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Crux Mathematicorum 37

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April 2018
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1

New Editor-in-Chief for CRUX with MAYHEM
We happily herald Shawn Godin of Cairine Wilson Secondary School, Orleans, ON, Canada as the next Editor-in-Chief of Crux Mathematicorum with Mathematical Mayhem eﬀective 21 March 2011. With this issue Shawn takes the reins, and though some of you know Shawn as a former Mayhem Editor let me provide some further background. Shawn grew up outside the small town of Massey in Northern Ontario. He obtained his B. Math from the University of Waterloo in 1987. After a few years trying to decide what to do with his life Shawn returned to school obtaining his B.Ed. in 1991. Shawn taught at St. Joseph Scollard Hall S.S. in North Bay from 1991 to 1998. During this time he married his wife Julie, and his two sons Samuel and Simon were born. In the summer of 1998 he moved with his family to Orleans where he has taught at Cairine Wilson S.S. ever since (with a three year term at the board oﬃce as a consultant). He also managed to go back to school part time and earn an M.Sc. in mathematics from Carleton University in 2002. Shawn has been involved in many mathematical activities. In 1998 he co-chaired the provincial mathematics education conference. He has been involved with textbook companies developing material for teacher resources and the web as well as writing a few chapters for a grade 11 textbook. He has developed materials for teachers and students for his school board, as well as the provincial ministry of education. Shawn is involved with the problem committees at the Centre for Education in Mathematics and Computing at the University of Waterloo where he currently works on the committee for the Fryer, Galois and Hypatia contests. He is a frequent presenter at local, regional and provincial math education conferences. In 1997 Shawn met Graham Wright, the former executive director of the CMS, at a conference. When Shawn moved to Ottawa Graham put him to work as Mayhem editor from 2001 to 2006 as well as helping with math camps in Ottawa. In his free time, Shawn enjoys hanging out with his wife and kids, reading (mainly science ﬁction and mathematics), and “playing” guitar. Shawn has already been receiving your submissions, and please send any future submissions to the addresses and e-mail addresses listed inside the back cover, with your full name and aﬃliation on each page. V´ aclav (Vazz) Linek, Herald and Guest Editor.

EDITORIAL
Shawn Godin
Hello CRUX with MAYHEM, it is great to be back! I always thought at some point I would come back to the editorial board, but I thought it would be when I retired and not as Editor-in-Chief. Sometimes life throws you a surprise and you have to roll with it. This is an interesting time for the journal. The readership is down a bit and Robert Woodrow is stepping down as the editor of the Olympiad corner after serving Crux for many, many years, so we are looking to make some changes. Ultimately, we want a publication that meets the needs and wants of its readers, so we will be looking to you for some feedback. In a future issue we will look to get some ideas from you. What do you like about CRUX with MAYHEM? What would you like changed? Are there things that we are not currently doing that we should be doing? Are there things that we are doing that we need to discontinue? Are there alternate formats that we should explore for CRUX with MAYHEM? We will need your input, so please start thinking so that you can give us some feedback later. It took quite a while for me to be oﬃcially appointed the Editor-in-Chief, and as a result we are a few months behind schedule. We will be doing our best to make up some of that time and get us closer to our real time line. To help facilitate this, all deadlines for solutions will appear as if we are on time. Having said that you will have some extra time before the material will be processed. We will post the status of the problems processing on the CMS web site, cms.math.ca/crux. I want to say a quick thank you to Vazz Linek, the rest of the members of the board and the staﬀ at the CMS for all their help getting me on my feet. The learning curve is steep but I am really looking forward to working on CRUX with MAYHEM. I know we can work together and continue the great work that Vazz, Jim, Bruce, Robert, Bill, Leo, Fred and all the other members of the CRUX with MAYHEM board past and present, have done. Let’s get started! Shawn Godin

SKOLIAD

No. 130

Lily Yen and Mogens Hansen
Please send your solutions to problems in this Skoliad by September 15, 2011. A copy of CRUX with Mayhem will be sent to one pre-university reader who sends in solutions before the deadline. The decision of the editors is ﬁnal.

Our contest this month is the Niels Henrik Abel Mathematics Contest, 2009– 2010, Second Round. Our thanks go to Øyvind Bakke, Norwegian University of Science and Technology, Trondheim, Norway, for providing us with this contest and for permission to publish it. We also thank Rolland Gaudet, University College of Saint Boniface, Winnipeg, MB, for translating this contest from English into French.

Niels Henrik Abel Mathematics Contest, 2009–2010 2nd Round
100 minutes allowed

A four-digit whole number is interesting if the number formed by the leftmost two digits is twice as large as the number formed by the rightmost two digits. (For example, 2010 is interesting.) Find the largest whole number, d, such that all interesting numbers are divisible by d.

2. A calculator performs this operation: It multiplies by 2.1, then erases all digits
to the right of the decimal point. For example, if you perform this operation on the number 5, the result is 10; if you begin with 11, the result is 23. Now, if you begin with the whole number k and perform the operation three times, the result is 201. Find k.

The pentagon ABCDE consists of a square, ACDE , with side length 8, and an isosceles triangle, ABC , such that AB = BC . The area of the pentagon is 90. Find the area of BEC .

In how many ways can one choose three diﬀerent integers between 0.5 and 13.5 such that the sum of the three numbers is divisible by 3?

5. 6.
a

If a and b are positive integers such that a3 − b3 = 485, ﬁnd a3 + b3 .
−2 2

.... . .. . ... . .. . . . . . . ..... . .. . . . ... . . .. . ... ... . . ... . . . . ... .. . . . . . .. C A .. . . . . . . . ..... . . . . . . . ..... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..... .. . . . . .. . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . ... . ..............................................

B .

If a and b are positive integers such that a3 + b3 = 2ab(a + b), ﬁnd b + a2 b−2 .

Let D be the midpoint of side AC in ABC . If ∠CAB = ∠CBD and the length of AB is 12, then ﬁnd the square of the length of BD .

8. If x, y, and z are whole numbers and xyz + xy +2yz + xz + x +2y +2z = 28
ﬁnd x + y + z . Henrik’s math class needs to choose a committee consisting of two girls and two boys. If the committee can be chosen in 3630 ways, how many students are there in Henrik’s math class?

10.

100!(1002 + 100 + 1). Find

Let S be 1!(12 + 1 + 1) + 2!(22 + 2 + 1) + 3!(32 + 3 + 1) + · · · +
S+1 . (As usual, k! = 1 · 2 · 3 · · · · · (k − 1) · k.) 101!

Concours Math´ ematique Niels Henrik Abel, 2009–2010 eme ronde 2i`
100 minutes sont accord´ ees

1. Un entier a ` quatre chiﬀres est int´ eressant si l’entier form´ e par les deux chiﬀres a ` l’extrˆ eme gauche est deux fois plus grand que l’entier form´ e par les deux chiﬀres a ` l’extrˆ eme droite. (Par exemple, 2010 est int´ eressant.) D´ eterminer le plus grand entier, d, tel que tous les nombres int´ eressants sont divisibles par d. 2. Une calculatrice eﬀectue cette op´ eration : elle multiplie par 2,1, puis elle eﬀace tous les chiﬀres a ` droite de la d´ ecimale. Par exemple, si on eﬀectue cette op´ eration a ` partir de 5, le r´ esultat est 10 ; a ` partir de 11, le r´ esultat est 23. Or, si on commence avec un entier k et qu’on eﬀectue cette op´ eration trois fois, le r´ esultat est 201. D´ eterminer k. 3. Le pentagone ABCDE consiste d’un carr´ e, ACDE , de cˆ ot´ es de longueur 8, puis d’un triangle isoc` ele, ABC , tel que AB = BC . La surface du pentagone est 90. D´ eterminer la surface de BEC . 4.
De combien de fa¸ cons pouvons-nous choisir trois entiers diﬀ´ erents entre 0, 5 et 13, 5, tels que la somme des trois entiers soit divisible par 3 ?
...... . . ... . ... . . . . . . ..... . . ... . ... . . . .. . ... .. ... . ... . . ... .. . . . . . A. C . . . . . . .... . . . . . . . . . . . . . ..... . . . . . . . . . . . .. . . . . . . . . . .... . . . .. . ........ . . . . . . . . . . . . . . . . . . . . . . .. . ...... . . . . . . . . . . . . . . .. . . . . ................................................

B .

5. Si a et b sont des entiers positifs tels que a3 − b3 = 485, d´ eterminer a3 + b3 .
a−2 b2 + a2 b−2 .

6. Si a et b sont des entiers positifs tels que a3 + b3 = 2ab(a + b), d´ eterminer 7. Soit D le mipoint du cˆ ot´ e AC
d´ eterminer x + y + z . dans ABC . Si ∠CAB = ∠CBD et si la longueur de AB est 12, d´ eterminer le carr´ e de la longueur de BD .

8. Si x, y et z sont des entiers et si xyz + xy + 2yz + xz + x + 2y + 2z = 28,

9. La classe de math´ ematiques d’Henri a besoin de choisir un comit´ e form´ e de deux
ﬁlles et de deux gar¸ cons. Si ce comit´ e peut ˆ etre form´ e de 3630 fa¸ cons, combien d’´ etudiants y a-t-il dans la classe de math´ ematiques d’Henri ?

10.

Soit S ´ egal a ` 1!(12 + 1 + 1) + 2!(22 + 2 + 1) + 3!(32 + 3 + 1) +
S+1

. (Comme d’habitude, k! = · · · + 100!(1002 + 100 + 1). D´ eterminer 101! 1 · 2 · 3 · · · · · (k − 1) · k.)

Next follow solutions to the selected problems from the 10th Annual Christopher Newport University Regional Mathematics Contest, 2009, given in Skoliad 124 at [2010:129–131]. (Note: Problems 1, 2, 3 and 4 ﬁrst appeared on the 2009 Calgary Junior Math Contest. – Ed.) Elves and ogres live in the land of Pixie. The average height of the elves is 80 cm, the average height of the ogres is 200 cm, and the average height of the elves and the ogres together is 140 cm. If 36 elves live in Pixie, how many ogres live there? ´ Solution by Lena Choi, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC. Let x be the number of ogres in Pixie. Then the total height of all the ogres is 200x, and the total height of all the elves is 36 · 80 = 2880. Therefore the total height of all the creatures in Pixie is 2880 + 200x. On the other hand, the average height of the 36 + x creatures in Pixie is 140, so their total height is 140(36 + x) = 5040 + 140x. Thus 2880 + 200x = 5040 + 140x, so x = 36.
Also solved by WEN-TING FAN, student, Burnaby North Secondary School, Burnaby, BC; GESINE GEUPEL, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany; and LISA WANG, student, Port Moody Secondary School, Port Moody, BC.

You are given a two-digit positive integer. If you reverse the digits of your number, the result is a number which is 20% larger than your original number. What is your original number? Solution by Gesine Geupel, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany.
1 x, Let x be the given two-digit number. Increasing x by 20%, that is, by 5 yields an integer, so x must be divisible by 5. Thus x ends in 0 or in 5. If the ones digit of x is 0, reversing the digits would decrease the number, so x must end in 5. If the tens digit is larger than 5, reversing the digits would again decrease the number. Thus only the numbers 15, 25, 35, 45, and 55 remain to be checked. Only 45 works out.

´ Also solved by LENA CHOI, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; WEN-TING FAN, student, Burnaby North Secondary School, Burnaby, BC; and LISA WANG, student, Port Moody Secondary School, Port Moody, BC. Alternatively, let 10a + b be the two-digit number, where a and b are digits. Increasing by 6 6 20% is the same as multiplying by 5 , so 10b +a = 5 (10a +b). Thus 5(10b +a) = 6(10a +b),

so 44b = 55a, so 4b = 5a. Clearly b is divisible by 5, and since b = 0 = a does not yield a two-digit number, b must be 5. Hence a = 4 and the number is 45.

Three squares are placed side-by-side inside a right-angled triangle as shown in the diagram. The side length of the smallest of the three squares is 16. The side length of the largest of the three squares is 36. What is the side length of the middle square?

Solution by Wen-Ting Fan, student, Burnaby North Secondary School, Burnaby, BC. Impose a coordinate system as in the ﬁgure. If the middle square has side length n, then the coordinates are as indicated. Since the slanted line passes through (0, 16), the equation of the line is y = mx + 16 for some slope, m.

..... ... .... . . . ......... . . . . ......... . . . . . . . . . . . . . . . . . . . . . . . . . (16 + n, 36) ............. . . . . . . . . . . . . . . . . . . . . . . ............................................................ . r . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (16 , n ) . . . . . . . ..... . . . . . . ......... .. . .. . . . . . r . .. ... .. . ................................. . .... . . . . . . . . . . . . . . . . . .............. ..... . . . (0, 16) . . . . . . . . . r . . ......................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . .. .. .. .. .. .. .. .. .. . .. .. .. . .. .. .. . .. .. .. .. . .. .. .. . .. .. ... . .. .. . .. .. .. . .. .. .. . .. .. .. . .. . .. .. .. . .. .. .. . .. .. .. .. . .. .. .. .. . .. .. .. . .. . .. .. .. .. .. .. . .. .. .. .. .. .. .. .. .. . .. .. .. .. .. .. .. . .. .. .. .. .. . .. .. .. . ........... ..... . ... . . . ... . . ... . . . . . ... . . ... ... . . ... . . ... . . ... .. . . . ... . . ... . . . . . . . . ... . . . . ... . . ... . ... . . . ... . ... . . ... . . . ... . ......... .. .. . .. .. .
Using the two other points yields n = 16m + 16 and 36 = m(16 + n) + 16 . Therefore 36 = m(16 + 16m + 16) + 16 = 16m2 + 32m + 16, so 1 5 or x = 2 . 0 = 16m2 + 32m − 20 = 4(2m + 5)(2m − 1), so x = − 2 1 In the ﬁgure, the slope is clearly positive, so m = 2 , and n = 16m + 16 = 24.
Also solved by LISA WANG, student, Port Moody Secondary School, Port Moody, BC. You can also solve this problem using similar triangles.

Friends Maya and Naya ordered ﬁnger food in a restaurant, Maya ordering chicken wings and Naya ordering bite-size ribs. Each wing cost the same amount, and each rib cost the same amount, but one wing was more expensive than one rib. Maya received 20% more pieces than Naya did, and Maya paid 50% more in total than Naya did. The price of one wing was what percentage higher than the price of one rib? Solution by Lisa Wang, student, Port Moody Secondary School, Port Moody, BC. Say Naya gets n pieces at $r each. Then Maya gets 1.2n pieces at, say, $w each. Then Naya pays $nr and Maya pays $1.2nw. Since Maya pays 50% more 1.5 than Naya, 1.2nw = 1.5nr , so w = 1 r = 1.25r , so one wing is 25% more .2 expensive than one rib.

´ Also solved by LENA CHOI, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; and WEN-TING FAN, student, Burnaby North Secondary School, Burnaby, BC.

A 9 × 12 rectangular piece of paper is folded once so that a pair of diagonally opposite corners coincide. What is the length of the crease? Solution by Wen-Ting Fan, student, Burnaby North Secondary School, Burnaby, BC. If you fold the paper as instructed and unfold it again, you obtain the ﬁgure below where the section outlined with thick lines used to overlap and the dashed line is the crease. The Pythagorean Theorem now yields that
12 − x x

92 + x2

81 + x2 = 144 − 24x + x2 24x = 63 63 x = 24

= (12 − x)2

... ......................................... .. ... ... ... ... ...12 − x 12 − x.... 9 ... ... ... ... ........................................ .. .
x 12 − x 12 − x x

Then redraw the diagram as in the ﬁgure below. Use the Pythagorean Theorem again:
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..

c2 = 92 + (12 − 2x)2 . Since x = . c = 45 4
63 , 24

c2 =

2025 , 16

12 − 2x

´ Also solved by LENA CHOI, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; and LISA WANG, student, Port Moody Secondary School, Port Moody, BC.

In calm weather, an aircraft can ﬂy from one city to another 200 kilometres north of the ﬁrst and back in exactly two hours. In a steady north wind, the round trip takes ﬁve minutes longer. Find the speed (in kilometres per hour) of the wind. Solution by the editors.

The airspeed of the plane is 400 = 200 kilometres per hour. Let w denote 2 the speed of the wind. Then, if you ﬂy with the wind, the ground speed is 200+ w; and if you ﬂy against the wind, the ground speed is 200 − w. Therefore, the plane 200 200 takes hours to ﬂy with the wind and hours to ﬂy against the wind. If you add these two expressions, you get the total time for the round trip,
200 + w 200 − w

but this time is given to be 2 +

5 60

hours, so 200 25 12 1 12 1 12

200

200 − w 200 + w 8(200 + w) + 8(200 − w) (200 − w)(200 + w) 3200

= = =

2002 − w2 w = ±40 .

Therefore the speed of the wind is 40 kilometres per hour.

A rectangular ﬂoor, 24 feet × 40 feet, is covered by squares of sides 1 foot. A chalk line is drawn from one corner to the diagonally opposite corner. How many tiles have a chalk line segment on them? Solution by Gesine Geupel, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany.

Since 24 = 3 and 40 = 5, consider instead the 3 × 5 rectangle on the left 8 8 in the ﬁgure. You can easily count that the diagonal crosses seven squares.

.................. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ........ ........ ................ . . . . . . . . . . ......... ......... .

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Now tile the 24 × 40 rectangle with 3 × 5 rectangles as in the righthand side of the ﬁgure. The diagonal of the 24 × 40 rectangle is also the diagonal of each of eight of the 3 × 5 rectangles. Therefore the diagonal crosses 56 of the 1 × 1 squares.

This issue’s prize of one copy of Crux Mathematicorum for the best solutions goes to Wen-Ting Fan, student, Burnaby North Secondary School, Burnaby, BC. As Skoliad editors we are quite pleased to see envelopes with “exotic” stamps in the mail, but receiving more Canadian solutions would be wonderful. The address is on the inside of the back cover. You do not have to solve the entire featured contest; a well-presented solution to a single problem is enough. The test for “well-presented” is that your classmates at school can understand it. You can send your solution(s) by mail or electronically—even if that means that we miss out on the exotic stamps.

Mayhem Problems
Please send your solutions to the problems in this edition by 15 August 2011. Solutions received after this date will only be considered if there is time before publication of the solutions. Each problem is given in English and French, the oﬃcial languages of Canada. In issues 1, 3, 5, and 7, English will precede French, and in issues 2, 4, 6, and 8, French will precede English. The editor thanks Jean-Marc Terrier of the University of Montreal for translations of the problems.

Note: As CRUX with MAYHEM is running behind schedule, we will accept solutions past the posted due date. Solutions will be accepted until we process them for publication. Currently we are delayed by about four months. Check the CMS website, cms.math.ca/crux, for our status in processing problems.

M470.

Proposed by the Mayhem Staﬀ

Vazz needs to buy desks and monitors for his new business. A desk costs $250 and a monitor costs $260. Determine all possible ways that he could spend exactly $10 000 on desks and monitors.

M471.

Proposed by the Mayhem Staﬀ

Square based pyramid ABCDE has a square base ABCD with side length 10. Its other four edges AE , BE , CE , and DE each have length 20. Determine the volume of the pyramid.

M472.

Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania Suppose that x is a real number. Without using calculus, determine the
2x2 − 8x + 17 and the minimum possible value of x2 − 4x + 7

maximum possible value of
x2 + 6x + 8 . x2 + 6x + 10

M473.

Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania Determine all pairs (a, b) of positive integers for which a2 + b2 − 2a + b = 5.

M474.

Proposed by Dragoljub Miloˇ sevi´ c, Gornji Milanovac, Serbia

Let a, b and x be positive integers such that x2 − bx + a − 1 = 0. Prove that a2 − b2 is not a prime number.

M475.
ON

Proposed by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,

Let x denote the greatest integer not exceeding x. For example, 3.1 = 3 and −1.4 = −2. Let {x} denote the fractional part of the real number x, that is, {x} = x − x . For example, {3.1} = 0.1 and {−1.4} = 0.6. Show that there exist inﬁnitely many irrational numbers x such that x · {x} = x . ................................................................. ´ M470. Propos´ e par l’Equipe de Mayhem Vazz doit acheter des pupitres et des ´ ecrans pour son nouveau commerce. Un pupitre coˆ ute 250$ et un ´ ecran 260$. Trouver de combien de mani` eres possibles il pourrait d´ epenser exactement 10 000$ en pupitres et ´ ecrans. ´ M471. Propos´ e par l’Equipe de Mayhem Une pyramide ABCDE a une base carr´ ee ABCD de cˆ ot´ e 10. Les quatre autres arˆ etes AE , BE , CE et DE sont toutes de longueur 20. Trouver le volume de la pyramide.

M472.

´ Propos´ e par Neculai Stanciu, Ecole secondaire George Emil Palade, Buz˘ au, Roumanie Supposons que x soit un nombre r´ eel. Sans utiliser le calcul diﬀ´ erentiel, d´ eterminer la valeur maximale possible de
x2 + 6x + 8 . x2 + 6x + 10 2x2 − 8x + 17 et la valeur minimale x2 − 4x + 7

possible de

M473.

´ Propos´ e par Neculai Stanciu, Ecole secondaire George Emil Palade, Buz˘ au, Roumanie D´ eterminer toutes a2 + b2 − 2a + b = 5. les paires (a, b) d’entiers positifs tels que

M474. Propos´ e par Dragoljub Miloˇ sevi´ c, Gornji Milanovac, Serbie
Soit a, b et x trois entiers positis tels que x2 − bx + a − 1 = 0. Montrer que a − b2 n’est pas un nombre premier.
2

M475. Propos´ e par Edward T.H. Wang, Universit´ e Wilfrid Laurier, Waterloo,
ON Notons x le plus grand entier n’exc´ edant pas x. Par exemple, 3,1 = 3 et −1,4 = −2. Notons {x} la partie fractionnaire du nombre r´ eel x, c.-` a-d, {x} = x − x . Par exemple, {3,1} = 0,1 et {−1,4} = 0,6. Montrer qu’il existe une inﬁnit´ e de nombres irrationnels x tels que x · {x} = x .

Mayhem Solutions
M432.
Proposed by the Mayhem Staﬀ. Determine the value of d with d > 0 so that the area of the quadrilateral with vertices A(0, 2), B (4, 6), C (7, 5), and D (d, 0) is 24. Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON. Let E = (0, 6), F = (7, 6), G = (7, 0) and Ω denote the area function. Then 1 (d × 2) = d; Ω(AOD ) = 2 1 Ω(BEA) = 2 (4 × 4) = 8; 3 (3 × 1) = 2 ; Ω(BF C ) = 1 2 1 5 (7−d). and Ω(CDG) = 2 (7−d)×5 = 2
å

........ .... .................. . . . . . . .... . . . . . . ... .. .... . . . . . . ............. A(0, 2) .. .. ............. ............. ... ....
E (0, 6)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

B (4, 6)

F (7, 6) C (7, 5)

G(7, 0)

D (d, 0)

Since Ω(OEF G) = 7 × 6 = 42, we have 24 = Ω(ABCD ) = 42 − d + 8 + 3 2 + 5 2

(7 − d) = 15 +

3 2

Solving we ﬁnd d = 6.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; NATALIA DESY, ´ student, SMA Xaverius 1, Palembang, Indonesia; SAMUEL GOMEZ MORENO, Universidad de Ja´ en, Ja´ en, Spain; AFIFFAH NUUR MILA HUSNIANA, student, SMPN 8, Yogyakarta, Indonesia; GEOFFREY A. KANDALL, Hamden, CT, USA; WINDA KIRANA, student, SMPN 8, Yogyakarta, Indonesia; JOSHUA LONG, Southeast Missouri State University, Cape ˇ ´ Gornji Milanovac, Serbia; RICARD Girardeau, MO, USA; DRAGOLJUB MILO SEVI C, ´ IES “Abastos”, Valencia, Spain; CAO MINH QUANG, Nguyen Binh Khiem High PEIRO, School, Vinh Long, Vietnam; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; NECULAI STANCIU, George Emil Palade Secondary School, Buz˘ au, Romania(2 solutions); and JOHN WYNN, student, Auburn University, Montgomery, AL, USA; Two incorrect solutions were received.

M433.

Proposed by Bruce Shawyer, Memorial University of Newfoundland, St. John’s, NL.

In triangle ABC , AB < BC , L is the midpoint of AC , and M is the midpoint of AB . Also, P is the point on LM such that M P = M A. Prove that ∠P BA = ∠P BC .

Solution by Souparna Purohit, student, George Washington Middle School, Ridgewood, NJ, USA.
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

It is well known that since M and L are the midpoints of AB and AC then BC M L so ∠P BC = ∠BP M . Also, since P M = AM = BM , ∆BM P is isosceles. Therefore ∠ABP = ∠BP M which, when combined with ∠P BC = ∠BP M , we conclude that ∠ABP = ∠P BC , as desired.
Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain(two solutions); GEORGE APOSTOLOPOULOS, Messolonghi, Greece; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; GEOFFREY A. KANDALL, Hamden, CT, USA; WINDA ˇ ´ Gornji KIRANA, student, SMPN 8, Yogyakarta, Indonesia; DRAGOLJUB MILOSEVI C, ´ IES “Abastos”, Valencia, Spain; BRUNO SALGUEIRO Milanovac, Serbia; RICARD PEIRO, ´ FANEGO, Viveiro, Spain; HUGO LUYO SANCHEZ, Pontiﬁcia Universidad Cat´ olica del Peru, Lima, Peru; NECULAI STANCIU, George Emil Palade Secondary School, Buz˘ au, Romania; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; and the proposer. One incorrect solution was received. Several readers pointed out that ∆AP B is right angled with the right angle at P .

M434.

Proposed by Heisu Nicolae, Pˆ ırjol Secondary School, Bac˘ au, Romania.

Determine all eight-digit positive integers abcdef gh which satisfy the relations a3 − b2 = 2, c3 − d2 = 4, 2e − f 2 = 7, and g 3 − h2 = −1. Solution by Arkady Alt, San Jose, CA, USA. Since 2 ≤ a3 ≤ 92 + 2 = 83 ⇔ 2 ≤ a ≤ 4 and a3 − 2 for such a can only be square for a = 3, then a = 3, b = 5. Since 4 ≤ c3 ≤ 92 + 4 = 85 ⇔ 2 ≤ c ≤ 4 and c3 − 4 for such c can only be square for c = 2 then c = 2, d = 2. Since 7 ≤ 2e ≤ 92 + 7 = 88 ⇔ 3 ≤ e ≤ 6 and 2e − 7 for such e can only be square for e = 3, e = 4 and e = 5 then (e, f ) = (3, 1), (4, 3), (5, 5). Since 0 ≤ g 3 ≤ 92 − 1 = 80 ⇔ 0 ≤ g ≤ 4 and g 3 + 1 for such g can only be square for g = 0 and g = 2 then (g, h) = (0, 1), (2, 3). Thus abcdef gh = 35 223 101, 35 224 301, 35 225 501, 35 223 123, 35 224 323, 35 225 523.
´ Also solved by RICHARD I. HESS, Rancho Palos Verdes, CA, USA; RICARD PEIRO, IES “Abastos”, Valencia, Spain; CAO MINH QUANG, Nguyen Binh Khiem High School, Vinh Long, Vietnam; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; and the proposer. Seven incomplete solutions were submitted. Most of the incomplete solutions missed the case where g = 0.

M435.

Proposed by Mih´ aly Bencze, Brasov, Romania.

Prove that
n

1+
k=1

1 k2

1 (k + 1)2

n(n + 2) n+1

Solution by Cao Minh Quang, Nguyen Binh Khiem High School, Vinh Long, Vietnam. For any k > 0 we have 1+ 1 k − 1 k+1
2

= 1+

1 k2

(k + k k+1 k(k + 1) 1 1 2 2 = 1+ 2 + + − k (k + 1)2 k(k + 1) k(k + 1) 1 1 = 1+ 2 + . k (k + 1)2 = then 1 k+1 + 1+ 1 − 1 + ··· + 1 + 1 n − 1 n+1 1 +
1 k

1 1)2

−

Hence S=

Èn

1 k2

+
1 k2

1 (k+1)2

k=1

1+
n

1 (k+1)2

−

1 k+1

. Therefore if we let

=
k=1

1+ 1+ 1 1

1 k 1

−

2 3 n(n + 2) = n+1− = , n+1 n+1 and we are done!

−

2 1

Also solved by ARKADY ALT, San Jose, CA, USA; MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; GEORGE APOSTOLOPOULOS, Messolonghi, ´ Greece; SAMUEL GOMEZ MORENO, Universidad de Ja´ en, Ja´ en, Spain; G.C. GREUBEL, Newport News, VA, USA; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; GEOFFREY A. KANDALL, Hamden, CT, USA; WINDA KIRANA, student, SMPN 8, ´ Yogyakarta, Indonesia; HUGO LUYO SANCHEZ, Pontiﬁcia Universidad Cat´ olica del Peru, Lima, Peru; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA; PEDRO HENRIQUE ´ IES “Abastos”, Valencia, Spain; O. PANTOJA, student, UFRN, Brazil; RICARD PEIRO, PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata ˇ ´ Gornji Milanovac, Serbia; NATALIA DESY, Roma, Rome, Italy; DRAGOLJUB MILO SEVI C, student, SMA Xaverius 1, Palembang, Indonesia; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; NECULAI STANCIU, George Emil Palade Secondary School, Buz˘ au, Romania; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; and the proposer.

M436.

Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania. Determine the smallest possible value of x + y , if x and y are positive integers with
x 2009 2008 < < . 2009 y 2010

Solution by Richard I. Hess, Rancho Palos Verdes, CA, USA.
d 1 1 d 1 1 < y− < 1 − 2010 so 2009 > y > 2010 . If Let y = x + d then 1 − 2009 y d = 1 there is no solution. If d = 2, y = 4019 is a solution so x = 4017 and x + y = 8036. If d > 2 then x, y > 6000 thus x + y > 12000. Therefore the minimum value of x + y is 8036.

´ Also solved by ARKADY ALT, San Jose, CA, USA; SAMUEL GOMEZ MORENO, Universidad de Ja´ en, Ja´ en, Spain; CAO MINH QUANG, Nguyen Binh Khiem High School, Vinh Long, Vietnam; DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; and the proposer. Five incorrect solutions were submitted. +1 < x < n then the solution with Alt, Manes and Wang proved in general that if nn +1 y n+2 the smallest sum corresponds to x = 2n + 1, y = 2n + 3 and thus x + y = 4n + 4. In general +c c for any two fractions of non-negative integers, in lowest terms, a < d the value a is called b b+d the mediant and it satisﬁes
a b

a+c b+d

c . d

If we also have bc − ad = 1 then the mediant is

the fraction with the lowest denominator in the interval

a
b

c d

M437. Proposed by Samuel G´ omez Moreno, Universidad de Ja´ en, Ja´ en, Spain.
Let x denote the greatest integer not exceeding x. For example, 3.1 = 3 and −1.4 = −2. Let {x} denote the fractional part of the real number x, that is, {x} = x− x . For example, {3.1} = 0.1 and {−1.4} = 0.6. Determine all rational numbers x such that x · {x} = x . Solution by David E. Manes, SUNY at Oneonta, Oneonta, NY, USA. The only rational number x such that x · {x} = x is x = 0. If n is an integer, then n · {n} = 0 = n = n has the only solution n = 0. Therefore, 0 is the only integer solution to the equation. Assume x is a rational number diﬀerent from an integer such that x x . Therefore, x x+1 = x x · {x} = x = x − {x}, then {x} = x+1 m 2 implies x = x (x + 1). Assume x = n where m and n are relatively prime integers and n > 1. Then m2 n2 = x m n +1 = x m+n n .

As a result, m2 = x (m + n) · n so that n is a divisor of m2 , a contradiction since m and n are relatively prime and n > 1 .
Also solved by ARKADY ALT, San Jose, CA, USA; RICHARD I. HESS, Rancho Palos ´ IES “Abastos”, Valencia, Spain; EDWARD T.H. WANG, Verdes, CA, USA; RICARD PEIRO, Wilfrid Laurier University, Waterloo, ON; and the proposer. Three incorrect solutions were submitted.

Problem of the Month
Ian VanderBurgh
Problems involving averages and their properties appear frequently on contests (Look at the solution to question 1 of Skoliad on page 5 – Ed.). This month and next, we will look at a few of these problems, at least one of which uses averages in a very subtle way. Problem 1 (2008 Small c Contest) The average of three numbers is 13. Two numbers are added to this list so that the average of all ﬁve numbers is 17. What is the average of the two new numbers? (A) 21 (B) 25 (C) 23 (D) 30 (E) 15 One of the things about average problems that I like is that there are really 1 things that you need to know about averages in order to be able only about 1 2 to do almost all of such problems. (That’s not to say that there isn’t a plethora of tricks of the trade that can be useful...) important things is how to calculate an average: add The ﬁrst of these 1 1 2 up the given numbers, count the given numbers, and divide the sum by the count to get the average. The extra 1 thing to remember is that the sum of the numbers 2 equals the count times the average. Expressing these facts algebraically, we see that if there are n numbers whose sum is S , then the average, a, satisﬁes the S equation a = . Rearranging this gives S = na. (I concede that occasionally n S as well.) we might use the fact that n = a Let’s solve Problem 1 using these properties and then look at our answer to see what we can observe. Solution to Problem 1. Since the average of the original three numbers is 13, then their sum is 3 × 13 = 39. Since the average of all ﬁve numbers is 17, then the sum of the ﬁve numbers is 5 × 17 = 85. The sum of the additional two numbers equals the sum of all ﬁve numbers minus the sum of the original three numbers, or 85 − 39 = 46. Therefore, the 46 average of these two numbers is = 23. 2 This problem is particularly nice, in my opinion, because it doesn’t require us to use any algebra. Let’s look at the data that we have: • the average of the ﬁrst 3 numbers is 13 • the average of all 5 numbers is 17 • the average of the last 2 numbers is 23

Do you notice anything about the position of the overall average relative to the averages of the ﬁrst and last numbers? You might have noticed that the overall average splits these averages in the ratio 4 : 6 which equals 2 : 3, which happens to be the ratio of the count of numbers in each partial average (arranged in reverse from what you might quickly guess). If this rule works in general, then if we had 5 numbers with average 22 and 3 numbers with average 46, the average of all 8 numbers should split 22 and 46 3 3 = of the way from 22 in the ratio 3 : 5. In other words, the average is 3+5 8 3 to 46, and so equals 22 + × (46 − 22) = 31. Try solving this problem using 8 the method that we used above to conﬁrm the answer. Putting this in a more general way, if m numbers have an average of a and n numbers have an average of b with a < b, then the average of the m + n numbers splits a and b in the ratio n : m (not m : n). Can you prove this? We’ll look at another problem next month where this approach is really useful. Problem 2 (2010 Pascal Contest) In the diagram, each of the ﬁve boxes is to contain a number. Each number in a bold outlined box must be the average of the number in the box to the left of it and the number in the box to the right of it. What is the value of x?
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q qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqq.q.q........................................ . q q . . q q q qq . . q q q . . q q q qq . . q q q q q q . x . 26 q . . q q q q . q q q q . . q . . qqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqq..qq........................................ .

(A) 28 (B) 30 (C) 31 (D) 32 (E) 34 Special cases often produce interesting facts. For example, if two numbers x+y = a or x + y = 2a. Try to use this x and y have an average of a, then 2 to solve the following problem algebraically. Solution to Problem 2. We label the numbers in the empty boxes as y and z , so the numbers in the boxes are thus 8, y, z, 26, x. Since the average of z and x is 26, then x + z = 2(26) = 52 or z = 52 − x. We rewrite the list as 8, y, 52 − x, 26, x.

Since the average of 26 and y is 52 − x, then 26 + y = 2(52 − x) or y = 104 − 26 − 2x = 78 − 2x. We rewrite the list as 8, 78 − 2x, 52 − x, 26, x. Since the average of 8 and 52 − x is 78 − 2x, then 8+(52 − x) = 2(78 − 2x) or 60 − x = 156 − 4x and so 3x = 96 or x = 32.

Especially while writing a contest, it’s very tempting to take the answer that we get and not think about it at all. But let’s actually take this a moment to use this answer and go back to the list in terms of x (written as 8, 78 − 2x, 52 − x, 26, x) and substitute to get the list 8, 14, 20, 26, 32.

Do you recognize what kind of sequence this list forms? This is an arithmetic sequence. (Look up this term if you’ve never seen it before.) Do you think that this is a coincidence? (Hint: The answer to this question is almost always no.) Let’s think about this by going back to the list 8, y, z, 26, x. Let’s avoid using algebra, but we’ll keep these labels to make things a little clearer. We are told that y is the average of 8 and z . The important fact to recognize here is that y is halfway between 8 and z . In other words, the diﬀerence y − 8 equals z − y . Similarly, z is the average of 26 and y , so 26 − z equals z − y . But there is a common diﬀerence in these two sentences! (And it’s no coincidence that I used the phrase common diﬀerence...) Since there is this common diﬀerence, then all three diﬀerences must be equal. Since 26 − 8 = 18, then each of these diﬀerences equals 18 ÷ 3 = 6, and so the numbers in the sequence are 8, 14, 20, 26, x. Can you extend this argument another step to explain why x = 32? So what is the connection between averages and arithmetic sequences? An arithmetic sequence is a sequence with the property that each term after the ﬁrst is the average of the term before and the term after. This is pretty neat, if you’ve never seen it before. One last thing to think about – the average is sometimes called the arithmetic mean. Coincidence?

Adams, Douglas (1952 - 2001) The ﬁrst nonabsolute number is the number of people for whom the table is reserved. This will vary during the course of the ﬁrst three telephone calls to the restaurant, and then bear no apparent relation to the number of people who actually turn up, or to the number of people who subsequently join them after the show/match/party/gig, or to the number of people who leave when they see who else has turned up. The second nonabsolute number is the given time of arrival, which is now known to be one of the most bizarre of mathematical concepts, a recipriversexcluson, a number whose existence can only be deﬁned as being anything other than itself. In other words, the given time of arrival is the one moment of time at which it is impossible that any member of the party will arrive. Recipriversexclusons now play a vital part in many branches of math, including statistics and accountancy and also form the basic equations used to engineer the Somebody Else’s Problem ﬁeld. The third and most mysterious piece of nonabsoluteness of all lies in the relationship between the number of items on the bill, the cost of each item, the number of people at the table and what they are each prepared to pay for. (The number of people who have actually brought any money is only a subphenomenon of this ﬁeld.) “Life, the Universe and Everything.” New York: Harmony Books, 1982.

THE OLYMPIAD CORNER
No. 291 R.E. Woodrow
This number we begin by looking at the ﬁles of solutions by readers to problems given in the February 2010 number of the Corner, and “A” problems proposed but not used at the 2007 IMO in Vietnam, given at [2010 : 18–19].

A2.

Let n be a positive integer, and let x and y be positive real numbers such that xn + y n = 1. Prove that
n k=1

1 + x2 k 1+ x4 k

n k=1

1 + y 2k 1+ y 4k

1 (1 − x)(1 − y )

Solved by Mohammed Aassila, Strasbourg, France; Arkady Alt, San Jose, CA, USA; Oliver Geupel, Br¨ uhl, NRW, Germany; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Geupel. From the identity 1 + x2 k 1+ x4 k + (1 − x3k )(1 − xk ) xk (1 + x4 k ) = 1 + x2 k 1+ x4 k + (1 + x4k ) − xk (1 + x2k ) xk (1 + x4 k ) = 1 xk

and the premise 0 < x < 1 we deduce that
n k=1

1 + x2 k 1 + x4 k
n k=1

<
k=1

1 xk

xn (1 − x)

1 − xn

(1)

Similarly we have

1 + y 2k 1 + y 4k

y n (1 − y )

1 − yn

(2)

The hypothesis xn + y n = 1 yields (1 − xn )(1 − y n ) xn y n = 1 − xn − y n + xn y n xn y n = 1. (3)

The desired inequality follows immediately from the relations (1), (2), and (3).

A3.

Find all functions f : R+ → R+ such that f (x + f (y )) = f (x + y ) + f (y )

for all x, y ∈ R+ . (Here R+ denotes the set of all positive real numbers.)

Solved by Mohammed Aassila, Strasbourg, France; and Michel Bataille, Rouen, France. We give Bataille’s version. We show that the unique solution is the function φ : x → 2x. Since 2(x + 2y ) = 2(x + y ) + 2y for all x, y ∈ R+ , this function φ is a solution. Conversely, let f be any solution and x, y be any positive real numbers. On the one hand, adding x on both sides of the given equation and taking the images under f , we successively have f (x + f (x + f (y ))) = f ((x + f (y )) + f (x + y )) = f (2x + y + f (y )) + f (x + y ) = f (2x + 2y ) + f (y ) + f (x + y ). On the other hand, using the given equation, we obtain f (x + f (x + f (y ))) = = f (2x + f (y )) + f (x + f (y )) f (2x + y ) + f (y ) + f (x + y ) + f (y ). (1)

It follows that f (2x + 2y ) = f (2x + y ) + f (y ). Now, suppose that 0 < a < b. We prove that f (a) < f (b).
−b , y = b − a in (1) and obtain • If b < 2a, we take x = 2a2 f (b) = f (a) + f (b − a), hence f (b) > f (a).

• If b > 2a, with x = f (b) > f (a) again.

b−2a , 2

y = a, (1) gives f (b) = f (a) + f (b − a) and = f a+
a 2

+a ) • If b = 2a, f (b) = f (2a) = f 2( a 2 2
a 2

a 2

f (a) + f > f (a). Thus, f is strictly increasing on (0, ∞) and, as such, is injective. In addition, if a, b are positive and distinct, say a < b, then a a f (a + b) = f (a) + f (b) (for f a + 2 · b− + f (a) = f 2a + 2 · b− ). 2 2 Lastly, let y > 0. Since f (y ) = y (otherwise f (x + f (y )) = f (x + y ) in contradiction with the given functional equation), we may write f (y + f (y )) = f (y ) + f (f (y )) as well as f (y + f (y )) = f (2y ) + f (y ). It follows that f (f (y )) = f (2y ) and since f is injective, f (y ) = 2y , as desired.

Next we look at the “C” problems proposed but not used at the 2007 IMO in Vietnam given at [2010: 19–20].

C2. A unit square is dissected into n > 1 rectangles such that their sides are parallel to the sides of the square. Any line, parallel to a side of the square and

intersecting its interior, also intersects the interior of some rectangle. Prove that one of the rectangles has no point on the boundary of the square. Solved by Oliver Geupel, Br¨ uhl, NRW, Germany. The proof is by contradiction. Assume the contrary and consider a counterexample where n is minimal. Let A be one of the vertices of the square, let E be the locus of the sides of the rectangles of the dissection, and let ABCD be the rectangle that contains the vertex A. Since the square is covered by the rectangles, at least one of the segments BC and DC has an extension in E beyond the point C . Without loss of generality assume that DC can be extended beyond C where the longest possible extension in E is up to a point E . Since the line DE intersects the interior of some rectangle, the point E is an interior point of the square. Since the square is covered by the rectangles, a segment EF orthogonal to CE where the points A and F are in the same half-plane relative to the line DE , also belongs to E . If there were a second rectangle beside ABCD which contains the side BC , then it could be glued together with the rectangle ABCD obtaining a counterexample with n − 1 rectangles, which contradicts our minimum hypothesis. Hence, there is a segment GH in E where G is an inner point of BC , and GH is parallel to CE . The rectangle with vertex C and sides on the lines CE and CG is now separated from the boundary of the square by the four segments HG, GC , CE and EF . This is a contradiction which completes the proof.
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C5. In the Cartesian coordinate plane let Sn = {(x, y) | n ≤ x < n + 1} for each integer n, and paint each region Sn either red or blue. Prove that any rectangle whose side lengths are distinct positive integers may be placed in the plane so that its vertices lie in regions of the same colour.
Solved by Oliver Geupel, Br¨ uhl, NRW, Germany. We generalize that any rectangle ABCD with distinct real sides AB = a and BC = b may be placed so that its vertices lie in regions of the same colour. Firstly consider the case where a ∈ / Z. Without loss of generality, we may assume that S0 and S1 have distinct colours. We place ABCD so that A and D have the common x-coordinate 1 − {a} while B and C have the

common x-coordinate a . We then translate the rectangle with oﬀset {a} in the positive x-direction. With this translation, A and D move from S0 to S1 , while B and D remain in S a . Consequently, in one of these two positions the vertices A, B, C, D lie in regions of the same colour. It remains to consider a, b ∈ Z. Let d = gcd(a, b), and let a0 , a1 , b0 , b1 be integers such that a = a0 d, b = b0 d, and a0 a1 + b0 b1 = 1. The proof is by contradiction. Suppose ABCD cannot be placed properly. Then S0 and Sa have distinct colours, and S0 and Sb have also distinct colours. By induction, S0 and Sua+vb , where u and v are integers, have distinct colours if and only if u + v is odd. By b0 a − a0 b = 0, we see that b0 − a0 is even. Since a0 and b0 are coprime, both are odd. Hence, a1 + b1 is odd, too. Thus, S0 and Sa1 a+b1 b = Sd have distinct colours. We conclude that S0 and S2d have the same colour. Assume a < b, which implies b0 ≥ 3. We pitch the rectangle so that the x-coordinates of A and D as well as the x-coordinates of B and C have distance 2d. It suﬃces to prove that the we can translate it so that A and B lie in regions of the same colour. If the angle between the x-axis and the line AB is ϕ and A and B are the projections of A and B , respectively, onto the x-axis, then we 2 a 2d = and A B = a cos ϕ = b2 / Z. By the ﬁrst obtain sin ϕ = 0 −4 ∈ b b0 b0 part of the proof, we can place A and B so that they lie in regions of the same colour. This completes the proof.
C
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C7.

A convex n-gon P in the plane is given. For every three vertices of P , the triangle determined by them is good if all its sides are of unit length. Prove that 2 n good triangles. P has at most 3

Comment by Mohammed Aassila, Strasbourg, France. This is not an original problem. It ﬁrst appeared in J. Pach and R. Pinchasi, How many unit triangles can be generated by n points in convex position?, American Math. Monthly 110 (5) 2003 : 40–406.

Next we move to the “G” problems proposed but not used at the 2007 IMO

in Vietnam, given at [1020 : 20–21].

G3.

Let ABC be a ﬁxed triangle, and let A1 , B1 , C1 be the midpoints of sides BC , CA, AB , respectively. Let P be a variable point on the circumcircle of ABC . Let lines P A1 , P B1 , P C1 meet the circumcircle again at A , B , C respectively. Assume that the points A, B , C , A , B , C are distinct, and that the lines AA , BB , CC form a triangle. Prove that the area of this triangle does not depend on P . Solved by Michel Bataille, Rouen, France. We denote by Γ the circumcircle of ABC and by A0 , B0 , C0 the points of intersection of the lines BB and CC , CC and AA , AA and BB , respectively. We will use areal coordinates relatively to (A, B, C ). The equation of the circle Γ is known to be a2 yz + b2 zx + c2 xy = 0 where A = BC , b = CA, c = AB . Let P (x0 , y0 , z0 ) and A (x , y , z ); note that the line AA has equation yz − zy = 0 and that (x0 , y0 , z0 ), (x , y , z ) are solutions to the system a2 yz + b2 zx + c2 xy = 0, x(y0 − z0 ) = x0 (y − z ). It readily follows that
y0 y , are solutions of an equation (with unknown U ) z0 z

(c2 x0 )U 2 + λU + (−b2 x0 ) = 0 for some real λ. From equation of AA is (c2 y0 )y + (b2 z0 )z = 0. Similarly, the equations of BB and CC are (c2 x0 )x + (a2 z0 )z = 0 and (b2 x0 )x + (a2 y0 )y = 0. Then, it is easily obtained that A0 (−a2 y0 z0 , b2 x0 z0 , c2 x0 y0 ), B0 (a2 y0 z0 , −b2 x0 z0 , c2 x0 y0 ),
y0 y b2 y z · = − 2 , we deduce = 2 so that the z0 z c − b2 z 0 c y0

Now, recall that if Mi (xi , yi , zi ) with xi + yi + zi = 1 for ¬ i = 1, 2, 3, the ¬ ratio ¬ x 1 y2 x 3 ¬ ¬ ¬ area(M1 M2 M3 ) y1 y2 y3 ¬ is the absolute value of the determinant ¬ ¬ ¬. Here, area(ABC ) ¬ z1 z2 z3 ¬ we have
area (A0 B0 C0 ) =| area (ABC )

C0 (a2 y0 z0 , b2 x0 z0 , −c2 x0 y0 ).

| where · b2 x0 z0 a2 y0 z0 − b2 x0 z0 + c2 x0 y0 1 −1 1 1 1 −1
¬ ¬ ¬ ¬. ¬ ¬

a2 y0 z0 −a2 y0 z0 + b2 x0 z0 + c2 x0 y0 c 2 x 0 y0

¬ ¬ −1 ¬ 1 · 2 ·¬ 2 2 a y0 z0 + b x0 z0 − c x0 y0 ¬ ¬ 1

Since P is on Γ, b2 x0 z0 + c2 x0 y0 = −a2 y0 z0 (for example) so that = (a2 y0 z0 )(b2 x0 z0 )(c2 x0 y0 ) ·4 (−2a2 y0 z0 )(−2b2 x0 z0 )(−2c2 x0 y0 )

and area (A0 B0 C0 ) = independent of P .

1 2

area (ABC ) ,

G5.

Triangle ABC is acute with ∠ABC > ∠ACB , incentre I , and circumradius R. Point D is the foot of the altitude from vertex A, point K lies on line AD such that AK = 2R, and D separates A and K . Finally, lines DI and KI meet sides AC and BC at E and F , respectively. Prove that if IE = IF then ∠ABC > 3∠ACB . Solved by Oliver Geupel, Br¨ uhl, NRW, Germany. The claim is false. We prove instead that ∠ABC ≤ 3∠ACB . We write a = BC , b = CA, c = AB , 2s = a + b + c, h = AD , α = ∠BAC , β = ∠ABC , γ = ∠ACB . We denote by r the inradius, by J the point on the segment AD such that DJ = 2r , and by M and N the perpendicular projections of I onto AC and BD , respectively.
A
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . ... . . . . ....... E J . M ... . .. . ... . h . I. . . . . . . ... . . . . .. . . . D. F. N a . . . . . . . . . . . . . . . . . . . . . .. . .
K

By standard formulas, we have AI 2 = (s − a)(s − b)(s − c) bc (s − a)bc r2 = · = , sin(A/2) s (s − b)(s − c) s 2Rh = bc, 4Rr = abc [ABC ] · [ABC ] s = abc s
AI

.
AK

We obtain AI 2 = 2R(h − 2r ) = AK · AJ ; hence = . By AJ AI ∠IAJ = ∠KAI , it follows that AIJ ∼ AKI . Thus, ∠AIJ = ∠AKI = ∠IKD . Recognizing the isosceles IJD , we deduce ∠AJI = 180◦ − ∠DJI = 180◦ − ∠IDJ = ∠IDK . We obtain AIJ ∼ IKD and consequently ∠DIK = ∠JAI = ∠BAI − ∠BAD = α 2 − (90◦ − β ) = β−γ 2 .

By IE = IF , the triangles EIM and F IN are congruent. The point N is an inner point of the segment CF . Now, if M is between C and E , then γ = 180◦ − ∠M IN = ∠DIN + ∠EIM > ∠DIK = ; consequently 2 β < 3γ . On the other hand, if the point E is between C and M , or E = M , then γ = 180◦ − ∠M IN = ∠DIK = done.
β−γ which implies β = 3γ . We are 2 β−γ

Next we move to the “N” problems proposed but not used at the 2007 IMO in Vietnam, given at [1020 : 21].

N1.

Find all pairs (k, n) of positive integers for which 7k − 3n divides k4 + n2 .

Solved by Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA. We will demonstrate that there is only one such pair, namely (k, n) = (2, 4). Suppose then that k, n are positive integers such that (7k − 3n ) | (k4 + n2 ); which means that k4 + n2 = r · (7k − 3n ), for some nonzero integer r k, n ∈ Z+ . (1)

First we show that both k and n must be even. We do so by ruling out the other three possibilities: both k and n being odd, k odd and n even, or (third possibility) k even and n odd. Possibility 1. both k and n are odd: k ≡ n ≡ 1 mod 2. We then have k4 ≡ n2 ≡ 1 (mod 8) (the square of an odd integer is congruent to 1 modulo 8. And so, k4 + n2 ≡ 1 + 1 ≡ 2 (mod 8) . Since k and n are both odd positive integers we also have k = 2m + 1, Thus 7k − 3n = 72m+1 − 32l+1 ≡ (72 )m · 7 − (32 )l · 3 ≡ 1 · 7 − 1 · 3 ≡ 7 − 3 ≡ 4 (mod 8) . (3) n = 2l + 1; where m, l are nonnegative integers. (2)

According to (3), 4 divides 7k − 3n . And by (2), the highest power of 2 dividing k4 + n2 ; is 21 = 2. This renders equation (1) contradictory or impossible. Hence possibility 1 is ruled out. Possibility 2. k is odd and n is even; k ≡ 1 (mod 2), n ≡ 0 (mod 2).

Clearly 7k − 3n ≡ 1 − 1 ≡ 0 (mod 2), while k4 + n2 ≡ 1 + 0 ≡ 1 (mod 2), which renders (1) contradictory modulo 2: the left-hand side is congruent to 0 modulo 2. So this possibility is ruled out as well. Possibility 3. k is even and n odd; k ≡ 0 (mod 2), n ≡ 1 (mod 2). Same argument as in Possibility 2; the left-hand side is congruent to 1 but the right-hand side is zero modulo 2. So this possibility is eliminated as well. We conclude that both positive integers k and n must be even: k = 2K and n = 2N, for some positive integers K and N. From (1) and (4) we obtain, 16K 4 + 4N 2 = k4 + n2 = r · (7K − 3N )(7K + 3N ). (5) (4)

According to (5), the positive integer 7K + 3N is a divisor of k4 + n2 . On the other hand, 75 = 16807 > 16 · 54 = 10, 000

(while 7K < 16K 4; for K = 1, 2, 3, 4). An easy induction shows that 7K > 16 · K 4 for K ≥ 5, we omit the details. Similarly we have 81 = 34 > 4 · 42 = 64 (while 3N < 4N 2 for N = 1, 2, 3). And an easy induction establishes that 3N > 4N 2 , for N ≥ 4. Therefore for K ≥ 5 and N ≥ 4 we have 7K + eN > 16K 4 + 4N 2 ; which implies that (7K + 3N ) · |7K − 3N | · |r | > 16K 4 + 4N 2 (6) since |r | · |7K − 3N | is a positive integer. Clearly (6) condtradicts (5). We have demonstrated that a necessary condition for (5) to hold true is K ≤ 4 and N ≤ 3; which means that there is only up to 12 possible pairs (K, N ) that may satisfy (5). We form the following table: K K K K K K K K K K K K = = = = = = = = = = = = 1, N 1, N 1, N 2, N 2, N 2, N 3, N 3, N 3, N 4, N 4, N 4, N = = = = = = = = = = = = 1 2 3 1 2 3 1 2 3 1 2 3 7K + 3N 10 16 34 52 58 76 346 352 370 2404 2410 2428 24 · K 4 + 22 · N 2 20 = 16 + 4 32 = 16 + 16 52 = 16 + 36 256 + 4 = 260 256 + 16 = 272 256 + 36 = 296 (16)(81) + 4 = 1300 (16)(81) + 16 = 1312 1296 + 36 = 1332 4096 + 4 = 4100 4096 + 16 = 4112 4096 + 36 = 4132

The above table shows that the only pairs (K, N ) for which 7K + 3N divides 24 · K 4 + 22 · N 2 are (K, N ) = (1, 1), (1, 2). However, the pair (1, 1)

does not satisfy (5) since 20 = r · 4 · 10, is impossible with r ∈ Z. The other pair, (K, N ) = (1, 2) does: 32 = r · (−2)(16), satisﬁed with r = −1. Thus (K, N ) = (1, 2) is the only pair; and so (k, n) = (2K, 2N ) = (2, 4).

N2.

Let b, n > 1 be integers. Suppose that for each k > 1 there exists an n integer ak such that b − an k is divisible by k . Prove that b = A for some integer A. Solved by Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA. To show that b is the nth power of an integer, it suﬃces to show that for every prime number p in the prime factorization of b, if pe is the power of p that appears in the prime factorization of b, then the exponent e is a multiple of n. We set b = pe · r , r a positive integer such that p does not divide r ; (r, p) = 1, e a positive integer. We apply the hypothesis of the problem with k = (pe )n = pen : k | b − an k means that there exists a positive integer λ such that b − an k = k · λ; and since b = pe · r , we obtain
e·n b − an · λ; k = p

e·n pe · r − a n · λ, k = p e, r, ak , λ positive integers such that (r, p) = 1

(1)

Since n > 1, it follows that 1 ≤ e < e·n Let pt , t a positive integer, be the highest power of p which divides ak : ak = pt · bk , t, bk positive integers such that (bk , p) = 1 From (1) and (3) we obtain
e·n pe · r − pn·t · bn · λ. k = p

(2)

(3)

(4)

Possibility 2. e < n · t holds.

Possibility 1. n · t < e, or,

We claim that (4) implies e = n · t. Indeed, if e = n · t, then either,

If Possibility 1 holds then, by (2) we have: 1 ≤ n · t < e , e · n. (5)

And so (4) implies that
c· n − n · t p(e−n·t) · r − bn · λ, k = p

(6)

which implies by (5) and (6) that p | bn k ; (since p is prime) p | bk , contrary to (3). If possibility 2 holds, then by (2), 1 ≤ e < min{e · n, n · t} = m And so, (4) implies,
e·n−e r − pn·t−e · bn ·λ k = p

(7) (8)

Thus (7) and (8) imply that p | r , contrary to (1). We have proved that e = n · t; and so b must be the nth power of an integer.

N4.

For every integer k ≥ 2, prove that 23k divides the number 2k+1 2k − 2k 2k−1

but 23k+1 does not. Comment by Mohammed Aassila, Strasbourg, France. This is not an original problem. It appeared ﬁrst in D.B. Fuchs and M.B. Fuchs, Arithmetic of binomial coeﬃcients, KVANT 6 (1970).

N5.

Find all surjective functions f : N → N such that for every m, n ∈ N and every prime p, the number f (m + n) is divisible by p if and only if f (m) + f (n) is divisible by p. (N is the set of all positive integers.) Solved by Oliver Geupel, Br¨ uhl, NRW, Germany. The identity f (n) = n is such a function. We prove that it is the unique solution. Suppose that f is any solution. We show by contradiction that f is injective. Assume the contrary. Consider the equivalence relation on N deﬁned by m ∼ n if and only if f (m) and f (n) have the same prime divisors. By assumption, there are numbers m < n such that f (m) = f (n). For every k ∈ N and every prime p, we have p | f (m+k) ⇔ p | f (m)+f (k) ⇔ p | f (n)+f (k) ⇔ p | f (n+k). Hence, for every k ∈ N it holds m + k ∼ n + k, i.e. each integer s > n is equivalent to s − (n − m). Let p be a prime such that the least number s with the property p | f (s) is greater than n. We have s ∼ s − (n − m), but p f (s − (n − m)), a contradiction. This completes the proof that f is injective. We show by contradiction that f (1) = 1.
¬ ¬ ¬
n

(1)

Suppose that there were a prime p such that p | f (1). Then, for every n ∈ N, we ¬ would have p ¬
k=1

f (1) and, by Mathematical Induction, p | f (n). Therefore, f

is not surjective, a contradiction, which completes the proof of (1).

We prove that for every m ∈ N it holds |f (m + 1) − f (m)| = 1 (2)

Suppose contrariwise that for any number m ∈ N there is a prime p such that p | f (m + 1) − f (m). Since f is surjective, there is a number n ∈ N such that p | f (m) + f (n). We obtain p | f (m + n) and p | f (m + 1) + f (n); hence p | f (m + n + 1). Thus, p | f (1), which contradicts (1). This proves that f (m + 1) = f (m) + 1. From (1) and (2) and the injectivity of f , it follows by Mathematical Induction that f (n) = n for every n ∈ N.

Next we turn to solutions to problems of the Bundeswettbewerb Mathematik 2006 given at [2010 : 22].

1. A circle is divided into 2n congruent sectors, n of them coloured black and the
remaining n sectors coloured white. The white sectors are numbered clockwise from 1 to n, starting anywhere. Afterwards, the black sectors are numbered counter clockwise from 1 to n, again starting anywhere. Prove that there exist n consecutive sectors having the numbers from 1 to n. Comment by Mohammed Aassila, Strasbourg, France. This problem appeared in the 20th Tournament of the Towns, Spring 1999, A-level, problem 4. The author is V. Proizvolov.

Let Q+ (resp. R+ ) denote the set of positive rational (resp. real) numbers. Find all functions f : Q+ → R+ that satisfy f (x) + f (y ) + 2xyf (xy ) = f (xy ) f (x + y ) for all x, y ∈ Q+ .

Solved by Michel Bataille, Rouen, France; and by Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA. We give Bataille’s solution. 1 We show that the only solution is x → 2 . x Let f satisfy the given functional equation, denoted by (E ) in what follows. 1 With x = y = 1, (E ) readily gives f (2) = 4 . Let a = f (1). With y = 1 in (E ), we obtain f (x) f (x + 1) = , (1) (2x + 1)f (x) + a which successively yields f (3) = f (2 + 1) =
1 . 4a2 + 5a + 7 1 and f (4) = f (3 + 1) = 4a + 5

1 . Comparing However, from (E ) with x = y = 2, we also deduce f (4) = 16 2 with (1), we see that 4a + 5a − 9 = 0, and, since a > 0, it follows that a = f (1) = 1.

Now, (1) rewrites as shows that

1 1 = + 2x + 1 and an easy induction f (x + 1) f (x)

1 1 = + 2nx + n2 f (x + n) f (x)
1

(2)

for all positive integers n and rationals x. Thus, f (n) = 2 for all n in N (x = 1 n in (2)) and 1 1 = + 2 + n2 . 1 f (1/n) f ( n + n) Comparing with the result given by (E ) with x = n and y = f and f to
1 n 1 , n

we have

1 n

− n2 −

1 n2

1 n

−1 =0

= n2 follows.
1 n

Lastly, if m, n ∈ N, taking x = m and y = 1 m2 + n2 + 2m n f m n
m n

in (E ) and using (2) lead + 2m n + m2
1 for all x2

= f =

m n

1 n2

and a short calculation gives f rational numbers x. numbers x, y .

n 2 . m

As a result, f (x) = 1 x2

Conversely, it is easily checked that x →

satisﬁes (E ) for all rational

3. The point P lies inside the acute-angled triangle ABC and C , A and B are the feet of the perpendiculars from P to AB , BC , CA. Find all positions of P such that ∠BAC = ∠B A C and ∠CBA = ∠C B A .
Solved by Titu Zvonaru, Com´ ane¸ sti, Romania.
.. ...... .. ....... . . .... . .... .. .... ... . .... . . . .... . . .... . . .... C ..... .... . . . .. ......... .... B . .. . .... ........... . . ..... . . . . . . .. ..... .............. P .......... . . . . . .... ......... ...... .... ...... . . . .... .. .. . . .. .... .. ... . . . . . . . .... . . ... . . . . . . . . . . . . . . .............. . . ........ . .. .... . . . . . ... . . . . . . . . . .... . . . . ........ . .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ........ ...... .... . . . . . . . . . . . . . . . . ........ .... ... . .. .... .......... . ........ .... .... .. . .. . ................................................... ....... ........... B ..... ........................ ................ ... . .. ................................. .. .... . .. ... C

Since the quadrilaterals P A BC and P A CB are cyclic, we have ∠BAC = ∠B A C = ∠B A P + ∠P A C

= ∠P CA + ∠P BA. It results that ∠BP C = 180◦ − ∠P BC − ∠P CB = 180◦ − (∠ABC − ∠P BA) − (∠BCA − ∠P CA)

= 180◦ − ∠ABC − ∠BCA + ∠P BA + ∠P CA = ∠BAC + ∠BAC = 2∠BAC,

hence ∠BP C = 2∠BAC . We deduce that the point P lies on an arc of a circle which passes through B and C . Similarly, we deduce that the point P lies on an arc of a circle which passes through A and C . It follows that there exists at most one point P satisfying the given condition. Since it is easy to see that the circumcentre satisﬁes the problem (the medial triangle is similar to the given triangle), we conclude that the desired point is the circumcentre.

And next we look at solutions for problems of the Bundeswettebwerb Mathematik 2007 given at [2010 : 22].

Show that one can distribute the integers from 1 to 4014 on the vertices and the midpoints of the sides of a regular 2007-gon so that the sum of the three numbers along any side is constant. Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA. Starting at any vertex number the vertices counterclockwise from v1 to v2007. Setting v2008 = v1 , label as mk the midpoint of side vk vk+1 . For k = 1, 2, . . . , 2007, assign the integer 2k − 1 to mk . For k = 1, 2, . . . , 1004, assign the integer 2k to v2009−2k , and for k = 1005, 1006, . . . , 2007, assign the integer 2k to v4016−2k. Then mj has the value 2j − 1, v2j has the value 4016 − 2j , and v2j +1 has the value 2008 − 2j . Hence, the sum on side v2j m2j v2j +1 is (4016 − 2j ) + (4j − 1) + (2008 − 2j ) = 6023, and the sum on side v2j −1 m2j −1 v2j is (20080(2j − 2)) + (4j − 3) + (4016 − 2j ) = 6023.

Each positive integer is coloured either red or green so that

(a) The sum of three (not necessarily diﬀerent) red numbers is red. (b) The sum of three (not necessarily diﬀerent) green numbers is green. (c) There is at least one green number and one red number.

Find all colourings that satisfy these conditions. Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and by Titu Zvonaru, Com´ ane¸ sti, Romania. We give Zvonaru’s response. Let R be the set of red numbers and G be the set of green numbers. Let a, b be two positive integers such that a = b, a ∈ R, b ∈ G. It is easy to see that a + a + a = 3a, and b + b + b = 3b, It results that R and G are inﬁnite sets. (1) b + b + 3b = 5b, b + b + 5b = 7b, . . . ∈ G. a + a + 3a = 5a, a + a + 5a = 7a, . . . ∈ R

To make a choice, we assume that 1 ∈ R: It follows that R contains all odd positive integers. If we suppose that the even integer 2k ∈ R, then an easy induction shows that every integer t, with t ≥ 2k belongs to R. We deduce that {t, t ≥ 2k} ⊂ R, hence the set G is ﬁnite — a contradiction with (1). It results that we have two possibilities: • (i) R is the set of all odd positive integers; G is the set of all even positive integers. • (ii) R is the set of all even positive integers; G is the set of all odd positive integers. Since 1 ∈ R and 2k ∈ R, then 1 + 1 + 2k = 2(k + 1) ∈ R and so on.

3. In triangle ABC the points E and F lie in the interiors of sides AC and BC (respectively) so that |AE | = |BF |. Furthermore, the circle through A, C and F and the circle through B , C and E intersect in a point D = C .
Prove that the line CD is the bisector of ∠ACB . Solved by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; and by Michel Bataille, Rouen, France. We give the solution of Amengual Covas.

32
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q F

Let the circle through B , C and E intersect AB at E = B . Let the circle through A, C , and F intersect AB at F = A. We denote by D the point where the line CD intersects AB . Observing that E , E , B and C are concyclic, as are F , F , A and C , we have AE · AC = AE · AB and BF · BC = BF · BA implying
AE · AB AE · AC = which simpliﬁes to BF · BC BF · BA

AC AE = BC BF

(1)

because AE = BF by hypothesis. Also, since E , E , B and C are concyclic, as are F , F , A, and C , then DD · D C = BD · D E and DD · D C = AD · D F implying AD · D F = BD · D E . Hence AD BD = = = = DE DF AD − D E

BD − D F AE BF AC BC

by (1).

By the converse of the internal angle bisector theorem, then, CD is the bisector of ∠ACB . That is, the line CD bisects ∠ACB .

Let a be a positive integer. How many nonnegative integers x satisfy x a
ë

x a+1

Solved by Michel Bataille, Rouen, France; Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Bataille. . We show that the required number is 2 By long division, we may write any nonnegative integer x as qa(a + 1) + r for some nonnegative integers q and r such that r < a(a +1). Using the fact that m + u = m + u for m ∈ Z and u ∈ R, the proposed equation becomes q+ Now,
r r ≥ , hence a a+1 r a a(a + 1)

r a

r a+1

(1)

Thus, the desired number is also the number of elements r A = {0, 1, 2, . . . , a(a + 1) − 1} such that r a
r a

≥

r a+1

and (1) cannot be satisﬁed if q ≥ 1. ∈ A with
î

r a+1

(2)

Any r ∈ A satisﬁes ka ≤ r < (k +1)a for some unique k ∈ {0, 1, 2, . . . , a} and then = k. Observing that such an r satisﬁes we see that this r is a solution if and only if
r ≥ k, that is r ≥ ka + k. As a+1 (k + 1)a r < < k + 1, a+1 a+1

a result, solutions r with ka ≤ r < (k + 1)a exist if and only if k ≤ a − 1, in which case the solutions are ka + k, ka + k + 1, . . . , ka + a − 1. Thus, for each k ∈ {0, 1, 2, . . . , a − 1}, we obtain (a − 1) − (k − 1) = a − k solutions and no other solution exists. In conclusion the number of solutions is (a − 0) + (a − 1) + · · · + (a − (a − 2)) + (a − (a − 1)) = a(a + 1) . 2

Next we move to the March 2010 number of the Corner and solutions to problems of the Republic of Moldova Selection tests for BMO 2007 and IMO 2007 given at [2010 : 81–83].

In triangle ABC the points M , N and P are the midpoints of the sides BC , AC and AB , respectively. The lines AM , BN and CP intersect the circumcircle of ABC at A1 , B1 and C1 , respectively. Prove that the area of the triangle ABC does not exceed the sum of the areas of the triangles BA1 C , AB1 C and AC1 B . Solved by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; by Michel Bataille, Rouen, France; by Prithwijit De, Homi Bhabha Centre for Science Education, Mumbai, India; and by Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of De.

Notation: [XY Z ] = Area of the triangle XY Z . ma = median on the side of a triangle with length a. We see at once that [BA1 C ] [ABC ] = M A1 AM , [AB1 C ] [ABC ] = N B1 BN , [AC1 B ] [ABC ] = P C1 CP .

Let BC = a, CA = b and AB = c. Chords AA1 and BC of the circumcircle of triangle ABC intersect at M . Therefore AM · M A1 = BM · M C . Also M . a and hence M A1 = is the midpoint of BC . Therefore BM = M C = 1 2 4AM Thus 2 M A1 a2 a . = = AM 4AM 2 2ma Similarly we can show that
N B1 = BN b 2mb
2

and

P C1 = CP

c 2mc

. Therefore ... (1)

[BA1 C ] [AB1 C ] [AC1 B ] 1 + + = [ABC ] [ABC ] [ABC ] 4 Recall that a2 b2 c2 Using these in (1) we obtain 1 4 where x = a2 m2 a + b2 m2 b + c2 m2 c = 1 9 = = = 4 9 4 9 4 9

a2 b2 c2 + + m2 m2 m2 b a c

2 2 (2(m2 b + mc ) − ma ) 2 2 (2(m2 c + ma ) − mb ) 2 2 (2(m2 a + mb ) − mc ).

2 x+y+z+

1 x

1 y

1 z

− 3 . . . , (2)

m2 a , m2 b

y =

m2 b , m2 c

and

z =
1 x

m2 c . m2 a ≥ 2, y +
1 y

1 1 +y +1 ) − 3 ≥ 9 because x + Now, 2(x + y + z + x z 1 z + z ≥ 2. Thus from (1) and (2) we can conclude that

≥ 2 and

[BA1 C ] + [AB1 C ] + [AC1 B ] ≥ [ABC ].

2. Let p be a prime number, p = 2, and m1 , m2 , . . . , mp positive consecutive integers, and σ a permutation of the set A = {1, 2, . . . , p}. Prove that the set A contains 2 distinct numbers k and l such that p divides mk · mσ(k) − ml · mσ(l) .
Solved by Prithwijit De, Homi Bhabha Centre for Science Education, Mumbai, India.

The residue classes modulo p of the p consecutive positive integers is C = {0, 1, 2, . . . , p − 1}. Let mk ≡ 0 (mod p) for some k ∈ A and σ (k) = k. Then there exists some l ∈ A, l = k such that σ (l) = k. Thus we obtain two distinct positive integers m and l such that p | (mk mσ(k) − ml mσ(l) ). Now suppose for some positive integer j ∈ A we have mj ≡ 0 (mod p) and σ (j ) = j . Consider the set A = A − {j }. Claim. There exist distinct positive integers k and l in A p | (mk mσ(k) − ml mσ(l) ). such that

To prove it assume on the contrary that no such pair of positive integers exists. Then {mr mσ(r) (mod p) : r ∈ A } = {1, 2, . . . , p − 1}. Now observe that mr mσ(r) ≡ (p − 1)! (mod p) . . . . (1)
r ∈A

Again observe that mr mσ(r) ≡ mr
r ∈A r ∈A

mσ(r)

(2) From (1) and (2) we conclude that (p − 1)! ≡ 1 (mod p) and this contradicts Wilson’s Theorem. Thus our assumption is wrong and the claim is correct.

r ∈A

≡ ((p − 1)!)2 (mod p)) . . . .

Inside the triangle ABC there exists a point T such that m(∠AT B ) = m(∠BT C ) = m(∠CT A) = 120◦ .

Prove that the Euler lines of the triangles AT B , BT C and AT C are concurrent. Solved by Titu Zvonaru, Com´ ane¸ sti, Romania.
A

Let M be the midpoint of BC , and let Ga be the centroid of the triangle BT C . Let A be the point on the other side of BC as A such that BA C is equilateral. It is known that the point T lies on AA . Since ∠BT C + ∠BA C = 180◦ , the quadrilateral BA CT is cyclic; it results that the circumcentre Oa of BA C is the circumcentre of BT C . We deduce that Oa Ga is the Euler line of BT C . Since 1 M Oa Ga M = = , we obtain that Oa Ga A T . Now, denoting G = Oa Ga ∩ AM , by similarity, GM O M 1 it results that = a = , hence G is the centroid of
GA Ga T 3 Oa A

ABC .

Oa A

M O

Similarly, we deduce that the Euler lines of CT A and AT B pass through the centroid of ABC , hence the three Euler lines are concurrent.

Determine the smallest positive integers m and k such that:

(a) there exist 2m + 1 consecutive positive integers whose sum of cubes is a perfect cube; (b) there exist 2k + 1 consecutive positive integers whose sum of squares is a perfect square. Solved by David E. Manes, SUNY at Oneonta, Oneonta, NY, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We use Manes’ solution. For part (a), m = 1 since there exist three consecutive positive integers whose cubes sum to a perfect cube; namely 33 + 43 + 53 = 63 . For part (b), we will show that k = 5. Let s(n, 2k + 1) = n2 + (n + 1)2 + · · · + (n + 2k)2 be the sum of the squares of 2k + 1 consecutive integers, the smallest of which is n. If k = 1, then s(n − 1, 3) = (n − 1)2 + n2 + (n + 1)2 = 3n2 + 2 ≡ 2 (mod 3) . Therefore, s(n − 1, 3) is not a perfect square since x2 is not congruent to 2 modulo 3 for any integer x. If k = 2, then s(n − 2, 5) = (n − 2)2 + (n − 1)2 + n2 + (n + 1)2 + (n + 2)2 = 5(n2 + 2) ≡ 2 or 3 (mod 4) . Thus, s(n − 2, 5) is not a perfect square since x2 ≡ 0 or 1 (mod 4) for every integer x. If k = 3, assume that s(n − 3, 7) = r 2 for some integer r . This equation reduces to 7(n2 + 4) = r 2 . Hence, 7 divides r so that r = 7t for some integer t. Therefore, n2 + 4 = 7t2 and so, n2 + 4 ≡ 0 (mod 7) or n2 ≡ 3 (mod 7), a contradiction since 3 is not a quadratic residue of 7. Thus, s(n − 3, 7) is not a perfect square. If k = 4, assume that s(n − 4, 9) = r 2 for some integer r . This equation reduces to 3(3n2 + 20) = r 2 . Therefore, 3 divides r so that r 2 = 9t2 for some integer t, whence 3n2 + 20 = 3t2 . Thus, 3 divides 20, a contradiction that shows s(n − 4, 9) is not a perfect square. If k = 5, assume that s(n − 5, 11) = m2 for some integer m. Then s(n−5, 11) = 11(n2 +10) = m2 implies that 11 divides m so that m2 = 112 t2 for some integer t. Therefore n2 + 10 = 11t2 or n2 − 1 ≡ 0 (mod 11). Thus, n ≡ ±1 (mod 11), and so n = 11j ± 1 for some integer j . Then s(n − 5, 11) = = 11(n2 + 10) = 11[(11j ± 1)2 + 10] 112 [11j 2 ± 2j + 1] = 112 [10j 2 + (j ± 1)2 ].

The problem now reduces to ﬁnding the smallest value of j so that 10j 2 +(j ± 1)2 is a perfect square. The value j = 1 is easily dispensed with. However, for j = 2, 10j 2 + (j + 1)2 = 72 and n = 11j + 1 = 23. Accordingly, s(18, 11) = 772 or 182 + 192 + 202 + · · · + 282 = 772 .

Let I be the incenter of triangle ABC and let R be the circumradius. Prove that AI + BI + CI ≤ 3R.

Solved by Arkady Alt, San Jose, CA, USA; Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; George Apostolopoulos, Messolonghi, Greece; Jos´ e Luis D´ ıazBarrero, Universitat Polit` ecnica de Catalunya, Barcelona, Spain; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the version of Amengual Covas. s=
a+b+c 2

We write a, b, c respectively for the lengths of the sides BC , CA, AB , and for the semiperimeter. Let r be the radius of the incircle.
A 2

Note that cos

may be expressed as

these and solving for AI , we get AI = and CI . → → Applying the Cauchy’s inequality − u · − v √ √ √ − → → − s −a s −b u = , , bc, ca, ab and v =
s s

− a) s −a , and also as s(s . Equating AI bc bc(s−a) , with symmetric results for BI s

s −c s

≤

− → u

, we now get
ab(s−c) s

− → v

with

AI + BI + CI

bc(s−a) −b) = + ca(s + s s √ ≤ ab + bc + ca

(1)
3 √ 3 R, 2

and s ≤ Using the relations ab + bc + ca = r 2 + 4Rr + s2 , r ≤ R 2 we have √ 2 R 2 R 3 3 ab + bc + ca ≤ = 9R2 + 4R · + R 2 2 2

(2)

By (1) and (2), we obtain the desired inequality. Since equality in (1) and (2) holds if and only if a = b = c, it holds in the required inequality if and only if ABC is equilateral. Comment. Also solved, by using the Erd˝ os-Mordell inequality, on page 38 of the book Experiences in Problem Solving: A W. J. Blundon Commemorative, Atlantic Provinces Council on the Sciences, Canada, (1994).

Let U and V be two points inside the angle BAC such that m(∠BAU ) = m(∠CAV ) .

Denote projections from U and V on the angle sides AC , AB as X1 , X2 and Y1 , Y2 respectively. Let W be the intersection of the lines X2 Y1 and X1 Y2 . Prove that U , V , W are collinear.

Solved by Titu Zvonaru, Com´ ane¸ sti, Romania.
A
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . .. . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Y X

Let the lines U V and AB intersect at M , and the lines U V and AC intersect at N . We denote α = ∠BAC, γ = ∠AN M, β = ∠AM N, ϕ = ∠BAV = ∠CAV.

Suppose that the lines X2 Y1 and M N intersect at the point W . We will prove that the points W , Y2 , X1 are collinear. By Menelaus’ theorem we obtain: WM WN · Y1 N Y1 A · X2 A X2 N = 1 ⇔ ⇔ WM WN WM · V N cos γ · AV cos ϕ = 1 (1)

AV cos ϕ M U cos β AV · M U cos β = WN AU · V N cos γ

By the converse of Menelaus’ theorem, we have to prove that WM WN · X1 N X1 A · Y2 A Y2 M = 1 ⇔ ⇔ WM WN WM · U N cos γ · AV cos(α − ϕ)

AU cos(α − ϕ) M V cos β AU · M V · cos β = . WN AV · U N · cos γ

By (1), it suﬃces to prove that AV · M U AU · V N = AU · M V ⇔ AU 2 AV
2

AV · U N

NV · V M

MU · UN

which is Steiner’s Theorem with respect to isogonal cevians.

Here a proof: We denote by [XY Z ] the area of MU NV · UN VM = = = [AM U ] · [AU N ]

XY Z . We have

[AV M ] [AN V ] AM · AU · sin ϕ

AN · AV sin(α − ϕ) AU 2 . AV 2

AU · AN sin(α − ϕ) AV · AM · sin γ

Let a1 , a2 , . . . , an (n ≥ 2) be real numbers in the interval [0, 1]. Let 3 3 S = a3 1 + a2 + · · · + an . Prove that
n i=1

ai 2n + 1 + S − a3 i

≤

1 3

Solved by Jos´ e Luis D´ ıaz-Barrero, Universitat Polit` ecnica de Catalunya, Barcelona, Spain; and by Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of D´ ıazBarrero. Since ai , 1 ≤ i ≤ n, lie in the interval [0, 1], then 1 − a3 i ≥ 0 for 1 ≤ i ≤ n, and n n ai ai ≤ . 3 2n + 1 + S − a i 2n + S i=1 i=1 So, it will suﬃce to prove that
n i=1

ai 2n + S

≤

1 3

or equivalently,
n

3(a1 + a2 + . . . + an ) ≤ 2n + S =

(1 + 1 + a3 i)
i=1

which trivially holds on account of AM-GM inequality. Indeed,
n i=1 n n

(1 + 1 + a3 i) ≥

3
i=1

1 · 1 · a3 i = 3

ai .
i=1

Equality holds when a1 = a2 = . . . = an = 1, and we are done.

10.

Find all polynomials f with integer coeﬃcients, such that f (p) is a prime for every prime p. Solved by David E. Manes, SUNY at Oneonta, Oneonta, NY, USA.

Note that the polynomial f (x) = x and the constant polynomials f (x) = c where c is a prime satisfy the requirements in the problem. They are the only such polynomials that do. Assume that f is a polynomial with integer coeﬃcients such that if p is a prime, then f (p) is a prime and also assume that f (x) = x and f (x) = c. Then there exists a prime π such that gcd(π, f (π )) = 1 since otherwise p divides f (p) for all primes p and it would follow that either f (x) = 0 or f (p) = p so that f (x) = x, both contradictions. By Dirichlet’s Theorem, there exist inﬁnitely many integers ni such that ni f (π ) + π is a prime, say ni f (π ) + π = qi , i = 1, 2, 3, . . . . Since qi − π divides f (qi ) − f (π ) we get that f (π ) divides f (qi ) for each i ≥ 1. Hence f (x) = f (π ), a contradiction. As a result, the only polynomials f with integer coeﬃcients for which f (p) is a prime for every prime p are f (x) = x and f (x) = c where c is a prime.

11.

Let ABC be a triangle with a = BC , b = AC , c = AB , inradius r and circumradius R. Let rA , rB and rC be the radii of the excircles of the triangle ABC . Prove that a2 2 rA − r rB rC + b2 2 rB − r rC rA + c2 2 rC − r rB rA = 4(R +3r ) .

Solved by Arkady Alt, San Jose, CA, USA; George Apostolopoulos, Messolonghi, Greece; Michel Bataille, Rouen, France; Geoﬀrey A. Kandall, Hamden, CT, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give Kandall’s write-up. Let K = [ABC ], s = 16K 2
a+b+c . 2

It is well known (Heron) that

= (a + b + c)(−a + b + c)(a − b + c)(a + b − c) = −(a4 + b4 + c4 ) + 2(a2 b2 + b2 c2 + a2 c2 )
abc

. and K = rs = rA (s − a) = rB (s − b) = rC (s − c) = 4R Let X = a2 We have a2 2 r − rA rB rC = a2 = 2(s − b) K s−b s−c − · · K s K K a2 a+b+c a−b+c 2· · sK 2 2 a−b+c a+b−c − · 2 2 1 (−3a4 + 3a2 b2 + 3a2 c2 + 2a2 bc). 4sK 2 rA − r rB rc = b2 2 rB − r rC rA + c2 2 rC − r rB rA .

Similarly, b2 c2 2 rB − r rC rA = = 1 4sK (−3b4 + 3a2 b2 + 3b2 c2 + 2ab2 c),

2 r − rC rB rA Consequently, X = 1 4Sk

1 (−3c4 + 3a2 c2 + 3b2 c2 + 2abc2). 4sK

(−3(A4 + B 4 + C 4 )

= = as required.

+6(A2 B 2 + B 2 C 2 + A2 C 2 ) + 2ABC (A + B + C )) 1 K 3 · 16K 2 + 2 · 4RK · 2 K r 4( r )K 4(R + 3r ),

12.

number r (n) of segments of length l is maximized. Prove that r (n) ≤ Solved by Oliver Geupel, Br¨ uhl, NRW, Germany.

Consider n distinct points in the plane n ≥ 3, arranged such that the
n2 . 3

We will apply Tur´ an’s Theorem: Let G be a simple graph with n vertices which does not contain a complete subgraph containing p vertices. Let r be the remainder of n modulo p. Then the number of edges of G is not greater than f (n, p) = (p − 2)n2 − r (p − 1 − r ) 2(p − 1) .

Consider the graph G whose vertices are the n given points and where two vertices P and Q are connected by an edge if and only if P Q = l. Clearly, G does not contain a complete subgraph with 4 vertices. By Tur´ an’s Theorem, the number of edges of G is not greater than f (n, 4) = 2n2 − r (3 − r ) 6 ≤ n2 3 .

Remark. Problem A-6 of the 49th William Lowell Putnam Competition (1978) asked for proving the inequality r (n) < 2n3/2 . This is a sharper upper bound for each n ≥ 36. See: The William Lowell Putnam Mathematical Competition problems and solutions: 1965-1984, ed. by G.L. Alexanderson, L.F. Klosinski, and L.C. Larson, MAA, 1986, p. 104f.

14.

Let b1 , b2 , . . . , bn (n ≥ 1) be nonnegative real numbers at least one of which is positive. Prove that P (X ) = X n − b1 X n−1 − · · · − bn−1 X − bn , has a single positive root p, which is simple, and that the absolute value of each root of P (X ) is not greater than p.

Solved by Jos´ e Luis D´ ıaz-Barrero, Universitat Polit` ecnica de Catalunya, Barcelona, Spain; and Oliver Geupel, Br¨ uhl, NRW, Germany. We give the solution by D´ ıazBarrero. We consider the continuous function A : (0, +∞) → R deﬁned by A(x) = Since A (x) = − b1 − 2b2 x3 b1 x + b2 x2 + ... + bn xn −1

< 0 for all x > 0, then A is a decreasing xn+1 continuous function. Furthermore, lim A(x) = −1 and lim A(x) = +∞. x2
x→+∞ x→0+

−. . . −

nbn

P (x) = 0 xn has only one positive root, say p, which is a zero of polynomial P . On the other hand, from A (p) < 0 follows P (p) > 0 and p is a simple zero. To see that all the zeros of P have modulus less than or equal to p we argue by contradiction. Assume that x0 is a zero of P and let |x0 | = α with α > p. Then A(α) < A(p) = 0 and P (α) > 0. On the other hand from n− 1 xn we have 0 = bn + bn−1 x0 + . . . + b1 x0 Therefore, on account of Bolzano’s theorem the equation A(x) = −
n− 1 −1 | ≤ bn + bn−1 |x0 | + . . . + b1 |xn | | xn 0 0 | = |bn + bn−1 x0 + . . . + b1 x0

from which follows P (α) = αn − b1 αn−1 − . . . − bn−1 α − bn < 0 Contradiction, and we are done. Comment. The ﬁrst part of the statement also follows immediately applying the well-known Descartes Rule of Signs.

15. A circle is tangent to the sides AB and AC of the triangle ABC and to its circumcircle at P , Q and R respectively. If P Q ∩ AR = {S } prove that m(∠SBA) = m(∠SCA).
Solved by Oliver Geupel, Br¨ uhl, NRW, Germany. Let Γ be the circle through A, B , and C , and let ∆ be the circle through P , Q, and R. Let D be the second intersection of ∆ and the line AR. Since AP is tangent to ∆, we have ∠P DR = ∠BP R. Since ∆ and Γ have a common tangent t at R, it holds ∠DP R = ∠(AR, t) = ∠ABR = ∠P BR. Hence, the triangles DP R and P BR are similar. It follows that ∠ARP = ∠P RD = ∠BRP , that is, the line RP is the internal bisector of ∠R in ABR.
. . . . . . . . . . . . . . . . . . . . . Γ . . . . . . . . . . . . .

∆

. ... .. .

.. . .. . .

Therefore,

AP BP

AR BR AR CR

Similarly,

AQ CQ

By AP = AQ, we obtain BP BR By PS QS we deduce that sin ∠P AS sin ∠QAS BP CQ By ∠BP S = ∠CQS , it follows that = CQ CR .

sin ∠BAR sin ∠CAR PS QS .

BR CR

BP S ∼

CQS . Consequently,

∠SBA = ∠SBP = ∠SCQ = ∠SCA, which completes the proof.

16. Prove that there are inﬁnitely many primes p for which there exists a positive
integer n such that p divides n! + 1 and n does not divide p − 1. Comment by David E. Manes, SUNY at Oneonta, Oneonta, NY, USA. A proof is by Paul Erd˝ os, (c.f. p. 558 of G.E. Hardy and M.V. Subbarao, “A modiﬁed Problem of Pillai and Some Related Questions”, The American Math. Monthly, Vol 109, pp. 554–559).

BOOK REVIEWS
Amar Sodhi
Alex’s Adventures in Numberland by Alex Bellos Bloomsbury Publishing, 2010 ISBN-13: 978-0747597162, hardcover, 448 pages, $30.95 Reviewed by Bruce Shawyer, Memorial University of Newfoundland, St. John’s, NL Alex’s Adventures in Numberland is a book about mathematics aimed at readers of all abilities. It was published in March 2010 in the UK by Bloomsbury. In the US the book appeared in June with the title Here’s Looking at Euclid, published by Free Press. The editions have diﬀerent covers. I have the UK edition (448 pages) and have not seen the US edition (320 + xi pages). Both are available in Canada via the usual retailers. The US edition of the book only contains a brief bibliography. Thanks to the wonders of the internet, the author has been able to make available on his WEB site (alexbellos.com) the complete chapter-by-chapter bibliography, with comments and suggested further reading. It is enhanced with links! Here are some of the diﬀerence between the two editions, gleaned from the WEB site: 1. The UK edition has a cartoon preceding each chapter. 2. Both the British and American publishers felt that their respective titles worked best for their respective audiences. 3. The American one is slightly shorter (the section on the British 50p piece is omitted, for example, since the shape means nothing in Arkansas or Wyoming) and it has less diagrams overall. 4. The British version also has a 12-page colour plate section. The author, Alex Bellos (with degrees in Mathematics and Philosophy) is a writer, broadcaster, football (soccer) lover and self-proclaimed math geek, with a colourful career including several books and short ﬁlms. The book consists of mathematics (real mathematics) for the lay person. But yet, it is mathematics for the mathematics student and for the mathematics teacher at all levels. It covers a very broad range of topics, which are all related in an engaging narrative style. For example, Bellos describes how Yorkshire shepherds count (his list agrees with the song by Jake Thackray); why business card origami is abhorrent to the Japanese; a mnemonic for the digits of π which is a remarkable modernist pastiche of Poe’s The Raven; and why we most commonly x as the name of a variable. We meet fanatics, crackpots, anthropologists and gurus as well as a few mathematicians (such as Aitken, Brahmagupta, Cantor, Descartes, Euler and Fibonacci).

All of the mathematics (with one exception) is well explained and correct. It is a pity that he has an error in the penultimate paragraph of the book. Here, he describes inﬁnite cardinals, and makes two claims: ﬁrst, that the number of curves in the plane is larger than c, the cardinality of the continuum (this is false if curves must be continuous; there are only c of them); and second, that nobody has been able to come up with a larger set (Cantor proved that the set of subsets of any set is larger than the original set - perhaps Alex meant a larger “naturally-occurring” set). Despite this, we have a book well worth reading. As I read it, I thought that it might well form the basis for an elementary course on the History of Mathematics. Now there is something that every student of mathematics should study. This book puts much of mathematics into context, and so, will encourage students to want to know more. Editor’s note: the North American edition, Here’s Looking at Euclid: A Surprising Excursion Through the Astonishing World of Math (ISBN 978-1-41658825-2) retails at $32.99.

Einstein, Albert (1879-1955) A human being is a part of the whole, called by us “Universe”, a part limited in time and space. He experiences himself, his thoughts and feelings as something separated from the rest, a kind of optical delusion of his consciousness. This delusion is a kind of prison for us, restricting us to our personal desires and to aﬀection for a few persons nearest to us. Our task must be to free ourselves from this prison by widening our circle of compassion to embrace all living creatures and the whole of nature in its beauty. Nobody is able to achieve this completely, but the striving for such achievement is in itself a part of the liberation and a foundation for inner security. In H. Eves “Mathematical Circles Adieu”, Boston: Prindle, Weber and Schmidt, 1977.

PROBLEMS
Solutions to problems in this issue should arrive no later than 1 August 2010. An asterisk ( ) after a number indicates that a problem was proposed without a solution. Each problem is given in English and French, the oﬃcial languages of Canada. In issues 1, 3, 5, and 7, English will precede French, and in issues 2, 4, 6, and 8, French will precede English. In the solutions’ section, the problem will be stated in the language of the primary featured solution. The editor thanks Jean-Marc Terrier of the University of Montreal for translations of the problems.

3601.

Proposed by Bill Sands, University of Calgary, Calgary, AB.

Suppose that b is a positive real number such that there are exactly two integers strictly between b and 2b, and exactly two integers strictly between 2b and b2 . Find all possible values of b.

3602.
Vietnam.

Proposed by Pham Van Thuan, Hanoi University of Science, Hanoi,

Prove that if ai > 0 for i = 1, 2, 3, 4, then 1
2 a2 + a2 i+1 + ai+2 cyclic i

≥

12 (a1 + a2 + a3 + a4 )2

3603.

Proposed by George Apostolopoulos, Messolonghi, Greece.

. Let D and E be points Let ABC be a given triangle and 0 < λ < 1 2 on AB such that AD = BE = λ · AB , and F , G points on AC such that AF = CG = λ · AC . Let BF ∩ CE = H and BG ∩ CD = I . Show that i) HI BC and 1 − 2λ BC . −λ+1

ii) HI =

λ2

3604.

Proposed by Michel Bataille, Rouen, France.
Ê1

Evaluate (x2 − x − 2)n dx lim Ê 10 . 2 n n→∞ 0 (4x − 2x − 2) dx

3605.

Proposed by Jos´ e Luis D´ ıaz-Barrero, Universitat Polit` ecnica de Catalunya, Barcelona, Spain. Let A(x) = xn + an−1 xn−1 + · · · + a1 x + 1 be a real polynomial with positive coeﬃcients and having all its zeros real. Prove that
n

A(1)A(2) · · · A(n) ≥ (n + 1)!

3606.

Proposed by V´ aclav Koneˇ cn´ y, Big Rapids, MI, USA.

Let ABC be a triangle with ∠A = 20◦ . Let BD be the angle bisector of ∠ABC with D on AC . If AD = DB + BC , determine ∠B .

3607.

Proposed by George Miliakos, Sparta, Greece.

Let c1 = 9, c2 = 15, c3 = 21, c4 = 25, . . ., where cn is the nth composite odd integer. Evaluate cn . lim n→∞ cn+1

3608.
Let

Proposed by Michel Bataille, Rouen, France. e1/x − 1

f (x) = (a) Show that for all x ∈ (0, ∞),

e1/(x+1)
Ö

−1

f (x) > (b) Prove or disprove: f (x) < for all x ∈ (1, ∞).

x+1 x

x+1 x−1

3609.
Italy.

Proposed by Panagiote Ligouras, Leonardo da Vinci High School, Noci,

Let r be a real number. and let D, E , and F be points on the sides BC, CA, and AB of a triangle ABC with BD DC = CE EA = AF FB = r.

The cevians AD, BE , and CF bound a triangle P QR whose area we denote by DEF ] equals 4. [P QR]. Find the value of r for which the ratio of the areas, [ [P QR]

3610.

Proposed by Peter Y. Woo, Biola University, La Mirada, CA, USA.

Let S = {2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, . . .} be the set of positive integers whose only prime divisors are 2 or 3. Let a1 = 2, a2 = 3, . . . , be the elements of S , with a1 < a2 < . . . .
∞ i=1

(i) Determine

1 ai

(ii)

For each positive integer n, let s(n) be the sum of all its divisors inn) cluding 1 and n itself. Prove s( < 3 for all members of S . n

3611. Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania.
Given x, y , and z are positive integers such that z (x + 1) x(y + 1) y (z + 1) , , and, x−1 y−1 z−1 are positive integers. Find the smallest positive integer N such that xyz ≤ N .

3612.

Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.

Find all nonconstant polynomials P such that P ({x}) = {P (x)}, for all x ∈ R, where {a} denotes the fractional part of a. .................................................................

3601.

Propos´ e par Bill Sands, Universit´ e de Calgary, Calgary, AB.

On suppose que b est un nombre r´ eel positif tel qu’il existe exactement deux entiers strictement compris entre b et 2b, de mˆ eme qu’exactement deux entiers strictement compris entre 2b et b2 . Trouver toutes les valeurs possibles de b.

3602.
Vietnam.

Propos´ e par Pham Van Thuan, Universit´ e de Science de Hano¨ ı, Hano¨ ı,

Montrer que si ai > 0 pour i = 1, 2, 3, 4, alors 1 a2 cyclique i + a2 i+1 + a2 i+2 ≥ 12 (a1 + a2 + a3 + a4 )2

3603.

Propos´ e par George Apostolopoulos, Messolonghi, Gr` ece.

. Soit D et E deux points sur Soit ABC un triangle et 0 < λ < 1 2 AB tels que AD = BE = λ · AB , F et G deux points sur AC tels que AF = CG = λ · AC . Soit BF ∩ CE = H et BG ∩ CD = I . Montrer que i) HI BC et

ii) HI =

λ2

1 − 2λ BC . −λ+1

3604.

Propos´ e par Michel Bataille, Rouen, France.

Calculer (x2 − x − 2)n dx lim Ê 10 . 2 n n→∞ 0 (4x − 2x − 2) dx
Ê1

3605.

Propos´ e par Jos´ e Luis D´ ıaz-Barrero, Universit´ e Polytechnique de Catalogne, Barcelone, Espagne. Soit A(x) = xn + an−1 xn−1 + · · · + a1 x +1 un polynˆ ome r´ eel a ` coeﬃcients positifs n’ayant que des racines r´ eelles. Montrer que
n

A(1)A(2) · · · A(n) ≥ (n + 1)!

3606.

´ . Propos´ e par V´ aclav Koneˇ cn´ y, Big Rapids, MI, E-U

Soit ABC un triangle avec ∠A = 20◦ . Soit BD la bissectrice de l’angle au sommet B avec D sur AC . Si AD = DB + BC , trouver ∠B .

3607.

Propos´ e par George Miliakos, Sparte, Gr` ece.

Soit c1 = 9, c2 = 15, c3 = 21, c4 = 25, . . ., o` u cn d´ esigne le ne entier impair non premier. Calculer
n→∞

lim

cn cn+1

3608.

Propos´ e par Michel Bataille, Rouen, France. e1/x − 1 . e1/(x+1) − 1
Ö

Soit f (x) =

(a) Montrer que pour tout x ∈ (0, ∞), f (x) >

x+1 x

(b)

Trouver si oui ou non, on a f (x) < pour tous les x ∈ (1, ∞) x+1 x−1

3609.

´ Propos´ e par Panagiote Ligouras, Ecole Secondaire L´ eonard de Vinci, Noci, Italie. Soit r un nombre r´ eel et D, E et F des points sur les cˆ ot´ es BC, CA et AB d’un triangle ABC avec BD DC = CE EA = AF FB = r.

Les c´ eviennes AD, BE et CF limitent un triangle P QR dont on d´ esigne l’aire DEF ] des aires est par [P QR]. Trouver la valeur de r pour laquelle le rapport [ [P QR] ´ egal a ` 4.

3610.

´ . Propos´ e par Peter Y. Woo, Universit´ e Biola, La Mirada, CA, E-U

Soit S = {2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, . . .} l’ensemble des entiers positifs dont les seuls diviseurs premiers sont 2 ou 3. Notons a1 = 2, a2 = 3, . . . les ´ el´ ements de S , avec a1 < a2 < . . . . (i) Trouver
∞ i=1

1 . ai

(ii)

Pour chaque entier positif n, soit s(n) la somme de tous ses diviseurs, y n) compris 1 et n lui-mˆ eme. Montrer que s( < 3 pour tous les ´ el´ ements de n S.

3611.

´ Propos´ e par Neculai Stanciu, Ecole secondaire George Emil Palade, Buz˘ au, Roumanie. On donne trois entiers positifs x, y et z tels que x(y + 1) y (z + 1) z (x + 1) , et x−1 y−1 z−1 sont des entiers positifs. Trouver le plus petit entier positif N tel que xyz ≤ N .

3612.

Propos´ e par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie.

Trouver tous les polynˆ omes non constants P tels que P ({x}) = {P (x)} pour tout x ∈ R, o` u {a} d´ esigne la partie fractionnaire de a.

Just a reminder, it makes it easier for us if problem proposals and solutions A are sent to us in electronic format. Material sent in TEX or L TEX is preferred, but we will also accept pdf, Microsoft Word as well as handwritten material (mailed or scanned). When sending electronic solutions, please name the ﬁles in a meaningful way to identify yourself and the problem. For example, if I was to submit a solution to problem 3603 from this issue, I would name it February 3603 Godin.tex. Please place each solution on its own separate sheet(s) with the problem number, your name and aﬃliation on each page. Multiple solutions on one page means we have to do lots of photocopying and it increases the chances that something will get overlooked or misﬁled. As always no problem is ever closed. We always accept new solutions and generalizations to past problems. Also, in the last issue [2010 : 545, 547], Chris Fisher published a list of unsolved problems from Crux. Below is a sample of one of these unsolved problems:

609

. [1981 : 49; 1982 : 27-28] Proposed by Ian June L. Garces, Ateneo de Manila University, The Philippines.

A1 B1 C1 D1 is a convex quadrilateral inscribed in a circle and M1 , N1 , P1 , Q1 are the mid-points of sides B1 C1 , C1 D1 , D1 A1 , A1 B1 , respectively. The chords A1 M1 , B1 N1 , C1 P1 , D1 Q1 meet the circle again in A2 , B2 , C2 , D2 , respectively. Quadrilateral A3 B3 C3 D3 is formed from A2 B2 C2 D2 as the latter was formed from A1 B1 C1 D1 , and the procedure is repeated indeﬁnitely. Prove that quadrilateral An Bn Cn Dn “tends to” a square as n → ∞. What happens if A1 B1 C1 D1 is not convex? Enjoy!

SOLUTIONS
No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems.

3501.

[2010 : 44, 46] Proposed by Hassan A. ShahAli, Tehran, Iran.

Let N be the set of positive integers, E the set of all even positive integers, and O the set of all odd positive integers. A set S ⊆ N is closed if x + y ∈ S for all distinct x, y ∈ S , and unclosed if x + y ∈ S for all distinct x, y ∈ S . Prove that if N is partitioned into A and B , where A is closed and nonempty, and B is unclosed and inﬁnite, then A = E and B = O . Solution by Harry Sedinger, St. Bonaventure University, St. Bonaventure, New York, USA. We prove ﬁrst that 1 ∈ B . Assume by contradiction that 1 ∈ A. Chose m, n ∈ B then m + n ∈ A. But since 1 ∈ A, it follows that all integers greater than m + n are in A, which contradicts B inﬁnite. We prove now that 2 ∈ A. Again assume by contradiction that 2 ∈ B . Chose n > 2, n ∈ B . Then 3 ∈ A, n + 1 ∈ A, n + 2 ∈ A and hence n + 4, n + 5 ∈ A. Then 2n + 5, 2n + 6, 2n + 7 ∈ A, and since 3 ∈ A, it follows that A contains all the integers greater than 2n + 5, which contradicts B inﬁnite. Now we show that 3 ∈ B . Assume by contradiction 3 ∈ A. Then 5 ∈ A, 7 ∈ A, 8 ∈ A, and then, since 2 ∈ A, it follows that A contains all the integers greater than 7. Again this contradicts B inﬁnite. Next we prove that neither A nor B contains two consecutive integers. If n, n + 1 are in A, then n ≥ 4, and since 2 ∈ A, it follows that A contains all the integers greater than n, a contradiction. Assume now that B contains two consecutive integers n, n + 1. Since B is inﬁnite, there exists m ∈ B ; m > n + 1. But then A contains the consecutive integers m + n, m + n + 1, a contradiction. Hence 1 ∈ B, 2 ∈ A and neither A nor B contains two consecutive integers. Then it follows by induction that 2k − 1 ∈ B, 2k ∈ A for all k ≥ 1.
ˇ Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; SEFKET ´ University of Sarajevo, Sarajevo, Bosnia and Herzegovina; ROY BARBARA, ARSLANAGI C, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, France; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; JOSEPH DiMURO, Biola University, La Mirada, CA, USA; IAN JUNE L. GARCES, Ateneo de Manila University, Quezon City, The Philippines; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; JOHAN GUNARDI, student, SMPK 4 BPK PENABUR, Jakarta, Indonesia; MICHAEL JOSEPHY, Universidad de Costa Rica, San Pedro, Costa Rica; KATHLEEN E. LEWIS, SUNY Oswego, Oswego, NY, USA; MISSOURI STATE UNIVERSITY PROBLEM SOLVING GROUP, Springﬁeld, MO, USA; JOEL SCHLOSBERG, Bayside, NY, USA; DIGBY SMITH, Mount Royal University, Calgary, AB; EDMUND SWYLAN, Riga, Latvia; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. There was one incorrect solution.

3502. [2010 : 44, 46]

Proposed by Jos´ e Luis D´ ıaz-Barrero, Universitat Polit` ecnica de Catalunya, Barcelona, Spain. Find all real solutions of the following system of equations x2 1 + x2 2 + x2 n + x2 2 + 21 x2 3 + 21 ··· x2 1 + 21 = = ··· = x2 2 + 77 , x2 3 + 77 , ··· x2 1 + 77 .

Similar solutions by Michel Bataille, Rouen, France ; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give Bataille’s write up. √ √ Let ui = x2 x + 77 − x + 21. Then, the equations are: i and f (x) = u1 = f (u2 ) ; u2 = f (u3 ) ; ... ; un−1 = f (un ) ; un = f (u1 ) .
1 f (x) = 2

We observe that f ([0, ∞)) ⊂ [0, ∞) and f (4) = 4.
√ 1 x+77

In addition

−

√ 1 x+21

, hence

|f (x)| <

1 1 ≤ √ . √ 2 x + 21 2 21

1 and f m denote the composition f ◦ f ◦ f ◦ ... ◦ f . Then Let k = 2√ 21 k < 1 and it follows by induction that

|(f m ) (x)| < km , for all x ∈ [0, ∞) and m positive integer. Let 1 ≤ i ≤ n. Then ui ∈ [0, ∞) and f n (ui ) = ui . We show that u i = 4. Assume by contradiction that ui = 4. Then by the Mean Value Theorem, there exists a c between ui and 4 so that |ui − 4| = |(f n )(ui ) − (f n )(4)| = |(f n ) (c)| |ui − 4| ≤ kn |ui − 4| . But this contradicts k < 1. This shows that ui = 4 for all i. Conversely u1 = u2 = .. = un = 4 is obviously a solution for our system. Thus, all the real solutions of the system are (±2, ±2, ..., ±2).

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; ROY BARBARA, Lebanese University, Fanar, Lebanon; BRIAN D. BEASLEY, Presbyterian College, Clinton, SC, USA; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; JOEL SCHLOSBERG, Bayside, NY, USA; ALBERT STADLER, Herrliberg, Switzerland; EDMUND SWYLAN, Riga, Latvia; and the proposer. One incomplete solution was submitted.

3503.

[2010 : 44, 47] Proposed by Bruce Shawyer, Memorial University of Newfoundland, St. John’s, NL. Given a triangle and the midpoints of its sides, with the use of a straight edge and only three uses of a pair of compasses, bisect all three angles of the triangle. Solution by Mohammed Aassila, Strasbourg, France. Soit ABC le triangle donn´ e ; I , J et K les milieux respectifs de BC , CA et AB . On va utiliser une seule fois le compas. On trace le cercle de centre K et de rayon KA = KB . Soit E le point d’intersection de la droite KJ avec ce cercle. Comme KE = KB , alors E est sur la bissectrice de l’angle ABC car KBE triangle isoc` ele et KJ parall` ele a ` BC . De mˆ eme, soit F le point d’intersection du cercle avec la droite KI , alors on a KF = KA, et donc F est sur la bissectrice de l’angle BAC car KAF triangle isoc` ele et KI est parall` ele a ` AC . Ces deux bissectrices se coupent en H , centre du cercle inscrit et donc sur la troisi` eme bissectrice CH . Les trois bissectrices sont donc AH , BH et CH .
Also solved by the following readers, with the number of uses of the compass indicated in parentheses: MICHEL BATAILLE, Rouen, France (2); OLIVER GEUPEL, Br¨ uhl, NRW, Germany (2); JOHAN GUNARDI, student, SMPK 4 BPK PENABUR, Jakarta, ´ ˇ Y, ´ Indonesia (3); GEOFFREY A. KANDALL, Hamden, CT, USA (2); V ACLAV KONE CN Big Rapids, MI, USA (2); MISSOURI STATE UNIVERSITY PROBLEM SOLVING GROUP, Springﬁeld, MO, USA (2); EDMUND SWYLAN, Riga, Latvia (1); PETER Y. WOO, Biola University, La Mirada, CA, USA (2); TITU ZVONARU, Com´ ane¸ sti, Romania (2); and the proposer (3). Swylan was the only other solver who used the compass just once. Bataille, the Missouri State University Problem Solving Group, and Woo all noted that by the Poncelet–Steiner theorem just one use of the compass suﬃces to carry out the construction.

3504.

Proposed by Mariia Rozhkova, Kiev, Ukraine.

Given triangle ABC , set Q = a cos2 A + b cos2 B + c cos2 C , and let ABC have area S and circumradius R. Prove that
S , with equality if and only if ABC is equilateral. R √ S 2 (b) Q ≤ if ABC is not obtuse, with equality if and only if ABC is an R

(a) Q ≥

isosceles right triangle.

Solution to part (a) by Michel Bataille, Rouen, France; solution to part (b) by the proposer, modiﬁed by the editor. (a) Let a = BC , b = CA, c = AB and let I and H be the incentre and orthocentre of ABC . Since 2sI = aA + bB + Cc, we have

55 − → − − → − − → − − → 2sHI = aHA + bHB + cHC , and so 4s2 HI 2 = a2 HA2 + b2 HB 2 + c2 HC 2 − − → − − → − − → − − → − − → − − → + 2abHA · HB + 2bcHB · HC + 2caHC · HA = a2 HA2 + b2 HB 2 + c2 HC 2 + ab(HB 2 + HA2 − c2 ) + bc(HC 2 + HB 2 − a2 ) + ca(HA2 + HC 2 − b2 ) = 2s(aHA2 + bHB 2 + cHC 2 − abc) . Now, if A is the midpoint of BC , then HA2 = (2OA )2 = 4R2 cos2 A and similar results hold for HB 2 and HC 2 . It follows that 2sHI 2 = 4R2 · Q − abc and sHI 2 abc + . Q= 4R2 2R2 Since = , we see that Q ≥ , with equality if and only if H = I , that 4R2 R R is, if and only if ABC is equilateral. (b) Substituting the well-known relations a = 2R sin A, b = 2R sin B , c = 2R sin C , S = 2R2 sin A sin B sin C into the inequality and canceling 2R from each side yields the equivalent inequality √ sin A cos2 A + sin B cos2 B + sin C cos2 C ≤ 2 sin A sin B sin C .
1 Assume that A ≥ B ≥ C and set φ = 2 (B + C ), ψ = π π ≤ φ ≤ , and the hypotheses we have 0 ≤ ψ ≤ 1 (B 2

abc

− C ). Then from

A = π − 2φ ,

B = φ+ψ,

C = φ−ψ.

In terms of φ and ψ the desired inequality takes the form sin 2φ cos2 2φ + cos2 (φ + ψ ) sin(φ + ψ ) + cos2 (φ − ψ ) sin(φ − ψ ) √ ≤ 2 sin 2φ sin(φ + ψ ) sin(φ − ψ ) . In this inequality make the replacements sin 2φ = 2 sin φ cos φ, cos 2φ = cos2 φ − sin2 φ, cos(φ ± ψ ) = cos φ cos ψ ∓ sin φ sin ψ , and sin(φ ± ψ ) = sin φ cos ψ ± cos φ sin ψ , then expand each side and cancel like terms, then cancel a common term sin φ > 0 from each side, then apply the identities sin2 x = 1 − cos2 x for x = φ, ψ . This yields the equivalent inequality + (1 − cos2 φ) cos ψ (1 − cos2 ψ ) − 2 cos2 φ(1 − cos2 ψ ) cos ψ √ √ ≤ 2(1 − cos2 φ) cos φ cos2 ψ − 2 cos3 φ(1 − cos2 ψ )
å

cos3 φ − (1 − cos2 φ) cos φ + cos2 φ cos3 ψ

Making these substitutions and simplifying yields the equivalent inequality √ √ x3 (2 + 2) + 4x2 y 3 − 3x2 y − 2xy 2 − x − y 3 + y ≤ 0 .

Set x = cos φ, y = cos ψ , so x ∈

√ è å√ è 1 2 2 , and y ∈ I = ,1 . 2 2 2

56 √ Let t = x 2. Then t and y are in the interval I , and we need to prove that √ 2+ 2 3 t f (t, y ) = √ + 2t2 y 3 − t2 y − ty 2 − √ − y 3 + y ≤ 0 . 2 2 2 2 We have
1 1 1 y + = (y − 1) y 2 − 2 2 2 √ 2 with equality only for y = 1 and y = . 2

f (1, y ) = y 3 − y 2 −

≤ 0,

y ∈I,

Next we prove that f (t, y ) ≤ f (1, y ). Let g (y ) = 4y 3 − 2y 2 − 3y + 1 and h(y ) = 4y 3 − 3y . We then have ä √ ç √ 2[f (t, y ) − f (1, y )] = (t − 1) t2 ( 2 + 1) + t( 2 + 1) + th(y ) + g (y ) . (1)

Since t − 1 ≤ 0, it suﬃces to prove that √ √ t2 ( 2 + 1) + t( 2 + 1) + th(y ) + g (y ) > 0 . Note that (1) follows from √ t( 2 + 1) + g (y ) ≥ 1 , √ t( 2 + 1) + h(y ) ≥ 1 ,

(2) (3)

2 √ √ 2 2 , so g (y ) is strictly increasing on I and g (y ) ≥ g 2 √ 2 √ √ 2 2 , so that t( 2 + 1) ≥ 1 + , and (2) follows. t≥ 2 2
2

since the left side of (1) is t times the left side of (3) plus the left side of (2). √ √ 1 + 10 1 ± 10 and < The quadratic equation g (y ) = 0 has roots y =

√ 2 − 2 = . Also, 2

Furthermore, h(y ) ≥ g (y ) for y ∈ I , since h(y ) = g (y ) + (2y 2 − 1) and 2y − 1 ≥ 0 on I , so (3) follows from (2). This completes the proof of the inequality in part (b). Equality holds in f (t, y ) − f (1, y√ ) ≤ 0 only for t = 1, and equality holds
2 . These cases correspond to a triangle 2 π π π with A = and B = C = , or to a degenerate triangle with A = B = 2 4 2

in f (1, y ) ≤ 0 only for y = 1 and y =

and C = 0.

ˇ Part (a) also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; SEFKET ´ University of Sarajevo, Sarajevo, Bosnia and Herzegovina; SCOTT BROWN, ARSLANAGI C, Auburn University, Montgomery, AL, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; JOE HOWARD, Portales, NM, USA; THANOS MAGKOS, 3rd High School of Kozani, Kozani, Greece; and the proposer. One incomplete solution to part (a), one incorrect solution to part (a), and three incomplete solutions to part (b) were received. The proposer said she was inﬂuenced by Crux problem 3167, which asked to show that a cos3 A+b cos3 B +c cos3 C ≤ abc/4R2 holds for non-obtuse triangles ABC . She indicated that the inequality in part (a) occurs in a diﬀerent form on p. 14 of V.P. Soltan and I. Majdan’s book Identities and Inequalities in a Triangle, Kishinev, 1982 (Russian), although the proof there is of a general nature and diﬀerent from the one she constructed.

3505. [2010 :

45, 47, 107, 109] Proposed by Yakub N. Aliyev, Qafqaz University, Khyrdalan, Azerbaijan. The circles Γ1 and Γ2 have a common centre O , and Γ1 lies inside Γ2 . The point A = O lies inside Γ1 ; a ray not parallel to AO that starts at A intersects Γ1 and Γ2 at the points B and C , respectively. Let tangents to corresponding circles at the points B and C intersect at the point D . Let E be a point on the line BC such that DE is perpendicular to BC . Prove that AB = EC if and only if OA is perpendicular to BC . Solution by Roy Barbara, Lebanese University, Fanar, Lebanon. Let A be the projection of O on the line = BC . A and A lie on the same side of B on (because they are interior points of Γ1 ), so we have OA ⊥ ⇔ A = A ⇔ AB = A B. (1)

Since ∠OBD = ∠OCD = 90◦ , B and C lie on the circle whose diameter is OD , which we will denote by Γ3 ; let M be its centre. Since M B = M C , the projection of M on is the midpoint N of BC . The lines OA , M N , and DE are parallel (they are all perpendicular to ). Since OM = M D it follows that A N = N E . Hence, A B = A N − BN = N E − N C = EC . That is, A B = EC. (2)

Using (2) and (1) we conclude that AB = EC if and only if AB = A B if and only if OA ⊥ .

ˇ Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; SEFKET ´ University of Sarajevo, Sarajevo, Bosnia and Herzegovina; MICHEL ARSLANAGI C, BATAILLE, Rouen, France; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; PRITHWIJIT DE, Homi Bhabha Centre for Science Education, Mumbai, India; RICHARD EDEN, student, Purdue University, West Lafayette, IN, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; JOHAN GUNARDI, student, SMPK 4 BPK PENABUR, Jakarta, Indonesia; JOHN G. HEUVER, Grande Prairie, AB; JOEL SCHLOSBERG, Bayside, ¨ NY, USA; MIHA¨ I STOENESCU, Bischwiller, France; PETER Y. WOO, Biola University, La Mirada, CA, USA; KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA; and the proposer.

3506.
Brazil.

[2010 : 45, 47] Proposed by Pedro Henrique O. Pantoja, student, UFRN,

Prove that Q(n) + Q n2 + Q n3 is a perfect square for inﬁnitely many positive integers n that are not divisible by 10, where Q(n) is the sum of the digits of n. Solution by the Missouri State University Problem Solving Group, Springﬁeld, MO, USA. More generally, we will show that the following result holds: If Q(n, b) denotes the sum of the digits of n in base b, then Q(n, b) + Q(n2 , b) + Q(n3 , b)

is a perfect square for inﬁnitely many positive integers n that are not divisible by b. Let a be the square-free part of b − 1, k = a 2 with ∈ N, and n = bk − 1. Now n consists of k b − 1’s when written in base b and hence Q(n) = k(b − 1). Using the binomial theorem, it is easy to see that the base b representation of n2 consists of k −1 b−1’s, one b−2, k −1 0’s, and one 1 (hence Q(n2 , b) = k(b−1)) and n3 consists of k − 1 b − 1’s, one b − 3, k − 1 0’s, one 2, and k b − 1’s (hence Q(n3 , b) = 2k(b − 1)). Therefore Q(n, b) + Q(n2 , b) + Q(n3 , b) = 4k(b − 1) = 4a(b − 1) 2 , but since a is the square-free part of b − 1, a(b − 1) is a perfect square and we’re done.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; ROY BARBARA, Lebanese University, Fanar, Lebanon; BRIAN D. BEASLEY, Presbyterian College, Clinton, SC, USA; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; JOSEPH DiMURO, Biola University, La Mirada, CA, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; JOHAN GUNARDI, student, SMPK 4 BPK PENABUR, Jakarta, Indonesia; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; PETER HURTHIG, Columbia College, Vancouver, BC; MICHAEL JOSEPHY, Universidad de Costa Rica, Costa Rica; R. LAUMEN, Deurne, Belgium; ALBERT STADLER, Herrliberg, Switzerland; EDMUND SWYLAN, Riga, Latvia; LI ZHOU, Polk Community College, Winter Haven, FL, USA; and the proposer. The other submitted solutions were similar and they can be summarized by the following 2 sequences nk = 2 × 10k + 7 (k ≥ 2), nk = 7 × 10k + 2 (k ≥ 2), nk = 10k − 1 (k ≥ 1), k k nk = 10 + 17 (k ≥ 4), and nk = 18 × 10 + 18 (k ≥ 4).

3507.

[2010 : 45, 47] Proposed by Pham Huu Duc, Ballajura, Australia.

Let a, b, and c be positive real numbers. Prove that a(b + c) + a2 + bc ≤ b(c + a) + b2 + ca c(a + b) c2 + ab 1 1 1 + + . a+b b+c c+a

2(a + b + c)

Solution by Joe Howard, Portales, NM, USA, modiﬁed by the editor. By the Cauchy-Schwarz inequality, we have that a(b + c)
cyc 2

+ bc

≤ 2(a + b + c)

a
cyc

+ bc

Thus, it suﬃces to show that a
cyc

a2 + bc

≤

1
cyc

a+b

Without loss of generality, let a ≥ b ≥ c. Then (a − c)(b − c) ≥ 0, so c c 1 c2 + ab ≥ ac + bc and c2 + ≤ ac+ = a+ . Therefore, it now suﬃces to ab bc b show that b 1 1 a + 2 ≤ + 2 a + bc b + ac b+c a+c

Since this inequality holds for a = b, we can assume that a > b ≥ c. This simpliﬁes to a2 c2 + b2 c2 + a3 b + ab3 + ab2 c + a2 bc ≤ 2abc2 + a4 + b4 + a3 c + b3 c c2 (a − b)2 + b2 c(a − b) + b3 (a − b) ≤ a3 (a − b) + a2 c(a − b) Since a − b > 0, we obtain c2 (a − b) ≤ a3 − b3 + c(a2 − b2 ) c2 (a − b) ≤ (a − b)(a2 + ab + b2 ) + c(a − b)(a + b) c2 ≤ a2 + ab + b2 + c(a + b) The result follows.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; and the proposer.

3508. [2010 : 45, 47] Proposed by Hung Pham Kim, student, Stanford University, Palo Alto, CA, USA.
Let a, b, c, d be nonnegative real numbers such that a + b + c + d = 4. Prove that √ √ √ √ √ a bc + b cd + c da + d ab ≤ 2 1 + abcd . Solution by the proposer. Let (x, y, z, a permutation √of (a, b, c, d) such that x ≥ y ≥ z ≥ t. √ √ t) be√ We clearly have x ≥ y ≥ z ≥ t and √ Ô Ô √ xyz ≥ xyt ≥ xzt ≥ yzt, and therefore, by the Rearrangement Inequality, we have √ Ô √ √ √ √ √ Ô x xyz + y xyt + z xzt + t yzt √ √ √ √ √ √ √ √ ≥ a abc+ b bcd+ c cda+ d dab. It remains to prove that √ √ Ô √ √ √ √ √ Ô x xyz + y xyt + z xzt + t yzt ≤ 2(1 + abcd), or √ Ô Ô √ √ ( xy + zt)( xz + yt) ≤ 2(1 + xyzt). Since uv ≤
1 2

u2 + v 2 , it is enough to prove that

√ Ô Ô √ √ ( xy + zt)2 + ( xz + yt)2 ≤ 4(1 + xyzt),

or xy + zt + xz + yt ≤ 4, which is equivalent to (x + t)(y + z ) ≤ 4. This is clearly true by the AM–GM Inequality, since x + y + z + t = 4, and we are done.
There were 2 incomplete solutions.

3509. [2010 : 45, 48] Proposed by Hung Pham Kim, student, Stanford University, Palo Alto, CA, USA.
Let a, b, and c be nonnegative real numbers such that a + b + c = 3. For each positive real number k, ﬁnd the maximum value of a2 b + k b2 c + k c2 a + k .

The proposer’s submitted solution is distributed among several other solutions to several other problem proposals, while all of the other submitted solutions to this problem were either incomplete or incorrect. The editor has therefore elected to leave this problem open until a correct and complete “one piece” solution is received.

3510. [2010 : 45, 48] Proposed by Cosmin Pohoat ¸˘ a, Tudor Vianu National College, Bucharest, Romania.
Let d be a line exterior to a given circle Γ with centre O . Let A be the orthogonal projection of O on the line d, M be a point on Γ, and X , Y be the intersections of Γ, d with the circle Γ of diameter AM . Prove that the line XY passes through a ﬁxed point as M moves about Γ. I. Solution by Johan Gunardi, student, Jakarta, Indonesia, modiﬁed by the editor. SMPK 4 BPK PENABUR,

When M lies on OA the lines OA and XY coincide, so that a ﬁxed point would necessarily lie on OA. For any position of M on Γ oﬀ OA let Q denote the intersection of XY and OA. We must prove that the position of Q is independent of the choice of M . Let P be the centre of Γ ; we ﬁrst show that OP QX is cyclic. To that end, note that ∠OP M is both an exterior angle of ∆AOP and half the apex angle of the isosceles triangle P XM . Therefore, ∠OAP + ∠P OA = ∠OP M = 90◦ − ∠AM X. (1)

Also, because M Y ⊥ Y A and Y A ⊥ OA, it follows that M Y ||OA and, thus, ∠OAP = ∠Y M A. From (1) therefore, ∠OAP + ∠P OA − ∠OAP = 90◦ − ∠AM X − ∠Y M A that is, ∠P OA = 90◦ − ∠Y M X. (2)

But ∠P OQ = ∠P OA and, because the angle at the center P of Γ is twice 1 ∠Y P X . Note also that the corresponding inscribed angle at M , ∠Y M X = 2 1 90◦ − 2 ∠Y P X = ∠P XY = ∠P XQ. Consequently, equation (2) becomes ∠P OQ = ∠P XQ, and we conclude that O, P, Q, X are concyclic, as desired. This now implies that ∠OQP = ∠OXP ; moreover, because triangles P XM and OM X are isosceles we have ∠OXP = ∠P M O . Deﬁne R to be the point where the line parallel to P Q through M meets OA. We have ∠ORM = ∠OQP = ∠P M O = ∠AM O . Therefore, triangles M RO and AM O are similar and we OR = OM ; hence have OM OA OR = OM 2 OA ,

so that R is a ﬁxed point. But because P is the midpoint of AM and P Q||M R, Q must be the midpoint of AR. We conclude that line XY passes through the midpoint of the ﬁxed segment AR as M moves about Γ. II. Solution by Li Zhou, Polk Community College, Winter Haven, FL, USA. We do not need the assumption that d is exterior to Γ so long as it is neither tangent to Γ nor passing through O . Let p be the power of A with respect to Γ. Note that p is negative if A is inside Γ. Let i be the transformation deﬁned by i(U ) = V if and only if A, U, V are collinear and as signed lengths, AU · AV = p. [Editor’s comment. Zhou called this transformation an inversion. When A is outside Γ, then i is indeed an inversion in the circle with centre A that is orthogonal to Γ; when A is inside Γ, i is the commutative product of a halfturn about A and inversion in the circle with centre A whose diameter is the chord intercepted from d by Γ.] Note that d and Γ are invariant under i. Let M = i(M ), X = i(X ), and Y = i(Y ). Then i(Γ ) is the line passing through M , X , Y and i(XY )is the circle passing through A, X , Y . Since AX ⊥ M X , we have AM ⊥ M X , whence M X is a diameter of Γ. Let B be the point symmetrical with A about O . Then the triangles OBX and OAM are congruent, which implies that ∠OBP = ∠OAM . This last angle equals the directed angle between the lines Y A and Y X (because corresponding sides are perpendicular). We conclude that B is on the circle through A, X , and Y , whence i(B ) is the ﬁxed point through which XY passes.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece(2 solutions); MICHEL BATAILLE, Rouen, France; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; JOEL SCHLOSBERG, Bayside, NY, USA; PETER Y. WOO, Biola University, La Mirada, CA, USA; and TITU ZVONARU, Com´ ane¸ sti, Romania; and the proposer. There was one incorrect submission. The ﬁnal equation of solution I implies that R is the inverse of A with respect to Γ, something Woo proved in his solution using projective geometry. Many of the other solutions easily solved the problem with the help of coordinates or trigonometry.

3511.

[2010 : 46, 48] Proposed by Pham Van Thuan, Hanoi University of Science, Hanoi, Vietnam. Let a, b, c, and d be nonnegative real numbers. Prove that a2 + b2 + c2
cyclic

≤

1 (a + b + c + d)8 . 64

Solution by George Apostolopoulos, Messolonghi, Greece. We write f (a, b, c, d) =
É

a2 + b2 + c2 , and without loss of generality

cyclic

we assume that a ≥ b ≥ c ≥ d. Since b2 + c2 + d2 ≤ b + c2 + d2 + a2 ≤ a + a2 + b2 + c2 ≤ a + c+d 2 c+d 2 c+d 2
2

,
2

+ b+

c+d 2

and a2 + c2 + d2 ≤ a + we obtain f (a, b, c, d) ≤ f a + and y = b + c+d 2
2

+ b+

c+d 2

inequality follows from an application of the AM–GM inequality: x +y
2 2

c+d , so that x + y = a + b + c + d. 2 ¡2 1 We now need to prove that x2 + y 2 x2 y 2 ≤ (x + y )8 . However, this 64

c+d c+d c+d ,b+ , 0, 0 . Let x = a + 2 2 2

xy =

1 2

x +y

(2xy ) ≤

1 2

x2 + y 2 + 2xy 2

1 8

(x + y ) ,

and the proof is complete. Equality holds precisely when two of a, b, c, d are equal and the remaining two are zero.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece (second solution); ˇ ´ University of Sarajevo, Sarajevo, Bosnia and Herzegovina; OLIVER SEFKET ARSLANAGI C, GEUPEL, Br¨ uhl, NRW, Germany; PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU, Com´ ane¸ sti, Romania; and the proposer.

3512.

[2010 : 46, 48] Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.

Let α be a real number and let p ≥ 1. Find
n n→∞

lim

np + (α − 1)kp−1 np − kp−1

k=1

Solution by Albert Stadler, Herrliberg, Switzerland. Let pn =
n É np + (α − 1)kp−1 . Then np − kp−1 k=1 n

ln pn =
k=1 n

1 + (α − 1) 1−
k n

k n
p− 1

p− 1

1 n

= −

k=1 n

ln 1 + (α − 1) ln 1 − k n

k n ·

p− 1

1 n (1)

p− 1

1 n

k=1

For each k = 1, 2, . . . , n, we have, as n → ∞ ln 1 + (α − 1) ln 1 − k n k n
p− 1 p− 1

· ·

1 n 1 n

= (α − 1) =− k n

k n ·

p− 1

1 n

+O 1 n2 .

1 n2

(2) (3)

p− 1

1 n

From (1), (2), and (3) we have
n

ln pn = α
k=1 n È n=1

k n

p− 1

1 +O n

1 n2

(4)

Note that the sum

Sn =

k n

p− 1

1 is a Riemann sum for the function n

f (x) = xp−1 over the interval [0, 1]. Hence,
1 n→∞

lim Sn =
0

xp−1 dx =

1 p

(5)

From (4) and (5) we have lim ln pn =
n→∞

α , hence lim pn = eα/p . n→∞ p

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; MICHEL BATAILLE, Rouen, France; PAUL BRACKEN, University of Texas, Edinburg, TX, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; JOE HOWARD, Portales, NM, USA; ANASTASIOS KOTRONIS, Heraklion, Greece; and the proposer. Two incorrect solutions were submitted.

3513.

[2010 : 46, 48] Proposed by Hassan A. ShahAli, Tehran, Iran.

Let α and β be positive real numbers, and r be a positive rational number. Prove that there exist inﬁnitely many integers m and n such that mα nβ = r,

where x is the greatest integer not exceeding x. Solution by Oliver Geupel, Br¨ uhl, NRW, Germany, modiﬁed by the editor. We show that the statement is false. It is well known (see D.O. Shklarsky, N.N. Chentzov, and I.M. Yaglom, The USSR Olympiad Problem Book: selected problems and theorems of elementary mathematics, Dover, New York, 1993, 1 1 Problem 108) that if α and β are positive irrational numbers such that + = 1,
α β

then all positive integers appear with no duplications in the two sequences α , 2α , 3α , . . . and β , 2β , 3β , . . . . [Ed.: These are known in the literature as complementary Beatty sequences.] √ √ 1 1 If we take r = 1, α = 2, and β = 2 + 2, then clearly + = 1, so
α β

there are no positive integers m and n such that

mα nβ

= 1. It is also clear that

m = 0, n = 0, and if m and n are of opposite signs, then mα = nβ . Finally, suppose m < 0 and n < 0. Using the trivial fact that x + −x = −1 for all reals x which are not integers, we see that the sequences −α , −2α , −3α , . . . and −β , −2β , −3β , . . . contain all nonnegative integers with no duplications. Hence it is again impossible for mα = nβ to hold, and our proof is complete.
Two incorrect solutions were submitted.

Crux Mathematicorum
with Mathematical Mayhem
edacteurs: Bruce L.R. Shawyer, James E. Totten, V´ aclav Linek Former Editors / Anciens R´

Crux Mathematicorum
Founding Editors / R´ edacteurs-fondateurs: L´ eopold Sauv´ e & Frederick G.B. Maskell Former Editors / Anciens R´ edacteurs: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Mathematical Mayhem
Founding Editors / R´ edacteurs-fondateurs: Patrick Surry & Ravi Vakil Former Editors / Anciens R´ edacteurs: Philip Jong, Jeﬀ Higham, J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia, Shawn Godin, Jeﬀ Hooper, Ian VanderBurgh

SKOLIAD

No. 131

Lily Yen and Mogens Hansen
Please send your solutions to problems in this Skoliad by October 15, 2011. A copy of CRUX with Mayhem will be sent to one pre-university reader who sends in solutions before the deadline. The decision of the editors is ﬁnal. Our contest this month is the National Bank of New Zealand Junior Mathematics Competition, 2010. Our thanks go to Warren Palmer, Otago University, Otago, New Zealand for providing us with this contest and for permission to publish it. La r´ edaction souhaite remercier Rolland Gaudet, de Coll` ege universitaire de Saint-Boniface, Winnipeg, MB, d’avoir traduit ce concours.

Concours math´ ematique de la Banque Nationale de la Nouvelle Z´ elande, 2010 Dur´ ee : 1 heure

1. Raymonde tient un s´ eminaire a ` son lieu de travail. Elle d´ esire
cr´ eer un anneau ininterrompu de tables identiques en formes de polygones r´ eguliers. (Dans un polygone r´ egulier, les cˆ ot´ es sont de longueurs ´ egales et les angles sont de mesures ´ egales. Les carr´ es et les triangles ´ equilat´ eraux sont r´ eguliers.) Chaque table doit avoir deux cˆ ot´ es complets qui co¨ ıncident avec les cˆ ot´ es d’autres tables, tel qu’illustr´ ea ` droite dans le cas du carr´ e ombr´ e. Raymonde a l’intention d’´ etaler du mat´ eriel a ` l’int´ erieur de l’anneau, de fa¸ con a ` ce que ce mat´ eriel soit visible par chacun. (Si vous ne pouvez pas nommer une forme, simplement en fournir le nombre de cˆ ot´ es. Par exemple, si vous croyez que la forme a 235 cˆ ot´ es, mais n’en connaissez pas le nom, simplement l’appeler un 235-gone ; noter que ceci fait partie de la r´ eponse a ` aucune des questions ci-bas.) 1. En premier lieu, Raymonde d´ ecide d’utiliser des tables carr´ ees identiques. Quel est le nombre minimal de tables carr´ ees a ` placer les unes contre les autres, de fa¸ con a ` ce qu’il y ait un espace vide au centre ? 2. Si Raymonde utilise le nombre minimal de tables carr´ ees, quelle est la forme de l’espace vide au centre ? 3. Raymonde consid` ere maintenant utiliser des tables en formes d’octagones (huit cˆ ot´ es chacune). (a) Quel est le nombre minimal de tables octagonales que doit utiliser Raymonde aﬁn qu’il y ait un espace vide au centre de l’enclos ? (b) Quel est le nom de la forme de l’espace vide au centre ? Si vous n’en connaissez pas le nom, il suﬃt d’en donner le nombre de cˆ ot´ es. ` part les carr´ 4. A es et les octagones, y a-t-il d’autres formes de tables possibles ? Si oui, les nommer. Sinon, indiquer qu’il n’y en a pas.

2. Une horloge analogue aﬃche le temps a ` l’aide de deux aiguilles. Chaque heure,
l’aiguille des minutes tourne par 360 degr´ es, tandis que l’aiguille des heures (qui est plus courte que l’aiguille des minutes) tourne par 360 degr´ es sur une p´ eriode de 12 heures. Deux exemples suivent.

3h00

6h00

1. Dessiner une horloge qui aﬃche 9h00. Assurez-vous que l’aiguille des heures est plus courte que celle des minutes. 2. Quel est l’angle entre les deux aiguilles a ` 3h00 et aussi a ` 9h00 ? 3. Quel temps est aﬃch´ e par l’horloge qui suit, a ` l’heure et a ` la minute pr` es ?

4. Quel est l’angle entre les aiguilles aux moments suivants ? (a) 1h00. (b) 2h00. (c) 1h30. ` quel moment, a 5. A ` la minute pr` es, entre 7h00 et 8h00, les aiguilles sont-elles a ` la mˆ eme position ?

3. Un entier a ` six chiﬀres “abcdef ” est form´ e en utilisant les chiﬀres 1, 2, 3, 4, 5, et 6, une et une seule fois chacun, de fa¸ con a ` ce que “abcdef ” est un multiple de 6, “abcde” est un multiple de 5, “abcd” est un multiple de 4, “abc” est un multiple de 3 et “ab” est un multiple de 2.
1. D´ eterminer une solution “abcdef .” Montrer votre travail. 2. La solution que vous avez obtenue est-elle unique (la seule possible) ? Si oui, expliquer bri` evement pourquoi. Sinon, donner une deuxi` eme solution.

Un rectangle 3 × 2 est divis´ e en six carr´ es ´ egaux, chacun contenant une bibitte. Lorsqu’une cloche sonne, chaque bibitte saute soit horizontalement soit verticalement pour atterrir dans un carr´ e voisin ; les bibittent ne peuvent pas sauter diagonalement ; aussi, elles doivent rester a ` l’int´ erieur du rectangle ; on ne peut pas savoir d’avance o` u le bibittes sauteront ; enﬁn, toute bibitte doit changer de carr´ e, aucune ne pouvant rester au mˆ eme carr´ e qu’avant. Comme exemple de fa¸ con de repr´ esenter ceci, le sextuplet (1, 1, 1, 1, 1, 1) d´ enoteune situation o` u chaque carr´ e contient une bibitte, que ce soit au d´ epart ou que ce soit a ` un moment plus tard, comme ¸ ca pourrait bien se produire. Deux

bibittes pourraient bien se retrouver dans le mˆ eme carr´ e. Le 2 2 1 sextuplet (2, 2, 1, 0, 0, 1) repr´ esente la situation donn´ ee par le 0 0 1 diagramme a ` droite ; il y a plusieurs successions de sauts pouvant donner ceci. Le premier chiﬀre repr´ esente un coin, le second un carr´ e au milieu d’un cˆ ot´ e, et ainsi de suite. 1. Quel est le nombre moyen de bibittes par carr´ e du rectangle 3 × 2, sans savoir o` u les bibittes sauteront ? ` partir de la situation initiale d’une bibitte par carr´ 2. A e, est-ce possible que trois bibittes se retrouvent dans le mˆ eme carr´ e, apr` es un seul son de cloche ? Si vous croyez que oui, ´ ecrire un sextuplet ordonn´ e comme les deux cihaut pour indiquer comment ceci pourrait se produire. Sinon, expliquer bri` evement pourquoi. 3. Pour un rectangle 3 × 2, avec une situation initiale d’une bibitte par carr´ e, il n’est certainement pas possible que quatre bibittes se retrouvent dans le mˆ eme carr´ e, apr` es un seul son de cloche. Donner la taille du plus petit rectangle pour lequel aurait ´ et´ e possible. ` partir d’une situation initiale avec une bibitte dans chaque carr´ 4. A e, il est impossible d’avoir cinq bibittes dans un carr´ e, apr` es un seul son de cloche, quel que soit la taille du rectangle. Dans quelques mots, expliquer pourquoi. 5. Pour le cas 3 × 2, avec une bibitte dans chaque carr´ e au d´ epart, combien de sextuplets non uniques, comme (1, 1, 1, 1, 1, 1), sont possibles apr` es un seul son de cloche ? Vous n’avez pas besoin d’en fournir la liste, bien que vous pourriez le faire.

Pania et Rangi font leur entraˆ ınement physique en courrant une fois par semaine autours des deux enclos situ´ es a ` le ferme de leur p` ere, a ` Kakanui ; ils vont de A a ` B a ` C a ` D , puis de retour a ` A. (Voir le sch´ ema.) En ligne directe de Aa ` C , la distance est de 6250 m` etres. AB est plus court que BC .

1. Si ABC est rectangle en ratio 3 : 4 : 5 avec l’angle rectangle a ` B, d´ eterminer les longueurs de ses cˆ ot´ es. 2. Si ABC est rectangle en ratio 3 : 4 : 5, avec l’angle rectangle a ` B, d´ eterminer la mesure de l’angle ∠CAB , a ` une d´ ecimale pr` es. 3. L’angle a ` B est eﬀectivement un angle rectangle, et AB et BC sont de longueurs enti` eres en m` etres, mais, cette fois-ci, les cˆ ot´ es ne sont pas en ratio 3 : 4 : 5. D´ eterminer les valeurs possibles de AB et BC . 4. L’angle a ` D n’est pas rectangle, mais ´ egale plutˆ ot 40◦ . CD est de longueur 600 m` etres. Utiliser cette information pour d´ eterminer la longueur de AD .

Indication : Dans tout triangle XY Z , les r` egles suivantes tiennent : loi de sinus : loi de cosinus :
y z x = = , sin X sin Y sin Z

x2 = y 2 + z 2 − 2yz cos X ,

o` u le cˆ ot´ e x est oppos´ ea ` l’angle X , le cˆ ot´ e y est oppos´ ea ` l’angle Y et le cˆ ot´ ez est oppos´ ea ` l’angle Z .

National Bank of New Zealand Junior Mathematics Competition, 2010
One hour allowed
Rebecca is holding a seminar at the place at which she works. She wants to create an unbroken ring of tables, using a set of identical tables shaped like regular polygons. (In a regular polygon, all sides have the same length, and all angles are equal. Squares and equilateral triangles are regular.) Each table must have two sides which completely coincide with the sides of other tables, such as the shaded square table seen to the right. Rebecca plans to put items on display inside the ring where everyone can see them. (If you cannot name a shape in this question, just give the number of sides. For example, if you think the shape has 235 sides, but don’t know the name, just call it a 235-gon—that isn’t an answer to any of the parts.) 1. Rebecca ﬁrst decides to use identical square tables. What is the minimum number of square tables placed beside each other so that there is an empty space in the middle? 2. If Rebecca uses the minimal number of square tables, what shape is left bare in the middle? 3. Rebecca considers using octagon (eight sides) shaped tables. (a) What is the minimum number of octagonal tables which Rebecca must have in order for there to be a bare space in the middle so that the tables form an enclosure? (b) What is the name given to the bare shape in the middle? If you can’t name it, giving the number of sides will be suﬃcient. 4. Apart from squares and octagons, are there any other shaped tables possible? If there are any, name one. If there isn’t, say so.

An analogue clock displays the time with the use of two hands. Every hour the minute hand rotates 360 degrees, while the hour hand (which is shorter than the minute hand) rotates 360 degrees over a 12-hour period. Two example times are shown below:

3 o’clock

6 o’clock

1. Draw a clock face which shows 9 o’clock. Make sure the hour hand is shorter than the minute hand. 2. What is the angle between the two hands at both 3 o’clock and 9 o’clock? 3. What time to the closest hour (and minute) does the following clock face show?

4. What is the angle between the two hands at the following times? (a) 1 o’clock. (b) 2 o’clock. (c) Half past one. 5. At what time (to the nearest minute) between 7 and 8 o’clock do the hands meet?

3. A six-digit number “abcdef ” is formed using each of the digits 1, 2, 3, 4, 5, and 6 once and only once so that “abcdef ” is a multiple of 6, “abcde” is a multiple of 5, “abcd” is a multiple of 4, “abc” is a multiple of 3, and “ab” is a multiple of 2.
1. Find a solution for “abcdef .” Show key working. 2. Is the solution you found unique (the only possible one)? If it is, brieﬂy explain why. If it isn’t, give another solution.

A 3 × 2 rectangle is divided up into six equal squares, each containing a bug. When a bell rings, the bugs jump either horizontally or vertically (they cannot jump diagonally and they stay within the rectangle) into a square adjacent to their previous square in any direction, although you cannot know in advance which exact square they will jump into. Every bug changes square; no bug stays put. As an example, the ordered sextuplet (1, 1, 1, 1, 1, 1) (where this represents the result, not the movement) represents the situation where every bug jumped so that each square still had one bug in it (it could happen). Alternatively, two

bugs could also land in the same square. An example (not 2 2 1 the only way this could happen) of this might be represented 0 0 1 by (2, 2, 1, 0, 0, 1) —see the diagram to the right. The ﬁrst number in the sextuplet represents a corner square, the second represents a square on the middle of a side, and so on. 1. What is the average number of bugs per square in the 3 by 2 rectangle no matter how the bugs jump? 2. From the initial situation of one bug in every square, is it possible for three bugs to end up in the same square if the bell rings only once? If you think it is, write an ordered sextuplet like the two above where this could happen. If you think it can’t happen, brieﬂy explain why not. 3. From the initial situation of one bug in every square, it is certainly not possible in a 3 × 2 rectangle for four bugs to end up in the same square if the bell rings only once. Write down the dimensions of the smallest rectangle for which it would be possible. 4. From the initial situation of one bug in every square, ﬁve bugs can never end up in the same square if the bell rings only once, no matter the size of the rectangle. In a few words, explain why not. 5. In the 3 × 2 case, how many non-unique sextuplets (like (1, 1, 1, 1, 1, 1)) are possible from the initial situation of one bug in every square, if the bell rings only once? You do not have to list them, although you might like to.

Pania and Rangi exercise weekly by running around two paddocks on their father’s farm near Kakanui from A to B to C to D and then back to A (see the diagram). In a direct line from A to C , the distance is 6250 m. AB is shorter than BC .

1. If ABC is a right angled triangle in the ratio of 3 : 4 : 5, with B at the right angle, ﬁnd the lengths of the sides. 2. If ABC is a right angled triangle in the ratio of 3 : 4 : 5, with B at the right angle, ﬁnd the size of ∠CAB to one decimal place. 3. The angle at B is in fact a right angle, and AB and BC are whole metres in length, but the sides are not in the ratio of 3 : 4 : 5. Find possible lengths for AB and BC . 4. The angle at D is not a right angle but is 40◦ , and CD is 600 m. Use this information to ﬁnd the length of AD . Hint: In any triangle XY Z , the following rules apply:

71
x y z = = , sin X sin Y sin Z

Sine Law: Cosine Law:

x2 = y 2 + z 2 − 2yz cos X ,

where side x is opposite to angle X , side y is opposite to angle Y , and side z is opposite to angle Z .

Next follow solutions to the Baden-W¨ urttemberg Mathematics Contest, 2009, given in Skoliad 125 at [2010:194–196].

Find all natural numbers n such that the sum of n and the digit sum of n is 2010.

Solution by Natalia Desy, student, SMA Xaverius 1, Palembang, Indonesia. Consider the n-digit number abcd; that is, n = 1000a + 100b + 10c + d. Then the digit sum of n is a + b + c + d, and the condition is that 1000a + 100b + 10c + d + a + b + c + d = 2010, so 1001a + 101b + 11c + 2d = 2010. Since all variables represent digits, a = 1 or a = 2. If a = 2, the condition is that 2002 + 101b + 11c + 2d = 2010, so 101b + 11c + 2d = 8. Since all variables represent digits, b and c must both be zero, and thus d = 4. Hence n = 2004. If a = 1, the condition is that 1001 + 101b + 11c + 2d = 2010, so 101b + 11c + 2d = 1009. Again, all variable represent digits, so 11c + 2d ≤ 11 · 9 + 2 · 9 = 117, so 101b ≥ 892, so b = 9. With a = 1 and b = 9, the condition is that 1001 + 909 + 11c + 2d = 2010, so 11c + 2d = 100. Since d is a digit, 2d ≤ 18, so 11c ≥ 82, so c ≥ 8 because c also is a digit. If c = 9, then 2d = 100 − 11c = 1, which is impossible. If c = 8, then 2d = 100 − 11c = 12, so d = 6. Hence n = 1986. Only two natural numbers satisfy the condition, namely 2004 and 1986.
´ Also solved by LENA CHOI, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; GESINE GEUPEL, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany; and RICHARD I. HESS, Rancho Palos Verdes, CA, USA.

A regular 18-gon can be cut into congruent pentagons as in the ﬁgure below. Determine the interior angles of such a pentagon.

´ Solution by Lena Choi, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC.
ca c b d e b d c e b d b e c a d a d c e e b e c d b c b d e a a a a b a b a e c c e a b b aaa e d b c e b d c e e b c d a d a e c b b d d b e e c d c b ac ac d d e ca e d d

Label the angles of the congruent pentagons as in the ﬁgure. In the centre of the ﬁgure, you can see that 6a = 360◦ , so a = 60◦ . The angle sum of an 18-gon is (18 − 2) · 180◦ = 2880◦ , so each interior 1 · 2880◦ = 160◦ . In the ﬁgure, the interior angles of angle in a regular 18-gon is 18 the 18-gon are e, a + c, and 2d. Thus e = 160◦ , a + c = 160◦ , and 2d = 160◦ , so c = 100◦ and d = 80◦ . In the ﬁgure you will also ﬁnd that 2b + d = 360◦ , so 2b = 280◦ , so b = 140◦ . The angles of the pentagon are (a, b, c, d, e) = (60◦ , 140◦ , 100◦ , 80◦ , 160◦ ).
Also solved by VINCENT CHUNG, student, Burnaby North Secondary School, Burnaby, BC; NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia; RICHARD I. HESS, ´ Rancho Palos Verdes, CA, USA; and ROWENA HO, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC.

In the ﬁgure on the right, ABE is isosceles with base AB , ∠BAC = 30◦ , and ∠ACB = ∠AF C = 90◦ . Find the ratio of the area of ESC to the area of ABC . A

C E S F B

Solution by Vincent Chung, student, Burnaby North Secondary School, Burnaby, BC. Since ∠BAC = 30◦ and ∠ACB = 90◦ , ∠ABC = 60◦ ; that is ABC is a 30◦ –60◦ –90◦ triangle. Assume without loss of generality that AB = 2, √ √ √ 3·1 = 23 . AC = 3, and BC = 1. Then the area of ABC is 2 Since ABE is isosceles and ∠BAC = 30◦ , ∠ABE = 30◦ and ∠AEB = 120◦ . Thus ∠CES = 60◦ . Since ∠ABE = 30◦ and ∠AF C = 90◦ ,

∠BSF = 60◦ . Thus ∠CSE = 60◦ , and ESC is equilateral. Since ∠ABC = 60◦ and ∠ABE = 30◦ , ∠CBE = 30◦ . As ∠ACB = 90◦ , it follows that ∠BEC = 60◦ , and BCE is a 30◦ –60◦ –90◦ 2 1 triangle. Since BC = 1, BE = √ and CE = √ . 3 3 Since ESC is equilateral with side
×
1 √ , 3

its height is 1 12
Ö

1 √ 3

−

2 3
1 √

1 √

=
·
1 2

1 3

−

1 4

1 2

3 . Therefore the area of ESC is 3 = 12 2 The ratio of the area of ESC to the area of √ √ 3 3 1 1 : 2 = 12 : 2 = 1 : 6. 12

√

ABC is, then,

´ Also solved by LENA CHOI, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia; RICHARD ´ I. HESS, Rancho Palos Verdes, CA, USA; ROWENA HO, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; and KENRICK TSE, student, Point Grey Secondary, Vancouver, BC.

Given two nonzero numbers z1 and z2 , let zn be

z1 , z2 , z3 , . . . form a sequence. Prove that if you multiply any 2009 consecutive terms of the sequence, then the product is itself a member of the sequence. Solution by Kenrick Tse, student, Point Grey Secondary, Vancouver, BC. Using the recursive deﬁnition of zn , z3 =
z2 , z1 z z5 = 4 = z3 z z7 = 6 = z5 z /z z3 1 = 2 1 = , z2 z2 z1 1/z2 z z z6 = 5 = = 1, z4 1/z1 z2 z z1 z8 = 7 = = z2 , . . . z6 z1 /z2

zn−1 for n > 2. Then zn−2

z4 =

1/z1 1 = , z2 /z1 z2 z1 /z2 = z1 , 1/z2

Since z7 = z1 and z8 = z2 , the sequence will repeat with period six: z1 , z2 ,

is itself one of the ﬁrst six terms. Therefore, whenever zn is a member of the sequence, then so is its reciprocal, z1 . n The product of the ﬁrst six terms is 1. Therefore the product of any six consecutive terms is 1. Since 2010 = 6 · 335, it follows that the product of any 2010 terms is 1. Therefore the product of any 2009 consecutive terms is the reciprocal of the next term, but this reciprocal is itself a member of the sequence as noted above.
´ Also solved by LENA CHOI, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; JONATHAN FENG, student, Burnaby North Secondary School, Burnaby, BC; ´ RICHARD I. HESS, Rancho Palos Verdes, CA, USA; and ROWENA HO, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC.

z 1 1 z1 , , , z1 , z2 , 2 , . . . . Note that the reciprocal of each of the ﬁrst six terms z1 z2 z2 z1

z2 , z1

Let ABC be an isosceles triangle such that ∠ACB = 90◦ . A circle with centre C cuts AC at D and BC at E . Draw the line AE . The perpendicular to AE through C cuts the line AB at F , and the perpendicular to AE through D cuts the line AB at G. Show that the length of BF equals the length of GF . Solution by Kenrick Tse, student, Point Grey Secondary, Vancouver, BC.

1 . Since The slope of the line AE is − r AE is perpendicular to CF , the slope of CF must then be r . (The product of the slopes C of perpendicular lines is −1.) Therefore the equation of CF is y = rx. Now, F is the intersection point of AB and CF , so it is the intersection of y = 1 − x and y = rx. Solving these two equations simultaneously 1 yields that rx = 1 − x, so x = r+1 and, thus, y = rx =

Impose a coordinate system such that C = (0, 0), A = (0, 1), B = (1, 0), D = (0, r ), and E = (r, 0), where r is the radius of the circle. Then the line AB has the equation y = 1 − x.

A G D F

r . r +1

That is,

F =

1 , r r +1 r +1

and, thus, y = 1 − x =

Similarly, the slope of DG is also r , so the equation of DG is y = rx + r . 1 −r Intersection with AB , that is y = 1 − x, yields that rx + r = 1 − x, so x = r +1
r +1 r +1

You can now calculate the distance from B to F using the Pythagorean
2 2 2 2
2

−

1 −r r +1

2r . r +1

That is, G =

1 −r , 2r r +1 r +1

r 1 r r r − 1 + r+1 − 0 = r− , Theorem: |BF |2 = r+1 + r+1 = (r2 +1 +1)2 √ 2 2 r 1 r −r r so |BF | = r+1 − r+1 + r2 − r+1 = 2. Likewise, |GF |2 = 1 r +1 +1 √ 2 2 −r r r 2r 2 + r+1 = (r+1)2 , so |GF | is also r+1 2. Thus |BF | = |GF |. r +1

Also solved by GEOFFREY A. KANDALL, Hamden, CT, USA, who uses this geometric approach: Let H be the point of intersection between CF and A AE , and let the line through E parallel to AB meet CF x at I . Since |CD | = |CE | and ABC is isosceles, you w G can label the lengths as in the ﬁgure. The problem now is to show that y = z . D y Since CH ⊥ AE , ACE ∼ AHC . Therefore |AC | |AH | F = |AC | , so (r + w )2 = |AC |2 = |AE | · |AH |. |AE | r Again, since so r2 = =
(r +w)2 r2

ACE |CE |2

∼ = =

CHE ,

|CE | |AE |

|HE | , |CE |

I C r E w

z B

|AE | · |HE |.

Hence
|AF | |IE |

|AE |·|AH | |AE |·|HE |

|AH | . |HE |

Since EI CIE ∼
(r +w) r2
2

AF ,
r +w . r

HAF ∼ =
|CE | |CB |

HEI , =
r . r +w

CF B , so =

|EI | |F B |

|AH | (r +w)2 so = |HE| = . r2 |AF | |AF | x+y Therefore z = |F B | = |IE|

Moreover, ·
|IE | |F B |

r r +w

Since DG
y r = w , so x x y x+y w = r+ . z r

CF ,
w , r

ADG ∼ +1 = =
w r x+y , z

ACF , so
x+y y

Thus

so x y x+y y

+ 1, so

x+y +r y r = ww , so 1 + x = 1+ w , so x w+r . The previous paragraph shows that r

and it follows that y = z .

We leave it to the reader to judge whether this geometric argument is preferable to the analytic geometry of the ﬁrst solution.

6. A gaming machine randomly selects a divisor of 20092010 and displays its ones
digit. Which digit should you gamble on? ´ Solution by Lena Choi, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC. Since 2009 = 72 · 41, 20092010 = 74020 · 412010. Thus any divisor of 20092010 has the form 7a · 41b , where a and b are integers such that 0 ≤ a ≤ 4020 and 0 ≤ b ≤ 2010.

Note that the ones digit of a product depends only on the ones digits of the factors, and use x ≡ y to mean that x and y have the same ones digit. Then 41b ≡ 1 for all 2011 possible values of b. Moreover, 7a · 41b ≡ 7a · 1 = 7a , so you just have to study the powers of 7: 70 = 1, 71 = 7, 72 = 49 ≡ 9, 73 = 72 · 7 ≡ 9 · 7 = 63 ≡ 3, and 74 = 73 · 7 ≡ 3 · 7 = 21 ≡ 1. Since the ones digit reached 1 again, the sequence of ones digits now repeats with period four. That is, 7a 7a 7a 7a ≡ ≡ ≡ ≡ 3 9 7 1 if if if if a a a a = = = = 3, 2, 1, 0, 7, 10, . . . , 6, 9, . . . , 5, 8, . . . , 4, 7, . . . , 4019 4018 4017 4016, 4020 .

Thus 7a has ones digit 1 for 1006 values of a, while 7a only has ones digit 7, 9, or 3 for 1005 values of a each. Hence the ones digit 1 occurs slightly more often among the divisors of 20092010 than any other digit, so you should gamble on the digit 1.
Also solved by RICHARD I. HESS, Rancho Palos Verdes, CA, USA.

This issue’s prize of one copy of Crux Mathematicorum for the best solutions goes to Kenrick Tse, student, Point Grey Secondary, Vancouver, BC. We look forward to receiving our readers’ solutions to our featured contest.

MATHEMATICAL MAYHEM
Mathematical Mayhem began in 1988 as a Mathematical Journal for and by High School and University Students. It continues, with the same emphasis, as an integral part of Crux Mathematicorum with Mathematical Mayhem. The interim Mayhem Editor is Shawn Godin (Cairine Wilson Secondary School, Orleans, ON). The Assistant Mayhem Editor is Lynn Miller (Cairine Wilson Secondary School, Orleans, ON). The other staﬀ member is Monika Khbeis (Our Lady of Mt. Carmel Secondary School, Mississauga, ON).

Mayhem Problems
Veuillez nous transmettre vos solutions aux probl` emes du pr´ esent num´ ero avant le 15 septembre 2010. Les solutions re¸ cues apr` es cette date ne seront prises en compte que s’il nous reste du temps avant la publication des solutions. Chaque probl` eme sera publi´ e dans les deux langues oﬃcielles du Canada (anglais et fran¸ cais). Dans les num´ eros 1, 3, 5 et 7, l’anglais pr´ ec´ edera le fran¸ cais, et dans les num´ eros 2, 4, 6 et 8, le fran¸ cais pr´ ec´ edera l’anglais. La r´ edaction souhaite remercier Jean-Marc Terrier, de l’Universit´ e de Montr´ eal, d’avoir traduit les probl` emes.

M476.

´ Propos´ e par l’Equipe de Mayhem.

On d´ eﬁnit comme s(n) la somme des chiﬀres de l’entier positif n. Par exemple, s(2011) = 2 + 0 + 1 + 1 = 4. Trouver le nombre d’entiers positifs de quatre chiﬀres n avec s(n) = 4.

M477.

´ Propos´ e par Neculai Stanciu, Ecole secondaire George Emil Palade, Buz˘ au, Roumanie.

Soit m un param` etre entier tel que l’´ equation x2 − mx + m + 8 = 0 ait une racine enti` ere. Trouver la valeur du param` etre m.

M478.

´ Propos´ e par l’Equipe de Mayhem.

On consid` ere l’ensemble des points (x, y ) du plan tels que x2 + y 2 − 22x − 4y + 100 = 0 . Soit P le point de cet ensemble pour lequel est maximal. D´ eterminer la distance x de P a ` l’origine.
y

M479.

´ Propos´ e par Neculai Stanciu, Ecole secondaire George Emil Palade, Buz˘ au, Roumanie. (a) Trouver le plus grand entier positif n pour lequel 3n est un diviseur de A. (b) Trouver le nombre de z´ eros en queue de la repr´ esentation de A en base 10. Soit A = 1 · 2 · 3 · · · · · 2011 = 2011!.

M480. M481.
ON.

Propos´ e par Dragoljub Miloˇ sevi´ c, Gornji Milanovac, Serbie.

Soit x, y et k trois nombres positifs tels que x2 + y 2 = k. Trouver la valeur minimale possible de x6 + y 6 en fonction de k. Propos´ e par Edward T.H. Wang, Universit´ e Wilfrid Laurier, Waterloo,

On suppose que a, b et x sont des nombres r´ eels avec ab = 0 et a + b = 0.
sin4 x cos4 x 1 sin6 x cos6 x Si + = , trouver la valeur de + en fonction de 3 a b a+b a b3

a et b.

.................................................................

M476.

Proposed by the Mayhem Staﬀ

Deﬁne s(n) to be the sum of the digits of the positive integer n. For example, s(2011) = 2 + 0 + 1 + 1 = 4. Determine the number of four-digit positive integers n with s(n) = 4.

M477.

Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania

Let m be an integer parameter such that the equation x2 − mx + m +8 = 0 has one integer root. Determine the value of the parameter m.

M478.

Proposed by the Mayhem Staﬀ x2 + y 2 − 22x − 4y + 100 = 0 .

Consider the set of points (x, y ) in the plane such that
y

Let P be the point in this set for which is the largest. Determine the distance x of P from the origin.

M479.

Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania (a) Determine the largest positive integer n for which 3n divides exactly into A. (b) Determine the number of zeroes at the end of the base 10 representation of A. Let A = 1 · 2 · 3 · · · · · 2011 = 2011!.

M480.

Proposed by Dragoljub Miloˇ sevi´ c, Gornji Milanovac, Serbia

Let x, y , and k be positive numbers such that x2 + y 2 = k. Determine the minimum possible value of x6 + y 6 in terms of k.

M481.
ON

Proposed by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,

Suppose that a, b, and x are real numbers with ab = 0 and a + b = 0. If and b.
sin4 x cos4 x 1 sin6 x cos6 x + = , determine the value of + in terms of a 3 a b a+b a b3

Mayhem Solutions
M438.
Proposed by the Mayhem Staﬀ.

Find all pairs of real numbers (x, y ) such that x2 + (y 2 − y − 2)2 = 0 . Solution by Allen Zhu, Conestoga High School, Berwyn, PA, USA. For x2 + (y 2 − y − 2)2 = 0 to be true with x, y ∈ R, both x = 0 and y − y − 2 = (y − 2)(y + 1) = 0 must hold, so the solutions are: (0, 2), (0, −1).
2

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; ADAMAS AQSA F.S., student, SMA Kharisma Bangsa, Indonesia; JACLYN CHANG, student, University of Calgary, Calgary, AB; NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; ANTONIO LEDESMA ´ LOPEZ, Instituto de Educaci´ on Secundaria No. 1, Requena-Valencia, Spain; PEDRO ´ IES “Abastos”, HENRIQUE O. PANTOJA, student, UFRN, Brazil; RICARD PEIRO, Valencia, Spain; PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; CAO MINH QUANG, Nguyen Binh Khiem High School, Vinh Long, Vietnam; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; NECULAI STANCIU, George Emil Palade Secondary School, Buz˘ au, Romania; GUSNADI WIYOGA, student, SMPN 8, Yogyakarta, Indonesia; and KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA.

M439.

Proposed by Eric Schmutz, Drexel University, Philadelphia, PA, USA.
1 1 1 + = . log2 x log5 x 100

Determine the positive integer x for which

Solution by Gusnadi Wiyoga, student, SMPN 8, Yogyakarta, Indonesia. have that Note that since loga b · logb a = 1 then loga b =
1 log2 x 1 . logb a

Consequently, we

= logx 2 and

1 log5 x

= logx 5. Thus, = 1 100 1

logx 2 + logx 5

logx 10 = x
1 100

100 = 10

x So, the positive integer x is 10100.

= 10100

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; G.C. GREUBEL, Newport News, VA, USA; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; YOUNGHUAN JUNG, The Wood´ L OPEZ ´ lands School, Mississauga, ON; MAR ´ IA ASCENSI ON CHAMORRO, I.B. Leopoldo ˇ ´ Gornji Milanovac, Serbia; NATALIA Cano, Valladolid, Spain; DRAGOLJUB MILO SEVI C, DESY, student, SMA Xaverius 1, Palembang, Indonesia; PEDRO HENRIQUE O. PANTOJA, ´ IES “Abastos”, Valencia, Spain; PAOLO student, UFRN, Brazil; RICARD PEIRO, PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; CAO MINH QUANG, Nguyen Binh Khiem High School, Vinh Long, Vietnam; ´ RODR´ ALEJANDRO S. CONCEPCION IGUEZ, student, University of Las Palmas de Gran Canaria; NECULAI STANCIU, George Emil Palade Secondary School, Buz˘ au, Romania; KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA; and ALLEN ZHU, Conestoga High School, Berwyn, PA, USA.

M441.

Proposed by Katherine Tsuji and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON. What is the maximum number of non-attacking kings that can be placed on an n × n chessboard? (A “king” is a chess piece that can move horizontally, vertically, or diagonally from one square to an adjacent square.) Solution by Bruno Salgueiro Fanego, Viveiro, Spain. We will denote by (i, j ) the square situated on row i and column j , where 1 ≤ i, j ≤ n. Case I: If n is even In each of the n rows, we can place at most n kings, because between any two 2 neighbouring kings there must exist at least one free square. Analogously, in each of the n columns we can place at most n kings too. If we place a king at the 2 squares (i, j ), where i, j are odd numbers then we have n kings in each column 2 and in each row that are non-attacking. Thus, in the n × n chessboard with n even, we can place a maximum of
n 2

n 2

n 2 2

kings.

Case II: If n is odd In each of the n rows we can place a maximum number of n+1 non-attacking kings. 2 Similar to above, if we want to have the maximum number of non-attacking kings,

then we must place a king at the squares (i, j ), where i, j are odd numbers giving n+1 kings in each column and in each row that are non-attacking. Thus, in the 2 n × n chessboard with n odd, we can place a maximum of kings.
n+1 2

n+1 2

n+1 2 2

Also solved by JACLYN CHANG, student, University of Calgary, Calgary, AB; GESINE GEUPEL, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany; RICHARD ´ I. HESS, Rancho Palos Verdes, CA, USA; ANTONIO LEDESMA LOPEZ, Instituto de Educaci´ on Secundaria No. 1, Requena-Valencia, Spain; GUSNADI WIYOGA, student, SMPN 8, Yogyakarta, Indonesia; ALLEN ZHU, Conestoga High School, Berwyn, PA, USA; and the proposers.

M443.

Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania.

Let x denote the greatest integer not exceeding x. For example, 3.1 = 3 and −1.4 = −2. Let {x} denote the fractional part of the real number x (that is, {x} = x − x ). For example, {3.1} = 0.1 and {−1.4} = 0.6. Find all positive real numbers x such that 2x + 3 x+2
ê í ë

2x + 1 x+1
í

14 9

Solution by Adamas Aqsa F.S., student, SMA Kharisma Bangsa, Indonesia.
ê ê í ê

x+3 x+1 1 1 First, note that 2 = 2 − x+2 . Similarly, 2 = 2 − x+1 . x+2 x+1 1 1 Since x is a positive real number, 0 < x+1 < 1 and 0 < x+2 < 1, which

implies that 1 < 2 −
1 x+1

1 x+2

1 < 2− < 2, then that means that 2 − = 1. Now, assembling the values we obtained plus the deﬁnition of {n}, we have 2x + 3 2x + 1 14 + = x+2 x+1 9 ë î ë î 2x + 3 2x + 3 2x + 1 14 − + = x+2 x+2 x+1 9 2x + 3 14 −1+1 = x+2 9 14 2x + 3 = x+2 9 9(2x + 3) = 14(x + 2) 4x = 1 1 x = 4
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; RICHARD I. HESS, Rancho Palos Verdes, CA, ´ L OPEZ ´ USA; MAR ´ IA ASCENSI ON CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain;

< 2. This means that 2 −
ê
1 x+1

1 x+2

= 1. Also, as

´ ´ IES SAMUEL GOMEZ MORENO, Universidad de Ja´ en, Ja´ en, Spain; RICARD PEIRO, “Abastos”, Valencia, Spain; PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; NECULAI STANCIU, George Emil Palade Secondary School, Buz˘ au, Romania; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; GUSNADI WIYOGA, student, SMPN 8, Yogyakarta, Indonesia; KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA; and ALLEN ZHU, Conestoga High School, Berwyn, PA, USA.

M444.
Ô

Proposed by Jos´ e Luis D´ ıaz-Barrero, Universitat Polit` ecnica de Catalunya, Barcelona, Spain. Let a and b be real numbers. Prove that a2 + b2 + 6a − 2b + 10 +
Ô

√ a2 + b2 − 6a + 2b + 10 ≥ 2 10 .

Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON. Let A = (−3, 1), B = (3, −1) and P = (a, b). Then
Ô Ô

a2 + b2 + 6a − 2b + 10 =

(a + 3)2 + (b − 1)2 = P A, (a − 3)2 + (b + 1)2 = P B, √ √ 40 = 2 10.

a2 + b2 − 6a + 2b + 10 = AB =

(−6)2 + 22 =

Hence the given inequality follows from the triangle inequality, P A + P B ≥ AB
Ô

a2 + b2 + 6a − 2b + 10 +

√ a2 + b2 − 6a + 2b + 10 ≥ 2 10.

Note that equality holds if and only if P is on the line segment, l, connecting A and B . Since the slope of l is − 1 , its equation is y = − 1 x. Hence, equality 3 3 holds if and only if a = −3b, where −1 ≤ b ≤ 1.

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; SHAMIL ASGARLI, student, Burnaby South Secondary School, Burnaby, BC; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; MAR ´ IA ´ ´ ASCENSI ON L OPEZ CHAMORRO, I.B. Leopoldo Cano, Valladolid, Spain; SAMUEL ´ GOMEZ MORENO, Universidad de Ja´ en, Ja´ en, Spain; PEDRO HENRIQUE O. PANTOJA, student, UFRN, Brazil; PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli ´ IES “Abastos”, Valencia, Spain; studi di Tor Vergata Roma, Rome, Italy; RICARD PEIRO, CAO MINH QUANG, Nguyen Binh Khiem High School, Vinh Long, Vietnam; NECULAI STANCIU, George Emil Palade Secondary School, Buz˘ au, Romania; GUSNADI WIYOGA, student, SMPN 8, Yogyakarta, Indonesia; ALLEN ZHU, Conestoga High School, Berwyn, PA, USA; and the proposer.

Problem of the Month
Ian VanderBurgh
Last month, we talked about averages and looked at a couple of related problems. This month, we’ll continue by looking at two more problems on this topic. Problem 1 (2010 Sun Life Financial Canadian Open Mathematics Challenge) On a calculus exam, the average of those who studied was 90% and the average of those who did not study was 40%. If the average of the entire class was 85%, what percentage of the class did not study? Solution 1 to Problem 1. Let x be the number of people who studied for the exam and let y be the number of people who did not study. We assume without loss of generality that the exam was out of 100 marks. Since the average of those who studied was 90%, then those who studied obtained a total of 90x marks. Since the average of those who did not study was 40%, then those who did not study obtained a total of 40y marks. Since the overall x+40y average was 85%, then 90x = 85. Therefore, 90x + 40y = 85x + 85y or +y 5x = 45y or x = 9y . Therefore, x : y = 9 : 1 = 90 : 10. This means that 10% of the class did not study for the exam. This is a good solution, but doesn’t take advantage of what we looked at last month related to weighted averages. Let’s try this approach. Solution 2 to Problem 1. The combined average (85%) splits the two partial averages (40% and 90%) in the ratio 45 : 5 or 9 : 1. This means that the number of people in the two categories must be in the inverse ratio, or 1 : 9. This is the ratio of the number of students who did not study to the number of students who did study. This ratio is equivalent to 10 : 90. Therefore, the percentage of the class that did not study is 10%. That was much easier, wasn’t it? Here is a second problem for this month involving averages. Problem 2 (2010 Cayley Contest) Connie has a number of gold bars, all of diﬀerent weights. She gives the 24 lightest bars, which weigh 45% of the total weight, to Brennan. She gives the 13 heaviest bars, which weigh 26% of the total weight, to Maya. She gives the rest of the bars to Blair. How many bars did Blair receive? (A) 14 (B) 15 (C) 16 (D) 17 (E) 18 This problem is one of my favourites from the past couple of years. One reason that I like this problem is that it’s not at all obvious that there is enough information to solve the problem. In fact, a couple of people involved in the contest

creation process were convinced that there was something missing! However, there is enough information to solve Problem 2. Give it a try before reading on! Solution to Problem 2. Connie gives 24 bars that account for 45% of the total weight to Brennan. Thus, each of these 24 bars accounts for an average of 45 % = 15 % = 1.875% of the total weight. 24 8 Connie gives 13 bars that account for 26% of the total weight to Maya. Thus, each of these 13 bars accounts for an average of 26 % = 2% of the total weight. 13 Since each of the bars that she gives to Blair is heavier than each of the bars given to Brennan (which were the 24 lightest bars) and is lighter than each of the bars given to Maya (which were the 13 heaviest bars), then the average weight of the bars given to Blair must be larger than 1.875% and smaller than 2%. Note that the bars given to Blair account for 100% − 45% − 26% = 29% of the total weight. If there were 14 bars accounting for 29% of the total weight, 29 the average weight would be 14 % ≈ 2.07%, which is too large. Thus, there must be more than 14 bars accounting for 29% of the total weight. If there were 15 bars accounting for 29% of the total weight, the average % ≈ 1.93%, which is in the correct range. If there were 16 weight would be 29 15 bars accounting for 29% of the total weight, the average weight would be 29 %≈ 16 1.81%, which is too small. The same would be true if there were 17 or 18 bars. Therefore, Blair must have received 15 bars. When we read this problem for the ﬁrst time, the fact that averages might enter in is not clear. But averages are useful in a pretty natural way, and perhaps in a “real life” way too. Often, using an average is a great way of estimating the size of objects in a collection, and that’s exactly what we’ve done here. We don’t know the actual sizes of the gold bars, but we can estimate and compare by using averages. As one ﬁnal note on this problem, can you ﬁnd the piece of information in this problem that seemed important, but was never used?

THE OLYMPIAD CORNER
No. 292 R.E. Woodrow
We begin the section of solutions from our readers with the ﬁle of solutions to problems of the Thai Mathematical Olympiad Examinations 2006, Selected problems, given at [2010 : 83–84].

Suppose f : R → R is a function satisfying f (x2 + x + 3) + 2f (x2 − 3x + 5) = 6x2 − 10x + 17

for all real x. Find f (85). Solved by Arkady Alt, San Jose, CA, USA; David E. Manes, SUNY at Oneonta, Oneonta, NY, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We use an edited version of the solution of Alt. Let p (x) = x2 + x + 3, q (x) = x2 − 3x + 5. Since p (1 − x) = (1 − x)2 + (1 − x) + 3 = x2 − 3x + 5 = q (x) , q (1 − x) = p(1 − (1 − x)) = p(x) and then 6 (1 − x) − 10 (1 − x) + 17 = 6x2 − 2x + 13
2

6x2 − 2x + 13 = f (p (1 − x)) + 2f (q (1 − x)) = f (q (x)) + 2f (p (x)) and from the system of equations f (p (x)) + 2f (q (x)) = 6x2 − 10x + 17 2f (p (x)) + f (q (x)) = 6x2 − 2x + 13 we obtain 3f (p (x)) = 2 (2f (p (x)) + f (q (x))) − (f (p (x)) + 2f (q (x))) = 6x2 + 6x + 9 ⇐⇒ f (p (x)) = 2p(x) − 3.
11 11 then for any y ≥ there is x such that 4 4 11 we have p (x) = y and, therefore, for any y ≥ 4

= 2 6x2 − 2x + 13 − 6x2 − 10x + 17

Since p (x) = x2 + x + 3 ≥

f (y ) = f (p (x)) = 2p(x) − 3 = 2y − 3. Hence, f (85) = 167.

Evaluate

8000 k=84

k 84

8084 − k 84

Solved by Arkady Alt, San Jose, CA, USA; Oliver Geupel, Br¨ uhl, NRW, Germany; David E. Manes, SUNY at Oneonta, Oneonta, NY, USA; and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON. We give Wang’s combinatorial argument. More generally, we consider the sum S (n, d) = where n, d ∈ N such that d ≤ show that S (n, d) =
n+1 ¡ n+1 ¡
2d+1 n− d k=d

k d

n−k d

n . We use a 2-way counting argument to 2

so 2d+1 is the number of (2d + 1)-subsets of T . On the other hand, each (2d + 1)-subset of T is completely determined by ﬁrst choosing a positive integer k to be the middle number, d ≤ k ≤ n − d and then select any d numbers from the k-subset {0, 1, 2, . . . , k − 1} and any d numbers from the (n − k)subset {k + 1, k + 2, . . . , n}. This procedure which is possible since d ≤ k and d ≤ n − k would yield all the (2d + 1)-subsets of T . Hence S (n, d) =
n− d k=d

. Let T = {0, 1, 2, . . . , n}. Then |T | = n + 1

k d

n−k d

n+1 2d + 1
8085¡
169

follows. In particular, the value of the given summation is S (8084, 84) =

Find all integers such n that n2 + 59n + 881 is a perfect square.

Solved by David E. Manes, SUNY at Oneonta, Oneonta, NY, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution by Zonvaru. If n2 + 59n + 881 is a perfect square, then 4(n2 + 59n + 881) is a perfect square. We have k2 ⇔ k2 = 4n2 + 4 · 59n + 3524 = 4n2 + 4 · 59n + 592 + 3524 − 592

⇔ k2

= (2n + 59)2 + 43 ⇔ (k − 2n − 59)(k + 2n + 59) = 43.

Since k, n are integers and 43 is prime, we have the following possibilities (i) (ii) k − 2n − 59 = 1 k + 2n + 59 = 43 k − 2n − 59 = −1 k + 2n + 59 = −43 ⇔ k = 22, n = −19; ⇔ k = −22, n = −40;

(iii) (iv)

k − 2n − 59 = 43 k + 2n + 59 = 1 k − 2n − 59 = −43 k + 2n + 59 = −1

⇔ k = 22, n = −40; ⇔ k = −22, n = −19.

For n = −19 and n = −40, n2 + 59n + 881 = 112 . It results that n2 + 59n + 881 is a perfect square for n = −19 and n = −40.

Find the least positive integer n such that √ n+1 3z − zn − 1 = 0 Solved by Arkady Alt, San Jose, CA, USA. Let z = cos ϕ + i sin ϕ, ϕ ∈ [0, 2π ). Since 1 + cos nϕ + i sin nϕ = 2 cos2
nϕ nϕ nϕ + 2i cos sin 2 2 2 nϕ nϕ nϕ cos = 2 cos + i sin 2 2 2

has a complex root z with |z | = 1.

then √ n+1 3z = 1 + cos nϕ + i sin nϕ √ n+1 nϕ nϕ nϕ ⇐⇒ 3z + i sin = 2 cos cos
2 2 2

yields

¬ ¬ √ ¬ n+1 ¬ nϕ nϕ nϕ ¬ ¬ = 2¬ 3 ¬z cos + i sin ¬cos ¬
√ ¬ ¬ ¬ ¬ √ 3 nϕ ¬ nϕ ¬ ¬ ¬ ⇐⇒ 3 |z |n+1 = 2 ¬cos = ¬cos ¬ ⇐⇒ ¬ 2 2 2

2 2 2 √ ¬ ¬ 3 nϕ ¬ 3 1 ¬ 2 nϕ ⇐⇒ = ¬cos = 2 cos ⇐⇒ cos nϕ = . ¬ ⇐⇒ 2 2 2 2 2 √ √ 1 1 3 3 If cos nϕ = then sin nϕ = ± and z n = ± i. 2 2 2 2 √ √ 3 i 3 1 i 3 Since z n + 1 = ± , z n+1 = z · z n = z ± then 2 2 2 2 √ √ √ n+1 √ 1 i 3 3 i 3 3z = z n + 1 ⇐⇒ 3z = ± ± 2 2 2 2 √ √ 1 3 3 1 ± i = ± i ⇐⇒ z 2 2 2 2 π π π π ⇐⇒ z cos ± + i sin ± = cos ± + i sin ± 3 3 6 6 π π ⇐⇒ z = cos ∓ + i sin ∓ . 6 6

87
π nπ

Hence, ± = ∓ + 2kπ ⇐⇒ n = ±12k − 2, k ∈ Z and, therefore, 3 6 the smallest positive integer n satisfying this equation is n = 10. So, a necessary condition is n ≥ 10. Let z = cos z
10

π π + i sin and n = 10 then 6 6

z 11 and √ 3z n+1

√ 10π 10π 5π 5π 1 i 3 = cos + i sin = cos + i sin = − , 6 6 3 3 2 2 √ 11π 11π 3 1 = cos + i sin = − i, 6 6 2 2 √ √ 3 3 1 3 = − i=1+ − = 1 + zn . 2 2 2 2

Thus, the least positive integer n such that root with |z | = 1 is 10.

√

3z n+1 − z n − 1 = 0 has a complex

Let pk denote the kth prime number. Find the remainder when
2550 k=2

p kk is divided by 2550.

p4 − 1

Solved by Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA. First we factor 2550 into prime powers: 2550 = (255) · (10) = 5 · 51 · 10 = 5 · 3 · 17 · 2 · 5. So, we easily see that 2550 = 2 · 3 · 52 · 17. (1) First, note that p2 = 3, p3 = 5, and p7 = 17. We will ﬁrst ﬁnd the congruence p4 − 1 p4 − 1 p4 − 1 classes the three integers p22 , p33 , and p77 ; belong to modulo 2550. We start with p22 Clearly
p4 − 1

= 33

−1

= 380 . and 380 ≡ 1 (mod 2) . (2)

380 ≡ 0 (mod 3)

By Fermat’s Little Theorem, 316 ≡ 1 (mod 17); and so 380 = (316 )5 ≡ 15 ≡ 1 (mod 17) . Consider 380 modulo 52 = 25. First 38 ≡ 34 So that, 380 = (38 )10 ≡ (11)10 ≡ (112 )5 ≡ (121)5 ≡ (−4)5 ≡ −24 ≡ 1 (mod 25) ≡ (−4)4 · (−4) ≡ (256)(−4) ≡ 6 · (−4)
2

(3)

≡ (81) ≡ (6) ≡ 36 ≡ 11 (mod 25) .

(4)

Altogether we have, from (2), (3), (4), that 380 ≡ 1 (mod ¡ 2) , 380 ≡ 1 (mod 17) , 2 3 ≡ 1 mod 5 and 380 ≡ 0 (mod 3)
80

(5)

Since 2, 17, 52 are pairwise ¡ relatively prime; (5) shows that ¡ 380 − 1 ≡ 0 mod 2 · 52 · 17 ; 380 ≡ 1 mod 2 · 52 · 17 . Thus, 380 = 2 · 52 · 17 · k + 1, for some positive integer k. Since k = 3 · k + r where r ∈ {0, 1, 2} and k ∈ Z+ , and since 380 ≡ 0 (mod 3) and 42 · 52 · 17 ≡ 2 · 1 · 2 ≡ 4 ≡ 1 (mod 3) we see that 380 = 2 · 3 · 52 · 17 · k + 2 · 52 · 17 · r + 1 ≡ r + 1 ≡ 0 (mod 3)

so r = 2. Thus, 380 = = 2 · 3 · 52 · 17 · k + 2 · 52 · 17 · 2 + 1 2550 · k + 1700 + 1 = 2550 · k + 1701
p4 − 1

we have shown that 380 = 1701 (mod 2550) p2 = 3 , Next, consider p3 5624 5624 5624 5624
p4 3 −1

p2 2
−1

≡ 1701 (mod 2550) .
2

(6)

= 55

= 5624 = 5(5

−1)(5 +1)

= 524·26. Clearly

= 1 (mod 3) (since 52 ≡ 1 (mod 3)); ≡ 0 mod 52 ; ≡ 1 (mod 2) ; and = 524·26 = 58·2·3·13 = (516 )39 ≡ 139 ≡ 1 (mod 17)

by Fermat’s Theorem. We see that 5624 ≡ 1 (mod 3), 5624 ≡ 1 (mod 2), 5624 ≡ 1 (mod 17) which implies that 5624 ≡ 1 (mod 2 · 3 · 17) so 5624 = 2 · 3 · 17 · l + 1 for some positive integer l. Observe that 2 · 3 · 17 ≡ 6(−8) ≡ −48 ≡ −(−2) ≡ 2 (mod 25). And so, 5624 = 2 · 3 · 17 · l + 1 ≡ 2l + 1 (mod 25). But 5624 ≡ 0 (mod 25); and so we must have 2l + 1 ≡ 0 (mod 25) ⇔ l ≡ 12 (mod 25); l = 25 · L + 12; for some L ∈ Z+ . Thus, 5624 = 2 · 3 · 17 · (25L + 12) + 1 = 2 · 3 ·ßÞ 52 · 17 ·L + 2 · 3 · 17 · 12 + 1.
2550

And so, 5624 ≡ 2 · 3 · 17 · 12 + 1 ≡ 1225 (mod 2550) p3 = 5 , p3 3
p4 − 1

≡ 1225 (mod 2550) .
2

(7)

Next consider p7 = 17. We have, p7 7
p4 − 1

= 1717

−1

= 17(17

−1)(172 +1)

= 1716·18·290.

And 1716·18·290 17 And also, 1716·18·290
16·18·290

≡ 1 (mod 3) (since 172 ≡ 1 (mod 3)) ≡ 0 (mod 17)

≡ 1 (mod 2)

and

Consider 1716·18·290 modulo 52 = 25. Observe that, 172 = 289 ≡ 275 + 14 ≡ 14 (mod 25); or equivalently, 172 ≡ −11 (mod 25) 1716 ≡ (172 )8 ≡ (−11)8 ≡ [(−11)2 ]4 ≡ (121)4 ≡ (−4)4 ≡ 256 ≡ 1 (mod 25) .

And thus 1716·18·290 ≡ [(17)16 ]18·290 ≡ 118·290 ≡ 1 (mod 25). We have 16·18·290 16·18·290 ·290 17 ≡ 1 (mod 3), and 1716·18 ≡ 1 ¡ ≡ 1 (mod 2), 17 ¡ 2 16·18·290 2 mod 5 , which implies that 17 ≡ 1 mod 2 · 3 · 5 ; and so 1716·18·290 = 2 · 3 · 52 · m + 1; for some m ∈ Z+ . And since 2 · 3 · 52 ≡ 2 · 3 · 25 ≡ 2 · 3 · 8 ≡ 2 · 3 · 8 ≡ 48 ≡ −3 (mod 17). We see that we must have 2 · 3 · 52 · m + 1 ≡ 0 (mod 17); −3m + 1 ≡ 0 (mod 17); 3m ≡ 1 (mod 17); m ≡ 6 (mod 17). So that m = 6 + 17 · M ; for some M ∈ Z+ . Altogether, 1716·18·290 = 2 · 3 · 52 · (6 + 17M ) + 1 = 2 · 3 ·ßÞ 52 · 17 ·M + 2 · 3 · 52 · 6 + 1 = 2550M + 901
2550

We have shown that, p7 = 17, p7 7
p4 − 1

≡ 901 (mod 2550)
p4 − 1

(8)

We now come to the last part of the problem by considering pkk ; where k ≥ 2 and k = 2, 3, 7. In other words, pk = 3, 5, 17; and pk > 2. Since pk is odd, we have p2 k ≡ 1 (mod 8); from which it follows (just write 2 pk = 8λ + 1; and square both sides) that p4 k ≡ (mod 16) . By Fermat’s Little Theorem, we also have (since pk = 5), p4 k ≡ 1 (mod 5) . From (9) and (10) it follows that p4 k −1 4 pk − 1 Thus, pk
p4 k −1

(9)

(10)

≡ 0 (mod 16 · 5) ; = 80t, for some positive integer t.

(11)

t = p80 k ; which implies

t t p80 ≡ 1 (mod 2) , p80 ≡ 1 (mod 3) , k k and (by Fermat’s Little Theorem since 80 is divisible by 16) t p80 ≡ 1 (mod 17) . k

(12)

t Now, consider p80 modulo 25. Since pk is not divisible by 5; we have k pk = 5 · q + r ; where q is a positive integer and r = 1, 2, 3, or 4. Consider 20 p20 k = (5q + r ) . Since both 20 and 5q are divisible by 5; it is clear that in the binomial expansion (5q + r )20 every term, except for the last one; is divisible by 20 25; thus p20 ≡ r 20 mod 25 k = (5q + r ) When r = 1, r 20 ≡ 1 (mod 25) When r = 2, r 20 ≡ 220 ≡ (26 )3 · 22 ≡ (64)3 · 22 ≡ (−11)3 · 22 ≡ (−11)2 · (−11) · 22 ≡ (121)(−11)(22 ) ≡ (−4)(−11) · 4 ≡ (−16)(−11) ≡ 9(−11) ≡ −99 ≡ 1 (mod 25). When r = 3; p20 ≡ 320 ≡ (34 )5 ≡ (81)5 ≡ (−6)5 ≡ (−6)2 · (−6)2 · (−6) ≡ (36)(36) · (−6) ≡ (11)(11) · (−6) ≡ (121)(−6) ≡ (−4)(−6) ≡ 24 ≡ −1 (mod 25). When r = 4; r 20 ≡ (44 )5 ≡ (256)5 ≡ 15 ≡ 1 (mod 25). We see that 20 in all cases; r 20 ≡ ±1 (mod 25). Therefore p20 ≡ r 20 ≡ ±1 k = (5q + r ) (mod 25). And so,

≡ (±1)4 ≡ 1 (mod 25) p80 k
t p80 ≡ 1 (mod 25). From (13) and (12), it is clear that for k = 2, 3, 7 k t p80 = p kk k p4 − 1

(13)

≡ 1 mod 2 · 3 · 52 · 17 ≡ 1 (mod 2550)

(14)

Thus, in the sum k=2 pkk ; every term with k = 2, 3, 7; is congruent to 1 modulo 2550 by (14). There are (2550 − 2) + 1 − 3 = 2550 − 4 such terms. We have
2550 k=2

È2550

p4 − 1

p kk

p4 − 1

≡ p2 2

p4 − 1

+ p3 2

p4 − 1

+ p7 7

p4 − 1

+ (2550 − 4) · 1
0 (mod 2550)

≡ 1701 + 1225 + 901 +

2550

−4

· 1 by (6), (7), and (8)

≡ 3823 ≡ 2550 + 1273 ≡ 1273 (mod 2550) Conclusion: The remainder is 1273.

A triangle has perimeter 2s, inradius r , and the distance from its incenter to the vertices are sa , sb and sc . Prove that 3 4 + r sa + r sb + r sc ≤ s2 12r 2 .

Solved by Arkady Alt, San Jose, CA, USA; George Apostolopoulos, Messolonghi, Greece; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the write-up by Zvonaru. We have sa = that 3 + sin 4 inequalities
A 2

+ sin A 2

r r , sb = sin B , sin A 2 2 2 B C s + sin ≤ . 2 2 12r 2

sc =

r sin C 2

, and we have to prove

The last inequality follows by known 3 2

sin

+ sin

B 2

+ sin

C 2

≤

(item 2.9 in [1])

[1] 0. Bottema, Geometric Inequalities, Groningen, 1969.

and 27r 2 ≤ s2 (Item 5.11 in [1]).

Find all primes p such that

2p−1 −1 p

is a perfect square.

Solved by David E. Manes, SUNY at Oneonta, Oneonta, NY, USA; and Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA. We give Mane’s version. The only such primes are p = 3 and p = 7. Assume that some integer n. Then 2 p− 1 − 1 = 2
p−1 2

2p−1 −1 p

= n2 for

p−1 2

−1

= pn2 .

then 2 2 = 2(2x2 + 2x + 1). Since 2x2 + 2x + 1 is odd, it follows that it can p−1 be a power of 2 only if 2x2 + 2x + 1 = 1 when x = 0. Therefore 2 2 − 1 = 1 p−1 1 = 1 or p = 3. Accordingly, the only values of p for which 2 p −1 is so that p− 2 a perfect square are p = 3 and p = 7.

e1 2e2 er The prime factorization of pn2 is pe p2 · · · p2 1 p2 r , where the exponent e and the primes p, p1 , p2 , . . . , pr are all odd integers. Then p−1 p−1 p−1 p−1 gcd 2 2 + 1, 2 2 − 1 = 1 implies that one of the terms 2 2 +1 or 2 2 −1 is equal to pe a2 and the other is equal to b2 for some integers a and b. Assume p−1 p−1 2 2 + 1 = b2 and let b = 2x + 1 so that 2 2 + 1 = 4x2 + 4x + 1. Therefore p−1 2 2 = 4x(x + 1). The product x(x + 1) is a power of 2 only if x = 1, and so p−1 p−1 1 = 3 or p = 7. If 2 2 − 1 = b2 = 4x2 + 4x + 1, 2 2 = 23 . Therefore, p− 2
p−1

Next we turn to reader’s solutions to problems of the 14th Turkish Mathematical Olympiad 2006 given at [2010: 84–85]. Let A1 , B1 and C1 be the feet of the altitudes belonging to the vertices A, B and C , in acute triangle ABC , respectively, and let OA , OB and OC be the incenters of the triangles AB1 C1 , BC1 A1 and CA1 B1 , respectively. Let TA , TB and TC be the points of tangency of the incircle of the triangle ABC to the sides BC , CA and AB , respectively. Show TA OC TB OA TC OB is a regular hexagon. Solved by Titu Zvonaru, Com´ ane¸ sti, Romania.

Let I be the incentre of triangle ABC and r be the inradius. The triangles AB1 C1 and ABC are similar, with AB1 B1 C 1 AC1 = = = cos A. AC AB BC The points A, OA , I are collinear, and we have AI = r sin
A 2

C B1 TB I OA A TC C1 B

AOA =

r sin
A 2

cos A.

We deduce that OA I = r sin
A 2

−

r sin
A 2

cos A =

r · 2 sin2 sin
A 2

A 2

,
A 2

hence OA I = 2r sin A . In IOA TC we have ITC = r , OA I = 2r sin 2 A ◦ ∠OA ITC = 90 − 2 . By the Law of Cosines, we obtain:
2 OA TC

and

= =

r 2 + 4r 2 sin2 r 2 + 4r 2 sin
2

A 2 A 2

− 2r · 2r sin − 4r 2 sin
2

A 2

cos 90◦ −

A 2

= r2.

It results that OA TC = r and all sides of TA OC TB OA TC OB are equal to r. But this hexagon has no equal angles; From the isosceles triangle OA TC I we have A ∠IOA TC = ∠OA ITC = 90◦ − . 2 It follows that ∠TB OA TC = 180◦ −A = B +C , and similarly ∠TC OB TA = A + C , ∠TA OC TB = A + B . We also obtain that ∠OA TC I = 180◦ − 2 90◦ − A 2 = A,

hence ∠OA TC OB = A + B , ∠OB TA OC = B + C , ∠OC TB OA = A + C .

Next we turn to solutions to problems of the Turkish Team Selection Test for IMO 2007 given at [2010: 85].

An airline company is planning to run two-way ﬂights between some of the six cities A, B , C , D , E and F . Determine the number of ways these ﬂights can be arranged so that it is possible to travel between any two of these six cities using only the ﬂights of this company.

Solved by Oliver Geupel, Br¨ uhl, NRW, Germany. For positive integers n1 , . . . , np let dn1 ,...,np denote the number of labeled graphs consisting of p disjoint connected subgraphs with n1 , . . . , np vertices. We are to determine d6 . The result will be d6 = 26704. To begin with, note that d1 = d2 = 1, d3 = 4. The total number of labeled graphs with 4 vertices is
4 64 = 2(2) = d1,1,1,1 + d2,1,1 + d2,2 + d3,1 + d4 4 1 4 2 4 = d4 d2 d2 d + d3 d1 + d4 1 + 1 + 2 2 2 2 3 = 1 + 6 + 3 + 16 + d4 = 26 + d4 ;

hence d4 = 38. The total number of labeled graphs with 5 vertices is
5 1024 = 2(2) = d1,1,1,1,1 + d2,1,1,1 + d2,2,1 + d3,1,1 + d3,2 + d4,1 + d5 5 1 5 3 5 5 5 =1+ + + d3 + d3 + d4 + d5 2 2 2 2 3 3 4 = 1 + 10 + 15 + 40 + 40 + 190 + d5 = 296 + d5 ;

thus d5 = 728. Finally, the total number of labeled graphs with 6 vertices is
6 32768 = 2(2) = d1,1,1,1,1,1 + d2,1,1,1,1 + d2,2,1,1 + d2,2,2 + d3,1,1,1 + d3,2,1 + d3,3 + d4,1,1 + d4,2 + d5,1 + d6 6 1 6 4 1 6 4 6 6 3 =1+ + + + d3 + d3 2 2 2 2 6 2 2 3 3 2 1 6 2 6 6 6 + d3 + d4 + d4 + d5 + d6 2 3 4 4 5 = 6064 + d6 ;

consequently d6 = 26704. Remark. The problem is well-known. A diﬀerent approach is given in Herbert S. Wilf’s book Generating Functionology, Second edition, Academic Press, 1994, page 87, formula (3.10.2).

Let a, b, c be positive real numbers such that a + b + c = 1. Prove that 1 2c 2 + 2c + 1 bc + 2a 2 + 2a + 1 ca + 2b2 + 2b ≥ 1 ab + bc + ca .

ab +

Solved by Arkady Alt, San Jose, CA, USA; Jos´ e Luis D´ ıaz-Barrero, Universitat Polit` ecnica de Catalunya, Barcelona, Spain; and Michel Bataille, Rouen, France. We give the version of D´ ıaz-Barrero.

Multiplying numerator and denominator of the RHS by ab + bc + ca yields 1 ab + 2c 2 + 2c + 1 bc + 2a 2 + 2a + 1 ca + 2b2 + 2b ≥ ab + bc + ca (ab + bc + ca)2

Now we claim that 1 ab + 2c2 + 2c ≥ ab (ab + bc + ca)2

Indeed, the preceding inequality is equivalent to (ab + bc + ca)2 ≥ ab(ab + 2c2 + 2c) which after some algebraic computations becomes a2 b2 + b2 c2 + c2 a2 + 2abc(a + b + c) ≥ ab(ab + 2c2 + 2c), and taking into account the constraint, we get a2 b2 + b2 c2 + c2 a2 + 2abc ≥ ab(ab + 2c2 + 2c) Canceling terms, we obtain b2 c2 + c2 a2 ≥ 2abc2 or equivalently √ b2 c2 + c2 a2 ≥ a2 b2 c4 = abc2 2 which holds on account of AM-GM inequality. Likewise, 1 bc + 2a2 + 2a and ≥ bc (ab + bc + ca)2

1 ca ≥ . ca + 2b2 + 2b (ab + bc + ca)2

Adding the preceding three inequalities the statement follows. Equality holds when a = b = c = 1/3, and we are done.

Now we turn to solutions to the Estonian Team Selection Contest 2007 given at [2010: 149].

Let D be the foot of the altitude of triangle ABC drawn from vertex A. Let E , F be the points symmetric to D with respect to the lines AB , AC , respectively. Let triangles BDE , CDF have inradii r1 , r2 and circumradii R1 , R2 , respectively. If SK denotes the area of ﬁgure K , prove that |SABD − SACD | ≥ |r1 R1 − r2 R2 |.

Solved by Oliver Geupel, Br¨ uhl, NRW, Germany; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give Geupel’s solution. Without loss of generality let AD = 1. Let denote ϕ1 = ∠BAD and ϕ2 = ∠CAD , where ϕ1 , ϕ2 ∈ [0, π/2). It holds BD = BE = tan ϕ1 and DE = 2 sin ϕ1 . Hence SABD = r1 R1 = 2(DE + BD + BE ) DE · BD · BE
1 tan ϕ1 and 2

sin ϕ1 tan2 ϕ1 2(sin ϕ1 + tan ϕ1 )

1 2

1 cos ϕ1

−1 .

have the same sign. It therefore suﬃces to prove

1 1 1 tan ϕ2 and r2 R2 = − 1 . We therefore have 2 2 cos ϕ ¬ ¬2 ¬ 1 1 1 ¬ ¬ to prove that |tan ϕ1 − tan ϕ2 | ≥ ¬ ¬ cos ϕ1 − cos ϕ2 ¬. Since tan x and cos x are 1 1 − both increasing for 0 < x ≤ π/2, the terms tan ϕ1 − tan ϕ2 and cos ϕ1 cos ϕ2

Similarly, SACD =

tan ϕ1 − tan ϕ2 ≥

1 cos ϕ1

−

1 cos ϕ2

(1)

under the hypothesis ϕ1 ≥ ϕ2 . But the function f (x) = This implies (1) and the proof is complete.

1 − tan x is cos x sin x − 1 decreasing for 0 ≤ x < π/2 as can be seen from the derivative f (x) = . cos2 x

Let n be a natural number, n ≥ 2. Prove that if some positive integer b then n is prime.

bn − 1 b− 1

is a prime power for

Solved by Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA. Assume that,
bn − 1 b− 1

First, we will treat the case in which p equals 2. This is done in Case 1. In Case 2, p ≥ 3; the proof splits into three subcases. Case 1. p = 2. We have, by (1), b
bn − 1 = b− 1 n− 1 n− 2

= pk ; bn − 1 = pk · (b − 1) where p is a prime number, n, b, k positive integers such that n ≥ 2 and b ≥ 2

(1)

2k ; or equivalently + · · · + b + 1 = 2k .

(2)

It follows from (2) that b must be odd and n even; since there are n terms, with each term being an odd number; in the sum bn−1 + bn−2 + · · · + b + 1. Thus n = 2l where l is a positive integer. (3)

By (2) and (3) we have, b2l − 1 = 2k · (b − 1) ⇔ (bl − 1)(bl + 1) = 2k · (b − 1). (4)

Since b is odd; the numbers bl − 1 and bl + 1 are consecutive even integers; their greatest common divisor is 2: (bl − 1, bl + 1) = 2. If l = 1, then n = 2 · 1 = 2, which is a prime (and b = 2k − 1). If l ≥ 2; then bl − 1 = (b − 1) · (bl−1 + bl−2 + · · · + b + 1). By (4) and (6) we obtain, (b − 1)(bl−1 + bl−2 + · + b + 1) · (bl + 1) = (bl−1 + bl−2 + · + b + 1) · (bl + 1) = 2k (b − 1); 2k , (6) (5)

(7)

since b is greater than 1; in fact b ≥ 3, since b is odd; and l ≥ 2. Each of the factors on the lefthand side of (7) is greater than 2. Thus (7) implies that bl−1 + bl−2 + · · · + b + 1 = 2k1 (8) and bl + 1 = 2k2 ; with k1 , k2 positive. Also, bl − 1 = (b − 1)(bl−1 + bl−2 + · · · + b + 1) = 2k1 · (b − 1). (9)

Case 2. p is an odd prime, p ≥ 3.

Since k1 ≥ 2, k2 ≥ 2; (8) and (9) show that 4 must be a common divisor of bl − 1 and bl + 1, contrary to (5). Recall from number theory that if p is an odd prime and a an integer not divisible by p, then the order of a modulo p is the least positive integer k such that ak ≡ 1 (mod p). When a ≡ 1 (mod p); then obviously the order of a modulo p equals 1. Otherwise the order of a is ≥ 2. The following Lemma is well-known in number theory and easily provable by using the division algorithm. Lemma 1. Let p be an odd prime; and b a positive integer not divisible by p, such that b ≡ 1 (mod p). If m is a positive integer, such that bm ≡ 1 (mod p), then m is divisible by d, d | m, where d is the order of b modulo p; d ≥ 2. Since by Fermat’s (Little) Theorem, bp−1 ≡ 1 (mod p). We have the following corollary of Lemma 2.

Lemma 2. Let p be an odd prime; b ∈ Z+ , b ≡ 0, 1 (mod p). Then the order of b modulo p is a divisor d ≥ 2 of p − 1. Back to the problem. Subcase 2a. Assume that b ≡ 1 (mod p).

We will prove that this case is impossible, it leads to a contradiction regardless of whether n is a prime or not. From b ≡ 1 (mod p) and (1) we have
−1 bn−1 + bn−2 + · · · + b + 1 = pk = bb− 1 and b = p · t + 1, for some positive integer t.
n

(10)

Since b ≡ 1 (mod p); bn−1 ≡ bn−2 ≡ · · · ≡ b ≡ 1 (mod p). And so by (10) we have, 1 + 1 + · · · + 1 ≡ 0 (mod p) ;
ßÞ

Clearly, since n ≥ 3 and b = p · t + 1, we have

terms n ≡ 0 (mod p) and so n ≥ 3.
n

(11)

bn−1 + bn−2 + · · · + b2 + b +1 ≥ b2 + b +1 = (p · t +1)2 +(p · t +1)+1 > p2 , which shows that we must have k ≥ 3 in (11). Now, bn − 1 = pk · (b − 1) ⇒ (p · t + 1)n − 1 = pk · (pt). Expanding with the binomial expansion yields (pt) +
n

n 1

(pt)

n− 1

+ ··· +

n n−1

(pt) + 1 − 1 = pk+1 · t.

(12)

Let pf be the highest power of p dividing n (see (11)); and pe be the highest power of p dividing t. Then n = n1 · pf , t = t1 · pe , f ≥ 1; and e ≥ 0 and n1 · t1 ≡ 0 (mod p). The highest power p dividing the left hand side of (12); is the highest n ¡ power of p dividing the term n− p · t = n · p · t = n1 · t1 · pf +e+1 . The right 1 k+1 k+1+e hand side is p ·t = p . Thus, we must have 1 + f + e = k + 1 + e; and so f = k; n = n1 · pk . If we look at the term (pt)n (left hand side of (12); k k (pt)n = pn · tn = pn1 ·p · tn1 ·p > pk+1 · t, since t ≥ 1 and n1 pk ≥ pk > k +1, in view of p ≥ 3 and k ≥ 3. We have a contradiction to (12). In subcases 2b and 2c below; we argue by contradiction. We assume that b ≡ 1 (mod p) and n to be a composite number ≥ 4, and we show that this leads to a contradiction. Observe that a composite number is either a prime power with exponent at least 2; or otherwise it has two distinct prime bases in its prime factorization. Subcase 2b. Assume that b ≡ 1 (mod p); and n ≥ 4 has at least two prime bases in its prime factorization into prime powers. This then implies that we can write, n = n1 · n2 ≥ 4 (actually 6) with 1 < n1 , n2 < n; where n1 , n2 are relatively prime positive integers; (n1 , n2 ) = 1. (13)

In other words, n can be written as a product of two relatively prime proper positive divisors. We have, bn1 n2 − 1 = pk · (b − 1); (bn1 )n2 − 1 = pk · (b − 1); (bn2 − 1)[(bn1 )n2 −1 + · · · + bn1 +1] = pk · (b − 1); (b − 1) · (bn2 −1 + · · · + b + 1)[(bn1 )n2 −1 + · · · + bn1 + 1] = pk · (b − 1); (bn1 −1 + b + 1) · [(bn1 )n2 + · · · + bn1 + 1] = pk (14)

Each factor on the left hand side of (14) is greater than 2. Clearly then (14) implies that each factor must be a power of p, so that, bn1 −1 + · · · + b + 1 = pk1 k1 a positive integer. Similarly, by factoring bn − 1 = bn1 n2 − 1 = (bn2 )n1 − 1 = (bn2 − 1)[(bn2 )n1 −1 + · · · + bn2 + 1] and using similar reasoning; we obtain, bn2 −1 + · · · + b + 1 = pk2 for some positive integer k2 From (15) and (16) we get bn1 − 1 = pk1 · (b − 1) and bn2 − 1 = pk2 · (b − 1) (17) (16) (15)

Thus, (17) ⇒ (bn1 ≡ 1 (mod p) and bn2 ≡ 1 (mod p)). By Lemma 1, both n1 and n2 must be divisible by the order ρ of b modulo p; since b ≡ 1 (mod p); we have ρ ≥ 2. And 2 ≤ ρ | n1 and ρ | n2 , a contradiction of (n1 , n2 ) = 1 in (13). Since j ≥ 2, bn − 1 is divisible by bq − 1; and hence by bq − 1 as well. Indeed, j 2 j −2 2 j −2 bq − 1 = bq ·q − 1 = (bq )q −1 bn − 1 = (bq )q = (bq
2 j −2

Subcase 2c. n = q j , j ≥ 2, where q is a prime number.
2

(bq − 1), if j = 2 2 j −2 2 − 1) · [(bq )q −1 + · · · + (bq ) + 1], if j ≥ 3.

−1

(18)

In either (j = 2 or j ≥ 3) case, from bn − 1 = pk · (b − 1); after the cancelation of the factor b − 1 from both sides of the last equation; it follows that bq−1 + bq−2 + · · · + b + 1 must be a power of p. Indeed, since bq − 1
2

= (bq )q − 1 = (bq − 1)[(bq )q−1 + · · · + bq + 1]

= (b − 1)(bq−1 + · · · + b + 1)[(bq )q−1 + · · · + bq + 1],

thus, bq−1 + · · · + b + 1 = pλ , for some positive integer λ. And hence bq − 1 = pλ · (b − 1) ⇒ bq ≡ 1 (mod p) . (19) By Lemma 1, (19) implies that the order d of b modulo p must divide q : 2 ≤ d | q . But q is a prime and therefore it follows that d = q . The order of b modulo p must equal the prime q . Hence, by Lemma 2 it follows that p − 1 ≡ 0 (mod q ) . (20)

Recall from (18) that bq − 1 is a factor of bn − 1. bn − 1 = (bq − 1) · N, N a positive integer and bn − 1 = pk (p − 1) Thus, ((bq )q − 1) · N = pk · (b − 1); (bq − 1) · N = pk · (b − 1);
2 2

(bq − 1)
ßÞ

[(bq )q−1 + · · · + bq + 1] = pk · (b − 1);

(b−1)(bq −1 +···+b+1)

So So each factor (on the left hand side of (21) must be a power of p: In particular (bq )q−1 + · · · + bq + 1 = pw , w ≥ 1. By (19) and (22) we deduce that, 1 + 1 + · · · + 1 ≡ 0 (mod p) ⇒ q ≡ 0 (mod p) ;
ßÞ
q

(bq−1 + · · · + b + 1)[(bq )q−1 + · · · bq + 1] = pk .

(21) (22)

(23)

and since p and q are both primes; (23) implies p = q . But (20) then implies p − 1 ≡ 0 (mod p), an impossibility. 2 D F C

In square ABCD the points E and F are chosen in the interior of sides BC and CD , respectively. The line drawn from F perpendicular to AE passes through the intersection point G of AE and BD . A point K is chosen on F G such that |AK | = |EF |. Find ∠EKF . A

K G E B

Solved by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; Michel Bataille, Rouen, France; Geoﬀrey A. Kandall, Hamden, CT, USA; Bruce Shawyer, Memorial University of Newfoundland, St. John’s, NL; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Amengual Covas.

100

D 45◦

F 45◦

K E G A B

With right angles at D and G, AGF D is a cyclic quadrilateral, and in its circumcircle, ∠GF A = ∠GDA = ∠BDA = 45◦ Thus, right-triangle AGF is isosceles with AG = GF . Since we also have AK = EF , right triangles AGK and F GE are congruent (side-angle-side) with KG = GE . Hence KGE is isosceles right-angled. Thus, ∠EKF = 180◦ −∠GKE = ◦ 180 − 45◦ = 135◦ .

Find all continuous functions f : R → R such that for all reals x, y ∈ R, f (x + f (y )) = y + f (x + 1) .

Solved by Michel Bataille, Rouen, France. The functions x → x + 1 and x → −x + 1 are solutions (readily checked). We show that there are no other solutions. Let f : R → R be a continuous function satisfying the given functional equation that we will denote by (E ). Taking x = −1 in (E ) yields f (f (y ) − 1) = y + f (0) for all y ∈ R. An immediate consequence is f (y ) = f (y ) =⇒ y = y and f is injective. Letting y → 0 and using the continuity of f gives f (x + f (0)) = f (x +1), hence f (0) = 1 and so f (f (y ) − 1) = y + 1 for all y ∈ R. Using (E ), it follows that f (x + y ) = f ((x − 1)+(y +1)) = f (x − 1+ f (f (y ) − 1)) = f (y ) − 1+ f (x). Thus, the continuous function g : x → f (x) − 1 satisﬁes the Cauchy functional equation g (x + y ) = g (x) + g (y ). It is known that g must be a linear function x → ax and so f must be x → ax + 1. Returning to (E ), we must have a(x + ay + 1) + 1 = y + a(x + 1) + 1 for all x, y that is, a2 y = y for all y . Hence a = 1 or a = −1 and the conclusion follows.

101

Next we turn to solutions from our readers to problems of the Russian Mathematical Olympiad 2007, 10th grade, given at [2010: 150–151]. an , let m = min{a0 , a0 +a1 , . . . , a0 +a1 +· · ·+an }. Prove that P (x) ≥ mxn for all x ≥ 1. Solved by Oliver Geupel, Br¨ uhl, NRW, Germany. By the deﬁnition of m, it holds a 0 − m ≥ 0, a 0 + a 1 − m ≥ 0, ..., a 0 + a 1 + . . . + a n − m ≥ 0.

2. (A. Khrabrov) Given a polynomial P (x) = a0xn + a1 xn−1 + · · · + an−1x +

Hence, we have for x ≥ 1 P (x) = a0 xn + a1 xn−1 + · · · + an = a0 (xn − xn−1 ) + (a0 + a1 )(xn−1 − xn−2 ) + · · · + (a0 + a1 + · · · + an−1 )(x − 1) + (a0 + a1 + · · · + an ) + (a0 + a1 + a2 )(xn−2 − xn−3 )

+ mxn ≥ mxn .

= (a0 − m)(xn − xn−1 ) + (a0 + a1 − m)(xn−1 − xn−2 ) + · · · + (a0 + a1 + · · · + an − m)(x − 1) + (a0 + a1 + · · · + an − m)

3. (V. Astakhov) In an acute triangle ABC , BB1 is a bisector. Point K is chosen on the smaller arc BC of the circumcircle, such that B1 K and AC are perpendicular. Point L is chosen on line AC such that BL and AK are also perpendicular. Line BB1 meets the smaller arc AC at point T . Prove that points K , L, T are collinear.
Solved by Oliver Geupel, Br¨ uhl, NRW, Germany. Fixing the typo in the problem, we substitute the hypothesis B1 K ⊥ AC by the condition B1 K ⊥ BC . Let D be the point of intersection of the lines BC and B1 K , and let the lines AK and BL intersect at the point E . Since the inscribed angles ∠CBK = ∠DBK and ∠CAK = ∠LAE have equal size, the right triangles DBK and EAL are similar. Hence, ∠BKB1 = ∠BKD = ∠ALE = ∠BLB1 , that is, the points B , B1 , K , and L are concyclic. B K

A T

B1 E D C L

102

Thus, ∠BB1 D = ∠BB1 K = ∠BLK = ∠KLE, which implies that the right triangles BB1 D and KLE are similar. Therefore, ∠T KA = ∠T BA = ∠T BC = ∠B1 BD = ∠LKE = ∠LKA. Consequently, the point L is on the line T K , which completes the proof.

6. (S. Berlov) Two circles ω1 and ω2 intersect at points A and B . Let P Q and RS be the segments of common tangents to these circles (points P and R lie on ω1 , while points Q and S lie on ω2 ). Ray RB intersects ω2 again at point W . If RB P Q, ﬁnd the ratio RP/BW .
Solved by Geoﬀrey A. Kandall, Hamden, CT, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give Kandall’s solution. Q P ω1 O1 O2 y B y N W

ω2

x R M

S Let O1 (O2 ) be the centre of ω1 (ω2 ). Extend P O1 (QO2 ) to meet RW at M (N ). Since ∠O1 P Q and ∠O2 QP are right angles and RW P Q, it follows that ∠O1 M N and ∠O2 N M are right angles (so P QN M is a rectangle) and M (N ) is the midpoint of RB (BW ). Let RB = 2x, BW = 2y . Then RS = P Q = M N = x + y . Since RS 2 = RW · RB , we have (x + y )2 = 2(x + y ) · 2x. Dividing by x + y , we obtain x + y = 4x, that is y = 3x. Therefore,
RB x 1 = = . BW y 3

103

BOOK REVIEWS
Amar Sodhi
The Calculus of Friendship by Steven Strogatz Princeton University Press, 2009 ISBN 978-0-691-13493-2 hardcover, 166 + xii pp. US$19.95 Reviewed by Georg Gunther, Sir Wilfred Grenfell College (MUN), Corner Brook, NL New books and new cars have this in common: one cannot wait to open them. Their lure is irresistible. If chemists could ever be bothered to synthesize the smell of a new car, undoubtedly L’eau d’Auto would soon become the fragrance of choice. Cracking open the cover of a new book carries a similar appeal: there is a promise of things to come, an anticipation of words to be read, a sense of excitement. And the magic happens when the book delivers on this promise, when the words capture and enthrall. The Calculus of Friendship is such a book. It is a remarkable story, based upon a thirty-year correspondence between the author and his former high-school mathematics teacher. The letters, many of which are reproduced in the book, are all about calculus. It is the passion and love for this discipline that connects these two men as they move through their respective lives, and it is calculus, with its unparalleled ability to provide deep and profound metaphors, that provides the connecting context. “Some books are to be tasted ”, wrote Francis Bacon, “others to be swallowed, and some few to be chewed and digested ”. This book certainly falls into Bacon’s third category. The author, Steven Strogatz, is the Jacob Gould Schurman Professor of Applied Mathematics at Cornell University. He has contributed widely to the study of synchronization in dynamical systems. In addition to writing numerous mathematical papers and books, he is the author of the best selling Sync: the emerging science of spontaneous order (2003). The book being reviewed here is written with a deceptive simplicity and grace of style that draws the reader eﬀortlessly from page to page. It provides a rare glimpse into the mind and heart of a top-notch mathematician. Always fascinating, at times deeply profound, it escapes being maudlin by the sheer simplicity of its prose, the elegance of its mathematics and the, at times, brutal self-appraisal of the author. There is a great deal of fascinating and exciting mathematics in this book. Interesting problems are tossed back and forth between former teacher and former student in a wonderful point and counterpoint of question and answer, leading to yet more questions, more avenues of investigation. Reading these letters, one is struck with the exuberance of the intellectual exercises being posed, with the thrill of the exploration, and the sheer fun of doing mathematics.

104

The book is superbly organized. The successive chapters follow the chronology of two lives, from 1974 to the present; each explores one of the mathematical topics raised in the course of the correspondence between these two men. In addition to including the actual text of the original letters, the author provides some of the mathematical background, as well as placing these discussions within a personal context. The mathematical content of this book is far from trivial. It covers diverse topics, from chase problems, discussion of irrationality, subtleties of inﬁnite series, explorations of randomness, all the way to chaos theory and questions about bifurcation. This book is well worth reading for its mathematical content alone. However, this book is so much more than simply a book about calculus. It is a testimonial to the bonds of friendship and to the complex and ever evolving connections between teacher and student. In mathematics, ﬁxed-point theorems play a central role. The major focus of Strogatz’s mathematical work is the science of synchrony: how spontaneous order can arise within inherently chaotic systems. In The Calculus of Friendship, the author explores one more facet of this theme: “like calculus itself, this book is an exploration of change. It’s about the transformation that takes place in a student’s heart, as he and his teacher reverse roles, as they age, as they are buﬀeted by life itself. Through all these changes, they are bound together by a love of calculus. For them it is more than a science. It is a game they love playing together – so often the basis of friendship between men – a constant while all around them is ﬂux ”. Towards the end of this book, Steven Strogatz acknowledges that his former mentor taught him “something profoundly mathematical, about how to live ”; the lesson is to “balance the inevitable against the unforeseeable, the two sides of change in this world. The orderly and the chaotic. The changes that calculus can tame, and the ones it cannot. He confronts them all, and not, like Zeno, with his mind alone but also with his heart ”. “The Calculus of Friendship ” is recommended for all. Practicing mathematicians will ﬁnd much that might be new to them, as well as meeting up with many old friends. Students will get glimpses into numerous directions that will compel their interest. Non-mathematicians can skip the hard stuﬀ, the mathematical equations, and ﬁnd a diﬀerent set of valuable truths in the vastly more complicated rules that lie at the heart of human relationships.

105

Crux Chronology
J. Chris Fisher 1 Highlights
March 1975, Eureka. Crux Mathematicorum with Mathematical Mayhem began life as the journal Eureka. Early in 1975, six members of the Carleton-Ottawa Mathematics Association (COMA) met privately and decided to launch Eureka to provide a forum for the exchange of mathematical information, especially interesting problems and solutions, among the members of the mathematical community in the Ottawa region, students and teachers alike. [1975 : 1] Three of the six were from Algonquin College: L´ eo Sauv´ e, who served as the ﬁrst editor (until 1986); Fred G.B. Maskell, COMA’s secretary-treasurer who became the ﬁrst managing editor (through 1984); and H.G. Dworschak. The other three were Viktors Linis of the University of Ottawa, R. Duﬀ Butterill of the Ottawa Board of Education, and Richard J. Semple of Carleton University. While Dworschak and Linis provided some support and numerous problems, it was L´ eo and Fred who provided the the energy and dedication required to turn their modest local venture into a journal that within two years developed an international following, a following that included some of the world’s ﬁnest mathematicians. In the words of his friend and colleague Kenneth S. Williams [1987 : 240-242], “L´ eo’s dedication and hard work, his broad knowledge and love of mathematics, his careful eye for detail, all enabled Crux to grow from a four-page problem sheet to the international mathematical problem-solving journal that it is today.” Many of Crux’s faithful contributors became L´ eo’s friends; their tributes to him can be found in the issue dedicated to him [1986 : 163-168]. March 1978, Crux Mathematicorum. Beginning with volume 4 number 3, the name of the journal changed to Crux Mathematicorum. After 32 issues had been published under the name Eureka, it was discovered that there was a journal Eureka published once a year by the Cambridge University Mathematical Society. Sauv´ e chose the new name: it is an idiomatic Latin phrase meaning a puzzle or problem for mathematicians [1978 : 89-90]. January 1979, the Olympiad Corner. Murray S. Klamkin initiated the Olympiad Corner [1979 : 12] to “provide, on a continuing basis, information about mathematical contests taking place in Canada, the U.S.A., and internationally.” It would also provide “practice sets of problems on which interested students could test and sharpen their mathematical skills and thereby possibly qualify to participate in some Olympiad.” Klamkin served as editor of the ﬁrst 80 columns, through December, 1986.

106

December 1984, Fred. Frederick G. B. Maskell (1904-1985) steps down as managing editor [1984 : 340], and died a few months later [1985 : 14, 34]. He was replaced by Kenneth S. Williams. October 1 1985, the CMS. For nearly eleven years Crux was published by Algonquin College and sponsored by the Carleton-Ottawa Mathematics Association. In 1979 The Canadian Mathematical Olympiad Committee and the Carleton University Mathematics Department added their support, joined later by the University of Ottawa Mathematics Department. In March, 1985, the Canadian Mathematical Society was asked to assume responsibility for its publication; that organization called for suggestions concerning the future of the journal and began the search for a new editor. [1985 : 100] On October 1, 1985, Crux became an oﬃcial publication of the Canadian Mathematical Society. [1985 : 234, 236] The October 1985 issue was dedicated to Philip Kileen, President of Algonquin College, who strongly supported Crux during its ﬁrst eleven years. For all future editors, the mathematics department of the current editor’s university lent support to the journal. February 1986, G.W. (Bill) Sands. Starting with Volume 12, number 2, Bill Sands (University Calgary) became the second editor (through 1995) . [1986 : 17, 37] He announced [1986 : 65] that the September issue would be dedicated to L´ eo Sauv´ e. January 1987 Robert Woodrow. The Olympiad Corner’s Murray Klamkin was replaced by Robert E. Woodrow [1986 : 263; 1987 : 2-3, 34], but Klamkin continued making valuable contributions to Crux until his death in 2004. June 19, 1987, L´ eo. L´ eo Sauv´ e died (Dec. 12, 1921 - June 19, 1987). [1987 : 240-242] Not only was L´ eo the consummate scholar, but he injected into each page of Crux a spicy liveliness that will probably never be matched. When a problem failed to attract solutions to his liking, he would write his own, assuming the guise of his contributor persona, Gali Salvatore (Salvatore = Sauv´ e). If there happened to be a gap he could not ﬁll, he would attach an editorial comment in which he criticized Salvatore’s proof; see [1984 : 31] where he attacked the solution (his own!) to problem 783, complaining that the whole argument rested on a formula that came without a proof “presumably because he (or she: is Gali a man’s name or a woman’s?) felt it was ‘easy.’ ” Edith Orr (= editor) was another persona who was able to make comments that an editor could not; her poetry was sometimes so racy that it would not have been allowed to pass through the mail had the postal inspectors thought to look closely at a math journal! See her comment on lambs [1983 : 211-212], which L´ eo, wearing his editor’s hat, dismissed as “scrofulous.” January 1988, Colour. Starting with volume 14, each issue came with a coloured front and back 1 cover; although it continued to be printed on 8 2 × 11 three-holed paper and

107

stapled together, it began to take on a more professional look. January 1989, Kenneth Williams. Kenneth Williams steps down from his position of managing editor and his role as technical editor. [1988 : 300-301] His duties were taken over by Graham P. Wright, the executive director of the Canadian Mathematical Society, together with members of his staﬀ; Wright served (except for a short period between 1999 and 2000 when the position was ﬁlled by Robert Quackenbush) until his retirement in May, 2009. January 1991, The Editorial Board. After ﬁve years of performing all the editorial duties with only occasional help from others, Sands organized the ﬁrst formal editorial board. Robert Woodrow was promoted to joint editor, and six others agreed to form the board. The board members will be listed later. January 1995, New format and Skoliad. Starting with volume 21, the appearance was changed to its current 10-inch format with purple covers. Also, the Skoliad Corner was inaugurated with its simpler “Pre-Olympiad” problem sets, with Robert Woodrow as its ﬁrst editor [1995 : 5]. The name was suggested by Richard Guy, who had searched a map of the Mount Olympus area, ﬁnding the mountain Scollis, and then making a portmanteau of this with scholar and Olympiad. At about 1/3 the height of Mount Olympus, Mount Scollis seems the appropriate metaphor for a junior Olympiad. It was later learned that skolion, an unrelated ancient Greek word, referred to songs sung by invited guests at banquets in ancient Greece extolling the virtues of the gods or heroic men! It seems to have come from the word for crooked, which seems appropriate for a problem section: short, diverse, and slightly twisted entertainments. January 1996, Bruce L.R. Shawyer. Bruce Shawyer (Memorial University of Newfoundland) became the third editor starting with volume 22 [1995 : 354; 1996 : 1]; he served until December 2002. Colin Bartholomew, also of Memorial University, served as assistant editor for one year, after which Clayton Halfyard took over that position from 1997 through 2002. With the new editor, the journal went from 10 issues of 36 pages (360 pages per volume) to 8 issues of 48 pages (384 pages per volume). Although the January and June issues were dropped, there were 24 extra pages per year, allowing increased eﬃciency of printing and a decrease in mailing costs. The journal went on-line for subscribers later that year [1996 : 289]. The Academy Corner. Shawyer produced the Academy Corner during his tenure as editor. It dealt with problem solving at the undergraduate level. [1996 : 28] It ended with column 49 when he stepped down in 2002. [2002 : 480] February 1997, Crux Mathematicorum with Mathematical Mayhem. Mathematical Mayhem had been founded in 1988 by Ravi Vakil and Patrick Surry, two Canadian IMO alumni, as a journal of high-school and college level

108

mathematics written by and for students. [1996 : 337; 1997 : 1-2, 30-31] When it amalgamated with Crux in volume 23, it brought additional high-school level material and gained, in return, a wider exposure. It was agreed that the new journal would continue the volume numbering of Crux as well as maintain its general external appearance. The number of pages per issue jumped from 48 to 64. The then current Mayhem editor Naoki Sato and assistant editor Cyrus Hsai continued in their positions for four years, through December 2000. Other editors and staﬀ are listed below. January 2002, fran¸ cais. The statement of all problems would henceforth appear in both English and French. Jean-Marc Terrier has been translating from the start; other translators, serving for various periods, have been Hidemitsu Sayeki, Martin Goldstein, and Rolland Gaudet. January 2003, Jim Totten. James Edward Totten (University College of the Cariboo, renamed Thompson Rivers University in 2005) became the fourth editor starting with volume 29. [2002 : 287-288; 2003 : 1] His plan was to step down in June of 2008, after serving the ﬁnal six months as co-editor, but he died on March 9, 2008 in his 61st year (born August 9, 1947). His Assistant editor was Bruce Crofoot. January 2006, Mayhem on-line. The Mayhem portion of the journal became open to the public on the internet starting with volume 32. Currently, all issues of Crux with Mayhem become free to the public after ﬁve years, while the Mayhem portion is always available for free. January 2008, V´ aclav (Vazz) Linek. Vazz Linek (University of Winnipeg) became co-editor of the journal, then became the journal’s ﬁfth editor starting in July. [2007 : 449; 2008 : 193-194] His assistant editor is Jeﬀ Hooper. September 2009. Johan Rudnick became the executive director of the Canadian Mathematical Society and, thereby, the new managing editor of Crux.

2 Crux Editors
Editors-in-chief L´ eo Sauv´ e G.W. Sands G.W. Sands and Robert Woodrow Bruce L.R. Shawyer James Totten V´ aclav (Vazz) Linek and James Totten Vazz Linek Shawn Godin March 1975 through January 1986 February 1986 through December 1990 January 1991 through December 1995 January 1996 through December 2002 January 2003 through May 2008 January 2008 through May 2008 June 2008 through March 2011 April 2011 to present

109

Managing Editors Frederick G.B. Maskell Kenneth S. Williams Graham P. Wright Robert Quackenbush Graham P. Wright Johan Rudnick Olympiad Corner Murray S. Klamkin Robert E. Woodrow

March 1975 through November 1984 December 1984 through December 1988 January 1989 through September 1999 October 1999 through December 2000 January 2001 through August 2009 September 2009 to present

1979 through 1986 1987 through 2010 1991 to present 1991 through 2003 1991 through September 1999 1991 through 1998 1991 through 1994 1993 through 2010 1994 1994 through 2002 (when he became editor-in-chief) 1995 through 1999 1998 through 2002∗ 1999 through 2001 2000 through 2007 2000 through 2009 2002-2009 (book review editor until 2008) 2003 through 2005 2003 to present 2006 through 2009 2008 through 2010 2009 to present 2009 to present 2010 2010 2011 to present 2011 to present 2011 to present 2011 to present

Crux Editorial Board J. Chris Fisher Richard Guy Denis Hanson (articles) Andy Liu (book reviews) Richard Nowakowski Edward T.H. Wang Rod De Peiza Jim Totten Catherine Baker Loki J¨ orgenson Alan Law (book reviews) Bruce Gilligan (articles) Iliya Bluskov John Grant McLoughlin Richard (Rick) Brewster Bruce Shawyer (editor-at-large) Maria Torres James Currie (articles) Amar Sodhi (book reviews) Nicolae Strungaru Jonatan Aronsson Dzung Minh Ha Robert Craigen Robert Dawson (articles) Chris Grandison Cosmin Pohoata
∗

Although he was responsible for the on-line edition starting in 1996, he was a board member

for only ﬁve years.

Skoliad Corner Robert Woodrow Shawn Godin Robert Bilinski V´ aclav Linek Lily Yen and Mogens Hansen

1995 through May 2001 September 2001 through 2004 2005 through 2008 February 2009 March 2009 to present

110

3 Mathematical Mayhem
Mayhem editors Before joining Crux: Patrick Surry, Ravi Vakil, Philip Jong, Jeﬀ Higham, J.P. Grossman, Andre Chang, and Naoki Sato. After the amalgamation: Naoki Sato 1997 through 2000 Shawn Godin 2001 through 2006 Jeﬀ Hooper 2007 Ian VanderBurgh 2008 through 2010 Mayhem assistant editors Cyrus Hsai 1997 Chris Cappadocia 2001 John Grant McLoughlin 2003 Jeﬀ Hooper 2006 Ian VanderBurgh 2007 Lynn Miller 2011 through 2000 through 2002 through 2005

to present

Mayhem editorial staﬀ (various terms) Richard Hoshino, Wai Ling Yee, Adrian Chan, Jimmy Chui, David Savitt, Donny Cheung, Paul Ottaway, Larry Rice, Dan MacKinnon, Ron Lancaster, Eric Robert, Monika Khbeis, Mark Bredin.

4 Special Issues and Articles
Special Issues 3:4 April 1977, The Gauss bicentennial Issue [1976 : 131, 161] 3:10 December 1977, Special Morley Issue; see also [1978 : 33-34, 132; 1978 : 304-305] 12:7 September 1986, Issue dedicated to L´ eo Sauv´ e [1986 : 65] 27:2 March 2001, Murray Klamkin 80th Birthday Issue [2001 : 65-85] 31:5 September 2005, Issue dedicated to Murray S. Klamkin [2004 : 361] 35:5 September 2009, Issue dedicated to James Edward Totten [2008 : 193] Encyclopedic Articles L´ eo Sauv´ e, The Celebrated Butterﬂy Problem. [1976 : 2-5] L´ eo Sauv´ e, The Steiner-Lehmus Theorem, and Charles W. Trigg, A Bibliography of the Steiner-Lehmus Theorem. [1976 : 19-24, 191-193]

5 Contributors and Friends
Honours and Awards Murray Klamkin (University of Alberta), M.A.A. Distinguished Service Award [1988 : 33]; David Hilbert International Award [1992 : 224] Marcin E. Kuczma (University of Warsaw), David Hilbert International Award [1992 : 224]

111

Ronald Dunkley (University of Waterloo), Order of Canada [1996 : 106] Andy Liu (University of Alberta), David Hilbert International Award [1996 : 201]; Outstanding University Professor [1999 : 65-66] Bruce Shawyer (Memorial University of Newfoundland), Adrien Pouliot Award [1998 : 19] Francisco Bellot Rosado (Institute Emilio Ferrari), Erd¨ os Prize [2000 : 320] Contributor Proﬁles R. Robinson Rowe [1977 : 92,184,248; 1978 : 6,63,129,189] Kestraju Satyanarayana [1981 : 294] Jack Garfunkel [1990 : 318] Leon Bankoﬀ [1995 : 292] Jordi Dou [2002 : 56; 2006 : 65] K.R.S. Sastry [2006 : 2] Toshio Seimiya [2000 : 114; 2006 : 129] Christopher J. Bradley [2006 : 257] D.J. Smeenk [2006 : 353] Michel Bataille [2007 : 1] Richard K. Guy [2007 : 65] Walther Janous [2007 : 385] Peter Y. Woo [2008 : 1] Arkady Alt [2010 : 65] John G. Heuver [2010 : 193] Death Notices Richard J. Sempel, Carleton University (1930-1977) [1977 : 188] R. Robinson Rowe (1896-1978) [1978: 152] Herman Nyon, Paramaribo, Surinam (? - 1982) [1982 : 268] Viktors Linis, University of Ottawa (1916-1983) [1983 : 192] Kestraju Satyanarayana (1897-1985) [1985 : 268] Geoﬀrey James Butler, University of Alberta (1944-1986) [1986 : 203] Samuel L. Greitzer, Rutgers University (1905-1988) [1988 : 161-162] Charles Trigg, San Diego, CA (1898-1989) [1989 : 224] Jakob T. Groenman, Arnhem, The Netherlands [1989 : 96] W.J. Blundon, Memorial University of Newfoundland (1916-1990) [1990 : 160] Jack Garfunkel (1910-1990) [1991 : 64] Oene Bottema, Delft, The Netherlands (1901-1992) [1993 : 32] Peter Joseph O’Halloran, University of Canberra (1931-1994) [1994 : 248-249] P´ al Erd¨ os (1913-1996) [1996 : 339] Leon Bankoﬀ, Los Angeles, CA (1908-1997) [1997 : 145] Jessie Lei, University of Toronto (1980-2000) [2000 : 1-2] Herta Freitag, Roanoke, VA (1908-2000) [2000 : 535] H.S.M. Coxeter, University of Toronto (1907-2003) [2003 : 232] Murray S. Klamkin, University of Alberta (1921-2004) [2004 : 361] Robert Barrington Leigh, University of Toronto (1986-2006) [2006 : 453] Jordi Dou (1911-2007) [2007 : 457] James Edward Totten, Thompson Rivers University (1947-2008) [2008 : 194]

112

PROBLEMS
Toutes solutions aux probl` emes dans ce num´ ero doivent nous parvenir au plus tard le 1er septembre 2011. Une ´ etoile ( ) apr` es le num´ ero indique que le probl` eme a ´ et´ e soumis sans solution. Chaque probl` eme sera publi´ e dans les deux langues oﬃcielles du Canada (anglais et fran¸ cais). Dans les num´ eros 1, 3, 5 et 7, l’anglais pr´ ec´ edera le fran¸ cais, et dans les num´ eros 2, 4, 6 et 8, le fran¸ cais pr´ ec´ edera l’anglais. Dans la section des solutions, le probl` eme sera publi´ e dans la langue de la principale solution pr´ esent´ ee. La r´ edaction souhaite remercier Jean-Marc Terrier, de l’Universit´ e de Montr´ eal, d’avoir traduit les probl` emes.

´ Propos´ e par Neculai Stanciu, Ecole secondaire George Emil Palade, Buz˘ au, Roumanie. R´ esoudre le syst` eme d’´ equations x(y + 1) x−1 = 7, y (z + 1) y−1 = 5 , et z (x + 1) z−1 = 12 .

3613.

o` u x, y et z sont des entiers positifs.

3614.

Propos´ e par Neven Juriˇ c, Zagreb, Croatie. ` A partir des d´ ecimales cons´ ecutives de 1 on obtient l’ensemble des points 7 A(1, 4), B (4, 2), C (2, 8),. . . dans le plan. Montrer que tous ces points appartiennent a ` une mˆ eme ellipse. Calculer l’aire de cette ellipse.

3615.
Vietnam.

Propos´ e par Pham Van Thuan, Universit´ e de Science de Hano¨ ı, Hano¨ ı,

Montrer que si x, y, z ≥ 0 et x + y + z = 1, alors √ xy z + xy +√ zx 1 +√ ≤ . x + yz y + zx 2 yz

3616.

Propos´ e par Dinu Ovidiu Gabriel, Valcea, Roumanie.
æ

Calculer L = o` u k ∈ R. lim n
2k

arctan(nk ) nk

n→∞

−

arctan(nk + 1) nk + 1

3617.

Propos´ e par Michel Bataille, Rouen, France.

Soit r un nombre rationnel positif. Montrer que si r r est rationnel, alors r est un entier.

113

3618.

Propos´ e par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie.

Soit α > 3 un nombre r´ eel. Trouver la valeur de
∞ ∞

n (n + m)α

n=1 m=1

3619.
que

Propos´ e par Pham Kim Hung, ´ etudiant, Universit´ e de Stanford, Palo ´ . Alto, CA, E-U Soit a, b et c trois nombres r´ eels non n´ egatifs tels que a + b + c = 3. Montrer (a2 b − c)(b2 c − a)(c2 a − b) ≤ 4(ab + bc + ca − 3a2 b2 c2 ) .

3620.

Propos´ e par John G. Heuver, Grande Prairie, AB.

Soit P un point int´ erieur du t´ etra` edre ABCD et d´ esignons par A ,B , C et D les points d’intersection des droites AP , BP , CP et DP avec les faces correspondantes oppos´ ees. On a alors AP BP CP DP PA PB PC PD = 3+2 AP + BP + CP + DP

PB PC PD AP CP AP DP + + + PA PB PA PC PA PD BP CP BP DP CP DP + + + . PB PC PB PD PC PD

PA AP BP

3621.
que 27 128

Propos´ e par Titu Zvonaru, Com´ ane¸ sti, Roumanie.

Soit a, b et c trois nombres r´ eels non n´ egatifs avec a + b + c = 1. Montrer 4 1+a 4 1+b 4 1+c 3 ab + bc + ca

[(a − b)2 +(b − c)2 +(c − a)2 ]+

≤

3622

. Propos´ e par George Tsapakidis, Agrinio, Gr` ece.

On donne un quadrilat` ere ABCD . (a) Trouver des conditions n´ ecessaires et suﬃsantes sur les cˆ ot´ es et les angles de ABCD pour qu’il existe un point int´ erieur P tel que deux droites perpendiculaires issues de P divisent le quadrilat` ere ABCD en quatre quadrilat` eres d’aire ´ egale. (b) D´ eterminer P .

114

3623.

Propos´ e par Michel Bataille, Rouen, France.

Soit z1 , z2 , z3 , z4 quatre nombres complexes distincts mais de mˆ eme module, α = |(z3 − z2 )(z3 − z4 )|, β = |(z1 − z2 )(z1 − z4 )| et u( ) = α(z1 − z4 ) + β (z3 − z4 ) α(z1 − z2 ) + β (z3 − z2 ) .

Montrer que u(+1) ou u(−1) est un nombre r´ eel.

3624.

Propos´ e par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie.
∞ n=1

Calculer la somme (−1)n−1 n 1− 1 2 + 1 3 − ··· + (−1)n+1 n .

3625.
Vietnam.
Ö

Propos´ e par Pham Van Thuan, Universit´ e de Science de Hano¨ ı, Hano¨ ı,

Soit a, b et c trois nombres r´ eels positifs. Montrer que a a+b + b b+c
Ö

c c+a

≤ 2

abc (a + b)(b + c)(c + a)

.................................................................

3613. Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania.
Solve the system of equations x(y + 1) x−1 = 7, y (z + 1) y−1 = 5 , and z (x + 1) z−1 = 12 .

Where x, y, and z are positive integers.

3614.

Proposed by Neven Juriˇ c, Zagreb, Croatia.

Taking consecutive decimal digits of 1 the set of points A(1, 4), B (4, 2), 7 C (2, 8),. . . in the plane is obtained. Prove that all these points belong to the same ellipse. Compute the area of the ellipse.

3615.
Vietnam.

Proposed by Pham Van Thuan, Hanoi University of Science, Hanoi,

Prove that if x, y, z ≥ 0 and x + y + z = 1, then √ xy z + xy +√ yz x + yz zx 1 +√ ≤ . y + zx 2

115

3616.

Proposed by Dinu Ovidiu Gabriel, Valcea, Romania.
æ é

Compute L = where k ∈ R. lim n
2k

arctan(nk ) nk

n→∞

−

arctan(nk + 1) nk + 1

3617. 3618.

Proposed by Michel Bataille, Rouen, France.

Let r be a positive rational number. Show that if r r is rational, then r is an integer. Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.
∞ ∞

Let α > 3 be a real number. Find the value of n . (n + m)α

n=1 m=1

3619.
CA, USA. that

Proposed by Hung Pham Kim, student, Stanford University, Palo Alto,

Let a, b, and c be nonnegative real numbers such that a + b + c = 3. Prove (a2 b − c)(b2 c − a)(c2 a − b) ≤ 4(ab + bc + ca − 3a2 b2 c2 ) .

3620.

Proposed by John G. Heuver, Grande Prairie, AB.

Let P be an interior point in tetrahedron ABCD and let AP , BP , CP , and DP meet the corresponding opposite faces in A , B , C , and D then AP BP CP DP PA PB PC PD = 3+2 AP PA BP + PB + AP BP + PA PB BP AP + PB PA CP BP + PC PB CP DP + PC PD CP AP DP + PC PA PD DP CP DP + . PD PC PD +

3621.

Proposed by Titu Zvonaru, Com´ ane¸ sti, Romania.

Let a, b, and c be nonnegative real numbers with a + b + c = 1. Prove that 27 4 4 4 3 [(a − b)2 +(b − c)2 +(c − a)2 ]+ + + ≤ . 128 1+a 1+b 1+c ab + bc + ca

116

3622

. Proposed by George Tsapakidis, Agrinio, Greece.

Let ABCD be a quadrilateral. (a) Find suﬃcient and necessary condition on the sides and angles of ABCD , so that there is an inner point P such that two perpendicular lines through P divide the quadrilateral ABCD into four quadrilaterals of equal area. (b) Determine P .

3623.

Proposed by Michel Bataille, Rouen, France.

Let z1 , z2 , z3 , z4 be distinct complex numbers with the same modulus, α = |(z3 − z2 )(z3 − z4 )|, β = |(z1 − z2 )(z1 − z4 )| and u( ) = α(z1 − z2 ) + β (z3 − z2 ) α(z1 − z4 ) + β (z3 − z4 ) .

Prove that u(+1) or u(−1) is a real number.

3624.

Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.
∞ n=1

Calculate the sum (−1)n−1 n 1− 1 1 (−1)n+1 + − ··· + 2 3 n .

3625.
Vietnam.
Ö

Proposed by Pham Van Thuan, Hanoi University of Science, Hanoi,

Let a, b, and c be positive real numbers. Prove that a + a+b b + b+c
Ö

c ≤ 2 c+a

abc . (a + b)(b + c)(c + a)

117

SOLUTIONS
No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems.

3514.

[2010 : 107, 109] Proposed by Michel Bataille, Rouen, France.

Let m be a positive real number and a, b, c real numbers such that a(a − b) + b(b − c) + c(c − a) = m . What is the range of ab(a − b) + bc(b − c) + ca(c − a)? Solution by Albert Stadler, Herrliberg, Switzerland. We set x = b − a, Then x + y + z = 0, −xy − yz − zx = a(a − b) + b(b − c) + c(c − a) , y = c − b, z = a − c. (1)

xyz = ab(a − b) + bc(b − c) + ca(c − a) ,

and so (w − x)(w − y )(w − z ) = w3 − mw − xyz . Thus, the range consists of all real numbers n for which w3 − mw − n has three real roots (if the roots x, y , z are real, then (1) can be solved for the corresponding real numbers a, b, c). A cubic equation has three real roots if and only if its discriminant D is not positive. Here the discriminant is D = and only if |n| ≤ 2
m 3
3/2

−m 3

−n 2

, so D ≤ 0 holds if
3/2

Therefore, the range is the interval −2

m 3

3/2

m 3

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; ROY BARBARA, Lebanese University, Fanar, Lebanon; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; PRITHWIJIT DE, Homi Bhabha Centre for Science Education, Mumbai, India; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. One incomplete solution was submitted.

118

3515.

[2010 : 107, 110] Proposed by Jos´ e Luis D´ ıaz-Barrero and Josep Rubi´ o-Masseg´ u, Universitat Polit` ecnica de Catalunya, Barcelona, Spain. Let x, y , and z be positive real numbers. Prove that
{x}2 x + y z
2

{y }2 y 2 + z x

{z }2 z 2 + x y

≥

x2 + y 2 + z 2 , x+y+z

where a is the greatest integer not exceeding a, and {a} = a − a . Solution by Salem Maliki´ c, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina. Note ﬁrst that for all positive reals a, b, and c we have by the Cauchy–Schwarz Inequality that {c }2 a which implies that {c }2 a Hence, {x}2 {y }2 {z }2 x 2 y 2 z 2 + + + + + y z z x x y 2 2 2 2 x y z x + y2 + z2 > + + = . x+y+z x+y+z x+y+z x+y+z
Also solved by ARKADY ALT, San Jose, CA, USA; GEORGE APOSTOLOPOULOS, ˇ ´ University of Sarajevo, Sarajevo, Bosnia and Messolonghi, Greece; SEFKET ARSLANAGI C, Herzegovina; MICHEL BATAILLE, Rouen, France; OLEH FAYNSHTEYN, Leipzig, Germany; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; JOHAN GUNARDI, student, SMPK 4 BPK PENABUR, Jakarta, Indonesia; PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; ALBERT STADLER, Herrliberg, Switzerland; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposers. The featured solution shows that the inequality is strict. The proofs given by most solvers are similar to the one featured above. obtained the stronger result in which the right side of the inequality is replaced by by proving that
x2 y +z 3 2

c b

(a + b) ≥ ({c} + c )

= c2 ,

c b

≥

c2 a+b

c2 a+b+c

Faynshteyn
+y +z · xx +y +z
2 2 2

y z +x

z2 x+y

≥

3 2

x +y +z x+y +z

. He also claimed the following

generalization without proof:

cyclic

{x}2n + y

x 2n z

≥

3 22n−1

x2 n + y 2 n + z 2 n . x+y+z

119

3516.

[2010 : 107, 110] Proposed by J´ anos Bodn´ ar, Budapest, Hungary.

Let P and Q be interior points of triangle ABC . Let AA , BB , and CC be three concurrent cevians through P . The line through A parallel to AQ intersects the lines BQ and CQ at points L and M , respectively. The line through B parallel to BQ intersects the lines CQ and AQ at points M and N , respectively. The line through C parallel to CQ intersects the lines AQ and BQ at points N and L , respectively. Is it true that triangles LM N and L M N have the same area? Solution by Shailesh Shirali, Rishi Valley School, India. The point P plays no role in the result: A , B , and C can be any points on the lines BC, CA, and AB , respectively; furthermore, Q can be any point except a vertex in the plane of ∆ABC . But even these points play no essential role. The result is a special case of a known theorem; for triangles LM N and L M N to have the same area, all that is required is that the six vertices satisfy L L||M N , LM ||N N, and M M ||N L ,

which they do by deﬁnition. In terms of neutral letters, Theorem. If A, B, C, D, E, F are six points in an aﬃne plane such that AB ||DE , BC ||EF , and CD ||F A, then triangles ACE and BDF have equal areas and the same orientation. It is easier to ﬁnd a proof for the theorem than a reference. We use vectors with an arbitrary point taken to be the origin. Since AB ||DE we have A − B × D − E = 0. Similarly, B − C × E − F = 0 and C − D × F − A = 0. Expanding these cross products we get, A×D−B×D−A×E+B×E = 0,

B × E − C × E − B × F + C × F = 0, C × F − D × F − C × A + D × A = 0.

Subtracting the second relation from the sum of the ﬁrst and third we get, A × C + C × E + E × A = B × D + D × F + F × B. The ﬁnal equation tells us that ∆ACE and ∆BDF have equal areas and the same orientation.
Also solved by Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; MICHEL BATAILLE, Rouen, France; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; and PETER Y. WOO, Biola University, La Mirada, CA, USA.

120

3517.[2010 : 108, 110]

Proposed by V´ aclav Koneˇ cn´ y, Big Rapids, MI, USA.

Is there a scalene triangle ABC for which there exists a point P in the plane of ABC such that, for each line through P , the sum of the squares of the distances of A, B , and C to is constant? Solutions by George Apostolopoulos, Messolonghi, Greece; Roy Barbara, Lebanese University, Fanar, Lebanon; Michel Bataille, Rouen, France; Oliver Geupel, Br¨ uhl, NRW, Germany; and the proposer. Yes, there are scalene triangles with the required property. In Cartesian coordinates we take P to be the origin. The line through P with normal unit vector (cos φ, sin φ) is described by the equation x cos φ + y sin φ = 0, whence the square of the distance from a point X = (x0 , y0 ) to that line is dX = (x0 cos φ + y0 sin φ)2 . The examples given by our ﬁve correspondents are listed alphabetically by last name; for each example the vertices are given in the ﬁrst three columns and their squared distances to x cos φ + y sin φ = 0 in the next three, followed in the ﬁnal column by the constant sum of those three squared distances. A (0, 5) (1, 0) (1, −2) (1, 0) (0, 5) B (3, 0) (2, 0) (2, 2) (0, 1 ) 2 (−3, 0) C (4, 0) √ (0, 5) (2, −1) (0, 23 ) (4, 0)
√

dA 25 sin2 φ cos2 φ (cos φ− 2 sin φ)2 cos2 φ 25 sin2 φ

dB 9 cos2 φ 4 cos2 φ (2 cos φ+ 2 sin φ)2 1 sin2 φ 4 9 cos2 φ

dC 16 cos2 φ 5 sin2 φ (2 cos φ− sin φ)2 3 sin2 φ 4 16 cos2 φ

sum 25 5 9 1 25

Also solved by ALBERT STADLER, Herrliberg, Switzerland. There was one incorrect submission. Stadler also took P to be the origin, but he used complex numbers a, b, and c to represent the vertices A,B , and C . He proved that the sum of the squares of the distances from A,B , and C to a line through 0 is constant if and only if a2 + b2 + c2 = 0; in other words, given vertices corresponding to the complex numbers a and b, the third vertex can be either square root of −(a2 + b2 ). This, of course, produces all triangles, even degenerate, with the required property with respect to the origin. The proposer was motivated by problem 3291 [2007 : 485, 487; 2008 : 492-494], which applied to isosceles triangles.

121

3518. [2010 : 108, 110] Proposed by Yakub N. Aliyev, Qafqaz University, Khyrdalan, Azerbaijan.
Prove that if n and i are integers with 0 ≤ i ≤ n, then Ô 1 Ci 0 < 1 − 2i+1 − 10(2i+1)n · 102n − 1 < , 2 102n where {x} denotes the fractional part of the real number x and Ci is the ith 1 2i , i ≥ 0. Catalan number, Ci = i+1 i Solution by Oliver Geupel, Br¨ uhl, NRW, Germany. The case n = 0 is immediate; so assume that n > 0. We will use the 1/2 ¡ 2i+1 standard formula Ci = (−1)i i · 2 and the inequality Ci < 22i . +1
Èi
2n(i+1−k)

Let s = 102n(i+1) − k=1 Ck−1 · 10 22k−1 . It is well-known that the Catalan number Ck−1 is an integer for each integer k > 0. Moreover, for 1 ≤ 2n(i+1−k) k ≤ i ≤ n, it holds that 2k − 1 < 2n ≤ 2n(i + 1 − k); hence 10 22k−1 is an integer. Consequently, s ∈ Z. (1) Let t = that 0<t< 1 102n
∞ k=1 1 102n

È∞

Ck−1 k=i+2 22k−1

k−i−2 1 . 102n

By the binomial series, we have

∞ C k− 1 1 1/2 = − (−1)k 2 k − 1 2 n 2 10 k=1 k √ 1− 1−1 1 = = . 102n 102n

(2)

Moreover, 22 i 1 Ci (3) 0 < 2i+1 + t < 2i+1 + = 1 . 2 2 2 √ Let u = 10(2i+1)n · 102n − 1. Using the Binomial series, we obtain
Ö

u = 10
i

2n(i+1)

1− 1/2 k

1 102n

= 10

2n(i+1)

∞

k=0

(−1)k 1/2

1/2 k

1 102nk

= −

k=0 ∞

(−1)k

102n(i+1−k) − (−1)i 1/2 k 1 102n(k−i−1)

i+1

k=i+2

(−1)k−1 Ci − t.

=s−

22i+1

Ci By (1) and (3), we obtain {u} = 1− 22 i+1 − t. The conclusion follows immediately from (2).

Also solved by ALBERT STADLER, Herrliberg, Switzerland; and the proposer.

122

3519. [2010 : 108, 110] Proposed by Nguyen Duy Khanh, student, Hanoi University of Science, Hanoi, Vietnam.
Two triangles ABC and A B C have areas S and S , respectively. Let wa , wb , wc be the lengths of the internal angle bisectors of ABC to the sides BC , AC , AB , respectively, and deﬁne wa , wb , wc similarly. Prove or disprove that √ wa wa + wb wb + wc wc ≥ 3 3SS . Solution by Oliver Geupel, Br¨ uhl, NRW, Germany. We show that the inequality is not valid in general. As usual, let a = BC , b = CA, c = AB . Let ABC be a triangle with a = 3 and c = b + 1 > 2. Using a well-known formula for the length of an angle bisector, we obtain wa = and similarly wb = 2 6(b + 1)(b + 2) b+4 , wc = 2 3b(b + 2) b+3 . bc 1 − a2 (b + c)2 = 2 (b − 1)b(b + 1)(b + 2) 2b + 1 ,

Let A B C be a triangle with B C = c, C A = b, A B = a. We have wa = wc , wb = wb , wc = wa . If p denotes the common semiperimeter of both triangles, then we obtain S=S = p(p − a)(p − b)(p − c) = 2(b − 1)(b + 2) .

We will prove that for suﬃciently large b, √ wa wa + wb wb + wc wc < 3 3SS . By substituting the relations above, we obtain 8b(b + 2) 3(b − 1)(b + 1) (b + 3)(2b + 1) + 24(b + 1)(b + 2) (b + 4)2 6(b − 1)(b + 2) . (1)

< 3

√ √ 8 3 · b = 4 3b, while the right As b → ∞, the left side of (1) is asymptotic to 2 √ √ 2 √ 2

side of (1) is asymptotic to 3 6b. Since (4 3) = 48 < 54 = (3 6) , the right side eventually exceeds the left side. Indeed, with a calculator we ﬁnd that this occurs when b = 13. This completes the proof.

No other solutions were received. Geupel remarked that a related inequality appears in CRUX Problem 2029 [1995 : 91, 129–32], namely √ wb wc + wc wa + wa wb ≥ 3 3S , which is true.

123

3520. [2010 : 108, 110] Proposed by Ricardo Barroso Campos, University of Seville, Seville, Spain.
Construct a triangle ABC such that the line through the incentre and the circumcentre is parallel to the side AB . I. Solution by Michel Bataille, Rouen, France. Construct ∆ABC given its circumcentre O , circumradius R, and ∠C . Suppose that we have constructed a scalene triangle ABC such that the line through O and the incentre I is parallel to AB . Since I is interior to the triangle, O must be on the same side of AB as C . For C the midpoint of AB it follows that ∠C OA = ∠BCA (= ∠C ). Observing that OC = d(O, AB ) = d(I, AB ) = r (the inradius) and OA = R (the circumradius), we deduce that r = cos C ; from Euler’s inequality 2r < R, so that ∠C is necessarily between R 60◦ and 90◦ . Moreover, if M is the point where the ray [OC ) intersects the circumcircle Γ, M C is the angle bisector of ∠C and, from a familiar theorem, M B = M A = M I . This analysis of the ﬁgure leads to the following three-step construction: • Construct an angle equal to the given ∠C with vertex at O , and the point A on one of its legs such that OA = R. Let C be the orthogonal projection of A onto the other leg, and let M be the point where the circle with centre O and radius OA intersects that leg. Denote the circle by Γ and the point where Γ intersects AC by B . • Since M O = AO , when 90◦ > ∠M OA > 60◦ we have M A > M O , so that the circle with centre M , radius M A, intersects the line through O parallel to AC in two points. Let I be one of those points. • Let C be the second point of intersection of the line M I with Γ. Then M C is the angle bisector of ∠BCA (since M is the midpoint of the arc AB of Γ that does not contain C ), and I is the incentre of ∆ABC (since it is the point on the angle bisector for which M I = M A = M B ). The triangle ABC satisﬁes the requirement. The construction shows that there is at most one triangle with OI ||AB for the given O, R, and ∠C , and it exists if and only if 60◦ < ∠C < 90◦ . II. Composite of solutions by Mohammed Aassila, Strasbourg, France; John G. Heuver, Grande Prairie, AB; and Ricard Peir´ o, IES “Abastos”, Valencia, Spain. Construct ∆ABC given O, I , and r . Should such a triangle exist, we let F be the orthogonal projection of I onto AB ; then IF = r . Consider the right triangle OIF ; since by Euler’s formula OI 2 = R2 − 2Rr , we deduce that OF 2 = R2 − 2Rr + r 2 , whence OF = R − r . It follows that if the point P is at a distance r on the line OF beyond F , OP = R. The construction then proceeds in four steps: • Draw the circle with centre I and radius r (which will become the incircle). Construct the perpendicular to IO at I , and denote by F one of the points

124

where it intersects the circle. Construct the line perpendicular to IF at F ; call it . (The line will become AB .) • Draw the circle with centre F and radius r = F I ; denote by P the point where it meets the line OP on the other side of F from O . • Draw the circle with centre O and radius OP . (This will be the circumcircle because OP = R.) Call the points A and B where it meets . • Deﬁne E to be the second point where the circle with centre A and radius AF intersects the incircle, and deﬁne C to be the second point where AE intersects the circumcircle. ABC is the required triangle. (Because by construction, AF is tangent to the incircle and AE = AF , it follows that AE must also be tangent to the incircle. BC is also tangent to the incircle because we constructed the length R so that OI 2 = R2 − 2Rr .) The construction shows that the required triangle exists and is unique for any given pair of points O and I , and positive length r .
Also solved by MOHAMMED AASSILA, Strasbourg, France(a second solution); GEORGE APOSTOLOPOULOS, Messolonghi, Greece; ROY BARBARA, Lebanese University, Fanar, ´ IES Alvarez ´ Lebanon(2 solutions); FRANCISCO JAVIER GARC´ IA CAPITAN, Cubero, Priego ´ ˇ Y, ´ de C´ ordoba, Spain; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; V ACLAV KONE CN Big Rapids, MI, USA; SHAILESH SHIRALI, Rishi Valley School, India; D.J. SMEENK, Zaltbommel, the Netherlands; ALBERT STADLER, Herrliberg, Switzerland; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. There was one incomplete submission. Because the problem called for the construction of just one example of the required triangle without specifying what we are given, our solvers provided a wide variety of constructions, several of which were as simple as our featured solutions. Most were based on properties of triangles for which OI ||AB ; such properties have been the subject of numerous problems in CRUX with MAYHEM such as 659 [1982 : 215-216], which was based on Problem 758 in Mathematics Magazine 43:5 (Nov. 1970) pages 285-286. Here are three of them: (1) Construct the triangle given ∠A and side length c; Smeenk and Woo both used the property cos A + cos B = 1. (2) Construct the triangle given side lengths a and b; Aassila, Apostolopoulos, Garcia Capit´ an, and Shirali all obtained a formula for c in terms of a and b: c= ab +

a4 − a2 b 2 + b 4 . a+b

(For example, one can apply the cosine law to the formula used in (1).) Shirali observed parenthetically that this formula implies that there are no integer-sided triangles with the required property (because the expression under the radical sign has an integer square root only if a = b = 0). (3) Construct the triangle given R and the distance dA from O to the side BC ; Geupel noted that Carnot’s theorem, dA + dB + dC = R + r [Nathan Altshiller Court, College Geometry, page 83, Section 146] becomes simpler when dC = r —the circle whose diameter is OC contains the projections of O onto BC and AC , which are easily constructed using the distances dA and R − dA .

125

3523. [2010 : 109, 111] Proposed by Slavko Simic, Mathematical Institute SANU, Belgrade, Serbia.
Let f : R → R be a continuously diﬀerentiable function. functional equation f (x) + f (y ) − 2f x+y Solve the

= p(x − y ) f (x) − f (y ) , 2 where p is a real parameter independent of x, y . Partial solution by the proposer, modiﬁed by the editor. We show that if f has a continuous second derivative, then (i) f (x) = ax + b if p =
1 , 4 1 , 4

(ii) f (x) = ax3 + bx2 + cx + d if p =

where a, b, c, and d are arbitrary constants. Diﬀerentiating the given equation with respect to x, we obtain f (x) − f 1 2 x+y 2 = p f (x) − f (y ) + (x − y )f (x) . (1)

Next, diﬀerentiating (1) with respect to y , we obtain f x+y 2 = p f (x) + f (y ) .
1 2

(2)

Setting y = x in (2) we then have 2p −

f (x) = 0.

1 If p = , then f (x) = 0, from which (i) follows. 4 1 If p = , then (2) becomes 4

f That is, g

x+y 2 x+y

1 f (x) + f (y ) . 2 1

g (x) + g (y ) , (3) 2 2 where g = f , which is a variant of Cauchy’s equation also known as Jensen’s equation. It is a well-known fact in the theory of functional equations that the only solution of (3) is g (x) = αx + β for some constants α and β . From this, (ii) follows immediately.
Also solved by MICHEL BATAILLE, Rouen, France, under the same stronger assumption on f . Three other solutions were submitted which were either incomplete or partially incorrect. The problem remains open if we assume only that f ∈ C 1 (−∞, ∞). That the only solution of (3) above is g(x) = αx + β for some constants α, β when g is continuous or monotone can be found in Funtional Equations and How to Solve Them, by C.G. Small, Springer Problem Books in Mathematics, 2007. In case (i) both sides of the given a(x + y )(x − y )2 + 1 b(x − y )2 . equation vanish, while in case (ii) both sides equal 3 4 2

126

3524.

[2010 : 109, 111] Proposed by Titu Zvonaru, Com´ ane¸ sti, Romania.

Let a1 , a2 , . . . , an+1 be positive real numbers satisfying the condition an+1 = min{a1 , a2 , . . . , an+1 }. Prove that
+1 +1 +1 an + an + · · · + an 1 2 n+1 − (n + 1)a1 a2 · · · an+1

− n(a1 − an+1 )(a2 − an+1 ) · · · (an − an+1 )] .

≥ (n + 1)an+1 [(a1 − an+1 )n + (a2 − an+1 )n + · · · + (an − an+1 )n

Solution by George Apostolopoulos, Messolonghi, Greece, expanded by the editor. Let t = an+1 and let xi = ai − an+1 for i = 1, 2, . . . , n, then we have that t > 0 and xi ≥ 0 for all i. We can now write the inequality as
n n i=1

(t + xi )n+1 + tn+1 − (n + 1)t
n

(t + xi )
i=1 n

≥ (n + 1)t

i=1

xn i −n

xi
i=1

(1)

remaining coeﬃcients, ti+1 for i = 1, 2, . . . , n − 1, are equal to 0 for the RHS so it now suﬃces to show that the remaining coeﬃcients of the LHS are non-negative, that is
n j =1

Consider (1) as a polynomial of t. Then, it suﬃces to show that, for each degree of t, the coeﬃcient of the LHS is not less than the coeﬃcient of the n+1 RHS. The coeﬃcients of t0 and follow immediately. The coeﬃcient of Èn Én t 1 n t is (n + 1)( i=1 xi − n i=1 xi ) for both sides of the inequality. The

n+1 i+1

n− i xj − (n + 1)

n− i 1≤j1 <j2 <···<jn−i ≤n k=1

xjk ≥ 0 .

(2)

By the AM-GM Inequality, we have that (n + 1)
n− i 1≤j1 <j2 <···<jn−i ≤n k=1

Én − i

k=1

xjk ≤

1 n− i

Èn−i

k=1 n

n− i xj . Thus, k n− i xj .

xjk ≤

n+1 n−1 n−i i

(3)

i=1

From (2) and (3), it suﬃces to show that n+1 i+1 n+1 i+1 and we are done. n+1 ≥ n n+1 n−1 . i n−i n−1 i ≥

From the deﬁnition of binomial coeﬃcients, we have that = n−ii+1 n+1 n−1 n−i i

Also solved by Michel Bataille, Rouen, France; Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and the proposer.

127

3525.

[2010 : 109, 111] Proposed by an anonymous proposer. √ √ √ a+b a + b − ab Let 0 ≤ a, b ≤ 1. Prove that ≤
1 + ab

Solution by George Apostolopoulos, Messolonghi, Greece. √ √ For convenience, take α = a and β = b. Then we need to show that for all 0 ≤ α, β ≤ 1 we have α2 + β 2 1 + α2 β 2 or equivalently (α + β − αβ ) This can be factored as αβ (1 − α)(1 − β )[α2 β 2 − αβ (α + β ) − αβ + 2] ≥ 0 . Since αβ (1 − α)(1 − β ) ≥ 0, it suﬃces to show that α2 β 2 − αβ (α + β ) − αβ + 2 ≥ 0 . But
2

≤ (α + β − αβ )2 ,

1 + α2 β 2 − α2 + β 2 ≥ 0 .

α2 β 2 − αβ (α + β ) − αβ + 2 ≥ α2 β 2 − 3αβ + 2 = (1 − αβ )(2αβ ) ≥ 0 .
ˇ ´ Also solved by ARKADY ALT, San Jose, CA, USA; SEFKET ARSLANAGI C, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, France; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; KEE-WAI LAU, Hong Kong, China; PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; PRITHWIJIT DE, Homi Bhabha Centre for Science Education, Mumbai, India; ALBERT STADLER, Herrliberg, Switzerland; DANIEL TSAI, student, Taipei American School, Taipei, Taiwan; PETER Y. WOO, Biola University, La Mirada, CA, USA; KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA; and the proposer.

128

3526.[2010 : 109, 111]

Proposed by Cao Minh Quang, Nguyen Binh Khiem High School, Vinh Long, Vietnam. Let a, b, and c be positive real numbers. Prove that a ≥ 1. a2 + 2(b + c)2 cyclic

Solution by Oliver Geupel, Br¨ uhl, NRW, Germany. By H¨ older inequality we have
2

a
cyclic

a2 + 2(b + c)2

a(a2 + 2(b + c)2 )
cyclic

≥ (a + b + c)3 .

Thus we only need to show that (a + b + c)3 ≥ or equivalently that a2 b + a2 c + ab2 + ac2 + b2 c + bc2 ≥ 6abc . a(a2 + 2(b + c)2 ) ,
cyclic

But this is immediate from the AM-GM inequality. This completes the proof. Equality holds if and only if a = b = c
Also solved by ARKADY ALT, San Jose, CA, USA; GEORGE APOSTOLOPOULOS, ˇ ´ University of Sarajevo, Sarajevo, Messolonghi, Greece(2 solutions); SEFKET ARSLANAGI C, Bosnia and Herzegovina; MICHEL BATAILLE, Rouen, France; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; PEDRO HENRIQUE O. PANTOJA, student, UFRN, Brazil; PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; ALBERT STADLER, Herrliberg, Switzerland; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer.

Crux Mathematicorum
with Mathematical Mayhem
Former Editors / Anciens R´ edacteurs: Bruce L.R. Shawyer, James E. Totten, V´ aclav Linek

Crux Mathematicorum
Founding Editors / R´ edacteurs-fondateurs: L´ eopold Sauv´ e & Frederick G.B. Maskell Former Editors / Anciens R´ edacteurs: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

129

EDITORIAL
Shawn Godin
Hello again CRUX with MAYHEM readers. Is it April already? I just wanted to thank you for your patience as I ease into my role as editor of this great publication many months later than I would have liked. You may (or may not) have noticed some subtle changes already. For one thing, we have a new font that is a lot more user friendly from my end. We have also changed the package that we use to create diagrams so it is much easier to create nicer looking diagrams. We are in the process of going through a couple of years of proposals to decide which ones we will use in the journal. The hope is that you will hear about the fate of your proposals soon after you have submitted them. It is the plan that in the next couple of months we will review the problems that have already be submitted in 2011. After that point, as we continue to work our way through the older problems still in our banks, problem proposers will be informed of the status of their proposal in a much more timely fashion. We will also look into classifying problems so we can inform proposers which areas already have a long queue of problems waiting to appear and which areas have a shortage. We are still behind publication schedule and will be for the remainder of 2011, but we should be closer to our real schedule going into 2012 (so you should get the February issue before July!). We will continue to use the “old” deadlines in print, but deadlines will be extended until we are ready to work on the material. A CRUX with MAYHEM page has been created on Facebook at www.facebook.com/pages/ Crux-Mathematicorum-with-Mathematical-Mayhem/152157028211955. Updates to our progress will be posted on the site. Messages informing you when certain problems have been sent to the editors for moderation will appear to give you a better idea of what the “real” deadline for the problems will be. We will also give you some hints as to what will appear in future issues. As I mentioned in an earlier editorial we are looking to make some changes. With increases in publishing and mailing costs we are exploring alternate ways to deliver CRUX with MAYHEM. Also, in an increasingly digital world, we are looking into ways that we can better interact with our readers. We hope that, sometime in the not too distant future, CRUX with MAYHEM readers can submit problem proposals and solutions online. You would then be able to check the status of your proposal to see if it has been accepted and when it is projected to appear. We are preparing a short online survey to get some feedback on various parts of CRUX with MAYHEM and to get your input into some of the proposed changes. Check the Facebook page for details and please take a few minutes to give us your opinion.

130

SKOLIAD

No. 132

Lily Yen and Mogens Hansen
Please send your solutions to problems in this Skoliad by December 15, 2011. A copy of CRUX with Mayhem will be sent to one pre-university reader who sends in solutions before the deadline. The decision of the editors is ﬁnal. Our contest this month is the Maritime Mathematics Competition, 2010. Our thanks go to David Horrocks, University of Prince Edward Island, for providing us with this contest and for permission to publish it.

Maritime Mathematics Competition, 2010
2 hours allowed

The Valhalla Winter Games are held in February, and the closing ceremonies are on the last day of the month. The ﬁrst Valhalla Winter Games were held in the year 750, and since that year, they have been held every ﬁve years. How many times have the closing ceremonies been held on February 29th ? Note that year Y is a leap year if exactly one of the following conditions is true: (a) Y is divisible by 4 but Y is not divisible by 100. (b) Y is divisible by 400.

A triangle with vertices A(0, 0), B (3, 4), and C (2, c) has area 5. Find all possible values of the number c.

Let f (x) = ax2 + bx + c, where a, b, and c are real numbers. Assume that f (0), f (1), and f (2) are all integers. (a) Prove that f (2010) is also an integer. (b) Decide if f (2011) is an integer.

If x is a real number, let x denote the largest integer which is less than or equal to x. For example, 7.012 = 7. If n is any positive integer, ﬁnd a (simple) formula for ë î 2n 2(n + 1) 2(n + 2) + + . 3 3 3 (a) If a is a positive number, prove that a+ 1 a ≥ 2. 1 b 1 ab

(b) If a and b are both positive numbers, prove that a+ 1 a +b+ + ≥ 4. 5 .
1 x

You may assume without proof that f (x) = x + for x ≥ 1.

is an increasing function

131

6 √.

A hole in a concrete wall has the shape of a semi-circle with a radius of 2 metres. A utility company wants to place one large circular pipe or two smaller circular pipes of equal radius through the hole to supply water to Watertown. If they want to maximise the amount of water that could ﬂow to Watertown, should they use one pipe or two pipes, and what size pipes(s) should they use?

Concours de Math´ ematique des Maritimes, 2010
2 heures a permis

1. Les Jeux d’hiver de Valhalla de d´ eroulent en f´ evrier, les c´ er´ emonies de clˆ oture
se tenant let dernier jour du mois. Les premier Jeux d’hiver de Valhalla se sont d´ eroul´ es en l’an 750 et les Jeux prennent place depuis a ` tous cinq ans. Combien de fois le c´ er´ emonies de clˆ oture ont-elle eu lieu le 29 f´ evrier ? Rappelons qu’une ann´ ee Y est une ann´ ee bissextile si l’une des conditions suivantes est satisfaite : (a) Y est divisible par 4 mais n’est pas divisible par 100. (b) Y est divisible par 400.

2. Si les sommets d’un triangle d’aire 5 sont A(0, 0), B (3, 4) et C (2, c), trouver
toutes les valeurs possibles de c.

3. Soit f (x) = ax2 + bx + c o` u a, b, et c sont des nombres r´ eel. Supposons que
f (0), f (1), et f (2) sont des entier. (a) Montre que f (2010) est aussi un entier. (b) D´ eterminer si f (2011) est un entier.

4. Si x est une nombre r´ eel, d´ enotons par x le plus grand entier inf´ erieur ou ´ egal a ` x. Par exemple, 7.012 = 7. Si n est un entier positif quelconque, trouver une formule (simple) pour
ë

2n 3

2(n + 1) 3

2(n + 2) 3

5. (a) Si a est un nombre r´ eel strictement positif, montre que
a+ 1 a ≥ 2.

(b) Si a et b sont des nombres r´ eels strictement positifs, montre que a+ 1 a +b+ 1 b + 1 ab ≥ 4. 5 .
1 x

Vous pouvez supposer sans preuve que la fonction f (x) = x + pour x ≥ 1.

est croissante

132

demi-cercle de rayon 2 m` etres. La ville veut se fournir de l’eau au moyens de tuyaux pass´ es a ` travers le trou. Pour maximiser le d´ ebit de l’eau, est-il pr´ ef´ erable d’utiliser en seul grand tuyau circulaire ou deux tuyaux circulaires plus petits de mˆ eme rayon, et quelle est la dimension des tuyaux a ` utiliser ?

6. Un mur de b´ eton dans e d’un trou en forme de √ la ville de Watertown est perc´

Next a comment from a reader about a problem whose solutions appeared in Skoliad 126 at [2010 : 262–263].

4. The diagram shows three squares and
angles x, y , and z . Find the sum of the angles x, y , and z . x y z Comment by Solomon W. Golomb, University of Southern California, Los Angeles, CA, USA. Problem 4. on pages 262–263 of the September 2010 issue of CRUX with MAYHEM asks for the sum of the three angles x, y and z , in the ﬁgure below, consisting of three adjacent unit squares, and three diagonals. Two proofs that x + y + z = 90◦ , one trigonometric and one geometric, are given. A B C D

x E F

y G

z H

Years ago, an equivalent problem (same diagram, and to show that z = x + y ) appeared in Martin Gardner’s “Mathematical Games” column in Scientiﬁc American, and drew a great deal of reader response. However, the most elegant solution, by the late Leon Bankoﬀ, has never been previously published. It uses no trigonometry, and needs no additional lines to be drawn. Bankoﬀ’s keen observation was that the triangles DEG and DGF are similar! (They have ∠DGF is common, and the including sides are in the √ same ratio, √ √ namely √ |DG| : |GF | = 2 : 1 = 2, and |EG| : |GD | = 2 : 2 = 2.) Hence angle x = ∠DEG in the larger triangle equals ∠GDF = x in the smaller triangle. A B C x x E F y zz G H D

So we have that the three angles of triangle DGF are x, y , and 90◦ + z , and must sum to 180◦ . Hence x + y + z = 90◦ .

133

Next follow solutions to the Swedish Junior High School Mathematics Contest, Final Round, 2009/2010, given in Skoliad 125 at [2010 : 259–260].

A 2009 × 2010 grid is ﬁlled with the numbers 1 and −1. For each row, calculate the product of the entries in that row. Do likewise for the columns. Show that the sum of all the row products and all the column products cannot be zero. ´ Solution by Rowena Ho, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC.

Each of the 2009 row products and each of the 2010 column products is either 1 or −1. For the sum of a collection of 1’s and −1’s to be zero, you must have equally many of each, but this is impossible since you have 4019 numbers. Thus the sum of all the row products and all the column products cannot be zero.
´ Also solved by LENA CHOI, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; MONICA HSIEH, student, Burnaby North Secondary School, Burnaby, BC; and JULIA PENG, student, Campbell Collegiate, Regina, SK.

The square ABCD has side length 6. The point P splits side AB such that |AP | : |P B | = 2 : 1. A point Q inside the square is chosen such that |AQ| = |P Q| = |CQ|. Find the area of CP Q. Solution by Monica Hsieh, student, Burnaby North Secondary School, Burnaby, BC.

Impose a coordinate system so that A = (0, 6), P B B = (6, 6), C = (6, 0), and D = (0, 0). Since A |AP | : |P B | = 2 : 1, P = (4, 6). Since |AQ| = |P Q|, Q is on the line given by x = 2. Since |AQ| = |CQ|, Q is on the line through B and D , y = x. Thus Q Q = (2, 2). ·2 = 6, the area of Now, the area of CDQ is 62 2·6 ·4 C = 8, the D BCP is 2 = 6, the area of AP Q is 42 6·2 2 area of ADQ is 2 = 6, and the area of ABCD is 6 = 36. Therefore the area of CP Q is 36 − 6 − 6 − 8 − 6 = 10.
´ Also solved by LENA CHOI, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; and RICHARD I. HESS, Rancho Palos Verdes, CA, USA.

The product of three positive integers is 140. Determine the sum of the three integers if the second integer is seven times the ﬁrst one. Solution by Gesine Geupel, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany. Say the three integers are a, 7a, and b. Then 7a2 b = 140. If a = 1, then b = 20, and the sum of the three positive integers is a +7a + b = 1+7+20 = 28. If a = 2, then b = 5, and the sum is 2 + 14 + 5 = 21. If a = 3, then b = 20 , 9

134

Also solved by RICHARD I. HESS, Rancho Palos Verdes, CA, USA; and MONICA HSIEH, student, Burnaby North Secondary School, Burnaby, BC. Rather than going through so many cases, it may be easier to consider that if 7a2 b = 140, then a2 b = 20 = 22 · 5. The only squares that divide 22 · 5 are 12 and 22 , so a = 1 or a = 2.

5 , which is not an integer. If a ≥ 5, which is not an integer. If a = 4, then b = 4 4 then b ≤ 5 , so b cannot be a positive integer. Hence, the sum is either 21 or 28.

4. Five points are placed at the intersections of a rectangular grid. Then the midpoint of each pair of points is marked. Prove that at least one of these midpoints lands on an intersection point of the grid. ´ Solution by Lena Choi, student, Ecole Dr. Charles Best Secondary School, ´ Coquitlam, BC; Rowena Ho, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; and Monica Hsieh, student, Burnaby North Secondary School, Burnaby, BC. Impose a coordinate system such that (0, 0) is a grid point and the size of the grid is 1. Then the ﬁve points all have integer coordinates and the midpoints are grid points if and only if they, too, have integer coordinates.
1 (x1 + x2 ), 2 (y1 + y2 ) . The midpoint between (x1 , y1 ) and (x2 , y2 ) is 1 2 Therefore a midpoint is a grid point if and only if x1 + x2 and y1 + y2 are both even. Note that “even plus even is even” and “odd plus odd is even” while the sum of two numbers of opposite parity is odd. In terms of parity of their coordinates, the grid points fall into four classes: (even, even), (even, odd), (odd, even), and (even, odd). Since you have ﬁve grid points, two must be in the same class. Say (x1 , y1 ) and (x2 , y2 ) are in the same class. Then x1 and x2 have the same parity, so x1 + x2 is even. Likewise, y1 + y2 is even, so the midpoint between (x1 , y1 ) and (x2 , y2 ) is a grid point.

Also solved by RICHARD I. HESS, Rancho Palos Verdes, CA, USA.

Points K and L on segment AM are placed such that |AK | = |LM |. Place the points B and C on one side of AM and point D on the other side of AM such that |BK | = |KM |, |CM | = |KL|, and |DL| = |LM |, and such that BK , CM , and DL are all perpendicular to AM . Prove that ABCD is a square.

B C L K D

Solution by Julia Peng, student, Campbell Collegiate, Regina, SK. Note that |AL| = |AK | + |KL| = |LM | + |KL| = |KM | = |BK | and that |AK | = |LM | = |DL|. Since ABK and DAL are both right-angled, |AB | = |AD | and ABK ∼ = DAL. Let O be the point on BK such that KM CO is a rectangle. Let P be the intersection of CO and the extension of DL. Then OP LK is a rectangle. Moreover, |KL| = |CM | = |P L|, so OP LK is a square.

135

Now, |DP | = |DL| + |LP | = |AK | + |KL| = |AL| and |P C | = |OC | − |OP |
A a b

a b O P

b a C

L b a D

= |KM | − |KL|

= |LM | = |DL| .

Again, both CDP and DAL are right-angled, so CDP ∼ = DAL. Finally, |OC | = |KM | = |AL| and |BO | = |BK | − |OK | = |AL| − |KL| = |AK |, so BCO ∼ = ABK since both triangles are right-angled. Since ABK ∼ = BCO ∼ = CDP ∼ = DAL, the sides of ABCD are all equal and the angles are equal as labelled. Since the angle sum in BAK is 180◦ , a + b = 90◦ , so ∠DAB = 90◦ , and ABCD is a square.
Several solvers showed that ABCD is a rhombus.

6. Let N be a positive integer. Ragnhild writes down all the divisors of N other than 1 and N . She then notes that the largest divisor is 45 times the smallest one. Which positive integers satisfy this condition?
´ Solution by Lena Choi, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC. The smallest possible proper divisor of N would be 2. If 2 is a divisor, then 2 · 45 = 90 would be the largest proper divisor. Since N must be the product of its largest and smallest proper divisors, N = 2 · 90 = 180. If 3 is the smallest proper divisor, then 3 · 45 = 135 is the largest proper divisor, and N = 3 · 135 = 405. If m > 3 and m is the smallest proper divisor, then 45m is the largest proper divisor, and N = 45m2 . But 45m2 is divisible by 3 since 45 is divisible by 3. Thus m is not the smallest proper divisor after all. This contradiction shows that the smallest divisor is at most 3. Hence N = 180 or N = 405.
Also solved by RICHARD I. HESS, Rancho Palos Verdes, CA, USA; ROWENA HO, ´ student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; and MONICA HSIEH, student, Burnaby North Secondary School, Burnaby, BC.

This issue’s prize of one copy of Crux Mathematicorum for the best solutions goes to Monica Hsieh, student, Burnaby North Secondary School, Burnaby, BC. We hope that our readers will enjoy the featured contest and share their solutions.

136

MATHEMATICAL MAYHEM
Mathematical Mayhem began in 1988 as a Mathematical Journal for and by High School and University Students. It continues, with the same emphasis, as an integral part of Crux Mathematicorum with Mathematical Mayhem. The interim Mayhem Editor is Shawn Godin (Cairine Wilson Secondary School, Orleans, ON). The Assistant Mayhem Editor is Lynn Miller (Cairine Wilson Secondary School, Orleans, ON). The other staﬀ members are Ann Arden (Osgoode Township District High School, Osgoode, ON) and Monika Khbeis (Our Lady of Mt. Carmel Secondary School, Mississauga, ON).

Mayhem Problems
Please send your solutions to the problems in this edition by 15 November 2011. Solutions received after this date will only be considered if there is time before publication of the solutions. Each problem is given in English and French, the oﬃcial languages of Canada. In issues 1, 3, 5, and 7, English will precede French, and in issues 2, 4, 6, and 8, French will precede English. The editor thanks Rolland Gaudet, Universit´ e de Saint-Boniface, Winnipeg, MB, for translating the problems from English into French.

M482.

Proposed by the Mayhem Staﬀ.

Using four sticks with lengths of 1 cm, 2 cm, 3 cm, and 5 cm, respectively, you can measure any integral length from 1 cm to 10 cm. Note that a stick may only be used once in a particular measurement, so the 1 cm, 2 cm, and 3 cm sticks could be used to measure 6 cm, but not the 3 cm stick twice. (a) Find a set of ten stick lengths that can be used to represent any integral length from 1 cm to 100 cm. (b) What is the fewest number of sticks that are needed to represent any integral length from 1 cm to 100 cm?

M483.

Proposed by Bruce Shawyer, Memorial University of Newfoundland, St. John’s, NL.

Triangle ABC has ∠BAC = 90◦ . The feet of the perpendiculars from A to the internal bisectors of ∠ABC and ∠ACB are P and Q, respectively. Determine the measure of ∠P AQ.

137

M484.

Proposed by Dragoljub Miloˇ sevi´ c, Gornji Milanovac, Serbia.

Solve the equation x2 + 4 x x−2
2

= 45 .

M485.
ON.

Proposed by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,

Prove that

n k=1

n k

1 n!

n k=1

kk (n − k)!

for all n ∈ N.

M486.

Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania. How many distinct numbers are in the list 12 − 1 + 4 12 + 1 , 22 − 2 + 4 22 + 1 , 32 − 3 + 4 32 + 1 , ..., 20112 − 2011 + 4 20112 + 1 ?

M487. Proposed by Samuel G´ omez Moreno, Universidad de Ja´ en, Ja´ en, Spain.
Let m be a positive integer. Find all real solutions to the equation
Õ

m + ···

√

x = x,

in which the integer m occurs n times. ................................................................. ´ M482. Propos´ e par l’Equipe de Mayhem. ` l’aide de baguettes de longueurs 1 cm, 2 cm, 3 cm et 5 cm, on peut A mesurer toute longueur enti` ere de 1 cm a ` 10 cm. Noter qu’une baguette oeut ˆ etre utilis´ ee une seule fois lors d’une mesure ; par exemple, les baguettes de 1 cm, 2 cm et 3 cm peuvent ˆ etre utilis´ ees pour mesurer 6 cm, tandis qu’on ne peut pas utiliser deux baguettes de 3 cm pour mesurer 6 cm. (a) D´ eterminer un ensemble de dix baguettes de longueurs enti` eres pouvant mesurer toute longueur enti` ere de 1 cm a ` 100 cm. (b) Quel est le plus petit nombre de baguettes permettant de mesurer toute longueur enti` ere de 1 cm a ` 100 cm ?

138

M483.

Propos´ e par Bruce Shawyer, Universit´ e Memorial de Terre-Neuve, St. John’s, NL.

Le triangle ABC est tel que ∠BAC = 90◦ . Les pieds des perpendiculaires de A jusqu’aux bissectrices internes des angles ∠ABC et ∠ACB sont P et Q respectivement. D´ eterminer la mesure de ∠P AQ.

M484.

Propos´ e par Dragoljub Miloˇ sevi´ c, Gornji Milanovac, Serbie.

R´ esoudre l’´ equation x2 + 4 x x−2
2

= 45 .

M485.
ON.

Propos´ e par Edward T.H. Wang, Universit´ e Wilfrid Laurier, Waterloo,

D´ emontrer que

n k=1

n k

1 n!

n k=1

kk (n − k)!

pour tout n ∈ N.

M486.

´ Propos´ e par Neculai Stanciu, Ecole secondaire George Emil Palade, Buz˘ au, Roumanie. Combien de nombres distincts y a-t-il dans la liste 12 − 1 + 4 12 + 1 22 − 2 + 4 22 + 1 32 − 3 + 4 32 + 1 20112 − 2011 + 4 20112 + 1

, ...,

M487.

Propos´ e par Samuel G´ omez Moreno, Universit´ e de Ja´ en, Ja´ en, Espagne.

Soit m un enti` ere positif. D´ eterminer toutes les solutions r´ eelles a ` l’´ equation
Õ

m + ···

√

x = x,

dans laquelle l’enti` ere m a lieu n fois.

Mayhem Solutions
M440.
Proposed by the Mayhem Staﬀ. In trapezoid ABCD , AB is parallel to DC and AD is perpendicular to AB . If AB = 20, BC = 5x, CD = x2 + 3x, and DA = 3x, determine the value of x.

139

Solution by Geoﬀrey A. Kandall, Hamden, CT, USA. There are two cases to consider. Case I: x2 + 3x > 20. Let P be the foot of the perpendicular from B to DC . Then BP = 3x and, according to Pythagoras, P C = 4x. Therefore, A 20 B x2 + 3x = 20 + 4x, (x − 5)(x + 4) = 0, x = 5 (since x > 0). Case II: x2 + 3x < 20. Proceeding as in Case I, we obtain (x2 + 3x) + 4x = 20, x2 + 7x − 20 = 0, √ −7 + 129 x= . 2 A 3x D C 20 5x 4x P x2 + 3x 3x B x2 − x − 20 = 0, 3x D
2

3x 4x P x + 3x

5x C

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; SCOTT BROWN, Auburn University, Montgomery, AL, USA; RICHARD I. HESS, Rancho Palos Verdes, CA, ´ USA; ANTONIO LEDESMA LOPEZ, Instituto de Educaci´ on Secundaria No. 1, Requenaˇ ´ Gornji Milanovac, Serbia; RICARD PEIRO, ´ Valencia, Spain; DRAGOLJUB MILO SEVI C, IES “Abastos”, Valencia, Spain; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; and NECULAI STANCIU, George Emil Palade Secondary School, Buz˘ au, Romania. Eight incorrect solutions were received. Most of the incorrect solutions neglected one of the cases.

M442.

Proposed by Carl Libis, Cumberland University, Lebanon, TN, USA.
¾ ¿

Consider the square array 1 n+1 . . . 2 n+2 . . . ··· ··· ··· n−1 n 2n − 1 2n . . . . . . 2 n − 1 n2

(n − 1)n + 1 (n − 1)n + 2

formed by listing the numbers 1 to n2 in order in consecutive rows. Determine the sum of the numbers on each diagonal. How does this sum compare to the “magic constant” that would be obtained if the n2 entries were rearranged to form a magic square?

140

Soluci´ on de Ricard Peir´ o, IES “Abastos”, Valencia, Spain. Los elementos de la diagonal principal son: 1, n + 2, 2n + 3, . . . , (n − 1)n + n. La suma es: D1 (n) = 1 + (n + 2) + (2n + 3) + · · · + [(n − 1)n + n] = = n(n + 1) +n (n − 1)n 2 = [1 + 2 + 3 + · · · + n] + n[1 + 2 + 3 + · · · + (n − 1)] 2 n3 + n

. 2 Los elementos de la diagonal secundaria son: n, 2n − 1, 3n − 2, . . . , n · n − (n − 1). La suma es: D2 (n) = n + (2n − 1) + (3n − 2) + · · · + [n · n − (n − 1)] = n[1 + 2 + 3 + · · · + n] − [1 + 2 + 3 + · · · + (n − 1)] =n = n(n + 1) 2 − (n − 1)n 2

. 2 La constante m´ agica de un cuadrado m´ agico n × n es: 1 M (n) = (1 + 2 + 3 + · · · + n2 ) n 1 n2 (n2 + 1) = n 2 3 n +n = = D1 (n) = D2 (n). 2
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; JACLYN ´ CHANG, student, University of Calgary, Calgary, AB; SAMUEL GOMEZ MORENO, Universidad de Ja´ en, Ja´ en, Spain; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; ´ ANTONIO LEDESMA LOPEZ, Instituto de Educaci´ on Secundaria No. 1, Requena-Valencia, Spain; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; NECULAI STANCIU, George Emil Palade Secondary School, Buz˘ au, Romania; GUSNADI WIYOGA, student, SMPN 8, Yogyakarta, Indonesia; and ALLEN ZHU, Conestoga High School, Berwyn, PA, USA.

n3 + n

M445.

Proposed by the Mayhem Staﬀ.

The lines with equations y = x + 1, y = mx − 1, and y = −4x + 2m pass through the same point. Determine all possible values for m. Solution by Aﬁﬀah Nuur Mila Husniana, student, SMPN 8, Yogyakarta, Indonesia. Given are three linear equations y = x + 1, y = mx − 1, (1) (2) (3)

y = −4x + 2m.

141

From (1) and (2) we have x(1 − m) = −2 −2 x= 1−m From (2) and (3) we have x(m + 4) = 2m + 1 2m + 1 x= m+4 From (4) and (5) we have −2 2m + 1 = 1−m m+4 −2(m + 4) = (1 − m)(2m + 1) 2m2 − 3m − 9 = 0 (2m + 3)(m − 3) = 0 mx − 1 = −4x + 2m (5) x + 1 = mx − 1 (4)

−2m − 8 = 2m − 2m2 + 1 − m

So the possible values for m are m = − 3 or m = 3. 2

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; ALPER CAY and LOKMAN GOKCE, Geomania Problem Group, Kayseri, Turkey; DAVINIA CERVERA GARC´ IA, Club Math´ ematique de l’Instituto de Ecuaci´ on Secundaria No. 1, Requena-Valencia, Spain; MUHAMMAD HAFIZ FARIZI, student, SMPN 8, Yogyakarta, Indonesia; G.C. GREUBEL, Newport News, VA, USA; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; WINDA KIRANA, student, SMPN 8, Yogyakarta, Indonesia; SALLY LI, student, Marc Garneau Collegiate Institute, Toronto, ON; DEBRA A. OHL, student, Angelo State University, San Angelo, TX, USA; KONSTANTINOS AL. NAKOS, Agrinio, Greece; RICARD ´ IES “Abastos”, Valencia, Spain; NECULAI STANCIU, George Emil Palade Secondary PEIRO, School, Buz˘ au, Romania; GUSNADI WIYOGA, student, SMPN 8, Yogyakarta, Indonesia; and KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA. Two incorrect solutions were submitted.

M446.

Proposed by J. Walter Lynch, Athens, GA, USA.

Let a, b, and c be positive digits. Suppose that b equals the product of a, b, and c, and ac = a + b + c. Determine a, b, and c. (Here ab is the two-digit positive integer with tens digit a and units digit b.) Solution by Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA. We are given 10a + c = a + b + c b = a·b·c

142

Since a, b, c are positive digits, then 1 ≤ a, b, c ≤ 9. Since b = 0 then b = a · b · c gives a · c = 1; which implies that a = 1 = c. From 10a + c = a + b + c, then we have 10 · 1 + 1 = 1 + b + 1; hence b = 11 − 2 = 9. Therefore a = 1, b = 9, c = 1.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; ALPER CAY and LOKMAN GOKCE, Geomania Problem Group, Kayseri, Turkey; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; AFIFFAH NUUR MILA HUSNIANA, student, SMPN 8, Yogyakarta, Indonesia; YOUNGHUAN JUNG, The Woodlands School, Mississauga, ON; WINDA KIRANA, student, SMPN 8, Yogyakarta, Indonesia; DONGCHAN LEE, University of Toronto, Toronto, ON; SALLY LI, student, Marc Garneau Collegiate Institute, Toronto, ON; MUHAMMAD ROIHAN MUNAJIH, student, SMPN 8, Yogyakarta, Indonesia; ALEECE NALBANDIAN, California State University, Fresno, CA, USA; DEBRA A. OHL, student, Angelo State ´ IES “Abastos”, Valencia, Spain; University, San Angelo, TX, USA; RICARD PEIRO, ´ PLANELLS CARCEL, ´ ANDRES Club Math´ ematique de l’Instituto de Ecuaci´ on Secundaria No. 1, Requena-Valencia, Spain; NECULAI STANCIU, George Emil Palade Secondary School, Buz˘ au, Romania; GUSNADI WIYOGA, student, SMPN 8, Yogyakarta, Indonesia; and INGESTI BILKIS ZULFATINAAS, student, SMPN 8, Yogyakarta, Indonesia. One incorrect solution was submitted.

M447.
Azerbaijan.

Proposed by Yakub N. Aliyev, Qafqaz University, Khyrdalan,

Let ABCD be a parallelogram. The sides AB and AD are extended to points E and F (respectively) so that E , C , and F all lie on a straight line. Prove that BE · DF = AB · AD . Solution by George Apostolopoulos, Messolonghi, Greece.
BC BE = . Since ABCD DC DF AD BE = , hence is a parallelogram we know BC = AD and DC = AB . So AB DF

The triangles BCE and F DC are similar, so

BE · DF = AB · AD .

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; ALPER CAY and LOKMAN GOKCE, Geomania Problem Group, Kayseri, Turkey; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; AFIFFAH NUUR MILA HUSNIANA, student, SMPN 8, Yogyakarta, Indonesia; WINDA KIRANA, student, SMPN 8, Yogyakarta, Indonesia; DONGCHAN LEE, University of Toronto, Toronto, ON; DEBRA A. OHL, student, Angelo State University, San Angelo, TX, USA; PEDRO HENRIQUE O. PANTOJA, student, UFRN, ´ IES “Abastos”, Valencia, Spain; JORGE SEVILLA LACRUZ, Brazil; RICARD PEIRO, Club Math´ ematique de l’Instituto de Ecuaci´ on Secundaria No. 1, Requena-Valencia, Spain; NECULAI STANCIU, George Emil Palade Secondary School, Buz˘ au, Romania; LOU VANG, California State University, Fresno, CA, USA; KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA; and INGESTI BILKIS ZULFATINAAS, student, SMPN 8, Yogyakarta, Indonesia. One incorrect solution was submitted.

M448.

Proposed by the Mayhem Staﬀ.

A polyhedron with exactly m + n faces has m faces that are quadrilaterals and n faces that are triangles. Exactly four faces meet at each vertex. Prove that n = 8.

143

Solution by Dongchan Lee, University of Toronto, Toronto, ON. The number of faces is F = m + n. Since there are 4 vertices and 4 edges in a quadrilateral, and 3 vertices and 3 edges in a triangle, then the total +3n . The total number of vertices will be number of edges would be E = 4m2 4m+3n since it is given that exactly four faces meet at each vertex. Using V = 4 Euler’s polyhedron formula, which says that the sum of the number of faces and the number of vertices is equal to the number of edges plus two, F + V = E + 2, 4m + 3n 4m + 3n = + 2. m+n+ 4 2 Solving the equation, we get n = 8.
Also solved by ALPER CAY and LOKMAN GOKCE, Geomania Problem Group, ´ Kayseri, Turkey; JORGE ARMERO JIMENEZ, Club Math´ ematique de l’Instituto de Ecuaci´ on Secundaria No. 1, Requena-Valencia, Spain; RICHARD I. HESS, Rancho Palos Verdes, CA, ´ USA; SALLY LI, student, Marc Garneau Collegiate Institute, Toronto, ON; RICARD PEIRO, IES “Abastos”, Valencia, Spain; and NECULAI STANCIU, George Emil Palade Secondary School, Buz˘ au, Romania.

M449.

Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania. Let E (x) =
4x . 4x + 2

(a) Prove that E (x) + E (1 − x) = 1. (b) Find the value of E
1 2010

2 2010

+ ··· + E

2008 2010

2009 . 2010

Solution by Aﬁﬀah Nuur Mila Husniana, student, SMPN 8, Yogyakarta, Indonesia. (a) From the given equation E (x) = E (x) + E (1 − x) = =
4x 4x +2

so, + 41 −x

4x + 2 41 −x + 2 4x (41−x + 2) + 41−x (4x + 2)

(4x + 2)(41−x + 2) 4 + 2(4x ) + 4 + 2(41−x ) = 4 + 2(4x ) + 2(41−x ) + 4 8 + 2(4x ) + 2(41−x ) = 8 + 2(4x ) + 2(41−x ) =1 [Ed. – Note that E (1 − x) = immediately.]
41−x × 1 4 −x + 2
4x 2 4x 2

2 and the conclusion follows 2 + 4x

144

(b) From (a) we know that E (x) + E (1 − x) = 1, thus E E 1 2010 2 +E 2009 2010 2008 =E 1 2010 2 +E 1−

1 2010 2

+E =E +E 1− =1 2010 2010 2010 2010 and so on. Then we have 1 2 2008 2009 1005 +E +...+E +E = 1004×1+E E 2010 2010 2010 2010 2010 Since E
1005 2010

42 4
1 2

Therefore the sum is 1004.5.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; ALPER CAY and LOKMAN GOKCE, Geomania Problem Group, Kayseri, Turkey; CHAO-PING CHEN, Henan Polytechnic University, Jiaozuo City, China and Mih´ aly Bencze, Lajos Aprily Highschool, Brasov, Romania; DIANA DOMINGUEZ, California State University, Fresno, CA, USA; MUHAMMAD HAFIZ FARIZI, student, SMPN 8, Yogyakarta, Indonesia; PABLO PARDAL GARCES, Club Math´ ematique de l’Instituto de Ecuaci´ on Secundaria No. 1, RequenaValencia, Spain; G.C. GREUBEL, Newport News, VA, USA; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; WINDA KIRANA, student, SMPN 8, Yogyakarta, Indonesia; DONGCHAN LEE, University of Toronto, Toronto, ON; SALLY LI, student, Marc Garneau Collegiate Institute, Toronto, ON; KONSTANTINOS AL. NAKOS, Agrinio, Greece; CARLOS TORRES NINAHUANCA, Lima, Per´ u; PEDRO HENRIQUE O. PANTOJA, student, UFRN, Brazil; ´ IES “Abastos”, Valencia, Spain; PAOLO PERFETTI, Dipartimento di RICARD PEIRO, Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; GUSNADI WIYOGA, student, SMPN 8, Yogyakarta, Indonesia; KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA; and INGESTI BILKIS ZULFATINAAS, student, SMPN 8, Yogyakarta, Indonesia;

, then E

1005 2010

1 . 2

M450.
Waterloo, ON.

Proposed by Edward T.H. Wang, Wilfrid Laurier University,

Prove that if n is an odd positive integer, then nn+2 + (n + 2)n is divisible by 2(n + 1). Solution by Osman Ekiz, Eskisehir, Turkey. From the Binomial Theorem, (n + 2)n can be written as: (n + 2)n = (n + 1) + 1 = n 0
n

(n + 1)n +

n 1

(n + 1)n−1 + · · · +

n n−1

(n + 1) + 1.

Since n is an odd number we can also write:

nn+2 = (n + 1 − 1)n+2 n+2 n+2 = (n + 1)n+2 − (n + 1)n+1 + · · · 0 1 n+2 + (n + 1) − 1. n+1

145

If we add the two expansions together, the constant terms cancel each other. Therefore, we have:
å

(n + 2)n + nn+2 =

n n (n + 1)n + (n + 1)n−1 + · · · 0 1 n n+2 (n + 1) + (n + 1)n+2 + n−1 0 è n+2 n+2 (n + 1)n+1 + · · · + (n + 1) − 1 n+1

Since all of the terms have a factor of n + 1, then (n + 2)n + nn+2 is divisible by n + 1 and we can rewrite the expression as:
å

(n + 2)n + nn+2 = (n + 1)

n 0 + −

(n + 1)n−1 + n n−1 n+2 1 +

1 n+2 0

(n + 1)n−2 + ... (n + 1)n+1 n+2 n+1
è

(n + 1)n + · · · +

+2 . Since nn = n and n = n + 2, then we have −1 n+1 2n + 2 which is an even number. n+1

Now we must prove that the expression in the square brackets above is an even number. Since we know that n + 1 is an even number, all of the terms with ¡ a factor of n + 1 are also even. Then we are left with only two terms, nn and −1
n+2¡ ¡ ¡
n ¡ n− 1

n+2¡
n+1

Hence (n + 2)n + nn+2 is divisible by 2(n + 1).
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; ADAM GREGSON, teacher, University of Toronto Schools, Toronto, ON; G.C. GREUBEL, Newport News, VA, USA; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; ANTONIO LEDESMA ´ LOPEZ, Instituto de Educaci´ on Secundaria No. 1, Requena-Valencia, Spain; DONGCHAN LEE, University of Toronto, Toronto, ON; SALLY LI, student, Marc Garneau Collegiate ´ IES “Abastos”, Valencia, Spain; NECULAI Institute, Toronto, ON; RICARD PEIRO, STANCIU, George Emil Palade Secondary School, Buz˘ au, Romania; and KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA. Note that since n is odd

(n + 2)n − 1 = (n + 2)n−1 + (n + 2)n + · · · + 1 is also odd. Thus their sum is even. n+1

nn+2 + 1 = nn+1 − nn + nn−1 − · · · + 1 is odd. Also n+1

146

Problem of the Month
Ian VanderBurgh
Problems involving probability can be very interesting and can lead to lots of discussion and debate. (If you don’t believe me about debate, try looking up the Monty Hall Problem.) These problems also give lots of opportunity for both creative solutions and plausible incorrect solutions. Here are two problems involving probability that have very diﬀerent ﬂavours. Problem 1 (2011 Euclid Contest) Three diﬀerent numbers are chosen at random from the set {1, 2, 3, 4, 5}. The numbers are arranged in increasing order. What is the probability that the resulting sequence is an arithmetic sequence? This type of problem is pretty standard. We are given a set of objects and a property. We need to determine the probability that a randomly chosen object from the set has the desired property. Often, the most direct approach is to count the total number of objects in the set and to count the number of objects in the set that have the desired property. The probability that we are after is the second number divided by the ﬁrst number. Here is an illustration of this technique. Solution to Problem 1. We consider choosing the three numbers all at once. We list the possible sets of three numbers that can be chosen: {1, 2, 3} {1, 2, 4} {1, 2, 5} {1, 3, 4} {1, 3, 5} {1, 4, 5} {2, 3, 4} {2, 3, 5} {2, 4, 5} {3, 4, 5} We have listed each in increasing order because once the numbers are chosen, we arrange them in increasing order. There are 10 sets of three numbers that can be chosen. Of these 10, the 4 sequences 1, 2, 3 and 1, 3, 5 and 2, 3, 4 and 3, 4, 5 are arithmetic sequences. Therefore, the probability that the resulting sequence is an arithmetic sequence is 2 4 or 5 . 10 Sometimes, a good problem solving strategy helps us think about a problem in a more straightforward way. The problem itself isn’t any easier, but it might be easier to attack. Separating the problem into counting these two diﬀerent sets of objects makes this easier to approach. When this contest was marked, the most popular incorrect answer to this 4 1 problem was 60 (or its reduced form of 15 ). Can you ﬁgure out what mistake might lead to this answer? The next problem is also about probability, but seems very diﬀerent at ﬁrst. Problem 2 (2011 Euclid Contest) A 75 year old person has a 50% chance of living at least another 10 years. A 75 year old person has a 20% chance of living at least another 15 years. An 80 year old person has a 25% chance of living at least another 10 years. What is the probability that an 80 year old person will live at least another 5 years?

147

This is a really interesting problem that has a good “real life context”. The data given is close to the actual data for Canadian adults. One approach is to use the given probabilities directly. Solution 1 to Problem 2. Suppose that the probability that a 75 year old person lives to 80 is p, the probability that an 80 year old person lives to 85 is q , and the probability that an 85 year old person lives to 90 is r . We want to the determine the value of q . For a 75 year old person to live at least another 10 years, they must live another 5 years (to age 80) and then another 5 years (to age 85). The probability of this is equal to pq . We are told in the question that this is equal to 50% or 0.5. Therefore, pq = 0.5. For a 75 year old person to live at least another 15 years, they must live another 5 years (to age 80), then another 5 years (to age 85), and then another 5 years (to age 90). The probability of this is equal to pqr . We are told in the question that this is equal to 20% or 0.2. Therefore, pqr = 0.2. Similarly, since the probability that an 80 year old person will live another 10 years is 25%, then qr = 0.25. 0. 2 pqr = = 0. 4. Since pqr = 0.2 and pq = 0.5, then r = pq 0. 5 0.25 qr = = 0.625. Since qr = 0.25 and r = 0.4, then q = r 0. 4 Therefore, the probability that an 80 year old person will live at least another 5 years is 0.625, or 62.5%. A second approach is actually to use the “count the objects” method that we discussed earlier. You might wonder what the set of objects is. Here is one way to do this. Solution 2 to Problem 2. Consider a population of 100 people, each of whom is 75 years old and who behave according to the probabilities given in the question. Each of the original 100 people has a 50% chance of living at least another 10 years, so there will be 50% × 100 = 50 of these people alive at age 85. Each of the original 100 people has a 20% chance of living at least another 15 years, so there will be 20% × 100 = 20 of these people alive at age 90. ) chance that an 80 year old person will live at Since there is a 25% (or 1 4 least another 10 years (that is, to age 90), then there should be 4 times as many of these people alive at age 80 than at age 90. Since there are 20 people alive at age 90, then there are 4 × 20 = 80 of the original 100 people alive at age 80. In summary, of the initial 100 people of age 75, there are 80 alive at age 80, 50 alive at age 85, and 20 people alive at age 90. Because 50 of the 80 people alive at age 80 are still alive at age 85, then the probability that an 80 year 50 old person will live at least 5 more years (that is, to age 85) is 80 = 5 , or 62.5%. 8 That works pretty well doesn’t it? It is always fascinating to me when mathematics becomes so connected to real life. Problem 2 is related to an area of mathematics called actuarial science, which has lots of applications to things like insurance and pensions. If you are interested in the idea of applying mathematics to the ﬁnancial industry, check out this ﬁeld!

148

THE OLYMPIAD CORNER
No. 293 R.E. Woodrow and Nicolae Strugaru
In this issue we begin a transition in the Corner. Problems editor Nicolae Strungaru, from Grant MacEwan University in Edmonton, has agreed to take over from Robert Woodrow who has been the editor of the Corner since 1987. Robert’s dedication to CRUX with MAYHEM over the years is greatly appreciated and he will be sorely missed. Material from Robert will continue to appear in CRUX with MAYHEM as we wrap up the solutions to the last sets of problems he published. The format of the Corner is changing slightly. It will still consist of problems from Olympiads from around the world, but, rather than printing the contests in their entirety, each column will consist of 10 questions, in both English and French, selected from diﬀerent contests. The origin of the question will be revealed when the solutions are published. We will have the same time lines as we do with the CRUX problems. Solutions will be due six months from the issue date and will appear in the same issue number of the next volume, one year later. The ﬁrst set of new Olympiad Corner problems is below, please send your solutions to Nicolae by email (preferred) at: [email protected] or by mail to Nicolae Strungaru Department of Mathematics and Statistics Grant MacEwan University Edmonton, AB Canada T5J 4S2 Enjoy the new Corner!

149

The solutions to the problems are due to the editor by 1 January 2012.

OC1. OC2.

Find all positive integers w, x, y and z which satisfy w! = x! + y ! + z !. Suppose that f is a real-valued function for which f (xy ) + f (y − x) ≥ f (y + x)

for all real numbers x and y . (a) Give a nonconstant polynomial that satisﬁes the condition. (b) Prove that f (x) ≥ 0 for all real x.

OC3.

Let ABCD be a convex quadrilateral with ∠CBD = 2∠ADB, ∠ABD = 2∠CDB and AB = CB.

Prove that AD = CD .

OC4.

Consider 70-digit numbers n, with the property that each of the digits 1, 2, 3, . . . , 7 appears in the decimal expansion of n ten times (and 8, 9 and 0 do not appear). Show that no number of this form can divide another number of this form.

OC5.

Suppose that the real numbers a1 , a2 , . . . , a100 satisfy a1 ≥ a2 ≥ · · · ≥ a100 ≥ 0, and a1 + a2 ≤ 100 a3 + a4 + · · · + a100 ≤ 100.

2 2 Determine the maximum possible value of a2 1 + a2 + · · · + a100 , and ﬁnd all possible sequences a1 , a2 , . . . , a100 which achieve this maximum.

OC6. In the diagram, ABCD is a square, with U and V interior points of the sides AB and CD respectively. Determine all the possible ways of selecting U and V so as to maximize the area of the quadrilateral P U QV .
A P U D

V Q B C

150

OC7.

Let n be a natural number such that n ≥ 2. Show that 1 n+1 1+ 1 3 + ··· + 1 2n − 1 > 1 n 1 2 + 1 4 + ··· + 1 2n .

OC8. For each real number r let Tr be the transformation of the plane that takes the point (x, y ) into the point (2r x, r 2r x + 2r y ). Let F be the family of all such transformations i.e. F = {Tr : r ∈ R}. Find all curves y = f (x) whose graphs remain unchanged by every transformation in F . OC9.
A deck of 2n + 1 cards consists of a joker and, for each number between 1 and n inclusive, two cards marked with that number. The 2n + 1 cards are placed in a row, with the joker in the middle. For each k with 1 ≤ k ≤ n, the two cards numbered k have exactly k − 1 cards between them. Determine all the values of n not exceeding 10 for which this arrangement is possible. For which values of n is it impossible?

OC10.

The number 1987 can be written as a three digit number xyz in some base b. If x + y + z = 1 + 9 + 8 + 7, determine all possible values of x, y, z, b. .................................................................

OC1.

Trouver tous les entiers positifs w, x, y et z qui satisfont w! = x! + y ! + z !.

OC2. Supposer que f

est une fonction a ` valeurs r´ eelles qui satisfait f (xy ) + f (y − x) ≥ f (y + x)

pour tous nombres r´ eels x et y . (a) Donner un polynˆ ome non constant qui satisfait cette condition. (b) Montrer que f (x) ≥ 0 pour tout nombre r´ eel x.

OC3. On consid` ere un quadrilat` ere convexe ABCD dans lequel
∠CBD = 2∠ADB, ∠ABD = 2∠CDB et D´ emontrer que AD = CD . AB = CB.

OC4. Consid´ erer les nombres n a ` 70 chiﬀres avec la propri´ et´ e que chacun des
chiﬀres 1, 2, 3, . . . , 7 apparaˆ ıt dix fois dans l’expansion d´ ecimale de n (et que 8, 9 et 0 n’y apparaissent pas). Montrer qu’aucun nombre de cette forme ne peut ˆ etre divis´ e par un autre nombre de la mˆ eme forme.

151

OC5. Supposons que les nombres r´ eels a1 , a2 , . . . , a100 satisfont aux conditions suivantes
a1 ≥ a2 ≥ · · · ≥ a100 ≥ 0, a1 + a2 ≤ 100

a3 + a4 + · · · + a100 ≤ 100.

2 2 D´ eterminer la valeur maximale possible de a2 1 + a2 + · · · + a100 , et trouver toutes les suites possibles a1 , a2 , . . . , a100 pour lesquelles ce maximum est atteint.

OC6. Sur le diagramme ci-dessous, ABCD est un carr´ e sur lequel on choisit des points U et V int´ erieurs aux cˆ ot´ es AB et CD respectivement. D´ eterminer toutes les fa¸ cons possibles de choisir U et V de telle sorte que la surface du quadrilat` ere P U QV soit maximale.
A P U D

V Q B C

OC7. Soit n un nombre naturel tel que n ≥ 2. Montrer que
1 n+1 1+ 1 3 + ··· + 1 2n − 1 > 1 n 1 2 + 1 4 + ··· + 1 2n .

OC8. Soit F la famille des transformations F

= {Tr : r ∈ R} o` u Tr transforme le point (x, y ) en le point (2r x, r 2r x +2r y ). Trouver toutes les courbes y = f (x) dont le graphe est invariant pour chacune des transformations de F .

OC9. Un jeu de 2n +1 cartes contient un joker et, pour chaque nombre entier de
1a ` n inclusivement, 2 cartes marqu´ ees de ce num´ ero. Les 2n + 1 cartes sont alors align´ ees avec le joker au milieu. De plus, pour chaque nombre entier k avec 1 ≤ k ≤ n, les deux cartes num´ erot´ ees k ont exactement k − 1 autres cartes entre elles. Trouver toutes les valeurs de n ne d´ epassant pas 10 pour lesquelles cet arrangement soit possible. Maintenant, pour quelles valeurs de n est-ce impossible ?

OC10. Le nombre 1987 s’´ ecrit a ` trois chiﬀres, xyz , dans une certaine base b.
Si x + y + z = 1 + 9 + 8 + 7, d´ eterminer toutes les valeurs possible de x, y, z et b.

152

First we look at solutions from the ﬁles to the 20th Korean Mathematical Olympiad, given at [2010: 152–153].

Triangle ABC is acute with circumcircle Γ and circumcentre O . The circle Γ has centre O , is tangent to O at A and to the side BC at D , and intersects the lines AB and AC again at E and F , respectively. The lines OO and EO intersect Γ again at A and G, respectively. The lines BO and A G intersect at H . Prove that DF 2 = AF · GH . Solved by Titu Zvonaru, Com´ ane¸ sti, Romania. T A Γ Γ E B D A O O H F G

C Since O E = O G and O A = O A , the quadrilateral AEA G is a parallelogram, hence A G AE (1) The triangles O AE and OAB are isosceles. It follows that ∠O EA = ∠EAO = ∠BAO = ∠ABO, hence EO BO. (2) By (1) and (2) we deduce that BHGE is a parallelogram; thus HG = BE and we have to prove that DF 2 = AF · BE (3)

Let a = BC , b = CA, c = AB . If b = c, then D is the midpoint of BC , AD is a diameter of Γ , BE = CF and DF ⊥ AC . It is easy to see that, in ADC with DF ⊥ AC , the equation (3) is true. We may assume that b > c, and we denote by T the intersection of the line BC with the tangent to Γ at A. Using the power of point T with respect to Γ, we obtain T A2 = T B · T C ⇔ T A2 = T B (T B + a),

153

and applying the Law of Cosines in

⇔ T B 2 + aT B = T B 2 + c2 + 2c · T B · cos B, hence TB =

T A2 = T B 2 + AB 2 − 2T B · AB · cos ∠ABT

ABT (with ∠ABT = 180◦ − B ), we have

c2 . a − 2c cos B Since O A = O D , then T D = T A and, using again the Law of Cosines, we deduce that TB + a = c2 a − 2c cos B +a = bc c2 + a2 − 2ac cos B a − 2c cos B c2 − = = b2 a − 2c cos B a − 2c cos B c(b − c) . ,

BD = T D − T B = Denoting α =

it results that BD = cα, DC = bα. Using the power of points B and C with respect to circle Γ , we get: hence BE = cα2 , CF = bα2 , AF = b(1 − α2 ). BE · BA = BD 2 , CF · CA = CD 2 ,

a(b − c) a(b − c) b−c a , we have α = 2 = 2 = ; a − 2c cos B a − 2ac cos B b − c2 b+c

1 − 2c cos B

a − 2c cos B

The equality (3) is equivalent to:

DF 2 = AF · BE ⇔ DC 2 + CF 2 − 2DC · CF · cos C = AF · BE ⇔ b2 α2 + b2 α4 − 2b2 α3 cos C = bcα2 (1 − α2 ) ⇔ b + bα2 − 2bα cos C = c(1 − α2 ) ⇔ b − c + (b + c) · which is true (by the Law of Cosines in a2 (b + c)2 − 2b · a

b+c

cos c = 0

⇔ b2 − c2 + a2 − 2ab cos c = 0. ABC ).

Find all triplets (x, y, z ) of positive integers satisfying 1 + 4x + 4y = z 2 .

Solved by Michel Bataille, Rouen, France; Prithwijit De, Homi Bhabha Centre for Science Education, Mumbai, India; and Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA. We give the solution of De. Observe that z is odd and is at least 3. Let z = 2m + 1 where m is a positive integer. Then the equation reduces to 4x−1 + 4y−1 = m(m + 1) Assume that x ≥ y and rewrite (1) as 4y−1 (4x−y + 1) = m(m + 1) (1)

(2)

Observe that gcd(m, m + 1) = 1. Therefore either

154

(a) m = 4y−1 , m + 1 = 4x−y + 1; or (b) m + 1 = 4y−1 , m = 4x−y + 1. If (a) holds then x = 2y − 1 and z = 22y−1 + 1. If (b) holds then we obtain 2 2 y − 3 − 2 2 x− 2 y − 1 = 1 (3)

5 , 2) which is inadmissible because x is not an The solution of (3) is (x, y ) = ( 2 integer. Hence the solution set in positive integers of this equation is

{(2k − 1, k, 22k−1 + 1) : k ∈ Z + } ∪ {(k, 2k − 1, 22k−1 + 1) : k ∈ Z + }. where Z + is the set of positive integers.

Find all pairs (p, q ) of primes such that pp + q q + 1 is divisible by pq .

Solved by Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA. Suppose that (p, q ) is such a pair of primes. Then, pp + q q + 1 = kpq, for some positive integer k p and q are primes. (1)

Since equation (1) is symmetric with respect to p and q ; and since p = q (by inspection, p = q would imply p | 1). There is no loss of generality in assuming p < q. (2) We distinguish between two cases: Case 1, in which p and q are both odd primes; and Case 2 wherein, p = 2 and q is an odd prime. Case 1. p and q are both odd primes. Thus, by (2) we must have 3 ≤ p < p + 2 ≤ q, 5 ≤ q. (3)

We will prove that no primes satisfying (1) and (3) exist. We make use of the concept of the order of a positive integer a modulo an odd prime r . If a and r are relatively prime, then the order of a modulo r is the smallest positive integer n such that an ≡ 1 (mod r ). When a ≡ 1 (mod r ), the order of a is equal to 1. Otherwise, it is some positive integer. The order of a exists, since by Fermat’s Little Theorem, we know that ar−1 ≡ 1 (mod r ). (Thus the set of all natural numbers n such that an ≡ 1 (mod r ) is nonempty). The following lemma is well-known in elementary number theory, we state it without proof. Lemma 1. Let r be an odd prime, and a a positive integer not divisible by r , and let n be the order of a modulo r . Then, if m is a positive integer such that am ≡ 1 (mod r ), n is a divisor of m.

155

From (1) it follows that, q q ≡ −1 (mod p)

⇒ q 2q ≡ (−1)2 ≡ 1 (mod p) .

(4)

Let n be the order of q modulo p. By (4) and Lemma 1, it follows that n is a divisor of 2q , which means that n = 1, 2, q , or 2q . If n = 1; then q ≡ 1 (mod p); q = 1 + pl, for some positive integer l, and going back to (1) we have pp + (1 + p · l)q + 1 = k · p · q (5)

It is evident from the binomial expansion of (1 + p · l)2 , that (5) implies 2 + λp = kpq , for some positive integer λ, which is impossible since this last equation implies p | 2; we know that p ≥ 3. Next, consider the case in which the order n (of q modulo p) is q or 2q . We know from Fermat’s Little Theorem that q p−1 ≡ 1 (mod p) . By Lemma 1, the order n (= q or 2q ) must divide p − 1. Since p − 1 is even and is q odd; we see that in either case 2q must divide p: therefore p−1 = p = 2q · t, for some positive integer t.

2qt + 1 > q,

which contradicts (3). There remains only one possibility to consider: the order n (of q modulo p) is equal to 2. q 2 ≡ 1 (mod p) ⇔ (q − 1)(q + 1) ≡ 0 (mod p)

⇔ q ≡ ±1 (mod p) ( since p is prime).

(6)

The case q ≡ 1 (mod p) has already been examined above (this was done in the case order n = 1). So, then suppose that q ≡ −1 (mod p), q = p · v − 1, v ∈ Z, v ≥ 2 We go back to (1) and this time we work modulo q : pp ≡ −1 (mod q ) ⇒ p2p ≡ 1 (mod q ) which implies by Lemma 1 that the order f of p modulo q must be a divisor of 2p. Thus, f = 1, 2, p, or 2p. Once again, by Fermat’s Little Theorem, we know that the order f must divide q − 1 by virtue of pq−1 ≡ 1 (mod q ). Hence, q−1 =f ·u q = f · u + 1, where f = 2, p, or 2p. (8) (7)

156

Note that the possibility f = 1 is ruled out: if f = 1 then p ≡ 1 (mod q ) which implies (since both p and q are positive and ≥ 3) that p > q ; contrary to (3). If f = p or 2p, then combining (7) with (8) yields p · v − f · u = 2, which implies (since f = p or 2p) that p divides 2; an impossibility since p ≥ 3. Finally suppose that f = 2. Then, p2 ≡ 1 (mod q ) ⇔ (p − 1)(p + 1) ≡ 0 (mod q ) ; and since q is a prime, we must have either p = 1 + q · w or p = −1 + q · w for some positive integer w which again contradicts the conditions in (3); for either possibility implies p > q (note that in either case, w ≥ 2). It is now clear that there are no odd primes p and q which satisfy (1). Case 2. p = 2 and q is an odd prime. From (1) we have, 22 + q q + 1 = 2kq ; 5 = q · (2k − q q−1 ). (9)

Equation (9) clearly shows that q | 5; and since q is a prime; we must have q = 5 = 313. and 2k − q q−1 = 1 so 2k = 54 + 1, thus k = 626 2 Conclusion: Taking into account symmetry, there exist exactly two pairs with the problem’s property: (p, q ) = (2, 5), (5, 2).

5. For the vertex A of ABC , let A be the point of intersection of the angle bisector at A with side BC , and let A be the distance between the feet of the perpendiculars from A to the lines AB and C , respectively. Deﬁne B and C similarly, and let be the perimeter of ABC . Prove that
A B C 3

≤

1 64

Solved by Arkady Alt, San Jose, CA, USA; Michel Bataille, Rouen, France; Prithwijit De, Homi Bhabha Centre for Science Education, Mumbai, India; Geoﬀrey A. Kandall, Hamden, CT, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give Bataille’s solution. We adopt the standard notations for the elements of ∆ABC and denote the orthogonal projections of A onto AB, AC by H, K , respectively. Since the line segment HK is a chord subtending ∠BAC of the circle with diameter AA , we have A = HK = AA sin A. As it is well-known, the length of the bisector is given by AA = =
2bc cos(A/2) so we obtain b+c

2bc sin(A/2) b+c

· 2 cos2 (A/2) =

2bc sin(A/2) b+c

· 1+

b2 + c2 − a2 2bc

It quickly follows that
A

2(s − a) sin(A/2) b+c

157

With similar results for
A B C 3

and

we ﬁnally have

sin(A/2) sin(B/2) sin(C/2) · 8(s − a)(s − b)(s − c) . (b + c)(c + a)(a + b)

From the following known formulas: rs = s(s − a)(s − b)(s − c), sin(A/2) sin(B/2) sin(C/2) = abc = 4rRs, and ab + bc + ca = s2 + r 2 + 4rR we ﬁrst deduce (b + c)(c + a)(a + b) = (a + b + c)(ab + bc + ca) − abc = 2s(s2 + r 2 + 4rR) − 4rRs = 2s(s2 + r 2 + 2rR) r3 Rs2 + Rr 2 + 2rR2 and then
A B C 3

r , 4R

(1)

By AM-GM, we have s 3 = (s − a) + (s − b) + (s − c) 3 ≥
3

(s − a)(s − b)(s − c) =

√ 3 r2s

so that s2 ≥ 27r 2. Recalling Euler’s inequality R ≥ 2r , we obtain Rs2 + Rr 2 + 2rR2 ≥ 54r 3 + 2r 3 + 8r 3 = 64r 3 and from (1),
A B C 3

≤

1 . 64

Next we turn to the 2006/2007 British Mathematical Olympiad, Round 1, given at [2010: 153].

Find four prime numbers less than 100 which are factors of 332 − 232 .

Solved by Arkady Alt, San Jose, CA, USA; Geoﬀrey A. Kandall, Hamden, CT, USA; Henry Ricardo, Tappan, NY, USA; Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give Ricardo’s write-up. Using the familiar ‘diﬀerence of squares’ identity repeatedly, we can write
4

332 − 232 = 5

(32 + 22 ).
k=1

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For k ≤ 2, the factors 32 + 22 are less than 100 and it is easy to pick out 5, 13 (when k = 1), and 97 (when k = 2) as prime factors. Now we observe that 332 = 916 ≡ 1 (mod 17) and 232 = 416 ≡ 1 (mod 17) by Fermat’s Little Theorem. Thus 332 − 232 ≡ 0 (mod 17) and 17 is the fourth prime factor we seek.

In the convex quadrilateral ABCD , points M , N lie on the side AB such that AM = M N = N B , and points P , Q lie on the side CD such that CP = P Q = QD . Prove that Area of AM CP = Area of M N P Q = 1 Area of ABCD . 3

Solved by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; by Geoﬀrey A. Kandall, Hamden, CT, USA; by Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and by Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Amengual Covas. A D M Q N P C N M Q P C A D

Figure 1

Figure 2

Since CP = P Q, we have (ﬁgure 1) Area of Since AM = M N , we have Area of Hence, Area of that is, Area of AM CP = Area of M N P Q Now, since the areas of triangles with equal altitudes are proportional to the bases of the triangles, we have (ﬁgure 2) Area of AM C = 1 3 (Area of ABC ) CP M + Area of AM P = Area of P QM + Area of MNP AM P = Area of MNP CP M = Area of P QM

159

and Area of Hence, Area of that is, Area of AM CP = and we are done. AM C +Area of CP A = CP A =

1 3 1 3

(Area of

CDA)

(Area of

ABC + Area of

CDA)

1 (Area of ABCD ) 3

The number 916238457 is an example of a nine-digit number which contains each of the digits 1 to 9 exactly once. It also has the property that the digits 1 to 5 occur in their natural order, while the digits 1 to 6 do not. How many such numbers are there? Solved by Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA. x1 x2 x3 x4 x5 x6 x7 x8 x9 1 2 3 4 5 6 7 8 9 {x1 , x2 , x3 , x4 , x5 , x6 , x7 , x8 , x9 } = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Note that there are exactly 9! = 1 · 2 · 3 · 4 · 5 · 6 · 7 · 8 · 9 nine-digit numbers with distinct nonzero digits. Let S be the set of all nine-digit numbers with distinct nonzero digits and such that the digits 1 to 5 occur in their natural order, and m = n(S ) = cardinality of the set S . 1 2 3 4 5 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 1 2 3 4 5 6 7 8 9

Let S1 be the set of all nine-digit numbers with distinct nonzero digits and such that the digits 1 to 6 occur in their natural order, and m1 = n(S1 ) = cardinality of the set S1 . Let S2 be the set of all nine-digit numbers with distinct nonzero digits and such that the digits 1 to 5 occur in their natural order; but the (numbers) digits 1 to 6 do not occur in their natural order; and m2 = n(S2 ) = cardinality of the set S2 . Then S = S1 ∪ S2 and S1 ∩ S2 = ∅. Therefore, n(S ) = m = m2 = n(S1 ) + n(S2 ); m1 + m2 ; m − m1 . (1)

160

To calculate m, observe that any ﬁve of the positions 1 through 9 may be chosen; for any such choice, the numbers 1 to 5 are placed in their natural order on those positions. Moreover, for each such choice of ﬁve positions; there are 4! ways to place the remaining numbers 6 to 9, on the remaining four positions. Hence, 9! 9! 9 = = 6 · 7 · 8 · 9. m = (4!) · = 4! 4!5! 5! 5 Similarly, m1 = (3!) Hence by (1) m2 = m − m1 = 6·7·8·9−7·8·9 = (7 · 8 · 9)(6 − 1) = 7 · 8 · 9 · 5 = 2520 9 6 = 3! 9! 3!6! = 9! 6! = 7 · 8 · 9.

Conclusion: There are exactly 2520 such numbers.

Two touching circles S and T share a common tangent which meets S at A and T at B . Let AP be a diameter of S and let the tangent from P to T touch it at Q. Show that AP = P Q. Solved by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; Geoﬀrey A. Kandall, Hamden, CT, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the two solutions of Amengual Covas.

Solution 1. Let O and O be the centers of S and P T , respectively. Then the line OO , joining the S centers of the touching circles, goes through the point of contact D . Now, P A is perpendicular to AB , Q O as is the radius O B to the point of D T contact with AB . Thus P A and O B are parallel and the alternate O angles P OD and BO D are equal. But triangles P OD and BO D are isosceles, and since their vertical A B angles are equal, so are their base angles. Therefore ∠ODP = ∠O DB , and D lies on P B . Next, diameter P A subtends a right angle at D , making AD the altitude to the hypotenuse in right-triangle ADP . By a standard mean proportion, then, we have P A2 = P D · P B (1)

161

On the other hand, the power of P with respect to T is P Q2 and also P D · P B ; hence, P Q2 = P D · P B (2) By (1) and (2), P Q2 = AP 2 . It follows that P Q = AP , as desired. Solution 2. Let R and r be the radii of circles S and T , respectively. Let O be the center of T and denote by C the foot of the perpendicular from O to AP . By the Pythagorean theorem, applied to right triangles P QO and P CO , P Q2 + QO that is, A B P Q2 + QO 2 = P C 2 + AB 2 √ and since AB = 2 Rr (for a proof, see e.g. Japanese Temple Geometry Problems, by H. Fukagawa and D. Pedoe, Canada, 1989, Example 1.1 on p. 3), P C = 2R − r , QO = r , we have P Q2 + r 2 = (2R − r ) + 4Rr . Hence P Q2 = = = 4R2 (2R)2 AP 2
2 2

P S 2R − r r O Q T

= PO 2 = P C 2 + CO

C r

It follows that P Q = AP , as desired.

For positive real numbers a, b, c prove that (a2 + b2 )2 ≥ (a + b + c)(a + b − c)(b + c − a)(c + a − b) .

Solved by Michel Bataille, Rouen, France; Geoﬀrey A. Kandall, Hamden, CT, USA; Giulio Loddi, High School student, Cagliari, Italy; Henry Ricardo, Tappan, NY, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give Loddi’s solution. We will treat a and b like constants and the right-hand side as a function of c: f (c) = = = = (a + b + c)(a + b − c)(b + c − a)(c + a − b) [(a + b)2 − c2 ] · [c2 − (a − b)2 ] −c4 + c2 [(a − b)2 + (a + b)2 ] − (a − b)2 (a + b)2 −c4 + 2c2 (a2 + b2 ) − (a2 − b2 )2

162

Let us ﬁnd the maximum of f (c). Diﬀerentiating with respect to c yields: f (c) = −4c3 + 4c(a2 + b2 ) = 4c[−c2 + a2 + b2 ]. The derivative of f (c) is √ zero when c = 0 or when c2 = a2 + b2 . 2 2 2 f (0) = −( a − b ) and f ( a2 + b2 ) = (a2 + b2 )2 − (a2 − b2 )2 = 4a2 b2 , so √ f (0) < f ( a2 + b2 ). We can guess that there is a maximum when c2 = a2 + b2 . But f ( a2 + b2 ) = 4a2 b2 ≥ −c4 + 2c2 (a2 + b2 ) − (a2 − b2 )2 = f (c) when c4 − 2c2 (a2 + b2 )+(a2 − b2 )2 +4a2 b2 ≥ 0. By computing the discriminant of this quadratic (in c2 ): = 4(a2 + b2 )2 − 4(a2 − b2 )2 − 16a2b2 = 16a2 b2 − 16a2 b2 = 0, √ so f ( a2 + b2 ) − f (c) ≥ 0 for all c. Finally, (a2 + b2 )2 ≥ 4a2 b2 follows from (a2 − b2 )2 ≥ 0 and thus LHS ≥ 4a2 b2 = f (
Ô Ô

Ed. – Note that f (c) = − c2 − (a2 + b2 ) + 4a2 b2 which yields the same result. √ 6. Let n be an integer. Show that, if 2 + 2 1 + 12n2 is an integer, then it is a perfect square. Solved by Arkady Alt, San Jose, CA, USA; Michel Bataille, Rouen, France; Henry Ricardo, Tappan, NY, USA; and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON. We use Bataille’s write-up. √ Suppose that 2 + 2 1 + 12n2 is a positive integer m. Then (m − 2)2 = 4(1 + 12n2 ) so that 1 + 12n2 must be a perfect square, say 1 + 12n2 = a2 where a is a positive integer. It follows that a2 − 3(2n)2 = 1
2 2

a2 + b2 ) ≥ f (c) = RHS
¡2

∀c > 0.

(1)

and the pair (a, 2n) is a solution to the Fermat equation x − 3y = 1 with x ≥ 1 and y even. It is well-known that the solutions √ k in nonnegative √ to this equation 3 = (2+ 3) , k = 0, 1, 2, . . .. integers are the√ pairs (xk , yk ) such that x + y k√ √ k Since xk+1 + 3yk+1 = (xk + yk 3)(2 + 3) the sequences (xk ), (yk ) are given by the recursion xk+1 = 2xk + 3yk , yk+1 = xk + 2yk and x0 = 1, y0 = 0. Using induction, it √ is easy to see that √ yk is even if and only if k is even. Note also that 2xk = (2 + 3)k + (2 − 3)k . Returning to (1) and assuming that n ≥ 0 without lost of generality, we must have a√ = xk and 2n = √ yk for some even k. Setting k = 2 , we ﬁrst deduce 2a = (2 + 3)2 + (2 − 3)2 and then √ √ √ √ m = 2 + 2a = (2 + 3)2 + (2 − 3)2 + 2(2 + 3) (2 − 3) √ √ 2 = (2 + 3) + (2 − 3) = x2 , a perfect square.

163

Next up are solutions to problems of the 2006/2007 British Mathematical Olympiad, Round 2, given at [2010: 154].

Triangle ABC has integer-length sides, and AC = 2007. The internal bisector of ∠BAC meets BC at D . Given that AB = CD , determine AB and BC . Solved by Michel Bataille, Rouen, France; Geoﬀrey A. Kandall, Hamden, CT, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We use Kandall’s version. B a−c c D a c

2007

Let BC = a, AB = c, so that BD = a − c. Note that 2007 = 32 · 223 (prime factorization). Since AD is an angle bisector, we have a−c c = c 2007 . (1)

Thus, c2 = 32 · 223(a − c). Both 3 and 223 divide c, so c = 3 · 223k (k a positive integer). From (1), c2 2007 + c = 223(k2 + 3k).

a =

Since a < c + 2007 (triangle inequality), we have 223(k2 + 3k) < 3 · 223k + 9 · 223, which reduces easily to k2 < 9. Thus, k = 1 or k = 2. If k = 1, then c = 3 · 223, a = 4 · 223, so c + a = 7 · 223 < 2007, which violates the triangle inequality. 2230. Therefore, k = 2, which means that c = 6·223 = 1338 and a = 10·223 =

164

Show that there are inﬁnitely many pairs of positive integers (m, n) such that m+1 n + n+1 m

is a positive integer. Solved by Prithwijit De, Homi Bhabha Centre for Science Education, Mumbai, India.
+1 +1 + nm . Observe that f (1, 1) = 4. Let f (m, n) = mn We claim that there are inﬁnitely many pairs of positive integers (m, n) such that f (m, n) = 4. f (m, n) = 4 implies m2 + (1 − 4n)m + n2 + n = 0. Viewing this as a quadratic in m and solving we get √ (4n − 1) ± 12n2 − 12n + 1 . (1) m = 2

Observe that t = 12n2 − 12n + 1 is odd and m is an integer if and only if t is a perfect square. So let t = p2 , for some positive integer p, then p2 = 3 q 2 − 2 (2)

where q = 2n − 1. If (p, q ) satisﬁes (2) then both p and q must be odd. Equation (2) is satisﬁed by (p1 , q1 ) = (1, 1) and if the positive integral pair (pk , qk ) satisﬁes (2) then so does (pk+1 , qk+1 ) where pk+1 qk+1 = 2 p k + 3 qk = p k + 2 qk .

Observe that {pk } and {qk } are increasing sequences and pk > qk for k > 1. Now deﬁne qk + 1 , nk = 2 2 2 2 for k ≥ 1. Observe that both mk and nk are positive integers as qk and qk+1 are odd positive integers. The set S = {(mk , nk ) : k ≥ 1} is an inﬁnite set (because {qk } is an increasing sequence) and consists of pairs of positive integers satisfying f (m, n) = 4. Thus we have produced inﬁnitely many pairs of positive integers (m, n) for which m+1 n is a positive integer. + n+1 m mk = (2nk − 1) + 12n2 k − 12nk + 1 = 2 qk + 1 + p k = qk+1 + 1

165

Let ABC be an acute-angled triangle with AB > AC and ∠BAC = 60◦ . Denote the circumcentre by O and the orthocentre by H and let OH meet AB at P and AC at Q. Prove that P O = HQ. Note: The circumcentre of triangle ABC is the centre of the circle which passes through the vertices A, B and C . The orthocentre is the point of intersection of the perpendiculars from each vertex to the opposite side. Solved by Michel Bataille, Rouen, France; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give Bataille’s solution. A

O H

Q C

A B Note that from AB > AC , we have C = ∠ACB > B = ∠ABC , hence 2C > B + C = 180◦ − A = 120◦ and so C > 60◦ . Also, since ∆ABC is acute-angled, AH = 2OA where A is the midpoint of BC , that is AH = 2R cos A = R (denoting the circumcentre by H ). It follows that ∆OAH is isosceles with AO = AH . Moreover, we have ∠HAC = 90◦ − C 1 (180◦ − ∠BOA) = 1 (180◦ − 2C ) = 90◦ − C as well. It and ∠P AO = 2 2 follows that the angle bisectors of ∠BAC and ∠OAH are the same line . Now, if ρ denotes the reﬂection in , the image ρ (OH ) of the line OH is OH itself (since OH ⊥ ) and the image ρ (AB ) is AC . As a result, the image of P , the intersection of OH and AB is Q, the intersection of OH and AC . Finally, ρ (P ) = Q, ρ (O ) = H and so P O = QH .

166

Next we move to the May 2010 number of the Corner and solutions from our readers to problems of the XV Olymp´ ıada Matem´ atica Rioplatense, Nivel 2, given at [2010; 214].

1. Let ABC be a right triangle with right angle at A. Consider all the isosceles triangles XY Z with right angle at X , where X lies on the segment BC , Y lies on AB , and Z is on the segment AC . Determine the locus of the medians of the hypotenuses Y Z of such triangles XY Z .
Solved by Oliver Geupel, Br¨ uhl, NRW, Germany; and Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA. We give the solution of Geupel.

A Z E Y M F

Let AEDF be the square where D , E , and F are on the segments BC , AB , and AC , respectively. We prove that D = X and that the locus of the midpoints M of segments Y Z is the line segment EF . The end point E and accordingly F is included if and only if AB ≥ AC and AB ≤ AC , respectively. Consider XY Z as described in the problem. By ∠Y AZ = ∠Y XZ = 90◦ , the quadrilateral AY XZ is cyclic. From the condition XY = XZ , we see that ∠XAY = ∠XAZ = 45◦ ; hence X = D . By ∠DM Y = ∠DEY = 90◦ , the quadrilateral DM EY is cyclic. Thus, ∠DEM = ∠DY M = 45◦ . Consequently, M is on EF . Vice versa, let M lie on EF . The cases M = E and M = F are possible if and only if AB ≥ AC and AB ≤ AC , respectively. Let us suppose that M = E, F . The perpendicular to DM through M cuts AB and AC at Y and Z , respectively. By ∠DM Y = ∠DEY = 90◦ , the quadrilateral DM EY is cyclic; hence ∠DY M = ∠DEM = 45◦ . Similarly ∠DZM = 45◦ . Consequently, XY Z is an isosceles right triangle, which completes the proof.

167

A ﬁnite number of (possibly overlapping) intervals on a line are given. If the rightmost 1/3 of each interval is deleted, an interval of length 31 remains. If the leftmost 1/3 of each interval is deleted, an interval of length 23 remains. Let M and m be the maximum and minimum of the lengths of an interval in the collection, respectively. How small can M − m be? Solved by Oliver Geupel, Br¨ uhl, NRW, Germany. The solution is 24. Consider the intervals [0, 33], [19, 28], and [25, 34]. If the rightmost 1/3 of each interval is deleted, then the union of the resulting intervals is the interval [0, 31] with length 31. If the leftmost 1/3 of each interval is deleted, then the union of the resulting intervals is [11, 34] with length 23. We have M = 33, m = 9; therefore M − m = 24. We prove that generally M − m ≥ 24. Let [a, b] be the minimal closed interval that contains all the given intervals. If the rightmost 1/3 of each interval is deleted, then an interval [a, r ] of length 31 remains. Thus, r − a = 31. At least one of the intervals with right end point in the interval [r, b] will be reduced by a segment not greater than b − r . The length of such an interval is not greater than 3(b − r ), which implies that m ≤ 3(b − r ). If the leftmost 1/3 of each interval is deleted, then an interval [ , b] of length 23 remains. Thus, b − = 23. The initial collection of intervals contains an interval with left bound a. Its left bound after the deletion of the left 1/3 is not less than . Hence, its length is not less than 3( − a), which implies that M ≥ 3( − a). We conclude M − m ≥ 3( − a) − 3(b − r ) = 3 [(r − a) − (b − )] = 3(31 − 23) = 24, which completes the proof.

168

BOOK REVIEWS
Amar Sodhi
Pythagoras’ Revenge: A Mathematical Mystery by Arturo Sangalli Princeton University Press, 2009 ISBN: 978-0-6910-4955-7, cloth, 188 + xviii pp. US$24.95 Reviewed by Mark Taylor, Halifax, N.S. Dr. Jule (formerly Jules) Davidson teaches group theory and non-Euclidean geometry at Indiana State University. He has reached the age of 34 and is becoming increasingly bored with the routine of academic life. On top of that he has “all but given up hope of becoming a famous mathematician”. Presumably in an eﬀort to complete the work for his Fields Medal before time runs out, Jule spends his evenings visiting canyousolveit.com, a website devoted to math puzzles. His attention is caught by the following problem: A group of twelve baseball players put their caps in a bag. After the caps are well shuﬄed, each player picks one at random. (1) Calculate the probability that none of the players will pick up his own hat; (2) What is this probability if there are inﬁnitely many players in the group? In the preface to the book, Sangalli states that one of his aims is “to reach those who usually shun mathematics”. The question above certainly has the potential to fulﬁll this aim. The problem can be explained in such a way that readers who may not understand the mathematical meaning of “probability” or “inﬁnitely many” can come to believe they understand what the question asks. Unfortunately the calculation of probabilities can be diﬃcult to understand even when the problem is simple to pose – the Monty Hall and the Birthday problems are two good examples. Jule’s answers are 0.3679 and 1/e for parts (1) and (2) respectively. A hint to the solution is provided in appendix 1. It uses the Inclusion-Exclusion Principle. This may have the eﬀect of damaging some readers for life or at least cause them to shun the appendices; the latter being unfortunate because the appendices also include Inﬁnitely Many Primes, A Simple Visual Proof of the Pythagorean Theorem and some paragraphs on Perfect, Triangular and Square Numbers all of which should be accessible to anyone with a Grade 10 education. Jule’s solution to the hat problem results in an invitation to compete for the “opportunity to help solve a 2,500 - year - old enigma”. Indiana Jule is quick to seize the opportunity and, after passing a number of tests, joins an esoteric group seeking the reincarnation of Pythagoras. The genesis of the Pythagorean cult and its beliefs are explained in Chapter 8 where we also learn Pythagoras prophesied his own reincarnation, and left instructions in a secret document which was to be guarded through the generations until he reappeared on earth. When questioned how the custodian of the secret document would recognize the reincarnation,

169

Pythagoras states, “He will be an extraordinary gifted man, eminently versed in the secrets of Number, of whom many wonderful things will be persistently related ”. Chapter 9 introduces Norton Thorp who at the age of 9 months spoke in complete sentences. When Norton was barely ﬁve an incredible thing happened to him. His guardian aunt, Therese, put him to bed and then sat down with her erstwhile lover Morris to a meal that consisted of “an assortment of dips that included grilled eggplant and lemon puree, a spread made from feta cheese spiced with chili pepper and garlic, and meat cooked in tomato and red wine sauce. An entre of burghul and potato cakes with lamb and apricot ﬁlling was followed by the main course: swordﬁsh baked in a lemon and paprika sauce and served on a bed of pilaf rice”. Just before dessert was to be served, the couple heard a piano playing. The young Norton, who had never had a piano lesson in his life, was playing the third movement from Mozart’s piano sonata in A major, K 331, with all the skill of a concert pianist! Ten years after the piano incident, Thorp is subject to another supernatural incident. This time he writes an excerpt from The Odyssey in an ancient Greek script. The search for the reincarnation of Pythagoras is paralleled in the book by a search for a copy of a manuscript by Pythagoras (no such manuscript was believed to exist). Of course, the latter search, driven by Professor Elmer Galway of Oriel College, Oxford, results in the unearthing of Pythagoras’ secret document. The discovery takes place in Rome and the chapter heading is the old proverb - to think that if the manuscript had been left in Greece the heading might have read “All Roams lead to Rhodes”. I should mention that Jule has a twin sister, Johanna Davidson, who has a Ph.D. in computer science. Johanna’s purpose in the book seems to be to link Jule to Norton Thorp, and to facilitate a discourse on randomness. Unfortunately she also aﬀords Sangalli the opportunity to display his inadequacies as a writer. I shudder, or perhaps cringe is a better word, to recall his description of Johanna. I can only assume that the material on random numbers had numbed the editor’s brain to such an extent that the paragraphs immediately following failed to register. You may detect a certain lack of enthusiasm for the book on my part. Such an observation is correct. However, I think Sangalli has produced the basis for what could be a commercially successful screenplay. His ingenious twists would transfer to ﬁlm without diﬃculty and the pace could make more acceptable the irrationalities within the plot because there would be little time for reﬂection. Of course, the mathematics would have to be toned down; perhaps reduced to the standard esoteric scribblings, with plenty of subscripts, superscripts, multiple integrals, Greek letters and tensor products. Yes, my advice would be to wait for the ﬁlm.

170

PROBLEMS
Solutions to problems in this issue should arrive no later than 1 November 2011. An asterisk ( ) after a number indicates that a problem was proposed without a solution. Each problem is given in English and French, the oﬃcial languages of Canada. In issues 1, 3, 5, and 7, English will precede French, and in issues 2, 4, 6, and 8, French will precede English. In the solutions’ section, the problem will be stated in the language of the primary featured solution. The editor thanks Jean-Marc Terrier of the University of Montreal for translations of the problems.

3626.
that

Proposed by Thanos Magkos, 3rd High School of Kozani, Kozani, Greece.

Let x, y , and z be positive real numbers such that x2 + y 2 + z 2 = 3. Prove 1 + x2 z+2 + 1 + y2 x+2 + 1 + z2 y+2 ≥ 2.

3627.

Proposed by Jos´ e Luis D´ ıaz-Barrero, Universitat Polit` ecnica de Catalunya, Barcelona, Spain. Find all quadruples a, b, c, d of positive real numbers that are solutions to the system of equations a +b +c +d = 4, 1 1 1 1 + 12 + 12 + 12 (1 + 3abcd) = 16 . a12 b c d

3628.

Proposed by George Apostolopoulos, Messolonghi, Greece.

Let a, b, c and r be the edge-lengths and the inradius of a triangle ABC . Find the minimum value of the expression E= a2 b2 a+b−c + b2 c2 b+c−a + c2 a2 c+a−b r −3 .

3629.

Proposed by Michel Bataille, Rouen, France.

Find the greatest positive integer m such that 2m divides 2011(2013
2016

−1 )

− 1.

3630.
CA, USA.

Proposed by Hung Pham Kim, student, Stanford University, Palo Alto,

Let a, b, and c be nonnegative real numbers such that a + b + c = 3. Prove that ab(b + c) bc(c + a) ca(a + b) + + ≤ 2. 2+c 2+a 2+b

171

3631.

Proposed by Michel Bataille, Rouen, France.

Let {xn } be the sequence satisfying x0 = 1, x1 = 2011, and xn+2 = 2012xn+1 − xn for all nonnegative integer n. Prove that
2 2 2 (2010 + x2 n + xn+1 )(2010 + xn+2 + xn+3 ) 2 (2010 + x2 n+1 )(2010 + xn+2 )

is independent of n.

3632.
Italy.

Proposed by Panagiote Ligouras, Leonardo da Vinci High School, Noci,

Let k be a real number such that 0 ≤ k ≤ 56. Prove that the equation below has exactly two real solutions: (x − 1)(x − 2)(x − 3)(x − 4)(x − 5)(x − 6) = k(x2 − 7x) + 720 .

3633.

Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.

Let g1 (x) = x and for natural numbers n > 1 deﬁne gn (x) = xgn−1 (x) . Let f : (0, 1) → R be the function deﬁned by f (x) = gn (x), where n = Determine lim f (x) or prove it does not exist.
x→0+

1 . x

For example, f

1 3

1 1 3 3

. Here

denotes the ﬂoor of a.

3634.

Proposed by Michel Bataille, Rouen, France.

ABC is an isosceles triangle with AB = AC . Points X , Y and Z are on − → − → − → rays AC , BA and AC respectively with AZ > AC and AX = BY = CZ . (a) Show that the orthogonal projection of X onto BC is the midpoint of Y Z . (b) If BZ and Y C intersect in W , show that the triangles CY A and CW Z have the same area.

172

3635.

Proposed by Mehmet Sahin, Ankora, Turkey.

Let ABC be an acute-angled triangle with circumradius R, inradius r , semiperimeter s, and with points A ∈ BC, B ∈ CA, and C ∈ AB arranged so that ∠ACC = ∠CBB = ∠BAA = 90◦ . Prove that: (a) |BC ||CA ||AB | = abc; (b) (c) |AA | |BB | |CC | = tan A tan B tan C ; |BC | |CA | |AB | Area(A B C ) 4R2 = 2 − 1. Area(ABC ) s − (2R + r )2 Proposed by Pham Van Thuan, Hanoi University of Science, Hanoi,

3636.
Vietnam.

Let a, b, c, and d be nonnegative real numbers such that a + b + c + d = 2. Prove that ab(a2 + b2 + c2 )+ bc(b2 + c2 + d2 )+ cd(c2 + d2 + a2 )+ da(d2 + a2 + b2 ) ≤ 2 .

3637.

Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.
∞ n=1

Let x be a real number with |x| < 1. Determine (−1)n−1 n ln(1 − x) + x + x2 2 + ··· + xn n .

.................................................................

3626.
que

eme Propos´ e par Thanos Magkos, 3i` -Coll` ege de Kozanie, Kozani, Gr` ece.

Soit x, y et z trois nombres r´ eels positifs tels que x2 + y 2 + z 2 = 3. Montrer 1 + y2 1 + z2 1 + x2 + + ≥ 2. z+2 x+2 y+2

3627.

Propos´ e par Jos´ e Luis D´ ıaz-Barrero, Universit´ e Polytechnique de Catalogne, Barcelone, Espagne.

Trouver tous les quadruplets a, b, c, d de nombres r´ eels positifs qui sont solutions du syst` eme d’´ equations a +b +c +d = 4, 1 1 1 1 + 12 + 12 + 12 (1 + 3abcd) = 16 . 12 a b c d

173

3628.

Propos´ e par George Apostolopoulos, Messolonghi, Gr` ece.

Soit a, b, c les longueurs des cˆ ot´ es d’un triangle ABC et r le rayon de son cercle inscrit. Trouver la valeur minimale de l’expression E= a2 b2 a+b−c + b2 c2 b+c−a + c2 a2 c+a−b r −3 .

3629.

Propos´ e par Michel Bataille, Rouen, France.

Trouver le plus grand entier positif m tel que 2m divise 2011(2013
2016

−1 )

− 1.

3630.

Propos´ e par Pham Kim Hung, ´ etudiant, Universit´ e de Stanford, Palo ´ . Alto, CA, E-U Soit a, b et c trois nombres r´ eels non n´ egatifs tels que a + b + c = 3. Montrer que bc(c + a) ca(a + b) ab(b + c) + + ≤ 2. 2+c 2+a 2+b

3631.

Propos´ e par Michel Bataille, Rouen, France.

Soit {xn } une suite satisfaisant x0 = 1, x1 = 2011 et, pour tout entier non n´ egatif n, xn+2 = 2012xn+1 − xn . Montrer que
2 2 2 (2010 + x2 n + xn+1 )(2010 + xn+2 + xn+3 ) 2 (2010 + x2 n+1 )(2010 + xn+2 )

est ind´ ependant de n.

3632.

´ Propos´ e par Panagiote Ligouras, Ecole Secondaire L´ eonard de Vinci, Noci, Italie. Soit k un nombre r´ eel tel que 0 ≤ k ≤ 56. Montrer que l’´ equation ci-dessous poss` ede exactement deux solutions r´ eelles : (x − 1)(x − 2)(x − 3)(x − 4)(x − 5)(x − 6) = k(x2 − 7x) + 720 .

3633.

Propos´ e par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie.

Soit g1 (x) = x et, pour les nombres naturels n > 1, on d´ eﬁnit gn (x) = xgn−1 (x) . Soit f : (0, 1) → R la fonction d´ eﬁnie par f (x) = gn (x), o` u n = Trouver la limite lim f (x) ou montrer qu’elle n’existe pas.
x→0+

1 . Par exemple, f x

1 3

1 1 3 3

. Ici, a d´ enote la partie enti` ere de a.

174

3634.

Propos´ e par Michel Bataille, Rouen, France.

Soit ABC un triangle isoc` ele avec AB = AC . On choisit respectivement − → − → − → trois points X , Y et Z sur les rayons AC , BA et AC avec AZ > AC et AX = BY = CZ . (a) Montrer que la projection orthogonale de X sur BC est le point milieu de Y Z. (b) Si BZ et Y C se coupent en W , montrer que les triangles CY A et CW Z ont la mˆ eme aire.

3635.

Propos´ e par Mehmet Sahin, Ankora, Turkey.

Soit ABC un triangle acutangle, r le rayon de son cercle inscrit, R le rayon de son cercle circonscrit, s son demi-p´ erim` etre. Soit de plus les points A ∈ BC, B ∈ CA et C ∈ AB arrang´ es de telle sorte que ∠ACC = ∠CBB = ∠BAA = 90◦ . Montrer que : (a) |BC ||CA ||AB | = abc ; (b) (c) |BC | |CA | |AB | Aire(A B C ) Aire(ABC ) = |AA | |BB | |CC | = tan A tan B tan C ; 4R2 s2 − (2R + r )2 − 1.

3636.
Vietnam.

Propos´ e par Pham Van Thuan, Universit´ e de Science de Hano¨ ı, Hano¨ ı,

Soit a, b, c et d des nombres r´ eels non n´ egatifs tels que a + b + c + d = 2. Montrer que ab(a2 + b2 + c2 )+ bc(b2 + c2 + d2 )+ cd(c2 + d2 + a2 )+ da(d2 + a2 + b2 ) ≤ 2 .

3637.

Propos´ e par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie.

Soit x un nombre r´ eel avec |x| < 1. D´ eterminer
∞ n=1

(−1)n−1 n ln(1 − x) + x +

x2 2

+ ··· +

xn n

175

SOLUTIONS
No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems.

3521. [2010 : 108, 111] Proposed by Dorin M˘ arghidanu, Colegiul Nat ¸ional “A.I. Cuza”, Corabia, Romania.
index k let ek = Let x1 , x2 , . . . , xn be real numbers in the interval [e, ∞) and for each
x1 + x2 + · · · + xk . Prove that xk

e2 en 1 xe ≥ nx1 + (n − 1)x2 + · · · + 2xn−1 + xn . 1 + x2 + · · · + xn

Solution by Richard Eden, student, Purdue University, West Lafayette, IN, USA. It is easy to show that f (x) = x1/x is decreasing on [e, ∞). [Ed.: f (x) = exp(ln(x)/x) and d2 /dx2 ln(x)/x = (1 − ln(x))/x2 ≤ 0 for x ∈ [e, ∞), so ln(x)/x decreases on the given interval and so does f (x).] For each index k, xk so
k xe k = xk

1/xk

≥ (x1 + x2 + · · · + xk )
(x1 +x2 +···+xk )/xk

1/(x1 +x2 +···+xk )

≥ x1 + x2 + · · · + xk .

Therefore,
n
k xe k

k=1

≥ (x1 ) + (x1 + x2 ) + · · · + (x1 + x2 + · · · + xn ) = nx1 + (n − 1)x2 + · · · + 2xn−1 + xn .

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece (2 solutions); ARKADY ALT, San Jose, CA, USA; MICHEL BATAILLE, Rouen, France; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; ALBERT STADLER, Herrliberg, Switzerland; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. Most solvers used a similar approach to that of the featured solution, except Curtis, who used Bernoulli’s inequality. Geupel noted that problem 3521 is equivalent to Mathematics Magazine problem 1794 by the same proposer.

176

3522. [2010 : 108, 111] Proposed by Dorin M˘ arghidanu, Colegiul Nat ¸ional “A.I. Cuza”, Corabia, Romania.
If a, b, c, and d are positive real numbers satisfying abcd = 1, prove that a 1+ b
cd

b 1+ c

c 1+ d

d 1+ a

16 2 + b2 + c 2 + d 2 a ≥ 2

Solution by Michel Bataille, Rouen, France. The inequality is equivalent to cd ln 1 + a b + da ln 1 + ≥ b c 16 + ab ln 1 + · ln 2, x ln(1 + x), f 2 c d
2

c d

+ bc ln 1 +

d a

a2 + b2 + c2 + d2

or, taking abcd = 1 into account and setting f (x) = (a2 + b2 + c2 + d2 ) 1 a f 2 a b
2

b f b2 c

1 c

1 d

f 2

d a

≥ 16 ln 2. By the Cauchy-Schwarz Inequality, the left side L satisﬁes L ≥ M 2 , where M = f (a/b) + f (b/c) + f (c/d) + f (d/a), √ so it suﬃces to show that M ≥ 4 ln 2. Now, let x1 = ln(a/b), x2 = ln(b/c), x3 = ln(c/d), x4 = ln(d/a), and for real x, let g (x) = f (ex ) = ex/2 (ln(1 + ex ))1/2 . Then g (x) = ex/2 (ln(1 + ex )) · ln(1 + ex )2 4 è 2ex ln(1 + ex ) ex ex x + + 2 ln(1 + e ) − . (ex + 1)2 ex + 1 ex + 1
x x

−3/2

e e > ex (as it follows So g (x) > 0 certainly holds, since 2 ln(1 + ex ) − ex +1 +1 u from ln(1 + u) > 1+u for positive u). Thus, g is convex on R, and from Jensen’s inequality we obtain

g (x1 ) + g (x2 ) + g (x3 ) + g (x4 ) ≥ 4g

x1 + x2 + x3 + x4 4

Since x1 + x2 + x3 + x4 = 0, we have the desired result √ 1/2 M ≥ 4 · e0 · ln(1 + e0 ) = 4 ln 2.
Also solved by Albert Stadler, Herrliberg, Switzerland; and the proposer. Two incorrect solutions were submitted. Each of the two incorrect submissions used Bernoulli’s inequality, but overlooked the fact that (1 + x)r < 1 + rx for x > 0 and 0 < r < 1.

177

3527.
that

[2010 : 171, 173] Proposed by Hung Pham Kim, student, Stanford University, Palo Alto, CA, USA. Let a, b, and c be nonnegative real numbers such that a + b + c = 3. Prove a2 b +
cyclic

3 2

b2 c +

3 2

≤

75 4

Solution by George Apostolopoulos, Messolonghi, Greece. First we will prove that if a, b, c are nonnegative real numbers satisfying a + b + c = 3, then ab2 + bc2 + ca2 ≤ 4 − abc . (1)

We will use the Rearrangement Inequality to prove this. Let (x, y, z ) be a permutation of (a, b, c) such that x ≥ y ≥ z . Since xy ≥ xz ≥ yz , we have ab2 + bc2 + ca2 = b · ab + c · bc + a · ac ≤ x · xy + y · xz + z · yz

= y (x + z )2 − xyz = y (x + z )2 − abc .

It suﬃces to show y (x + z )2 ≤ 4, which follows from the AM-GM Inequality: 2y (x + z )2 = 2y (x + z )(x + z ) ≤ 2y + (x + z ) + (x + z ) 3
3

2(x + y + z ) 3

= 8;

and (1) is established. Now a2 b +
cyclic

3 2

b2 c +

3 2

≤

75 4

⇐⇒ rA + 3B − 12 ≤ 0 ,

where r = abc, A = ab2 + bc2 + ca2 , and B = a2 b + b2 c + c2 a. By the AM–GM Inequality we have r ≤ 1. Also, by Schur’s inequality, (a + b + c)3 ≥ 3abc + 4ab(a + b) + 4bc(b + c) + 4ca(c + a) , from which we obtain A + B ≤ Finally, we have rA + 3B − 12
27 − 3r . From (1) we have A ≤ 4 − r . 4

= (r − 3)A + 3(A + B ) − 12 27 − 3r ≤ (r − 3)(4 − r ) + 3 − 12 4 −4r 2 + 19r − 15 15 = = −(r − 1) r − 4 4

≤ 0,

178

< 0. which holds because 0 ≤ r ≤ 1 and r − 4 Equality holds if and only if (a, b, c) = (1, 1, 1) or (a, b, c) is a permutation of (0, 1, 2).
˘ Bucharest, Romania; KEE-WAI LAU, Hong Kong, Also solved by MARIAN DINCA, China; PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; STAN WAGON, Macalester College, St. Paul, MN, USA; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer.

3528. [2010 : 171, 173] Proposed by Hiroshi Kinoshita and Katsuhiro Yokota, Tokyo, Japan.
The incircle of triangle ABC touches the sides BC , AC , AB at the points A , B , C , respectively. Let ρ, ra , rb , rc denote the inradii of the triangles A B C , AB C , BC A , CA B , respectively, and let r be the inradius of the triangle ABC . Prove that r =
1 (ρ + ra + rb + rc ) . 2

ˇ Similar solutions by Sefket Arslanagi´ c, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; Oliver Geupel, Br¨ uhl, NRW, Germany; Salem Maliki´ c, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; and the proposers. The desired equation is a consequence of two known theorems. Problem 1.1.4, page 3 of H. Fukagawa and D. Pedoe, Japanese Temple Geometry Problems, says that if the tangents at points B and C of a circle Γ intersect in a point A, then the incentre of ∆AB C lies on Γ. Since the simple proof is not given there, we shall prove it here. Let the line joining A to the centre of Γ meet the circle at P ; we shall show that P is the incentre of ∆AB C . Since AC is tangent to Γ at C , we have ∠P C A = ∠P A C for any point A on the arc B C opposite P . But P is the midpoint of arc B C so that ∠P A C = ∠P A B = ∠P C B , whence C P bisects ∠AC B . Since P also lies on the bisector of ∠B AC , it must be the incentre of ∆AB C , as claimed. To set up the second theorem, let Ia , Ib , Ic , and I be the incentres of the triangles AB C , A BC , A B C , and ABC , respectively. We have seen that Ia , Ib , and Ic lie on the circumcircle of ∆A B C , which we again call Γ; note that Γ has centre I and radius r . If da , db , and dc are the distances from I to the sides B C , C A , and A B , respectively, then r = da + ra = db + rb = dc + rc . Carnot’s theorem applied to ∆A B C with its circumradius r and inradius ρ says that da + db + dc = r + ρ. (See, for example, Nathan Altshiller Court, College Geometry, page 83.) It follows that 3r = (da + ra ) + (db + rb ) + (dc + rc ) = r + ρ + ra + rb + rc , which is equivalent to the desired result.
Also solved by ARKADY ALT, San Jose, CA, USA; MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; GEORGE APOSTOLOPOULOS, Messolonghi, Greece;

179

MICHEL BATAILLE, Rouen, France; MIHAELA BLANARIU, Columbia College Chicago, Chicago, IL, USA; RICHARD EDEN, student, Purdue University, West Lafayette, IN, USA; ´ ˇ Y, ´ Big Rapids, MI, USA; JOHN G. HEUVER, Grande Prairie, AB; VACLAV KONECN KEE-WAI LAU, Hong Kong, China; PRITHWIJIT DE, Homi Bhabha Centre for Science Education, Mumbai, India; JOEL SCHLOSBERG, Bayside, NY, USA; ALBERT STADLER, Herrliberg, Switzerland; PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU, Com´ ane¸ sti, Romania; and the proposers (a second solution). Geupel referred to the result as ”well known”, and provided the reference http://www.artofproblemsolving.com/Forum/viewtopic.php?t=275874. The solution on that web page is essentially the same as our featured solution. Most of the other submissions were based on formulas equivalent to Carnot’s theorem which, applied to ∆ABC , become A B C r = cos A + cos B + cos C − 1 = 4 sin sin sin . R 2 2 2

3529.

[2010 : 171, 174] Proposed by Michel Bataille, Rouen, France.

Let A be a point on a circle Γ with centre O and t be the tangent to Γ at A. Triangle P OQ is such that P is on Γ, Q is on t, and ∠P OQ = 90◦ . Find the envelope of the perpendicular to AP through Q as P OQ varies. I. Solution by the proposer. We shall see that the envelope is the parabola, minus its vertex, with focus t P O and directrix t. Point P can be any point of Γ except for A and its diametrically opposed point. We denote by the perpendicular to AP through P Q, and by U the point of intersection of Γ with the line AP . Since OP = OA, O A ∠OP A = 90◦ and so lines OP and must intersect, say at R. Lastly, let H be the projection of R onto t. From the U Q ﬁgure we see that ∠OP A = ∠OAP = ∠AQU = ∠RQH , while ∠OQR = 90◦ − ∠QRP = ∠RP U = ∠OP A; it follows that ∠OQR = ∠RQH . Thus, H R the right triangles ROQ and RHQ are congruent; so RO = RH , whence, R must lie on the parabola P with focus O , directrix t. As the perpendicular bisector of OH , is the tangent to P at R. Conversely, let be the tangent to P at a point R of P distinct from its vertex, and suppose that it meets t at Q. Let the perpendicular to through A intersect Γ again at P and at U . We show that ∠QOP = 90◦ . As above, let H be the projection of R onto t. Since, using directed angles here, ∠HOQ = ∠QHO = ∠QAU , we have ∠OP U = ∠OP A = ∠P AO = ∠OQR, hence 180◦ = ∠U QO +∠OQR = ∠U QO +∠OP U . Thus, P, O, Q, U are concyclic and the claim follows since ∠P U Q = 90◦ .

180

Comment. Note that this problem oﬀers an alternative construction for the points and tangents of a parabola using the circle centred at the focus and tangent to the directrix. II. Solution by Oliver Geupel, Br¨ uhl, NRW, Germany. Consider Cartesian coordinates (x, y ) with O = (0, 0) and A = (1, 0), and let P = (cos ϕ, sin ϕ). Then for ϕ = 0, π , we have Q = (1, − cot ϕ), and . the perpendicular to AP has slope tan ϕ 2 Firstly, take 0 < ϕ < π . The family of perpendiculars to AP through Q as ϕ varies between 0 and π is given implicitly by U (x, y, ϕ) = 0, where U (x, y, ϕ) = (x − 1) tan ϕ − y − cot ϕ. 2

According to H. v. Mangoldt and K. Knopp, Einf¨ uhrung in die H¨ ohere Mathematik, Vol. 2, 10th ed. (S. Hirzel Verlag Leipzig, 1957) Paragraph 176, for the existence of an envelope one must check that the partial derivatives Ux , Uy , Uϕ , Uϕx , Uϕy , and Uϕϕ are continuous on the domain of deﬁnition (which is easily veriﬁed here); cos(ϕ/2) 1 moreover, −Uϕϕ = sin2 = 0 and Ux Uϕy − Uy Uϕx = 2 cos 2 ϕ = 0, ϕ sin(ϕ/2) as required (where we inserted the values of x and y that we obtain below into the second derivatives). The theorem implies that the equation of the envelope in terms of x and y can be found by solving simultaneously the pair of equations Uϕ = 0, From 0 = Uϕ = (x − 1) · we solve for x: x−1= − so that x= 1 U = 0. 1 2 cos2 ϕ 2 2 cos2
ϕ 2

1 sin2 ϕ

sin2 ϕ ϕ 2

2 Plugging this into the equation U = 0 yields y = − cot ϕ − 1 sin ϕ
1 (1 2

1 − cot2

= − cot − y 2 ), or (y < 0),

ϕ 2

Combining these results, we obtain x =

y 2 = −2x + 1

which is the lower branch of the parabola with focus O and directrix t. For −π < ϕ < 0, we obtain the upper branch of the same parabola. The desired envelope is thus the parabola with the exception of its vertex.
Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; GEORGE APOSTOLOPOULOS, Messolonghi, Greece; CHIP CURTIS, Missouri Southern

181

´ ˇ Y, ´ Big Rapids, MI, USA; ALBERT State University, Joplin, MO, USA; VACLAV KONECN STADLER, Herrliberg, Switzerland; and PETER Y. WOO, Biola University, La Mirada, CA, USA. Two submissions dealt instead with a related problem. All the correct solutions used the same argument as in our solution II, except for solution I. Geupel’s version stood out, however, because he provided the required details (and, consequently, was the only person other than the proposer to explicitly exclude the parabola’s vertex from the envelope). The two submissions that went astray determined the locus of the point U (in the notation of solution I) rather than the envelope of the lines QU . It turns out that the locus is a right strophoid—the pedal curve of a parabola with respect to the point of intersection of its axis and directrix (that is, the locus of the points where a tangent to the parabola meets the perpendicular dropped to it from the pedal point A). It is the curve in the 3 and a loop that is tangent to the parabola accompanying ﬁgure with a vertical asymptote x = 2 1 at its vertex ( 2 , 0). In terms of the coordinatization of solution II, the locus satisﬁes y 2 = (x − 1)2 1 + 2(x − 1) 1 − 2(x − 1) .

3530. [2010 : 171, 174] Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.
Let f : [0, 1] → R be an integrable function which is continuous at 1. Let k be a ﬁxed positive integer, and let
1

an =
0

f (x) (1 + xn )(1 + xn+k )

dx .

Find L = lim an and lim n(L − an ).
n→∞ n→∞

Solution by Oliver Geupel, Br¨ uhl, NRW, Germany. We shall show that L=
0 1

f (x)dx 1 2

(1)

and
n→∞

lim n(L − an ) =

+ ln 2 f (1) .

(2)

(x) The sequence fn (x) = (1+xn f of integrable functions converges )(1+xn+k ) to f (x) almost everywhere in [0, 1] as n → ∞. Moreover, |fn (x)| ≤ |f (x)| for all n. By Lebesgue’s Dominated Convergence Theorem, we obtain 1 1 1

L = lim

n→∞

fn (x)dx =
0

0 n→∞

lim fn (x)dx =
0

f (x)dx ,

thus proving the limit (1). We will ﬁrst prove the limit (2) for f ∈ C 1 and subsequently for general integrable functions f . Let us assume that f ∈ C 1 and make use of the following lemma [1].

182

Lemma If g : [0, 1] → R is a continuous function such that limn→0+ and is ﬁnite, then for any function f : [0, 1] → R of class C 1 ,
1 n→∞

g (x) x

exists

lim n
0

f (x)g (xn )dx = f (1)

1 0

g (x) x

dx .

We have n(L − an ) = n 1 dx n )(1 + xn+k ) (1 + x 0 1 xn (2 + xn ) f (x) =n dx (1 + xn )2 0 1 xn (1 − xk ) f (x) −n dx . (1 + xn )2 (1 + xn+k ) 0 f (x) 1 −
x(2+x) , (1+x)2 1

By applying the lemma with g (x) =
1 n→∞

we obtain
1

lim n
å
0

f (x)

xn (2 + xn ) (1 + xn )2 1 1+x

dx = f (1)
0

g (x) x

è1

= f (1) ln(1 + x) −

=
0

1 2

+ ln 2 f (1) .

Moreover, f is bounded, and for 0 ≤ x < 1, it holds that limn→∞ nxn = 0. Hence, for each x ∈ [0, 1],
n→∞

lim nf (x)

xn (1 − xk ) = 0. (1 + xn )2 (1 + xn+k )

By Lebesgue’s Dominated Convergence Theorem, we obtain
1 n→∞

lim n
0

f (x)

xn (1 − xk ) = 0. (1 + xn )2 (1 + xn+k )

This proves (2) for f ∈ C 1 . Finally, let us drop the hypothesis on f and assume that f is an integrable function which is continuous at 1. By the continuity, for each ﬁxed > 0 there exists a number δ > 0 such that |f (x) − f (1)| < whenever 1 − δ ≤ x < 1. We have
1

n(L − an ) = n

dx (1 + + xn+k ) 1 −δ xn (1 + xk + xn+k ) =n dx f (x) (1 + xn )(1 + xn+k ) 0 1 xn (1 + xk + xn+k ) +n f (x) dx . (1 + xn )(1 + xn+k ) 1 −δ
0

f (x) 1 −

xn )(1

183

By Lebesgue’s Dominated Convergence Theorem, lim n
1 −δ 0

n→∞

f (x)

xn (1 + xk + xn+k ) dx = 0 . (1 + xn )(1 + xn+k )

since the integrand is bounded and pointwise convergent to 0. Hence,
1 n→∞

lim n(L − an ) = lim n
n→∞

f (x)
1 −δ

xn (1 + xk + xn+k ) (1 + xn )(1 + xn+k )

dx .

We have that f (1) − < f (x) < f (1) + for 1 − δ ≤ x < 1. Since the relation (2) holds for the constant functions f1 (x) = f (1) − and f2 (x) = f (1) + , we conclude that 1 + ln 2 (f (1) − ) ≤ lim n(L − an ) ≤ n→∞ 2 This holds for each > 0. Consequently, lim n(L − an ) = 1 2 + ln 2 f (1) . 1 + ln 2 (f (1) + ) . 2

n→∞

Also solved by George Apostolopoulos, Messolonghi, Greece (part 1 only); Paolo Perfetti, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy (part 1 only); Albert Stadler, Herrliberg, Switzerland; and the proposer. Two incorrect solutions to part 2 were received. References [1] T.-L. T. R˘ adulescu, V. D. R˘ adulescu, T. Andreescu, Problems in Real Analysis, Springer, 2009, § 9.5.12., page 348.

3531. [2010 : 172, 174] Proposed by K.S Bhanu, Institute of Science, Nagpur, India, and M.N. Deshpande, Nagpur, India.
Let a, b be positive integers. On the real line, A stands at −a and B stands at b. A fair coin is tossed, and if it shows heads then A moves one unit to the right, while if it shows tails then B moves one unit to the left. The process stops when A or B reaches the origin. Let PA (a, b) be the probability that A reaches the origin before B , and deﬁne PB (a, b) similarly. Prove that E (a, b) = 2aPA (a + 1, b) + 2bPB (a, b + 1) , where E (a, b) is the expected number of tosses before the process terminates. Solution by the Joel Schlosberg, Bayside, NY, USA. Consider a specﬁc a, b. For 0 ≤ k ≤ b − 1 let Ak be the event that A reaches the origin before B after k tosses show tails; for 0 ≤ k ≤ a − 1, let Bk be the event È that B reaches the origin before A after k tosses show heads. Then È b−1 a− 1 PA (a, b) = k=0 P (Ak ) and PB (a, b) = k=0 P (Bk ).

184

Since Ak occurs if and only if there are a + k tosses, of which the ﬁrst a + k − 1 contain a − 1 heads and k tails and the (a + k)th toss is heads, P (Ak ) = a+k−1 k
b−1 k=0

1 2

a+k−1

1 2

= 1 2k

1 2a+k

a+k−1 k

PA (a, b) = Similarly,

P (Ak ) =

1 2a

b−1 k=0

a+k−1 k

P (Bk ) =
a− 1

1 2b+k

b+k−1 k
a− 1 k=0

1 P (Bk ) = b PB (a, b) = 2 k=0 Furthermore, E (a, b) =
b−1 k=0 b−1 k=0

1 b+k−1 2k k

(a + k)P (Ak ) + a+k k a+k k +
a− 1 k=0

a− 1 k=0

(b + k)P (Bk ) b+k k b+k k + 2b 2b+1
a− 1 k=0

a 2a+k 1 2k

b 2b+k
a− 1 k=0

a 2a

b−1 k=0

b 2b

1 2k

2a 2a+1

b−1 k=0

1 2k

(a + 1) + k − 1 k

1 2k

(b + 1) + k − 1 k

= 2aPA (a + 1, b) + 2bPB (a, b + 1)
Also solved by George Apostolopoulos, Messolonghi, Greece; Victor Arnaiz and Pedro A. Castillejo, students, Universidad Complutense de Madrid, Madrid, Spain; Keith Ekblaw, Walla Walla, WA, USA; Oliver Geupel, Br¨ uhl, NRW, Germany; Kathleen E. Lewis, SUNY Oswego, Oswego, NY, USA; Albert Stadler, Herrliberg, Switzerland; and the proposer.

3532.
France.

Correction. [2010 : 172, 174, 239] Proposed by Michel Bataille, Rouen,

Let triangle ABC have circumradius R, inradius r , and let δa , δb , δc be the distances from the centroid to the sides BC , CA, AB , respectively. Prove that Ö √ √ √ √ δa + δb + δc R r ≤ ≤ . 3 2

185

Solution by Richard Eden, student, Purdue University, West Lafayette, IN, USA. Let of ABC have centroid P , area F , and semiperimeter s. Since the area
2F F 2F 2F aδa = , so δa = . Similarly, δb = 2 3 3a 3b

BP C is one-third of F , then

and δc = . 3c Using the Harmonic Mean – Root Mean Square Inequality, and the fact that F = rs, we have 3
2F + 3a 2F + 3b 2F 3c 3a 3b 3c + + 2F 2F 2F ,

≤

or 3 ≤ √ √ √ δa + δb + δc which is the desired ﬁrst inequality. To prove the second inequality, we write δa = = so that 2F bc sin A = 3a 3a (2R sin B )(2R sin C ) sin A 3 · 2R sin A
Ö Ö

s F

1 r

2R sin B sin C 3

2R √ 2R sin B + sin C δa = sin B sin C ≤ 3 3 2 √ √ with similar statements for δb and δc . Adding these together yields
Ô

δa +

δb +

δc ≤

2R 3

(sin A + sin B + sin C ) .

The sine function is concave on (0, π ), so by Jensen’s Inequality we have √ sin A + sin B + sin C A+B+C π 3 ≤ sin = sin = . 3 3 3 2 Therefore, √ √ √ δa + δb + δc ≤
√ 2R 3 3 · = 3 3 2 R , from which the 2

second inequality follows immediately. In both inequalities, equality holds if and only if

ABC is equilateral.

Also solved by ARKADY ALT, San Jose, CA, USA; GEORGE APOSTOLOPOULOS, ˇ ´ University of Sarajevo, Sarajevo, Bosnia and Messolonghi, Greece; SEFKET ARSLANAGI C, Herzegovina (two solutions); MIHAELA BLANARIU, Columbia College Chicago, Chicago, IL, USA; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; PRITHWIJIT ˘ Bucharest, DE, Homi Bhabha Centre for Science Education, Mumbai, India; MARIAN DINCA,

186

Romania; OLEH FAYNSHTEYN, Leipzig, Germany; OLIVER GEUPEL, Br¨ uhl, NRW, ´ ˇ Y, ´ Big Rapids, MI, USA; JOEL SCHLOSBERG, Bayside, NY, Germany; VACLAV KONECN USA; ALBERT STADLER, Herrliberg, Switzerland; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. It is worth noting that the proposed inequality implies Euler’s inequality, 2r ≤ R.

3533. [2010 : 172, 174] Proposed by Cao Minh Quang, Nguyen Binh Khiem High School, Vinh Long, Vietnam.
Let a, b, c be positive real numbers such that a + b + c = 1. Let m and n be positive real numbers satisfying 6m ≤ 5n. Prove that ma + nbc b+c + mb + nca c+a + mc + nab a+b ≤ 3m + n 2 .

Solution to a corrected version of the problem by the proposer. We prove instead that under the assumption that a, b, c, m, n are all positive real numbers with a + b + c = 1 and 6m ≥ 5n. By the AM–GM inequality , we have ma + nbc 9(ma + nbc)(b + c) + ≥ 3(ma + nbc) . b+c 4 Adding the above inequality to its two cyclic variants and using Schur’s inequality 1 abc ≥ [4(ab + bc + ca) − 1], we obtain
9 mb + nca mc + nab 3m + n ma + nbc + + ≥ b+c c+a a+b 2

ma + nbc

b+c c+a ≥ 3m + 3n(ab + bc + ca) − 9(ma + nbc)(1 − a)

mb + nca

mc + nab a+b 9(mb + nca)(1 − b) 9(mc + nab)(1 − c)

3 = m+ 4 3 = m+ 4

4 4 4 ç 1ä 3n(ab + bc + ca) + 9m(a2 + b2 + c2 ) + 27nabc 4 1 [3n(ab + bc + ca) + 9m(1 − 2ab − 2bc − 2ca) + 27nabc] 4 3 1 ≥ 3m − n + (15n − 18m)(ab + bc + ca) 4 4 3 1 3m + n ≥ 3m − n + (15n − 18m)(a + b + c)2 = . 4 12 2

Counterexamples to the original problem given by ARKADY ALT, San Jose, CA, ˇ ´ University of Sarajevo, Sarajevo, Bosnia and Herzegovina; USA; SEFKET ARSLANAGI C, DIONNE BAILEY, ELSIE CAMPBELL, and CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; PAOLO

187

PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; and PETER Y. WOO, Biola University, La Mirada, CA, USA. 1 1 Geupel and Perfetti gave the counterexample a = b = 4 ,c= 2 , m = 1, n = 2 to the original problem. The proposer’s original wording of the problem was almost identical to the corrected version of the problem stated above, except that m, n were separately given as negative in the proposer’s version.

3534.

[2010 : 172, 175] Proposed by Mih´ aly Bencze, Brasov, Romania.

Let x1 , x2 , . . . , xn be positive real numbers, where n ≥ 2, and let α ≥ 1. Prove that
n n α

(n − 1) ≥

α−1 k=1

xα k

xk
k=1 α n n +1 xα k k=1 k=1 α−1

2α
1≤i<j ≤n

xi xk

+ (n − 1)α−1

Solution by Michel Bataille, Rouen, France.
n

By homogeneity, we may suppose that proved becomes
n k=1 α

xk = 1. The inequality to be

(n − 1) that is,

α−1 k=1

xα k

≥

2
1≤i<j ≤n n

xi xj

+ (n − 1)α−1
α

+1 xα k k=1

(n − 1) On the one hand,

α−1 k=1

(1 −

xk )xα k

≥

xi xj
i=j

(1)

xi xj =
i=j i=1

xi
1≤j ≤n,j =i n

=
i=1

xi (1 − xi )

and on the other hand, since inequality yields
n i=1

(1 − xi ) = n − 1 and x → xα is convex, Jensen’s
n

(n − 1)α−1

k=1

(1 − xk )xα k

≥ (n − 1)α−1 · (n − 1)
n α

(1 − xk )xk n−1

k=1

=
i=1

xi (1 − xi )

The inequality (1) immediately follows.

188

Also solved by RICHARD EDEN, student, Purdue University, West Lafayette, IN, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; ALBERT STADLER, Herrliberg, Switzerland; and the proposer.

3535. [2010 : 172, 175] Proposed by Walther Janous, Ursulinengymnasium, Innsbruck, Austria.
Let a, b, and c be positive real numbers and let α ≥ 0. Prove that a2 + bc b+c
α

b2 + ca c+a

c2 + ab a+b

≥ 31−α (a + b + c)α .

Solution by Albert Stadler, Herrliberg, Switzerland. We will prove that the statement is false for 0 < α ≤ 0.13432, however it hold true for α ≥ 0.5. The open question is what is the smallest positive α for which the inequality holds true. We will prove ﬁrst the following result: If the inequality holds for some α = A and p is any real number so that p > 1 then the inequality also holds for α = Ap: Indeed, by the H¨ older inequality
Ap Ap Ap
1 p

a2 + bc b+c ≥ a2 + bc b+c

+
A

b2 + ca c+a b2 + ca c+a

+
A

c2 + ab a+b c2 + ab a+b

31 − p
A

Thus, since the inequality holds for α = A, we get
Ap Ap Ap
1 p

a2 + bc b+c

b2 + ca c+a

c2 + ab a+b

31 − p

≥ 31−A (a + b + c)A , or equivalently a2 + bc b+c
Ap

b2 + ca c+a

c2 + ab a+b

≥ 31−Ap (a + b + c)Ap ,

which completes our claim. Problem 3437 [2009 : 174, 176; 2010 : 190-191], showed that the statement holds true for α = 0.5, thus by our argument it holds true for α ≥ 0.5.

189

An easy computation shows that the statement fails for a = 1.6 · 10−6 ; b = 0.73 ; c = 0.049 and α = 0.13432. Thus, again by the above argument, the inequality cannot hold for 0 < α ≤ 0.13432. The question of ﬁnding the smallest positive α for which the inequality holds true is still open.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; and PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy. There was also one incorrect solution. The original problem was an “open problem”, the star was missing. The question of ﬁnding the smallest α is still open. All solvers showed that the inequality holds if α is not small (≥ 0.5 for Stadler, Apostolopoulos and Geupel, ≥ 1 for Perfetti) and fails for α small enough. Geupel and Apostolopoulos proved that there exists a small α for which there is a counterexample and Perfetti found that α = 0.1 works.

3536. [2010 : 173, 175] Proposed by Samuel G´ omez Moreno, Universidad de Ja´ en, Ja´ en, Spain.
Find all positive integers n and k such that the equation {x2n } = {x} has 2010 roots inside the interval [k, k + 1), where x is the greatest integer not exceeding x and {x} = x − x . Solution by George Apostolopoulos, Messolonghi, Greece, modiﬁed by the editor. Observe that {x2n } = {x} if and only if f (x) = x2n − x is an integer. Since x ∈ [k, k + 1) and k ≥ 1, we see that f (x) = x(x2n−1 − 1) is an increasing function of x on this interval, as it is the product of two nonnegative and increasing functions. Therefore, since f is continuous, the number of roots of the original equation is the number of integers in the interval [f (k), f (k +1)) = [k2n − k, (k +1)2n − k − 1), which is (k + 1)2n − k2n − 1. Thus, we seek all positive integers k, n such that (k + 1)2n − k2n − 1 = 2010, or (k +1)2n − k2n = 2011. Now, 2011 is a prime number, so by factoring a diﬀerence of squares we deduce that (k +1)n −kn = 1 and (k +1)n +kn = 2011, and hence kn = 1005 = 3 · 5 · 67. Finally, since 1005 contains primes which only divide it to the ﬁrst power, we see that (k, n) = (1005, 1) is the only solution.
Also solved by VICTOR ARNAIZ and PEDRO A. CASTILLEJO, students, Universidad Complutense de Madrid, Madrid, Spain; MICHEL BATAILLE, Rouen, France; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; JOEL SCHLOSBERG, Bayside, NY, USA; and the proposer.

190

3537.

[2010 : 173, 175] Proposed by Marian Marinescu, Monbonnot, France.

Let f : [0, 1] → R be continuous, and let g : [0, 1] → R be monotonic and diﬀerentiable with g (0) = 0. Prove that there is a number 0 < a < 1 such that
a 1 a

f (x)g (x) dx =
0 0

f (x) dx
0

g (x) dx

Solution by Albert Stadler, Herrliberg, Switzerland. Let h(x) = f (x) − 0 f (t)dt. Then h is continuous on [0, 1] and Ê1 0 h(x)dx = 0. It suﬃces to prove that there exists an α ∈ (0, 1) such that
α

Ê1

h(x)g (x)dx = 0 .
0

By eventually replacing g with −g , without loss of generality we can assume g is non-decreasing. Ê Êx x Let F (x) := 0 h(t)dt and H (x) = 0 h(t)g (t)dt. Then F, H are continuously diﬀerentiable on [0, 1]. We need to prove that H (α) = 0 for some 0 < α < 1. Lets note that if F is constant then F ≡ 0, and hence h ≡ 0, which solves the problem. So we can assume that F is not constant. Also, if g is constant, then g ≡ 0, and the problem is also trivial. Thus we can also assume that g is not constant. Integration by parts yields H (x) = F (t)g (t)|x 0 −
x x 0 x

F (t)g (t)dt = F (x)g (x) −
x 0

F (t)g (t)dt
0

(1)

=
0

F (x)g (t) − F (t)g (t)dt =

[F (x) − F (t)]g (t)dt

Since F is continuous on [0, 1] it attains an absolute max and and absolute min. Also, since F (0) = F (1) = 0, and F is not constant, at most one of them can occur at an endpoint of the interval. We break now the problem in two cases: First case: F attains one of the extremum only at the end points of the interval [0, 1]. By eventually replacing h by −h, we can assume in this case without loss of generality that F attains its absolute minimum only at 0 and 1. This means that F (x) > 0 for all x ∈ (0, 1). Let a be so that F (a) is the absolute max of F on [0, 1]. Since F is not constant, we get that F (a) > 0 and a ∈ (0, 1). Since g is non-constant and non-decreasing, and F (t) > 0 for all t ∈ (0, 1) we get
1 1

H (1) =
0

[F (1) − F (t)]g (t)dt = −

F (t)g (t)dt < 0 ,
0

191

and H (a) =

a 0

[F (a) − F (t)]g (t)dt ≥ 0 .

If H (a) = 0 we are done, while if H (a) > 0, by the Intermediate Value Theorem we have H (c) = 0 for some c ∈ (a, 1). Second case: F attains both the absolute maximum and the absolute minimum inside (0, 1). of F . Let a, b ∈ (0, 1) so that F (a), F (b) are the absolute maxima and minima

Then, since g (t) ≥ 0 we have H (a) ≥ 0 and H (b) ≤ 0. Then, by the Intermediate Value Theorem, H (c) = 0 for some c in the closed interval deﬁned by a and b. Since a, b ∈ (0, 1), we get that c ∈ (0, 1), which completes the proof.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer .

3538.
Israel.

[2010 : 173, 175] Proposed by Victor Oxman, Western Galilee College,

In the plane you are given a triangle ABC with its internal angle bisector BD , a point E on the side BC such that ED is the bisector of angle AEC , and the circumcircle of the triangle ABC (but not its centre). Construct the centre of that circle using only a straightedge. [Ed.: The Poncelet–Steiner Theorem says that given a circle with its centre, we can carry out all the ruler-compass constructions in the plane of that circle by straightedge only. See Crux problems 2694, 2695, and 2696 [2002 : 553-557 ].] Combination of solutions by Michel Bataille, Rouen, France and Peter Y. Woo, Biola University, La Mirada, CA, USA. The construction is based on classical properties of complete quadrangles. As B shown in the ﬁgure, we deﬁne F to A D be the intersection of the given lines P AB and ED , and consider the quadrangle BF CD . The diagonal points B C A = F B ∩ CD and E = BC ∩ F D O R E determine a diagonal AE which meets the other two sides in points P on C Q BD and Q on F C that are separated F Γ harmonically by the two diagonal points. Thus, the lines BP (= BD ) and B BQ are separated harmonically by BE (= BC ) and BA. Because BP

192

bisects ∠EBA, it follows that BQ ⊥ BP . Extending BQ and BP to the second points B and B where these lines meet the given circumcircle, call it Γ, B B must be a diameter of Γ. Apply the same argument to the quadrangle BQCA: F = AB ∩ QC and E = BC ∩ QA are diagonal points which separate harmonically the points R, where F E meets the side BQ, and D , where it meets AC . Thus AF and AE separate harmonically AR and AD . Because BD and ED bisect angles at B and E in triangle ABE (internally or externally depending on where E lies on the line BC ), D must be a tritangent centre, whence AD is a bisector of ∠A and, therefore, AD ⊥ AR. Extending AD and AR to the points C and C where they meet Γ, we see that CC is another diameter of Γ. The centre of Γ is then the point where CC intersects B B . Note that to construct the centre we had to draw only the ﬁve new lines: F C, BQ, B B , AR, and CC (and to extend some of the given segments). The construction works as long as E is a point on the line BC distinct from B and C (not just restricted to the segment BC as required by the statement of the problem).
Also solved by the proposer. Bataille contributed the idea of using harmonic conjugates, while Woo contributed the observation that AD is a bisector of ∠A in ∆ABE . From that bisector property we see that E can be located as the point where the line BC meets the reﬂection of the line AB in AD . This implies that ∠BAC + ∠EAC = 180◦ so that, as observed by the proposer, E is interior or exterior to the segment BC according as ∠BAC is obtuse or acute. When that angle is 90◦ , then E = B and our construction for the centre of Γ fails; nevertheless, the centre is still easily constructed by straightedge using a somewhat modiﬁed procedure.

Crux Mathematicorum
with Mathematical Mayhem
edacteurs: Bruce L.R. Shawyer, James E. Totten, V´ aclav Linek Former Editors / Anciens R´

Crux Mathematicorum
Founding Editors / R´ edacteurs-fondateurs: L´ eopold Sauv´ e & Frederick G.B. Maskell Former Editors / Anciens R´ edacteurs: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

193

EDITORIAL
Shawn Godin
Over the last few years the number of people subscribing to CRUX with MAYHEM has been declining while the publication and shipping costs have been increasing. As a result, the CMS is looking for ways to revitalize the journal and so we are conducting a reader survey. We want to hear from the readers about which parts of the journal they like, and which parts should we should consider retiring. Also, there are a number of new features that we are considering for which we need your input. It is vital that we hear from you, the readers, so that we can deliver a journal that best suits your needs. The survey should take about ﬁve minutes to complete, so please take some time to give us some feedback. You can access the survey through our Facebook page: www.facebook.com/pages/ Crux-Mathematicorum-with-Mathematical-Mayhem/152157028211955, or you can access it directly at: http://www.surveymonkey.com/s/W982CPC We have been working our way through the vast backlog of problem proposals. We have just started going through the proposals received in 2011. The hope is, that we will get through the most recent year within the next couple of months. That way, when a proposal gets submitted, the author will know its fate within a reasonable amount of time. We are also looking into an online system for problem proposal and solution submission that will streamline the process. After we get up to date on the current year, we will continue with the proposals from previous years. A few years back, former editor-in-chief V´ aclav (Vazz) Linek suggested the categorization of Crux problems. This will allow us to deliver a more diverse set of problems and to let the readers know what our needs are. We will group the problems into the following areas: Algebra and Number Theory, Logic, Calculus, Combinatorics, Inequalities (including Geometric Inequalities), Geometry, Probability, and Miscellaneous Problems. Currently we have a large number of inequalities, enough to last us for years to come. Historically, we do receive a good supply of geometry questions. We are in real need of good questions from the other areas. Please continue to send us your good problems, especially from those other areas.

194

SKOLIAD

No. 133

Lily Yen and Mogens Hansen
Please send your solutions to problems in this Skoliad by February 15, 2012. A copy of CRUX with Mayhem will be sent to one pre-university reader who sends in solutions before the deadline. The decision of the editors is ﬁnal. Our contest this month is selected problems from the 21st Transylvanian Hungarian Mathematical Competition, 9th Form, 2011. Our thanks go to Dr. Mih´ aly Bencze, president of The Transylvanian Hungarian Competition, Brasov, Romania, for providing us with this contest and for permission to publish it. La r´ edaction souhaite remercier Rolland Gaudet, de Coll` ege universitaire de Saint-Boniface, Winnipeg, MB, d’avoir traduit ce concours.

La 21e comp´ etition math´ ematique hongroise-transylvanienne, 2011
9e classe, Probl` emes choisis

1. D´ emontrer que si a, b, c, et d sont des nombres r´ eels, alors
a + b + c + d − a2 − b2 − c2 − d2 ≤ 1 .

2. Comparons les deux nombres suivants,
A= 22 ßÞ
·2 ··

and B =

33 ßÞ
c

·3 ··

;

2011 copies de 2
c

2010 copies de 3
)

lequel est le plus ´ elev´ e, A ou B ? (Noter que ab ´ egale a(b

et non (ab )c .)

3. D´ eterminer toutes les solutions en entiers naturels a ` chacune des ´ equations.
a. 20x2 + 11y 2 = 2011. b. 20x2 − 11y 2 = 2011.

4. Dans le parallelogramme ABCD, on a ∠BAD

= 45◦ et ∠ABD = 30◦ . D´ emontrer que la distance de B a ` la diagonale AC est 1 |AD |. 2

5. Quelle est la prochaine ann´ ee avec quatre vendredi 13 ?

195

21st Transylvanian Hungarian Mathematical Competition, 2011
Selected problems for the 9th form

Prove that if a, b, c, and d are real numbers, then a + b + c + d − a2 − b2 − c2 − d2 ≤ 1 .

Compare the following two numbers, A= 2 ßÞ
2·
2 ··

and B =

3 ßÞ
c

3·

3 ··

;

2011 copies of 2
c

2010 copies of 3

which is larger, A or B ? (Note that ab equals a(b ) , not (ab )c .)

Find all natural number solutions to each equation: a. 20x2 + 11y 2 = 2011.

b. 20x2 − 11y 2 = 2011. In the parallelogram ABCD , ∠BAD = 45◦ and ∠ABD = 30◦ . Show that the distance from B to the diagonal AC is 1 |AD |. 2

What is the next year with four Friday the 13ths?

Next follow solutions to the British Columbia Secondary School Mathematics Contest 2010, Junior Final Round, Part B, given in Skoliad 127 at [2010 : 353 – 354].

1a.

Find the sum of all positive integers less than 2010 for which the ones digit is either a ‘3’ or an ‘8’. Solution by Szera Pinter, student, Moscrop Secondary School, Burnaby, BC. The positive numbers with ones digit 3 or 8 form an arithmetic sequence with common diﬀerence 5, namely 3, 8, 13, 18, . . . , 2008. The formula for the nth term of an arithmetic sequence with ﬁrst term a and common diﬀerence d is tn = a + d(n − 1). In the problem, a = 3, d = 5, and the last term is 2008, so 2008 = 3 + 5(n − 1), thus n = 402, whence the sequence has 402 terms. The formula for the sum of an arithmetic sequence is S=
(ﬁrst term) + (last term) number · of terms 2

The sum in the problem is therefore S =

3 + 2008 · 402 = 404 211. 2

196

´ Also solved by LENA CHOI, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia; DAVID FAN, student, Campbell Collegiate, Regina, SK; GESINE GEUPEL, student, Max Ernst ´ Gymnasium, Br¨ uhl, NRW, Germany; and ROWENA HO, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC. Solvers who are not familiar with arithmetic sequences can easily ﬁnd the sum using Gauss’s trick as explained in the comments to the solution to Problem 7 in Skoliad 129 given at [2010 : 485].

1b. Two cans, X and Y , both contain some water. From X Tim pours as much water into Y as Y already contains. Then, from Y he pours as much water into X as X already contains. Finally, he pours from X into Y as much water as Y already contains. Each can now contains 24 units of water. Determine the number of units of water in each can at the beginning.
Solution by David Fan, student, Campbell Collegiate, Regina, SK. Let x be the number of units of water originally contained in can X , and let y be the number of units in can Y . The following table then shows the contents in each can as water is poured back and forth: X x x−y 2(x − y ) = 2x − 2y 2x − 2y − (3y − x) = 3x − 5y Y y 2y 2y − (x − y ) = 3y − x 2(3y − x) = 6y − 2x

Thus 3x − 5y = 24 and 6y − 2x = 24. Solving these two simultaneous equations yields that x = 33 and y = 15.
´ Also solved by LENA CHOI, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia; GESINE GEUPEL, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany; ROWENA HO, student, ´ Ecole Dr. Charles Best Secondary School, Coquitlam, BC; and JULIA PENG, student, Campbell Collegiate, Regina, SK. Since the two cans together in the end hold 48 units of water, you know from the outset that y = 48 − x. Whether using this fact saves eﬀort is a matter of taste.

2. The area of

AP E shown in the diagram is 12. Given that |AB | = |BC | = |CD | = |DE |, determine the sum of the areas of all the triangles that appear in the diagram.

A B C D E ´ Solution by Janice Lew, student, Ecole Alpha Secondary School, Burnaby, BC. Since |AB | = |BC | = |CD | = |DE |, AP B , BP C , CP D , and DP E all have the same base and height, so they have the same area. Since AP E has area 12 each of AP B , BP C , CP D , and DP E have area 3.

197

The following table now lists all the triangles in the diagram and their areas: AP B , BP C , CP D , and AP C , BP D , CP E AP D and BP E AP E Grand total DP E Area 3 6 9 12 Total 12 18 18 12 60

Also solved by NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia; DAVID FAN, student, Campbell Collegiate, Regina, SK; GESINE GEUPEL, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany; and SZERA PINTER, student, Moscrop Secondary School, Burnaby, BC.

3. Given that P QRS is a square and that ABS is an equilateral triangle (see the diagram), ﬁnd the ratio of the area of AP S to the area of ABQ.
Let a denote the side length of the square, let x be |P A|, and let y = |√ BR|. By the Pythagorean Theorem, | SA | = a2 + x2 and Ô 2 2 |SB | = a + y . Since ABS is equilateral, |SA| = |SB |, so a2 +x2 = a2 +y 2 , so x2 = y 2 , so x = y . Using the Pythagorean Theorem on AQB yields that |AB |2 = (a − x)2 + (a − y )2 , but |AB |2 = |SA|2 = a2 + x2 and x = y , so
2 2 2 2 2

S R Solution by Natalia Desy, student, SMA Xaverius 1, Palembang, Indonesia.
P x A a−x Q

a−y B y

S
2

a + x = 2(a − x) = 2(a − 2ax + x ) = 2a − 4ax + 2x2 . Thus 2ax = a2 − 2ax + x2 = (a − x)2 . 1 ax, and the area of ABQ is Now, the area of AP S is 2 1 1 1 2 ( a − x )( a − y ) = ( a − x ) = · 2 ax = ax . Hence the ratio of the 2 2 2 1 1 two areas is 2 ax : ax = 2 : 1 = 1 : 2.
´ Also solved by LENA CHOI, student, Ecole Dr. Charles Best Secondary School, ´ Coquitlam, BC; and ROWENA HO, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC.

Find the ﬁve distinct integers for which the sums of each distinct pair of integers are the numbers 0, 1, 2, 4, 7, 8, 9, 10, 11, and 12. ´ Solution by Lena Choi, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC. If the ﬁve numbers are a, b, c, d, and e, and a < b < c < d < e, then a + b < a + c, and all the other sums are larger. Thus a + b = 0 and a + c = 1,

198

so b = −a and c = b + 1. Moreover, c + e < d + e and all the other sums are smaller. Thus c + e = 11 and d + e = 12, so d = c + 1 = b + 2. That is, b, c, and d are consecutive integers. Since c + e = 11, it follows that b + e = 10. Since b, c, and d are consecutive integers, so are b + c, b + d, and c + d. Among the given sums, only 4, 7, 8, and 9 remain. Thus b + c = 7, b + d = 8, and c + d = 9. Finally, a + e = 4. To summarise, a + b = 0, a + c = 1, a + d = 2, a + e = 4, b + c = 7, b + d = 8, b + e = 10, c + d = 9, c + e = 11, and d + e = 12. Since a + b = 0 and a + c = 1, 2a + b + c = 1. But b + c = 7, so 2a = −6, so a = −3. Therefore b = 3, c = 4, d = 5, and e = 7.
Also solved by DAVID FAN, student, Campbell Collegiate, Regina, SK; GESINE GEUPEL, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany; ROWENA HO, student, ´ Ecole Dr. Charles Best Secondary School, Coquitlam, BC; and JULIA PENG, student, Campbell Collegiate, Regina, SK. Placing the ten sums in a chart like this a+b a+c a+d a+e b+c b+d b+e

c+d c+e

d+e

can make the bookkeeping easier. Note that the sums increase as you move down or right in the diagram.

A rectangle contains three circles, as in the diagram, all tangent to the rectangle and to each other. The height of the rectangle is 4. Determine the width of the rectangle. Solution by Julia Peng, student, Campbell Collegiate, Regina, SK.

Since the height of the rectangle is 4, the radius of the large circle is 2, and both the small circles have radius 1. Now connect the centres of the three circles. 4 This forms an isosceles triangle with sides 3 and base 2. By√ the Pythagorean the height of this triangle √ Theorem, √ is 32 − 12 = 8 = 2 2. The distance from the base of the triangle to the right-hand edge of the rectangle is 1, and the distance from the left vertex of the triangle √ of the rectangle is 2, so the width of the rectangle is √ to the left edge 2 + 2 2 + 1 = 3 + 2 2.
´ Also solved by LENA CHOI, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia; ROWENA ´ HO, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; and JANICE LEW, ´ student, Ecole Alpha Secondary School, Burnaby, BC.

This issue’s prize of one copy of Crux Mathematicorum for the best solutions goes to Natalia Desy, student, SMA Xaverius 1, Palembang, Indonesia. We wish our readers the best of luck solving our featured contest.

199

MATHEMATICAL MAYHEM
Mathematical Mayhem began in 1988 as a Mathematical Journal for and by High School and University Students. It continues, with the same emphasis, as an integral part of Crux Mathematicorum with Mathematical Mayhem. The interim Mayhem Editor is Shawn Godin (Cairine Wilson Secondary School, Orleans, ON). The Assistant Mayhem Editor is Lynn Miller (Cairine Wilson Secondary School, Orleans, ON). The other staﬀ members are Ann Arden (Osgoode Township District High School, Osgoode, ON) and Monika Khbeis (Our Lady of Mt. Carmel Secondary School, Mississauga, ON).

Mayhem Problems
Veuillez nous transmettre vos solutions aux probl` emes du pr´ esent num´ ero avant le 15 f´ evrier 2012. Les solutions re¸ cues apr` es cette date ne seront prises en compte que s’il nous reste du temps avant la publication des solutions. Chaque probl` eme sera publi´ e dans les deux langues oﬃcielles du Canada (anglais et fran¸ cais). Dans les num´ eros 1, 3, 5 et 7, l’anglais pr´ ec´ edera le fran¸ cais, et dans les num´ eros 2, 4, 6 et 8, le fran¸ cais pr´ ec´ edera l’anglais. La r´ edaction souhaite remercier Rolland Gaudet, Universit´ e de Saint-Boniface, Winnipeg, MB, d’avoir traduit les probl` emes.

M488.

´ Propos´ e par l’Equipe de Mayhem.

Un triangle a les sommets (x1 , y1 ), (x2 , y2 ) et (x3 , y3 ). (a) Si x1 < x2 < x3 et y3 < y1 < y2 , d´ eterminer la surface du triangle. (b) D´ emontrer que si on laisse tomber les conditions sur x1 , x2 , x3 , y1 , y2 et y3 , alors l’expression que vous avez fournie en (a) donne soit la surface, soit −1 fois la surface.

M489.
que

´ Propos´ e par Neculai Stanciu, Ecole secondaire George Emil Palade, Buz˘ au, Roumanie. D´ emontrer que si m et n sont des entiers positifs relativement premiers tels m 1+ 1 2 + 1 3 + ··· + 1 2010 = n,

alors 2011 divise n.

200

M490.

Propos´ e par Johan Gunardi, ´ etudiant, SMPK 4 BPK PENABUR, Jakarta, Indon´ esie. Pour n entier positif, soit S (n) la somme des chiﬀres dans l’expression d´ ecimale (base 10) de n. Soit m un entier positif donn´ e ; d´ emontrer qu’il existe n entier positif tel que m =
S (n2 ) . S (n)

M491.
ON.

Propos´ e par Edward T.H. Wang, Universit´ e Wilfrid Laurier, Waterloo,

Soient a, b et c des constantes, pas n´ ecessairement distinctes. R´ esoudre l’´ equation ci-bas : (x − a)2 + (x − b)2 + (x − c)2 = 1.

(x − a)2 − (b − c)2

(x − b)2 − (c − a)2

(x − c)2 − (a − b)2

M492.

Propos´ e par Pedro Henrique O. Pantoja, ´ etudiant, UFRN, Br´ esil.

D´ emontrer que
2009 k=0

(k + 1)![6k (6k + 11) − k − 1] = 2011! 62010 − 1 .

M493.

´ Propos´ e par Neculai Stanciu, Ecole secondaire George Emil Palade, Buz˘ au, Roumanie. D´ eterminer tous les entiers positifs x qui satisfont a ` l’´ equation
ä√ ç √ x+ x+1 x+ x ä√ ç + ä√ ç = 1, 4x + 1 + 4022 4x + 2 + 4022

o` u [x] est la partie enti` ere de x.

M494.

Propos´ e par Dragoljub Miloˇ sevi´ c, Gornji Milanovac, Serbie.

eterminer la valeur minimum ¬ Soit ¬z un nombre complexe tel que |z | = 2. D´ 1¬ ¬ de ¬z − ¬. z .................................................................

M488.

Proposed by the Mayhem Staﬀ.

A triangle has vertices (x1 , y1 ), (x2 , y2 ), and (x3 , y3 ). (a) If x1 < x2 < x3 and y3 < y1 < y2 , determine the area of the triangle. (b) Show that, if the conditions on x1 , x2 , x3 , y1 , y2 , and y3 are dropped, the expression from (a) gives either the area or −1 times the area.

201

M489.

Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania. Prove that if m and n are relatively prime positive integers such that m 1+ 1 1 1 + + ··· + 2 3 2010 = n,

then 2011 divides n.

M490. Proposed by Johan Gunardi, student, SMPK 4 BPK PENABUR, Jakarta,
Indonesia. For any positive integer n, let S (n) denote the sum of the digits of n (in base 10). Given a positive integer m, prove that there exists a positive integer n such that m =
S (n2 ) . S (n)

M491.
ON.

Proposed by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,

Let a, b, and c be given constants, not necessarily distinct. Solve the equation below: (x − b)2 (x − c)2 (x − a)2 + + = 1. (x − a)2 − (b − c)2 (x − b)2 − (c − a)2 (x − c)2 − (a − b)2

M492.

Proposed by Pedro Henrique O. Pantoja, student, UFRN, Brazil.
2009 k=0

Prove that (k + 1)![6k (6k + 11) − k − 1] = 2011! 62010 − 1 .

M493.

Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania. Find all positive integers x that satisfy the equation ä√ ç √ x+ x+ x+1 x ä√ ç + ä√ ç = 1, 4x + 1 + 4022 4x + 2 + 4022

where [x] is the integer part of x.

M494.

Proposed by Dragoljub Miloˇ sevi´ c, Gornji Milanovac, Serbia.

Let ¬ ¬ z be a complex number such that |z | = 2. Find the minimum value of 1¬ ¬ ¬z − ¬. z

202

Mayhem Solutions
M451.
Proposed by the Mayhem Staﬀ. Square ABCD has side length 6. Point P is inside the square so that AP = DP = 5. Determine the length of P C . Solution by Florencio Cano Vargas, Inca, Spain. We draw the line parallel to BC that passes through P . Let F be the intersection point of this line with DC . Since P A = P D , the point P lies on the perpendicular = 3. Triangle bisector of AD and then P F = AD 2 DP F is a right-angled triangle, so: DF =
Ô

A 5

6 5 P 3

F C

DP 2 + P F 2 = 4 .
Ô Ô

B 32 + 22 = √ 13 .

Moreover, triangle CF P is also right angled, hence: PC = P F 2 + CF 2 =

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; JORGE ´ ARMERO JIMENEZ, Club Math´ ematique de l’Instituto de Ecuaci´ on Secundaria No. 1, Requena-Valencia, Spain; SCOTT BROWN, Auburn University, Montgomery, AL, USA; ALPER CAY and LOKMAN GOKCE, Geomania Problem Group, Kayseri, Turkey; JACLYN CHANG, student, University of Calgary, Calgary, AB; NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia; GERHARDT HINKLE, Student, Central High School, Springﬁeld, MO, USA; AFIFFAH NUUR MILA HUSNIANA, student, SMPN 8, Yogyakarta, Indonesia; HEEYOON KIM, Conyers, GA, USA; WINDA KIRANA, student, SMPN 8, Yogyakarta, Indonesia; SALLY LI, student, Marc Garneau Collegiate Institute, Toronto, ON; JAGDISH MADNANI, Bangalore, India; MITEA MARIANA, No. 2 Secondary School, ˇ ´ Gornji Milanovac, Serbia; RICARD Cugir, Romania; DRAGOLJUB MILO SEVI C, ´ IES “Abastos”, Valencia, Spain; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; PEIRO, LU´ IS SOUSA, ISQAPAVE, Angola; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; LEI WANG, Missouri State University, Springﬁeld, MO, USA; GUSNADI WIYOGA, student, SMPN 8, Yogyakarta, Indonesia; and KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA.

M452.

Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania.
Õ

a+ (a) Suppose that x = Prove that x2 − a = x.

√

a + · · · for some real number a > 0.

(b) Determine the integer equal to
30 +

Ô Õ √ 30 + 30 + · · · − 42 + Ô √ 6 + 6 + 6 + ···

42 +

√

42 + · · · .

203

Solution by Pedro Henrique O. Pantoja, student, UFRN, Brazil. (a) By squaring both sides of the equation we have,
Õ

x =a+

a+
Õ

√ a + · · · ⇒ x2 − a = x .

√ (b) From part (a) we have that 30 + 30 + 30 + · · · is the positive root of the equation x2 − x − 30 = 0. We can factor the equation to get (x − 6)(x + 5) = 0, so x = −5 or x = 6. Since x is positive, then
Õ

30 + Similarly, we get
Õ

30 +

√

30 + · · · = 6 .

6+ Hence
Õ Õ

√ 6 + 6 + · · · = 3 and

42 +

√ 42 + · · · = 7 .

30 +

√ Õ 30 + 30 + · · · − 42 + √ 6 + 6 + 6 + ···

42 +

√ 6 42 + · · · = − 7 = −5 . 3

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; SCOTT BROWN, Auburn University, Montgomery, AL, USA; FLORENCIO CANO VARGAS, Inca, Spain; ALPER CAY and LOKMAN GOKCE, Geomania Problem Group, Kayseri, Turkey; JACLYN CHANG, student, University of Calgary, Calgary, AB; NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia; MARC FOSTER, student, Angelo State University, San Angelo, TX, USA; G.C. GREUBEL, Newport News, VA, USA; GERHARDT HINKLE, Student, Central High School, Springﬁeld, MO, USA; AFIFFAH NUUR MILA HUSNIANA, student, SMPN 8, Yogyakarta, Indonesia; HEEYOON KIM, Conyers, GA, USA; WINDA KIRANA, student, SMPN 8, Yogyakarta, Indonesia; SALLY LI, student, Marc Garneau Collegiate Institute, ´ IES “Abastos”, Valencia, Spain; PAOLO PERFETTI, Toronto, ON; RICARD PEIRO, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; ´ PLANELLS CARCEL, ´ ANDRES Club Math´ ematique de l’Instituto de Ecuaci´ on Secundaria No. 1, Requena-Valencia, Spain; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; LU´ IS SOUSA, ISQAPAVE, Angola; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; LEI WANG, Missouri State University, Springﬁeld, MO, USA; GUSNADI WIYOGA, student, SMPN 8, Yogyakarta, Indonesia; KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA; and the proposer. One incorrect solution was received.

M453.
Azerbaijan.

Proposed by Yakub N. Aliyev, Qafqaz University, Khyrdalan,

Let ABCD be a parallelogram. Sides AB and AD are extended to points E and F so that E , C , and F lie on a straight line. In problem M447, we saw that BE · DF = AB · AD . Prove that √ √ Ô AE + AF ≥ AB + AD .

204

Solution by George Apostolopoulos, Messolonghi, Greece. We have (BE − DF )2 ≥ 0 ⇐⇒ BE 2 + DF 2 − 2BE · DF ≥ 0 ⇐⇒ (BE + DF )2 ≥ 4BE · DF

⇐⇒ BE 2 + DF 2 + 2BE · DF − 4BE · DF ≥ 0 ⇐⇒ (BE + DF )2 ≥ 4AB · AD √ ⇐⇒ BE + DF ≥ 2 AB · AD

√ ⇐⇒ (AE − AB ) + (AF − AD ) ≥ 2 AB · AD √ ⇐⇒ AE + AF ≥ AB + AD + 2 AB · AD √ √ √ 2 2 ⇐⇒ AE + AF ≥ AB + AD + 2 AB · AD √ √ Ô ⇐⇒ ( AE + AF )2 ≥ ( AB + AD )2 So √ AE + AF ≥ √ AB + √ AD .

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; FLORENCIO CANO VARGAS, Inca, Spain; ALPER CAY and LOKMAN GOKCE, Geomania Problem Group, Kayseri, Turkey; HEEYOON KIM, Conyers, GA, USA; WINDA KIRANA, student, SMPN 8, Yogyakarta, Indonesia; SALLY LI, student, Marc Garneau ˇ ´ Gornji Milanovac, Serbia; Collegiate Institute, Toronto, ON; DRAGOLJUB MILO SEVI C, ´ IES PEDRO HENRIQUE O. PANTOJA, student, UFRN, Brazil; RICARD PEIRO, “Abastos”, Valencia, Spain; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; JORGE SEVILLA LACRUZ, Club Math´ ematique de l’Instituto de Ecuaci´ on Secundaria No. 1, Requena-Valencia, Spain; LU´ IS SOUSA, ISQAPAVE, Angola; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; LEI WANG, Missouri State University, Springﬁeld, MO, USA; GUSNADI WIYOGA, student, SMPN 8, Yogyakarta, Indonesia; KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA; and the proposer.

M454.

Proposed by Pedro Henrique O. Pantoja, student, UFRN, Brazil.

Determine all real numbers x with 16x + 1 = 2x + 8x . Solution by Marc Foster and Travis B. Little, students, Angelo State University, San Angelo, TX, USA. The equation may be re-written in the following ways: 16x + 1 = 8x + 2x , 2x 8x + 1 = 8x + 2x , (2x − 1)(8x − 1) = 0 . This implies that either 2x = 1 or 8x = 1. In either case, the solution is x = 0, which is the solution to the original equation.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; FLORENCIO CANO VARGAS, Inca, Spain; ALPER CAY and LOKMAN GOKCE, Geomania Problem Group, Kayseri, Turkey; NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia; HEEYOON KIM, Conyers, GA, USA; WINDA KIRANA, student, SMPN 8, Yogyakarta,

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˜ Indonesia; LUIZ ERNESTO LEITAO, Par´ a, Brazil; SALLY LI, student, Marc Garneau Collegiate Institute, Toronto, ON; MITEA MARIANA, No. 2 Secondary School, Cugir, ´ Romania; ARTURO PARDO PEREZ, Club Math´ ematique de l’Instituto de Ecuaci´ on ´ IES “Abastos”, Valencia, Spain; Secundaria No. 1, Requena-Valencia, Spain; RICARD PEIRO, PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; LU´ IS SOUSA, ISQAPAVE, Angola; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; LEI WANG, Missouri State University, Springﬁeld, MO, USA; GUSNADI WIYOGA, student, SMPN 8, Yogyakarta, Indonesia; KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA; and the proposer. Most of the other solvers converted everything to base 2 and solved the related equation y4 + 1 = y + y3 .

M455.
Romania.

Proposed by Gheorghe Ghit ¸a ˘, M. Eminescu National College, Buz˘ au,

Suppose that n is a positive integer. (a) If the positive integer d is a divisor of each of the integers n2 + n + 1 and 2n3 + 3n2 + 3n − 1, prove that d is also a divisor of n2 + n − 1. (b) Prove that the fraction
2n3 n2 + n + 1 is irreducible. + 3n2 + 3n − 1

Solution by Florencio Cano Vargas, Inca, Spain. (a) We can write: n2 + n − 1 = (2n3 + 3n2 + 3n − 1) − 2n(n2 + n + 1) Then if d divides 2n3 + 3n2 + 3n − 1 and n2 + n + 1, then it also divides n2 + n − 1. (b) Let d be a positive common divisor of the numerator and denominator. From (a) we know that d also divides n2 + n − 1. But n2 + n + 1 and n2 + n − 1 only diﬀer by two units, so then d ≤ 2. To discard the possibility d = 2, we note that n2 + n + 1 = n(n + 1) + 1 is always odd since either n or n + 1 is even. Then the only possibility that remains is d = 1 and the fraction is irreductible.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; ALPER CAY and LOKMAN GOKCE, Geomania Problem Group, Kayseri, Turkey; JACLYN CHANG, student, University of Calgary, Calgary, AB (part (a) only); SALLY LI, student, Marc Garneau ´ Collegiate Institute, Toronto, ON; ANTONIO LEDESMA LOPEZ, Instituto de Educaci´ on Secundaria No. 1, Requena-Valencia, Spain; MITEA MARIANA, No. 2 Secondary School, Cugir, Romania; MISSOURI STATE UNIVERSITY PROBLEM SOLVING GROUP, Springﬁeld, MO, USA; PEDRO HENRIQUE O. PANTOJA, student, UFRN, Brazil; RICARD ´ IES “Abastos”, Valencia, Spain; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; PEIRO, LU´ IS SOUSA, ISQAPAVE, Angola; LEI WANG, Missouri State University, Springﬁeld, MO, USA; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA; and the proposer. One incorrect solution was received.

M456.

Proposed by Mih´ aly Bencze, Brasov, Romania.

Let f and g be real-valued functions with g an odd function, f (x) ≤ g (x) for all real numbers x, and f (x + y ) ≤ f (x) + f (y ) for all real numbers x and y . Prove that f is an odd function.

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Solution by Paolo Perfetti, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy. From f (x) ≤ g (x) for any x, and from the oddness of g (x) we get f (−x) ≤ g (−x) = −g (x) ≤ −f (x) . Moreover, we have f (0) = f (x + (−x)) ≤ f (x) + f (−x) =⇒ f (−x) ≥ f (0) − f (x) , and f (x + 0) ≤ f (0) + f (x) =⇒ f (0) ≥ 0 . Since f (0) ≥ 0, then from (1) we can conclude that f (−x) ≥ −f (x). Then we have −f (x) ≤ f (−x) ≤ −f (x) . Hence f (x) is an odd function.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; FLORENCIO CANO VARGAS, Inca, Spain; JAVIER GARC´ IA CAVERO, Club Math´ ematique de l’Instituto de Ecuaci´ on Secundaria No. 1, Requena-Valencia, Spain; ALPER CAY and LOKMAN GOKCE, Geomania Problem Group, Kayseri, Turkey; SALLY LI, student, Marc Garneau Collegiate Institute, Toronto, ON; MISSOURI STATE UNIVERSITY PROBLEM SOLVING GROUP, Springﬁeld, MO, USA; PEDRO HENRIQUE O. PANTOJA, student, UFRN, Brazil; ´ IES “Abastos”, Valencia, Spain; BRUNO SALGUEIRO FANEGO, Viveiro, RICARD PEIRO, Spain; LU´ IS SOUSA, ISQAPAVE, Angola; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; LEI WANG, Missouri State University, Springﬁeld, MO, USA; KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA, USA; and the proposer.

(1)

Problem of the Month
Ian VanderBurgh
Here is a neat problem that has an easy-to-understand and appreciate reallife context and leads us to a good discussion of two diﬀerent methods of solution. Problem (2010 AMC 8) Everyday at school, Jo climbs a ﬂight of 6 stairs. Jo can take the stairs 1, 2, or 3 at a time. For example, Jo could climb 3, then 2, then 1. In how many ways can Jo climb the stairs? (A) 13 (B) 18 (C) 20 (D) 22 (E) 24 I can just picture hundreds of Grade 8 students trying this after writing this contest in 2010... There are actually quite a lot of ways of climbing the stairs! The ﬁrst step (ahem, no pun intended) is to read and understand the problem. As part of this, we should ﬁddle to see what ways we can ﬁnd. We are looking for ways of adding combinations of 1s, 2s and 3s to get 6. Try playing with this for a couple of minutes!

207

What combinations did you get? Some that you might get include 3 + 3, 3 + 1 + 1 + 1, and 2 + 3 + 1. One question that we should immediately ask is whether re-arranging the order of a given sum makes a diﬀerence. Does it? Yes – for example, 2 + 3 + 1 (Jo takes 2 steps, then 3 steps, then 1 step) is diﬀerent from 2 + 1 + 3 (2 steps, then 1 step, then 3 steps) which are both diﬀerent than 3 + 1 + 2, and so on. Can you ﬁnd more ways to re-arrange this particular sum? The sum 3 + 1 + 1 + 1 can also be re-arranged in a number of ways. How many can you ﬁnd? So it looks as if there are now two sub-problems – ﬁnding the diﬀerent combinations of 1s, 2s and 3s that give 6, and then ﬁguring out the number of ways in which we can re-arrange each of these combinations. Let’s get a handle on the second sub-problem ﬁrst. To do this, we’ll consider a slightly diﬀerent context: How many diﬀerent “words” can be made from the letters of AAAAB, AAAC, AABB, and ABC? (By a “word” in this case, we mean a rearrangement of the letters; it doesn’t actually have to form a real word!) In each case, we could exhaustively list out the possibilities or look for a diﬀerent approach: • AAAAB: 5 words List : AAAAB, AAABA, AABAA, ABAAA, BAAAA Alternate approach : If we start with the four As (AAAA), there are then ﬁve possible positions for the B: either before the ﬁrst A or after each of the four As. Thus, there are ﬁve words. • AAAC: 4 words List : AAAC, AACA, ACAA, CAAA Alternate approach : Can you modify the previous argument to ﬁt this case? • AABB: 6 words List : AABB, ABAB, ABBA, BAAB, BABA, BBAA Alternate approach : While there are good ways to count the words in this case using more advanced mathematics like combinatorics, actually coming up with a simple explanation of why the answer is 6 without actually just doing it isn’t that easy. Here’s one try. Suppose that the word starts with A. Put in the A and the two Bs to get ABB; the remaining A can go in three places (right before the ﬁrst B or after either B). So there are 3 words beginning with A. Can you see why there are also 3 words beginning with B? • ABC: 6 words List : ABC, ACB, BAC, BCA, CAB, CBA Alternate approach : There are 3 possibilities for the ﬁrst letter; for each of these, there are 2 possibilities for the second letter (all but the letter we already chose); the last letter is then completely determined. This tells us that there are 3 × 2 × 1 = 6 possible words. Now let’s combine this information about re-arrangements with a systematic way of ﬁnding the diﬀerent combinations.

208

Solution 1. Let’s ﬁnd the possible combinations in an organized way. We’ll start by assuming that the order of steps doesn’t actually matter, and then incorporate the order at the end. Coming up with a good way to ﬁnd all of the combinations might require a bit of ﬁddling. One good method to use is to organize these by the number of 1s. Can there be six 1s? Yes: 1 + 1 + 1 + 1 + 1 + 1 = 6. Can there be ﬁve 1s? No: if there are ﬁve 1s, then Jo has only 1 step left for which she needs another 1. Can there be four 1s? Yes: if there are four 1s, then Jo has 2 steps left, which must be taken up by a 2. (It can’t be two 1s since we’re only allowed four 1s.) This gives us 1 + 1 + 1 + 1 + 2. Can there be three 1s? Yes: if there are three 1s, then Jo has 3 steps left. We can’t divide the 3 into two pieces without using a 1, so the only way is 1 + 1 + 1 + 3. Can there be two 1s? Yes: if there are two 1s, then Jo has 4 steps left. To avoid using a 1, this 4 must be 2 + 2. This gives us 1 + 1 + 2 + 2. Can there be one 1? Yes: with 5 steps left and not using a 1, the remaining 5 must be 2 + 3. This gives us 1 + 2 + 3. Can there be zero 1s? Yes. If there are no 2s, then there are only 3s, so we have 3 + 3. If there is a 2, then Jo has 4 steps left, which must be 2 + 2 since no 1s are used. In this case, we have 3 + 3 or 2 + 2 + 2. So ignoring order, the possibilities are (i) 1 + 1 + 1 + 1 + 1 + 1, (ii) 1 + 1 + 1 + 1 + 2, (iii) 1 + 1 + 1 + 3, (iv) 1 + 1 + 2 + 2, (v) 1 + 2 + 3, (vi) 3 + 3, and (vii) 2 + 2. The combinations in (i), (vi) and (vii) can’t be re-arranged in any other order. That gives us 1 way in each case. How can we re-arrange the sums in (ii), (iii), (iv), and (v)? There are 5 ways of arranging the sum in (ii). This is because this sum can be related to the word AAAAB from before, with each A representing a 1 and B representing the 2. Each re-arrangement of the sum in (ii) is the same as one of the words that we talked about earlier. Since there were 5 words, then there are 5 ways of arranging the sum. Can you see how to relate the sums in (iii), (iv) and (v) to the words earlier? Try this out! You’ll ﬁnd that the sum in (iii) can be arranged in 4 ways, and the sum in each of (iv) and (v) can be arranged in 6 ways. Therefore, there are 1 + 5 + 4 + 6 + 6 + 1 + 1 = 24 ways that Jo can climb the stairs. While there was a fair bit of work required to actually make that solution work, we didn’t have to do anything really hard. But, we had to be very, very careful. Also, this method might not “scale up” very well to a larger number of steps, because of the number of cases that we had to consider. Let’s switch gears. Sometimes looking at smaller cases helps in one of two ways: either by showing us a pattern that might continue or more directly by allowing us to capitalize on these smaller cases.

209

What do smaller cases look like here? They are cases with fewer stairs. Let’s try a few: • With 1 stair, there is only 1 way for Jo to climb. • With 2 stairs, there are 2 ways: 1 + 1 and 2. • With 3 stairs, there are 4 ways: 1 + 1 + 1, 1 + 2, 2 + 1, and 3. • With 4 stairs, there are 7 ways: 1 + 1 + 1 + 1, 1 + 1 + 2, 1 + 2 + 1, 2 + 1 + 1, 1 + 3, 3 + 1, and 2 + 2. Do you notice anything about the number of ways in these four cases? Do you think that it is a coincidence that 1+2+4 = 7? In other words, is it a coincidence that the sum of the numbers of ways for 1, 2 and 3 stairs gives us the number of ways for 4 stairs? Solution 2. We have seen that with 1, 2 and 3 stairs, there are 1, 2 and 4 ways, respectively. If Jo is to climb 4 stairs, then she starts by climbing 1 stair (leaving 3) or by climbing 2 stairs (leaving 2) or by climbing 3 stairs (leaving 1). If she starts by climbing 1 stair, then the number of ways that she can ﬁnish climbing is the number of ways to climb the remaining 3 stairs. In other words, the number of ways that she can climb the stairs staring with 1 stair is equal to the number of ways in which she can climb 3 stairs. (There are 4 ways to do this.) If she starts by climbing 2 stairs, then the number of ways that she can ﬁnish climbing is the number of ways to climb the remaining 2 stairs. In other words, the number of ways that she can climb the stairs staring with 2 stairs is equal to the number of ways in which she can climb 2 stairs. (There are 2 ways.) Similarly, the number of ways of climbing starting with 3 stairs is equal to the number of ways of climbing the remaining 1 stair. (There is 1 way.) Therefore, the number of ways of climbing 4 stairs equals the sum of the number of ways of climbing 3, 2 and 1 stairs, or 4 + 2 + 1 = 7. What happens with 5 stairs? In this case, Jo starts with 1, 2 or 3 stairs, leaving 4, 3 or 2 stairs. Using a similar argument, the total number of ways of climbing 5 stairs equals the sum of the number of ways of climbing 4, 3 and 2 stairs, or 7 + 4 + 2 = 13. Continuing along these lines, for 6 stairs, the total number of ways will equal the sum of the number of ways of climbing 5, 4 and 3 stairs, or 13 + 7 + 4 = 24. We’ve just used a method called recursion in this second solution. This can be a very powerful approach in cases where it will work. Recursion is particularly useful in ﬁelds like computer science. I’ll leave you with a challenge. Can you determine the number of ways that Jo could go up the stairs if there were 10 stairs? Which approach do think that you’d want to use?

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THE OLYMPIAD CORNER
No. 294 R.E. Woodrow and Nicolae Strugaru
The problems from this issue come from the selection tests for the Balkan, Indian and Slovenian IMO teams and the Singapore Mathematical Olympiad. Our thanks go to Adrain Tang for sharing the material with the editor.
Toutes solutions aux probl` emes dans ce num´ ero doivent nous parvenir au plus tard le 15 mars 2012. Chaque probl` eme sera publi´ e dans les deux langues oﬃcielles du Canada (anglais et fran¸ cais). Dans les num´ eros 1, 3, 5 et 7, l’anglais pr´ ec´ edera le fran¸ cais, et dans les num´ eros 2, 4, 6 et 8, le fran¸ cais pr´ ec´ edera l’anglais. Dans la section des solutions, le probl` eme sera publi´ e dans la langue de la principale solution pr´ esent´ ee. La r´ edaction souhaite remercier Jean-Marc Terrier, de l’Universit´ e de Montr´ eal, d’avoir traduit les probl` emes.

OC11.

Etant donn´ e deux sous-ensembles non vides A, B ⊆ Z , on d´ eﬁnit A + B et A − B par A + B = {a + b | a ∈ A, b ∈ B } , A − B = {a − b | a ∈ A, b ∈ B } . Dans ce qui suit, on travaille avec des sous-ensembles ﬁnis non vides de Z. Montrer qu’on peut recouvrir B par au plus
|A + B | tel que | A| |A + B | translat´ es de A − A, c.-` a-d. | A|

qu’il existe X ⊆ Z avec |X | ≤ B⊆

x∈X

(x + (A − A)) = X + A − A .

OC12. Soit k un entier positif plus grand que 1. Montrer que pour tout entier
non n´ egatif m, il existe k entiers positifs n1 , n2 , . . . , nk , tels que
2 2 m+k n2 . 1 + n2 + · · · + nk = 5

OC13. Soit ABC un triangle acutangle et soit D un point sur le cˆ ot´ e AB . Le
cercle circonscrit du triangle BCD coupe le cˆ ot´ e AC en E . Le cercle circonscrit du triangle ADC coupe le cˆ ot´ e BC en F . Soit O le centre de gravit´ e du triangle CEF . Montrer que les points D et O et les centres de gravit´ e des triangles ADE , ADC , DBF et DBC sont cocycliques et que la droite OD est perpendiculaire a ` AB .

211

OC14. Soit an , bn , n
b1 = 0 et, pour n ≥ 1,

= 1, 2, . . . deux suites d’entiers d´ eﬁnis par a1 = 1, an+1 = 7an + 12bn + 6 , bn+1 = 4an + 7bn + 3 .

Montrer que a2 erence de deux cubes cons´ ecutifs. n est la diﬀ´

OC15. Une r` egle de longueur

a k ≥ 2 marques distantes de ai unit´ es d’une des extr´ emit´ es avec 0 < a1 < · · · < ak < . La r` egle est appel´ ee r` egle de Golomb si les longueurs mesurables grˆ ace aux marques de la r` egle sont des entiers cons´ ecutifs commen¸ cant avec 1, et telles que chaque longueur soit mesurable entre une unique paire de marques sur la r` egle. Trouver toutes les r` egles de Golomb.

OC16. Etant donn´ e a1
montrer que

≥ 1 et ak+1 ≥ ak + 1 pour tout k = 1, 2, . . . , n,

2 3 3 a3 1 + a2 + · · · + an ≥ (a1 + a2 + · · · + an ) .

OC17.

Montrer que les sommets d’un pentagone convex ABCDE sont cocycliques si et seulement si on a d(E, AB ) · d(E, CD ) = d(E, AC ) · d(E, BD ) = d(E, AD ) · d(E, BC ) .

OC18. Etant donn´ e n nombres complexes a1 , a2 , . . . , an , non n´ ecessairement
k k distincts, et des entiers positifs distincts k, l tels que ak 1 , a2 , . . . , an et l l al , a , . . . , a sont deux collections de nombres identiques, montrer que chaque 1 2 n aj , 1 ≤ j ≤ n, est une racine de l’unit´ e.

OC19. Il y a eu 64 participants dans un tournoi d’´ echecs. Chaque paire a jou´ e
une partie qui s’est termin´ e soit par un gagnant ou par un match nul. Si une partie s’´ etait termin´ ee par un match nul, alors chacun des 62 participants restants gagnait contre au moins un des deux joueurs. Il y a eu au moins deux parties avec match nul dans ce tournoi. Montrer qu’on peut aligner tous les participants sur deux rangs de sorte que chacun d’eux a gagn´ e contre celui qui se trouve juste derri` ere lui.

OC20. Etant donn´ e un entier n ≥ 2, trouver la valeur maximale que la somme
x1 + x2 + · · · + xn puisse atteindre lorsque les xi prennent toutes les valeurs positives sujettes aux conditions x1 ≤ x2 ≤ · · · ≤ xn et x1 + x2 + · · · + xn = x1 x2 · · · xn . .................................................................

212

OC11.

For non-empty subsets A, B ⊆ Z deﬁne A + B and A − B by

A + B = {a + b | a ∈ A, b ∈ B } , A − B = {a − b | a ∈ A, b ∈ B } . In the sequel we work with non-empty ﬁnite subsets of Z. Prove that we can cover B by at most there exists X ⊆ Z with |X | ≤ B⊆
|A + B | such that | A| |A + B | translates of A − A, i.e. | A|

x∈X

(x + (A − A)) = X + A − A .

OC12.

Let k be a positive integer greater than 1. Prove that for every nonnegative integer m there exist k positive integers n1 , n2 , . . . , nk , such that
2 2 m+k n2 . 1 + n2 + · · · + nk = 5

Let ABC be an acute triangle and let D be a point on the side AB . The circumcircle of the triangle BCD intersects the side AC at E . The circumcircle of the triangle ADC intersects the side BC at F . Let O be the circumcentre of triangle CEF . Prove that the points D and O and the circumcentres of the triangles ADE , ADC , DBF and DBC are concyclic and the line OD is perpendicular to AB .

OC13.

OC14.

Let an , bn , n = 1, 2, . . . be two sequences of integers deﬁned by a1 = 1, b1 = 0 and for n ≥ 1, an+1 = 7an + 12bn + 6 , bn+1 = 4an + 7bn + 3 .

Prove that a2 n is the diﬀerence of two consecutive cubes.

OC15.

A ruler of length has k ≥ 2 marks at positions ai units from one of the ends with 0 < a1 < · · · < ak < . The ruler is called a Golomb ruler if the lengths that can be measured using the marks on the ruler are consecutive integers starting with 1, and each such length be measurable between a unique pair of marks on the ruler. Find all Golomb rulers.

OC16.

Given a1 ≥ 1 and ak+1 ≥ ak + 1 for all k = 1, 2, . . . , n, show that
3 3 2 a3 1 + a2 + · · · + an ≥ (a1 + a2 + · · · + an ) .

OC17.

Prove that the vertices of a convex pentagon ABCDE are concyclic if and only if the following holds d(E, AB ) · d(E, CD ) = d(E, AC ) · d(E, BD ) = d(E, AD ) · d(E, BC ) .

213

OC18. If a1, a2 , . . . , an are n non-zero complex numbers, not necessarily k k distinct, and k, l are distinct positive integers such that ak 1 , a2 , . . . , an and l l l a1 , a2 , . . . , an are two identical collections of numbers. Prove that each aj , 1 ≤ j ≤ n, is a root of unity. OC19.
There were 64 contestants at a chess tournament. Every pair played a game that ended either with one of them winning or in a draw. If a game ended in a draw, then each of the remaining 62 contestants won against at least one of these two contestants. There were at least two games ending in a draw at the tournament. Show that we can line up all the contestants so that each of them has won against the one standing right behind him.

OC20.

Given an integer n ≥ 2, determine the maximum value the sum x1 + x2 + · · · + xn may achieve, as the xi run through the positive integers subject to x1 ≤ x2 ≤ · · · ≤ xn and x1 + x2 + · · · + xn = x1 x2 · · · xn .

We now turn to solutions of the II International Zhautykov Olympiad in Mathematics given at [2009 : 376–377].

2. The points K and L lie on the sides AB and AC , respectively, of the triangle ABC such that BK = CL. Let P be the point of intersection of the segments BL and CK , and let M be an inner point of the segment AC such that the line M P is parallel to the bisector of the angle ∠BAC . Prove that CM = AB .
Solved by Michel Bataille, Rouen, France; Geoﬀrey A. Kandall, Hamden, CT, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Bataille. We introduce the glide-reﬂection (or reﬂection) g such that g(B ) = C A and g(K ) = L. The axis of g is parallel to the angle bisector m of ∠BAC and passes through the midM points X and W of BC and KL (see ﬁgure). We observe that the line ALC is W K a transversal of ∆BKP (with A on L BK, L on BP and C on KP ). From P a well-known theorem, the midpoints X C m of the “diagonals” AP, LK, CB are B collinear. It follows that the midpoint of AP is on the axis XW of g. Thus, the lines m and M P , which are parallel to the axis of g, are equidistant of this axis. As a result, g(m) = M P . Since in addition the image under g of the line AB = BK is the line AC = CL, we obtain g(A) = M . Recalling that g preserves distances, AB = CM follows.

214

Let a, b, c, and d be real numbers such that a + b + c + d = 0. Prove that (ab + ac + ad + bc + bd + cd)2 + 12 ≥ 6(abc + abd + acd + bcd) .

Solved by Arkady Alt, San Jose, CA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Zvonaru. Since d = −a − b − c, the given inequality is equivalent to ⇔ (a2 + b2 + c2 + ab + bc + ca)2 + 12 [ab + ac + bc − (a + b + c)2 ]2 + 12 ≥ 6abc − 6(a + b + c)(ab + bc + ca) + 6(a + b + c)(ab + bc + ca) − 6abc ≥ 0.

Because (a + b + c)(ab + bc + ca) − abc = (a + b)(b + c)(c + a), we obtain 1 4 [(a + b)2 + (b + c)2 + (c + a)2 ] + 12 + 6(a + b)(b + c)(c + a) ≥ 0.
a+b , 2

Denoting z =

b+c , 2

c+a , 2

we have to prove that

(x2 + y 2 + z 2 )2 + 24xyz + 48 ≥ 0. By AM-GM Inequality, we have (x2 + y 2 + z 2 )2 ≥ 9|xyz |4/3 and because 24xyz ≥ −24|xyz |, it suﬃces to prove that 9t4 − 24t3 + 48 ≥ 0, where t = |xyz |1/3 . This is true, since 9t4 − 24t3 + 48 = 3(t − 2)2 (3t2 + 4t + 4) ≥ 0. Next we turn to problems of the 50th Mathematical Olympiad of the Republic of Moldova given at [2009 : 377–378].

Let a, b, and c be the side lengths of a right triangle with hypotenuse of length c, and let h be the altitude from the right angle. Find the maximum value c+h of .
a+b

Solved by Arkady Alt, San Jose, CA, USA; Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; Michel Bataille, Rouen, France; Geoﬀrey A. Kandall, Hamden, CT, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Bataille. Let Q = We have
c+h . a+b

We show that the maximum value of Q is Q = c2 + hc (a + b)c = a2 + b2 + ab . √ (a + b) a2 + b2

√ 2 . 4

215
√ √

Clearly, Q = 3 4 2 when a = b. To prove that Q ≤ 3 4 2 in any case, we rewrite this inequality successively as Ô √ 4(a2 + b2 + ab) ≤ 3 2(a + b) a2 + b2 16(a2 + b2 )2 + 16a2 b2 + 32ab(a2 + b2 ) ≤ 18(a2 + b2 )(a2 + b2 + 2ab) 16a2b2 ≤ 2(a2 + b2 )2 + 4ab(a2 + b2 ).
2 2 2 2 2

(1)
2 2 2 2

Now, a +b ≥ 2ab > 0, hence 2(a +b ) ≥ 8a b and 4ab(a +b ) ≥ 8a b so that (1) certainly holds. This completes the proof.

The quadrilateral ABCD is inscribed in a circle. The tangents to the circle at A and C intersect at a point P not on BD and such that P A2 = P B · P D . Prove that BD passes through the midpoint of AC . Solution by Michel Bataille, Rouen, France. Let B be the second point of Γ A intersection of P D with the circle Γ = m (ABCD ). Note that B = B (since D B P is not on BD ) and that P B · P D P is the power of P with respect to Γ. Q 2 Thus P B · P D = P A = P B · P D , B O so that P B = P B . If O is the centre of Γ, we also have OB = OB , hence the line OP is the perpendicular C S bisector of BB . It follows that OP is the bisector of the angle ∠BP D . Now, let m be the perpendicular to OP at P . The lines OP and m are the two bisectors of the lines BP, BD , hence (P D, P O, P B, m) is a harmonic pencil of lines and P O and m meet BD at points Q and S which are conjugate with respect to Γ. As a result, m is the polar of Q since it passes through S and is perpendicular to OQ. Finally, Q is on the polar AC of P and so is the common point of OP and AC , that is, the midpoint of AC . This completes the proof.

Triangle ABC is isosceles with AC = BC and P is a point inside the triangle such that ∠P AB = ∠P BC . If M is the midpoint of AB , prove that ∠AP M + ∠BP C = 180◦ .

216

Solution by Geoﬀrey A. Kandall, Hamden, CT, USA. Let θ = ∠P AB = ∠P BC , ϕ = ∠P AC = ∠P BA, ψ = ∠M P B , 2γ = ∠ACB . Then θ + ϕ + γ = 90◦ .
C 2γ P
γ

90 ψ ψ

◦

−

ϕ θ A M

ϕ θ ϕ θ

Rotate ACP about C counterclockwise through an angle 2γ to the position BCP and draw P P . We have CP = CP , P A = P B , ∠P BC = ϕ, and ∠CP P = 90◦ − γ . Extend P M past M its own length to a point N and draw BN . Then AM P ∼ = BM N , so N B = P A and ∠N BM = θ . It now follows that P BP ∼ = N BP , so ∠P P B = ψ . Consequently, ∠AP M + ∠BP C = [180◦ − (θ + ϕ) − ψ ] + [(90◦ − γ ) + ψ ] = 180◦ .

The interior angles of a convex octagon are all equal and all side lengths are rational numbers. Prove that the octagon has a centre of symmetry. Solution by Geoﬀrey A. Kandall, Hamden, CT, USA. Our notation is evident from the ﬁgure. The interior angle sum of the octagon ABCDEF GH is 1080◦ , hence each interior angle is 135◦ and each exterior angle 45◦ . Thus, the octagon can be enclosed in a rectangle, as shown. √ √ u+w t 2 = v + 2, Consequently, s + r+ 2 2 √ √ that is s + a 2 = v + b 2, where s, v , a, b are rational numbers.
H G u √ 2 P w F v E
w 2 r 2

√ 2 A r

t 2

√

t C D

√ v −s If a = b, then 2 a would be rational, which is not so. Thus, a = b, −b hence s = v . Therefore, ABEF is a parallelogram and its diagonals AE and BF bisect each other at P , which is the centre of symmetry of ABEF .

u 2

√

217

Similarly, this same point P is the centre of symmetry of parallelogram BCF G, and so on around the octagon. Thus, P is the centre of symmetry of the octagon.

Let M = {x2 + x | x is a positive integer}. For each integer k ≥ 2 prove that there exist a1 , a2 , . . . , ak , bk in M such that a1 + a2 + · · · + ak = bk . Solution by Geoﬀrey A. Kandall, Hamden, CT, USA. The proof is by induction on k. Note that x2 + x = x(x + 1), so each element of M is even. Since 12 + 30 = 42 and 12 = 3 · 4, 30 = 5 · 6, 42 = 6 · 7 ∈ M , the desired result holds for k = 2. Now the inductive step. Suppose that for some k ≥ 2 we have a1 + a2 + · · · + ak = bk , where a1 , a2 , . . . , ak , bk ∈ M . Since bk is even, we have bk = 2c for some positive integer c. Moreover, bk ≥ a1 + a2 ≥ 4, so c ≥ 2. Let ak+1 = (c − 1)c ∈ M . Then a1 + a2 + · · · + ak + ak+1 = 2c + (c − 1)c = c(c + 1) ∈ M, and the induction is complete.

Next we turn to solutions from our readers to problems of the Republic of Moldova Mathematical Olympiad Second and Third Team Selection Tests given at [2009 : 378–379].

Let a, b, c be the side lengths of a triangle and let s be the semiperimeter. Prove that a (s − b)(s − c) bc + b (s − c)(s − a) ac + c (s − a)(s − b) ab ≥ s.

Solution by Arkady Alt, San Jose, CA, USA. Let x := s − a, y := s − b, z := s − c then x, y, z > 0, a = y + z , b = z + x, c = x + y , s = x + y + z and the original inequality becomes
Ö

(y + z )
cyc

yz (x + y ) (z + x)

≥ x + y + z,

where x, y, z > 0. Since
Ö

(y + z )
cyc

yz (x + y ) (z + x)

=
cyc

(y + z )

yz (x + y ) (z + x)

(x + y ) (z + x)

218

and by Cauchy and AM-GM Inequalities (y + z ) yz (x + y ) (z + x) ≥ (y + z ) = = √ 2 yz ) √ √ (y + z ) yz (x + yz ) √ x (y + z ) yz + (y + z ) yz √ √ 2x yz yz + (x + y )yz 2xyz + (y + z ) yz yz (x +

≥ = then (y + z )
cyc

= yz ((x + y ) + (x + z ))

yz (x + y ) (z + x)

(x + y ) (z + x)

≥ =

yz ((x + y ) + (x + z ))
cyc

(x + y ) (z + x) yz yz + z+x x+y yz yz + z+x x+y cyc zx x+y +
cyc

cyc

=
cyc

yz x+y

=
cyc

zx + yz x+y

=
cyc

z (x + y ) x+y

= x + y + z.

The point P is in the interior of triangle ABC . The rays AP , BP , and CP cut the circumcircle of the triangle at the points A1 , B1 , and C1 , respectively. Prove that the sum of the areas of the triangles A1 BC , B1 AC , and C1 AB does not exceed s(R − r ), where s, R, and r are the semiperimeter, the circumradius, and the inradius of triangle ABC , respectively. Solution by Titu Zvonaru, Com´ ane¸ sti, Romania. We will prove the statement of the problem for the points A1 , B1 , C1 such that A1 belongs to arc BC which does not contain point A, and similarly for B1 and C1 . Let [XY Z ] be the area of XY Z . We denote a = BC , b = CA, c = AB . Let M be the midpoint of arc BC which contains the point A1 (which does not contain the point A). It is easy to see that [A1 BC ] ≤ [M BC ]. We have ∠M BC = ∠M CB = ∠M AC =
A . 2 A

(1)
B A1 M C

219

By the Law of Sines, we obtain BM = 2R sin [BM C ] = BM · BC · sin ∠M BC 2

A . It follows that 2

= aR sin2

A 2

(2)

By (1) and (2), it follows that [A1 BC ] + [B1 AC ] + [C1 AB ] B C A + bR sin2 + cR sin2 ≤ aR sin2 2 2 2 aR(1 − cos A) + bR(1 − cos B ) + cR(1 − cos C ) = 2 1 R = R(a + b + c) − (a cos A + b cos B + c cos C ) 2 2 R2 = sR − (2 sin A cos A + 2 sin B cos B + 2 sin C cos C ) 2 R2 = sR − (sin 2A + sin 2B + sin 2C ) 2 R2 R2 abc = sR − · 4 sin A sin B sin C = sR − ·4· 2 2 8R3 abc = sR − = sR − [ABC ] = sR − sr = s(R − r ) . 4R The equality holds if and only if AA1 , BB1 , CC1 are the bisectors of ABC .

Let a, b, and c be positive real numbers such that abc = 1. Prove that a+3 (a + 1)2 + b+3 (b + 1)2 + c+3 (c + 1)2 ≥ 3.

Solved by Arkady Alt, San Jose, CA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Zvonaru. We have 2(a + 3) (a + 1)2 −2 = = Denoting x = 2a + 6 − 2a 2 − 4a − 2 (1 + a)2 1 − 2a + a2 + 3(1 − a2 ) = (1 + a)2
1−a , y = 1+a 1−b , and z = 1+b

1−a 1+a

+3·

1−a . 1+a

1−c , it results that 1+c

x, y, z ∈ [−1, 1] and we have to prove that x2 + y 2 + z 2 + 3(x + y + z ) ≥ 0. (1)

220

Since abc = 1, we obtain x+y+z = (1 − a)(1 + b + c + bc) + (1 − b)(1 + a + c + ac) (1 + a)(1 + b)(1 + c) + = = a + b + c − ab − bc − ca (1 − c)(1 + a + b + ab) (1 + a)(1 + b)(1 + c)

(1 + a)(1 + b)(1 + c) (1 − a)(1 − b)(1 − c) = − = −xyz. (1 + a)(1 + b)(1 + c)

(1 + a)(1 + b)(1 + c) −1 + a + b(1 − a) + c(1 − a) − bc(1 − a)

Thus, the inequality (1) is equivalent to x2 + y 2 + z 2 ≥ 3xyz, which is true because by AM-GM Inequality and since x, y, z ∈ [−1, 1] we have x2 + y 2 + z 2 ≥ 3
3

x2 y 2 z 2 ≥ 3xyz.

Next we move to the Russian Mathematical Olympiad 2007, 11th Grade, given at [2010: 151–152].

(N. Agakhanov) The product f (x) = cos x cos 2x cos 3x . . . cos 2k x is written on the blackboard (k ≥ 10). Prove that it is possible to replace one “cos” by “sin” such that the product obtained f1 (x) satisﬁes the inequality |f1 (x)| ≤ 3 · 2−1−k for all real k. Solved by Oliver Geupel, Br¨ uhl, NRW, Germany. We prove that it suﬃces to replace the factor “cos 3x” by “sin 3x” whenever k ≥ 2. We start with the identity sin x · cos x cos 2x cos 4x · · · cos 2k x = sin 2k+1 x 2k+1 (k ≥ 0). (1)

We prove (1) by induction. It is obvious for k = 0. Assume that it holds for some ﬁxed integer k ≥ 0. Then sin x · cos x cos 2x cos 4x · · · cos 2k x cos 2k+1 x = = sin 2k+1 x 2k+1 sin 2k+2 x 2k+2 · cos 2k+1 x ,

221

which completes the induction and therefore the proof of (1). Moreover, we have that sin 3x = sin x cos 2x + cos x sin 2x = sin x(1 − 2 sin2 x) + 2 sin x(1 − sin2 x) = 3 sin x − 4 sin3 x. Putting this all together, we conclude that |f1 (x)| ≤ |3 sin x − 4 sin3 x| · |cos x cos 2x cos 4x · · · cos 2k x|
¬ ¬ ¬ sin 2k+1 x ¬ ¬ ¬ −1 −k . ≤ 3·¬ ¬≤3·2 ¬ 2k+1 ¬

= |3 − 4 sin2 x| · |sin x| · |cos x cos 2x cos 4x · · · cos 2k x|

(A. Polyansky) The incircle of a triangle ABC touches sides BC , AC , AB at points A1 , B1 , C1 , respectively. Segment AA1 intersects the incircle again at point Q. Line is parallel to BC and passes through A. Lines A1 C1 and A1 B1 intersect at points P and R, respectively. Prove that ∠P QR = ∠B1 QC1 . Solved by Geoﬀrey A. Kandall, Hamden, CT, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give Kandall’s presentation. Observe that 1 A1 C1 = ∠BA1 C1 = ∠BC1 A1 , 2 ∠BA1 C1 = ∠AP C1 , ∠BC1 A1 = ∠P C1 A ∠A1 QC1 =

so ∠AP C1 = ∠P C1 A. Therefore, ∠AP C1 is supplementary to ∠AQC1 , so quadrilateral P AQC1 is cyclic. Consequently, ∠P QA = ∠P C1 A = ∠A1 QC1 and similarly, ∠RQA = ∠A1 QB1 . Thus, by angle addition, ∠P QR = ∠B1 QC1 .

A Q C1

Next we look at solutions to problems of the XV Olymp´ ıada Matem´ atica Rioplatense, Nivel 2, given at [2010; 214] that we started last issue.

4. Let a1, a2 , . . . , an be positive numbers. The sum of all the products ai aj with i < j is equal to 1. Show that there is a number among them such that the √ sum of the remaining numbers is less than 2.

222

Solved by Oliver Geupel, Br¨ uhl, NRW, Germany; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give Zvonaru’s write-up. We prove by induction that
à
n− 1 n n n

ai aj =
i=1

ai
j =1

aj
j =i

(1)

i=1 j =i+1

For n = 2 we have 2a1 a2 = a1 a2 + a2 a1 , which is true. For n = 3 we have 2(a1 a2 + a1 a3 + a2 a3 ) = a1 (a2 + a3 ) + a2 (a1 + a3 ) + a3 (a1 + a2 ), which is also true. Suppose that (1) is true for some n = k, k > 1. We have to prove that
à
k k+1 k+1 k+1

2
i=1 j =i+1

ai aj =
i=1

ai
j =1

aj
j =i

(2)

We have
k k+1

2
i=1 j =i+1

ai aj

= 2

k−1

ai aj + 2
à
i=1 k k

ai ak+1
k

i=1 j =i+1 k

=
i=1

ai
j =1

aj
j =i k

+
i=1

ai ak+1 + ak+1
j =1

àà
k+1

=
i=1

ai
j =1

aj + ak+1
j =i k+1

+ ak+1
j =1

aj
j =k+1

à
k+1

=
i=1

ai
j =1

aj
j =i k+1

+ ak+1
j =1

aj
j =k+1

k+1

=
i=1

ai
j =1

aj
j =i

Denoting s = a1 + a2 + · · · + an and using (1) we obtain 2 = a1 (s − a1 ) + a2 (s − a2 ) + · · · + an (s − an ). √ If there is i such that s − ai < 2, we are done. (3)

223 √

Suppose that for i = 1, 2, . . . , n, s − ai ≥ 2 =

2. Using (3) we deduce that

a1 (s − a1 ) + a2 (s − a2 ) + · · · + an (s − an ) √ ≥ 2(a1 + a2 + · · · + an ), √ hence a1 √ + a2 + · · · + an ≤ 2. It follows that, for example, √ a1 + a2 + · · · + an−1 < 2, a contradiction with the inequality s − an ≥ 2. Next we move to solutions to problems of the XV Olymp´ ıada Matem´ atica Rioplatense 2006, Nivel 3, given at [2010; 215–216].

(a) For each k ≥ 3, ﬁnd a positive integer n that can be represented as the sum of exactly k mutually distinct positive divisors of n. (b) Suppose that n can be expressed as the sum of exactly k mutualy distinct positive divisors of n for some k ≥ 3. Let p be the smallest prime divisor of n. Show that 1 p + 1 p+1 + ··· + 1 p+k−1 ≥ 1.

Solved by Oliver Geupel, Br¨ uhl, NRW, Germany; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give Geupel’s write-up. (a) Consider n = 2k−2 · 3 with the divisors dj = 2k−2−j · 3 (1 ≤ j ≤ k − 2), dk−1 = 2, dk = 1. We have
k

dj = 3
j =1

k−2 j =1

2k−2−j + 3 = 3(2k−2 − 1) + 3 = 2k−2 · 3 = n.
k

(b) Let d1 > d2 > . . . > dk be divisors of the integer n such that
j =1

dj = n.

By d1 < n we have n d1 n d2 n n dk

n ≥ p. d1 is an increasing sequence of integers, we also have n d3 ≥ p + 2, . . . , n dk ≥ p + k − 1.

Since

, ...,

d2 Consequently,

≥ p + 1,

1 1 1 1 + + ··· + ≥ p p+1 p+k−1 n which completes the proof.

d j = 1,
j =1

224

2. Let ABCD be a convex quadrilateral such that AB = AD and CB = CD. The bisector of ∠BDC cuts BC at L, and AL cuts BD at M , and it is known that BL = BM . Determine the value of 2∠BAD + 3∠BCD .
Solved by Geoﬀrey A. Kandall, Hamden, CT, USA; and Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA. We give Kandall’s solution. A
1 θ 2

B
90 −

M
1 ϕ 2

D
1 ϕ 45 − 4

1 ϕ 2

C Let ∠BAD = θ and ∠BCD = ϕ (in degrees). We want to determine 2θ + 3ϕ.
1 The line AC bisects ∠BAD and ∠BCD ; ∠CBD = 90 − 2 ϕ, 1 1 1 1 1 ∠CDL = 2 (90 − 2 ϕ) = 45 − 4 ϕ. From BLM , ∠BLM = 2 (90 + 2 ϕ) = 1 ϕ; from DLC , ∠DLC = 180 − ϕ − (45 − 4 ϕ) = 135 − 3 ϕ. 45 + 1 4 4 3 1 1 Consequently, ∠ALD = 180 − (45 + 4 ϕ) − (135 − 4 ϕ) = 2 ϕ = ∠ACD .

Therefore, quadrilateral ALCD is cyclic, so ∠DLC = ∠DAC , that is, 3 1 135 − 4 ϕ= 2 θ . It follows easily that 2θ + 3ϕ = 540.

The acute triangle ABC with (AB = AC ) has circumcircle Γ, circumcentre O and orthocentre H . The midpoint of BC is M and the extension of the median AM intersects Γ at N . The circle of diameter AM intersects again Γ at A and P. Show that the lines AP , GC , and OH are concurrent if and only if AH = HN . Solved by Michel Bataille, Rouen, France; and Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA. We give Bataille’s solution. Let V denote the point diametrically opposite to A on Γ. Then, V C ⊥ CA, hence BH V C . Similarly CH V B and it follows that BHCV is a parallelogram with centre M . As a result, the line HM = HV meets Γ again at U such that AU ⊥ U M . Therefore U is also on the circle with diameter AM and U = P .

225

A U P O H J B M N V C Γ

Now, let AP intersect the line BC at J . The point H is also the orthocentre of ∆AJM (since lines AH and M P are altitudes). Thus, the line OH passes through J if and only if OH ⊥ AN . But OA = ON , hence OH is perpendicular to AN if and only if OH is the perpendicular bisector of the line segment AN that is, if and only if HA = HN . The result follows.

Next we turn to solutions from our readers to problems of the 21 Olimpiada Iberoamericana de Matematico, Guayoquil, given at [2010; 216–217]. In the scalene triangle ABC , with ∠BAC = 90◦ , the tangent line to the circumcircle at A intersects the line BC at M . Let S and R be the points where the incircle of ABC touches AC and AB respectively. The line RS intersects the line BC at N . The lines AM and SR meet at U . Show that triangle U M N is isosceles. Solved by George Apostolopoulos, Messolonghi, Greece; Michel Bataille, Rouen, France; Geoﬀrey A. Kandall, Hamden, CT, USA; Zelator; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give Bataille’s solution. A S R U N M B I O C

Let I denote the incentre of ∆ABC . The quadrilateral ARIS is a rectangle (because ∠ARI = ∠ASI = ∠RAI = 90◦ ), even a square since in addition IR = IS . It follows that ∠ARS = 45◦ .

226

Observing that γ = ∠ACB and ∠U AB subtend the same arc of the circumcircle, we have ∠N U M = ∠AU R = 180◦ − γ − 135◦ = 45◦ − γ. From ∠U M N = 180◦ − ∠AM B = ∠M AB + ∠M BA = γ + (180◦ − (90◦ − γ )) = 90◦ + 2γ we deduce ∠U N M = 180◦ − (45◦ − γ ) − (90◦ + 2γ ) = 45◦ − γ. (2) (1)

From (1) and (2), ∠N U M = ∠U M N and so ∆U M N is isosceles with MN = MU.

Let a1 , a2 , . . . , an be real numbers. Let Èd be the diﬀerence between the smallest and the largest of them, and let s = i<j |ai − aj |. Show that 4 and determine the conditions under which equality holds in each equality. Solved by Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give Zvonaru’s version. Without loss of generality we may assume that a1 ≥ a 2 ≥ · · · ≥ an , and È we denote d(a1 , . . . , an ) = a1 − an and s(a1 , . . . , an ) = i<j |ai − aj |. For n = 2, we have s = d and both inequalities are equalities. For n = 3, we have s = 2d; the left inequality is equality and the right inequality is true. Assume that n ≥ 4. We have s(a1 , . . . , an ) =
n− 1 n

(n − 1)d ≤ s ≤

n2 d

i=1 j =i+1

(ai − aj )
n− 2 n− 1

= a 1 − a 2 + a 2 − a n + · · · + a 1 − a n− 1 + a n− 1 − a n +a1 − an +
i=2 j =i+1

(ai − aj )

= (n − 1)d(a1 , . . . an ) + s(a2 , . . . , an−1 ). Since s(a2 , . . . ak−1 ) ≥ 0, it results that the left inequality is true. The equality holds if and only if n = 2 or n = 3 or n ≥ 4 and a2 = · · · = an−1 (that is, s(a2 , . . . , an−1 ) = 0). For the right inequality we proceed by induction. If n = 2, 3, the inequality is true. Suppose that the inequality is true for any K ≤ n and we have to prove that (n + 1)2 s(a1 , . . . , an+1 ) ≤ d(a1 , . . . , an+1 ). 4

227

Since it is obvious that d(a2 , . . . , an ) ≤ d(a1 , . . . , an+1 ), we obtain s(a1 , . . . , an+1 ) = nd(a1 , . . . , an+1 ) + s(a2 , . . . , an ) (n − 1)2 ≤ nd(a1 , . . . , an+1 ) + d(a2 , . . . , an ) 4 (n − 1)2 d(a1 , . . . , an+1 ) ≤ nd(a1 , . . . , an+1 ) + 4 4n + (n − 1)2 = d(a1 , . . . , ak+1 ) 4 (n + 1)2 = d(a1 , . . . , an+1 ) n and the induction is complete. The equality holds if and only if n = 2 or a1 = a2 = · · · = an .

The numbers 1, 2, 3, . . . , n2 are placed in the cells of an n × n board, one number per cell. A coin is initially placed in the cell containing the number n2 . The coin can move to any of the cells which share a side with the cell it currently occupies. First, the coin travels from the cell containing the number 1 to the cell containing the number n2 , using the smallest possible number of moves. Then the coin travels from the cell containing the number 1 to the cell containing the number 2 using the smallest number of moves, and then from the cell containing the number 3, and continuing until the coin returns to the initial cell, taking a shortest route each time it travels. The complete trip takes N steps. Determine the smallest and largest possible values of N . Solved by Oliver Geupel, Br¨ uhl, NRW, Germany.

According to http://www.imomath.com/othercomp/Ib/IbMO06.pdf the trip should be n2 → 1 → 2 → . . . → n2 − 1 → n2 . We will show that min N is n2 if n is even and n2 + 1 if n is odd. Moreover, we prove that max N is n3 − 2 for even n and n3 − n for odd n. Since we must enter each of the n2 cells, we have min N ≥ n2 . By colouring the cells alternately black and white like a chess board, we see that we can return to a cell of the colour of the initial cell only after an even number of steps; hence min N ≥ n2 + 1 if n is odd. Examples of trips that reach these bounds are given in Figures 1 and 2, for even and odd n, respectively. In each step one edge is passed. Thus, the total number of steps is equal to the number of passed edges, counted with the multiplicity of crossings. Consider a k × n rectangle consisting of the k leftmost or rightmost columns or of the k uppermost or lowermost rows of the board, where 1 ≤ k ≤ n/2. The interior edge of this rectangle can be passed not more than 2kn times, speciﬁcally, twice for each cell, that is on entering and on leaving the rectangle.

228

Figure 1 If n is odd, then
(n−1)/2 k=1

Figure 2

N ≤4

2kn = (n − 1)n(n + 1) = n3 − n.

If n is even, the board consists of four quadrants, where we can alternate between the upper-left and the lower-right quadrant as well as between the upperright and the lower-left quadrant. We must at least twice change the pair of quadrants in order to return to the inital cell thus losing two crossings. Therefore,
(n−2)/2 k=1

N ≤4

2kn + 2n2 − 2 = n3 − 2.

Examples of arrangements that reach these bounds are given in Figures 3 and 4, for n = 2m and n = 2m + 1, respectively.

odd numbers odd numbers
1,...,2m2 −1

odd numbers
2m2 +2m+1,..., 4m2 +4m−1

odd numbers
2m2 +1,...,4m2 −1

1,..., 2m +2m−1
2

even numbers
2m2 +2,...,4m2

even numbers even numbers
2,...,2m2 2m2 +2m+2,..., 4m2 +4m

even numbers
2,..., 2m2 +2m

Figure 3

Figure 4

229

Determine all pairs (a, b) of positive integers such that 2a + 1 and 2b − 1 are relatively prime and a + b divides 4ab + 1. Solved by David E. Manes, SUNY at Oneonta, Oneonta, NY, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give Zelator’s write-up. Suppose that a and b are positive integers such that a + b | 4ab + 1 and gcd(2a + 1, 2b − 1) = 1. Then 4ab + 1 = k · (a + b) , (1) for some positive integer k. From (2a + 1)(2b + 1) = 4ab + 2(a + b) + 1 and (1), we obtain (2a + 1)(2b + 1) = (k + 2) · (a + b) . (2) Observe that a + b is relatively prime to 2a + 1. Indeed, if d is the greatest common divisor of a + b and 2a + 1, then d must be a divisor of any linear combination (with integer coeﬃcients) of a + b and 2a + 1. In particular, d must divide 2(a + b) − (2a + 1) = 2b − 1. Thus d | 2a + 1 and d | 2b − 1 and so, by the coprimeness condition in (1), it follows that d = 1. We have shown that, gcd(a + b, 2a + 1) = 1. (3)

Euclid’s lemma in number theory postulates that if an integer divides the product of two other integers, and it is relatively prime to one of those (two) integers, then it must divide the other one. Clearly then, by Euclid’s lemma, (3), and (2), it follows that a + b must divide 2b + 1. Thus there exists a positive integer, m, such that 2b + 1 = m · (a + b) . We rewrite (4) in the form, m · a + (m − 2) · b = 1 . (5) (4)

Clearly, if m ≥ 2, the lefthand side of (5) is greater than 1, since a and b are positive integers. Therefore m = 1, which yields, a − b = 1 and a = b + 1. We have proven that if two positive integers a and b satisfy conditions (1), then a = b + 1. The converse is also true. Indeed if a = b + 1, then 4ab + 1 = 4(b + 1)b + 1 = (2b + 1)2 = (a + b)2 , which shows that a + b divides 4ab + 1. Moreover, we have 2a + 1 = 2(b + 1) + 1 = 2b + 3, which is relatively prime to 2b − 1, since if D = gcd(2b − 1, 2b + 3). Then D | [(2b + 3) − (2b − 1)] = 4. But D is odd, since 2b − 1 and 2b + 3 are. Thus D = 1. Conclusion: The pairs (a, b) = (b + 1, b), where b can be any positive integer, are the solution to this problem.

230

5. The circle Γ is inscribed in quadrilateral ABCD with AD and CD tangent to Γ, at P and Q, respectively. If BD intersects Γ at X and Y and M is the midpoint of XY , prove that ∠AM P = ∠CM Q.
Solved by Oliver Geupel, Br¨ uhl, NRW, Germany.
B G Q R M T O R Q A P D F E C C

We argue in terms of projective geometry, assuming that parallel lines meet at a point of inﬁnity. Let O be the midpoint of Γ, let E be the pole of BD with respect to Γ, and let BC and AB be tangent to Γ at points R and S , respectively. We assume without loss of generality that the points C and E are on the same side of the line BD . Lemma 1. The lines AC , P Q, and RS meet at E . Moreover, BD ⊥ EM , and the points E , M , and O are collinear. Proof. Let P Q meet AC at E1 , and let RS and AC meet at E2 . By P D = P Q and Menelaus’ Theorem, for ACD and the line P Q it holds E1 A E1 C Analogously, E2 A E2 C · RC SA = 1. · QC PA = E1 A E1 C · QC QD · PD PA =1

Since QC = RC and P A = SA, we obtain E1 A E1 C = E2 A E2 C

and consequently E1 = E2 . The point E1 lies on the polar P Q of point D . Hence, D is on the polar of E1 . Similarly, B is also on the polar of E1 . Thus, BD is the polar of E1 , that is, E1 coincides with the pole E of BD , and BD ⊥ EM . Thus, E , M , and O are collinear. 2 Let P , Q , R , and S be the reﬂections of P , Q, R, and S in the line EM .

231

Lemma 2. The lines P Q , P Q, RS , and R S meet at M . Proof. Let U and V be the intersection of BD and P Q and the intersection of BD and P Q , respectively. Let the line U V meet P Q at point W . By Menelaus’ Theorem, for EU V and the line P Q it holds PE PU · WU WV · QV QE = 1;

hence, by Q V = QU and Q E = QE , PE PU : QE QU = WV WU .

On the other hand, since E and U are harmonic conjugates with respect to the points P and Q, we have P E QE : = 1. P U QU Consequently, W = M , that is, M lies on P Q . Analogously, M lies on P Q, RS , and R S . 2 We are now prepared to prove that ∠AM P = ∠CM Q. Since EM bisects ∠P M P , the orthogonal line BD bisects the complementary angle P M Q, that is, ∠DM P = ∠DM Q. (1)

Let C be the reﬂection of C in the line EM . Since the tangents of Γ in Q and R meet at C , we see that the tangents in Q and R meet at C . Let the tangents to Γ in P and R meet at point F , and let the tangents in S and Q meet at point G. Consider the circumscribed hexagon AP F C Q G. By Brianchon’s Thereom, the lines AC , P Q , and F G are concurrent. Next, consider the circumscribed hexagon F R C GSA. By Brianchon’s Theorem, the lines AC , R S , and F G are concurrent. Therefore, the four lines AC , F G, P Q , and R S are concurrent. By Lemma 2, their intersection is the point M . We deduce that the points A, C , and M are collinear. Let T be the intersection of AB and EM . Then, ∠AM T = ∠C M E = ∠CM E . (2) By (1) and (2), we obtain ∠AM P = 90◦ − ∠DM P − ∠AM T = 90◦ − ∠DM Q − ∠CM E = ∠CM Q, which completes the proof.

That completes the material for this number of the Corner.

232

BOOK REVIEWS
Amar Sodhi
Icons of Mathematics: An Exploration of Twenty Key Images by Claudi Alsina & Roger B. Nelsen Dolciani Mathematical Exposition #45 The Mathematical Association of America, 2011 ISBN: 978-0-88385-352-8 (print), 978-0-88385-986-5 (electronic) Hard cover, 327+xvii pages, US$69.95 Reviewed by Edward J. Barbeau, University of Toronto, Toronto, ON The author of a geometry book directed at a general audience has a daunting task. For about 2500 years, amateur and professional mathematicians in Europe and Asia have uncovered an abunadance of fascinating results about simple geometric ﬁgures, particularly circles and triangles, and the end is apparently not in sight. Many diﬀerent techniques have been developed to establish them, each shining its own particular light and allowing for diﬀerent insights and connections to be made. How can one choose among such a wealth of facts and arguments and arrive at a book that is readable, representative, focussed and compact? The authors of the book under review use twenty diagrams, or “icons”, as an organizing principle. Many of them have a particular historical or cultural signiﬁcance, such as the Yin-Yang symbol (a circle bisected by two semi-circles of half the diameter), the “windmill” diagram used by Euclid in his proof of Pythagoras’ theorem (I:47) (a right triangle with squares constructed outwards on the sides), the three circles of a standard Venn diagram, and a trapezoid ﬁgure used by President Garﬁeld in his proof of Pythagoras’ theorem. Others are just representative ﬁgures to introduce a theme. The prerequisites for this book are modest. The reader should have a secondary background in Euclidean geometry and trigonometry, along with an active imagination. Most of the results are proved in an informal way; the chief tools are standard Euclidean arguments, algebraic manipulation and proofs without words that involve dissections and shifting ﬁgures around. A small number of propositions are mentioned without proof, and some are listed as challenges at the end of each chapter, with solutions provided in an appendix. Each chapter is lightened by digressions on mathematical personalities and artefacts, history and background, and occurrences of geometrical ﬁgures in ordinary life. For example, on page 203, we ﬁnd photographs of star-shaped badges with 5, 6, 7 and 8 points worn by law enforcement oﬃcers along with speculation about their origin. Two pages later, we learn about the role of star polygons in the design of columns for a modern church in Barcelona by Antoni Gaudi (1852-1926). This book need not be read straight through. Since the twenty chapters are generally self-contained, readers can forage at random and follow something that catches their fancy. CRUX with MAYHEM readers will ﬁnd a lot of familiar material along with results and arguments that will be new to them. The scope of

233

the book can be suggested by a list of some topics that make an appearance. Apart from standard material on triangles, circles and inequalities, the authors touch on Dido’s isoperimetric problem, Euclid’s construction of the regular solids, reptiles, cevians, the butterﬂy theorem, conic sections, Reuleux polygons, star polygons, self-similarity, spirals, the Monge sponge and Sierpinski carpet and tilings. The treatment is light but satisfying, and readers wanting more are directed to other resources. There are some themes that recur throughout the book, such as tilings and dissections, inscribed and escribed ﬁgures and cevians. The most prominent of these is the Pythagorean theorem, which is treated in many places. In the opening chapter, Euclid’s diagram is generalized to the Vecten conﬁguration (squares escribed on an arbitrary triangle) which leads to an insightful proof of the Cosine Law and the solution to two problems by the American puzzler, Sam Loyd (1841-1911). The second chapter uses the icon of one square inscribed inside another for an ancient Chinese proof of the Pythagorean theorem and continues to get a diagrammatic proof of standard inequalities. The trapezoidal diagram used by U.S. President Garﬁeld (1831-1881) to prove the theorem is pressed into service for some trigonometric equations. The chapter on similar ﬁgures provides the occasion to prove the Pythagorean as well as the Reciprocal Pythagorean theorem (that, if a2 + b2 = c2 , then (1/a)2 + (1/b)2 = (1/h)2 where h is the altitude to the hypotenuse). The right triangle has a chapter of its own, in which Pythagorean triples are characterized and the Pythagorean relation is used to derive some inequalities. Another characterization of Pythagorean triples is found using two overlapping squares of areas a2 and b2 inside a square of area c2 = a2 + b2 . An arrangement of tatami rectangular mats in a square is behind a proof of the Pythagorean theorem credited to Bhaskara (1114-1185). In the ﬁnal chapter, inﬁnitely many demonstrations of the theorem can be had by laying a hypotenuse grid atop a plane tiling of two diﬀerent squares. This is an enjoyable and useful book that provides a lot of material that could be presented to secondary students. The challenges at the end of the chapter are generally accessible and often require some insight to obtain an elegant solution. I found very little to quibble with. The approach to the proof of the result on page 133 (the locus of the centres of circles touching an ellipse and passing through a point inside the ellipse) seems backwards when a direct assault is just as easy. In Challenge 3.8, the diagram needs to be justiﬁed by the result that if ABCD is a trapezoid with AB and DC both perpendicular to BC , E is the midpoint of BC and ∠AED = 90◦ , then ED bisects angle ADC . This book suggests how geometry might be rehabilitated in the school curriculum. Now that geometry is no longer seen as the centrepiece of the syllabus and a prerequisite for later mathematics, it can be taught as an important milestone in human intellectual history, a celebration of human ingenuity, a more natural approach to proof and a source of personal pleasure. This book is a ﬁne vehicle to carry out such a course.

234

PROBLEMS
Toutes solutions aux probl` emes dans ce num´ ero doivent nous parvenir au plus tard le 15 mars 2012. Une ´ etoile ( ) apr` es le num´ ero indique que le probl` eme a ´ et´ e soumis sans solution. Chaque probl` eme sera publi´ e dans les deux langues oﬃcielles du Canada (anglais et fran¸ cais). Dans les num´ eros 1, 3, 5 et 7, l’anglais pr´ ec´ edera le fran¸ cais, et dans les num´ eros 2, 4, 6 et 8, le fran¸ cais pr´ ec´ edera l’anglais. Dans la section des solutions, le probl` eme sera publi´ e dans la langue de la principale solution pr´ esent´ ee. La r´ edaction souhaite remercier Jean-Marc Terrier, de l’Universit´ e de Montr´ eal, d’avoir traduit les probl` emes.

3638.

Propos´ e par Michel Bataille, Rouen, France.

Etant donn´ e un triangle ABC , on arrange respectivement les points D, E, F sur les droites BC ,CA,AB , de sorte que BD : DC = λ : 1 − λ, CE : EA = µ : 1 − µ, AF : F B = ν : 1 − ν.

Montrer que DEF est le triangle p´ edal du triangle ABC si et seulement si (2λ − 1)BC 2 + (2µ − 1)CA2 + (2ν − 1)AB 2 = 0.

3639.

Propos´ e par Pham Kim Hung, ´ etudiant, Universit´ e de Stanford, Palo ´ . Alto, CA, E-U Soit a, b et c trois nombres r´ eels non n´ egatifs tels que a + b + c = 3. Montrer que b2 c c2 a a2 b + + ≤ 1. a+b+1 b+c+1 c+a+1

3640.

Propos´ e par Roy Barbara, Universit´ e Libanaise, Fanar, Liban. √ 3 On consid` ere la fonction f (x) = − 4x6 + 6x3 + 3.

(a) Trouver les points ﬁxes de f (x), s’il y en a. (b) Trouver les points p´ eriodiques de p´ eriode 2 de f (x), s’il y en a. (c) Montrer que x = −1 est l’unique nombre r´ eel tel que x et f (x) sont tous deux des entiers.

3641.

Propos´ e par Jos´ e Luis D´ ıaz-Barrero, Universit´ e Polytechnique de Catalogne, Barcelone, Espagne. Soit 0 ≤ x1 , x2 , . . ., xn < π/2 n nombres r´ eels. Montrer que 1 n
n 1/2

sec(xk )
k=1

1−

1 n

sin(xk )
k=1

≥ 1.

235

3642.

Propos´ e par Michel Bataille, Rouen, France.

Evaluer
2 n 0 (2x − 5x − 1) dx . lim Ê 1 2 n n→∞ 0 (x − 4x − 1) dx

Ê1

3643.
Vietnam.

Propos´ e par Pham Van Thuan, Universit´ e de Science des Hano¨ ı, Hano¨ ı,

Soit u et v deux nombres r´ eels positifs. Montrer que 1 8 17 − 2uv u2 + v2 ≤
3

u v

v u

≤

(u + v )

1 u

1 v

Pour chaque in´ egalit´ e, d´ eterminer quand il y a ´ egalit´ e.

3644.

Propos´ e par George Apostolopoulos, Messolonghi, Gr` ece.

On op` ere une trisection des cˆ ot´ es AB et AC du triangle ABC avec les points D , E et F , G respectivement, de telle sorte que AE = ED = DB et AF = F G = GC . La droite BF coupe respectivement CD , CE aux points K , L, tandis que BG coupe CD , CE en N , M respectivement. Montrer que : (a) KM est parall` ele a ` BC ; (b) Aire(KLM ) =
5 Aire(KLM N ). 7

3645.

Propos´ e par Jos´ e Luis D´ ıaz-Barrero et Juan Jos´ e Egozcue, Universit´ e Polytechnique de Catalogne, Barcelone, Espagne. Soit a, b et c trois nombres positifs tels que a2 + b2 + c2 + 2abc = 1. Montrer que a
cyclique

1 b

−b

1 c

−c

> 2.

3646.

Propos´ e par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie.

Soit α ≥ 0 et soit β un nombre positif. Trouver la limite
n

L(α, β ) =

n→∞

lim

1+
k=1

kα nβ

−n

236

3647.

´ Propos´ e par Panagiote Ligouras, Ecole Secondaire L´ eonard de Vinci, Noci, Italie. Montrer que dans un triangle ABC dont les rayons des cercles exinscrits sont ra , rb et rc , on a (ra + rb )(rb + rc )
cyclique

≥ 9,

o` u AB = c, BC = a et CA = b.

3648.
S (S −4) 4

Propos´ e par Michel Bataille, Rouen, France.

Trouver tous les nombres r´ eels x, y, z tels que xyz = 1 et x3 + y 3 + z 3 = y y x z z o` uS= y +x+z +y +x +x . z Propos´ e par Pham Van Thuan, Universit´ e de Science des Hano¨ ı, Hano¨ ı,

3649.
Vietnam.

Soit a, b et c trois nombres r´ eels positifs et soit k = (a + b + c) Montrer que (a3 + b3 + c3 ) 1 a3 + 1 b3 + 1 c3 ≥ k3 − 15k2 + 63k − 45 4
k−5±

1 1 1 + + a b c

et que l’´ egalit´ e a lieu si et seulement si (a, b, c) = ou une quelconque de ses permutations.

√ 2 k − 10k + 9 , 1, 1 4

3650.

Propos´ e par Mehmet Sahin, Ankara, Turquie.

Soit ABC un triangle acutangle avec A ∈ BC , B ∈ CA et C ∈ AB arrang´ es de telle sorte que ∠ACC = ∠CBB = ∠BAA = 90◦ . Montrer que (a) |BC ||CA ||AB | = abc; (b) AA BB CC BC CA AB (c) A(ABC ) A(A B C ) =1+ 4R2 (2R + r )2 − s2 . = tan(A) tan(B ) tan(C );

237

3638.

Proposed by Michel Bataille, Rouen, France.

Let ABC be a triangle and let points D, E, F lie on lines BC , CA, AB , respectively, such that BD : DC = λ : 1 − λ, CE : EA = µ : 1 − µ, AF : F B = ν : 1 − ν.

Show that DEF is a pedal triangle with regard to ∆ABC if and only if (2λ − 1)BC 2 + (2µ − 1)CA2 + (2ν − 1)AB 2 = 0.

3639.
CA, USA.

Proposed by Hung Pham Kim, student, Stanford University, Palo Alto,

Let a, b, and c be nonnegative real numbers such that a + b + c = 3. Prove that b2 c c2 a a2 b + + ≤ 1. a+b+1 b+c+1 c+a+1

3640.

Proposed by Roy Barbara, Lebanese University, Fanar, Lebanon. √ Consider the function f (x) = − 3 4x6 + 6x3 + 3.

(a) Find the ﬁxed points of f (x), if any. (b) Find the periodic points with period 2 of f (x), if any. (c) Prove that x = −1 is the unique real number such that x and f (x) are both integers.

3641.

Proposed by Jos´ e Luis D´ ıaz-Barrero, Universitat Polit` ecnica de Catalunya, Barcelona, Spain. Let 0 ≤ x1 , x2 , . . . , xn < π/2 be real numbers. Prove that 1 n
n

sec(xk )
k=1

1−

1 n

1/2

sin(xk )
k=1

≥ 1.

3642.

Proposed by Michel Bataille, Rouen, France.
Ê1

Evaluate
2 n 0 (2x − 5x − 1) dx lim Ê . 1 2 n n→∞ 0 (x − 4x − 1) dx

238

3643.
Vietnam.

Proposed by Pham Van Thuan, Hanoi University of Science, Hanoi,

Let u and v be positive real numbers. Prove that 1 8 17 − 2uv u2 + v2 ≤
3

u v

v u

≤

(u + v )

1 u

1 v

For each inequality, determine when equality holds.

3644.

Proposed by George Apostolopoulos, Messolonghi, Greece.

We trisect the sides AB and AC of triangle ABC with the points D , E and F , G respectively such that AE = ED = DB and AF = F G = GC . The line BF intersects CD , CE in the points K , L respectively, while BG intersects CD , CE in N , M respectively. Prove that: (a) KM is parallel to BC ; (b) Area(KLM ) =
5 Area(KLM N ). 7

3645.

Proposed by Jos´ e Luis D´ ıaz-Barrero and Juan Jos´ e Egozcue, Universitat Polit` ecnica de Catalunya, Barcelona, Spain. Let a, b, and c be positive numbers such that a2 + b2 + c2 + 2abc = 1. Prove that 1 1 −b − c > 2. a b c cyclic

3646.

Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.

Let α ≥ 0 and let β be a positive number. Find the limit
n

L(α, β ) =

n→∞

lim

1+
k=1

kα nβ

−n

3647.
Italy.

Proposed by Panagiote Ligouras, Leonardo da Vinci High School, Noci,

Show that in triangle ABC with exradii ra , rb and rc , (ra + rb )(rb + rc )
cyclic

≥ 9,

where AB = c, BC = a, and CA = b.

239

3648.

Proposed by Michel Bataille, Rouen, France.
S (S −4) 4

Find all real numbers x, y, z such that xyz = 1 and x3 + y 3 + z 3 = y z z where S = x +x +y +y +x +x . y z z

3649.
Vietnam.

Proposed by Pham Van Thuan, Hanoi University of Science, Hanoi,

Let a, b, and c be three positive real numbers and let k = (a + b + c) Prove that (a3 + b3 + c3 ) 1 a3 + 1 b3 + 1 c3 ≥ k3 − 15k2 + 63k − 45 4
k−5±

1 a

1 b

1 c

, or

and equality holds if and only if (a, b, c) = any of its permutations.

√ 2 k − 10k + 9 , 1, 1 4

3650.

Proposed by Mehmet Sahin, Ankara, Turkey.

Let ABC be an acute-angled triangle with A ∈ BC , B ∈ CA, and C ∈ AB arranged so that ∠ACC = ∠CBB = ∠BAA = 90◦ . Prove that (a) |BC ||CA ||AB | = abc; (b) AA BB CC BC CA AB (c) A(ABC ) A(A B C ) =1+ 4R2 (2R + r )2 − s2 . = tan(A) tan(B ) tan(C );

240

SOLUTIONS
No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems.

3526.

[2010 : 109, 111] Proposed by Cao Minh Quang, Nguyen Binh Khiem High School, Vinh Long, Vietnam. Let a, b, and c be positive real numbers. Prove that a
cyclic

a2 + 2(b + c)2

≥ 1.

Solution by Oliver Geupel, Br¨ uhl, NRW, Germany. By H¨ older inequality we have
2

a
cyclic

+ 2(b +

c)2

a(a2 + 2(b + c)2 )
cyclic

≥ (a + b + c)3 .

Thus we only need to show that (a + b + c)3 ≥ or equivalently that a2 b + a2 c + ab2 + ac2 + b2 c + bc2 ≥ 6abc . But this is immediate from the AM-GM inequality. This completes the proof. Equality holds if and only if a = b = c.
Also solved by ARKADY ALT, San Jose, CA, USA; GEORGE APOSTOLOPOULOS, ˇ ´ University of Sarajevo, Messolonghi, Greece (2 solutions); SEFKET ARSLANAGI C, Sarajevo, Bosnia and Herzegovina; MICHEL BATAILLE, Rouen, France; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; PEDRO HENRIQUE O. PANTOJA, student, UFRN, Brazil; PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; ALBERT STADLER, Herrliberg, Switzerland; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer.

a(a2 + 2(b + c)2 ) ,
cyclic

241

3539.

[2010 : 239, 241] Proposed by Jos´ e Luis D´ ıaz-Barrero, Universitat Polit` ecnica de Catalunya, Barcelona, Spain and Pantelimon George Popescu, Bucharest, Romania. Let A and B be 2 × 2 square matrices with real entries. Prove that the equations det(xA ± B ) = 0 have all of their roots real if and only if [trace(AB ) − trace(A)trace(B )]2 ≥ 4 det(A) det(B ) . Solution by the Henry Ricardo, Tappan, NY, USA. The result å is false èwithout further å conditions è on the matrix A. For example, 1 0 1 1 and B = , we ﬁnd that det(xA ± B ) = letting A = 0 0 −1 0 å è x ± 1 ±1 = 1, so that det(xA ± B ) = 0 has no roots despite the det ∓1 0 fact that [trace(AB ) − trace(A)trace(B )]2 = 0 = 4 det(A) det(B ).

If A is a 2 × 2 matrix with real entries, we can use the easily proved result (see Fact 4.9.3 in Matrix Mathematics (Second Edition) by Dennis S. Bernstein, Princeton University Press, 2009) that det(A + B ) − det(A) − det(B ) = trace(A)trace(B ) − trace(AB ) .

Thus, using basic properties of the determinant and trace, det(xA ± B ) = det(A)x2 ± (trace(A)trace(B ) − trace(AB ))x + det(B ) . Now if A is nonsingular, the equations det(xA ± B ) = 0 have all of their roots real if and only if [trace(AB ) − trace(A)trace(B )]2 ≥ 4 det(A) det(B ).

ˇ Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; SEFKET ´ University of Sarajevo, Sarajevo, Bosnia and Herzegovina; DIONNE ARSLANAGI C, BAILEY, ELSIE CAMPBELL, and CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; MURIEL BAKER, student, Southeast Missouri State University, Cape Girardeau, Missouri, USA; ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, France; BRIAN D. BEASLEY, Presbyterian College, Clinton, SC, USA; PAUL BRACKEN, University of Texas, Edinburg, TX, USA; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; HTET NAING LIN, Southeast Missouri State University, Cape Girardeau, Missouri, USA; ´ student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; DAVID SALEM MALIKIC, E. MANES, SUNY at Oneonta, Oneonta, NY, USA; JAMES MEYER, student, Southeast Missouri State University, Cape Girardeau, Missouri, USA; CRISTINEL MORTICI, Valahia University of Tˆ argovi¸ ste, Romania; JOHN POSTL, St. Bonaventure University, St. Bonaventure, NY, USA; HANNAH PREST, student, Southeast Missouri State University, Cape Girardeau, Missouri, USA; JAMES REID, student, Angelo State University, San Angelo, TX, USA; DIGBY SMITH, Mount Royal University, Calgary, AB; CULLAN SPRINGSTEAD, Southeast Missouri State University, Cape Girardeau, Missouri, USA; ALBERT STADLER, Herrliberg, Switzerland; ELIZABETH WAMSER, student, Southeast Missouri State University, Cape Girardeau, Missouri, USA; HAOHAO WANG and JERZY WOJDYLO, Southeast Missouri State University, Cape Girardeau, Missouri, USA; BRENT WESSEL, student, Southeast Missouri State University, Cape Girardeau, Missouri, USA; DANIEL WINGER, student, St. Bonaventure University, Allegany, NY, USA; PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU, Com´ ane¸ sti, Romania; and the proposer. As the matrices were all 2 × 2 almost all the other solutions were by direct computation. The proposer was the only other solver to use properties of determinants and the trace. The featured solution was the only solution to note the condition that A needed to be nonsingular.

242

3540.

[2010 : 239, 241] Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.

Triangle ABC has semiperimeter s and area F . A square P QRS with side length x is inscribed in ABC with P and Q on BC , R on AC , and S on AB . Similarly y and z are the sides of squares two vertices of which lie on AC and AB , respectively. Prove that √ s(2 + 3) −1 −1 −1 . x +y +z ≤ 2F Combination of solutions by John G. Heuver, Grande Prairie, AB and the proposer. As usual, the side lengths of ∆ABC will be denoted by a, b, and c, and the altitudes by ha , hb , and hc . By the similarity of triangles RAS and CAB we have
h −x x = a , so that a ha

x= Similarly,

aha 2F a + ha = , or x−1 = . a + ha a + ha 2F and z −1 = c + hc . 2F

b + hb 2F This allows us to deduce that y −1 =

x−1 + y −1 + z −1 =

2s + ha + hb + hc 2F

The desired conclusion follows from the familiar inequality √ ha + hb + hc ≤ s 3; see, for example, [1] page 60, formula 6.1 or 6.2. Equality holds if and only if ∆ABC is equilateral.
Also solved by ARKADY ALT, San Jose, CA, USA; GEORGE APOSTOLOPOULOS, ˇ ´ University of Sarajevo, Sarajevo, Bosnia Messolonghi, Greece; SEFKET ARSLANAGI C, and Herzegovina; ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, France; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; RICHARD EDEN, student, Purdue University, West Lafayette, IN, USA; OLEH ´ FAYNSHTEYN, Leipzig, Germany; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; VACLAV ˇ Y, ´ Big Rapids, MI, USA; ALBERT STADLER, Herrliberg, Switzerland; PETER KONECN Y. WOO, Biola University, La Mirada, CA, USA; and TITU ZVONARU, Com´ ane¸ sti, Romania. Heuver used the Cauchy-Schwarz to deduce that ha + hb + hc ≤ Ô Inequality √ Ô 2 2 2 2 2 3 h2 h2 a + hb + hc , and then applied a + hb + hc ≤ s , which is (9.8) on page 201 of [2]. Zvonaru ﬁnished his solution with [1].
1 a

1 b

1 c

√

≤

3 , 2r

which is item 5.22 on page 54 of

References [1] O. Bottema et al., Geometric Inequalities, Groningen, 1969 [2] D.S. Mitrinovi´ c et al., Recent Advances in Geometric Inequalities, Kluwer Academic Publishers, 1989

243

3541.

[2010 : 240, 242] Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.

Triangle ABC has circumcentre O , circumradius R, orthocentre H , side lengths a, b, c, and altitudes AD , BE , CF , where points D , E , F lie on the sides BC , AC , AB , respectively. The Euler line of triangle ABC intersects BC in P and AC in Q, and the quadrilateral ABP Q is cyclic. Show that a2 + b2 = 6R2 , and express the length of P Q in terms of a, b, c. Solution by the proposer. Because ABP Q is cyclic, ∠CP Q = ∠BAQ = ∠BAC . Since DE joins the feet of two altitudes, we also have ∠BAC = ∠CDE , whence P Q||DE . OH is (by assumption) the same line as P Q. Because P lies on BC , we have ∠ACH = ∠OCP so that ACF and P CO are similar right triangles; that is, CO ⊥ P Q. In the right triangle HOC , CH 2 = OH 2 + OC 2 = OH 2 + R2 . (1)

Formula 5.8(1) on page 50 of O. Bottema et al., Geometric Inequalities, Groningen, 1969 says that OH 2 = 9R2 − (a2 + b2 + c2 ). (2) Moreover, in any triangle CH = 2R cos C (see, for example Roger A. Johnson Advanced Euclidean Geometrey, Paragraph 252(e), page 163), so that equations (1) and (2) gives us 9R2 − (a2 + b2 + c2 ) + R2 = 4R2 cos2 C. Using the sine law, we replace c by 2R sin C in the last equation to get a2 + b2 = 6R2 , as desired. To determine the length of P Q we ﬁrst recall that ∠CP Q = ∠BAC , which implies that the triangles P QC and ABC are similar. Because CO and CF are PQ = AB , or (using the deﬁnition of corresponding altitudes, we deduce that CO CF sine, the sine law, and the equation from the previous paragraph), PQ = Rc b sin A = 2R2 c b · 2R sin A = 2R2 c ba = c(a2 + b2 ) 3ab .

Also solved by OLIVER GEUPEL, Br¨ uhl, NRW, Germanyand PETER Y. WOO, Biola University, La Mirada, CA, USA. Although the proposer stated his problem correctly, the published statement of the problem [2010 : 240, 242] contained two errors. Both Geupel and Woo reported the errors, but Woo ﬁgured out how to correct them, and he provided a complete solution. One of the editor’s errors is worth describing: The proposer correctly called the quadrilateral ABP Q inscriptible to mean that it can be inscribed in a circle. Unfortunately the word is ambiguous in English, as is its more common version inscribable—it can mean “permitting something to be inscribed in it” as in, “This book is inscribable.” (that is, it is possible to write an inscription in the book), as well as meaning capable of being inscribed in something, as in, “A rectangle is the only parallelogram that is inscribable in a circle.” One should restrict the use of the word to situations where the

244

context is clear (that is, when it is clear which object is being inscribed where); otherwise it is preferable to use the word cyclic (a cyclic polygon) or concyclic (a concyclic set of points). A similar error in CRUX with MAYHEM a few years ago brought forth a similar editorial comment [1997 : 530-531].

3542 . [2010 : 240, 242] Proposed by Cosmin Pohoat ¸˘ a, Tudor Vianu National College, Bucharest, Romania.
The mixtilinear incircles of a triangle ABC are the three circles each tangent to two sides and to the circumcircle internally. Let Γ be the circle tangent to each of these three circles internally. Prove that Γ is orthogonal to the circle passing through the incentre and the isodynamic points of the triangle ABC . [Ed.: Let ΓA be the circle passing through A and the intersection points of the internal and external angle bisectors at A with the line BC . The isodynamic points are the two points that ΓA , ΓB , and ΓC have in common.] No solutions to this problem were submitted. remains open. Problem 3542 therefore

3543.

[2010 : 240, 242] Proposed by Mehmet Mehmet S ¸ ahin, Ankara, Turkey.

Triangle ABC has inradius r , circumradius R, and angle bisectors [AD ], [BE ], [CF ], where points D , E , F lie on the sides BC , AC , AB , respectively. Let R be the circumradius of triangle DEF . Prove that R ≤
R4 . 16r3

A combination of solutions by Arkady Alt, San Jose, CA, USA and Michel Bataille, Rouen, France. We will prove the inequality R ≤
R , which implies the required inequality, 2
1 (a 2

since

R R4 ≤ is equivalent to Euler’s inequality, 2r ≤ R. 2 16r3

Let a = BC , b = CA, c = AB , s = area of the enclosed ﬁgure. Since

+ b + c), and let [·] denote the

follows that

BD DC a ca ab = = , we have BD = and CD = . c b c+b b+c b+c bc ab bc ca Similarly, AE = , CE = , AF = , BF = , and it a+c a+c a+b a+b

245

[BDF ] = [CED ] =

1 2

, 4R(b + c)(a + c) ab2 c2 . [AF E ] = 4R(a + b)(a + c)
abc , a straightforward calculation yields 4R

b+c a+b a2 b2 c

· sin B =

a2 bc2 4R(b + c)(a + b)

Using [ABC ] =

[EDF ] = [ABC ] − [BDF ] − [CED ] − [AF E ] = From the Law of Cosines,

(abc)2 . 2R(a + b)(b + c)(c + a)

EF 2 =

(bc)2 2(bc)2 (abc)2 + − · cos A = · Ka , (a + c)2 (a + b)2 (a + b)(a + c) (a + b)2 (a + c)2 (bc)2
1 [(a + c)2 + (a + b)2 − 2(a + b)(a + c) cos A]. Now, a2

where Ka =

Ka =

ç 1 ä 2 a + 2 a (1 − cos A )( a + b + c ) a2 8s =1+ sin2 (A/2) a 8s(s − b)(s − c) =1+ abc r 2 s2 2rs 8 · =1+ . =1+ 4Rrs s − a R(s − a)

1 ä 2 2a + b2 + c2 + 2ab + 2ac a2 ç − 2a2 cos A − 2ab cos A − 2ac cos A − 2bc cos A

As a result, EF = abc

√ abc Ka , and similarly (a + b)(a + c)
Ô

FD =

(a + b)(b + c)

Kb ,

DE =

abc (a + c)(b + c)

Kc ,

where Kb = 1 +

From these results, we obtain

2rs 2rs and Kc = 1 + . R(s − b) R(s − c)

246

R =

EF · F D · DE 4[EDF ]

Rabc 2(a + b)(b + c)(c + a)

Ka Kb Kc .

Using abc = 4Rrs and (a + b)(b + c)(c + a) = 2s(s2 + r 2 + 2rR), it readily follows that R = rR2 s2 + r2 + 2rR
Ô

Ka Kb Kc ,

and we will have proved R ≤
Ô

R if we can show that 2

Ka Kb Kc ≤

R s2 + r 2 + 2rR s2 + r 2 · = 1 + . 2 rR2 2rR

(1)

With this aim, we compute Ka Kb Kc as follows Ka Kb Kc = 1 + where S= T = 1 s−a + 1 1 s−b + + 1 s−c = 1 r + 4R rs , 1 (s − c)(s − a) = 1 r2 , 2rs R S+ 4r 2 s2 R2 T + 8r 3 s3 R3 (s − a)(s − b)(s − c) .

(s − a)(s − b)

(s − b)(s − c)

r 2 s = (s − a)(s − b)(s − c) . This yields Ka Kb Kc = 9 + 2r R + 4s2 R2 + 8rs2 R3 .

√ Now, from the well-known inequalities 2r ≤ R and 2s ≤ 3R 3, we obtain Ka Kb Kc ≤ 9 + 1 + 27 + 27 = 64, and so Ka Kb Kc ≤ 8 . √ But we also have s ≥ 3r 3, hence 2rR 2r · 2r and (1) directly follows from (2) and (3). 1+ s2 + r 2 ≥1+ 27r 2 + r 2 = 8,
Ô

(2)

(3)

Also solved by OLEH FAYNSHTEYN, Leipzig, Germany; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; and the proposer.

247

3544.

[2010 : 240, 242] Proposed by Mehmet S ¸ ahin, Ankara, Turkey.

Triangle ABC has excentres Ia , Ib , Ic and Ha , Hb , Hc are the orthocentres of triangles Ia BC , Ib CA, Ic AB , respectively. Prove that Area(Ha CHb AHc B ) = 2Area(ABC ) . Similar approaches by the solvers listed below with an asterisk. Let I be the incenter of ABC . Since Ha B and IC are each perpendicular to Ia C , Ha B IC . Similarly, Ha C IB , so Ha BIC is a parallelogram. Similarly, Hb CIA and Hc AIB are parallelograms. ∠BAC , ∠Ia BC = Since ∠BIa C = 90◦ − 1 2 ∠ ABC , and ∠ Ia CB = 90◦ − 90◦ − 1 2 1 ∠ACB , Ia BC is acute, so Ha is inside 2 Ia BC . Similarly, Hb is inside Ib CA and Hc is inside Ic AB . Since A, B and C lie on segments Ib Ic , Ic Ia and Ia Ib respectively and I is inside ABC , Ha CHb AHc B is convex and I is in its interior. Therefore, Ic

Hc I

A Hb C Ib

B Ha

Area(Ha CHb AHc B ) = Area(Ha BIC ) + Area(Hb CIA) + Area(Hc AIB ) = 2Area(BIC ) + 2Area(CIA) + 2Area(AIB ) = 2Area(ABC ) .
Solved by ∗ ARKADY ALT, San Jose, CA, USA; ∗ GEORGE APOSTOLOPOULOS, Messolonghi, Greece; MICHEL BATAILLE, Rouen, France; ∗ EMMANUEL LANCE CHRISTOPHER, Ateneo de Manila University, The Philippines; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; ∗ RICHARD EDEN, student, Purdue University, West Lafayette, IN, USA; OLEH FAYNSHTEYN, Leipzig, Germany; ∗ OLIVER GEUPEL, Br¨ uhl, NRW, Germany; ∗ JOEL SCHLOSBERG, Bayside, NY, USA; ∗ PETER Y. WOO, Biola University, La Mirada, CA, USA; ∗ TITU ZVONARU, Com´ ane¸ sti, Romania; and the proposer.

3545.

[2010 : 240, 242] Proposed by Michel Bataille, Rouen, France.

Given a line and points A and B with A ∈ / and B ∈ , ﬁnd the locus of points P in their plane such that P A + QB = P Q for a unique point Q of . Solution by Oliver Geupel, Br¨ uhl, NRW, Germany. Proposition 1 If ⊥ AB , then L consists of the line parallel to through A with the exception of the point A, and the mid-perpendicular line of the segment AB .

248

Proof: We consider Cartesian (x, y )-coordinates such that, without loss of generality, A = (2, 0), B = (0, 0), and is the line x = 0. For any points P = (p, r ), Q = (0, y ), the condition P A + QB = P Q is equivalent to (p − 2)2 + r 2 + |y | = p2 + (r − y )2 . (1)

After squaring both sides and erasing equal terms on both sides, this successively becomes equivalent to (p − 2)2 + r 2 + y 2 + 2|y | (p − 2)2 + r 2 = p2 + (r − y )2 , and [y < 0 ∧ (r − s)y = 2(p − 1)] ∨ [y ≥ 0 ∧ (r + s)y = 2(p − 1)] , (2)

where s = (p − 2)2 + r 2 . We consider the three cases p = 1, p = 2, and p ∈ / {1, 2} in succession. If p = 1, then, by the deﬁnition of s, r − s and r + s are both nonzero. Hence, y = 0 is the unique solution of the logical expression (2), that is, the mid-perpendicular of AB belongs to the locus L. Let p = 2. If r < 0, then the ﬁrst alternative of the condition (2) yields the solution y = 1/r , while the second alternative is contradictory. Similarly, if r > 0, there is the unique solution y = 1/r . If r = 0 then the condition (2) is contradictory. Therefore, the parallel to through A with the exception of the point A is included in L. Finally, let p ∈ / {1, 2}. By (r − s)(r + s) = r 2 − s2 < 0, so numbers 2(p − 1)/(r − s) and 2(p − 1)/(r + s) have opposite signs. Therefore there cannot be exactly one value of y for which (2) is true. Thus, the number of points Q with the property (1) is either 0 or 2, that is, P ∈ / L. Proposition 2 Assume that and AB are not perpendicular. Let g be the perpendicular to through B . Let π denote the parabola with directrix g and focus A. Let R be the region that is bounded by π and that contains the point A, where the boundary π belongs to R. Let m be the mid-perpendicular of the segment AB . Then m is tangential to π and the locus L consists of the region R and the line m with the exception of their tangential point. Proof: We consider Cartesian (x, y )-coordinates such that, without loss of generality, A = (a, 2), a > 0, B = (0, 0), and is the line x = 0. For any points P = (p, r ), Q = (0, y ), the condition (1) is equivalent to (p − a)2 + (r − 2)2 + |y | = p2 + (y − r )2 . (3)

After squaring both sides, this becomes equivalent to [y < 0 ∧ (r − s)y = t] ∨ [y ≥ 0 ∧ (r + s)y = t],

249

where s =

(p − a)2 + (r − 2)2 and t = pa + 2r − 2 − a2 /2. The parabola 1 π is given by the equation 4(y − 1) = (x − a)2 . We have t = P · A − A2 = 2 1 P − A · A. Hence, the condition t = 0 holds if and only if P is on m. It is 2 easy to check that m touches π at the point (0, a2 /4 + 1) on the line . We consider in succession the three cases where P is above π , on π , and below π . If P is above π , then (r − s)(r + s) = r 2 − s2 = [P g ]2 − [P A]2 > 0. Hence the numbers t/(r − s) and t/(r + s) have the same sign. Thus, exactly one of the two alternatives (3) is satisﬁable, and the solution is unique. Therefore, the interior of R is included in L. Let P be on π . Then r − s = [P g ] − [P A] = 0, r + s > 0, r = (p − a)2 /4 + 1, t = p2 /2. If p = 0 then t > 0 holds. Hence t/(r + s) > 0. Thus, the ﬁrst alternative in (3) is contradictory, while the second alternative yields a unique solution. Otherwise, if p = 0, then each y < 0 satisﬁes the condition (3). Therefore, the parabola π with the exception of its tangential point with m belongs to L. Finally, let P be below π . We consider the situations for t = 0 and t = 0 in succession. Firstly, let t = 0. By (r − s)(r + s) = r 2 − s2 = [P g ]2 − [P A]2 < 0, the numbers t/(r − s) and t/(r + s) have opposite signs. Therefore, either both alternatives in (3) are satisﬁable or both are contradictory. Thus, the number of points Q with the property (1) is either 0 or 2, that is, P ∈ / L. Finally, let t = 0. Then P is on m so P A = P B . The condition (1) is therefore equivalent to P B + BQ = P Q, which is satisﬁed if and only if the points B , P , and Q are collinear where B belongs to the line segment P Q. The unique point Q with this property is Q = B . We have shown that the line m with the exception of its tangential point with the parabola π is included in the locus L.
Also incompletely solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; and the proposer whose solutions were relatively short and more geometric.

3546.

[2010 : 240, 242] Proposed by Michel Bataille, Rouen, France.

Let n be a positive integer. Prove that (n 2) 0 <
k=0

(−1)k n+k

n ¡
2

≤

1 nn

Solution by the proposer. The inequality holds for n = 1, so we can assume that n ≥ 2. Let N =

n¡
2

250

and let S denote the central sum,
N

S=
k=0 1

(−1)

N k

x
0

n+k−1

dx =
0

n− 1 k=0

N k

(−x)k dx

=
0

xn−1 (1 − x)N dx .

It follows that S > 0. For k = 1, 2, . . . , n − 1 and by the AM-GM inequality, (1 − x)xk = kk (1 − x) · ≤ kk x k k k k+1 (1 − x) + k(x/k) k+1 · x ··· x

kk (k + 1)k+1

Since N = 1 + 2 + · · · + (n − 1), (1 − x)n−1 xN = [(1 − x)x] · [(1 − x)x2 ] · · · [(1 − x)xn−1 ] ≤ Thus, S=
0

22
1

22 33

···

(n − 1)n−1 nn

xN (1 − x)n−1 dx ≤

1 , nn

which completes the proof.
Also solved by Oliver Geupel, Br¨ uhl, NRW, Germany; and Albert Stadler, Herrliberg, Switzerland.

3547. [2010 : 241, 243] Proposed by Jos´ e Luis D´ ıaz-Barrero, Universitat Polit` ecnica de Catalunya, Barcelona, Spain.
Triangle ABC has perimeter equal to 1, inradius r , circumradius R, and side lengths a, b, c. Prove that a b c √ +√ +√ ≥ 1−a 1−c 1−b
abc

2 . 1 + 4r (r + 4R)

Solution by Richard Eden, student, Purdue University, West Lafayette, IN, USA. and r = (s − a)(s − b)(s − c)/s, where s is The relations R = 4rs the semiperimeter, are well-known. Since s = 1/2, then 16Rr = 8abc and 4r 2 = 8(1/2 − a)(1/2 − b)(1/2 − c) = (1 − 2a)(1 − 2b)(1 − 2c). Therefore, 1 + 4r (r + 4R) = 1 + (1 − 2a)(1 − 2b)(1 − 2c) + 8abc = 4(ab + bc + ca) − 2(a + b + c) + 2 = 4(ab + bc + ca)

251

The inequality to prove is then 2(ab + bc + ca) √ The function f (x) = 1/ x is convex for x > 0. Since a + b + c = 1, we can think of a, b, c as weights and apply Jensen’s inequality, b c a +√ +√ = af (b + c) + bf (a + c) + cf (a + b) √ a+c b+c a+b ≥ f (a(b + c) + b(a + c) + c(b + c)) = f (2(ab + bc + ca) 1 = , 2(ab + bc + ca) which completes the proof.
Also solved by Arkady Alt, San Jose, CA, USA; George Apostolopoulos, Messolonghi, ˇ Greece; Sefket Arslanagi´ c, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; Michel Bataille, Rouen, France; Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Charles R. Diminnie, Angelo State University, San Angelo, TX, USA; Oliver Geupel, Br¨ uhl, NRW, Germany; Salem Maliki´ c, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; Albert Stadler, Herrliberg, Switzerland; Peter Y. Woo, Biola University, La Mirada, CA, USA; and the proposer.

√

a b+c

+√

b a+c

+√

c a+b

≥

3548. [2010 : 241, 243] Proposed by Pham Van Thuan, Hanoi University of Science, Hanoi, Vietnam.
Let x, y , and z be nonnegative real numbers. Prove that x2 − xy + y 2 ≤ x + y + z + x2 + y 2 + z 2 − xy − yz − zx .

cyclic

I. Solution by the proposer, modiﬁed slightly by the editor. Due to complete symmetry we may assume without loss of generality that x = min{x, y, z }. Then
Ô

x2 − xy + y 2 = x2 − xz + z 2 =

x(x − y ) + y 2 ≤ y x(x − z ) + z 2 ≤ z .

and

Hence it suﬃces to show that x2 + y 2 + z 2 − xy − yz − zx . (1) √ Ô Since y 2 − yz + z 2 − x = (y − z )2 + yz − x ≥ x2 − x = 0, we may square both sides of (1) to obtain x2 + y 2 + z 2 − yz − 2x y 2 − yz + z 2 ≤ x2 + y 2 + z 2 − xy − yz − zx y 2 − yz + z 2 − x ≤

252

or x(y + z ) ≤ 2x y 2 − yz + z 2 . (2) Squaring both sides, (2) is equivalent, in succession, to x2 (y 2 + 2yz + z 2 ) ≤ 4x2 (y 2 − yz + z 2 ) 3x2 (y − z )2 ≥ 0 .

3x2 y 2 − 6x2 yz + 3x2 z 2 ≥ 0 The proof is complete.

II. Solution by Albert Stadler, Herrliberg, Switzerland, expanded by the editor. We show that the proposed inequality follows from the result below, known as Hlawka Inequality [see D.S. Mitrinovi´ c, J.E. Peˇ cari´ c and A.M. Fink, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, 1993; p.521]: If V is an inner product space, then for all a, b, c ∈ V , ||a + b|| + ||b + c|| + ||c + a|| ≤ ||a|| + ||b|| + ||c|| + ||a + b + c|| , where || · || denotes the norm induced by the inner product. If we let V c = z −1 ,− 2 Furthermore,
√ 3 2

, 23 , and = R2 and set a = x(1, 0), b = y − 1 2 √ √ Ô , then ||a|| = x2 = x, ||b|| = y 2 = y , ||c|| = z 2 = z .
2

√

¬¬ ¬¬ √ ¬¬ ¬¬ 1 3 1 ¬¬ ¬¬ ||a + b|| = ¬¬ x − y, y ¬¬ = x− y ¬¬ ¬¬ 2 2 2 ¬¬ ¬¬ √ ¬¬ ¬¬ 1 3 ¬¬ ¬¬ (y − z ) ¬¬ ||b + c|| = ¬¬ − (y + z ), ¬¬ ¬¬ 2 2 Ö

3 4

y2 =

x2 − xy + y 2 ,

=
¬¬ ¬¬ ¬¬ ||c + a|| = ¬¬ ¬¬

1 3 (y + z )2 + (y − z )2 = y 2 − yz + z 2 , 4 4 ¬¬ √ ¬¬ Ô 1 3 1 2 3 ¬¬ x − z, − z ¬¬ = x − z + z 2 = x2 − zx + z 2 , ¬¬ 2 2 2 4
¬¬ ¬¬ ¬¬ ¬¬ ¬¬

and

¬¬ √ ¬¬ 1 3 ¬¬ ||a + b + c|| = ¬¬ x − (y + z ), (y − z ) ¬¬ 2 2

= = The result follows.

x−

1 2

(y + z )

3 4

(y − z )2

x2 + y 2 + z 2 − xy − yz − zx .

253

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; MICHEL BATAILLE, Rouen, France; RICHARD EDEN, student, Purdue University, West Lafayette, IN, USA; OLEH FAYNSHTEYN, Leipzig, Germany; KEE-WAI LAU, Hong Kong, China; PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; and PETER Y. WOO, Biola University, La Mirada, CA, USA. As usual, Stan Wagon veriﬁed the validity of the inequality by using “FindInstance” (in 13 seconds).

3549. [2010 : 241, 243] Proposed by Hung Pham Kim, student, Stanford University, Palo Alto, CA, USA.
Let a, b,¡and c be nonnegative ¡ ¡real numbers such that a + b + c = 3. Prove that 1 + a2 b 1 + b2 c 1 + c2 a ≤ 5 + 3abc.

Solution by Arkady Alt, San Jose, CA, USA. Note ﬁrst that the given inequality is equivalent to a2 b + b2 c + c2 a + abc(ab2 + bc2 + ca2 ) + a3 b3 c3 ≤ 4 + 3abc . We ﬁrst establish the following lemma: Lemma If a, b, and c are nonnegative real numbers, then a2 b + b2 c + c2 a + abc ≤ 4 27 (a + b + c)3 . (2) (1)

Proof: Due to the cyclic symmetry of a, b, and c in (2) we may assume, without loss of generality, that c = min{a, b, c}. We consider two cases separately: Case (i). Suppose b ≤ a. By the AM-GM Inequality, we have 1 4 (a + b + c)3 = 27 2 Since b(a + c)2 − (a2 b + b2 c + c2 a + abc) = abc + bc2 − b2 c − c2 a = c(a − b)(b − c) ≥ 0 we have a2 b + b2 c + c2 a + abc ≤ b(a + c)2 and (2) follows from (3) and (4). Case (ii). Suppose b > a. We have 2(a2 b + b2 c + c2 a + abc) =
cyclic

2b + 2(a + c) 3

≥

1 2b(a + c)2 = b(a + c)2 . (3) 2

= c(ab + bc − b2 − ca)

(4)

(a2 b + ab2 ) + 2abc +
cyclic

(a2 b − ab2 ) (5)

= (a + b)(b + c)(c + a) − (a − b)(b − c)(c − a) .

254

By the AM-GM Inequality we have (a + b)(b + c)(c + a) ≤ = so 4 27 (a + b + c)3 ≥ 1 2 (a + b)(b + c)(c + a) 1 (a − b)(b − c)(c − a) 2 (a + b) + (b + c) + (c + a) 3 8 (a + b + c)3 27
3

= a2 b + b2 c + c2 a + abc + ≥ a2 b + b2 c + c2 a + abc since (a − b)(b − c)(c − a) ≥ 0. This completes the proof of the lemma.

Since a + b + c = 3, (2) becomes a2 b + b2 c + c2 a ≤ 4 − abc and since 4 − abc is invariant under the interchanging of a and b, we have max{a2 b + b2 c + c2 a, ab2 + bc2 + ca2 } ≤ 4 − abc. Therefore, a2 b + abc
cyclic cyclic

ab2 + a3 b3 c3 ≤ (1 + abc)(4 − abc) + a3 b3 c3 .
3

Finally, since abc ≤ 4 + 3abc −

a+b+c 3

= 1 we have

a2 b + abc
cyclic cyclic

ab2 + a3 b3 c3

= a2 b2 c2 − a3 b3 c3 = a2 b2 c2 (1 − abc) ≥ 0 which establishes (1) and completes the proof.

≥ 4 + 3abc − (1 + abc)(4 − abc) − a3 b3 c3

ˇ Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; SEFKET ´ University of Sarajevo, Sarajevo, Bosnia and Herzegovina; PAOLO ARSLANAGI C, PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; HAOHAO WANG and JERZY WOJDYLO, Southeast Missouri State University, Cape Girardeau, Missouri, USA; and the proposer. Stan Wagon gave his usual veriﬁcation using “FindInstances”.

255

3550.
Romania.

[2010 : 241, 243] Proposed by Ovidiu Furdui, Campia Turzii, Cluj,

Find the sum
∞ ∞ n+m

n=1 m=1

(−1)n+m

ln 2 −

1 n+m+i

i=1

Solution by Oliver Geupel, Br¨ uhl, NRW, Germany. Let
m

am = ln 2 − and bn = We have
2m ∞ m=1

1 m+i

i=1

(−1)m+n am+n .

am = ln 2 −
1

(−1)i−1 i
2m 1 0

i=1

=
0 1

1 1+x 1 1+x

dx − dx −

i=1 1 0

(−1)i−1 xi−1 dx
1

=
0

1 − x2 m 1+x

dx =
0

x2 m 1+x

dx ;

hence
m i=1

(−1)n+1 an+1 = =

1 0 1 0

1 1+x

(−x ) 1+x

i=1 2 n+1

(−1)n+i x2(n+i) dx · 1 − (−x2 )m dx 1 + x2

Then for all 0 ≤ x ≤ we have 0 ≤ |fm (x)| ≤ 2f (x). Also, on [0, 1) fm (x) → f (x) pointwise. Then, by the Lebesgue Dominated Convergence Theorem we have 1 (−x2 )n+1 1 1 − (−x2 )m (−x2 )n+1 bn = lim · dx = dx . 2 m→∞ 0 1+x 1 + x2 0 (1 + x)(1 + x ) It follows that
N 1

(−x2 )n+1 (−x2 )n+1 1 − (−x2 )m · and f ( x ) = . Let fm (x) = 1+x 1 + x2 (1 + x)(1 + x2 )

bn =
i=1 0 1

1 (1 + x)(1 + x2 )
4

N n=1

(−x2 )n+1 dx

=
0

x 1 − (−x2 )N · dx (1 + x)(1 + x2 ) 1 + x2

256

Applying again the Lebesgue Dominated Convergence Theorem we get:
∞ i=1 1

bn =
0

x4 (1 + x)(1 + x2 )2
1

= =

1 2(1 + x) 8 1 + x2 5 ln 2 − π 8

+ 3 ln(1 + x2 ) + 2 ln(1 + x) − 4 arctan(x)

Thus, the sum is

5 ln 2−π . 8

Also solved by PAUL BRACKEN, University of Texas, Edinburg, TX, USA; PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; ALBERT STADLER, Herrliberg, Switzerland; and the proposer .

Crux Mathematicorum
with Mathematical Mayhem
Former Editors / Anciens R´ edacteurs: Bruce L.R. Shawyer, James E. Totten, V´ aclav Linek

Crux Mathematicorum
Founding Editors / R´ edacteurs-fondateurs: L´ eopold Sauv´ e & Frederick G.B. Maskell Former Editors / Anciens R´ edacteurs: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

257

EDITORIAL
Shawn Godin
Hello CRUX with MAYHEM readers, welcome to the September 2011 issue. I must apologize up front for the ongoing delays. The ﬁrst contract between the CMS and the Ottawa District School Board started in March of 2011 left me with some catching up to do. The contract ran out in January and the negotiations took longer than any of the parties involved thought they would. As a result, for two months I was back to my “day job” full time. Fortunately, things are ﬁnally straightened out and we are moving forward with the 2011 volume of CRUX with MAYHEM . We have been going through the backlog of proposals and should be up to date with those within a couple of months. With respect to problem proposals, for the last couple of years the majority of the proposals submitted are inequalities. As a result, we have many proposals in the bank. We seem to receive the right amount of geometry problems to keep our problem sets going, so keep those questions coming. Currently we are very short on problems from number theory, algebra, calculus and combinatorics. Please consider sending us your nice proposals from these areas. As always, proposals sent to CRUX with MAYHEM should be original. If you have sent a proposal to another publication, do not send it to us unless the problem has been rejected, or you have withdrawn it and vice versa. Currently we construct the problem sets by looking at problems that have been proposed by readers and trying to select a set of problems from a number of diﬀerent topics that will be appealing to the readers. We have no idea how the readers really feel about these problems (other than the fact that if we receive a number of solutions to a problem from the readers, those readers must have enjoyed it). We have started another series of short surveys where you can give us feedback on the problems we propose in Crux . Each survey asks you to rate each problem (and it isn’t necessary to rate them all) from -5 (meaning you hated the problem) to 5 (meaning you loved the problem). A score of 0 means you are indiﬀerent. You can ﬁnd the addresses of the surveys for the problems from the 2011 volume on our Facebook page. The survey for the problems from this issue can be found at: http://www.surveymonkey.com/s/LK998Z9 Thank you to the people who have taken the time to answer our recent survey on-line. We have compiled your input and will be looking into making some changes. We are currently searching for columnists to start up some regular columns on problem-solving. This issue contains the ﬁrst installment of a semiregular column Recurring Crux Conﬁgurations. The column explores ideas that have resurfaced in Crux problems over the years and the ﬁrst entry was written by editorial board member J. Chris Fisher, I hope you enjoy it.

258

A major change that we will be making, starting in volume 38, is separating the Mathematical Mayhem and Skoliad Corner material from Crux. Mayhem has been a part of Crux since volume 23 in 1997. It will continue to exist, separately, in electronic format only. We would like to publish articles more regularly in Crux, but we have been receiving many articles lately that are inappropriate. Crux is a problem-solving journal at the secondary and university undergraduate levels; as such, articles should reﬂect the scope of the journal. Some possible categories that articles could fall into: 1. Articles about problem solving methods or techniques. An example of an article of this type is Summations according to Gauss by Gerhard J. Woeginger which appears in this issue. Articles about a theorem that is known, but maybe not well known to all the readership of Crux are also acceptable. 2. Articles inspired by a particular problem. These could include generalizations of problems that have appeared in CRUX with MAYHEM or elsewhere. An example of an article of this type is A Generalization of Mayhem Problem M396 Involving Pythagorean Triangles [2010 : 540–544] by Konstantine Zelator . 3. Articles that would be of interest to problem solvers. These could include articles surrounding a problem that could have been one of the Crux numbered problems (although, may have been a bit too involved). An example of an article of this type is A Nest of Euler Inequalities by Luo Qi which appears in this issue. Crux is not a research journal. Articles concerning “serious” mathematics will not be considered for publication (i.e. they will be rejected outright). Crux is not primarily aimed at teachers (although many of the readership are teachers). Articles concerning pedagogy will not be considered for publication. We welcome articles from the readers. Articles should be 2 to 8 pages in length. Longer articles may be considered if they can logically be separated into multiple parts. Please send your appropriate articles to our articles editor: Robert Dawson, Department of Mathematics and Computing Science, Saint Mary’s University, Halifax, NS, Canada, B3H 3C3, or by email to [email protected]. Shawn Godin

259

SKOLIAD

No. 134

Lily Yen and Mogens Hansen
Please send your solutions to problems in this Skoliad by July 15, 2012. A copy of CRUX with Mayhem will be sent to one pre-university reader who sends in solutions before the deadline. The decision of the editors is ﬁnal. Our contest this month is the Baden-W¨ urttemberg Mathematics Contest, 2010. Our thanks go to the Landeswettbewerb Mathematik Baden W¨ urttemberg for providing us with this contest and for permission to publish it. We also thank Rolland Gaudet, Universit´ e de Saint-Boniface, Winnipeg, MB, for translating the contest.

Baden-W¨ urttemberg Mathematics Contest, 2010 1. 2. 3.
Sonja has nine cards on which the nine smallest two-digit prime numbers are printed. She wants to order these cards in such a way that neighbouring cards always diﬀer by a power of 2. In how many ways can Sonja order her cards? A 50 cm by 30 cm by 28 cm box contains wooden blocks that all measure 10 cm by 9 cm by 7 cm. At most how many blocks can ﬁt in the box? Explain how to ﬁt that many blocks into the box. Five distinct positive numbers are given. Forming all possible sums of two of these numbers you obtain seven diﬀerent sums. Show that the sum of the ﬁve original numbers is divisible by 5.

4. Three squares are arranged as in the ﬁgure. Show that the two shaded triangles
have the same area.

Triangle △ABC is isosceles and ∠ACB = 90◦ . The point D is on the line AC beyond C , and the point E is on the line CB beyond B . Show that |CD | = |CE | if line BD is perpendicular to line AE .
D C

B E

260

The product of three positive integers is three times as large as their sum. Find all such triples.

Concours math´ ematique Baden-W¨ urttemberg 2010 1. Sonya dispose de neuf cartes sur lesquelles sont imprim´ es les neuf plus petits
nombres premiers ` a deux chiﬀres. Elle voudrait ordonner ses cartes de fa¸ con ` a ce que les cartes voisinantes diﬀ` erent toujours par une puissance de 2. De combien de mani` eres est-ce que Sonya peut ordonner ses cartes ?

2. Une boˆ ıte de taille 50 cm par 30 cm par 28 cm contient des blocs en bois, chacun de taille 10 cm par 9 cm par 7 cm. Au plus, combien de tels blocs entrent dans la boˆ ıte ? Expliquer comment faire entrer ce nombre de blocs dans la boˆ ıte. 3. Cinq nombres positifs distincts vous sont donn´ es. En formant toutes les sommes
possibles de deux de ces nombres, on constate qu’on obtient sept sommes distinctes. D´ emontrer que la somme des cinq nombres originaux est divisible par 5.

4. Trois carr´ es sont dispos´ es tel qu’illustr´ e par la ﬁgure. D´ emontrer que les deux
triangles ombrag´ es ont la mˆ eme surface.

5. Le triangle △ABC est isoc` ele et ∠ACB = 90◦ . Le point D se trouve sur la
ligne AC , au-del` a de C , et le point E se trouve sur la ligne CB , au-del` a de B . D´ emontrer que |CD | = |CE | si la ligne BD est perpendiculaire ` a la ligne AE .
D C

B E

6. Le produit de trois entiers positifs est trois fois aussi grand que leur somme.
D´ eterminer tout tel triplet.

261

Next follow solutions to the Mathematics Association of Quebec Contest, Secondary level, 2010, given in Skoliad 128 at [2010:417–419].

An alphametic is a small mathematical puzzle consisting of an equation in which the digits have been replaced by letters. The task is to identify the value of each letter in such a way that the equation comes out true. Diﬀerent letters have diﬀerent values, diﬀerent digits are represented by diﬀerent letters, and no number begins with a zero. For example, the alphametic PAPA + PAPA = MAMAN has the solution P = 7, A = 5, M = 1, and N = 0, yielding 7575 + 7575 = 15150. Find the solution to this “reversing” alphametic: NOMBRE × 3 5 = ERBMON .

´ Solution by Lena Choi, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC.
3 If NOMBRE × 5 = ERBMON, then NOMBRE × 3 = ERBMON × 5, so NOMBRE is divisible by 5. Therefore E = 0 or E = 5. However, ERBMON does not begin with a zero, so E = 5. Thus ERBMON > 500 000, so NOMBRE = 5 × ERBMON > 833 333. 3 Therefore N = 8 or N = 9. When multiplying integers, the ones digit of the result depends only on the ones digits of the factors. Therefore, the ones digit of NOMBRE × 3 is the ones digit of E × 3 which is 5, since E = 5. Now, NOMBRE × 3 = ERBMON × 5, so the ones digit of ERBMON × 5 is also 5. Therefore, the ones digit of N × 5 must be 5, so N cannot be 8. Thus N = 9. You now know that the ones digit of NOMBRE × 3 is 9. Note that the ones 5 digit of NOMBRE ÷ 5 equals the ones digit of RE ÷ 5. Therefore the ones digit 3 of NOMBRE × 5 equals the ones digit of RE × 3 , which, thus, must be 9. Since 5 E = 5, either R = 1 or R = 6. If R = 1, then NOMBRE = ERBMON × 5 ≤ 3 5 = 866 665, contradicting the fact that N = 9. Thus R = 6. 519 999 × 3 Now, NOMBRE = ERBMON × 5 ≥ 560 000 × 5 > 933 333, so O ≥ 3. 3 3 Moreover, ERBMON is divisible by 3, so E + R + B + M + O + N is divisible by 3. Thus NOMBRE is divisible by 3, and, hence, ERBMON = NOMBRE × 3 5 is divisible by 9. Therefore, N + O + M + B + R + E = 9 + O + M + B + 6 + 5 = O + M + B + 20 is divisible by 9. You must now ﬁnd three digits, (O, M, B), from {0, 1, 2, 3, 4, 7, 8} subject to the two conditions that O ≥ 3 and that O + M + B + 20 is divisible by 9. Only ten triples satisfy these conditions: (3, 0, 4), (3, 4, 0), (4, 0, 3), (4, 1, 2), (4, 2, 1), (4, 3, 0), (7, 1, 8), (7, 8, 1), (8, 1, 7), and (8, 7, 1). 3 Of these only one satisfy the further condition that NOMBRE × 5 = ERBMON, namely (O, M, B) = (3, 4, 0).

934 065 ×

3 = 560 439 5

3 The solution 934 065 × 5 = 560 439 was also found by ROWENA HO, student, ´ ´ Ecole Dr. Charles Best Secondary School, Coquitlam, BC; JANICE LEW, student, Ecole Alpha

262

Secondary School, Burnaby, BC; and SZERA PINTER, student, Moscrop Secondary School, Burnaby, BC.

2. Find all polynomials of the form p(x) = x3 + mx +6 whose roots are integers.
Solution by Billy Suandito, Palembang, Indonesia. All integer roots of p(x) must be factors of 6 (for the reason why, see the editors’ note below ). Thus, the possible integer roots are 1, 2, 3, 6, −1, −2, −3, and −6. If 1 is a root, then 0 = p(1) = 13 + m + 6 = 7 + m, so m = −7. Thus p(x) = x3 − 7x + 6 = (x − 1)(x2 + x − 6) = (x − 1)(x + 3)(x − 2). Thus p(x) has three integer roots if m = −7. If 2 (or −3) is a root, you ﬁnd the above example again. If 3 is a root, then 0 = p(3) = 33 + 3m + 6 = 33 + 3m, so m = −11. Thus p(x) = x3 − 11x + 6 = (x − 3)(x2 + 3x − 2), but the roots of x2 + 3x − 2 are not integers. If you continue checking the possible roots, 6, −1, −2, and −6, in a similar manner, you will ﬁnd that p(x) fails to have integer roots in each case except m = −7. Thus p(x) = x3 − 7x + 6 is the only solution.

The solution p(x) = x3 − 7x + 6 was also found by WEN-TING FAN, student, Burnaby North Secondary School, Burnaby, BC; and LISA WANG, student, Port Moody Secondary School, Port Moody, BC. Alternatively, say the three roots are a, b, and c. Then x3 + mx + 6 = (x − a)(x − b)(x − c) = x3 − (a + b + c)x2 + (ab + ac + bc)x − abc. Thus a + b + c = 0 and abc = −6. It follows that exactly one of the roots is negative. Since the roots are integers, they must be factors of 6. The only possibility is, then, that the roots are 1, 2, and −3. Thus the only solution is p(x) = (x − 1)(x − 2)(x + 3) = x3 − 7x + 6 as above. Note, as in the previous paragraph, that if the leading coeﬃcient of a polynomial is 1, then the next-to-leading coeﬃcient is the negative of the sum of the roots and the constant term is, apart from a sign, the product of the roots. This observation often comes in handy in contests.

A line is located at units from the centre of a 2 circle of radius 1, separating it into two parts. What is the area of the smaller part?

√

Solution by Lisa Wang, student, Port Moody Secondary School, Port Moody, BC. Let O be the centre of the circle and A and C be the endpoints of the chord. Let B be the point on AC that √ is closest to O . Then |OB | = 22 , ∠OBA = 90◦ , and B is the midpoint of AC . By the Pythagorean Theorem, Õ |AB | = 12 − (
√ 2 2 ) 2

A 1 O
√ 2 2

√ |AC | = 2, so the area of △AOC is 1 · |AC | · |OB | = 2 √ C √ 1 2 1 · 2 · 2 = 2. 2 Moreover, △AOB is isosceles and ∠AOB = 45◦ . By symmetry, ∠COB = 45◦ , so ∠AOC = 90◦ , whence sector AOC is a quarter

1−

1 2

√ 2 . 2

Therefore

263

1 circle. Thus the area of sector AOC is 4 π 12 = π . 4 It follows that the area of the shaded segment is

´ Also solved by LENA CHOI, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; VINCENT CHUNG, student, Burnaby North Secondary School, Burnaby, BC; ´ ROWENA HO, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; and BILLY SUANDITO, Palembang, Indonesia.

π 4

−

1 2

≈ 0.285.

The ﬁgure shows a map of a city. In how many ways can you travel along the roads of the city from point A to point B if you can only travel east and south (right and down in the ﬁgure)? B Solution by Vincent Chung, student, Burnaby North Secondary School, Burnaby, BC. Consider this simpler map. Surely, you can A C D E 1 1 1 arrive at C in only one way, as indicated. To reach D , you must come from C , so you can arrive at D in I only one way. Similarly at E and F . You can reach F 1 G 2 H 3 4 G either from C or from F , so you can arrive at G in two ways. You can reach H either from D or from 1 3 6 10 G, so you can arrive at H in 1 + 2 = 3 ways. Now J K L M continue in this way: if you can reach X from Y and Z , then the number of paths to X is the sum of the number of paths to Y and the number of paths to Z . Using this method of counting paths in the original map yields a total of 220 paths from A to B :
A 1 1 1 1 1 2 3 4 5 5 1 3 6 1 4 10 10 15 15 10 20 35 50 10 30 65 115 10 40 105 220

´ Also solved by LENA CHOI, student, Ecole Dr. Charles Best Secondary School, ´ Coquitlam, BC; ROWENA HO, student, Ecole Dr. Charles Best Secondary School, ´ Coquitlam, BC; JANICE LEW, student, Ecole Alpha Secondary School, Burnaby, BC; and BILLY SUANDITO, Palembang, Indonesia.

264

5. (a) How many zeroes are at the right-hand end of the number 1 × 2 × 3 × · · · × 52?
´ Solution by Janice Lew, student, Ecole Alpha Secondary School, Burnaby, BC. The number 1 × 2 ×· · ·× 52 is written more compactly as 52!. Let S denote the set {1, 2, 3, . . . , 52}. Half of the numbers in S are even, so 52! is divisible by 226 . One quarter of the numbers in S are divisible by 4. These contribute an extra 52 = 13 copies of 2. Since 52 = 6.5, the numbers in S that are divisible 4 8 by 8 contribute a further 6 copies of 2. Three of the numbers in S are divisible by 16, and one is divisible by 32. Thus 52! is divisible by 226+13+6+3+1 = 249 but not by any higher power of 2. Likewise, 52! is divisible by 510+2 = 512 but no higher power of 5. Since 10 = 2 × 5, it follows that 52! is divisible by 1012 but no higher power of 10. Thus 52! ends in exactly 12 zeroes.
´ Also solved by LENA CHOI, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; VINCENT CHUNG, student, Burnaby North Secondary School, Burnaby, BC; ´ ROWENA HO, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; and SZERA PINTER, student, Moscrop Secondary School, Burnaby, BC.

(b) What is the rightmost nonzero digit of 1 × 2 × · · · × 52? (For example, the rightmost nonzero digit of 1 × 2 × · · · × 12 = 479 001 600 is 6.) Solution by the editors. In Part (a), our solver began factoring 52! into primes. Now complete the process: Since 52 ≈ 17.3, 52 ≈ 5.8, and 52 ≈ 1.9, it follows that 52! is 3 9 27 17+5+1 23 divisible by 3 = 3 . Likewise, 52! is divisible by 77+1 = 78 , 114 , 134 , 173 , 192 , 232 , and once each by 29, 31, 37, 41, 43, and 47. Thus 52! = 249 · 323 · 512 · 78 · 114 · 134 · 173 · 192 · 232 · 29 · 31 · 37 · 41 · 43 · 47. The rightmost non-zero digit of 52! equals the ones digit of 52! 1012 = 237 · 323 · 78 · 114 · 134 · 173 · 192 · 232 · 29 · 31 · 37 · 41 · 43 · 47 .

Let a ≡ b denote that the whole numbers a and b have the same ones digit. Since the ones digit of a product depends only on the ones digits of the factors, 237 = 22 · (25 )7 = 22 · (32)7 ≡ 22 · 27 = 29 = 512 ≡ 2. Likewise, 323 = 33 · (34 )5 = 33 · (81)5 ≡ 33 · 1 = 27 ≡ 7, and 78 = (74 )2 = (2401)2 ≡ 1, and 114 ≡ 1, and 134 ≡ 34 = 81 ≡ 1, and 173 ≡ 73 = 343 ≡ 3, and 192 ≡ 92 = 81 ≡ 1, and 232 ≡ 32 = 9, and, of course, 29 ≡ 9, 31 ≡ 1, 37 ≡ 7, 41 ≡ 1, 43 ≡ 3, and 47 ≡ 7. Thus 52! ≡ 2 · 7 · 1 · 1 · 1 · 3 · 1 · 9 · 9 · 1 · 7 · 1 · 3 · 7 = 500094 ≡ 4 . 1012 Therefore the rightmost non-zero digit of 52! is 4.

265

Juliette and Philippe play the following game: At the beginning of the game, each corner of a square is covered with a number of chips. In turn, each player chooses one side of the square and removes as many chips as (s)he wants from the endpoints of that side provided (s)he takes at least one chip. It is not necessary to remove the same number of chips from each endpoint. The player who removes the last chip wins. At the beginning of the game on the square ABCD there are 10 chips on corner A, 11 chips on B , 12 chips on C , and 13 chips on D . If Juliette begins, how should she play?

´ Solution by Lena Choi, student, Ecole Dr. Charles Best Secondary School, ´ Coquitlam, BC; Rowena Ho, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; and Szera Pinter, student, Moscrop Secondary School, Burnaby, BC. Let a, b, c and d denote the number of chips on A, B , C , and D , respectively. Strategy: Juliette should play to ensure that a = c and b = d. If Philippe receives a square with a = c and b = d, then he must remove at least one chip, and he cannot remove chips from both ends of a diagonal. Therefore he will always pass a square to Juliette with a = c and/or b = d. If Juliette receives a square with a = c and/or b = d, then for each diagonal she should choose the endpoint with the larger number of chips. Then she should choose the side that connects those two endpoints and remove chips until a = c and b = d. Thus Juliette is always able to follow the strategy above. Eventually, a = c = 0 and b = d = 0, and Juliette wins.

7. Find all functions F : R → R such that F (x) + xF (−x) = 1 for all real numbers x. Solution by Billy Suandito, Palembang, Indonesia.
If F (x) + xF (−x) = 1 for all values of x, then the equation also holds for −x; that is, F (−x) − xF (x) = 1. Multiplying this last equation by x yields that xF (−x) − x2 F (x) = x. Subtract this from the original equation to get
1−x . 1 + x2

that F (x) + x2 F (x) = 1 − x, so (1 + x2 )F (x) = 1 − x, so F (x) =

This issue’s prize of one copy of Crux Mathematicorum for the best ´ solutions goes to Lena Choi, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC. We invite the reader to submit solutions to one or more of our problems.

266

MATHEMATICAL MAYHEM
Mathematical Mayhem began in 1988 as a Mathematical Journal for and by High School and University Students. It continues, with the same emphasis, as an integral part of Crux Mathematicorum with Mathematical Mayhem. The interim Mayhem Editor is Shawn Godin (Cairine Wilson Secondary School, Orleans, ON). The Assistant Mayhem Editor is Lynn Miller (Cairine Wilson Secondary School, Orleans, ON). The other staﬀ members are Ann Arden (Osgoode Township District High School, Osgoode, ON), Nicole Diotte (Windsor, ON), Monika Khbeis (Our Lady of Mt. Carmel Secondary School, Mississauga, ON) and Daphne Shani (Bell High School, Nepean, ON).

Editorial
Shawn Godin
Hello Mayhem readers. The CMS and the editors of CRUX with MAYHEM have been working on some changes for the journal. Over the last few months, we have set up a page on Facebook to communicate with our readers. We have surveyed the readers of CRUX with MAYHEM and have listened to their comments. One change that we are planning directly impacts the readers of Mayhem. As of volume 38, Mathematical Mayhem will no longer be part of Crux Mathematicorum. It will continue to exist, but only on the web. The numbering and dates of the issues will revert back to the days before it joined Crux Mathematicorum. Mathematical Mayhem will appear on line 5 times a year: September, November, January, March and May. The last volume of Mathematical Mayhem to appear as a stand alone was volume 8 [1995-1996]. The next stand alone volume starting in September 2012 will be volume 24 [20122013]. We will also be expanding and adding features that will be of interest to mathematics students and teachers at the high school level. The ﬁrst issue will be here before you know it, so get ready! Also, this issue marks the last installment of the Problem of the Month by Ian VanderBurgh. Ian has been writing this column since September 2004 [2004 : 264-265] (57 columns!). We will miss Ian’s column, it has been a great part of Mayhem. All the best Ian and thanks for sharing your passion of mathematical problem-solving with us. Shawn Godin

267

Mayhem Problems
Please send your solutions to the problems in this edition by 1 August 2012. Solutions received after this date will only be considered if there is time before publication of the solutions. Each problem is given in English and French, the oﬃcial languages of Canada. In issues 1, 3, 5, and 7, English will precede French, and in issues 2, 4, 6, and 8, French will precede English. The editor thanks Rolland Gaudet, Universit´ e de Saint-Boniface, Winnipeg, MB, for translating the problems from English into French.

M495.

Proposed by the Mayhem Staﬀ.

All possible lines are drawn through the point (0, 0) and the points (x, y ), where x and y are whole numbers with 1 ≤ x, y ≤ 10. How many distinct lines are drawn?

M496.

Proposed by Sally Li, student, Marc Garneau Collegiate Institute, Toronto, ON. Show that if we write the numbers from 1 to n around a circle, in any order, then, for all x = 1, 2, . . . , n, we are guaranteed to ﬁnd a block of x consecutive x(n + 1) numbers that add up to at least . Here ⌈y ⌉ is the ceiling function, 2 that is, the least integer greater than or equal to y . So ⌈6.2⌉ = 7, ⌈π ⌉ = 4, ⌈−8.3⌉ = −8 and ⌈10⌉ = 10.

M497.
b

Proposed by Pedro Henrique O. Pantoja, student, UFRN, Brazil.

Find all integers a, b, c where c is a prime number such that ab + c and a − c are both perfect squares.

M498.

Proposed by Bruce Shawyer, Memorial University of Newfoundland, St. John’s, NL. Right triangle ABC has its right angle at C . The two sides CB and CA are of integer length. Determine the condition for the radius of the incircle of triangle ABC to be a rational number.

M499.

Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania. Two circles of radius 1 are drawn so that each circle passes through the centre of the other circle. Find the area of the goblet like region contained between the common radius, the circumferences and one of the common tangents as shown in the diagram to the right.

268

M500.

Proposed by Edward T.H. Wang and Dexter S.Y. Wei, Wilfrid Laurier University, Waterloo, ON. Let N denote the set of natural numbers. (a) Show that if n ∈ N, there do not exist a, b ∈ N such that where [a, b] denotes the least common multiple of a and b. [a, b] = n, a+b

(b) Show that for any n ∈ N, there exists inﬁnitely many triples (a, b, c) of [a, b, c] natural numbers such that = n, where [a, b, c] denotes the least a+b+c common multiple of a, b and c. .................................................................

M495.

´ Propos´ e par l’Equipe de Mayhem.

Toutes les droites possibles sont trac´ ees ` a partir du point (0, 0) et des points (x, y ), o` u x et y sont des entiers tels que 1 ≤ x, y ≤ 10. Combien de droites distinctes ont ´ et´ e trac´ ees ?

M496.

Propos´ e par Sally Li, Institut coll´ egial Marc Garneau, Toronto ON.

D´ emontrer que si on place tous les entiers de 1 ` a n autour d’un cercle dans un ordre quelconque alors, pour tout x = 1, 2, . . . , n, on pourra certainement trouver x(n + 1) un bloc de x nombres cons´ ecutifs dont la somme sera d’au moins , 2 ⌈y ⌉ d´ esignant la partie “plafond” de y , c’est-` a-dire le plus petit entier plus grand ou ´ egal ` a y . Ainsi ⌈6.2⌉ = 7, ⌈π ⌉ = 4, ⌈−8.3⌉ = −8 and ⌈10⌉ = 10.

M497.

Propos´ e par Pedro Henrique O. Pantoja, ´ etudiant, UFRN, Br´ esil.

D´ eterminer tous les entiers a, b et c o` u c est un nombre premier tel que ab + c et ab − c sont tous les deux des carr´ es d’entiers.

M498.

Propos´ e par Bruce Shawyer, Universit´ e Memorial de Terre-Neuve, St. John’s, NL. Le triangle rectangle ABC a l’angle droit ` a C ; les deux cˆ ot´ es CB et CA sont de longueurs enti` eres. D´ eterminer la condition pour que le rayon du cercle inscrit du triangle ABC soit un nombre rationnel.

M499.

´ Propos´ e par Neculai Stanciu, Ecole secondaire George Emil Palade, Buz˘ au, Roumanie. Deux cercles de rayon 1 sont trac´ es de fa¸ con ` ce que chacun passe par le centre de l’autre. a D´ eterminer la surface de la r´ egion en forme de gobelet qui se trouve entre le rayon commun, les circonf´ erences et une des tangentes communes, tel qu’illustr´ ea ` droite.

269

M500.

Propos´ e par Edward T.H. Wang et Dexter S.Y. Wei, Universit´ e Wilfrid Laurier, Waterloo, ON. Soit N l’ensemble des nombres naturels. (a) D´ emontrer que si n ∈ N alors il n’existe aucun a, b ∈ N tels que o` u [a, b] d´ enote le plus petit commun multiple de a et b. [a, b] a+b = n,

(b) D´ emontrer que si n ∈ N alors il existe un nombre inﬁni de triplets (a, b, c) [a, b, c] d’entiers naturels tels que = n, o` u [a, b, c] d´ enote le plus petit a+b+c commun multiple de a, b et c.

Mayhem Solutions
M457.
Proposed by the Mayhem Staﬀ. Suppose that A is a digit between 0 and 9, inclusive, and that the tens digit of the product of 2A7 and 39 is 9. Determine the digit A. Solution by Florencio Cano Vargas, Inca, Spain. We write 2A7 = 2 · 102 + A · 10 + 7 and 39 = 3 · 10 + 9. Multiplying both numbers and grouping we get: 2A7 · 39 = 8 · 103 + 3A · 102 + (9A + 7) · 10 + 3 . The condition stated in the problem implies that 9A + 7 ≡ 9 (mod 10) which implies that 9A ≡ 2 (mod 10). Hence, the solution is A = 8.

Also solved by JACLYN CHANG, student, University of Calgary, Calgary, AB; NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia; WINDA KIRANA, ˜ student, SMPN 8, Yogyakarta, Indonesia; LUIZ ERNESTO LEITAO, Par´ a, Brazil; TRAVIS ´ B. LITTLE, students, Angelo State University, San Angelo, TX, USA; RICARD PEIRO, IES “Abastos”, Valencia, Spain; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; GUSNADI WIYOGA, student, SMPN 8, Yogyakarta, Indonesia; INGESTI BILKIS ZULPATINA, student, SMPN 8, Yogyakarta, Indonesia;

M458.

Proposed by the Mayhem Staﬀ.

Convex quadrilateral ABCD has AB = AD = 10 and BC = CD . Also, AC is perpendicular to BD , with AC and BD intersecting at P . If BP = 8 and CD = CP + 2, determine the area of quadrilateral ABCD . Solution by Ingesti Bilkis Zulpatina, student, SMPN 8, Yogyakarta, Indonesia. From the properties which are written above, ABCD is surely a kite since AB = AD , BC = CD , and AC ⊥ BD .

270

Using the Pythagorean theorem: AP 2 = AB 2 − BP 2
2

A 10 B 8 P 10 D

AP = 36 ∴ AP = 6 and BP 2 + CP 2 = CB 2

64 + CP 2 = CP 2 + 4CP + 4 60 = 4CP CP = 15 Hence AC = AP + P C = 6 + 15 = 21 . Thus the area of quadrilateral ABCD is [ABCD ] = AC × BD 2 =

CP + 2

21 × 16 2

= 168

square units.
Also solved by SCOTT BROWN, Auburn University, Montgomery, AL, USA; FLORENCIO CANO VARGAS, Inca, Spain; JACLYN CHANG, student, University of Calgary, Calgary, AB; NATALIA DESY, student, SMA Xaverius 1, Palembang, Indonesia; WINDA KIRANA, student, SMPN 8, Yogyakarta, Indonesia; MITEA MARIANA, No. 2 ´ IES “Abastos”, Valencia, Spain; Secondary School, Cugir, Romania; RICARD PEIRO, BRUNO SALGUEIRO FANEGO, Viveiro, Spain; and GUSNADI WIYOGA, student, SMPN 8, Yogyakarta, Indonesia.

M459.

Proposed by Neven Juriˇ c, Zagreb, Croatia.

Determine whether or not it is possible to create a collection of ten distinct subsets of S = {1, 2, 3, 4, 5, 6} so that each subset contains three elements, each element of S appears in ﬁve subsets, and each pair of elements from S appears in two subsets. Solution by Jaclyn Chang, student, University of Calgary, Calgary, AB. It is possible to create ten distinct subsets of S = {1, 2, 3, 4, 5, 6} such that each subset contains three elements, each element of S appears in ﬁve subsets, and each pair of elements from S appears in two subsets. Each of the following distinct subsets contains three elements of S : S 1 = {1, 2, 3}, S 2 = {1, 2, 4}, S 3 = {1, 3, 5}, S 4 = {1, 4, 6}, S 5 = {1, 5, 6}, S 6 = {2, 3, 6}, S 7 = {2, 4, 5}, S 8 = {2, 5, 6}, S9 = {3, 4, 5}, S10 = {3, 4, 6}. Each element of S appears in ﬁve subsets of S : Element 1 in S1 , S2 , S3 , S4 ,S5 ; Element 2 in S1 , S2 , S6 , S7 , S8 ; Element 3 in S1 , S3 , S6 , S9 , S10 ; Element 4 in S2 , S4 , S7 , S9 , S10 ; Element 5 in S3 , S5 , S7 , S8 , S9 ; Element 6 in S4 , S5 , S6 , S8 , S10 . Each pair of elements from S appears in two subsets of S :

271

Also solved by ALEX SONG, Detroit Country Day School, Detroit, MI, USA, and EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON. One incomplete solution was received.

{1, 2} {2, 4} {4, 5} {1, 6}

in in in in

S1 , S2 , S7 , S4 ,

S2 ; S7 ; S9 ; S5 ;

{2, 3} {3, 6} {1, 5} {3, 4}

in in in in

S1 , S6 , S3 , S9 ,

S6 ; S10 ; S5 ; S10 ;

{3, 5} {1, 4} {2, 6} {5, 6}

in in in in

S3 , S2 , S6 , S5 ,

S9 ; S4 ; S8 ; S8 .

{1, 3} in S1 , S3 ; {2, 5} in S7 , S8 ; {4, 6} in S4 , S10 ;

M460.
and K =

Proposed by Dragoljub Miloˇ sevi´ c, Gornji Milanovac, Serbia. √ a+b , G = ab, 2

Let a and b be positive real numbers. Deﬁne A =
a 2 + b2 . 2

(c) G + K ≤ 2A, and (d) G4 + K 4 ≥ 2A4 . (a) By direct computation we get G2 + K 2 = ab + = (b) Since (a − b)4 ≥ 0 we get

Prove that (a) G2 + K 2 = 2A2 , (b) A2 ≥ KG,

Solution by Jaclyn Chang, student, University of Calgary, Calgary, AB. a2 + b2 2
2

a2 + 2ab + b2 2
2

(a + b) a+b =2 2 2

= 2A2 .

⇒ a4 + 4a3 b + 6a2 b2 + 4ab2 + b4 ≥ 8a3 b + 8ab2 ⇒ ⇒ (a + b)4 16 a+b 2 ≥
2

a4 − 4a3 b + 6a2 b2 − 4ab3 + b4 ≥ 0 a3 b + ab3 ⇒ a2 a+b 2 + 2 b2
4

2 √ ≥ ab

≥

ab(a2 + b2 ) 2

⇒ A2 ≥ KG . ≥ GK . Thus, using

[Ed.: Note that from the AM-GM inequality the result from (a) we get A =
2 2 2 G2 +K 2 2

G2 +K 2 2

(d) Since (a − b)4 ≥ 0 we have a4 − 4a3 b + 6a2 b2 − 4ab3 + b4 ≥ 0, hence a4 + 6a2 b2 + b4 ≥ 4a3 b + 4ab3 ⇒ ⇒ 2a4 + 12a2b2 + 2b4 ≥ 8 a4 + 2a2 b2 + b4 4 a4 + 6a2 b2 + b4 8 a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 8 ≥2 a+b 2
4

(c) We have(G + K ) = G + 2KG + K 2 , but from (b) we know that 2KG ≤ 2A2 , thus (G + K )2 ≤ G2 + K 2 + 2A2 . Using part (a) we can deduce (G + K )2 ≤ 4A2 = (2A)2 and therefore G + K ≤ 2A. 4a3 b + 4ab3 8

≥ GK .]

≥

⇒ a2 b2 +

⇒ G4 + K 4 ≥ 2A4 .

272

[Ed.: Note that from (a) and (b), we have G4 + K 4 = (G2 + K 2 )2 − ¡2 2K 2 G2 = 2A2 − 2K 2 G2 ≥ 4A4 − 2A4 = 2A4 .]
´ Also solved by MIHALY BENCZE, Brasov, Romania; PAUL BRACKEN, University of Texas, Edinburg, TX, USA; SCOTT BROWN, Auburn University, Montgomery, AL, USA(parts a, b, c); FLORENCIO CANO VARGAS, Inca, Spain; NATALIA DESY, ˜ student, SMA Xaverius 1, Palembang, Indonesia(parts a, c); LUIZ ERNESTO LEITAO, ´ IES “Abastos”, Valencia, Spain; PAOLO PERFETTI, Par´ a, Brazil(part a); RICARD PEIRO, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; GUSNADI WIYOGA, student, SMPN 8, Yogyakarta, Indonesia; INGESTI BILKIS ZULPATINA, student, SMPN 8, Yogyakarta, Indonesia; and the proposer.

M461.

Proposed by Landelino Arboni´ es, Colegio Marcelino Champagnat, Santo Dominigo, Dominican Republic.

A Champagnat number is equal to the sum of all the digits in a set of consecutive positive integers, one of which is the number itself. Thus, 42 is a Champagnat number, since 42 is the sum of all of the digits of 39, 40, 41, 42, 43, 44. Prove that there exist inﬁnitely many Champagnat numbers. Solution by the proposer. We prove that for any n > 6 there is at least one Champagnat number with n +1 digits. Indeed, consider the number 10n and suppose if is not a Champagnat number. Let kn be the greatest number such that the digital sum of the numbers 10n , 10n + 1, 10n + 2, . . . , 10n + kn is less than 10n . Consider now the number N equal to the digital sum ofnall the integers from 10n to 10n + kn + 1 inclusive. Now, since kn is at least 9(10 (since each of the numbers is less than 10n+1 − 1 n+1) which has a digital sum of 9(n + 1)) and N is at most 10n + 9(n + 1) (only if kn + 1 = 10n+1 − 1), then (if n > 6) N is one of the numbers between 10n and 10n + kn + 1 inclusive, and hence it is a Champagnat number being the sum of a set of consecutive numbers, one of which is itself.
No other solutions were received.

M462.

Proposed by Alex Song, Detroit Country Day School, Detroit, MI, USA and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON. Let ⌊x⌋ denote the greatest integer not exceeding x and let ⌈x⌉ denote the smallest integer greater than or equal to x. For example, ⌊3.1⌋ = 3, ⌈3.1⌉ = 4, ⌊−1.4⌋ = −2, and ⌈−1.4⌉ = −1. Determine all real numbers x for which ⌊x⌋⌈x⌉ = x2 . Solution by Ricard Peir´ o, IES “Abastos”, Valencia, Spain. If x = n ∈ Z, then ⌊x⌋ = n and ⌈x⌉ = n. Hence, ⌊x⌋⌈x⌉ = n2 = x2 for all x ∈ Z. If x ∈ Z and x > 0, then there exists n ∈ N ∪ {0} such that n < x < n + 1. We can then conclude that ⌊x⌋ = n and ⌈x⌉ = n + 1. Consequently, ⌊x⌋⌈x⌉ = n(n + 1) = x2 , hence x = n(n + 1). If x ∈ Z and x < 0, then there exists n ∈ N ∪ {0} such that −(n + 1) < x < −n. We can then conclude that ⌊x⌋ = −(n + 1) and ⌈x⌉ = −n. Consequently,

273

numbers for which ⌊x⌋⌈x⌉ = x2 is x = ±n or x = ± n(n + 1), n ∈ N ∪ {0}.
Also solved by PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; and the proposers. Three incomplete solutions were received.

⌊x⌋⌈x⌉ = n(n + 1) = x2 , hence x = − n(n + 1). Thus, the set of all real

Problem of the Month
Ian VanderBurgh
Many problems that appear on contests are word problems. problem that appeared last year on a Scottish competition: Here is a

Problem (2010-2011 Scottish Mathematical Challenge) Katie had a collection of red, green and blue beads. She noticed that the number of beads of each colour was a prime number and that the numbers were all diﬀerent. She also observed that if she multiplied the number of red beads by the total number of red and green beads she obtained a number exactly 120 greater than the number of blue beads. How many beads of each colour did she have? Often, the ﬁrst step with a word problem is to translate the words into mathematics. Since this problem is dealing with the numbers of red, green and blue beads, let’s assign a variable to each of these numbers – say, r , g and b, respectively. (We’ll write this up nicely in a minute.) These seem to be the relevant quantities. We are next told that each of these quantities is a prime number. Let’s make a mental note to come back to this, and keep reading. The fact that the product of the number of red beads with the sum of the numbers of red and green beads is 120 more than the number of blue beads translates into the equation r (r + g ) = 120 + b. Now, I seem to remember that usually when we have three variables, one equation is not enough to determine the values of the variables. (Often, we need three equations.) This is mildly concerning, but let’s persevere to see what happens. What information haven’t we used? We haven’t used the fact that each of r , g and b is a prime number. How can we use this information? Again, let’s back up half a step. What do we know about prime numbers? It’s good to check the deﬁnition ﬁrst: a prime number is a positive integer larger than 1 (remember, 1 is not prime) that has no positive divisors other than 1 and itself. Is there a “formula” for prime numbers? There isn’t a good one that we know. However, there are lots and lots of properties of prime numbers: all prime numbers other than 2 are odd, there are inﬁnitely many prime numbers, every prime number

274

greater than 3 is either one more or one less than a multiple of 6. . . The list goes on! Many mathematicians spend much of their professional lives investigating properties of prime numbers. Given such a vast number to choose from, how do we know what properties to use? Therein lies the essence of problem solving! Figuring this out is not always easy. Here’s a solution to the problem. Solution Suppose that r , g and b are the numbers of red, green and blue beads, respectively. We are told that each of r , g and b is a prime number and that r (r + g ) = 120 + b. Let’s focus on the fact that the only even prime number is 2 and on the parity of the two sides of the equation. (Remember, checking parity means checking to see if an integer is even or odd.) If both r and g are odd, then r + g is even, so the left side of the equation is even, which means the right side is even. If 120 + b is even, then b is even, which means that b = 2. In this case, r (r + g ) = 122. Since 122 = 2 × 61 and each of 2 and 61 is prime, then we must have r = 2 or r + g = 2. Neither of these is possible, since r cannot equal b (since r , b and g are all diﬀerent) and since r + g is at least 4. Also, since 2 is the only even prime number, then r and g can’t both be even, since then they would both be 2, which would contradict the given hypothesis that r , b and g are all diﬀerent. Therefore, r and g are even and odd in some order. In other words, one of r and g equals 2 and the other is an odd prime number. Which is which? If r = 2, then the equation becomes 2(2 + g ) = 120 + b. Since the left side is even, then the right side is even too, so again b = 2 which is impossible since our assumption is that r = 2. Thus, it must be the case that g = 2 and r is an odd prime. This gives r (r + 2) = 120 + b. So we’ve got one value (g = 2), but still have one equation and two unknowns. What to do? Let’s try solving for b, which gives b = r 2 + 2r − 120. At this point, it might occur to try to factor the right side to obtain b = (r + 12)(r − 10). How does this help? Since b is a prime number, then it can’t be factored in many ways! Aha – that is probably useful. If b is a prime number that is written as the product of two integers, then one of the factors is either 1 or −1. This gives us four possibilities to check (r + 12 equals 1 or −1 and r − 10 equals 1 or −1). The only one that yields a positive value of r that is a prime number is r − 10 = 1, giving r = 11. In this case, b = (11 + 12)(11 − 10) = 23, which is (thankfully) a prime number. Therefore, g = 2, r = 11 and g = 23. We can check that these satisfy the original hypotheses.

275

THE OLYMPIAD CORNER
No. 295 R.E. Woodrow and Nicolae Strugaru
The problems from this issue come from the Italian Team Selection Test, the British Mathematical Olympiad, the Macedonian Mathematical Olympiad, the China Western Mathematical Olympiad, the Austrian Mathematical Olympiad, the Olimpiadi Italiane della Matematica and the Chinese Mathematical Olympiad. Our thanks go to Adrian Tang for sharing the material with the editor.
The solutions to the problems are due to the editor by 1 August 2012. Each problem is given in English and French, the oﬃcial languages of Canada. In issues 1, 3, 5, and 7, English will precede French, and in issues 2, 4, 6, and 8, French will precede English. In the solutions’ section, the problem will be stated in the language of the primary featured solution. The editor thanks Jean-Marc Terrier of the University of Montreal for translations of the problems.

OC21. A sequence of real numbers {an} is deﬁned by a0 = 0, 1, a1 = 1 − a0,

and an+1 = 1 − an (1 − an ) for n = 1, 2, · · · . Prove that for any positive integer n, we have 1 1 1 a0 a1 · · · an = 1. + + ··· + a0 a1 an

OC22.

Consider a standard 8 × 8 chessboard consisting of 64 small squares coloured in the usual pattern, so 32 are black and 32 are white. A zig-zag path across the board is a collection of eight white squares, one in each row, which meet at their corners. How many zig-zag paths are there?

OC23.

Determine all nonnegative integers n such that n(n − 20)(n − 40)(n − 60) · · · r + 2009

is a perfect square where r is the remainder when n is divided by 20.

OC24. Let O be the circumcentre of the triangle ABC . Let K and L be the intersection points of the circumcircles of the triangles BOC and AOC with the bisectors of the angles at A and B respectively. Let P be the midpoint of KL, M symmetrical to O relative to P and N symmetrical to O relative to KL. Prove that KLM N is cyclic. OC25.
Show that the inequality 3n > (n!)4 holds for all positive integers n.
2

276

OC26.
satisfy

Find all functions f from the real numbers to the real numbers which f x3 + f y 3 = (x + y )(f x2 + f y 2 − f (xy ))

for all real numbers x and y .

OC27.

A natural number k is said to be n-squared if, for every colouring of the squares in a chessboard of size 2n × k with n colours, there are 4 squares with the same colour whose centres are the vertices of a rectangle with sides parallel to the sides of the chessboard. For any given n, ﬁnd the smallest natural number k which is n-squared. A ﬂea is initially at the point (0, 0) of the Euclidean plane. It then takes n jumps. Each jump is taken in any of the four cardinal directions (north, east, south or west). The ﬁrst jump has length 1, the second jump has length 2, the third jump has length 4, and so on, the nth jump has length 2n−1 . Prove that if we know the number of jumps and the ﬁnal position, we can uniquely determine the path the ﬂea took.

OC28.

OC29.
satisfying

Let n ≥ 3 be a given integer, and a1 , a2 , · · · , an be real numbers
n 1≤i<j ≤n

min

|ai − aj | = 1. Find the minimum value of

k=1

|ak |3 .

OC30.

Let P be an interior point of a regular n-gon A1 A2 · · · An . The lines Ai P meet A1 A2 · · · An at another point Bi , where i = 1, 2, · · · , n. Prove that
n i=1 n

P Ai ≥

P Bi .
i=1

.................................................................

et an+1 = 1 − an (1 − an ) pour n = 1, 2, · · · . Montrer que pour tout entier positif n, on a a0 a1 · · · an 1 a0 + 1 a1 + ··· + 1 an = 1.

OC21. On d´ eﬁnit une suite de nombres r´ eels {an } par a0 = 0, 1, a1 = 1 − a0 ,

OC22. On consid` ere un ´ echiquier standard 8 × 8 form´ e de 64 petits carr´ es de couleur et disposition usuelles, soit 32 blancs et 32 noirs. Un chemin en zigzag sur l’´ echiquier est une collection de huit carr´ es blancs, un par rang´ ee, se touchant ` a leurs coins. Combien de chemins en zigzag y-a-t’il ?

277

OC23. Trouver tous les entiers non n´ egatifs n tels que
n(n − 20)(n − 40)(n − 60) · · · r + 2009 soit un carr´ e parfait o` u r est le reste de la division de n par 20.

OC24. Soit O le centre du cercle circonscrit du triangle ABC . Soit K et L les points d’intersection respectifs des cercles circonscrits BOC et AOC avec les bissectrices des angles en A et B . Soit P le milieu de KL, M le sym´ etrique de O par rapport ` a P et N le sym´ etrique de O par rapport a ` KL. Montrer que les points KLM N sont cocycliques. OC25. Montrer que l’in´ egalit´ e 3n
positifs n.
2

> (n!)4 est valable pour tous les entiers

OC26. Trouver toutes les fonctions f d’une variable a ` valeurs r´ eelles et satisfaisant
f x3 + f y 3 = (x + y )(f x2 + f y 2 − f (xy )) pour tous les nombres r´ eels x et y .

OC27. Un nombre naturel k est appel´ e n-carr´ e si, pour tout coloriage des cases
d’un ´ echiquier de dimension 2n × k avec n couleurs, il y a 4 cases de mˆ eme couleur dont les centres sont les sommets d’un rectangle dont les cˆ ot´ es sont parall` eles ` a ceux de l’´ echiquier. Pour tout n donn´ e, trouver le plus petit nombre naturel k qui soit n-carr´ e.

OC28. On imagine une puce ` a l’origine (0, 0) du plan euclidien. La voil` a qui
eﬀectue n sauts. Chaque saut a lieu dans l’une quelconque des quatre directions cardinales (nord, est, sud ou ouest). Les longueurs des sauts cons´ ecutifs sont, dans l’ordre, de 1, 2, 4 et ainsi de suite, le n-i` eme saut ´ etant de 2n−1 . Montrer que si l’on connaˆ ıt le nombre de sauts et la position ﬁnale, on peut en d´ eduire univoquement le chemin suivi par la puce.

OC29. On donne un entier n
satisfaisant
1≤i<j ≤n

≥ 3 et soit a1 , a2 , · · · , an des nombres r´ eels
n k=1

min

|ai − aj | = 1. Trouver la valeur minimale de

|ak |3 .

OC30. Soit P un point int´ erieur d’un polygone r´ egulier A1 A2 · · · An ` a n cˆ ot´ es.
Les droites Ai P coupent ce polygone en un autre point Bi , o` u i = 1, 2, · · · , n. Montrer que
n i=1 n

P Ai ≥

P Bi .
i=1

278

Next we turn to our ﬁle of solutions from readers to problems of the Youth Mathematical Olympiad of the Asociaci´ on Venezolana de Competencias Matem´ aticas, 2006, given at [2009 : 380].

A positive integer has 223 digits and the product of these digits is 3446 . What is the sum of the digits? Solved by Geoﬀrey A. Kandall, Hamden, CT, USA; David E. Manes, SUNY at Oneonta, Oneonta, NY, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Manes. Let n be the positive integer with 223 digits. Then the maximum product of the digits of n is 9223 = 3446 when all of the digits of n are nines. Therefore, the digits of n consist of 223 nines and so, the sum of the digits is 9(223) = 2007.

Find all solutions of the equation m2 − 3m + 1 = n2 + n − 1, where m and n are positive integers. Solved by Michel Bataille, Rouen, France; Geoﬀrey A. Kandall, Hamden, CT, USA; David E. Manes, SUNY at Oneonta, Oneonta, NY, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Bataille. The solutions are the pairs (m, n) = (a +2, a) where a is a positive integer. The equation rewrites as (m− 3 )2 − 5 = (n+ 1 )2 − 5 or (m− 3 )2 −(n+ 1 )2 = 0 2 4 2 4 2 2 that is, (m + n − 1)(m − n − 2) = 0. Thus positive integers m, n are solutions if and only if m − n − 2 = 0. The result follows.

Deﬁne the sequence a1 , a2 , a3 , . . . as follows: let a1 = a2 = 1003; a3 = a2 − a1 = 0; a4 = a3 − a2 = −1003; and in general an+1 = an − an−1 for any n ≥ 2. Compute the sum of the ﬁrst 2006 terms of the sequence. Solved by Oliver Geupel, Br¨ uhl, NRW, Germany; Geoﬀrey A. Kandall, Hamden, CT, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution by Geupel. Let us compute the next items of the sequence: a5 = a4 − a3 = −1003 , a7 = a6 − a5 = 1003 , a6 = a5 − a4 = 0 ,
È6

a8 = a7 − a6 = 1003 .

We see that the sequence is periodic with period 6 and n=1 an = 0. Noticing that 2006 ≡ 2 (mod 6), we ﬁnd that the sum of the ﬁrst 2006 items is
2006

an = a1 + a2 = 2006.
n=1

279

AB − BC

Let ABC be an isosceles triangle with ∠B = ∠C = 72◦ . Find the value of BC . Hint: Consider the bisector CD of ∠ACB .

Solved by Geoﬀrey A. Kandall, Hamden, CT, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution by Kandall. Let CD be the bisector of ∠ACB , and let a = BC , b = AB = AC . We have ∠BAC = 36◦ and ∠BDC = 72◦ . Therefore, AD = CD = BC = a, so BD = b − a. Triangles ABC and CDB are similar, hence b a = b− ≡ λ. (Note that λ > 0.) a a
a 1 b We have λ = b− = a − 1 = λ − 1, hence a BC a 2 λ − λ − 1 = 0. Consequently, AB = b− = −BC a

A
36◦

a D b−a B
72◦ 72◦

λ=

√ 1+ 5 . 2

36◦ 36◦

To ﬁnish the material from the October 2009 ﬁles we turn to solutions from our readers to problems of the 42nd Mongolian Mathematical Olympiad, 10th Grade, given at [2009 : 380–381].

Let a, b, c, d, e, and f be positive integers satisfying the relation ab + ac + bc = de + df + ef , and let N = a + b + c + d + e + f . Prove that if N | (abc + def ), then N is a composite number. Solution by Jan Verster, Kwantlen University College, BC. Expanding the following polynomial, and using the relation ab + ac + bc = de + df + ef , we get (x + a)(x + b)(x + c) − (x − d)(x − e)(x − f )

= x3 + (a + b + c)x2 + (ab + ac + bc)x + abc = N x2 + abc + def − x3 + (d + e + f )x2 − (de + df + ef )x + def

Then, if we let x = d, say, this becomes (d + a)(d + b)(d + c) = N d2 + (abc + def ) Thus, if N |(abc + def ), we also have N |(d + a)(d + b)(d + c). Let p be a prime such that p|N . Then p must divide at least one of d + a, d + b or d + c. Thus p ≤ max(d + a, d + b, d + c) < N , so p is a proper factor of N , and N must be composite.

280

Now we turn to solutions from readers to problems of the Olympiade Suisse de math´ ematiques 2005, tour ﬁnal, given at [2009 : 82–83].

Soint ABC un triangle aigu. Soient M et N des points arbitraires sur les cˆ ot´ es AB et AC respectivement. Les cercles de diam` etre BN et CM se coupent en P et Q. Montrer que les points P , Q et l’orthocentre du triangle ABC se trouvent sur une droite. Solved by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; and Michel Bataille, Rouen, France. We give the write-up of Amengual Covas. Let Ω, Ω1 , and Ω2 the circles on H BC , BN , and CM as diameters, respectively. V Since Ω1 and Ω2 intersect at P and Q, U P the line P Q is the radical axis of Ω1 and A Ω2 . Let H denote the orthocenter of △ABC and let U and V be the feet of the altitudes from B and C respectively. Since ∠BU N = ∠BU C = 90◦ , both the circles Ω and Ω1 pass through U . Hence the line BU is the radical axis of Ω and Ω1 . M B Ω1 Q N C

Ω2 Similarly, the line CV is the radical axis Ω of Ω and Ω2 . Since BU and CV intersect at H , the orthocenter of △ABC is the radical center of Ω, Ω1 , and Ω2 . Hence H lies on the radical axis of Ω1 and Ω2 , that is, H lies on the line P Q. This completes the proof of the collinearity of P , Q and the orthocenter of △ABC . As shown in the proof, the condition △ABC acute is not necessary.

Next we ﬁnish up the solutions to problems of the 55th Czech and Slovak Mathematical Olympiad 2006 given at [2009: 81–82].

ˇ cek, P. Cal´ (J. Svrˇ abek) Solve in real numbers the system of equations tan2 x + 2 cot2 2y tan2 y + 2 cot2 2z tan2 z + 2 cot2 2x = 1, = 1, = 1.

(1)

Solved by Arkady Alt, San Jose, CA, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give Zelator’s write-up.

281

First note that the real numbers x = kπ ± π , y = mπ ± π , z = nπ ± π ; 4 4 4 where k, m, n are (arbitrary) integers; are solutions of the system (1). We see, by inspection, that any solutions must satisfy 0 ≤ tan2 x, tan2 y , 1 2 tan z ≤ 1 and 0 ≤ cot2 2y , cot2 2x, cot2 2z ≤ 2 . However, in fact none of tan x, tan y , tan z can be zero, since if tan x = 0; then, and only then x = tπ , for some t ∈ Z. But then 2x = 2tπ ; and so cot 2x would be undeﬁned. Thus, 0 < tan2 x , tan2 y , tan2 z ≤ 1 and 0 ≤ cot2 2y, cot2 2x, cot2 2z < 1 2 (2)

Next, observe if one of tan2 x, tan2 y , tan2 z is equal to 1; then all three of them are and the three cotangent terms are zero. Indeed, suppose that tan2 x = 1; then tan x = 1 or −1; which implies x = kπ ± π ; k ∈ Z. But then from the ﬁrst 4 equation in (1) we obtain cot 2y = 0; 2y = ρπ + π , ρ ∈ Z; y = ρ π+ π ; when 2 2 4 π π . ρ = even = 2λ; y = λπ + 4 ; and when ρ = odd = 2λ + 1; y = λπ + 34 3π π These two formulas, since 4 = π − 4 ; can be condensed into one: y = mπ ± π ; 4 m ∈ Z. Then tan2 y = 1; and from the second equation in (1), we obtain cot 2z = ; n ∈ Z. 0; and from which (via a similar argument) we obtain z = nπ ± π 4 2 2 2 We conclude that either tan x = tan y = tan z = 1, which produces the solutions x = kπ ± π , y = mπ ± π , z = nπ ± π . Or alternatively, 4 4 4 0 < tan2 x , tan2 y , tan2 z < 1 and 0 < cot2 2y , cot2 2x , cot2 2z <
1 2

(3)

Below, we shall ﬁnd all the real solutions to system (1) that satisfy the conditions in (3). Suppose (x1 , y1 , z1 ) is a solution satisfying (3). We put r1 = tan2 x1 , r2 = tan2 y1 , and r3 = tan2 z1 . So that by (3), 0 < r1 = tan2 x1 , r2 = tan2 x1 , r3 = tan2 z1 < 1 1 and 0 < cot2 2y1 , cot2 2x1 , cot2 2z1 < 2 Since (x1 , y1 , z1 ) satisﬁes the system (1) we have, tan2 x1 + 2 cot2 2y1 tan2 y1 + 2 cot2 2z1 tan2 z1 + 2 cot2 2x1 = = = 1 1 1 (5) (4)

Consider the ﬁrst equation in (12) and multiply across by tan2 2y1 in order to obtain, since cot2 2y1 · tan2 2y1 = 1, tan2 2y1 · tan2 x1 + 2 = tan2 2y1 . Applying the double-angle identity tan 2y1 = 2(1 − tan2 y1 )2 = 4 tan2 y1 .
2 tan y1 1−tan2 y1

gives 4 tan2 y1 tan2 x1 +

⇔ tan4 y1 + 2 tan2 y1 tan2 x1 − 4 tan2 y1 + 1 = 0;

282

2 (with r1 = tan2 x1 , r2 = tan2 y1 ) r2 + 2r2 r1 − 4r2 + 1 = 0. Working similarly with the second and third equations in (12) we altogether obtain 2 r2 + 2 r2 r1 − 4 r2 + 1 = 0 2 r3 + 2 r3 r2 − 4 r3 + 1 = 0 2 r1 + 2 r1 r3 − 4 r1 + 1 = 0

(6)

Adding the three equations in (6) yields (r1 + r2 + r3 )2 − 4(r1 + r2 + r3 ) + 3 = 0; or equivalently, [(r1 + r2 + r3 ) − 1][(r1 + r2 + r3 ) − 3] = 0 . (7)

By (4), it follows that 0 < r1 + r2 + r3 < 3; which implies in conjunction with (7) that r1 + r2 + r3 = 1 (8) Now, we go to (6) and we multiply the ﬁrst equation by r1 r3 , the second equation by r1 r2 ; and the third equation by r2 r3 , in order to obtain
2 2 r1 r3 r2 + 2 r2 r3 r1 − 4 r1 r2 r3 + r1 r3 2 2 r1 r2 r3 + 2 r1 r3 r2 − 4 r1 r2 r3 + 4 1 4 2 2 2 r2 r3 r1 + 2 r1 r2 r3 − 4 r1 r2 r3 + r2 r3

= 0 = 0 = 0 (9)

We add the three equations in (9) to get, r1 r2 r3 (r1 + r2 + r3 ) + 2r1 r2 r3 (r1 + r2 + r3 ) −12r1 r2 r3 + (r1 r3 + r1 r2 + r2 r3 ) = 0; and by (8) we arrive at r1 r3 + r1 r2 + r2 r3 = 9r1 r2 r3 = 9P (product). By the Arithmetic-Geometric Mean inequality we also have, r1 r3 + r1 r2 + r2 r3 ≥ 3 and by (10); 9P ≥ 3P 2/3 ⇔ P 1/3 ≥ or equivalently, P ≥ 1 27 .
3

(10)

(r1 r3 )(r1 r2 )(r2 r3 ); 1 3

;

(11)

283

Next, consider the cubic polynomial of degree three, f (t) = (t − r1 )(t − r2 )(t − r3 ); f (t) = t3 − (r1 + r2 + r3 )t2 + (r1 r2 + r2 r3 + r3 r1 )t − r1 r2 r3 ; and by (8) and (10); f (t) = t3 − t2 + 9P · t − P . (13) (12)

Next, we make use of the following lemma. The facts stated in the Lemma are well-known, standard material on cubic polynomial functions. Lemma 1. Suppose that g (t) is a cubic polynomial function of degree 3 with leading coeﬃcient a > 0; and let g ′ (t) be the derivative of g (t); g ′ (t) being a quadratic trinomial. (i) If the discriminant of g ′ (t) is negative; then the function g (t) has no critical numbers in its domain, and therefore it has no points of local maximum or local minimum. The function g (t) is increasing throughout R, and has only one inﬂection point. And the function g (t) has exactly one real root, and two conjugate complex roots. (ii) If the discriminant of g ′ (t) is zero; then the function g (t) has exactly one critical number ρ in its domain R. The point (ρ, g (ρ)) is both an inﬂection point and a critical point on the graph of g (t). The function g (t) increases throughout R, it has no points of local maximum or minimum. Furthermore g (t) has the form, g (t) = a(t − ρ)3 + κ, for some κ ∈ R. Thus, g (t) has a triple real root, the real number r =
3

−

κ a

+ ρ.

Consider the derivative of f (t) in (13): f ′ (t) = 3t2 − 2t + 9P . The discriminant of f ′ (t) is D = 4 − 4 · 3 · 9 · P = 4(1 − 27P ). By (11) it follows that D = 0 or D < 0. The second possibility, D < 0 is eliminated by Lemma 1(i), since according to (12), f (t) has three real roots. Thus we must have D = 0. By Lemma 1(ii), it follows that r1 = r2 = r3 ; and therefore from (8) we 1 obtain r1 = r2 = r3 = 3 . Back to (4): tan2 x1 = 1 1 ⇔ tan x1 = ± √ ; 3 3 x1 = kπ ± π ; 6 for some k ∈ Z .

Likewise, we obtain y1 = mπ ± and z1 = nπ ± π 6 π 6 , for m ∈ Z , for n ∈ Z .

284

Conversely, one can verify directly that if the triple (x1 , y1 , z1 ) has the above form; then it satisﬁes the system (1). Conclusion. The set of solutions S of system (1) is the union of two disjoint families or solution sets S1 , S2 ;

S = S1 ∪ S2 , where S1 = {(x, y, z ) | x, y, z ∈ R and such that x = kπ ± π , y = mπ ± π , 4 4 π z = nπ ± 4 }; where m, n, k can be any integers; and all eight combinations of signs are allowed. And S2 = {(x, y, z ) | x, y, z ∈ R and such that x = kπ ± π , 6 π y = mπ ± π , z = nπ ± } ; where m , n , k can be any integers; and all eight 6 6 combinations of signs are allowed.

7. (I. Voronovich) The point K (distinct from the orthocentre) lies on the altitude
CC1 of the acute triangle ABC . Prove that the feet of the perpendiculars from C1 to the segments AC , BC , BK , and AK lie on a circle. Solution by Michel Bataille, Rouen, France. We denote by D , U , V , E the feet of the perpendiculars from C1 to AC , AK , BK , BC , respectively, and by H the orthocentre of ∆ABC . First, we show that D, U, V, E are not collinear. Assuming the contrary, let BK meet AC at B1 . From Simson’s theorem, C1 lies on the circumcircle of ∆AKB1 . Since ∠AC1 K = 90◦ , it follows that AK is a diameter of the circle (AKB1 ) and so ∠KB1 A = 90◦ . Thus BB1 is the altitude from B of ∆ABC and K is its orthocentre, contradicting the hypothesis. C B1 U D C1 A

K V

If CA = CB , since the line CC1 is an axis of symmetry of the ﬁgure, DU V E is an isosceles trapezium and D, U, V, E are concyclic. From now on, we assume that CA = CB . Let X be the point of intersection of the lines DE and AB . Since the circle 2 with diameter CC1 passes through D and E , we have XD · XE = XC1 . On the other hand, since ∠XEB = ∠CED = ∠CC1 D = ∠DAX , the triangle XBE and XDA are similar.

285

X Y B

E C1 A

It follows that XD · XE = XA · XB and so
2 XA · XB = XC1 .

(1)

Now, if Y is the pole of the line CC1 with respect to the circle with diameter AB , points Y, C1 are harmonic conjugates with respect to A, B and so (1) characterizes X as being the midpoint of Y C1 . Reasoning in the same way with triangle KAB instead of CAB , we see that the line U V passes through X as well and 2 that XU · XV = XC1 = XD · XE. Since U, V, D, E are not collinear, the relation XU · XV = XD · XE implies that U, V, D, E are concyclic, as required. (E. Barabanov, V. Kaskevich, S. Mazanik, I. Voronovich) An equilateral triangle of side n is divided into n2 unit equilateral triangles by lines parallel to its sides. Determine the smallest possible number of small triangles that must be marked so that any unmarked triangle has at least one side in common with a marked triangle. Solution by Oliver Geupel, Br¨ uhl, NRW, Germany. Let Tn be the smallest number of marked triangles. We prove: Tn = n . 4 Since each marked triangle has at most three neighboring unmarked 2 . Vice versa, we show triangles, we have 3Tn ≥ n2 − Tn ; hence Tn ≥ n 4 that n2 Tn ≤ . (1) 4 The proof is by induction. The relation (1) holds for n ∈ {1, 2}; see Figure 1 for n = 2. Assume that(1) holds for each n ∈ {1, 2, . . . , N − 1}. Let ∆k denote an equilateral triangle of side length k. We make three cases. Figure 1 Case 1: N is even. The triangle ∆N can be covered by one ∆N −2 and N − 1 triangles ∆2 (Figure 2). By induction, TN ≤ TN −2 + (N − 1)T2 ≤ N 2
2
2

−1

+N −1 =

N2 4

286

∆2k+2 ∆ N −2 ∆2 ∆2 ... ∆2 ∆2 ∆2k ∆2k+2 ∆2k+2 Figure 3

∆2k+2 ∆2k

∆2k

Figure 2

Figure 4

Case 2: N = 4k + 3 (k ≥ 0). The triangle ∆N can be covered by one ∆2k and three ∆2k+2 (Figure 3). By induction, TN ≤ T2k + 3T2k+2 ≤ k2 + 3(k + 1)2 = (4k + 3)2 4 .

Case 3: N = 4k + 1 (k ≥ 1). The triangle ∆N can be covered by three ∆2k and one ∆2k+2 (Figure 4). By induction, TN ≤ 3T2k + T2k+2 ≤ 3k2 + (k + 1)2 = This completes the proof. (4k + 1)2 . 4

Next we look at readers’ solutions to problems given in the November 2009 number of the Corner and the 24th Iranian Mathematical Olympiad, First Round, given at [2009 : 435].

Given integers m > 2 and n > 2, prove there is a sequence of integers a0 , a1 , . . . , ak such that a0 = m, ak = n, and (ai + ai+1 ) | (ai ai+1 + 1) for each i = 0, 1, . . . , k − 1. Solution by Titu Zvonaru, Com´ ane¸ sti, Romania. Taking a0 = m, a1 = 1, a2 = 1, . . . , ak−1 = 1, ak = n we have a0 + a1 = m + 1; a1 + a2 = 2; . . . a0 a1 + 1 = m + 1 a1 a2 + 1 = 2 . . .

ak−2 + ak−1 = 2; ak−2 ak−1 + 1 = 2 ak−1 + ak = n + 1; ak−1 ak + 1 = n + 1 hence (ai + ai+1 ) | (ai ai+1 + 1) for each i = 0, 1, . . . , k − 1.

287

Find all two-variable polynomials p(x, y ) with real coeﬃcients such that p(x + y, x − y ) = 2p(x, y ) for all real numbers x and y .

Solution by Arkady Alt, San Jose, CA, USA. Note that p(2x, 2y ) = p((x + y ) + (x − y ), (x + y ) − (x − y )) = 2p(x + y, x − y ) = 4p(x, y ). Excluding the trivial case p(x, y ) ≡ 0 we assume further that p(x, y ) = 0. Since p(0, 0) = p(2 · 0, 2 · 0) = 4p(0, 0) then p(0, 0) = 0 and p(x, y ) is not constant, moreover p(x, 0) and p(0, y ) are not constants. Note that any such two-variable polynomial p(x, y ) can be represented in the form p(x, y ) = A(x) + B (y ) + xyC (x, y ), where deg A(x) = n > 0, deg B (y ) = m > 0, more precisely A(x) = p(x, 0) = an xn + an−1 xn−2 + . . . + a1 x, B (y ) = p(0, y ) = bm xm + bm−1 xm−1 + . . . + b1 x, where an = 0, bm = 0. Since p(2x, 2y ) = 4p(x, y ) then in particular for y = 0 and any real x we have p(2x, 0) = 4p(x, 0) if and only if 2n an = 4an , 2n−1 an−1 = 4an−1, . . . , 2a1 = 4a1 if and only if n = 2, k1 = 0, Similarly we obtain m = 2, b1 = 0. Thus, p(x, y ) = ax2 + xyC (x, y ) + by 2 . Since p(x, x) = x2 (a + b + C (x, x)) and p(2x, 2x) = 4p(x, x) then for x = 0 we have 4x2 (a + b + C (2x, 2x)) = 4x2 (a + b + C (x, x)) if and only if C (2x, 2x) = C (x, x) if and only if C (x, x) is constant. Indeed, since C (x, x) = c + c1 x2 + · · · + ck x2k then C (2x, 2x) = C (x, x) if and only if ci = 22i ci , i = 1, 2, . . . , k if and only if ci = 0, i = 1, 2, . . . , k. So, p(x, y ) = ax2 + cxy + by 2 and since p(x + y, x − y ) = 2p(x, y ) if and only if a(x + y )2 + c(x2 − y 2 ) + b(x − y )2 = 2ax2 + 2cxy + 2by 2 if and only if (b + c − a)x2 + (a − b − c)y 2 + 2(a − c − b)xy = 0 for any x, y then c = a − b. Therefore, p(x, y ) = ax2 + (a − b)xy + by 2 and all such two-variable polynomials p(x, y ) of the second degree satisfy p(x + y, x − y ) = 2p(x, y ). Indeed, p(x + y, x − y ) = ax2 + 2axy + ay 2 + bx2 − 2bxy + by 2 + (b − a)(x2 − y 2 ) = 2(ax2 + (a − b)xy + by 2 ) = 2p(x, y ).

Let ω1 and ω2 be two circles such that the centre of ω1 is located on ω2 . If the circles intersect at M and N , AB is an arbitrary diameter of ω1 , and A1 and B1 are the second intersections of AM and BN with the circle ω2 (respectively), prove that A1 B1 is equal to the radius of ω1 . Solution by Michel Bataille, Rouen, France. The following lemma will be used twice: Let C and C ′ be two circles intersecting at U, V . If points P, Q on C and P ′ , Q′ on C ′ are such that P, U, P ′ and Q, V, Q′ are collinear, then P Q and P ′ Q′ are parallel. Q′ Q C P U V C′ P′

288

Indeed, denoting by ∠(·, ·) the directed angle of lines, we have (modulo π ) ∠(P Q, P ′ Q′ ) = ∠(P Q, P U ) + ∠(P ′ U, P ′ Q′ ) = ∠(V Q, V U ) + ∠(V U, V Q′ ) = ∠(V Q, V Q′ ) = 0 where the second equality follows from the concyclicity of P , Q, U , V and P ′ , Q′ ,U ,V . A1 This lemma clearly implies that A1 B1 ω M and AB are parallel. Now, consider A B1 the inversion I in the circle ω1 . Since I(M ) = M and I(N ) = N , I A′ exchanges the line M N and the circle O1 ω2 . It follows that I(A1 ) = A′ is the ω1 B ω2 point of intersection of M N and O1 A1 . N We remark that A′ is also on the the circumcircle ω of ∆O1 AM , which is the image of AM under I. Applying again the lemma, but this time to the circles ω and ω1 , we obtain BN O1 A′ . As a result, we have A1 B1 O1 B and BB1 O1 A1 so that A1 B1 BO1 is a parallelogram and A1 B1 = O1 B , the radius of ω1 .

Next we look at solutions from the ﬁle for the 24th Iranian Mathematical Olympiad, Third Round, given at [2009 : 437–438].

Does there exist a sequence of positive integers a0 , a1 , a2 , . . . such that
n È

gcd(ai , aj ) = 1 whenever i = j , and for all n the polynomial irreducible in Z[x]? Solved by Mohammed Aassila, Strasbourg, France.

ai xi is

i=0

Consider, for example, an = pn where (pn ) is a sequence of prime numbers such that pn > p0 + p1 + · · · + pn−1 . Then, by this condition, all the roots of Èn i i=0 pi z lie inside the unit circle |z | < 1.

Triangle ABC is isosceles with AB = AC . The line ℓ passes through A and is parallel to BC . The points P and Q are on the perpendicular bisectors of AB and AC , respectively, and such that P Q ⊥ BC . The points M and N are on ℓ and such that ∠AP M and ∠AQN are right angles. Prove that
1 1 2 + ≤ . AM AN AB

Solved by Mohammed Aassila, Strasbourg, France; Michel Bataille, Rouen, France; and Oliver Geupel, Br¨ uhl, NRW, Germany. We give the solution by Bataille.

289

Q 1 AH 1 AH = and = . B C AM AP 2 AN AQ2 Now, let O denote the circumcentre of ∆ABC . Then, from ∠OP Q = ∠ABC = B and ∠P QO = ∠ACB = C = B (acute angles with perpendicular sides) we deduce OP = OQ and ∠P OA = B, ∠AOQ = 180◦ − B . Using the law of cosines and denoting by R the circumradius of ∆ABC , it follows that
AP 2 = OP 2 + R2 − 2R · OP · cos B

Let H be the orthogonal projection of P (or Q) onto ℓ. Since triangles ∆AP M and ∆AQN are rightangled at P and Q, respectively, we have AP 2 = AH · AM and AQ2 = AH · AN , hence

ℓ

= (OP − R)2 + 2R · OP (1 − cos B ) = (OP − R)2 + 4R · OP sin2 (B/2)

AQ2 = OP 2 + R2 + 2R · OP · cos B

= (OP − R)2 + 2R · OP (1 + cos B ) = (OP − R)2 + 4R · OP cos2 (B/2)

and so AP 2 ≥ 4R · OP sin2 (B/2), Observing that 1 AM + 1 AN AQ2 ≥ 4R · OP cos2 (B/2).

AH = sin ∠(P OA) = sin B , we obtain OP

= =

AH AP 2

2R sin(B/2) 1 2 = . 2R sin(B/2) cos(B/2) AB

AH AQ2

≤

cos(B/2)

sin(B/2) cos(B/2)

(since AB = 2R sin B by the law of sines). Clearly equality holds if and only if OP = OQ = R.

Find all polynomials p(x) of degree 3 such that for all nonnegative real numbers x and y p(x + y ) ≥ p(x) + p(y ) . Solved by Michel Bataille, Rouen, France; and Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA. We give the solution of Bataille. Let us say that p(x) is superadditive if p(x + y ) ≥ p(x) + p(y ) for all nonnegative x and y . We show that the superadditive polynomials of degree 3 are the polynomials ax3 + bx2 + cx + d

290

where c ∈ R, a > 0, d ≤ 0 and 8b3 ≥ 243da2 . Let p(x) = ax3 + bx2 + cx + d where a, b, c, d are real numbers and a = 0 and let δ (x, y ) = p(x + y ) − p(x) − p(y ) = 3axy (x + y ) + 2bxy − d. Clearly, p(x) is superadditive if and only if δ (x, y ) ≥ 0 for all x, y ≥ 0. First, assume that it is the case. Then, δ (0, 0) ≥ 0, hence d ≤ 0. Also, for x > 0, 3a + b x − d 2x3 = δ (x, x) 2x3 ≥0

so that 3a ≥ 0 (by letting x approach inﬁnity) and a > 0 is obtained. Lastly, 8b3 ≥ 243da2 holds if b ≥ 0 (since d ≤ 0) and if b < 0 results from 0≤δ −2b −2b , 9a 9a = 8b3 − 243da2 243a2 .

Conversely, assume that the conditions a > 0, d ≤ 0 and 8b3 ≥ 243da2 hold. Clearly, δ (x, y ) ≥ 0 for all x, y ≥ 0 if b ≥ 0. Suppose that b < 0. Observing that √ √ δ (x, y ) ≥ 3axy · 2 xy + 2bxy − d = φ( xy ) where φ(t) = 6at3 + 2bt2 − d, it suﬃces to show that φ(t) ≥ 0 for t ≥ 0. 2b Since the derivative of φ is given by φ′ (t) = 18at t + 9 , it is readily seen that a
3 2

2b −243da the minimum of φ on [0, ∞) is φ − , which is nonnegative by = 8b 243 9a a2 assumption. Thus, φ(t) ≥ 0 for t ≥ 0 and the proof is complete.

We continue with solutions from our readers to problems given in the May 2010 number of the Corner with the XVIII Olimpiada de Matem´ atica de Paises del Cono Sur given at [2010: 217–218].

Find all the pairs (x, y ) of nonnegative integers that satisfy x3 y + x + y = xy + 2xy 2 .

Solved by George Apostolopoulos, Messolonghi, Greece; Michel Bataille, Rouen, France; Oliver Geupel, Br¨ uhl, NRW, Germany; Geoﬀrey A. Kandall, Hamden, CT, USA; David E. Manes, SUNY at Oneonta, Oneonta, NY, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give Bataille’s write-up. The solutions are the pairs (0, 0), (1, 1), (2, 2). It is readily checked that these pairs are solutions. Conversely, let (x, y ) be any solution. Then we have y = x(y + 2y 2 − 1 − x2 y ) (1)

291

and x = y (x − x3 − 1 + 2xy ).
2 2

(2)

From (1), x divides y (since y + 2y − 1 − x y is an integer) and from (2), y divides x (since x − x3 − 1 + 2xy is an integer). It follows that |x| = |y | and since x, y are nonnegative integers, x = y . Now, from the equation, we must have x4 + 2x = x2 + 2x3 that is, x(x2 − 1)(x − 2) = 0, which implies x ∈ {0, 1, 2}. The result follows.

Given are 100 positive whole numbers whose sum equals their product. Determine the minimum number of occurrences of the number 1 among the 100 numbers. Solved by Henry Ricardo, Tappan, NY, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the version by Zvonaru. Let x100 ≤ x99 ≤ · · · ≤ x2 ≤ x1 be the given 100 positive integers and let kmin be the searched minimum number of 1’s. The equation x100 + x99 + · · · + x2 + x1 = x100x99 . . . x2 x1 can be rewritten as x100 x2 x1 + ··· + + = 1. x100 · x99 · · · x2 x1 x100 · x99 · · · x2 x1 x100 · x99 · · · x2 x1

x1 1 Since x100 ≤ x99 ≤ · · · ≤ x2 ≤ x1 , it results that x100 ·x99 ≥ 100 , ···x2 x1 that is x100 · x99 · · · x2 ≤ 100. We deduce that at most 6 numbers among x100 , x99 , · · · , x2 may be greater than 1, hence kmin ≥ 93. Suppose that kmin = 93. We must solve the equation

93 + x7 + x6 + x5 + x4 + x3 + x2 + x1 = x1 x2 x3 x4 x5 x6 x7 , where x1 ≥ x2 ≥ x3 ≥ x4 ≥ x5 ≥ x6 ≥ x7 ≥ 2. We have 93+7x1 ≥ 93+ x1 + x2 + x3 + x4 + x5 + x5 + x7 = x1 x2 x3 x4 x5 x6 x7 ≥ 26 x1 that is 57x1 ≤ 93, a contradiction with x1 ≥ 2. Suppose now that kmin = 94. As above, we must solve the equation 94 + 6x1 ≥ 94 + x6 + x5 + x4 + x3 + x2 + x1 = x1 x2 x3 x4 x5 x6 . We have 94 + 6x1 ≥ x1 + x2 + x3 + x4 + x5 + x6 = x1 x2 x3 x4 x5 x6 ≥ 32x1 , that is 26x1 ≤ 94.

292

If x1 = 2, then x2 = x3 = x4 = x5 = x6 and we have no solution (because 94 + 6 · 2 = 26 ). If x1 = 3, we have to solve the equation 97 + x6 + x5 + x4 + x3 + x2 = 3x2 x3 x4 x5 x6 . We deduce that 97 + 5x2 ≥ 97 + x2 + x3 + x4 + x5 + x6 = 3x2 x3 x4 x5 x6 ≥ 48x2 and it follows that 43x2 ≤ 97, hence x2 = 2. We obtain x2 = x3 = x4 = x5 = x6 = 2 and we have no solution (because 97 + 5 · 2 = 3 · 32). Since the equation 95 + x1 + x2 + x3 + x4 + x5 = x1 x2 x3 x4 x5 has solution x1 = x2 = x3 = 3, x4 = x5 = 2 (95 + 3 + 3 + 3 + 2 + 2 = 108, 3 · 3 · 3 · 2 · 2 = 108), it results that kmin = 95.

3. Let ABC be an acute triangle with altitudes AD, BE , CF , where D, E, F lie
on BC , AC AB , respectively. Let M be the midpoint of BC . The circumcircle of triangle AEF cuts the line AM at A and X . The line AM cuts the line CF at Y . Let Z be the point of intersection of AD and BX . Show that the lines Y Z and BC are parallel. Solved by Michel Bataille, Rouen, France; Oliver Geupel, Br¨ uhl, NRW, Germany; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give Geupel’s solution. Let α = ∠CAB , β = ∠ABC , and let H be the orthocentre of △ABC . Because of △ABD ∼ △AHF , we have AD : AB = AF : AH , so that AD · AH = AB · AF = AB · AC cos α. For the median line AM , we have 4AM = 2AB + 2AC − BC . By the Law of Cosines it now follows that B M D C
2 2 2 2

F X Y H Z E

BC 2 = 2AB 2 + 2AC 2 − BC 2 − 4AB · AC cos α = 4AM 2 − 4AD · AH ; hence BM 2 = AM 2 − AD · AH.

293

By ∠AEH = ∠AF H = 90◦ , the point H lies on the circumcircle of quadrilateral AEXF . Hence ∠AXH = ∠AF H = 90◦ . Thus △AHX ∼ △AM D , which implies that AH : AX = AM : AD and therefore AM 2 − BM 2 = AD · AH = AM · AX = AM (AM − M X ) = AM 2 − AM · M X .

We obtain BM 2 = AM · M X and therefore AM : BM = BM : XM . Noticing that ∠AM B = ∠BM X , we obtain △ABM ∼ △BXM . Hence, ∠Y XZ = ∠BXM = ∠ABM = β . Also, ∠Y HZ = ∠CHD = ∠DBA = β . We obtain ∠Y XZ = ∠Y HZ , so the quadrilateral HXY Z is cyclic, which establishes ∠HZY = 180◦ − ∠HXY = 90◦ and Y Z BC .

Some cells of a 2007 × 2007 table are coloured. The table is “charrua” if none of the rows and none of the columns are completely coloured. (a) What is the maximum number k of coloured cells that a charrua table can have? (b) For such k, calculate the number of distinct charrua tables that exist. Solved by Titu Zvonaru, Com´ ane¸ sti, Romania. (a) We will determine the minimum number k′ of uncoloured cells that a charrua table can have. It is easy to see that this minimum of uncoloured cells is 2007 (if k′ < 2007, then there is at least a row or a column which is completely coloured). An example of a charrua table with k′ = 2007: the uncoloured cells are c(1, 1), c(2, 2), . . . , c(2007, 2007). It results that the maximum k of coloured cells that a charrua table can have is 2007 × 2007 − 2007 = 2006 × 2007. (b) We can choose in 2007! ways the uncoloured cells, one in each row (but not two in the same columns).

Let ABCDE be a convex pentagon that satisﬁes the following:

(i) There is a circle Γ tangent to each of the sides. (ii) The lengths of the sides are all whole numbers. (iii) At least one of the sides of the pentagon has length 1. (iv) The side AB has length 2. Let P be the point of tangency of Γ with AB . (a) Determine the length of segments AP and BP . (b) Give an example of a pentagon satisfying the given conditions. Solved by Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We print Zvonaru’s solution.

294

(a) Let Q, R, S, T be the points of tangency of Γ with BC , CD , DE , EA respectively. Let AP = AT = x, BP = BQ = y , CQ = CR = z , DR = DS = t, ES = ET = e. We denote by Z the set of integers. Without loss of generality, we may assume that x ≥ y . We have x+y ∈Z y+z ∈Z z+t∈ Z t+e ∈Z ⇒ ⇒ ⇒

y T x A x P

B y

z C z R t

e E eQ t D

x + y − (y + z ) ∈ Z ⇒ x − z ∈ Z z−e ∈Z e−y ∈Z

(1) (2) (3)

e+x∈ Z x+y ∈Z

By (1), (2), and (3) we obtain x − z + e − y ∈ Z ⇒ x − y ∈ Z. It results that there is one integer m such that x−y x+y = m = 2.

It follows that x = 1 + m . Since 1 ≤ x < 2, we deduce that 1 ≤ 1 + m < 2 2 2 ⇔ 0 ≤ m < 2. If m = 0, then x = y = 1 and we obtain that z , e and t are positive integers; this leads to a contradiction with condition (iii). 3 If m = 1, then x = 2 ,y= 1 , hence AP = 3 , BP = 1 . 2 2 2 (b) Let A′ BCDEF be a regular hexagon with A′ B = 1. This hexagon has an inscribed circle (tangent to each of the sides). The line EF intersects the line A′ B at the point A. Since △AA′ F is equilateral, we have AB = AE = 2, BC = CD = DE = 1, hence the pentagon ABCDE satisﬁes the given conditions. E D

A′

Show that for each positive whole number n, there is a positive whole number k such that the decimal representation of each of the numbers k, 2k, . . . , nk contains all the digits 0, 1, 2, . . . , 9.

295

Solved by Oliver Geupel, Br¨ uhl, NRW, Germany. Let the decimal representation of the number k consist of the leading digit 1, followed by 99 blocks of length ⌊log n⌋ + 3 where the mth block is the decimal representation of the number m + 1 with leading zeros (m = 1, 2, . . . , 99): 1 0... 0002 0 . . . 0003 . . . 0 . . . 0099 0 . . . 0100 ßÞ ßÞ ßÞ ßÞ block
1

block

Note that the length of the decimal representation of the integer j (m + 1) is ⌊log(j (m + 1))⌋ + 1 ≤ ⌊log n + log 100⌋ + 1 ≤ ⌊log n⌋ + 3 (j = 1, . . . , n). The decimal representation of the number jk consists of the decimal representation of the number j followed by 99 blocks of length ⌊log n⌋ + 3 where the mth block is the decimal representation of the number j (m + 1) with leading zeros. It therefore suﬃces to prove that, for each j (j = 1, . . . , n), every digit occurs in the decimal representation of one of the numbers j, 2j, 3j, . . . , 100j . To this end, for any ﬁxed j consider the integer ℓ such that 10ℓ−1 ≤ j < 10ℓ . It holds 10ℓ+1 ≤ 100j and each of the nine intervals [10l , 2 · 10l ), [2 · 10l , 3 · 10l ), . . . , [9 · 10l , 10l+1 ) contains a number from the arithmetic sequence j, 2j, 3j, . . . , 100j . The leading digit of the numbers in the mth interval is m (m = 1, 2, . . . , 9). Thus, the digits 1, 2, . . . , 9 occur. The last digit of 100j is zero. This completes the proof.

Next we move to the ﬁle of solutions from our readers to problems given in the September 2010 number of the Corner and the 2007 Bulgarian National Olympiad at [2010: 274].

1. (Emil Kolev, Alexandar Ivanov) The quadrilateral ABCD is such that ∠BAD+
∠ADC > 180◦ and is circumscribed around a circle of center I . A line through I meets AB and CD at points X and Y , respectively. Prove that if IX = IY then AX · DY = BX · CY . Solved by Titu Zvonaru, Com´ ane¸ sti, Romania. Let M, N, P, Q be the projections of point I onto the sides AB , BC , CD , DA, respectively. Since IX = IY and IM = IP , the rightangled triangles IM X and IP Y are congruent, hence ∠IXM = ∠IY P . If Y lies between D and P , and M lies between A and X , then AB CD , a contradiction with ∠BAD + ∠AOC > 180◦ . We may assume that Y lies between C and P , and X lies between B and M . B N C I Y P D Q X M

296

We denote a = AQ = AM , b = BM = BN , c = CN = CP , d = DP = DQ, α = ∠BAD , β = ∠CBA, γ = ∠DCB , δ = ∠ADC , ϕ = ∠IXM = ∠IY P , m = XM = Y P , r = IM = IN = IP = IQ. IM r Since tan α = tan ∠M AI = AM = a , we have 2 ϕ= hence tan ϕ = − tan and tan ϕ = tan Since tan ϕ = 1 m hence m=
r , m

360◦ − α − δ 2

β+γ 2

r +r α+δ r (a + d) = − a r dr = 2 2 1− a · d r − ad

β+γ 2

r (b + c) bc − r 2

we deduce that a+d = b+c bc − r2 =

a+b+c+d bc − ad

− ad

bc − ad . a+b+c+d

(1)

It follows that AX · DY = BX · CY is equivalent to (a + m)(d + m) = (b − m)(c − m) ad + m(a + d) = bc − m(b + c) bc − ad m = , a+b+c+d

which is true by (1).

(Nikolai Nikolov, Oleg Mushkarov) Find the √ least natural number n for which √ π cos n cannot be expressed in the form p + q + 3 r, where p, q and r are rational numbers. Solved by Mohammed Aassila, Strasbourg, France. We will prove that the least positive integer is n = 7. Let ω = e 7 . We have 0 = ω 7 + 1 = (ω + 1)(1 − ω + ω 2 − · · · + ω 6 ), hence 0 = (ω 3 + ω −3 ) − π (ω 2 + ω −2 ) + (ω + ω ) − 1. Let x = ω + ω −1 = 2 cos n , then x veriﬁes 3 2 3 the equation 0 = (x − 3x) − (x − 2) + x − 1 = x − x2 − 2x + 1. Let π P (x) = x3 − x2 − 2x + 1. Note that the roots of P are 2 cos( π ), 2 cos( 37 ) 7 √ √ 5π 3 and 2 cos( 7 ). Assume, by contradiction, that x = P + q + r, then x is irrational since if x = a with gcd(a, b) = 1 then a3 − a2 b − 2ab2 + b3 = 0, b 3 hence b | a , and b = ±1 and similarly a = ±1, in conclusion x = ±1, which is false.
iπ

297

Case 1: r = 0. Then x2 − 2px + p2 − q = 0. By the Euclidean division we have P (X ) = (X − a)(X 2 − 2pX + p2 − q ) + (bX + c). Hence bx + c = 0. Since x is irrational, we have b = c = 0 and hence P (X ) = (X − a)(X 2 − 2pX + p2 − q ). Consequently, P has a rational root a. Impossible. Case 2: q = 0. We have (x − p)3 = r . Since X 3 − 3pX 2 + 3p2 X − p3 − r is not proportional to P (X ) (compare the coeﬃcients of X and of X 2 ), an Euclidean division yields that x is a root of a polynomial (not equal to 0) and with degree ≤ 2, hence we are in case 1. Case 3: q = 0 and r = 0. √ We have r = (x − p − q )3 . Like in case 2, there exists a polynomial √ A = 0 of degree ≤ 2 in Q[ q ][X ] such that A(x) = 0. If deg A = 1 then we are in case 1. If deg A = 2 by an Euclidean division we have P (X ) = A(X )(X − a) + B (X ). Since B (x) = 0, then thanks to case 1 we have B = 0, √ hence P has a for a root in Q + Q q . As in the ﬁrst case, we have a contradiction. Alternative solution: (in french and using higher algebra). Soit x = cos(π/7). Soit K = Q(x). Alors K est une extension galoisienne √ √ de degr´ e ϕ(14)/2 = 3 de Q. Si x ´ etait de la forme p + q + 3 r, alors en ´ elevant √ √ au cube x − p − q , on voit que q ∈ K . Comme [K : Q] est impair, il ne peut √ pas contenir d’extension quadratique de Q donc q est rationnel. Par cons´ equent, √ 3 x est de la forme p + y , o` u y = r. Comme K/Q est galoisienne, K contient jy . On v´ eriﬁe ensuite que K = Q[j, y ] est de degr´ e 6 sur Q. Contradiction.

((Emil Kolev, Alexandar Ivanov) Let k > 1 be an integer. A set of positive integers S is called good if all positive integers can be painted in k colors such that no element of S is a sum of two distinct numbers of the same color. Find the largest positive integer t for which the set S = {a + 1, a + 2, a + 3, . . . , a + t} is good for all positive integers a. Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA. We claim that the largest such t, which we denote tmax (k), is 2k − 2. Case 1. To show that tmax (k) ≥ 2k − 2, consider even and odd a separately. (a) If a = 2r − 1 and t = 2k − 2, then S = {2r, 2r + 1, . . . , 2r + 2k − 3}. We assign colours as follows:

298

Colour Colour Colour . . . Colour Colour

1 2 3 k−1 k

1, 2, 3, . . . , r r+1 r+2 r+k−2 r + k − 1, r + k , r + k + 1, . . . , 2r + 2k − 4

All other positive integers are assigned colour k. The sum of any two distinct integers of colour 1 is smaller than any element of S ; the sum of any two distinct integers of colour k is larger than any element of S . Thus, no element of S is a sum of two distinct integers of the same colour. (b) If a = 2r and t = 2k − 2, then S = {2r + 1, 2r + 2, . . . , 2r + 2k − 2}. We assign colours as before except that colour k is assigned to all integers from n + k − 1 through 2r + 2k − 3. Again, no element of S is the sum of two distinct integers of the same colour. Case 2. To show that tmax (k) ≤ 2k − 2, suppose t = 2k − 1, and set a = 2. Then S = {3, 4, 5, . . . , 2k + 1}. Consider the colours assigned to the integers 1, 2, 3, . . . , k +1. Suppose i and j are elements of this list with i < j , and assume i and j are assigned the same colour. Then 3 ≤ i + j ≤ k + (k + 1) = 2k + 1, so i + j ∈ S . Hence the set S is not good. Thus, the largest t is given by tmax (k) = 2k − 2.

(Oleg Mushkarov, Nikolai Nikolov) Find the least number m for which any ﬁve equilateral triangles with combined area m can cover an equilateral triangle of area 1. Solved by George Apostolopoulos, Messolonghi, Greece; and Oliver Geupel, Br¨ uhl, NRW, Germany. We give the version of Geupel. The answer is m = 2. For convenience of presentation, we rescale the given equilateral triangle T such that its side length is 1. We ﬁrstly prove that any ﬁve equilateral triangles T1 , . . . , T5 with side 2 lengths a1 ≤ · · · ≤ a5 and a2 1 + · · · + a5 = 2 can cover T . Since the case a5 ≥ 1 is obvious, we may assume a5 < 1. Then
2 2 (a3 + a4 )2 ≥ 3a2 3 + a4 ≥ 2 − a5 > 1 .

Let us place T3 , T4 , T5 in one of the angles of T each, as shown in ﬁgure 1. If T3 , T4 , and T5 cover T then we are done. Otherwise, there remains an uncovered equilateral triangle T ′ in the interior of T . Each point of T is covered by at most two of the triangles T3 , T4 , T5 , T ′ . Hence, the combined area of T1 and T2 is not less than twice the area of T ′ ; thus T2 covers T ′ . We have proved that m ≤ 2.

T5 T4 T′ T3

Figure 1

299

In order to show that m ≥ 2, for arbitrary ǫ > 0 we present equilateral 2 triangles T1 , . . . , T5 with side lengths a1 ≤ · · · ≤ a5 such that a2 1 + · · · + a5 > 2 − ǫ that cannot cover T . Let δ= Indeed,
2 a2 1 + · · · + a5 > 2 − 20δ = 2 − ǫ .

ǫ 20

a1 = a2 = a3 = δ ,

a 4 = a 5 = 1 − 5δ .

Let us pin congruent equilateral triangles U , V , W with side length 2δ to the vertices of T as shown in ﬁgure 2. On applying T3 and T4 to T , each of T3 and T4 can meet at most one of the triangles U , V , W . Hence, there is a triangle among U , V , W which is not met by T3 and T4 , say U has this property. But the combined area of T1 , T2 , T3 is less than the area of U , so T1 , T2 , T3 cannot cover U . This completes both our counterexample and the proof that m ≥ 2.

W Figure 2

Next we look at solutions for the 48th IMO Bulgarian Team, First Selection Test, given at [2010: 275].

The sequence {ai }∞ i=1 is such that a1 > 0 and an+1 =
√1 2n

an 1+a2 n

for n ≥ 1.

(a) Prove that an ≤

for n ≥ 2;
7 √ . 10 n

(b) Prove that there exists n such that an >

Solved by Arkady Alt, San Jose, CA, USA; and Chip Curtis, Missouri Southern State University, Joplin, MO, USA. We give the solution by Alt. Since an > 0, n ≥ 1 then an+1 =
an 1 ⇐⇒ 2 = 1 + a2 an+1 n 1 + an an
2

⇐⇒ where bn :=

1 1 1 = 2 + a2 n + 2 ⇐⇒ bn+1 = bn + 2 + b , a2 a n n n+1

1 , n ≥ 1 and we will prove: a2 n

a) bn ≥ 2n for n > 2; b) There is n such that bn <
100n . 49

300

a) Since bn+1 = bn + 2 +

1 bn

> bn + 2 and b2 ≥ 4, by induction bn ≥ 2n.
1 1 ≤ bn + 2 + ,n≥2 bn 2n

b) Since bn ≥ 2n for n ≥ 2 then bn+1 = bn + 2 + and, therefore, bn+1 − b2 = where hn =
n k=2 n

(bk+1 − bk ) ≤

2+
k=2

1 2n

= 2 (n − 1) +

1 (hn − 1) , 2

n È 1 . Thus, n k=1

bn+1 ≤ 2n − =⇒ bn < 2n +

1 hn + b2 , n ≥ 3 . 2

5 1 1 + hn + b2 < 2 (n + 1) + hn+1 + b2 , n ≥ 2 2 2 2

Note that hn <

√ 2n, n ∈ N. Indeed, by the Cauchy Inequality we have
n

h2 n ≤ n· and
n k=1

k=1

1 k2

1 <1+ k2

n k=2

1 =1+ (k − 1) k

n k=2

1 1 − k−1 k

=1+1−

1 < 2. n

√ n n Since 100 = 2n + 2 and hn < 2n then it suﬃces to prove that there 49 √ 49 √ √ 1 n is n such that 2 ⇐⇒ 49b2 < 2n 2n + b2 < 2 2n − 49 . 49 2 It is easy to see that the latter inequality holds for any n ≥ n0 = max
2 492 b2 2 51 , 2 8

Another variant of ending solution (b): Since 2n < bn then lim
n→∞

bn = 2 and, therefore, for any ε > 0 there is n0 ∈ N such that n bn 2 < 2 + ε ⇐⇒ bn < (2 + ε) n for all n > n0 (ε). In particular for ε = n 49 2 100n 2 we have bn < 2 + n= for all n > n0 . 49 49 49

1√ 2n + 2n + b2 1√ 2 < 2n + 2n + b2 and lim = 2 n→∞ 2 n

That completes the Corner for this issue.

301

BOOK REVIEWS
Amar Sodhi
Lobachevski Revisited by Seth Braver The Mathematical Association of America, 2011 ISBN: 978-0-88385-979-7 (e-book) 237 + x pages, US$35.00 (print on demand, US$95.00) Reviewed by J. Chris Fisher, University of Regina, Regina, SK During the 1820s Lobachevski (1792-1856) in Russia and Bolyai (1802-1860) in Hungary independently discovered non-Euclidean Geometry—the geometry in which there are two lines parallel to a given line through a given point not on that line. Because of the overwhelming importance of their ideas, it might be hard for us today to understand how their work could have been so thoroughly ignored by their contemporaries—it was only after their deaths that the mathematical world paid much attention to the subject, and several decades elapsed before the full implication of their achievements became appreciated. Whereas Bolyai was so discouraged that he gave up publishing mathematics, Lobachevski optimistically produced further Russian accounts of non-Euclidean geometry; when his fellow Russians failed to recognize the signiﬁcance of his work, he then published treatments of his theory in French and in German. His little German book of 1840, Geometrische Untersuchungen zur Theorie der Parallellinien, forms the core of the book under review here. Seth Braver’s translation, with the title shortened to The Theory of Parallels, appears twice: at the end as a 23-page appendix, and spread out over the ﬁrst 200 pages, printed in red type and supplemented by Braver’s introduction and notes, printed in black. The resulting “illuminated” Lobachevski is intended for “student, professional, [and] layman.” Braver’s book would certainly make a superb textbook for an undergraduate course in non-Euclidean geometry. His commentary provides the historic and philosophical background that explains the mathematical environment in which Lobachevski worked, as well as the signiﬁcance of his work. His mathematical ideas are clearly motivated, and the relevant achievements of predecessors and contemporaries are brieﬂy outlined. Explanations are provided to ﬁll in details that many of today’s students might otherwise ﬁnd diﬃcult; the commentary is informative and entertaining, which most students would appreciate. For example, when I taught such a course a few years back I was surprised when several students complained about the diagrams: in imaginary geometry (to use Lobachevski’s terminology), straight lines in diagrams are often represented by curves so as to avoid unwanted intersections. Braver points out that although space in imaginary geometry looks the same at every point, “it looks very diﬀerent at diﬀerent scales. On a tiny scale, it resembles Euclidean geometry, and serious deviations become noticeable only on a large, possibly astronomical, scale. Since similar ﬁgures do not exist in imaginary geometry, accurate scaled down drawings are impossible.” I wish I had thought of this explanation to give my students. In general, the author provides the student with good explanations of what is the same and

302

what is diﬀerent in this new geometry. He supplements Lobachevski’s proofs with further details and alternative arguments, together with related results and elegant arguments from Saccheri, Lambert, Legendre, Gauss, Bolyai, and others. A teacher using this book as a text, however, would have to provide details of Euclidean theorems and proofs, and perhaps examples of modern arguments involving betweenness axioms. Also, the book comes with no exercises. As usual, I disagree with the MAA’s pricing policy: the $90 US price tag on the printed edition seems designed to encourage the student to purchase the $35 e-book. According to the book representative who sent me the printed version, the electronic version is equally hard to navigate—the references are not linked so that there is no quick way to locate an item that has been cross-referenced. The book is less successful in its eﬀort to reach the professional. Lobachevski’s intended audience was the professional mathematician of the mid-nineteenth century. He wrote quite well; his arguments provide pretty much the same level of detail that would be expected by readers of the geometry problems in Crux, so not much help would be needed for any reader familiar with Euclid. The supplementary comments on philosophy and history have been taken almost entirely from standard sources that are readily available and are not entirely reliable. Whereas the light, whimsical tone of the commentary might be suitable for the undergraduate seeing the material for the ﬁrst time, I found many explanations lacking in depth. This shortcoming was most evident in the notes accompanying the preface where Lobachevski lists his preliminary theorems and claims that their “proofs present no diﬃculties.” Braver is content to denigrate the ﬁrst ﬁve propositions (“A Rough Start”, “[they] should be demoted to the status of descriptions (or axioms)”, “... if either he or Euclid had tried to prove it rigorously, they would have found the task impossible”, and on and on. Far from thinking Lobachevski’s arguments were faulty, I was struck by how much thought he put into the foundations, and how far his thoughts had advanced beyond Euclid. Compare Euclid’s second postulate, “A ﬁnite straight line may be extended continuously in a straight line.” with Lobachevski’s Paragraph 3, “By extending both sides of a straight line suﬃciently far, it will break out of any bounded region. In particular, it will separate a bounded plane region into two parts.” Surely, he wanted to make it clear that lines stretch to inﬁnity in both directions (which is not clear in Euclid, who demands only that his segments be extendable). We can also see here a primitive notion of separation decades before Pasch made the concept rigorous (in 1882). How more valuable it would have been had Braver discussed where the initial ﬁve “theorems” originated, why Lobachevski thought they were easily established and, most importantly, where and how he used them in subsequent proofs. At any rate, the reader does not have to be reminded every time Proposition 3 is invoked, that Lobachevski lacked 20th century tools. The author fails to mention why he felt the need for a new translation from the original German, which seems to be only superﬁcially diﬀerent from Bruce Halsted’s 1881 translation. But whichever translation you can get your hands on, any person who likes geometry should read The Theory of Parallels— the ﬁnal half-dozen propositions constitute some of the most clever and exciting

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mathematics ever conceived. That is where Lobachevski develops the geometry of the horosphere and then establishes that the geometry of the sphere is independent of the parallel postulate (that is, the geometry and trigonometry of a sphere is the same in imaginary geometry as in Euclidean geometry); these results lead him directly to the key formulas of imaginary geometry.

Unsolved Crux Problems
As remarked in the problem section, no problem is ever closed. We always accept new solutions and generalizations to past problems. Recently, Chris Fisher published a list of unsolved problems from Crux[2010 : 545, 547]. Below is a sample of two of these unsolved problems:

342⋆.

[1978 : 133, 297; 1980 : 319-22] Proposed by James Gary Propp, Great Neek, NY, USA. For ﬁxed n ≥ 2, the set of all positive integers is partitioned into the (disjoint) subsets S1 , S2 , . . ., Sn as follows: for each positive integer m, we have m ∈ Sk if and only if k is the largest integer such that m can be written as the sum of k distinct elements from one of the n subsets. Prove that m ∈ Sn for all suﬃciently large m. (If n = 2, this is essentially equivalent to Problem 226 [1977 : 205]).

1754⋆. [1992 : 175; 1993 : 151-2; 1994 : 196-9; 1995 : Walther Janous, Ursulinengymnasium, Innsbruck, Austria.

236-8] Proposed by

Let n and k be positive integers such È that 2 ≤ k < n, and let x1 , x2 , . . ., n xn be non-negative real numbers satisfying i=1 xi = 1. Prove or disprove that x1 x2 . . . xk ≤ max 1 kk , 1 nk−1 ,

where the sum is cyclic over x1 , x2 , . . ., xn . [The case k = 2 is known — see inequality (1) in the solution of CRUX 1662 [1992 : 188].] Good luck solving these problems. We would love to receive your solutions so that we could cross them oﬀ our list.

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RECURRING CRUX CONFIGURATIONS
J. Chris Fisher Triangles for which 2b2 = c2 + a2
Many conﬁgurations have popped up again and again in Crux geometry problems over the years. In this occasionally appearing column we will recall some of them—they might not be as common as a right triangle, but their properties are generally useful and often surprising. This month we investigate triangles with sides a, b, c that satisfy 2b2 = c2 + a2 . Following the suggestion of Leon Bankoﬀ 2 a2 we will call them root-mean-square triangles because b = c + is the root 2 mean square of a and c. These triangles have also been called automedian and self-median (see property 1 below), as well as quasi-isosceles (see problem 727 below). Leon Bankoﬀ reported [1978 : 13-16] that in a series of Mathesis articles [4], 69 properties of these triangles are listed, generally with an abundance of earlier references instead of proofs. He listed his favorite 10 properties and supplied the short proofs. Here we shall leave the proofs as exercises with hints where appropriate. For these ten properties we assume that ABC is a triangle whose sides satisfy 2b2 = c2 + a2 . (1) Let ma , mb , mc denote the medians to sides a, b, c, respectively, and let H, O , and G denote the orthocentre, circumcentre, and centroid. The symmedians ka , kb , and kc are the reﬂections of the medians in the respective angle b+c bisectors; they meet in the symmedian point K . As usual, let s = a+2 be the semiperimeter, and R denote the circumradius. Property 1. ma = 23 c, mb = 23 b, mc = 23 a. This property follows immediately by plugging (1) into the formula for the length of a median in terms of the sides. Note that the medians can be translated to form a triangle that is oppositely similar to triangle ABC ; this property is the source of the “automedian” terminology that was used throughout [4]. Perhaps in French the word sounds less asinine, but to me it does not convey the image of a triangle that is oppositely similar to the triangle formed by its three medians. K.R.S. Sastry favored the terminology self-median. He proved a converse in “The Triangle: A Parametric Description” [2004 : 497-501]: If b is the length of the middle side, then ∆ABC is (oppositely) similar to its medial triangle if and only √ 3 if b = 2 mb , if and only if (1) holds. Note that the ratio of similitude of corresponding sides tells us that the area of the triangle formed by the medians is √ 3/4 that of the original triangle. It is also seen that ma + mb + mc = 3s, a familiar property of the equilateral triangle, which is, of course, a special case of a root-mean-square triangle. Property 2. b2 = 2ca cos B. (This is the cosine law applied to (1).)
√ √ √

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Property 3. 2 cot B = cot C + cot A. (You can get started by applying the sine law to both sides of (1).) Property 4. 2 cos 2B = cos 2C + cos 2A. Property 5. b2 = AG2 + BG2 + CG2 . Property 6. 4·Area(ABC ) = b2 tan B . 2 2 Property 7. 2m2 b = mc + ma . 2 2 Property 8. 2BH = CH + AH 2 . Property 9. If B ′ and B ′′ are the third vertices of equilateral triangles constructed externally and internally on side AC , then ∠B ′ BB ′′ is a right angle. Hint. A 1796 theorem of Nicholas Fuss implies that B ′ B 2 + B ′′ B 2 = a + b2 + c2 [3, p. 220, par. 354c]; more easily than tracking down a reference, one can prove Fuss’s theorem by applying the cosine law to triangles BCB ′ and BCB ′′. b Property 10. m = 2 cos B . kb
2

Also in [1978 : 13], W.J. Blundon showed how to construct a triangle satisfying (1) given the segment AC : Construct the equilateral triangle ACD and let O be the midpoint of AC ; Property 1 tells us that a triangle ABC √ satisﬁes (1) if and only if OB = 23 b, whence the locus of B is the circle with centre O and radius OD . The editor L´ eo Sauv´ e added (on page 16) that if you want to generate speciﬁc examples of root-mean-square triangles with integer sides, you could use the theorem that all primitive solutions of 2b2 = c2 + a2 are given by a = u2 + 2uv − v 2 , b = u2 + v 2 , c = |u2 − 2uv − v 2 |, where u > v , with u and v relatively prime positive integers of diﬀerent parity [1], [2, pp. 435 ﬀ.]. Of course, your values of a, b, c must satisfy the triangle inequality. Another approach to this result was taken by K.R.S. Sastry in “Pythagoras Strikes Again!” [1998 : 276-280]; speciﬁcally, he proved that
2 2 If a0 , b0 , c0 are the sides of a right triangle in which c2 0 + a0 = b0 , a0 > c0 , and b0 > 2c0 , then a = a0 + c0 , c = a0 − c0 , and b = b0 are the sides of a triangle that satisﬁes c2 + a2 = 2b2 ; conversely, if the sides of a triangle satisfy a > b > c and c2 + a2 = 2b2 , then 1 a0 = 2 (a + c) and c0 = 1 (a − c) are the legs of a right triangle 2 whose hypotenuse is b, a0 > c0 , and b > 2c0 .

For a list of all 36 primitive root-mean-square triangles with perimeters less than 1000, see [1978 : 194]. The smallest (that is not equilateral) has side lengths 17, 13, 7, which comes from a 12-13-5 right triangle by Sastry’s theorem. The 1978 discussion of these triangles came out of the solution to problem 210 [1977 : 10, 160-164, 196-197; 1978 : 13-16, 193-194] (proposed by Murray S. Klamkin): P, Q, R denote points on the sides BC, CA, and AB , respectively, of a given triangle ABC ; determine all triangles ABC such that if BP BC = CQ CA = AR AB =k 1 = 0, , 1 , 2

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then P QR (in some order) is similar to ABC . The solution revealed that if the triangles are directly similar, then they must be equilateral. When a > b > c the value of k in one of the three resulting solutions has 2b2 − c2 − a2 in its denominator, whence root-mean-square triangles ABC would have only two oppositely similar triangles P QR instead of three, and those two triangles would be congruent. Problem 309 [1978 : 12, 200-202] (Proposed by Peter Shor). Let ABC be a triangle with a ≥ b ≥ c or a ≤ b ≤ c. Let the bisectors of ∠cma and ∠amc meet at R. Prove that (a) AR ⊥ CR if and only if 2b2 = c2 + a2 ; (b) if 2b2 = c2 + a2 , then R lies on mb . Is the converse of (b) true? Yes, the converse turns out to be true (when b is assumed to be the middle side). The featured solution of Daniel Sokolowsky made use of yet another interesting property: Theorem. If D and E are the midpoints of sides AB and BC , and G is the centroid, then 2b2 = c2 + a2 if and only if BDGE is a cyclic quadrilateral. Problem 313 (revised) [1978 : 35, 207-209] (Proposed by Leon Bankoﬀ). The sides of a nonequilateral triangle satisfy 2b2 = c2 + a2 if and only if GK (the join of the centroid and the symmedian point) is parallel to AC . Bankoﬀ seems not to have worried much about converses! Although his version of Problem 313 called for a proof of suﬃciency only, both featured solutions make clear that (1) is both necessary and suﬃcient for GK to be parallel to AC . He found the theorem (without a converse) in Mathesis, t. IX (1889) p. 208, where it was attributed to Lemoine (Mathesis, t. V (1885) p. 104). As for Bankoﬀ’s list of ten properties (reproduced above), although I did not carefully write down the proofs, I believe that all ten converses hold for triangles with a > b > c or c > b > a. Problem 383 [1978 : 250, 174-176] (Proposed by Daniel Sokolowsky). Let ma , mb , mc be respectively the medians AD, BE, CF of a triangle ABC with centroid G. Prove that (a) if ma : mb : mc = a : b : c, then ∆ABC is equilateral; b (b) if m = c , then either (i) b = c or (ii) quadrilateral AEGF is cyclic; mc b (c) if both (i) and (ii) hold in (b), then ∆ABC is equilateral. By the theorem from the proof of Problem 309 above, quadrilateral AEGF is cyclic in part (b) if and only if 2a2 = b2 + c2 . Part (b) should be compared with the quasi-isosceles property of Bottema and Groenman discussed in the next problem, number 727. Problem 727 [1982 : 78; 1983 : 115, 180-181] (Proposed by J.T. Groenman). Let tb and tc be the symmedians issued from vertices B and C of triangle ABC and terminating in the opposite sides b and c, respectively. Prove that tb = tc if and only if b = c. It turned out that this problem had already appeared several times elsewhere; instead of a solution, three references were provided: Amer. Math.

307

Monthly, 51 (Dec 1944) 590-591; Scripta Mathematica, 22 (1956) 102; and Mathematics Magazine, 40 (May 1967) 165, and 41 (Jan 1968) 48-49. There was also reference to a problem in the Pi Mu Epsilon Journal that claimed to prove that the analogous result holds also for exsymmedians. Not so, according to a paper by Groenman with O. Bottema in Nieuw Tijdschrift voor Wiskunde, 70 (1983) 143-151. For the details, recall that an exmedian is a line through a vertex parallel to the opposite side of the triangle, while an exsymmedian is the reﬂection of the exmedian in the external angle bisector. The Bottema-Groenman result says that if the exsymmedians from B and C are equal in length, then either b = c (and ∆ABC is isosceles), or 2a2 = b2 + c2 (and the triangle is a root-mean-square triangle). In the latter case they call the triangle quasi-isosceles. Warning: their article is written in Dutch. After Problem 727 (from 28 years ago!) it seems as if Sastry has singlehandedly kept the subject of root-mean-square triangles alive. Besides the Crux articles from 1998 and 2004 mentioned earlier, he published the article [5] and posed Problem 473 in [6]: Prove that in a scalene triangle ABC , the bisectors of ∠ABC and ∠AGC intersect on the side AC if and only if c2 + a2 = 2b2 . He is also responsible for Problem 2252 [1997 : 300; 1998 : 375-376] (Proposed by K.R.S. Sastry). Prove that the nine-point circle of a triangle trisects a median if and only if the side lengths of the triangle are proportional to the lengths of its medians in some order. This completes the list of results on root-mean-square triangles that I found in the problems pages of CRUX with MAYHEM. Triangles for which 2b = c + a and triangles for which 2B = C + A have likewise been enthusiastically investigated in this journal. We will discuss them in future columns.

References
[1] Leon Bankoﬀ, Solution to Problem 91 (A Long Probate), Mathematics Magazine, 41:5 (Nov-Dec 1968) 291-293. [2] Leonard Eugene Dickson, History of the Theory of Numbers, vol. II, Chelsea, 1952. [3] Roger A. Johnson, Modern Geometry, Houghton Miﬄin, 1929; reissued as Advanced Euclidean Geometry, Dover, 1960. [4] Mathesis: (1902) 205-208; (1903) 196-200, 226-230, 245-248; (1926) 68-69. [5] K.R.S. Sastry, Self-Median Triangles, Mathematical Spectrum, 22 (1989/90) 58-60. [6] K.R.S. Sastry, Problem 473, College Mathematics Journal, 23:2 (March 1992) 162; 24:2 (March 1993) 186-188.

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Summations according to Gauss
Gerhard J. Woeginger
A well-known anecdote relates that when Carl Friedrich Gauss (1777–1855) was only ten years old, his school teacher wanted to keep the pupils busy and asked them to add up all the integers from 1 to 100. Almost immediately Gauss placed his slate on the table and said “There it is.” The slate just contained the number 5050 without further calculations. When the teacher ﬁnally checked the results, Gauss’s slate was the only one with the correct answer. If there is any truth in this anecdote, then the young Gauss must have paired up the integers in the following (or some closely related) way. In his mind, he wrote down the summation twice: once in the standard fashion from left to right, and once he ﬂipped it around and wrote it from right to left. 1 + 2 + 3 + 4 + · · · · · · + 99 + 100

100 + 99 + 98 + 97 + · · · · · · + 2 + 1

This yielded 100 vertically aligned pairs, where the numbers in each pair added up to 101. Hence the sum of all listed numbers was 100 · 101, and as each number was listed twice the answer desired by Gauss’s teacher was 1 · 100 · 101 = 5050. 2 In this article we will discuss several related problems that all can be settled by this “write it once down left-to-right and once right-to-left” approach of Gauss. For warming up, the reader may want to generalize the above calculation to an arbitrary number of terms. Problem 1 Determine a closed form expression for 1 + 2 + 3 + 4 + · · · + n. Our next problem is a standard textbook exercise in the manipulation of sums of binomial coeﬃcients. Problem 2 Determine a closed form expression for the following sum S : S = 1· n n n n +2· +3· + · · · + (n + 1) · 0 1 2 n (2)

We present a solution that is based on the trick of Gauss. We write down the sum once again, but in reversed order with its terms taken from right to left: S = (n + 1) · n n +n· n n−1 + (n − 1) · n n−2 + ··· +1· n 0 (3)

This yields n + 1 vertically aligned pairs, where the kth pairs consists of the kth term in (2) and the kth term in (3). By applying the well-known relation
Copyright c 2011 Canadian Mathematical Society Crux Mathematicorum with Mathematical Mayhem, Volume 37, Issue 5

309

n ℓ up to

n n−ℓ n k−1

with ℓ = k − 1, we derive that the terms in the kth pair add n n−k+1 n k−1

k·

+ (n − k + 2) ·

= (n + 2) ·

(4)

Although the resulting value in the right hand side of (4) still depends on the parameter k, we have made substantial progress. By using (4), we see that the sum of all terms listed in (2) and (3) is 2S = (n + 2) · n 0 + n 1 + ··· + n n = (n + 2) · 2n .

Here we used the fact that the binomial coeﬃcients in the nth row of Pascal’s triangle add up to 2n . Hence the answer to this problem is S = (n + 2) · 2n−1 . The following problem can be settled by a very similar argument. The reader is encouraged to verify that the answer is Tn = n.
2n

Problem 3 For n ≥ 1 evaluate Tn =

k=0

k · cos

kπ 2n

Now let us turn to a summation problem from the 2000 Asian Paciﬁc Mathematical Olympiad (APMO’2000).
101

Problem 4 Compute the sum k = 0, . . . , 101.
k=0

x3 k 3x2 k − 3xk + 1

with xk = k/101 for

Every single term in this summation is bulky, and there are lots of bulky terms that must be added up. Computing this sum by hand is certainly not a good idea. Let us follow Gauss and let us pair up the terms in the given sum with the terms in reversed order: term 0 is paired with term 101, term 1 with term 100, term 2 with term 99, and so on. Observe that in every resulting pair we have xk + x101−k = k/101 + (101 − k)/101 = 1, which implies 3x2 k − 3xk + 1 = =
3 x3 = 3x2 101−k + (1 − x101−k ) 101−k − 3x101−k + 1.

(1 − xk )3 + x3 k

Adding up the term for xk and the term for x101−k then yields x3 k 3x2 k − 3xk + 1 = + 3x2 101−k − 3x101−k + 1 + (1 − xk )3 + x3 k x3 101−k x3 101−k = =
3 x3 k + (1 − xk )

x3 k (1 − xk )3 + x3 k

(1 − xk )3 + x3 k

= 1.

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Since altogether there are 102 pairs, we see that twice the value of the sum equals 102, and that hence the answer to the APMO problem is 51. The bulky summation can actually be performed in a routine fashion! The following problem was posed as problem A3 on the 1980 William Lowell Putnam Mathematics Competition.
π/2

Problem 5 Evaluate I =
x=0

1 + (tan x)

√ 2

The integrand in this problem looks horrible. A little bit of playing around soon conﬁrms our impression that it is hopeless to search for an antiderivative in closed form. But remember that a deﬁnite integral is really just some kind of fancy summation, as it is obtained by adding up lots of very small numbers. Hence let us try to apply the trick of young Gauss. We write the integral down once again, but we ﬂip it over so that this time we integrate from right to left. This ﬂipping over operation corresponds to the substitution y = π/2 − x with dy = −dx. We derive:
0

I =
y=π/2

−dy

π/2 √ 2

=
y=0

1 + (tan(π/2 − y ))

√

(5)

We recall the trigonometric identity tan(π/2 − α) = cot(α), and with its help we rewrite the integrand in (5) as 1 = 1 = (tan y ) (tan y )
√ √ 2

1 + (tan(π/2 − y )) Now (5) and (6) imply
π/2

√

1 + (cot y )

√ 2

. +1

(6)

=
x=0 π/2

π/2 √ 2

1 + (tan x) 1 + (tan x)

+
y=0

(tan y ) (tan y )
π/2 √

√

dy +1

√ √

2 2

=
x=0

1 + (tan x)

dx =
x=0

dx = π/2.

Therefore the answer to this Putnam problem is I = π/4. Note that the exponent √ 2 does not play any special role in our calculations. If we replace it by an arbitrary positive real number, the answer will still remain π/4. The same integration theme resurfaced seven years later as problem B1 on the 1987 Putnam exam. The reader should have little diﬃculty in ﬁnding the right substitution for the following problem, which leads to the answer J = 1.

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Problem 6 Evaluate J =
x=2

ln(9 − x) dx ln(9 − x) + ln(x + 3)

√ The occurrence of the function ln x √ in this problem is purely artiﬁcial. The solution virtually remains the same, if ln x is replaced by any integrable function f (x) for which f (9 − x) + f (x + 3) = 0 for 2 ≤ x ≤ 4. Finally we want to discuss problem A4 from the 1999 Putnam exam. Problem 7 Sum the series S =
∞ ∞

m=1 n=1

m2 n . 3m (n3m + m3n )

Let us ﬂip over the summation so that m becomes n and simultaneously n becomes m. Of course this does not change the value of the sum, and (similarly as before) we get 2S =
∞ ∞

m2 n 3m (n3m + m3n )

∞

n2 m 3n (m3n + n3m )

(7)

m=1 n=1

We pair up terms (similarly as before), and by some simple algebra derive m2 n 3m (n3m
∞

m3n )

n2 m 3n (m3n + n3m )

mn 3m 3n

(8)

The right hand side of (8) indicates that it might be useful to investigate the auxiliary sum T =
∞ m=1

m/3m . Since
∞ m=0

m=1

3T =

m 3m−1

m+1 3m

= T +

∞ m=0

1 3m

= T +

3 2

we conclude T = 3/4. By combining this with (7) and (8) we derive (similarly as before) 2S =
∞ ∞

mn 3m 3n

∞ m=1

m 3m

m=1 n=1

∞ n=1

n 3n

= T2 =

9 16

Thus the ﬁnal answer to our ﬁnal problem is S = 9/32.

Gerhard J. Woeginger Department of Mathematics and Computer Science TU Eindhoven P.O. Box 513, NL-5600 MB Eindhoven The Netherlands [email protected]

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A nest of Euler Inequalities
Luo Qi
Abstract For any given △ABC , we deﬁne the antipodal triangle. Repeating this construction gives a sequence of triangles with circumradii Rn and inradii rn obeying a generalized form of Euler’s inequality 2n Rn ≥ · · · ≥ 22 R2 ≥ 2R1 ≥ R0 ≥ 2r0 ≥ 22 r1 ≥ · · · ≥ 2n+1 rn , (n = 1, 2, · · · ), with equalities iﬀ △ABC is equilateral.

Key words: Euler inequality; antipodal triangle Let R, r be the radius of circumcircle and inscribed circle of a triangle; then R ≥ 2r , with equalities iﬀ the triangle is equilateral ([1], p.50). This is the famous Euler inequality. In this note, we are going to build a nest of Euler inequalities for a certain family of related triangle. Deﬁnition 1 If a vertex A of a triangle ABC and another point A′ on the perimeter divide the perimeter into two equal parts (that is, |AB | + |BA′ | = |AC | + |CA′ |) we call |A′ | the antipode of |A|, and the triangle △A′ B ′ C ′ of which three vertices are antipodes of A, B, C respectively the antipodal triangle of △ABC . Note that A′ is necessarily on the (non-extended) edge BC and in fact it is the point where that edge touches the appropriate escribed circle.[2] Thus, we can easily ﬁnd a way to draw an antipodal triangle △A′ B ′ C ′ of a given triangle △ABC . Lemma 1 Denote by a, b, c, a1 , b1 , c1 , s, s1 , A, A1 the sides, semiperimeters, and areas of △ABC and its antipodal triangle △A1 B1 C1 , and let R and r be the circumradius and inradius of △ABC . Then 1. |AB1 | = |BA1 | = s − c, |AC1 | = |CA1 | = s − b, |BC1 | = |CB1 | = s − a; 2. 3. A1 A

r ; 2R

a1 b1 c1 abc

≥

r (with equality iﬀ △ABC is equilateral ); 4R

4. 2s1 ≥ s( with equality iﬀ △ABC is equilateral ).
Copyright c 2011 Canadian Mathematical Society Crux Mathematicorum with Mathematical Mayhem, Volume 37, Issue 5

313

Proof (See the ﬁgure 1) (1) We have 1 (|AB | + |BC | + |CA|) − |AB |, 2 1 |BA1 | = (|AB | + |BC | + |CA|) − |AB | 2 |AB1 | = and so |AB1 | = |BA1 | = s − c. In the same way, we have |BC1 | = |CB1 | = s − a. A B1 C1 C |AC1 | = |CA1 | = s − b,

A1 Figure 1

(2) Denote by AAB1 C1 , ABA1 C1 , ACA1 B1 , the areas of △AB1 C1 , △BA1 C1 , △CA1 B1 . Because AAB1 C1 A ABA1 C1 A ACA1 B1 A we have A1 A = A − AAB1 C1 − ABA1 C1 − ACA1 B1 AAB1 C1 A ABA1 C1 = = = |AB1 | · |AC1 | (s − c)(s − b) = , |AB | · |AC | c·b |BA1 | · |BC1 | (s − c)(s − a) = , |BA| · |BC | c·a |CA1 | · |CB1 | (s − b)(s − a) = |CA| · |CB | b·a

= 1−

ACA1 B1 − − A A A (s − c)(s − b) (s − c)(s − a) (s − b)(s − a) = 1− − − c·b c·a b·a 2(s − a)(s − b)(s − c) = a·b·c

314

By Heron’s and other well-known formulas, A= Thus A1 A = 2(s − a)(s − b)(s − c) a·b·c = 2A2 s · 4A · R = A 2sR = r 2R s(s − a)(s − b)(s − c) = abc 4R = sr

(3) Using the law of sines on △AB1 C1 , we have a1 sin A = = ≥ s−b s−c s−b+s−c = = sin ∠AB1 C1 sin ∠AC1 B1 sin ∠AB1 C1 + sin ∠AC1 B1 a
∠AC1 B1 2 sin ∠AB1 C1 + cos 2 a a = A 2 cos A 2 sin π− 2 2 ∠AB1 C1 −∠AC1 B1 2

(with equality iﬀ ∠AB1 C1 = ∠AC1 B1 ). Therefore we have a1 sin A A ≥ = sin a 2 2 cos A 2 (with equality iﬀ ∠AB1 C1 = ∠AC1 B1 ). In the same way, we have b1 b B 2

≥ sin

(with equality iﬀ ∠BA1 C1 = ∠BC1 A1 ), and c1 c ≥ sin C 2

(with equality iﬀ ∠CB1 A1 = ∠CA1 B1 ). Multiplying these three inequalities, we obtain a1 b1 c1 A B C ≥ sin sin sin abc 2 2 2 (with equality iﬀ △ABC is equilateral). Also, because A= we have (2R)3 sin A sin B sin C 4R = 2Rr (sin A + sin B + sin C ) 2 . 1 2 ab sin C = abc 4R = r (a + b + c) 2

315

So r R = = = = = 2 sin A sin B sin C sin A + sin B + sin C 16 sin A sin B sin C cos 2 2 2
A cos B cos C 2 2 2 A+B A−B C 2 sin 2 cos 2 + 2 sin 2 cos C 2 16 sin A sin B sin C cos A cos B cos C 2 2 2 2 2 2 A−B C 2 cos C (cos + sin ) 2 2 2 B C A B 8 sin A sin sin cos cos 2 2 2 2 2 A+B B A B cos + sin sin + cos cos A 2 2 2 2 2 B C A 8 sin A sin sin cos cos B 2 2 2 2 2 B A B A B cos A cos + sin sin + cos cos − 2 2 2 2 2 2

sin

A 2

sin

B 2

= 4 sin Thus

A B C sin sin 2 2 2

a1 b1 c1 abc

≥

r 4R

(with equality iﬀ △ABC is equilateral). (4) Construct perpendiculars B1 E and C1 D to BC at E and D , respectively (See Figure 2). Then a1 ≥ |DE | (with equality iﬀ BC B1 C1 .) But |DE | = a − |BD | − |CE | = a − (s − a) cos B − (s − a) cos C , so a1 ≥ a − (s − a)(cos B + cos C ) with equality iﬀ BC A B1 C1 E C B1 C 1 .

B Similarly, we have:

A1 D Figure 2

b1 ≥ b − (s − b)(cos C + cos A) with equality iﬀ CA c1 ≥ c − (s − c)(cos A + cos B ) with equality iﬀ AB

C1 A1 , A1 B1 .

316

Adding up these three inequalities yields 2s1 ≥ ≥ = 2s − (a cos A + b cos B + c cos C ) 1 2s − (a cos B + a cos C + b cos A + b cos C + c cos A + c cos B ) 2 1 2s − (a + b + c) = s 2

Therefore 2s1 ≥ s with equality iﬀ △ABC is equilateral. Studying these two triangles we can also ﬁnd other interesting properties. The reader may verify that the antipodal triangle is “less equilateral” in that the ratio between longest and shortest side is always greater than in the original triangle. Theorem 1 Denote by R, R1 , r, r1 , the circumradii and inradii of △ABC and its antipodal triangle △A1 B1 C1 . Then 2R1 ≥ R ≥ 2r ≥ 4r1 , with equalities iﬀ △ABC is equilateral. Proof By (2) and (3) of lemma 1, we have r 2R = A1 A = a1 b1 c1 /4R1 abc/4R = Ra1 b1 c1 R1 abc ≥ R R1 · r 4R = r 4R1

So 2R1 ≥ R, with equality iﬀ △ABC is equilateral. By (2) and (4) of lemma 1, we have 1 4 ≥ r 2R = A1 A = r1 s1 rs ≥ r1 s1 r · 2s1 = r1 2r

and so 2r ≥ 4r1 with equality iﬀ △ABC is equilateral. Hence, we get 2R1 ≥ R ≥ 2r ≥ 4r1 , again with equalities iﬀ △ABC is equilateral. Using mathematical induction and the theorem we immediately get: Corollary 1 (See Figure 3) Let △A0 B0 C0 be given, and let R0 , R1 , · · · , Rn ; r0 , r1 , · · · , rn denote the circumradii and inradii of △A0 B0 C0 , △A1 B1 C1 , · · · , △An Bn Cn respectively, and △Ai Bi Ci is the antipodal triangle of △Ai−1 Bi−1 Ci−1 , (i = 1, 2, · · · ). Then 2n Rn ≥ · · · ≥ 22 R2 ≥ 2R1 ≥ R0 ≥ 2r0 ≥ 22 r1 ≥ · · · ≥ 2n+1 rn , with equalities iﬀ △A0 B0 C0 is equilateral.

317

A0 An B n− 1 Cn C0

Cn− 1 Bn B0

An−1 Figure 3

So, we build a nest of Euler inequalities. Open question: In [3] Yang derives Euler-type inequalities for tetrahedra in 3-dimensional space. Can we deﬁne antipodal points for a tetrahedron in such a way that the results of this paper generalize?

References
[1] O. Bottema et al. Geometric Inequalities, Wolter-Noordhoﬀ, Groningen, 1969. [2] Weisstein, Eric W. “Semiperimeter.” FromMathWord–A Wolfram Web Resource. http://mathworld.wolfram.com/Semiperimeter.html. Accessed 10 January 2012. [3] Yang Shiguo. Several improvements of tetrahedral Euler inequality. China collection of studies of elementary mathematics [M]. Zhengzhou province: Henan education press,1992.6. Mathematics Department, Guilin Normal College, Guilin,

Luo Qi Mathematics Department Guilin Normal College Guilin, Guangxi China [email protected]

318

PROBLEMS
Solutions to problems in this issue should arrive no later than 1 September 2012. An asterisk (⋆) after a number indicates that a problem was proposed without a solution. Each problem is given in English and French, the oﬃcial languages of Canada. In issues 1, 3, 5, and 7, English will precede French, and in issues 2, 4, 6, and 8, French will precede English. In the solutions’ section, the problem will be stated in the language of the primary featured solution. The editor thanks Jean-Marc Terrier of the University of Montreal for translations of the problems.

3650.

Replacement. Proposed by Michel Bataille, Rouen, France. Let ABC be a triangle and R, O , G and K its circumradius, circumcentre, centroid and Lemoine point, respectively. Prove that KA KB KC = CA · = AB · = GA GB GC

BC ·

3(R2 − OK 2 ).

Recall that a symmedian of a triangle is the reﬂection of the median from a vertex in the angle bisector of the same vertex. The Lemoine point of a triangle is the point of intersection of the three symmedians. Proposed by Hung Pham Kim, student, Stanford University, Palo Alto, CA, USA. Let a, b, and c be nonnegative real numbers such that a + b + c = 3. Prove that a2 b + b2 c + c2 a + abc + 4abc(3 − ab − bc − ca) ≤ 4 .

3651.

3652.

Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania. Let α and β be positive real numbers. Find the value of
n n→∞

lim

1+
k=1

kα nβ

Proposed by Peter Y. Woo, Biola University, La Mirada, CA, USA. Let O be the centre of a sphere S circumscribing a tetrahedron ABCD . Prove that: (i) there exists tetrahedra whose four faces are obtuse triangles; and (ii) if O is inside or on ABCD , then at least two faces of ABCD are acute triangles.

3653.

319

3654.

Proposed by Pham Van Thuan, Hanoi University of Science, Hanoi, Vietnam. Let a, b, c, and d be nonnegative real numbers such that a2 +b2 +c2 +d2 = 1. Prove that

3655.

a3 + b3 + c3 + d3 + abc + bcd + cda + dab ≤ 1 .

Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania. Calculate the integral
1 1

1 dxdy, 1 − xy 0 0 where {a} = a − ⌊a⌋ denotes the fractional part of a. x

3656.

Proposed by Michel Bataille, Rouen, France. Let AB be a ﬁxed chord of an ellipse that is not a diameter and let M N be a variable diameter. Show that the locus of the intersection of M A and N B is an ellipse with the same eccentricity as that of the original ellipse, and ﬁnd a geometrical description of its centre.

3657.

Proposed by Thanos Magkos, 3rd High School of Kozani, Kozani, Greece. Prove that for the angles of any triangle the following inequality holds cos2 A cos2 B cos2 C 1 + + ≥ . 2 2 2 1 + cos A 1 + cos B 1 + cos C 2

3658.

Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania. Let −π < θ0 < θ1 < · · · < θk < π and let aj , j = 0, 1, · · · , k, be complex numbers. Prove that if
k n→∞

lim

aj cos(θj n) = 0 ,
j =0

then aj = 0 for all j .

3659.

Proposed by Michel Bataille, Rouen, France. Let P be a point on a circle Γ with diameter AB . The tangent to Γ at P intersects the tangents at A and B in D and C , respectively. Let M be any point of the line BC and V the point of intersection of M D and BP . If the parallel to BC through V meets CD in U , show that the line M U is tangent to Γ.

Proposed by Dragoljub Miloˇ sevi´ c, Gornji Milanovac, Serbia. Triangle ABC has inradius r , circumradius R, and side lengths a, b, c. Prove that y+z x y for all positive real numbers x, y and z . · 1 a2 + z+x · 1 b2 + x+y z · 1 c2 ≥ 1 Rr ,

3660.

320

3661.
USA.

Proposed by Paul Yiu, Florida Atlantic University, Boca Raton, FL,

Consider a triangle ABC with the midpoints D , E , F of its sides BC , CA, AB . For an arbitrary point P , let X , Y , Z be the reﬂections of P in D , E , F respectively. Show that the lines AX , BY , CZ are concurrent.

3662.

Proposed by Michel Bataille, Rouen, France.

Let R denote the set of positive integers whose base ten expression is a single repeated digit (e.g. 5 ∈ R, 222 ∈ R, 88888 ∈ R). Let T (n) = (n − 2)2 + n2 + (n + 2)2 where n is a non-negative integer. (a) Find all even integers n such that T (n) ∈ R. (b) Find one odd integer n > 1 such that T (n) ∈ R. Extra credit will be given to anyone who ﬁnds more than one odd integer n > 1 such that T (n) ∈ R.

3663.

Proposed by Pedro Henrique O. Pantoja, student, UFRN, Brazil.

Let a, b, c be positive real numbers. Prove that
3

2a 4a + 4b + c

2b 4b + 4c + a

2c 4c + 4a + b

< 2.

.................................................................

3650.

Remplacement. Propos´ e par Michel Bataille, Rouen, France.

Soit ABC un triangle et soit respectivement R, O , G et K le rayon et le centre de son cercle circonscrit, son centre de gravit´ e et son point de Lemoine. Montrer que KA GA KB GB KC GC

BC ·

= CA ·

= AB ·

3(R2 − OK 2 ).

Rappelons qu’une sym´ ediane d’un triangle est la r´ eﬂexion d’une m´ ediane par rapport ` a la bissectrice issue d’un mˆ eme sommet. Le point Lemoine d’un triangle est le point d’intersection des trois sym´ edianes.

3651.

Propos´ e par Pham Kim Hung, ´ etudiant, Universit´ e de Stanford, Palo ´ . Alto, CA, E-U Soit a, b et c trois nombres r´ eels non n´ egatifs tels que a + b + c = 3. Montrer que a2 b + b2 c + c2 a + abc + 4abc(3 − ab − bc − ca) ≤ 4 .

321

3652.

Propos´ e par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie.

Soit α et β deux nombres r´ eels positifs. Trouver la valeur de la limite
n n→∞

lim

1+
k=1

kα nβ

´ . Propos´ e par Peter Y. Woo, Universit´ e Biola, La Mirada, CA, E-U Soit O le centre d’une sph` ere S circonscrite ` a un t´ etrah` edre ABCD . Montrer que (i) il existe des t´ etrah` edres dont les quatre faces sont des triangles obtusangles ; et (ii) si O est ` a l’int´ erieur ou sur ABCD , alors au moins deux faces de ABCD sont des triangles acutangles.

3653.

3654.
Vietnam.

Propos´ e par Pham Van Thuan, Universit´ e de Science des Hano¨ ı, Hano¨ ı,

Soit a, b, c et d quatre nombres r´ eels non n´ egatifs tels que a2 + b2 + c2 + d2 = 1. Montrer que

a3 + b3 + c3 + d3 + abc + bcd + cda + dab ≤ 1 .

3655.

Propos´ e par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie. Calculer l’int´ egrale
1 0 1

x
0

1 1 − xy

dxdy,

3656.

o` u {a} = a − ⌊a⌋ d´ enote la partie fractionnaire de a.

Propos´ e par Michel Bataille, Rouen, France. Dans une ellipse, on ﬁxe une corde AB qui ne soit pas un diam` etre et soit M N un diam` etre variable. Montrer que le lieu du point d’intersection de M A et N B est une ellipse de mˆ eme excentricit´ e que l’ellipse originale, et trouver une description g´ eom´ etrique de son centre.

3657.

eme Propos´ e par Thanos Magkos, 3i` -Coll` ege de Kozanie, Kozani, Gr` ece. Montrer qu’on a l’in´ egalit´ e suivante

cos2 A 1+ cos2 A

cos2 B 1+ cos2 B

cos2 C 1+ cos2 C

≥

1 2

pour les angles de n’importe quel triangle.

322

3658.

Propos´ e par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie. Soit −π < θ0 < θ1 < · · · < θk < π et soit aj , j = 0, 1, · · · , k, k nombres complexes. Montrer que si
k n→∞

lim

aj cos(θj n) = 0 ,
j =0

alors aj = 0 pour tout les j .

3659.

Propos´ e par Michel Bataille, Rouen, France. Soit P un point sur un cercle Γ de diam` etre AB . La tangente ` a Γ en P coupe respectivement les tangentes en A et B aux points D et C . Soit M un point quelconque sur la droite BC et V le point d’intersection de M D et BP . Si la parall` ele ` a BC passant par V coupe CD en U , montrer que la droite M U est tangente ` a Γ.

3660.

Propos´ e par Dragoljub Miloˇ sevi´ c, Gornji Milanovac, Serbie. Soit a, b, c les longueurs des cˆ ot´ es d’un triangle ABC , r le rayon de son cercle inscrit et R le rayon de son cercle circonscrit. Montrer que y+z x · 1 a2 + z+x y · 1 b2 + x+y z · 1 c2 ≥ 1 Rr ,

pour tous les nombres r´ eels positifs x, y et z . ´ . Propos´ e par Paul Yiu, Florida Atlantic University, Boca Raton, FL, E-U On consid` ere un triangle ABC et D , E , F les points milieu de ses cˆ ot´ es BC , CA, AB . Pour un point arbitraire P , soit X , Y , Z les r´ eﬂexions respectives de P par rapport ` a D , E , F . Montrer que les droites AX , BY , CZ sont concourantes.

3661.

3662.

Propos´ e par Michel Bataille, Rouen, France. Soit R l’ensemble des entiers positifs dont l’´ ecriture en base dix est la r´ ep´ etition d’un seul chiﬀre (p.ex. 5 ∈ R, 222 ∈ R, 88888 ∈ R). Soit T (n) = (n − 2)2 + n2 + (n + 2)2 o` u n est un entier non n´ egatif. (a) Trouver tous les entiers pairs n tels que T (n) ∈ R. (b) Trouver un entier impair n > 1 tel que T (n) ∈ R. On donnera un cr´ edit suppl´ ementaire ` a quiconque trouve plus d’un tel entier.

3663.

Propos´ e par Pedro Henrique O. Pantoja, ´ etudiant, UFRN, Br´ esil. Soit a, b, c trois nombres r´ eels positifs. Montrer que
3

2a 4a + 4b + c

2b 4b + 4c + a

2c 4c + 4a + b

< 2.

323

SOLUTIONS
No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems.

3224.

[2007 : 112, 115; 2008 : 121- 124] Proposed by J. Chris Fisher and Harley Weston, University of Regina, Regina, SK. Let A0 B0 C0 be an isosceles triangle whose apex angle A0 is not 120◦ . We deﬁne a sequence of triangles An Bn Cn in which △Ai+1 Bi+1 Ci+1 is obtained from △Ai Bi Ci by reﬂecting each vertex in the opposite side (that is, Bi Ci is the perpendicular bisector of Ai Ai+1 , and so forth). Prove that all three angles approach 60◦ as n → ∞. [Ed: This problem is a special case of an open problem described by Judah Schwartz in “Can technology help us make the mathematics curriculum intellectually stimulating and socially responsible?”, International Journal of Computers for Mathematical Learning, 4 (1999), pp. 99–119.]

II. Solution by Gr´ egoire Nicollier, University of Applied Sciences of Western Switzerland, Sion, Switzerland. In his paper [1] Nicollier provides another solution to our problem (which is restricted to isosceles triangles). In addition he resolves Schwartz’s open problem by describing those triangles for which iterating the reﬂection map produces a sequence of triangles whose limit is equilateral, whose limit is degenerate, and whose limit is neither equilateral nor degenerate.
References [1] Gr´ egoire Nicollier, Reﬂection triangles and their iterates, Forum Geometricorum, 12 (2012) 83–128.

3551.
Romania.

[2010 : 314, 316] Proposed by Ovidiu Furdui, Campia Turzii, Cluj,

Let p ≥ 2 be an integer. Find the product
∞ n=1

1 1+ √ p ⌊ n⌋

(−1)n−1

where ⌊a⌋ is the greatest integer not exceeding a. Solution by Joel Schlosberg, Bayside, NY, USA. [mp + 1, (m + 1)p − 1], so
p (m+1) È −1

There are an equal number of odd and even integers in the interval
n=mp

(−1)n−1 = (−1)m

−1

= (−1)m−1 .

324

Therefore,
N n=1

1 1 + ¤√ ¥ p n
√

(−1)n−1

⌊ p N ⌋−1 (m+1)p −1
m=1 n=mp

1 1 + ¤√ ¥ p n
(m+1)p −1

(−1)n−1

⌊ p N ⌋−1
m=1 √ p ⌊ N ⌋−1 m=1

√

n=⌊

√ p p N⌋

1 1 + ¤√ ¥ p n
N È √ p p N⌋

(−1)n−1

1 m

n=mp

(−1)n−1

1 í 1 + ê√ p N
ê√
p

n=⌊

(−1)n−1

m+1 m

(−1)m−1

1+O

1 N

Letting N → ∞ and using Wallis’ product,
∞ n=1

1 1 + ¤√ ¥ p n

(−1)n−1

∞ m=1

m+1 m · 4 3 · 4 5 ·

(−1)m−1

2 1

2 3

6 5

6 7

··· =

π 2

Also solved by OLIVER GEUPEL, Br¨ uhl, NRW, Germany; JOHN HAWKINS and ˇ DAVID R. STONE, Georgia Southern University, Statesboro, GA, USA; NEVEN JURI C, Zagreb, Croatia; PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; ALBERT STADLER, Herrliberg, Switzerland; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. One incorrect solution was submitted.

3552.

[2010 : 314, 316] Proposed by N. Javier Buitrago Aza, Universidad Nacional de Colombia, Bogota, Colombia. Let θ be a real number. Prove that
n− 1 k=0

2kπ −θ n 2kπ 3 + 2 cos −θ n

sin

2 + 4(−1)n sin2 5Fn

(−1)n n sin(nθ )

nθ 2

where Fn denotes the nth Fibonacci number. Solution by Michel Bataille, Rouen, France, modiﬁed and expanded by the editor. √ First, we have, by the well known Binet’s formula that 5Fn = αn − β n √ √ 1 2 where α = 2 (1 + 5) and β = 1 (1 − 5). Note that αβ = −1 so 5Fn = 2 2n 2n n 2 2 2 α + β − 2(−1) . Also, α + β = (α + β ) − 2αβ = 3.

325

Now, for real x, let z = e−ix . Then it is readily checked that sin x 3 + 2 cos x = 1 2i 1 = 2i 1 = 2i · 3 + eix + e−ix 2i 3 + z −1 + z 2i z 2 + 3z + 1 2 1 3z + 2 2 + (α + β 2 )z − 1 −1 = z 2 + 3z + 1 2i z 2 + (α2 + β 2 )z + 1 1 1 + −1 . 2 1+α z 1 + β2z
2πi

eix − e−ix

z −1 − z

1 − z2

Letting w = e− n and replacing x with 2kπ − θ , k = 0, 1, 2, . . . , n − 1, n −i( 2kπ −θ ) iθ k iθ − 2kπi n n we have z = e =e ·e = e · w . Hence, sin x 1 = 3 + 2 cos x 2i 1 1 + −1 1 + α2 eiθ wk 1 + β 2 eiθ wk . (1)

The identity below is known and can be veriﬁed easily by the method of partial fractions decomposition: 1 tn −1 = 1 n
n− 1 k=0

1 wk t −1

(2)

From (1) and (2) we have
n− 1 k=0

sin 3+

2kπ −θ n 2 cos 2kπ − n

θ + 1 (−β 2 eiθ )wk −1 +1

= = = =

−1 2i n 2i n 2i n 2i ·

n− 1 k=0

1 (−α2 eiθ )wk −1

1 1 + −1 n 2 n inθ 1 − (−1) α e 1 − (−1)n β 2n einθ 1 − (−1)n (α2n + β 2n )einθ + e2inθ e−inθ − einθ 1 − e2inθ

2 The result now follows by substituting α2n + β 2n = 2(−1)n + 5Fn and 2 nθ cos(nθ ) = 1 − 2 sin . 2

e−inθ − (−1)n (α2n + β 2n ) + einθ (−1)n n sin(nθ ) = 2n . α + β 2n − 2(−1)n cos(nθ )

Also solved by PAUL DEIERMANN, Southeast Missouri State University, Cape Girardeau, MO, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; JOEL SCHLOSBERG, Bayside, NY, USA; ALBERT STADLER, Herrliberg, Switzerland; and the proposer. Both Bataille and Geupel pointed out that this problem (proposed by the same person) has appeared as problem U173 in Mathematical Reﬂections, 2010, issue 5. The solution featured above is diﬀerent from the one that was published in issue 6.

326

3553.

[2010 : 314, 317] Proposed by Michel Bataille, Rouen, France.

Let A, B , and C be the angles of a triangle. Prove that B 2 C 2
2

sin A cos
cyclic

cos

≤

cos6
cyclic

A 2

Solution by Dionne Bailey, Elsie Campbell, and Charles R. Diminnie, Angelo State University, San Angelo, TX, USA. Using basic trigonometric identities and the fact that A + B + C = π , we obtain B 2 C 2 A cos B cos C = cos A cos C
å

sin A cos

cos A

= 2 sin = 1

= = = =

2 2 2 2 2 2 2 2 å A A+B+C A+B−C A−B+C cos sin + sin + sin 2 2 2 2 2 è A−B−C + sin 2 1 A π π π π cos sin + sin − C + sin − B − sin −A 2 2 2 2 2 2 1 A cos [1 + cos C + cos B − cos A] 2 2 å è 1 A C B A cos 2 cos2 + 2 cos2 − 2 cos2 2 2 2 2 2 å è A 2 C 2 B 2 A cos cos . + cos − cos 2 2 2 2

cos

sin

A+B

+ sin

A−B

It follows that B C cos 2 2
2

sin A cos

= cos6

A A B C + cos2 cos4 + cos4 2 2 2 2 B 2 B 2 C 2 A + 2 cos cos − 2 cos cos2 2 2 2 2 è A C − 2 cos2 cos2 , 2 2
C 2

with similar expressions for

sin B cos

cos

A 2 2

and

sin C cos

A 2

cos

B 2 . 2

327

Therefore, sin A cos
cyclic

B 2 A 2

cos

C 2

=
cyclic

cos6
å

− − − = ≤

cos
å

A 2 B 2 C 2

cos

B 2 C 2 A 2

− 2 cos

A 2 A 2 A

cos

B 2 B 2 B

cos

C 2 C 2 C

+ cos

A 2 B 2 C

cos

cos2
å

cos4 cos4

− 2 cos2 − 2 cos2

cos2

+ cos2

cos4

2 è A 2 è B

cos2 cos2 + cos2 cos4 2 2 2 2 2 å è2 B A A C cos6 − cos2 cos2 − cos2 2 2 2 2 cyclic cyclic cos6
cyclic

cos2

A . 2

cos2 A 2

Furthermore, since 0 < A , B, C < π , equality is attained if and only if 2 2 2 2 B C 2 2 = cos 2 = cos 2 ; that is, if and only if A = B = C = π , so that 3 the given triangle is equilateral.
Also solved by ARKADY ALT, San Jose, CA, USA; GEORGE APOSTOLOPOULOS, ˇ ´ University of Sarajevo, Sarajevo, Bosnia and Messolonghi, Greece; SEFKET ARSLANAGI C, Herzegovina; OLEH FAYNSHTEYN, Leipzig, Germany; OLIVER GEUPEL, Br¨ uhl, NRW, ´ student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; Germany; SALEM MALIKIC, ALBERT STADLER, Herrliberg, Switzerland; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer.

3554.

[2010 : 314, 317] Proposed by Pham Huu Duc, Ballajura, Australia.

Let a, b, and c be positive real numbers. Prove that √ √ √ Ö Ö a2 + bc b2 + ca c2 + ab a b c + + ≥ + + . b+c c+a a+b b+c c+a a+b Solution by Joe Howard, Portales, NM, USA. By symmetry we can assume that a ≥ b ≥ c > 0. Then (a − c)(b − c) ≥ 0 and thus c2 + ab ≥ c(a + b) . √ c2 + ab a+b c(a + b) a+b
Ö

Hence

≥

c a+b

328

To complete the inequality, we will prove that √ √
2

a2 + bc b+c

b2 + ac a+c

≥

a b+c

b a+b

First we will show that √ √ a2 + bc b2 + ac b+c a+c
Ö

≥

a b+c

b a+b

(1)

This simpliﬁes to c(a3 + b3 ) ≥ abc(a + b). Dividing by c(a + b) this inequality is equivalent to a2 − ab + b2 ≥ ab , or (a − b)2 ≥ 0 . Thus, (1) holds. Now we also show that a2 + bc (b + This simpliﬁes to a4 + b4 + a3 c + b3 c + 2abc2 ≥ a3 b + b3 a + a2 c2 + b2 c2 + a2 bc + ab2 c , or (a − b)2 (a2 + ab + b2 ) + c(a − b)[a(a − c) − b(b − c)] ≥ 0 . This last inequality is an immediate consequence of a ≥ b ≥ c > 0. This completes the proof.
ˇ Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; SEFKET ´ University of Sarajevo, Sarajevo, Bosnia and Herzegovina; OLIVER GEUPEL, ARSLANAGI C, ´ student, Sarajevo College, Sarajevo, Bosnia and Br¨ uhl, NRW, Germany; SALEM MALIKIC, Herzegovina; PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; ALBERT STADLER, Herrliberg, Switzerland; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer .

c)2

b2 + ac (a + c)2

≥

a b+c

b a+b

(2)

329

3555.

[2010 : 315, 317] Proposed by Vahagn Aslanyan, Yerevan, Armenia.

Let a and b be positive integers, 1 < a < b, such that a does not divide b. Prove that there exists an integer x such that 1 < x ≤ a and both a and b divide xφ(b)+1 − x, where φ is Euler’s totient function. Similar solutions by Oliver Geupel, Br¨ uhl, NRW, Germany and John Hawkins and David R. Stone, Georgia Southern University, Statesboro, GA, USA. We give Geupel the write up. We write a = a1 a2 ; b = b1 b2 where
m n
i pα i ; a2 = i pα i ; b1 =

m
i pβ i ; b2 =

n
i pβ i ,

a1 =
i=1

i=m+1

i=1

i=m+1

with distinct primes pi and nonnegative integers αi , βi satisfying αi ≥ βi for all 1 ≤ i ≤ m and αi < βi for all i > m. That is a1 is the product of prime powers that occur with at least the same exponent in a as in b, and b1 is the product of the corresponding powers in b. We prove that x = a1 works. The hypothesis a ∤ b yields x = 1. We know that gcd(b1 , b2 ) = 1, thus φ(b) = φ(b1 )φ(b2 ). Since gcd(x, b2 )=1, by Euler Theorem we have xφ(b) ≡ xφ(b2 ) Hence xφ(b)+1 ≡ x (mod a1 b2 ) . From the deﬁnition of a1 , a2 , b1 , b2 , it follows that b1 |a1 and a2 |b2 and thus, both a, b divide a1 b2 . Thus, both a, b divide xφ(b)+1 − x which completes the proof.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, France; JOSEPH ´ student, Sarajevo DiMURO, Biola University, La Mirada, CA, USA; SALEM MALIKIC, College, Sarajevo, Bosnia and Herzegovina; JOEL SCHLOSBERG, Bayside, NY, USA; ALBERT STADLER, Herrliberg, Switzerland; and the proposer . There was also an incorrect solution. Three solvers mentioned that this problem appeared simultaneously as Problem O170 in Mathematical Reﬂections 5 (2010). Geupel pointed that the solution featured in that journal is a wrong, since it uses the incorrect fact that gcd( and b are relatively prime [ED: a = 4, b = 6 a,b) is a counterexample].
φ(b1 )

≡ (1)φ(b1 ) ≡ 1 (mod b2 ) .

330

3557.

[2010 : 315, 317] Proposed by Paolo Perfetti, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy. Let {ak }∞ k=1 be a sequence of positive real numbers with ak+1 ≤
ak (p) . Let Sn = 1 − ak
2n n→∞ n È k=1 1/p ∞ È k=1

ak = 1 and

ap k
n

, and for p ≥ 1 prove that j 1/p ak+j aj Sk+j
(p) 1/n

lim

k 2

= 0.

k=n+1

j =1

[Ed.: The upper limit of summation is now correctly stated as 2n; our apologies for this error.] Solution by Albert Stadler, Herrliberg, Switzerland. We will only È use the condition that {ak }∞ k=1 is a sequence of positive numbers such that ak converges to a positive number c. The conditions that c = 1 and ak+1 ≤
k≥1

We ﬁrst note that by the AM–GM Inequality (a1 a2 · · · an )1/n ≤ a1 + a2 + · · · + an c ≤ . n n
1/n

ak are superﬂuous. 1 − ak

It follows that
2n k=n+1

k n

n j =1

j 1/p ak+j aj Sk+j
(p)

1/n

≤ 2n ·

c n

n n<k≤2n

max

j 1/p ak+j Sk+j
(p)

. (1)

j =1

Put
n (p) Tn = k=1 n (p) Mn

ap k, j 1/p ak+j Sk+j
(p) p/n n

n<k≤2n

max

j =1

n<k≤2n

max

jap k+j Tk+j
(p)

1/n

j =1

We also have
n 1/n n 1/n

j
j =1 n 1/n

≤ 1 n

n
j =1 n

= n,

(2)

ap k+j
j =1 (p)

≤

ap k+j ,
j =1 (p) (p)

(3)

Tk+n ≥ Tk+n−1 ≥ · · · ≥ Tk+1 ≥ Tk

(p)

(4)

331

where (3) holds by the AM–GM Inequality. Using (2), (3), and (4) we obtain
n È (p) Mn ≤

n<k≤2n

max

j =1

ap k+j
(p)

By assumption,

È
k≥1

ak converges, so there are only ﬁnitely many ak for which
È
k≥1

ak ≥ 1 and the other summands satisfy 0 ≤ ap k ≤ ak < 1. Thus, converges, say to σ . Put ǫk = σ −
n È (p) Mn ≤ k È j =1

ap k

ap j

∞ È j =k+1

ap j.

It then follows that

n<k≤2n

max

j =1

ap k+j ≤
n<k≤2n

(p) Tk

max

ǫk σ − ǫk

→ 0 as n → ∞ .

Finally, we conclude from (1) that
2n k=n+1

k n

n j =1

j 1/p ak+j aj Sk+j
(p)

1/n

→ 0 as n → ∞ ,
p

since the sum is nonnegative and bounded above by 2c n tends to inﬁnity.

Mn , which vanishes as

(p)

Also solved by Oliver Geupel, Br¨ uhl, NRW, Germany, and the proposer. Geupel also corrected the problem and removed the hypothesis ak+1 ≤ ak /(1 − ak ).

3558.

[2010 : 315, 317] Proposed by Johan Gunardi, student, SMPK 4 BPK PENABUR, Jakarta, Indonesia.

Given two distinct positive integers a and b, prove that there exists a positive integer n such that an and bn have diﬀerent numbers of digits. Solution by Skidmore College Problem Solving Group, Skidmore College, Saratoga Springs, NY, USA(abbreviated by editor). Without loss of generality we assume 0 < a < b. For each positive integer j let Ij denote the interval (ja, jb). It will be suﬃcient to show that there exist positive integers n and N such that 10N ∈ In , for then na will require at most N digits and nb more than N digits in their decimal representations. It is observed that Im and Im+1 will overlap for all suﬃciently large m. a Indeed, for any m > M = ⌈ b− ⌉ it is easy to see that m(b − a) > a, therefore a

332

(m + 1)a < mb and thus Im ∩ Im+1 = ((m + 1)a, mb) = ∅. Consequently
m>M

Im = (M + 1)a, lim mb = ((M + 1)a, ∞)
m→∞

. Now take any integer N such that 10N ∈ ((M + 1)a, ∞) and we will have that 10N ∈ In for some n > M as claimed.

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; ROY BARBARA, Lebanese University, Fanar, Lebanon(2 solutions); CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; JOSEPH DiMURO, Biola University, La Mirada, CA, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; KEE-WAI LAU, Hong Kong, China; KATHLEEN E. LEWIS, SUNY Oswego, Oswego, NY, USA; NORVALD MIDTTUN, Royal Norwegian Naval Academy, Sjøkrigsskolen, Bergen, Norway; JOEL SCHLOSBERG, Bayside, NY, USA; DIGBY SMITH, Mount Royal University, Calgary, AB; EDMUND SWYLAN, Riga, Latvia; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer who provided several solutions.

3559

⋆. [2010 : 315, 318] Proposed by Thanos Magkos, 3 Kozani, Kozani, Greece.
(b + c)2 4bc ≤ s2 3r (4R + r ) .

High School of

Let ABC be a triangle with side lengths a, b, c, inradius r , circumradius R, and semiperimeter s. Prove that

Solution I by Albert Stadler, Herrliberg, Switzerland, expanded slightly by the editor. Let F be the area of the triangle. We have the formulae F = So s2 3r (4R + r ) = s4 = s3 abc 4R = rs = s(s − a)(s − b)(s − c) .

3F (4Rs + F ) 3abcs + 3(s − a)(s − b)(s − c) 1 s2 = · 2 3 s − (a + b + c)s + ab + bc + ca 1 (a + b + c)2 = · 3 (a + b + c)2 − 2(a + b + c)2 + 4(ab + bc + ca) 1 a2 + b2 + c2 + 2ab + 2bc + 2ca = · . 3 −a2 − b2 − c2 + 2ab + 2bc + 2ca

333

Hence, the given inequality is equivalent in succession to 3(b + c)2 (−a2 − b2 − c2 + 2ab + 2bc + 2ca)

0 ≤ (3b2 + 10bc + 3c2 )a2 − 2(b + c)(3b2 + 2bc + 3c2 )a

≤ 4bc(a2 + b2 + c2 + 2ab + 2bc + 2ca) + (b + c)2 (3b2 − 2bc + 3c2 ) . (1)

Let f (a) be the quadratic polynomial in a on the right hand side of (1) and let ∆ denote its discriminant. Since the leading coeﬃcient of f (a) is clearly positive, to conclude that f (a) ≥ 0 it suﬃces to show that ∆ ≤ 0. Letting d = 3b2 + 3c2 we have ∆ = 4(b + c)2 [(d + 2bc)2 − (d − 2bc)(d + 10bc)] = 4(b + c)2 (4bcd + 4b2 c2 − 8bcd + 20b2 c2 ) = 4(b + c)2 (−4bcd + 24b2 c2 )

so the proof is now complete.

= −48bc(b + c)2 (b2 + c2 − 2bc) = −48bc(b + c)2 (b − c)2 ≤ 0

Solution II by Joe Howard, Portales, NM, USA. We prove the following extension: (b + c)2 4bc ≤ (a + b + c)3 27abc ≤ s2 3r (4R + r ) .

Let An and Gn denote the arithmetic and geometric means of n positive n numbers, respectively. Then it is known [1] that A is monotonically increasing. Gn
A2 3 The left inequality follows from G ≤ A . G3 2 To establish the right inequality we use the known formulae: a + b + c = 2s 2 +b+c)3 and abc = 4Rrs. Then (a27 ≤ 3r(4s is equivalent in succession to abc R+r )

8s3 s2 ≤ 27(4Rrs) 12Rs + 3r 2 24Rr + 6r 2 ≤ 27Rr 2r ≤ R which is the famous Euler’s Inequality.
ˇ Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; SEFKET ´ University of Sarajevo, Sarajevo, Bosnia and Herzegovina; ROY BARBARA, ARSLANAGI C, Lebanese University, Fanar, Lebanon; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; KEE-WAI LAU, Hong Kong, ´ student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; and China; SALEM MALIKIC, PETER Y. WOO, Biola University, La Mirada, CA, USA. Arslanagi´ c, Geupel and Maliki´ c all pointed out that equality holds if and only if the triangle is equilateral. The proposer remarked that the proposed inequality sharpens the result 3r (4R + r ) ≤ s2 by G. Colombier and T. Doucet (see item 5.5 on p.49 of [2]).

334

References [1] B. Arbed, From Tricks to Strategies for Problem Solving, Int. J. Math., Edu. Sci. Tech. 21(3), 1990; pp. 429 – 438 [2] O. Bottema et al., Geometric Inequalities, Groningen, 1969

3560.

[2010 : 315, 318] Proposed by Pham Van Thuan, Hanoi University of Science, Hanoi, Vietnam. Let x and y be real numbers such that x2 + y 2 = 1. Find the maximum value of ¬ ¬ f (x, y ) = |x − y | + ¬x3 − y 3 ¬ . Solution by Joel Schlosberg, Bayside, NY, USA. Since |xy | ≤
1 (x2 2

+ y2 ) = 1 2 ≥ 0,

1 , 2

x2 + xy + y 2 ≥

1 − 2xy ≥ 0,

and 2 + xy ≥ 0 .

Noting that |x3 − y 3 | = |x − y ||x2 + xy + y 2 | and using the AM-GM inequality we see that (f (x, y ))
2

= |x − y |2 1 + |x2 + xy + y 2 | = (1 − 2xy )(2 + xy )2 ≤

= (x2 − 2xy + y 2 )(1 + x2 + xy + y 2 )2 (1 − 2xy ) + 2(2 + xy ) 3
3

5 3

; for ,

that is, f (x, y ) ≤ {x, y } =

5 3/2 . 3

Since x2 + y 2 = 1, and f (x, y ) = or
5 3/2 . 3

5 3/2 3

√ √ 1+ 5 1− 5 √ , √ 2 3 2 3

{x, y } =

√ √ 1− 5 1+ 5 − √ , − √ 2 3 2 3

the maximum value of f (x, y ) is

Also solved by ARKADY ALT, San Jose, CA, USA(2 solutions); GEORGE ˇ ´ University of Sarajevo, APOSTOLOPOULOS, Messolonghi, Greece; SEFKET ARSLANAGI C, Sarajevo, Bosnia and Herzegovina; DIONNE BAILEY, ELSIE CAMPBELL, and CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, France; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; PAUL DEIERMANN, Southeast Missouri State University, Cape Girardeau, MO, USA; JOSEPH DiMURO, Biola University, La Mirada, CA, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; JOHN HAWKINS and DAVID R. STONE, Georgia Southern University, Statesboro, GA, USA; RICHARD ˇ Zagreb, Croatia; KEEI. HESS, Rancho Palos Verdes, CA, USA; NEVEN JURI C, WAI LAU, Hong Kong, China; NORVALD MIDTTUN, Royal Norwegian Naval Academy, Sjøkrigsskolen, Bergen, Norway; CRISTINEL MORTICI, Valahia University of Tˆ argovi¸ ste,

335

Romania; SKIDMORE COLLEGE PROBLEM SOLVING GROUP, Skidmore College, Saratoga Springs, NY, USA(3 solutions); DIGBY SMITH, Mount Royal University, Calgary, AB; ALBERT STADLER, Herrliberg, Switzerland; EDMUND SWYLAN, Riga, Latvia; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. There was one incorrect submission.

3561.

[2010 : 316, 318] Proposed by Mih´ aly Bencze, Brasov, Romania.

An n-sided polygon has perimeter k with k2 < 2n2 . Prove that some three consecutive vertices along the polygon form a triangle with area less than 1 unit. Solution by George Apostolopoulos, Messolonghi, Greece. If s1 , . . . , sn , sn+1 are the side lengths in cyclic order (with sn+1 = s1 ), Èn then si > 0 and k = i=1 si . A triangle formed by three consecutive vertices that determine an angle θi has an area Fi that satisﬁes Fi = si si+1 sin θi 2 ≤ si si+1 2 .

From this inequality together with the AM-GM inequality and k2 < 2n2 (given) we get
n n

Fi
i=1

≤ ≤ =

si si+1 2
Èn

1 2n
n

si
i=1 2

i=1

1 2n 1 2n k n

i=1

n
2n

k2 2n2

< 1.

We deduce that at least one Fi is less than 1, as desired.
ˇ ´ University of Sarajevo, Sarajevo, Bosnia and Also solved by SEFKET ARSLANAGI C, Herzegovina; ROY BARBARA, Lebanese University, Fanar, Lebanon; OLIVER GEUPEL, ´ student, Sarajevo College, Sarajevo, Bosnia and Br¨ uhl, NRW, Germany; SALEM MALIKIC, Herzegovina; JOEL SCHLOSBERG, Bayside, NY, USA; SKIDMORE COLLEGE PROBLEM SOLVING GROUP, Skidmore College, Saratoga Springs, NY, USA; ALBERT STADLER, Herrliberg, Switzerland; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. There was one incorrect submission. All the submitted solutions were quite similar, although most used an indirect argument.

336

3562.

[2010 : 316, 318] Proposed by Vahagn Aslanyan, Yerevan, Armenia.

Let p be a prime number. Prove that there exists a prime number q such that m m p | (q − 1) and with the property that q k | ap − bp whenever q k | ap − bp for positive integers a, b, m, k with a and b not divisible by q . Solution by Joel Schlosberg, Bayside, NY, USA, modiﬁed slightly by the editor. Since (p2 , p + 1) = 1, by Dirichlet’s Theorem, there exists inﬁnitely many primes in the arithmetic progression p + 1 + np2 , n = 0, 1, 2, . . . . Thus, there exists a prime q such that q = p + 1 + np2 or q − 1 = p + np2 for some n. Hence p | q − 1 but p2 ∤ q − 1. Suppose that a, b, m, k are positive integers such that q ∤ a, q ∤ b and m m q k | ap − bp . Then (ab−1 )p
m

≡ 1 (mod q k )

(1)

where φ denotes Euler’s totient function. By (1) and (2), the multiplicative order of ab−1 modulo q k divides (pm , (q − 1)q k−1 ) which equals p since p ∤ q , p | q − 1 and p2 ∤ q − 1. Therefore, (ab−1 )p ≡ 1 (mod q k ), from which ap ≡ bp (mod q k ) follows.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; JOSEPH DiMURO, Biola University, La Mirada, CA, USA; ALBERT STADLER, Herrliberg, Switzerland; and the proposer. Geupel remarked that this problem is closely related to problem 6 of the IMO 2003. [Ed: The IMO problems asked to show that for each prime p there exists a prime q such that np − p is not divisible by q for any positive integer n.]

where b−1 denotes the multiplicitive inverse of b modulo q k . Since bb−1 ≡ 1 (mod q k ), q k ∤ b−1 so (q k , ab−1 ) = 1. Hence, by Euler’s theorem, we have k−1 k (ab−1 )(q−1)q = (ab−1 )φ(q ) ≡ 1 (mod q k ) (2)

Crux Mathematicorum
with Mathematical Mayhem
Former Editors / Anciens R´ edacteurs: Bruce L.R. Shawyer, James E. Totten, V´ aclav Linek

Crux Mathematicorum
Founding Editors / R´ edacteurs-fondateurs: L´ eopold Sauv´ e & Frederick G.B. Maskell Former Editors / Anciens R´ edacteurs: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

337

SKOLIAD

No. 135

Lily Yen and Mogens Hansen
Please send your solutions to problems in this Skoliad by August 15, 2012. A copy of CRUX with Mayhem will be sent to one pre-university reader who sends in solutions before the deadline. The decision of the editors is ﬁnal. Our contest this month is the Calgary Mathematical Association 35th Junior High School Mathematics Contest, Part B, 2011. Our thanks go to Robert Woodrow and Bill Sands, both of University of Calgary, Alberta, for providing us with this contest and for permission to publish it.

L’Association math´ ematique de Calgary 35e Comp´ etition Junior de Math´ ematique Ronde ﬁnale, partie B, 2011. 1. Ariel a achet´ e une certaine quantit´ e d’abricots. 90% du poids d’un abricot est
constitu´ e d’eau. Elle s` eche les abricots jusqu’` a ce que 60% du poids d’un abricot soit constitu´ e d’eau. 15 kg se sont ainsi ´ evapor´ es. Quel ´ etait le poids initial des abricots (en kg) ?

2. Un groupe de dix amis vont ensemble au cin´ ema. Un autre groupe de neuf amis vont aussi au mˆ eme cin´ ema. Quatorze des ces 19 personnes ont achet´ e chacune en plus une confection r´ eguli` ere de popcorn. Au total il s’est av´ er´ e que le coˆ ut combin´ e du ticket de cin´ ema plus les popcorn ´ etait le mˆ eme pour chacun des deux groupes. Le prix du ticket de cin´ ema est 6$. Trouver tous les prix possibles d’une confection r´ eguli` ere de popcorn. 3. Dans la ﬁgure, |AB | = 6, |AC | = 6, et ∠BAC
est un angle droit. On tire deux arcs de cercle passant par B et C : un arc est centr´ e en A et l’autre est 6 6 un demi-cercle de diam` etre BC . Quelle est l’aire du triangle ABC ? Quelle est la longueur de BC ? B C Calculer l’aire comprise entre les deux arcs de cercle, c’est-` a-dire l’aire de la partie hachur´ ee de la ﬁgure. ´ 4. Etant donn´ e un rectangle diﬀ´ erent d’un carr´ e, un d´ ecoupage carr´ e consiste a ` d´ ecouper le rectangle en deux parties, dont une est un carr´ e (l’autre ´ etant un rectangle possiblement ´ egal a ` un carr´ e). Par exemple, un d´ ecoupage carr´ e d’un rectangle 2 × 7 donne un carr´ e 2 × 2 et un rectangle 2 × 5, comme montr´ e.
2 2 7 −→ 2 2 2 5 A

338

` chaque ´ On part d’un rectangle 40 × 2011. A etape, on eﬀectue un d´ ecoupage carr´ e de la partie rectangulaire qui n’est pas carr´ ee. On continue jusqu’` a ce que toutes les parties soient des carr´ es. Combien de carr´ es y a-t-il au total ? Cinq ´ equipes, A, B , C , D , et E , par´ equipe victoires d´ efaites nuls ticipent a ` un tournoi de hockey o` u chaque A 3 ´ equipe joue contre chacune autre exacteB 1 1 C 1 ment une fois. Tout match r´ esulte en une D 0 victoire pour une ´ equipe et une d´ efaite pour E 4 l’autre ou bien un match nul. Le tableau comportait a ` l’origine tous les r´ esultats du tournoi, mais quelques cases ont ´ et´ e eﬀac´ ees. En d´ epit de l’information manquante, l’issue de chaque match peut ˆ etre d´ etermin´ ee de fa¸ con unique. Pour chacun des dix jeux, de d´ eterminer qui a gagn´ e ou si elle ´ etait un match nul.

6. Soit ABC un triangle dont les longueurs des cˆ ot´ es sont donn´ ees par |AB | = 5, |AC | = 7, et |BC | = 8. Le point D appartenant au cˆ ot´ e AC est tel que |AB | = |CD |. On prolonge le cˆ ot´ e BA au-del` a de A jusqu’` a un point E tel que |AC | = |BE |. Appelons F le point d’intersection de la droite ED et du cˆ ot´ e BC .
E A D

Trouver les longueurs de AD , AE , BF , et F C .

Calgary Mathematical Association 35th Junior High School Mathematics Contest Final Round, Part B, 2011. 1.
Ariel purchased a certain amount of apricots. 90% of the apricot weight was water. She dried the apricots until just 60% of the apricot weight was water. 15 kg of water was lost in the process. What was the original weight of the apricots (in kg)?

A group of ten friends all went to a movie together. Another group of nine friends also went to the same movie together. Fourteen of these nineteen people each bought a regular bag of popcorn as well. It turned out that the total cost of the movie plus popcorn for one of the two groups was the same as for the other group. A movie ticket costs $6. Find all possible costs of a regular bag of popcorn.

339

3. In the diagram, |AB | = 6, |AC | = 6, and ∠BAC is a right angle. Two arcs are drawn: a circular arc with centre A and passing through B and C , and a semi-circle with diameter BC . What is the area of ABC ? What is the length of BC ? Find the area between the two arcs; that is, ﬁnd the area of the shaded region in the diagram.

A 6 B 6 C

4. Given a non-square rectangle, a square-cut is a cutting-up of the rectangle into two pieces, a square and a rectangle (which may or may not be a square). For example, performing a square-cut on a 2 × 7 rectangle yields a 2 × 2 square and a 2 × 5 rectangle, as shown.
2 2 5 2 7 −→ 2 2

You are initially given a 40 × 2011 rectangle. At each stage, you make a squarecut on the non-square piece. You repeat this until all pieces are squares. How many square pieces are there at the end?

5. Five teams, A, B , C , D, and E , participate Team Wins Losses Ties in a hockey tournament where each team plays A 3 against each other team exactly once. Each game B 1 1 C 1 either ends in a win for one team and a loss for D 0 the other, or ends in a tie for both teams. The E 4 table originally showed all of the results of the tournament, but some of the entries in the table have been erased. The result of each game played can be uniquely determined. For each of the ten games, determine who won or if it was a tie. 6.
A triangle ABC has sides |AB | = 5, |AC | = 7, and |BC | = 8. Point D is on side AC such that |AB | = |CD |. We extend the side BA past A to a point E such that |AC | = |BE |. Let the line ED intersect side BC at a point F .
E A D

Find the lengths of AD , AE , BF , and F C .

340

Next follow solutions to the City Competition of the Croatian Mathematical Society, 2010, secondary level, grade 1, given in Skoliad 129 at [2010:481–483].

Let n be a positive integer and a a non-zero real number. Reduce the fraction a3n+1 − a4 .

a2n+3 + an+4 + a5

Solution by Harris Lin, student, Killarney Secondary School, Vancouver, BC. If n = 1, the expression equals If n = 2, the expression equals
a4 − a4 a5 +a5 +a5

= 0.

a4 (a3 − 1) a7 − a4 = , a7 + a6 + a5 a5 (a2 + a + 1)

but a3 − 13 = (a − 1)(a2 + a + 1), so the expression equals
a4 (a − 1)(a2 + a + 1) a−1 = . a5 (a2 + a + 1) a

If n = 3, the expression equals
a4 (a6 − 1) a10 − a4 = , a9 + a7 + a5 a5 (a4 + a2 + 1)

but, again, a6 − 1 = (a2 )3 − 13 = (a2 − 1)((a2 )2 + (a2 ) + 1) = (a2 − 1)(a4 + a2 + 1) , so the expression equals
a4 (a2 − 1)(a4 + a2 + 1) a2 − 1 = . 5 4 2 a (a + a + 1) a

Now hazard the guess that a3n+1 − a4 a n− 1 − 1 = . a2n+3 + an+4 + a5 a The equation holds if and only if (a3n+1 − a4 )a = (an−1 − 1)(a2n+3 + an+4 + a5 ) , so a3n+2 − a5 = an−1a2n+3 + an−1 an+4 + an−1 a5 − a2n+3 − an+4 − a5 = a3n+2 + a2n+3 + an+4 − a2n+3 − an+4 − a5 = a3n+2 − a5 ,

but this is obviously true, so the guess is correct.

341

Find a positive integer which when multiplied by 9 gives an integer between 1100 and 1200, and when multiplied by 13 gives an integer between 1500 and 1600.

Solution by Gesine Geupel, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany. Since 122 · 9 = 1098 < 1100 and 124 · 13 = 1612 > 1600, the only integer that could satisfy the conditions is 123. Now, 123 · 9 = 1107 and 123 · 13 = 1599, so 123 does indeed satisfy the conditions.

3. Three circles, each with radius 2, are given in the plane such that the centre of each lies on the intersection of the other two. Determine the area of the intersection of the three disks bounded by those circles.
Solution by Kristian Hansen, student, Burnaby North Secondary School, Burnaby, BC.

The three centres form an equilateral triangle with side length 2. The intersection of the three disks consists of this triangle and three segments. √ congruent √ 2 − 12 = Using the Pythagorean Theorem, the triangle has height 2 3. Thus √ √ the area of the triangle is 2 2 3 = 3. Each of the segments are a part of a sector in a circle of radius 2, . Since the triangle is equilateral, the sector √ angle is 60◦ , so the area of the sector √ is 60 2 2 2 π 2 = π . The triangle has area 3 , so each segment has area π − 3. 360 3 3 √ √ √ 3) = 2 π − 2 3. Therefore the area of the intersection is 3 + 3( 2 π − 3

Consider the integer n. Let m be the integer obtained from n by removing its ones digit. If n − m = 2010, ﬁnd n. Solution by Gesine Geupel, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany. Introduce letters for the digits of n and m so the problem has the form
abcd − abc 2010 .

Since abc is at most 999, the digit a must be 2 or 3. However, if a = 3, then n − m > 3000 − 399 = 2601 > 2010, so a = 2. Now the problem has the form

342

2bcd − 2bc 2010 .

Therefore, b must be 2 or 3. If b = 3, then n−m > 2300−239 = 2061 > 2010, so b = 2, and the problem has the form
22cd − 22c 2010 .

Again, c must be 3 or 4, but if c = 4, then n−m > 2240−224 = 2016 > 2010, so c = 3, and the problem has the form
223d − 223 2010 .

Now d must be 3. Thus n = 2233.
Alternatively, let d denote the ones digit of n. Then n = 10m + d, so 2010 = n − m = 9m + d. Since 2010 ÷ 9 ≈ 233.3, m = 233. Finally, d = 2010 − 9m = 3, and n = 2233 as above.

A bag contains a suﬃcient number of red, white, and blue balls. Each child in a given group takes three balls at random from the bag. What is the smallest number of children in the group that ensures that two of them have taken the same combination of balls, that is, the same number of balls of each colour? Solution by Gesine Geupel, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany. The possible colour combinations are RRR, W W W , BBB , RRW , RW W , RRB , BBR, W W B , W BB , RW B . Since there are 10 colour combinations, 10 children could have all diﬀerent combinations, but with 11 children, at least two would have the same combination. Thus the answer is 11 children.

If a2 + 2b2 = 3c2 , prove that a+b b+c + b−a b−c · a + 2b + 3c a+c

is a positive integer. Solution by Harris Lin, student, Killarney Secondary School, Vancouver, BC. Note that a + b a + 2b + 3c a2 + 2ab + 3ac + ab + 2b2 + 3bc · = . b+c a+c ab + bc + ac + c2 Since a2 + 2b2 = 3c2 , this equals 3ab + 3ac + 3bc + 3c2 ab + bc + ac + c2 = 3(ab + ac + bc + c2 ) ab + bc + ac + c2 = 3.

343

Similarly, b−c · a + 2b + 3c a+c = ab + 2b2 + 3bc − ac − 2bc − 3c2 ab + bc − a2 − ac .

b−a

Since a2 = 3c2 − 2b2 , this equals ab + bc − ac − a2 Thus ab + bc − a2 − ac · = 1.

a+b b+c

b−a

b−c

a + 2b + 3c a+c

= 3 +1 = 4.

A right triangle, ABC , with legs of lengths 15 and 20 and the right angle at vertex B is congruent to a triangle, BDE , with the right angle at vertex D . The point C lies strictly inside the segment BD , and the points A and E are on the same side of the straight line BD . (a) Find the distance between points A and E . (b) Find the area of the intersection of ABC and BDE .

Solution by Kristian Hansen, student, Burnaby North Secondary School, Burnaby, BC. (a) The triangles must sit as in the diagram on the A E left. Since ∠ABC = ∠BDE = 90◦ , there F exists a point F on AB such that BDEF is a 20 15 rectangle. Now |AF | = 5 and |EF | = 20, so the Pythagorean Theorem yields that √ |AE |2 = 15 2 2 20 425 = B C D √ + 5 = 425. Thus |AE | = 20 5 17. (b) Now impose a coordinate system such that A = (0, 20), B = (0, 0), C = (15, 0), D = (20, 0), and E = (20, 15). Then the line through B 3 x= 4 x. Moreover, the line through A and and E has the equation y = 15 20 −20 C has the equation y = 15 x + 20 = − 4 x + 20. These two lines intersect 3 9+16 3 4 when 4 x = − 3 x + 20, thus 12 x = 20 ⇒ 25 x = 20, so x = 48 , and 12 5 3 36 y = 4x = 5 . Thus the intersection between ABC and BDE is itself a triangle with 1 height 36 and base 15. Therefore the desired area is 2 · 15 · 36 = 54. 5 5

344

Let p and q be diﬀerent odd prime numbers. Prove that the integer (pq + 1)4 − 1 has at least four diﬀerent prime divisors. Solution by the editors. Since a2 − b2 = (a − b)(a + b), (pq + 1)4 − 1 = = (pq + 1 − 1)(pq + 1 + 1)((pq )2 + 2pq + 1 + 1) = pq (pq + 2)(p2 q 2 + 2pq + 2) . (pq + 1)2 − 1 (pq + 1)2 + 1

Since p and q are odd, pq +2 is not divisible by either p or q . Let n denote pq +2, and note that n is odd. Then pqn + 2 = pq (pq + 2) + 2 = p2 q 2 + 2pq + 2, so (pq + 1)4 − 1 = pqn(pqn + 2) . Note that pqn + 2 is not divisible by p or q since they are odd. If n is not a power of a single prime, then n is divisible by at least two diﬀerent primes, and (pq + 1)4 − 1 is divisible by at least four diﬀerent primes. If n is a power of a prime, say n = r k where r is a prime and k is a positive integer, then r is odd because n is odd. Therefore pqn + 2 is not divisible by either of p, q , or r , so pqn + 2 contains a prime factor diﬀerent from p, q , and r . Thus (pq + 1)4 − 1 is divisible by at least four diﬀerent primes.
This problem was the Problem of the Month in the December 1999 issue of Crux Mathematicorum, [1999 : 495]. We encourage the reader to look up the solution at http://cms.math.ca/crux/

This issue’s prize of one copy of Crux Mathematicorum for the best solutions goes to Gesine Geupel, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany. We hope to receive more reader solutions to this issue’s featured contest.

345

MATHEMATICAL MAYHEM
Mathematical Mayhem began in 1988 as a Mathematical Journal for and by High School and University Students. It continues, with the same emphasis, as an integral part of Crux Mathematicorum with Mathematical Mayhem. The interim Mayhem Editor is Shawn Godin (Cairine Wilson Secondary School, Orleans, ON). The Assistant Mayhem Editor is Lynn Miller (Cairine Wilson Secondary School, Orleans, ON). The other staﬀ members are Ann Arden (Osgoode Township District High School, Osgoode, ON), Nicole Diotte (Windsor, ON), Monika Khbeis (Our Lady of Mt. Carmel Secondary School, Mississauga, ON) and Daphne Shani (Bell High School, Nepean, ON).

Mayhem Problems
Veuillez nous transmettre vos solutions aux probl` emes du pr´ esent num´ ero avant le 1 septembre 2012. Les solutions re¸ cues apr` es cette date ne seront prises en compte que s’il nous reste du temps avant la publication des solutions. Chaque probl` eme sera publi´ e dans les deux langues oﬃcielles du Canada (anglais et fran¸ cais). Dans les num´ eros 1, 3, 5 et 7, l’anglais pr´ ec´ edera le fran¸ cais, et dans les num´ eros 2, 4, 6 et 8, le fran¸ cais pr´ ec´ edera l’anglais. La r´ edaction souhaite remercier Rolland Gaudet, Universit´ e de Saint-Boniface, Winnipeg, MB, d’avoir traduit les probl` emes.

M501.

´ Propos´ e par l’Equipe de Mayhem.

Un grillage 4 par 4 est form´ e de chevilles amovibles se situant a ` un centim` etre l’une de l’autre, tel qu’illustr´ e dans le sch´ ema. Des ´ elastiques sont attach´ es aux chevilles de fa¸ con a ` former des carr´ es ; deux carr´ es diﬀ´ erents de taille 2 par 2 sont illustr´ es dans le sch´ ema. Combien de diﬀ´ erents carr´ es sont possibles ?

M502.

´ Propos´ e par l’Equipe de Mayhem.

Lors de leur dernier match de basket, Arianne, Bernard et Claudine ont compt´ e un total de 23 points entre eux. Chaque joueur a compt´ e au moins 1 point ; Claudine a compt´ e au moins 10 fois. De combien de fa¸ cons les 23 points peuvent-ils ˆ etre r´ epartis, s’il faut satisfaire aux conditions ci-haut ? Par exemple, 5 points par Arianne, 3 points par Bernard et 15 points par Claudine serait une possibilit´ e et 3 points par Arianne, 5 par Bernard et 15 par Claudine en serait une autre. Propos´ e par John Grant McLoughlin, Universit´ e du Nouveau-Brunswick, Fredericton, NB. ´ Ecrire n’importe quel nombre entier et, a ` la suite, y adjoindre son renversement. Par exemple, si on a ´ ecrit 13, on obtiendra 1331. Montrer que le r´ esultat de cette op´ eration est toujours divisible par 11.

M503.

346

M504.

Propos´ e par Bruce Shawyer, Universit´ e Memorial de Terre-Neuve, St. John’s, NL. ` l’int´ A erieur d’un triangle rectangle de cˆ ot´ es 3, 4 et 5, deux cercles ´ egaux sont trac´ es de fa¸ con a ` ce qu’ils soient tangents l’un a ` l’autre et ` a un des cˆ ot´ es. De plus, un des cercles est tangent a ` l’hypot´ enuse ; l’autre cercle de la paire est tangent a ` l’autre cˆ ot´ e. D´ eterminer les rayons des cercles dans les deux cas.

M505.

´ Propos´ e par Neculai Stanciu, Ecole secondaire George Emil Palade, Buz˘ au, Roumanie. D´ emontrer que pour tout entier positif n, les nombres A = 5n + 7 et B = 6n2 +17n +12 sont relativement premiers (i.e. n’ont aucun facteur commun sauf 1).

M506.

Propos´ e par John Grant McLoughlin, Universit´ e du Nouveau-Brunswick, Fredericton, NB ; et Bruce Shawyer, Universit´ e Memorial de Terre-Neuve, St. John’s, NL.

Nous tentons de cr´ eer un ensemble de nombres, tels que chacun est obtenu en prenant seulement ses propres chiﬀres et leur appliquant les quelconques op´ erations arithm´ etiques et/ou symboles qui vous sont familiers. Chaque expression doit inclure au moins un symbole/op´ eration ; le nombre de fois qu’un √ chiﬀre apparait est le mˆ eme que dans le nombre lui-mˆ eme. Par exemple, 1 = 1, 36 = 6 × 3! et 121 = 112 . Toute contribution valide sera reconnue. .................................................................

M501.

Proposed by the Mayhem Staﬀ.

A 4 by 4 square grid is formed by removable pegs that are one centimetre apart as shown in the diagram. Elastic bands may be attached to pegs to form squares, two diﬀerent 2 by 2 squares are shown in the diagram. How many diﬀerent squares are possible?

M502.

Proposed by the Mayhem Staﬀ.

At their last basketball game Alice, Bob and Cindy scored a total of 23 points between them. Each player got at least 1 point, and Cindy scored at least 10. How many diﬀerent ways could the 23 points been awarded to satisfy the conditions? For example: 5 points for Alice, 3 points for Bob, 15 for Cindy; and 3 points for Alice, 5 points for Bob, 15 for Cindy; are two diﬀerent possibilities.

M503.

Proposed by John Grant McLoughlin, University of New Brunswick, Fredericton, NB. Write any number and then follow that number by adjoining its reversal. For example, if you write 13 then you would get 1331. Show that the resulting number is always divisible by 11.

347

M504.

Proposed by Bruce Shawyer, Memorial University of Newfoundland, St. John’s, NL. Inside a right triangle with sides 3, 4, 5, two equal circles are drawn that are tangent to one another and to one leg. One circle of the pair is tangent to the hypotenuse. The other circle of the pair is tangent to the other leg. Determine the radii of the circles in both cases.

M505.

Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania. Prove that, for all positive integers n, the quantities A = 5n + 7 and B = 6n2 + 17n + 12 are coprime (i.e. have no common factors other than 1).

M506.

Proposed by John Grant McLoughlin, University of New Brunswick, Fredericton, NB; and Bruce Shawyer, Memorial University of Newfoundland, St. John’s, NL. We are trying to create a set of positive integers, that each can be formed using their own digits only, along with any mathematical operations and/or symbols that are familiar to you. Each expression must include at least one symbol/operation; the number of √ times a digit appears is the same as in the number itself. For example, 1 = 1, 36 = 6 × 3! and 121 = 112 . All valid contributions will be acknowledged.

Mayhem Solutions
M463.
√ The square ABCD has side length 2 2. A circle with centre A and radius 1 is drawn. A second circle with centre C is drawn so that it just touches the ﬁrst circle at point P on AC . Determine the total area of the regions inside the square but outside the two circles. Solution by Gloria Fang, student, University of Toronto Schools, Toronto, ON. From the Pythagorean Theorem we get √ √ √ AC = (2 2)2 + (2 2)2 = 8+8 = √ 16 = 4. Since CP is a radius of the larger circle, and AP = 1 we get CP = CA − P A = 4 − 1 = 3, therefore the radius of the large circle is 3. Using [A] to represent the area of ﬁgure A we get H B P √ 2 2 Proposed by the Mayhem Staﬀ.

A 1 J F

√ 2 2

348

[GDF ] = [F CG] − [F CD ] α · π (3)2 − = 2π √ 9 = α− 2 2

√ √ 32 − (2 2)2 · 2 2 2

where α = ∠F CG = cos−1 2 3 2 . Next, we have [P CDF ] = [P CG] − √ π [GDF ] = 98 − 9 α + 2. Thus [JP F ] = [ACD ] − [AP J ] − [P CDF ] = 2 √ √ π π 4− π − 98 −9 α + 2 = 4 − 2 − 54 +9 α. 8 2 2 By symmetry [IP E ] = [JP E ] thus the area of the regions inside the square but not inside a circle is 2[JP E ], or √ 5π + 9α = 8 − 2 2 − + 9 cos−1 8−2 2− 2 2 √ 5π √ 2 2 3 =0 ˙ .38 units.

√

´ IES Also solved by FLORENCIO CANO VARGAS, Inca, Spain; RICARD PEIRO, “Abastos”, Valencia, Spain; and BRUNO SALGUEIRO FANEGO, Viveiro, Spain. Four incorrect solutions were received.

M464.

Proposed by the Mayhem Staﬀ.

Let x be the greatest integer not exceeding x. √ For example, 3.1 = 3 and −1.4 = −2. Find all real numbers x for which x2 + 1 − 1 = 2. Solution by Florencio Cano Vargas, Inca, Spain. The given equation is equivalent to the inequalities: 2≤
Ô Ô

Also solved by GLORIA FANG, student, University of Toronto Schools, Toronto, ON; MUHAMMAD HAFIZ FARIZI, student, SMPN 8, Yogyakarta, Indonesia; MITEA MARIANA, ´ IES “Abastos”, Valencia, Spain; No. 2 Secondary School, Cugir, Romania; RICARD PEIRO, PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; and EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON. Two incorrect solutions were received.

√ √ √ √ and then, the result is the interval x ∈ (− 15, −2 2] ∪ [2 2, 15).

⇔ 9 ≤ x2 + 1 < 16 ⇔ 8 ≤ x2 < 15

x2 + 1 − 1 < 3 ⇔ 3 ≤

x2 + 1 < 4

M465. Proposed by Antonio Ledesma L´ opez, Instituto de Educaci´ on Secundaria
No. 1, Requena-Valencia, Spain. The integer 20114022 is divisible by 2011. Determine if there exists a positive integer that is divisible by 2011 and whose digits add to 2011. Solution by Bruno Salgueiro Fanego, Viveiro, Spain.

349

The answer is yes. The positive integer 201120112011 · · · 201110055, which contains 500 groups with digits 2011 and 10055 = 5 × 2011 as its ﬁnal digits, is divisible by 2011 because a direct division would show that it is equal to 2011 × 100010001 · · · 1000100005, where the last number of this product contains 499 groups with digits 1000 and the digits 100005 as its ﬁnal digits. Another calculation shows that the digits add to
(2 + 0 + 1 + 1) + (2 + 0 + 1 + 1) + · · · + (2 + 0 + 1 + 1) + (1 + 0 + 0 + 5 + 5) 500 times Þ ß = 4 + 4 + 4 + · · · + 4 +11 = 2011 .

Other solutions can be obtained in a similar way, for example 20112011 · · · 201112066, made up of 499 groups with digits 2011 and 12066 = 3 × 2011 as its ﬁnal digits.
Also solved by FLORENCIO CANO VARGAS, Inca, Spain; GLORIA FANG, student, University of Toronto Schools, Toronto, ON; SALLY LI, student, Marc Garneau ´ IES “Abastos”, Valencia, Spain; Collegiate Institute, Toronto, ON; RICARD PEIRO, EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; and the proposer.

M466.

Proposed by Pedro Henrique O. Pantoja, student, UFRN, Brazil.

Determine all pairs (m, n) of positive integers such that 2m − 2 = n!. I. Solution by Sally Li, student, Marc Garneau Collegiate Institute, Toronto, ON.
! ! The condition is equivalent to 2m−1 − 1 = n . Since n is non-zero, it must 2 2 be odd, thus n can only be 1, 2 or 3. Trying each case we ﬁnd the only solutions are m = n = 2 or m = n = 3.

II. Solution by Cao Minh Quang, Nguyen Binh Khiem High School, Vinh Long, Vietnam. We have 2m − 2 = n! or 2m = 2 + n!. If n ≥ 4, then 2 + n! ≥ 26, thus m > 4. Hence, 2m ≡ 0(mod 4) and n! + 2 ≡ 2(mod 4), a contradiction. Therefore, n ≤ 3. By direct calculation we get (m, n) = (3, 3) or (m, n) = (2, 2).
Also solved by FLORENCIO CANO VARGAS, Inca, Spain; GLORIA FANG, student, ´ IES “Abastos”, Valencia, Spain; University of Toronto Schools, Toronto, ON; RICARD PEIRO, BRUNO SALGUEIRO FANEGO, Viveiro, Spain; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; and the proposer. One incorrect solution was received.

M467.

Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania. Determine all real numbers x for which (x − 2010)3 + (2x − 2010)3 + (4020 − 3x)3 = 0 .

Solution by Jos´ e Luis D´ ıaz-Barrero, Universitat Polit` ecnica de Catalunya, Barcelona, Spain. First, we observe that the given equation can be written as (x − 2010)3 + (2x − 2010)3 = (3x − 4020)3 .

350

Putting a = x − 2010, b = 2x − 2010, and c = 3x − 4020, we have a + b = c and a3 + b3 = c3 from which immediately follows (a + b)3 = a3 + b3 or 3ab(a + b) = 0. To get the solutions we consider the following two cases: (i) a + b = 3x − 4020 = 0 from which we obtain the solution x = 1340; (ii) ab = (x − 2010)(2x − 2010) = 0 from which we obtain the solutions x = 2010 and x = 1005. Since the polynomial has degree three and we have found three solutions, on account of the Fundamental Theorem of Algebra, the given equation does not have more roots and we are done.
´ Also solved by MIHALY BENCZE, Brasov, Romania; FLORENCIO CANO VARGAS, Inca, Spain; CAO MINH QUANG, Nguyen Binh Khiem High School, Vinh Long, Vietnam; GLORIA FANG, student, University of Toronto Schools, Toronto, ON; WINDA KIRANA, student, SMPN 8, Yogyakarta, Indonesia; SALLY LI, student, Marc Garneau Collegiate ´ IES “Abastos”, Valencia, Spain; PAOLO Institute, Toronto, ON; RICARD PEIRO, PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; and the proposer. Bencze generalized the problem by writing it as (ax − b)3 + (cx − d)3 + (b + d − (a + b+d b , x2 = d and x3 = a . Thus c)x)3 = 0, with ac(a + c) = 0, which has solutions x1 = a c +c setting a = 1, b = 2010, c = 2 and d = 2010 yields the given problem and its solutions.

M468.
Romania.
3

Proposed by Gheorghe Ghit ¸a ˘, M. Eminescu National College, Buz˘ au,

Determine all pairs (p, q ) of prime numbers for which each of p + q , p + q 2 , p + q , and p + q 4 is a prime number. Solution by Florencio Cano Vargas, Inca, Spain. Let p, q be prime numbers. If we require that p + q > 2 has to be prime, either p = 2 or q = 2 to ensure that p + q is odd. Let us consider two cases separately: Case 1: q = 2 In this case, p must be an odd prime number p ≥ 3. Clearly, p = 3 is a solution since p + q = 5, p + q 2 = 7, p + q 3 = 11 and p + q 4 = 19 are prime numbers. For p > 3 we will study the values of the expressions modulo 3. Let p ≡ t(mod 3) where t = 1 or t = 2 since p > 3 is a prime. Then p + q ≡ t + 4 ≡ t + 1(mod 3)
2

p + q ≡ t + 2(mod 3)

and therefore, either p + q or p + q 2 is divisible by three, thus they both cannot be prime. Case 2: p = 2 In this case, q must be an odd prime number q ≥ 3. Clearly, q = 3 is a solution since p + q = 5, p + q 2 = 11, p + q 3 = 29 and p + q 4 = 83 are prime

351

numbers. For q > 3 we will study the values of the expressions modulo 3. Let q ≡ t(mod 3) where t = 1 or t = 2 since q > 3 is a prime, then p + q 2 ≡ 2 + t2 (mod 3) . p + q ≡ 2 + t(mod 3)

If t = 1, p + q is divisible by three and if t = 2, p + q 2 is divisible by three, therefore, either p + q or p + q 2 is divisible by three, thus they both cannot be prime. Summarizing, the only pairs (p, q ) of prime numbers which satisfy the condition of the problem are (2, 3) and (3, 2).
Also solved by CAO MINH QUANG, Nguyen Binh Khiem High School, Vinh Long, Vietnam; GLORIA FANG, student, University of Toronto Schools, Toronto, ON; MITEA ´ IES “Abastos”, MARIANA, No. 2 Secondary School, Cugir, Romania; RICARD PEIRO, Valencia, Spain; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; and the proposer.

M469. Proposed by Antonio Ledesma L´ opez, Instituto de Educaci´ on Secundaria
No. 1, Requena-Valencia, Spain. Prove that, for all real numbers x, we have 2sin x + 2cos x
2

≥ 22 −

√ 2

Solution by Gloria Fang, student, University of Toronto Schools, Toronto, ON. √ 2sin x + 2cos x By the AM-GM inequality we have that ≥ 2sin x · 2cos x , so 2sin x + 2cos x ≥ 4 · 2sin x · 2cos x thus 2sin x + 2cos x Using well known trigonometric identities we get sin x + cos x = = =
1 √ 2 2

≥ 22+sin x+cos x .

sin x +
1 √ 2 π 4

1 √ 2

cos x
π 4

cos √

sin x + sin
1 √ 2

cos x

Since −1 ≤ sin

π 4

π +x . 4 √ √ + x ≤ 1, we must have 2 sin π + x ≥ − 2. Thus 4 2 sin
2

2sin x + 2cos x

≥ 22+sin x+cos x = 22+

√ 2 sin( π 4 +x)

≥ 22 −

√ 2

´ Also solved by MIHALY BENCZE, Brasov, Romania; FLORENCIO CANO VARGAS, Inca, Spain; CAO MINH QUANG, Nguyen Binh Khiem High School, Vinh Long, Vietnam; ´ LUIS D´ JOSE IAZ-BARRERO, Universitat Polit` ecnica de Catalunya, Barcelona, Spain; SALLY ´ IES “Abastos”, LI, student, Marc Garneau Collegiate Institute, Toronto, ON; RICARD PEIRO, Valencia, Spain; PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; and the proposer. [Ed.: Note the solution could be shortened by noting that sin 2x = 2 sin(x) cos(x) ≤ √ 1 ⇒ (sin(x) + cos(x))2 ≤ 2 ⇒ 2 − 2 ≤ 2 + sin(x) + cos(x), from which the result follows.]

352

THE OLYMPIAD CORNER
No. 296 R.E. Woodrow and Nicolae Strungaru
The problems from this issue come from the Italian Team Selection Test, the British Mathematical Olympiad, the Macedonian Mathematical Olympiad, and the Chinese Mathematical Olympiad. Our thanks go to Adrian Tang for sharing the material with the editors.
Toutes solutions aux probl` emes dans ce num´ ero doivent nous parvenir au plus tard le 1 octobre 2012. Chaque probl` eme sera publi´ e dans les deux langues oﬃcielles du Canada (anglais et fran¸ cais). Dans les num´ eros 1, 3, 5 et 7, l’anglais pr´ ec´ edera le fran¸ cais, et dans les num´ eros 2, 4, 6 et 8, le fran¸ cais pr´ ec´ edera l’anglais. Dans la section des solutions, le probl` eme sera publi´ e dans la langue de la principale solution pr´ esent´ ee. La r´ edaction souhaite remercier Jean-Marc Terrier, de l’Universit´ e de Montr´ eal, d’avoir traduit les probl` emes.

OC31.

Trouver toutes les paires (p, q ) de nombres premiers tels que pq | (5p + 5q ).

OC32. Soit ABC un triangle acutangle avec ∠B = ∠C . Soit O le centre de
son cercle circonscrit et H son orthocentre. Montrer que le centre du cercle BOH se situe sur la droite AB .

OC33. Soit n et k deux entiers tels que n ≥ k ≥ 1. On consid` ere un cercle de

n ampoules ´ electriques, toutes ´ eteintes. A chaque tour, on peut changer le statut d’un ensemble quelconque de k ampoules cons´ ecutives. Parmi les 2n combinaisons possibles, combien peut-on en engendrer (a) si k est un premier impair ? (b) si k est un entier impair ? (c) si k est un entier pair ?

OC34. Soit m et n deux entiers avec 4 < m < n, et A1 A2 · · · A2n+1 un 2n + 1-gone r´ egulier. Soit P = {A1 , A2 , · · · , A2n+1 }. Trouver le nombre de m-gones convexes avec exactement deux angles droits internes et dont les sommets sont tous dans P . OC35. Trouver toutes les paires d’entiers (x, y) telles que
y 3 = 8x6 + 2x3 y − y 2 .

353

OC36. Soit a, b et c les longueurs des cˆ ot´ es oppos´ es respectivement aux angles ∠A, ∠B et ∠C d’un triangle obtusangle ABC . Montrer que
a3 cos A + b3 cos B + c3 cos C < abc .

OC37.

Trouver tous les entiers n rendant possible le coloriage de toutes les arˆ etes et diagonales d’un n-gone convexe avec n couleurs satisfaisant les conditions suivantes : (i) Chacune des arˆ etes et diagonales est colori´ ee par une seule couleur ; (ii) Pour tout ensemble de trois couleurs distinctes, il existe un triangle dont les sommets sont des sommets du n-gone et les trois arˆ etes sont respectivement colori´ ees par les trois couleurs de cet ensemble.

OC38. Soit a, b et c trois nombres r´ eels positifs tels que ab + bc + ca =
Montrer que l’in´ egalit´ e suivante est satisfaite : a a2 − bc + 1 + b b2 − ca + 1 + c c2 − ab + 1 ≥ 1 a+b+c .

1 . 3

´ OC39. Etant donn´ e un entier positif n, soit b(n) le nombre d’entiers positifs dont les repr´ esentations binaires apparaissent comme blocs d’entiers cons´ ecutifs dans la repr´ esentation binaire de n. Par exemple b(13) = 6, puisque 13 = 11012, qui contient comme blocs cons´ ecutifs les repr´ esentations binaires de 13 = 11012, 6 = 1102 , 5 = 1012 , 3 = 112 , 2 = 102 et 1 = 12 . Montrer que si n ≤ 2500, alors b(n) ≤ 39, et d´ eterminer les valeurs de n pour lesquelles on a ´ egalit´ e. et N les intersections de deux cercles Γ1 et Γ2 . Soit AB la tangente commune aux deux cercles, la plus rapproch´ ee de M , disons A ∈ Γ1 et B ∈ Γ2 . Soit respectivement C et D les points sym´ etriques de A et B par rapport a ` M . Soit respectivement E et F les intersections du cercle circonscrit de DCM et des cercles Γ1 et Γ2 . Montrer que les rayons des cercles circonscrits des triangles M EF et N EF sont d’´ egale longueur. .................................................................

OC40. Soit M

OC31. OC32.

Find all pairs (p, q ) of prime numbers such that pq | (5p + 5q ).

Let ABC be an acute-angled triangle with ∠B = ∠C . Let the circumcentre be O and the orthocentre be H . Prove that the centre of the circle BOH lies on the line AB .

354

OC33. Let n and k be integers such that n ≥ k ≥ 1. There are n light bulbs placed in a circle. They are all turned oﬀ. Each turn, you can change the state of any set of k consecutive light bulbs. How many of the 2n possible combinations can be reached
(a) if k is an odd prime? (b) if k is an odd integer? (c) if k is an even integer? Let m, n be integers with 4 < m < n, and A1 A2 · · · A2n+1 be a regular 2n + 1-gon. Let P = {A1 , A2 , · · · , A2n+1 }. Find the number of convex m-gons with exactly two acute internal angles whose vertices are all in P .

OC34.

OC35.

Find all pairs of integers (x, y ) such that y 3 = 8x6 + 2x3 y − y 2 .

OC36.

The obtuse-angled triangle ABC has sides of length a, b, and c opposite the angles ∠A, ∠B and ∠C respectively. Prove that a3 cos A + b3 cos B + c3 cos C < abc .

OC37.

Find all integers n such that we can colour all the edges and diagonals of a convex n-gon by n given colours satisfying the following conditions: (i) Every one of the edges or diagonals is coloured by only one colour; (ii) For any three distinct colours, there exists a triangle whose vertices are vertices of the n-gon and the three edges are coloured by the three colours, respectively.

Let a, b, c be positive real numbers such that ab + bc + ca = Prove the inequality: a a2 − bc + 1 + b b2 − ca + 1 + c c2 − ab + 1 ≥ 1 a+b+c .

OC38.

1 . 3

OC39. Given a positive integer n, let b(n) denote the number of positive integers whose binary representations occur as blocks of consecutive integers in the binary expansion of n. For example b(13) = 6 because 13 = 11012 , which contains as consecutive blocks the binary representations of 13 = 11012 , 6 = 1102 , 5 = 1012 , 3 = 112 , 2 = 102 and 1 = 12 . Show that if n ≤ 2500, then b(n) ≤ 39, and determine the values of n for which equality holds.

355

OC40. Let M and N be the intersection of two circles, Γ1 and Γ2 . Let AB be the line tangent to both circles closer to M , say A ∈ Γ1 and B ∈ Γ2 . Let C be the point symmetrical to A with respect to M , and D the point symmetrical to B with respect to M . Let E and F be the intersections of the circle circumscribed around DCM and the circles Γ1 and Γ2 , respectively. Show that the circles circumscribed around the triangles M EF and N EF have radii of the same length.
First we look at the remainder of the solutions for the 48th IMO Bulgarian Team, First Selection Test, given at [2010: 275] that we started last issue.

Let A1 A2 A3 A4 A5 be a convex pentagon such that the triangles A1 A2 A3 , A2 A3 A4 , A3 A4 A5 , A4 A5 A1 , A5 A1 A2 have the same area. Prove that there exists a point M such that the triangles A1 M A2 , A2 M A3 , A3 M A4 , A5 M A1 have the same area. Solved by Titu Zvonaru, Com´ ane¸ sti, Romania, shortened by the editor.

Triangles A5 A1 A2 and A1 A2 A3 have A4 the same base and area, whence A1 A2 ||A5 A3 . Similarly we deduce that each side of the given A A3 5 pentagon is parallel to a diagonal. If in a convex pentagon each side is parallel to a diagonal, then the ratio of a diagonal to the corresponding parallel side is the golden section √ 1+ 5 A1 A2 ϕ = . For a simple proof of this 2 result see the solution to problem M133 [2005: 278-279]. Recall that aﬃne transformations preserve the ratios of segment lengths along parallel lines, so that the aﬃne transformation that takes the ﬁrst three vertices of the regular pentagon A1 A2 A3 A4 A5 to A1 A2 A3 will take the vertex A4 (which is the point on the line through A1 parallel to A2 A3 for which the directed segment A1 A4 is ϕ times the length of the directed segment A2 A3 ) to the point A4 . Similarly, A5 is taken to A5 . Of course, the centre of gravity M of the regular pentagon has the property that the areas of the ﬁve triangles Ai M Ai+1 are equal. (In fact the triangles are congruent.) Our aﬃne transformation takes M to the centre of gravity M of the given pentagon. Because aﬃne transformations preserve ratios of areas, the areas of Ai M Ai+1 are equal, as required.

Prove that there are no distinct positive integers x and y such that x2007 + y ! = y 2007 + x! .

Solved by George Apostolopoulos, Messolonghi, Greece; and Oliver Geupel, Br¨ uhl, NRW, Germany. We give the solution of Geupel. We prove the more general result that for each integer n ≥ 2 such that n = 2m − 1 (m ∈ N) the function f : N → Z : f (x) = x! − xn is injective.

356

Firstly, we show that f is increasing for x ≥ 2n. Indeed, for x ≥ 2n we have x! ≥ (1 · x)(2 · (x − 1)) · · · (n · (x + 1 − n)) ≥ x · x · · · x = xn . Taking into account that 1 n 1 2n 1+ ≤ 1+ ≤ e, x 2n we obtain å è 1 n 1 n f (x + 1) = 1 + (x! − xn ) + x + 1 − 1 + x! x x æ é 1 2n ≥ f (x) + x + 1 − 1 + x! ≥ f (x) + (x + 1 − e)x! > f (x), 2n which completes the proof that f is increasing for x ≥ 2n. We prove our initial claim by contradiction. Assume that we have x ≤ y and f (x) = f (y ). For k ∈ Z and p prime, let dp (k) denote the greatest α ∈ Z such that pα | k. Let p be any prime divisor of x. Then, p | x!, p | xn , and p | y !. Hence, p | xn − x! + y !, that is p | y n and therefore p | y . From y ! − x! = x![(x + 1)(x + 2) · · · y − 1], we see that dp (x!) = dp (y ! − x!) = dp (y n − xn ) ≥ n. By
ë î

dp (x!) =

x p

x p2

+ ··· ≤ x

1 p

1 p2

+ ···

x p−1

we deduce x ≥ (p − 1)dp (x!) ≥ (p − 1)n. Because f (z ) is increasing for z ≥ 2n, we must have p = 2 and x = 2m . Thus, n ≤ 2m < 2n. From d2 (y ! − x!) = d2 (x!) = 2m 1 2 + 1 4 + ··· + 1 2m = 2m − 1

and n | d2 (y n − xn ), we conclude n | 2m − 1. Consequently n = 2m − 1, a contradiction.

4. Given a point P on the side AB of a triangle ABC , consider all pairs of points (X, Y ), X ∈ BC , Y ∈ AC such that ∠P XB = ∠P Y A. Prove that the mid-points of the segments XY lie on a straight line.
Solved by Titu Zvonaru, Com´ ane¸ sti, Romania. Let Q, R be the projections of P onto BC and AC , respectively, and let M be the midpoint of XY . C Y M

X Q P B

R A

357

We denote m = CR, n = CQ, P = RQ, α = ∠P XB = ∠P Y A. By Stewart’s Theorem we obtain QC 2 · RY − QY 2 · CR + QR2 · CY = CR · CY · Y R ⇔ m · QY 2 = n2 · RY + p2 (m − RY ) − m(m − RY ) · RY ⇔ ⇔ m · QY 2 = RY (n2 − p2 − m2 ) + mp2 + m · RY 2 m · QY 2 = RY (−2mp cos ∠CRQ) + mp2 + m · RY 2

Since the quadrilateral CRP Q is cyclic, we deduce cos ∠CRQ = cos ∠CP Q = It follows that QY 2 = −2p · and similarly RX 2 = −2p · By (1) and (2) we have QY 2 − RY 2
2 2

PQ CP

RX tan α CR

RX · RY · tan α CP

+ p2 + RY 2

(1)

RX · RY · tan α CP

+ p2 + QX 2 .

(2)

⇔ 2(RX + RY ) − XY ⇔ RM 2

= 2(QX 2 + QY 2 ) − XY 2 = QM 2 ,

= RX 2 − QX 2

hence the point M lies on the perpendicular bisector of QR.

The real numbers ai , bi , 1 ≤ i ≤ n, are such that
n n n

a2 i = 1,
i=1 i=1

b2 i = 1

and
i=1

ai bi = 0 .

Prove that
n 2 n 2

ai
i=1

+
i=1

≤ n.

Solved by George Apostolopoulos, Messolonghi, Greece; and Henry Ricardo, Tappan, NY, USA. We use the write-up of Apostolopoulos. Let x =
Èn
i=1

ai and y =

Èn

i=1

bi . By Cauchy-Schwarz Inequality we

358

have
n 2 n

(x2 + y 2 )2

=
i=1 n

(ai x + bi y )

≤n

(ai x + bi y )2
i=1

= n
i=1

2 2 2 (a2 i x + bi y + 2ai bi xy ) n n 2 a2 i +y i=1 2 i=1 n

= n x2

b2 i + 2xy
i=1

ai bi

= n(x2 + y ) so (x2 + y 2 )2 ≤ n(x2 + y 2 ), namely
n 2 n 2

x +y ≤n⇔

ai
i=1

+
i=1

≤ n.

For a ﬁnite set S denote by P (S ) the set of all subsets of S . The function f : P (S ) → R is such that f (X ∩ Y ) = min(f (X ), f (Y )) for any two subsets X, Y ∈ P (S ). Find the largest number of distinct values that f can take. Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA. Given S = {a 1 , a 2 , . . . , a n },

deﬁne, for i = 1, 2, . . . , n, Xi = S \{ai }. Set xi = f (Xi ), and deﬁne M = f (S ). This deﬁnes f on all other subsets of S since all others can be formed from intersections of these. Moreover, these other function values are in {x1 , x2 , . . . , xn , M }. Hence, f has at most n + 1 distinct values. In fact, each subset of S can be obtained in a unique way (apart from order) from intersections of the Xi and S ; hence, any set of choices of the xi and of M gives an allowable f . Accordingly, f can have n + 1 distinct values.

Next we turn to the solutions of problems of the 48th IMO Bulgarian Team, Second Selection Test, given at [2010; 276].

Find all positive integers m such that 2m αm − (α + β )m − (α − β )m 3α2 + β 2

359

is an integer for all integer values of α, β with αβ = 0. Solved by George Apostolopoulos, Messolonghi, Greece; Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Apostolopoulos. Denote x = α + β , y = α − β . Then (x + y )m − xm − y m (2α)m − (α + β )m − (α − β )m = , 3α2 + β 2 x2 + xy + y 2 where αβ = 0 ⇔ x2 = y 2 . Now we consider (x + y )m − xm − y m and x2 + xy + y 2 as polynomials with one variable x. Do the division and we get that (x + y )m − xm − y m = (x2 + xy + y 2 )f (x, y ) + g (y )x + h(y ), where f (x, y ), g (y ), h(y ) are polynomials with integer coeﬃcients. Suppose m satisﬁes that x2 + xy + y 2 | (x + y )m − xm − y m . Fix y , let x vary in the set of positive integers. We have x2 + xy + y 2 | g (y )x + h(y ). But for very large x, |x2 + xy + y 2 | > |g (y )x + h(y )|, then g (y )x + h(y ) = 0 ⇒ g (y ) = h(y ) = 0. We deduce that for every y , g (y ) = h(y ) = 0. Thus g and h are both zero polynomials. On the other hand if g and h are zero polynomials, it is clear that x2 + xy + y 2 | (x + y )m − xm − y m for x2 = y 2 . Thus m is valid ⇔ g ≡ h ≡ 0. π π Next let w = cos 23 + i sin 23 , then w2 + w + 1 = 0, w3 = 1. We claim that g ≡ h ≡ 0 ⇔ (w + 1)m − wm − 1 = 0. If g ≡ h ≡ 0, then (x + y )m − xm − y m = (x2 + xy + y 2 )f (x, y ). This holds for all x and y . Take x = wy , then x2 + xy + y 2 = 0 ⇒ (x + y )m − xm − y m = 0 ⇒ y m [(w + 1)m − wm − 1] = 0, choose y = 0 then (w + 1)m − wm − 1 = 0. If (w + 1)m − wm − 1 = 0 we take x = wy , then (x + y )m − xm − y m = 0 and x2 + xy + y 2 = 0. We deduce that g (y )wy + h(y ) = 0. Since g (y ) and h(y ) are real numbers and w is not, g (y )y must be zero. Then g (y ) ≡ 0 for all y . Finally h(y ) ≡ 0. Now we only need to ﬁnd m such that (w + 1)m − wm − 1 = 0 , w + 1 = −w2 , (−1)m w2m − wm − 1 = 0 . Case 1: If m ≡ 0 (mod 3), then (−1)m w2m = (−1)m , wm = 1, and (−1)m − 2 = 0, no solution. Case 2: If 3 m we have (−1)m ω 2m − ω m − 1 = 0 . Also, since ω is a third root of unity and 3 m, we have ω 2m + ω m + 1 = 0 . By adding these two relations we get ω 2m [1 + (−1)m ] = 0 . Thus (−1)m = −1 ⇒ 2 m .

360

Thus 3 m; 2 m ⇒ m = ±1 (mod 6) . Finally all valid m are m ≡ ±1 (mod 6).

3. Find all integers n ≥ 3 such that: for any two positive integers m < n − 1, r < n − 1 there exist m distinct elements of the set {1, 2, . . . , n − 1} whose sum is congruent to r modulo n.
Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA. We give the solution of Curtis. We claim that the n’s with the desired property are exactly the odd n ≥ 3. 1. Suppose that k is a positive integer, n = 2k + 2, m = 2k, r = k + 1, and Sn = {1, 2, . . . , n − 1}. For j ∈ Sn , let xj denote the sum of all elements of Sn except the j th . Then xj = (k + 1)(2k + 1) − j . Thus, xj ≡ r mod n ⇔ (k + 1)(2k + 1) − j ≡ (k + 1) mod 2k + 2 ⇔ 2(k + 1) divides 2k(k + 1) − j ⇔ (k + 1)(2k) ≡ j mod 2(k + 1)

Since 2(k + 1) divides 2k(k + 1), n must divide j . But 1 ≤ j ≤ n − 1, a contradiction. Hence, no even n has the desired property. 2. Now suppose that k is a positive integer and n = 2k + 1. Let r ∈ Sn . (a) Suppose m = 2l + 1, where l is a positive integer less than k. Let
A= {1, 2, 3, . . . , r − 1, r + 1, . . . , l + 1} {1, 2, 3, . . . , l} {1, 2, . . . , 2k − r, 2k + 2 − r, . . . , l + 1} if if if r≤l l + 1 ≤ r ≤ 2k − l r ≥ (2k + 1) − l.

Then
i∈A

[i + (n − i)] + r

(b) Suppose m = 2l and r ≥ 3. Let
A=

is a sum of m distinct elements of Sn . This sum is congruent to r mod n since i + (n − i) ≡ 0 mod n for each i.
{2, 3, . . . , r − 2, r, r + 1, . . . , l + 1} {2, 3, . . . , l} {2, 3, . . . , 2k + 1 − r, 2k + 3 − r, . . . , l + 1} if if if 3≤r ≤l l + 1 ≤ r ≤ 2k − l r ≥ (2k + 1) − l.

Then 1 + (r − 1) +
i∈A

[i + (n − i)]

is a sum of m distinct elements of Sn . The sum is congruent to r mod n.

361

(c) Suppose m = 2l and r = 1. Then
l+1

2 + 2k +
i=3

[i + (n − i)] ≡ 1 mod n.

(d) Suppose m = 2l ≤ 2k − 4 and r = 2. Then
l+2

3 + 2k +
i=4

[i + (n − i)] ≡ 2 mod n.

(e) Suppose m = 2l = 2k − 2 = n − 3 and r = 2. Since i=
i∈Sn

(n − 1)n 2

= kn ≡ 0 mod n,

i=1,2k−2

i∈Sn

i ≡ 0 − (2k − 1) ≡ 2 mod n.

Thus, for n = 2k + 1 ≥ 3, and 1 ≤ m, r < n − 1, there exist m distinct elements of Sn with sum congruent to r mod n.

Solve the system x2 + yu = x2 + yz = (x + u)n u4

where x, y and z are prime numbers and u is a positive integer. Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA. We give the solution by Curtis. From the second equation, yz = (u2 + x)(u2 − x). Since y and z are prime, the only factors of the left-hand side are 1, y , z , and yz . Thus there are at most four possibilities. 1. The case u2 + x = 1, u2 − x = yz is impossible since u and x are both positive. 2. If u2 + x = yz and u2 − x = 1, then x = (u + 1)(u − 1). Since x is prime, we must have u − 1 = 1 and u + 1 = x. Thus, u = 2 and x = 3. From u2 + x = yz , we obtain 7 = yz , which is impossible. 3. If u2 + x = y and u2 − x = z , the ﬁrst equation gives x2 + u(u2 + x) = (x + u)n , or x2 + u3 + ux = (x + u)n . If n ≥ 3, then (x + u)n ≥ (x + u)3 > x2 + u3 + ux. Hence any solution must have n < 3.

362

(a) If n = 1, then x2 + u3 + ux = x + u, which can be rewritten as x(x − 1) + u(u2 − 1) + ux = 0 which is impossible since the left-hand side is positive. (b) If n = 2, then x2 + u3 + ux = x2 + 2ux + u2, which can be rewritten as u[u(u − 1) − x] = 0. Since u > 0, we have x = u(u − 1). Since x is prime, u = 2 and x = 2, implying that y = 6 and z = 2. 4. If u2 + x = z and u2 − x = y , the ﬁrst equation gives x2 + u(u2 − x) = (x + u)n . As before, if n ≥ 3, the right-hand side is greater than the left-hand side. (a) If n = 1, then x2 + u(u2 − x) = x + u, which can be rewritten as x(x − 1) + u(u2 − x − 1) = 0. But u4 − x2 = yz ≥ 4 so that u4 ≥ x2 + 4 and u2 > x. Thus, u2 − x − 1 ≥ 0, and x(x − 1) + u(u2 − x − 1) > 0, a contradiction.

(b) If n = 2, then x2 + u(u2 − x) = (x + u)2 , which can be rewritten as u[u(u − 1) − 3x] = 0. Thus u(u − 1) = 3x and u − 1 ∈ {1, 3, x, 3x}. • If u − 1 = 1, then u = 2, and 3x = 2, which is impossible. • If u − 1 = 3, then u = 4, and x = 4, which is not a prime. • If u − 1 = x, then x(x + 1) = 3x, implying that x = 2. In this case, u = 3, y = 7, and z = 11. • If u − 1 = 3x, then u = 1, so that x = 0, a contradiction.

In summary, the only solutions are (x, y, z, u, n) ∈ {(2, 6, 2, 2, 2), (2, 7, 11, 3, 2)}.

And now we look at solutions to problems of the 2007 Mediterranean Mathematical Competition, given at [2010: 277].

Let x ≤ y ≤ z be real numbers, such that xy + yz + zx = 1. Prove that 1 1 xz < 2 . Is it possible to improve the value of the constant 2 ?

Solved by Mohammed Aassila, Strasbourg, France; Arkady Alt, San Jose, CA, USA; George Apostolopoulos, Messolonghi, Greece; Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Oliver Geupel, Br¨ uhl, NRW, Germany; Edward T.H. Wang and Dexter Wei, Wilfrid Laurier University, Waterloo, ON; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Wang and Wei.
1 If x = 0, then xz = 0 < 2 . If x > 0, then 2xz = xz + xz ≤ yz + zx < 1 1 xy + yz + zx < 1 so xz < 2 . Suppose x < 0. If z ≥ 0, then xz ≤ 0 < 2 so it suﬃces to consider the case when x ≤ y ≤ z < 0.

363

If ε ≥ 1 0<ε< 2 .

Set r = −z , s = −y , and t = −z . Then r , s, and t are all positive such 1 by what we showed above. that t ≤ s ≤ r and ts + sr + rt = 1. Hence, tr < 2 1 Then xz < 2 follows. is the best upper bound for xz by showing that for Now we prove that 1 2 any ε > 0, there exists x, y , z satisfying x ≤ y ≤ z , xy + yz + zx = 1 and 1 xz > 2 − ε.
1 , 2

simply take x = y = z =
√ 2ε 2

√ 3 . 3

Hence we may assume that

We set x = y =
2 3

and z =

z −ε √ . 2 2ε

2ε < 2 − ε or ε < which is true. ε ε Next, xy + yz + zx = x2 + 2xz = 2 + 2− = 1. Finally, xz > 1 −ε 2 2 2 −ε 1 1 ε 1 ε is equivalent, in succession, to 4 > 2 − ε; 2 − 4 > 2 − ε; 4 < ε which is clearly true and our proof is complete.

Then y ≤ z is equivalent to

The quadrilateral ABCD is convex and cyclic, and the diagonals AC and BD intersect at the point E . Given that AB = 39, AE = 45, AD = 60 and BC = 56, determine the length of CD . Solved by George Apostolopoulos, Messolonghi, Greece; Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We use the solution of Apostolopoulos. The triangles BEC and AED are similar, so BE AE = BC AD ⇒ BE = AE · BC AD = 45 · 56 60 = 42 . O B A

Also, the triangles AEB and CED are similar, so CE CD = ED = BE , namely CD = ED = CE = l > 0. AB AE 29 45 42 Using Ptolemy’s Theorem yields: AB · CD + AD · BC = AC · BD C

E D

⇒ 39(39l) + 60 · 56 = (45 + CE )(42 + ED ) ⇒ 392 l + 60 · 56 = (45 + 42l)(42 + 45l) ⇔ 315l2 + 378l − 245 = 0 , l=

7 15

or
7 15

l<0 = 18.2.

namely l =

7 , 15

thus CD = 39l = 39 ·

3. In the triangle ABC √ , the angle α = ∠A and the side a = |BC | are given. It is known that a = rR, where r is the inradius and R is the circumradius of ABC . Determine all such triangles, that is, compute the sides b and c of all such triangles.

364

Solved by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; George Apostolopoulos, Messolonghi, Greece; Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Amengual Covas. Let b = CA, c = AB , and s = √ From a = rR, we obtain
1 (a 2

+ b + c).

a2 R

.
abc , 4R

(1) and also as rs. Equating (2)

Note that the area of ABC may be expressed as those we get abc = 4sRr = 4sa2 . Thus bc = 4as.

− a) s − a) −a = s(s = s(4 = s4 . Since cos α = 2 cos2 A − 1, Then cos2 A 2 bc as a 2 s −3 a this is equivalent to cos α = 2a or s = a(3 + 2 cos α), so that (2) becomes

bc = 4a2 (3 + 2 cos α). We also have b+c = 2s − a = 2a(3 + 2 cos α) − a = a(5 + 4 cos α).

(3)

(4)

We solve the system (3) and (4) by considering b and c as roots of a quadratic equation with coeﬃcients determined by the product (3) and the sum (4) of the roots: x2 − a(5 + 4 cos α)x + 4a2 (3 + 2 cos α) = 0. (5) The requirement that b and c be sides of ABC forces the discriminant of (5) a2 [(5 + 4 cos α)2 − 16(3 + 2 cos α)] = a2 [(4 cos α + 1)2 − 24] to be non-negative. This is true only if cos α ≥ α ≤ 12◦ 54 15 . If this holds, the roots of (5) x= 1 2 a[5 + 4 cos α ±
√ 24−1 , 4

that is, only if

(4 cos α + 1)2 − 24] ABC .

yield the length of the sides CA and AB of

Let x > 1 be a noninteger number. Prove that x + {x} [ x] − [ x] x + {x} + x + [ x] {x} − {x} x + [ x] > 9 , 2

365

where [x] and {x} represents the integer and the fractional part of x. Solved by Mohammed Aassila, Strasbourg, France; George Apostolopoulos, Messolonghi, Greece; Michel Bataille, Rouen, France; Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Henry Ricardo, Tappan, NY, USA; Edward T.H. Wang and Kaiming Zhao, Wilfrid Laurier University, Waterloo, ON; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the improved results of Bataille.
{x} x {x} Let L = x+x − x+x + x+ − x+ . We show that L > 16 , {x} {x} x 3 slightly improving the proposed inequality. Using x + {x} = x, we calculate

x + {x} x+ x x2 + {x}2 + x + = x {x} {x} x and x x + {x} It follows that L= + x+ x {x} = (x + {x})(x + x ) x2 + {x}2 + x
2

x2 + {x}2 + x 2x2 + {x} x

Now, recalling that 2(a2 + b2 ) > (a + b)2 for distinct positive real numbers a, b, we see that 2x2 (x2 + {x}2 + x
2

2x2 (x2 + {x}2 + x 2 ) . {x} x (2x2 + {x} x )

) > x2 (2x2 + ({x} + x )2 ) = 3x4

(note that {x} ∈ [0, 1) and x ≥ 1 so that {x} = x ). In addition, from AM-GM, {x} x (2x2 + {x} x ) < and we ﬁnally deduce L > (3x4 ) · ({x} + x )2 4 · 2x2 + 16 3 ({x} + x )2 4 = 9x4 16

16 9x4

We return to the ﬁles of solutions from our readers and the 24th Balkan Mathematical Olympiad 2007 given at [2010: 277–278].

Let ABCD be a convex quadrilateral with AB = BC = CD , AC = BD and let E be the intersection point of its diagonals. Prove that AE = DE if and only if ∠BAD + ∠ADC = 120◦ . Solved by Mohammed Aassila, Strasbourg, France; Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; George Apostolopoulos, Messolonghi, Greece; Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Amengual Covas.

366

B
α y

C x β E y v D

x u

A Let ∠DAE = u, ∠EDA = v , ∠ABD = α and ∠ACD = β . Letting the base angles in isosceles triangles ABC and BCD be x and y , respectively, we have u + v = x + y in AED , BEC and x + α = y + β in ABE , CDE , respectively, because of the vertically opposite angles at E . Therefore, ∠BAD + ∠ADC = (x + u) + (v + y ) = (u + v ) + (x + y ) = 2(x + y ) We must have α = β . Suppose α = β : Then the condition x + α = y + β implies x = y and we would have ∠ABC = ∠BCD , making ABC and BCD congruent which contradicts the assumption AC = BD . Now, by the law of sines, AE sin α = = = so AE = DE ⇔ ⇔ sin α = sin β α + β = 180◦ (since a = β ) AB sin ∠BEA CD sin ∠DEC DE . sin β (∠BEA and ∠DEC are vertically opposite angles) (1)

⇔ ⇔

⇔

∠BAD + ∠ADC + x + y = 180◦ (since the angles of ABCD add up to 360◦ ) 1 ∠BAD + ∠ADC + (∠BAD + ∠ADC ) = 180◦ 2 ∠BAD + ∠ADC = 120◦

by (1)

Find all functions f : R → R such that for all x, y ∈ R f (f (x) + y ) = f (f (x) − y ) + 4f (x)y.

Solved by Mohammed Aassila, Strasbourg, France; and Chip Curtis, Missouri Southern State University, Joplin, MO, USA. We give the solution by Curtis.

367

Then

We claim that the only solutions are the 0 function and functions f (x) = x2 + b, where b is a real number. It is readily veriﬁed that these satisfy the functional equation, so we assume that f not identically 0 satisﬁes the functional z equation. Choose x ¯ ∈ R such that f (¯ x) = 0. Let z ∈ R, and set y ¯= .
8f (¯ x)

f (¯ x) +

z 8f (¯ x)

−f

f (¯ x) −

z 8f (¯ x)

z 2

Let x1 = f (¯ x) + y ¯ and x2 = f (¯ x) − y ¯. Then z = 2[f (x1 ) − f (x2 )]. With v = f (x1 ) − 2f (x2 ), we have z = f (x1 ) + v, so that f (z ) = f (f (x1 ) + v ) = f (f (x1 ) − v ) + 4f (x1 )v = f (2f (x2 )) + 4f (x1 )v. Let b = f (0), and in the functional equation, let x = x2 and y = f (x2 ). Then f (2f (x2 )) = b + 4[f (x2 )]2 . Thus, f (z ) = b + 4[f (x2 )]2 + 4f (x1 )v = b + 4[f (x2 )]2 + 4f (x1 )[f (x1 ) − 2f (x2 )] = b + z2 . = b + 4[f (x1 ) − f (x2 )]2

Find all positive integers n such that there is a permutation σ of the set {1, 2, . . . , n} such that the number below is a rational number
Õ

σ (1) +

σ (2) +

··· +

σ (n)

Ed.: A permutation of the set {1, 2, . . . , n} is a one-to-one function of this set to itself. Solved by Mohammed Aassila, Strasbourg, France; and George Apostolopoulos, Messolonghi, Greece. We give the version of Apostolopoulos. Suppose that for some n ∈ N∗ we have
Õ

σ (1) +

σ (2) +

··· +

σ (n) = v1 ∈ Q

368
Õ

then σ (1) +

σ (2) +

··· +

σ (n) is a rational number. Similar, for each
Ö Õ

k ∈ {1, 2, . . . , n} the number vk = is a rational number. Deﬁne αk =

σ (k) +
Õ

σ (k + 1) +

··· +

σ (n)

√ n + · · · + n, for each k ∈ N∗ . √ By an easy induction, we can prove k ∈ N∗ , as a √ that αk < n + 1, for each 2 result we will have v1 < αn < n + 1. If l > 0 is such that l ≤ n < (l + 1)2 then for some i ∈ {1, 2, . . . , n} we have σ (i) = l2 . n+ Case 1. i = n. Then we have
Õ

l< ⇒

l2 + l2 +

σ (i + 1) +
Õ

··· + ··· +
Õ

σ (n) <

√

n+1<l+2

σ (i + 1) + σ (i + 1) +

σ (n) = l + 1 σ (n) < √ n+1 < l + 2 ⇒ l < 1,

⇒ 2l + 1 = a contradiction. Case 2. i = n.

··· +

If l > 1, then (l2 − 1) ∈ {σ (1), σ (2), . . . , σ (n − 1)}. Suppose that j < n is such that σ (j ) = l2 − 1. Similarly to Case 1, we get
Õ

l< ⇒

l2 − 1 + l2 −1+

σ (j + 1) +
Õ

··· +

√ √ l2 < n + 1 < l + 2

σ (j + 1) + σ (j + 1) +

⇒ 2l + 2 =

√ · · · + l2 = l + 1 √ √ · · · + l2 < n + 1 < l + 2 ,

a contradiction. If l = 1, then n ∈ {1, 2, 3}. We conclude that for n = 1 and for n = 3 there are permutations that satisfy Õ the terminus relation. For n = 1, we have √ √ 1 = 1 and for n = 3, we have 2 + 3 + 1 = 2. But for n = 2 no such permutation exists. So n = 1, n = 3.

Next we turn to the Indian Team Selection Test 2007 given at [2010: 278– 279].

1. Let ABC be a triangle with AB = AC , and let Γ be its circumcircle. The incircle γ of ABC moves (slides) on BC in the direction of B . Prove that when γ touches Γ internally, it also touches the altitude through A.

369

Solved by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; Michel Bataille, Rouen, France; D.J. Smeenk, Zaltbommel, the Netherlands; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Smeenk. We denote Γ = Γ(O, R), γ = γ (I, r ). Line passes through I and is parallel to BC . D lies on , and ID = r . OD intersects Γ at E . It suﬃces to show that DE = r , and so OD = R − r . OI = R cos α − r , ID = r , and we are to show (R − r )2 = (R cos α − r )2 + r 2 As β = γ we have r (1) E B C A Γ γ O D I

= R(cos α + cos β + cos γ − 1) = R(− cos 2β + 2 cos β − 1) = 2R cos β (1 − cos β ). (2)

Substituting (2) in (1) we see that it holds and we are done.

Consider the quadratic polynomial p(x) = x2 + ax + b, where a, b are in the interval [−2, 2]. Determine the range of the real roots of p(x) = 0 as a and b vary over [−2, 2]. Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA. The polynomial p(x) has real roots if and only if its discriminant D = a2 − 4b is nonnegative. Set x+ = −a +
1 2 a , 4

√ a 2 − 4b 2

and

x− =

−a −

√ a 2 − 4b 2

Since −2 ≤ b ≤ − a 2

we have 0 ≤ a2 − 4b ≤ a2 + 8. Hence, √ a2 + 8 2 + and −a − √ a2 + 8 2 ≤ x− ≤ −a 2 .

≤ x+ ≤

−a +

√ −1 ≤ x+ ≤ 1 + 3. √ The function g (t) = 1 (−t − t2 + 8) is also continuous and decreasing on 2 [−2, 2]; hence, √ −1 − 3 ≤ x− ≤ 1. √ √ Thus the range of the real roots of p(x) is [−1 − 3, 1 + 3].

The function f (t) = hence

1 (−t 2

√ t2 + 8) is continuous and decreasing on [−2, 2];

370

Let triangle ABC have side lengths a, b, c; circumradius R, and internal angle bisector lengths wa , wb , wc . Prove that c2 + a2 a2 + b2 b2 + c2 + + > 4R . wa wb wc

Solved by George Apostolopoulos, Messolonghi, Greece; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Apostolopoulos.
2bc cos

It is well-known that wa = b2 + c2
cyclic

b+c

A 2 , and thus we have

> 4R ⇔ ⇔ ⇔

(b2 + c2 )(b + c)
cyclic

2bc cos

A 2

> 4R >2
A 2

(b2 + c2 )(b + c)
cyclic

4Rbc cos

A 2

(b2 + c2 )(b + c) sin
cyclic

2abc

> 1. A > 1. Let 2

Note that b2 + c2 ≥ 2bc, b + c > a. It suﬃces to prove that r be the inradius of the triangle. Then
cyclic

sin
cyclic

cos A + 1 +

r R

> 1. This inequality

holds for all triangles. Note that , , are also the three angles 2 2 2 π−A A of a triangle thus cos > 1. That is sin > 1. 2 2 cyclic cyclic

π−A π−B π−C

Show that in a non-equilateral triangle, the following are equivalent:

(a) The angles of the triangle are in arithmetic progression; (b) The common tangent to the nine-point circle and the in-circle is parallel to the Euler line. Solved by Michel Bataille, Rouen, France; Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Bataille. We show that both (a) and (b) are equivalent to: (c) one of the angles of the triangle is 60◦ . Let ABC be the given non-equilateral triangle with BC = a, CA = b, AB = c. If (c) holds, then the angles of ∆ABC are of the form 60◦ , 60◦ + α, ◦ C ◦ 60 − α, hence (a) holds. Conversely, if, say, B = A+ , then B = 1802−B , 2 hence B = 60◦ . Thus, (a) ⇐⇒ (c).

371

Now, let O, H, I, and N denote the circumcentre, the orthocentre, the incentre, and the centre of the nine-point circle, respectively. As usual, let R and s be the circumradius and the semi-perimeter of ∆ABC . It is well-known that the incircle is internally tangent to the nine-point circle. It follows that − → − → − − → − → − → 1− OH , (b) is equivalent to OH ⊥ IN . Since N I = OI − ON = OI − 2 2 − − → − → − → the condition OH ⊥ IN is equivalent to OH · OI = OH . Using 2sOI = 2 − → − → − → − − → − → − → − → aOA + bOB + cOC and OH = OA + OB + OC , a simple computation shows 2 − − → − → that OH · OI = OH is itself equivalent to 2 s = −a cos 2A − b cos 2B − c cos 2C − → − → (note that for example OA · OB = R2 cos 2C , C being acute or not). We successively rewrite (1) as a(1 + 2 cos 2A) + b(1 + 2 cos 2B ) + c(1 + 2 cos 2C ) = 0 sin A + 2 sin A cos 2A + sin B +2 sin B cos 2B + sin C + 2 sin C cos 2C = 0 sin 3A + sin 3B + sin 3C = 0 (the latter because 2 sin A cos 2A = sin 3A − sin A, etc.). As a result, (b) is equivalent to sin 3A + sin 3B + sin 3C = 0, or, with B C A cos 32 cos 32 = 0. Finally, the help of the familiar trig formulas, to 4 cos 32 3A 3B 3C ◦ (b) is equivalent to 90 = 2 or 2 or 2 and (b) ⇐⇒ (c). (1)

7. Let a, b, c be nonnegative real numbers such that a + b ≤ c +1, b + c ≤ a +1 and c + a ≤ b + 1. Prove that
a2 + b2 + c2 ≤ 2abc + 1 . Solved by Arkady Alt, San Jose, CA, USA; George Apostolopoulos, Messolonghi, Greece; Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We use the solution by Alt. First note that a, b, c ≤ 1. Indeed, c + a ≤ b + 1 ⇒ c + 2a ≤ a + b + 1 ≤ c + 1 + 1 ⇒ 2a ≤ 2 ⇒ a ≤ 1 , and, similarly, b, c ≤ 1. Let x := 1 − a, y := 1 − b, z := 1 − c. Then x, y, z ∈ [0, 1] , a = 1 − x, b = 1 − y, c = 1 − z a+b ≤c+1 b+c ≤a+1 c+a≤ b+1 ⇐⇒ z ≤x+y x ≤y+z y ≤z+x ⇐⇒ |x − y | ≤ z ≤ x + y,

372

and the original inequality becomes (1 − x) + (1 − y ) + (1 − z ) ≤ 2 (1 − x) (1 − y ) (1 − z ) + 1 ⇐⇒ x2 + y 2 + z 2 ≤ 2 (xy + yz + zx) − 2xyz ⇐⇒ (x − y ) ≤ 2z (x + y − xy ) − z 2 . ⇐⇒ x2 + y 2 − 2xy ≤ 2z (x + y − xy ) − z 2
2 2 2 2

(1)

Let f (z ) = 2z (x + y − xy ) − z 2 . Since z ∈ [|x − y | , x + y ] then min f (z ) = min {f (|x − y |) , f (x + y )}, and, therefore,
z

(1) ⇐⇒ (x − y ) ≤ min 2z (x + y − xy ) − z 2
z

⇐⇒ (x − y ) ≤ min {f (|x − y |) , f (x + y )} ⇐⇒ ⇐⇒ ⇐⇒ We have x2 + y 2 ≤ (x + y ) (x + y − xy ) ⇐⇒ (x + y ) xy ≤ 2xy (x − y ) ≤ f (|x − y |) 2 (x − y ) ≤ f (x + y )
2 2

(2)

(x − y ) ≤ 2 |x − y | (x + y − xy ) − (x − y ) 2 2 (x − y ) ≤ 2 (x + y ) (x + y − xy ) − (x − y ) (x − y ) ≤ |x − y | (x + y − xy ) x2 + y 2 ≤ (x + y ) (x + y − xy )
2

⇐⇒ 0 ≤ xy (2 − x − y )

and (x − y )2 ≤ |x − y | (x + y − xy ) ⇐⇒ 0 ≤ |x − y | (x + y − xy − |x − y |) . Since x, y ∈ [0, 1] then the inequality 0 ≤ xy (2 − x − y ) obviously holds

and

|x − y | ≤ x + y − xy ⇐⇒ xy − x − y ≤ x − y ≤ x + y − xy ⇐⇒ xy − x ≤ x −y ≤ y − xy ⇐⇒

0 ≤ x (2 − y ) . 0 ≤ y (2 − x)

373

8. Given a ﬁnite string S of symbols a and b, we write (S ) for the number of a’s in S minus the number of b’s. (For example, (abbabba) = −1.) We call a string S balanced if every substring (of consecutive symbols) T of S has the property −1 ≤ (T ) ≤ 2. (Thus abbabba is not balanced, as it contains the substring bbabb and (bbabb) = −3). Find, with proof, the number of balanced strings of length n.
Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Oliver Geupel, Br¨ uhl, NRW, Germany. We give Geupel’s solution. We prove that the number of balanced strings of length n is n + 1. If a string S is balanced, then it does not contain the string bb as a substring, because (bb) = −2. Moreover, S contains at most one substring of the form aa, and the remaining part of S must alternate between the symbols a and b. Conversely, any string with not more than one occurrence of the substring aa and alternating symbols in the remaining parts of the string is balanced. We have two such strings with no substring aa and n − 1 strings with exactly one substring aa. This completes the proof.

Deﬁne the functions f , g , h on Z × Z × Z as follows: f (x, y, z ) g (x, y, z ) h(x, y, z ) = (3x + 2y + 2z, 2x + 2y + z, 2x + y + 2z ) , = (3x + 2y − 2z, 2x + 2y − z, 2x + y − 2z ) , = (3x − 2y + 2z, 2x − y + 2z, 2x − 2y + z ) .

Given a primitive Pythagorean triplet (x, y, z ), with x > y > z , prove that (x, y, z ) can be uniquely obtained by repeated application of f , g , h to the triple (5, 4, 3). For example: (697, 528, 455) = f ◦ h ◦ g ◦ h(5, 4, 3). Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA. Representing f , g , and h by the matrices F , G, and H given by
¾ ¿ ¾ ¿ ¾ ¿

F =

3 2 2

2 2 1
¾

2 1 2

3 2 2
¿

2 2 1

−2 −1 −2

H =

3 2 2

−2 −1 −2 −2 1 −2
¿

2 2 1

their inverses are given by F −1 = 3 −2 −2 −2 2 1 −2 1 2
¾ ¾

, 3 2 −2

G −1 = −2 −1 2 −2 −2 1

3 −2 2
¿

−2 2 −1

H −1 =

Let (x, y, z ) be a primitive Pythagorean triple with x > y > z . Then there exist relatively prime opposite-parity integers s, t with s > t such that x = s2 + t2 and either y = 2st and z = s2 − t2 or y = s2 − t2 and z = 2st.

374

Case 1. Suppose ﬁrst that y = 2st and z = s2 − t2 . Then 2st > s2 − t2 , so that t2 > s2 − 2st = s(s − 2t) > t(s − 2t). This implies that 3t > s. We have x y = F −1 • F −1 z (3t > s > 2t); x y • G −1 z (2t > s > t); x • H −1 y z (2t > s > t).
¾ ¾ ¿ ¾ ¿ ¾

s2 + t2 2st s2 − t2 s2 + t2 2st s2 − t2 s2 + t2 2st s2 − t2

t2 + (s − 2t)2 t2 − (s − 2t)2 2t(s − 2t) t2 + (2t − s)2 t2 − (2t − s)2 2t(2t − s) t2 + (2t − s)2 2t(2t − s) t2 − (2t − s)2

= G −1

= H −1

Since s and t are relatively prime, s = 2t unless s = 2 and t = 1; if s > 2t, then the ﬁrst right-hand side is a primitive Pythagorean triple closer to the origin in Z × Z × Z than the original; if s < 2t, the second and third right-hand sides are primitive Pythagorean triples closer to the origin than the original. We note that with [x y z ]T = F −1 ([x y z ]T ), y > z if and only if t2 − (s − 2t)2 > 2t(s − 2t), which is equivalent to y > z . Likewise, with [x y z ]T = G−1 ([x y z ]T ), y > z if and only if t2 − (2t − s)2 > 2t(2t − s), which is√ equivalent to s2 − 6 st + 7t2 < 0. This inequality holds if and only √ if (3 − 2)t < s < (3 + 2)t. This also implies √ that with [x y z ]T = −1 T H ([x y z ] ), y > z if and only if s < (3 − 2)t. Hence, when y is even and z is odd, we ﬁnd the previous triple by √ applying F −1 if √ Pythagorean −1 −1 3t > s > 2t, G if 2t > s > (3 − 2)t, and H if (3 − 2 2)t > s > t. Case 2. Similarly, if y = s2 − t2 and z = 2st, then s2 − t2 > 2st, implying that s(s − 2t) > t2 > 0, so that s > 2t. We have x y • F −1 = F −1 z (3t > s > 2t);
¾ ¾ ¿ ¾ ¿ ¾ ¿

s2 + t2 s2 − t2 2st s2 + t2 s2 − t2 2st s2 + t2 s2 − t2 2st

t2 + (s − 2t)2 2t(s − 2t) t2 − (s − 2t)2 (s − 2t)2 + t2 2(s − 2t)t (s − 2t)2 − t2 (s − 2t)2 + t2 (s − 2t)2 − t2 2(s − 2t)t

• G −1 (s > 3t);
¾

x y z x y z

= G −1

• H −1 (s > 3t).

= H −1

375

Suppose that (s, t) = (2, 1). If s < 3t, then the ﬁrst right-hand side is a primitive Pythagorean triple closer to the origin than the original triple. If s > 3t, the second and third right-hand sides are primitive Pythagorean triples closer to the origin than the original triple. With [x y z ]T = F −1 ([x y z ]T ), y > z if and only if 2t(s − 2t) > t2 − (s √ − 2t)2 , which is equivalent to y > z . Similarly, y > z if and only if s < (3 + √ 2)t when [x y z ]T = G−1 ([x y z ]T ), and y > z if and only if s > (3 + 2)t, when [z y z ]T = H −1 ([x y z ]T ). Hence, if y is odd and z is√ even, we ﬁnd the previous triple by applying F −1 if √ −1 −1 if s > (3 + 2)t. 3t > s > 2t; G if (3 + 2)t > s > 3t, and H Thus, repeatedly applying F −1 , G−1 , and H −1 to a primitive Pythagorean triple yields a sequence of primitive Pythagorean triples, or equivalently a sequence of lattice points in the (s, t)-plane successively closer to the origin, terminating when s = 2, t = 1, which corresponds to the triple (5, 4, 3). At each step, there is a unique choice of F −1 , G−1 , or H −1 for which x > y > z . Reversing the process yields a unique path to (x, y, z ) from (5, 4, 3). Find all pairs of integers (x, y ) such that y 2 = x3 − p2 x, where p is a prime such that p ≡ 3 (mod 4). Solution based on an approach of George Apostolopoulos, Messolonghi, Greece, modiﬁed by the editor. cases. The equation can be rewritten as y 2 = (x − p)(x + p)x. There are two

11.

relatively prime. Their product is the integer
2 2 2

Case 1. p y , Then (p, (x − p)x(x + p)) = 1. When x is even, then x − p, x + p and x are pairwise relatively prime and so must all be squares. Since x is a multiple of 4 and p ≡ 3 (mod 4), it follows that x + p ≡ 3 (mod 4), and so cannot be square. Thus there are no solutions when x is even. p x+p When x is odd, then (x − p, x + p) = 2 so that x− , 2 and x are pairwise 2
2 y 2 , 2 2

so they are all squares. Let

Case 2. p | y . If y = 0, then (x, y ) = (p, 0), (−p, 0), (0, 0) are solutions. We show that there are no other solutions. Suppose that y = 0. Since p | y , then (x − p, x + p, x) = p and p2 | y . Dividing both sides of the equation by p3 , we have that pb2 = (a − 1)a(a + 1), where x = pa, y = p2 b. Since b = 0, a − 1 > 0.

1 and s = p− . x = r , x − p = 2s , x + p = 2t . Then t − s = p, t = p+1 2 2 Thus (p − 1)2 p2 + 1 r 2 = (x − p) + p = +p = , 2 2 whereupon p2 − 2√ r 2 = −1. The of this pellian equation are √ positive solutions √ given by pn + qn 2 = (1 + 2)(3 + 2 2)n for n a nonnegative integer. Since pn+1 = 6pn − pn−1 , it can be veriﬁed that pn ≡ (−1)n (mod 8). The ﬁrst few solutions are (1, 1), (7, 5), (41, 29), (239, 169). The equation is solvable if and only if n is odd and p = pn is a prime. For example, we obtain (p, x, y ) = (7, 25, 2 × 3 × 4 × 5) = (7, 25, 120) and (p, x, y ) = (239, 1692 , 2 × 119 × 120 × 169) = (239, 28561, 4826640).

376

The prime p divides exactly one of a − 1, a + 1, a. Since any pair of these three integers is co-prime, two of them must be squares that diﬀer by either 1 or 2. But this is possible only when a = 0 and a = 1, both of which are excluded. The result follows. Comment by the editor. The condition that p ≡ 3 (mod 4) seems artiﬁcial. When p is any odd prime and does not divide y , we deduce as above that x − p, x and x + p are all square. The only pair of squares that diﬀer by p are
2 p− 1 2 2

and p+1 . Thus there is no integer x for which all of x − p, x and x + p are 2 square and so there are no solutions when x is even and y is not a multiple of p. When x is odd, we can pursue the foregoing argument to ﬁnd solutions when p ≡ 1 (mod 8), such as (p, x, y ) = (41, 292 , 2 × 20 × 21 × 29) = (41, 841, 24360). The reader is invited to examine the case that p = 2.

12.

Find all functions f : R → R satisfying the equation f (x + y ) + f (x)f (y ) = (1 + y )f (x) + (1 + x)f (y ) + f (xy ) ,

for all x, y ∈ R. Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA. Substituting y = 1 into f (x + y ) + f (x)f (y ) = (1 + y )f (x) + (1 + x)f (y ) + f (xy ) gives f (x + 1) = 3f (x) + f (1)[1 + x − f (x)]. Substituting y = −1 into (1) gives f (x − 1) = f (−x) + f (−1)[1 + x − f (x)]. Substituting y = 0 into (1) gives f (0)[f (x) − x − 2] = 0. (4) (3) (2) (1)

Hence, either f (0) = 0 or f (x) = x + 2 for all real x. Substituting this last possibility into (1) yields the contradiction 2xy = 0 for all x, y ∈ R. Thus f (0) = 0. Setting x = 1 and y = −1 in (1) gives f (1)f (−1) = 2f (−1) + f (−1), so that f (−1)[3 − f (1)] = 0. Hence, either f (1) = 3 or f (−1) = 0.

We note for future reference that substituting, respectively, y = x and y = −x in (1) gives f (2x) + [f (x)]2 = 2(1 + x)f (x) + f (x2 ) (5)

377

and f (x)f (−x) = (1 − x)f (x) + (1 + x)f (−x) + f (−x2 ). (6) Case 1. If f (1) = 3, then setting y = 1 gives f (x + 1) + 3f (x) = 2f (x) + 3(1 + x) + f (x), which is equivalent to f (x + 1) = 3(1 + x). Thus, for all real x, f (x) = 3x. Case 2. Now suppose f (−1) = 0 and f (1) = 3. From (3), we have f (x − 1) = f (−x), so that, by replacing x with x + 1, f (x) = f (−x − 1). Set y = −x − 1 in (1). Then [f (x)]2 = −xf (x) + (1 + x)f (x) + f (−x − x2 ). Hence, In particular, if x = 1, we have f (−2) = [f (1)]2 − f (1). But from (7), we have f (−2) = f (1). Thus, f (1) = [f (1)]2 − f (1), so that f (1)[f (1) − 2] = 0. Therefore, either f (1) = 0 or f (1) = 2. Subcase (a). Suppose f (1) = 0. Equations (2) and (3) give f (x + 1) = 3f (x) and f (x − 1) = f (−x), respectively. Applying (8) with x = 1 yields f (2) = 0. Thus, substituting y = 2 in (1) gives f (x + 2) = 3f (x) + f (2x). On the other hand, applying (8) twice yields f (x + 2) = 9f (x). From (11) and (12), we have f (2x) = 6f (x). (13) (12) (11) (10) (9) (8) f (−x − x2 ) = [f (x)]2 − f (x). (7)

378

Replacing x with x − 1 in (8) gives f (x) = 3f (−x); combining with (9) gives f (−x) = Applying (14) and then (5) to (6) gives 1 [f (x)]2 = (1 − x)f (x) + 3 1 [f (x)]2 = (1 − x)f (x) + 3 1 1 (1 + x)f (x) + f (x2 ) 3 3 1 (1 + x)f (x) 3 1 + {f (2x) + [f (x)]2 − 2(1 + x)f (x)}. 3 (15) 1 3 f (x). (14)

Simplifying gives From (13) and (15), we obtain f (2x) = 2(2x − 1)f (x).

6f (x) = 2(2x − 1)f (x) 4(2 − x)f (x) = 0. Since f (2) = 0, this last equation implies that f (x) is identically 0. Subcase (b). Now suppose that f (1) = 2. From (2) and (3), f (x + 1) = f (x) + 2(1 + x) and Replacing x with x − 1 in (16) and combining with (17) gives f (−x) = f (x) − 2x. Applying (18) and then (5) to (6) gives f (x)[f (x) − 2x] = (1 − x)f (x) + (1 + x)[f (x) − 2x] + f (2x) [f (x)] − 2xf (x) = 2f (x) − 2x(1 + x) + f (2x) + [f (x)]2 0 = −2x(1 + x) + f (2x) − 2x2
2

(16) (17)

f (x − 1) = f (−x).

(18)

f (x)[f (x) − 2x] = (1 − x)f (x) + (1 + x)[f (x) − 2x] + f (x2 ) − 2x2 + [f (x)]2 − 2(1 + x)f (x) − 2x2 − 2(1 + x)f (x) − 2x2

f (2x) = 4x2 + 2x

f (2x) = (2x)2 + (2x) for all real x, so that f (x) = x2 + x. In summary, there are three possibilities, each of which is readily veriﬁed to satisfy (1).

379

• If f (1) = 3, then f (x) = 3x. • If f (−1) = 0 and f (1) = 0, then f (x) ≡ 0. • If f (−1) = 0 and f (1) = 2, then f (x) = x2 + x.

Next we turn to the ﬁles of readers’ solutions to problems given in the October 2010 number of the Corner and the Olimpiada Nacional Escolar de Matematica 2009, Level 1, given at [2010: 372–373]. If P , E , R and U represent digits diﬀerent from 0 and pairwise diﬀerent such that P ER + P RU + P U E + 2009 = P ERU , ﬁnd all the values that P + E + R + U can take. Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the version of Zvonaru. The given condition is equivalent to 100P + 10E + R + 100P + 10R + U + 100P + 10U + E + 2009 700P + 89E = = 1000P + 100E + 10R + U ; 2009 + U R.

Since 2009 + U R ≤ 2009 + 98 and 89E ≥ 89, we deduce that P ≤ 2. If P = 1, then we have 89E = 1309 + U R. We do not obtain a solution because 89E ≤ 89 · 9 = 801. If P = 2, then we obtain 89E = 609 + U R. Since 609 + U R ≥ 609 + 12 = 621 and 609 + U R ≤ 609 + 98 = 707, we deduce that 621 ≤ E ≤ 707 , 89 89 hence E = 7. It results that P ERU = 2741 and P + E + R + U = 14.

Andr´ es and Bertha play on a 4 × 4 table with tetrominos as shown.

Andr´ es begins the game placing 4 tetrominos of the same shape on the table without overlaps and leaving no empty space. Then Bertha must write on each square of the table one of the numbers 1, 2, 3 or 4 in such a way that each row and column has no two numbers repeated. Bertha wins if each tetromino on the chart covers 4 diﬀerent numbers. (a) Show that Bertha can always win the game. (b) Andr´ es ﬁlls the table with 4 tetrominos where at least 2 are diﬀerent. Is it true that in this situation, playing with the same rules, Bertha can always win?

380

Solved by Oliver Geupel, Br¨ uhl, NRW, Germany; and Gesine Geupel, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany (joint work). Let S1 , . . . , S7 denote the given shapes from left to right. The table can not be covered with 4 pieces of shape S3 or 4 pieces of shape S4 . The possible coverings with 4 pieces of the same shape S1 , S2 , S5 , S6 , S7 , respectively, are each similar to one of the assemblies shown in Figure 1 below. It is demonstrated in Figure 1 how Bertha can win the game (a) in each case. 3 4 2 1 4 1 3 2 2 3 1 4 1 2 4 3 1 2 3 4 3 4 1 2 2 1 4 3 4 3 2 1 3 4 1 2 2 1 3 4 1 2 4 3 4 3 2 1 4 3 1 2 1 2 3 4 2 1 4 3 3 4 2 1 1 2 3 4 2 3 4 1 3 4 1 2 4 1 2 3

Figure 1 Andr´ es can win the game (b) if he puts the covering in Figure 2. For the proof, we use coordinates as in Figure 2. For example, (A, d) denotes the lowerleft cell. With no loss of generality assume that Bertha writes the numbers in the upper-left piece according to Figure 2. In the lower-left piece, the number 1 can not occur in column A. Hence (B, c) = 1. Then (A, b) in the same piece cannot be 1. Also (A, b) cannot be 3 or 4, because both numbers occur in the same row. Thus (A, b) = 2. Also, (D, a) = 4 and (C, a) = 3. Now, the occurrence of 2 in the upper-right tetromino must be (D, c), with the consequence that there is no possible entry for (C, c). Therefore, Bertha loses the game (b). 1 2 3 4 a 2 3 4 b 1 2 c d A B C D Figure 2

To complete this number of the Corner we look at solutions to the Olimpiada Nacional Escolar de Matematica 2009, Level 3 given at [2010 : 373–374].

For each positive integer N let c(n) be the number of decimal digits of N . Let A be a set of positive integers such that if a and b are two distinct elements of A, then c(a + b) + 2 > c(a) + c(b). Find the largest number of elements that A can have. Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA. We give the presentation by Curtis.

381

Note that c(n) is a nondecreasing function, and that for each positive integer n, 10c(n)−1 ≤ n ≤ 10c(n) − 1. Hence, if a ≤ b, then c(a + b) ≤ c(2b) ≤ c[2 · (10c(b) − 1)] ≤ c(10c(b)+1 − 1) = c(b) + 1. c(a + b) ≤ 1 + max{c(a), c(b)}. Thus, If c(b) ≥ c(a), then 3 + c(b) ≥ c(a + b) + 2 > c(a) + c(b), so that c(a) ∈ {1, 2}. Similarly, if c(a) > c(b), then c(b) ∈ {1, 2}. If c(b) = c(a), then c(b) = c(a) ∈ {1, 2}. Write the elements of A as 1 ≤ x1 < x2 < x3 < · · · . If i < j , then xi < xj , implying that c(xi ) ≤ c(xj ), so that c(xi ) ∈ 1, 2. Hence, A has at most one element greater than or equal to 100, so A is ﬁnite, and #(A) ≤ 100. Write A = {xi }N i=1 . We have N ≤ 100. Suppose that i < j and c(xi ) = c(xj ) = 2. Then c(xi + xj ) > 2. In particular, the smallest two 2-digit numbers in A have sum at least 100. The table shows the maximum number of 2-digit numbers for various possible values of the smallest 2-digit number in A. Thus, A can have at most nine 1-digit xi min xj Max. num. of numbers, at most ﬁfty 2-digit numbers, and at 2-digit num. most one number with more than two digits, so 10 90 11 N ≤ 60. In fact, A can have 60 numbers. One 20 80 21 possible such set is 30 70 31 40 60 41 A = {1, 2, 3, . . . , 9, 50, 51, 52, . . . , 99, 100}. 49 51 50 It is readily veriﬁed that any distinct pair a, b ∈ 50 51 50 A satisﬁes c(a + b) + 2 > c(a) + c(b). 51 52 49 2. In a quadrilateral ABCD, a circle is drawn that is tangent to the sides AB , BC , CD and DA at the points M , N , P and Q respectively. Prove that if (AM )(CP ) = (BN )(DQ), then ABCD can be inscribed in a circle. Solved by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain. A
A 2

x = λt Q t

D t P

x = λt M
r = λt tan
A 2

r = z tan

C 2

y = λz z

z
C 2

y = λz N

382

We put AM = x, BN = y , CP = z and DQ = t. Then we have = y = λ, say, where λ is a real number. Thus, x = λt and xz = yt, that is, x t z y = λz . The tangents AM and AQ have the same length and similarly at the other vertices. Therefore AB BC CD = AM + M B = x + y = λ(t + z ) = BN + N C = y + z = (λ + 1)z = CP + P D = z + t

DA = DQ + QA = t + x = (λ + 1)t. By the law of cosines, applied to triangles ABD and BCD , we have BD 2 = [(λ + 1)t]2 + [λ(t + z )]2 − 2λ(λ + 1)t(t + z ) cos A BD 2 = [(λ + 1)z ]2 + (z + t)2 − 2(λ + 1)z (z + t) cos C or (λ + 1)2 (t2 − z 2 ) + (t + z )2 (λ2 − 1) = 2(λ + 1)(z + t)(λt cos A − z cos C ) that is (λ+1)(z +t)[(λ+1)(t−z )+(λ−1)(t+z )] = 2(λ+1)(z +t)(λt cos A−z cos C ) which simpliﬁes to λt − z = λt cos A − z cos C or λt(1 − cos A) = z (1 − cos C ), which is equivalent to A C 2λt sin2 = 2z sin2 . (1) 2 2 We also have A C λt tan = z tan (= r ) (2) 2 2 where r is the radius of the inscribed circle in quadrilateral ABCD . From (1) and (2) we obtain, by division, 2λt sin2
A 2 A 2

λt tan z tan tan A C C A ⇒ 2 sin cos = 2 sin cos 2 2 2 2 so sin A = sin C . Similarly, we ﬁnd sin B = sin D .

2z sin2

C 2 C 2

⇒

2 sin2

A 2 A 2

sin2 tan

C 2 C 2

(3) (4)

From (3), either A = C or A + C = 180◦ . From (4), either B = D or B + D = 180◦ .

383

Case (a). A + D = 180◦ or B + D = 180◦ . In this case ABCD is cyclic since opposite angles are supplementary. Case (b). A = C and B = D . Denote the center of the incircle of ABCD by O . Right-angled triangles AM O , CP O and BN O , DQO are congruent with x = z and y = t. Since xz = yt, we obtain x = y = z = t making OM , ON , OP , OQ the perpendicular bisectors of segments AB , BC , CA, AD , respectively. Thus, OA = OB = OC = OD making ABCD cyclic. (a) There are 8 points placed on a circle. We say that Juliana performs “operation T ” if she chooses 3 such points and paints the sides of the triangle they determine in such a way that each painted triangle has at most one vertex in common with a previously painted triangle. What is the greatest number of operations T that Juliana can make? (b) If in part (a), if you have 7 points instead of 8 points, then what is the greatest number of operations T Juliana can make? Solved by Oliver Geupel, Br¨ uhl, NRW, Germany. We interpret the problem so that each painted triangle must have at most one vertex in common with every previously painted triangle. We claim that the greatest number of operations T is 8 in part (a) and 7 in part (b). We start with the proof of part (a) and denote the points by the numbers 1, 2, . . . , 8. A possible sequence of triangles of length 8 is 123, 456, 167, 148, 268, 347, 358, 257. On the other hand, consider a sequence of operations T . By assumption, each painted edge belongs to only one painted triangle. Therefore, each point from {1, 2, . . . , 8} is adjacent with an even number of edges of painted triangles, hence, with not more than 6 edges. Then, the total number of painted edges is not greater than 8 · 6/2 = 24. Consequently, we have not more than 24/3 = 8 triangles, which completes the proof of part (a). It remains to prove part (b). Denote the points by the numbers 1, 2, . . . , 7. A possible sequence of length 7 of triangles is 123, 345, 367, 146, 157, 247, 256. On the other hand, we have not more than = 21 edges, hence not more than 2 21/3 = 7 steps, which completes the proof of part (b). 7

That completes the Corner for this issue.

384

BOOK REVIEWS
Amar Sodhi
Charming Proofs : A Journey Into Elegant Mathematics by Claudi Alsina and Roger B. Nelsen The Mathematical Association of America, 2010 ISBN: 978-0-88385-348-1, Hardcover, 295 + xxiv pages, US$59.95 Reviewed by R. P. Gallant, Grenfell Campus, Memorial University of Newfoundland, Corner Brook, NL Alsina and Nelsen have set out to collect some beautiful proofs in elementary mathematics. Although beauty is in the eye of the beholder, I think many will agree they have succeeded. The book includes chapters individually devoted to results on polygons, triangles, equilateral triangles, quadrilaterals, squares, curves, and results from three-dimensional geometry. Other chapters include ‘Adventures in Tiling and Coloring’, ‘Distinguished Numbers’, ‘Points in the Plane’, ‘A Garden of Integers’, and a ﬁnal chapter containing assorted results. Each chapter closes with a selection of 10–15 relevant problems for the reader to attempt. Solutions to these challenges are provided at the end of the book. The selection of content is ripe for supporting visuals, and indeed “Charming Proofs” is distinguished by its numerous (over 300(!)) diagrams. The authors have written several other books devoted to the use of diagrams in mathematics, and that experience shows in this book. This book is part of the Dolciani Mathematical Expositions series and as such is intended to be suﬃciently elementary for undergraduate and even some high school students. “Charming Proofs” hardly uses calculus, and even then only in a handful of places. The book should be fully accessible to serious mathematics undergraduates, and much will be accessible to talented high school students. I must mention an error. In a short discussion about types of proofs, the authors confuse the converse for the contrapositive (page xxii). You may feel the same pang of concern I did upon ﬁnding this doozy so early in the book, but I assure you that the very few mistakes I found (like the one on page 108 involving the perimeter of the Varignon parallelogram, as another example) are essentially typos and of no consequence if one is paying attention. Certainly anyone interested in elegant proofs should consider this book. Also, anyone interested in mathematical competitions should ﬁnd this a useful problem book, though in this case be mindful of the elementary nature and geometric emphasis of the book. The book contains results both familiar and less familiar, and should be attractive to inexperienced and experienced readers of both classes. In summary, “Charming Proofs” is a wonderful collection of elementary proofs and related problems and should ﬁnd its way onto the bookshelves of many.

385

RECURRING CRUX CONFIGURATIONS 2
J. Chris Fisher Triangles for which 2b = c + a
This month we explore triangles ABC whose sides a, b, c are in arithmetic progression; we shall see that with the triangle labeled so that b is the intermediate side, having the sides in arithmetic progression is equivalent to requiring ∠BIO = 90◦ , as well as IG parallel to AC , and many other noteworthy properties (where I, O , and G are the incentre, circumcentre, and centroid, respectively). As in last month’s column, I will supply statements, references, and occasional hints, leaving the proofs as exercises. Problem 268 [1977 : 190; 1978 : 78-79] (Proposed by Gali Salvatore = L´ eo Sauv´ e). Show that in ∆ABC with a ≥ b ≥ c, the sides are in arithmetic progression if and only if 2 cot B 2 = 3 tan C 2 + tan A 2 .

The featured solution made use of the half-angle identities tan A C s−b tan = 2 2 s and
cyclic

tan

A B tan 2 2

= 1,

where s is the semiperimeter (a + b + c)/2. One solver, Charles W. Trigg, added the comment that in triangles where c + a = 2b, seven other relationships “follow easily”: 1. 2. 3. 4. 5. 6. 7. cos C + cos A = 4 sin2 B . 2 a cos C − c cos A = 2(a − c). ca = 6Rr (where R and r are the circumradius and inradius). −3 b . cos A = 4c2 c B < 60◦ except when the triangle is equilateral. GI ||BC . In the special A = C + 90◦ , then a, b, c are in the ratio √ √case where √ ( 7 + 1) : 7 : ( 7 − 1).

The editor Sauv´ e remarked that copies of Trigg’s proofs were available from him on request; sadly both he and Trigg died long ago, so today it would probably be faster for the reader to discover the proofs for himself. Number 6, however, has appeared in this journal several times, twice as a “Klamkin Quickie” [1996 : 61] and [2001 : 79]. An even quicker proof appeared as D.L. MacKay’s solution to Problem E411 in [1]; namely, Prove that if the sides of a triangle form an

386

arithmetic progression the line joining the centroid to the incentre is parallel to one side: We have b − a = c − b if and only if s = 3b/2, and r= ∆ s = 2∆ 3b = h 3 ,

where ∆ is the area of ∆ABC and h is the altitude from B to CA. Thus the incentre and centroid are equidistant from the side BC . This proof (expanded somewhat) was reproduced as Problem 82 in [5]. A reﬁned version of the same problem had appeared a couple years earlier in [2]. For that version, recall that the Nagel point (the common intersection point of the lines joining a vertex to the point where the opposite excircle touches a side) lies on the line GI . It turns out that in a triangle whose sides are in arithmetic progression, the common diﬀerence |a − b| = |b − c| equals the distance from I to the Nagel point. The MathWorld web page has proposed calling GI the Nagel line, but the name seems not to have caught on yet. Another solver of Problem 268, Leon Bankoﬀ, submitted two striking results (which he saw—so he said— “out of the cornea of his eye”): i. If c + a = 2b, then cot
A B C k + cot = 2 cot ; 2 2 2

ii. If c2 + a2 = 2b2 , then cot C + cot A = 2 cot B . The ﬁrst follows from Problem 268, while the second is Property 3 from our previous column on root-mean-square triangles. Problem 2870 [2003 : 399; 2004 : 382-383] (Proposed by Toshio Seimiya). Given triangle ABC with incentre I , circumcentre O , and centroid G, suppose that ∠AIO = 90◦ . Prove that IG||BC . The featured solution also proved the converse (for scalene triangles); this result combined with Trigg’s Property 6 (above) implies that for scalene triangles, ∠BIO ≤ 90◦ if and only if 2b ≤ c + a, with equality holding only simultaneously. This version is Problem 1506 in [7], where there are two published solutions; it also appeared as problem 2 on the Second Hong Kong Mathematical Olympiad 1999, with a solution in CRUX with MAYHEM [2005 : 520-521]. Amengual Covas added three further references, namely [3], [4], [6] which, he said, “provide other relationships.” Problem 3197 [2006 : 516, 518; 2007 : 501-502] (Proposed by Paul Deiermann). If AB is a ﬁxed line segment, ﬁnd the triangle ABC which has maximum area among those which satisfy ∠AIO = π/2. What is this maximum area? Michel Bataille’s featured solution plugged 2a = b + c into Heron’s formula for the area and found that the maximum area is achieved for the triangle with sides √ 3+ 3 c = 1, a = , and b = 2a − 1. 3

387

Finally, triangles with sides in arithmetic progression are mentioned in a footnote to a problem dealing with lines through vertex B that are perpendicular to IO . Problem 2246 (reworded) [1997 : 244; 1998 : 318-319] (Proposed by D.J. Smeenk). Given the nonequilateral triangle ABC , suppose that the line through B that is perpendicular to OI intersects the bisector of ∠BAC at P , and that the line through P parallel to AC intersects BC at M . Show that I, G, and M are collinear. The problem breaks down, of course, should ∆ABC be equilateral (because then I and O would coincide). Note that BI ⊥ IO implies that P = I , whence the conclusion that G is on the line IM gives yet another proof (although convoluted) that BI ⊥ IO implies GI ||AC . A comment attached to the problem pointed out that for M to be deﬁned, IG and BC could not be parallel, which thus forbids AI ⊥ IO , whence the condition 2a = b + c should have been added to the statement of the problem.

References
[1] J.H. Butchart, Problem E411, American Mathematical Monthly, 47:10 (Dec. 1940) 708-709. (Appeared in 1940, page 175). [2] Walter B. Clarke, Problem 172, National Mathematics Magazine, 12:5 (Feb. 1938) 249-250. (Appeared in 1938, page 194.) [3] F. G.-M., Exercices de g´ eom´ etrie, 7th ed., pp. 464-465. [4] F. G.-M., Exercices de trigonom´ etrie, pp. 413-414, 481. [5] Ross Honsberger, Mathematical Morsels. Dolciani Mathematical Expositions No. 3, 1978, pages 209-210. [6] I. Shariguin, Problemas de geometr´ ıa Planimetr´ ıa, editorial Mir, 1989, pp. 96-97. [7] Wu Wei Chao, Problem 1506, Mathematics Magazine, 70:4 (Oct. 1997) 302303. (Appeared in October 1996.)

388

PROBLEMS
Toutes solutions aux probl` emes dans ce num´ ero doivent nous parvenir au plus tard le 1 octobre 2012. Une ´ etoile ( ) apr` es le num´ ero indique que le probl` eme a ´ et´ e soumis sans solution. Chaque probl` eme sera publi´ e dans les deux langues oﬃcielles du Canada (anglais et fran¸ cais). Dans les num´ eros 1, 3, 5 et 7, l’anglais pr´ ec´ edera le fran¸ cais, et dans les num´ eros 2, 4, 6 et 8, le fran¸ cais pr´ ec´ edera l’anglais. Dans la section des solutions, le probl` eme sera publi´ e dans la langue de la principale solution pr´ esent´ ee. La r´ edaction souhaite remercier Jean-Marc Terrier, de l’Universit´ e de Montr´ eal, d’avoir traduit les probl` emes.

3651. Correction. Propos´ e par Pham Kim Hung, ´ etudiant, Universit´ e de ´ . Stanford, Palo Alto, CA, E-U
Soit a, b et c trois nombres r´ eels non n´ egatifs tels que a + b + c = 3. Montrer que a2 b + b2 c + c2 a + abc + 4abc(3 − ab − bc − ca) ≤ 5 .

3664.
que

Propos´ e par Pham Kim Hung, ´ etudiant, Universit´ e de Stanford, Palo ´ . Alto, CA, E-U Soit a, b et c trois nombres non n´ egatifs tels que a + b + c = 3 Montrer

|(1 − a2 b)(1 − b2 c)(1 − c2 a)| ≤ 3|1 − abc| .

3665.

Propos´ e par Nguyen Thanh Binh, Hanoi, Vietnam.

Dans une quadrilat` ere cyclique ABCD , soit M le point d’intersection des diagonales AC et BD , et soit Q le point d’intersection de la droite passant par M et le point milieu de BC . Montrer que M Q est perpendiculaire a ` AD si et seulement si les cˆ ot´ es AD et BC sont parall` eles (en quel cas ABCD est un trap` ezo¨ ıde isoc` ele) ou les diagonales sont perpendiculaires (et alors on a la conﬁguration de Brahmagupta).

3666.

Propos´ e par Michel Bataille, Rouen, France.

Soit R l’ensemble des paires d’entiers relativement premiers et S = {(a, b) ∈ R : 5a + 42b ≡ 0 (mod 1789)} . Trouver une bijection explicit entre S et R − S .

389

3667.

´ . Propos´ e par Joe Howard, Portales, NM, E-U

On suppose que bi > 0 pour i = 1, 2, . . . , n ; n ≥ 3 ; et Montrer que
n i=1 n− 2 bi ≥ n i=1

Én
i=1

bi = 1.

n−1 2

3668.

Propos´ e par Neven Juriˇ c, Zagreb, Croatie.

On suppose que p, q et r sont trois nombres premiers distincts. Combien de 1 1 1 solutions en entiers positifs l’´ equation x +y = z (o` u y = pqr ) poss` ede-t-elle ?

3669.

Propos´ e par Michel Bataille, Rouen, France.

Soit ABC un triangle inscrit dans un cercle Γ et soit r un nombre r´ eel − − → − − → − → − → (r = 0, 1). On d´ eﬁnit les points D , E , F par BD = r BC , CE = r CA, − → − → AF = r AB . Le cercle Γ coupe encore DA en D1 et la parall` ele a ` BC par D1 en B . On construit les points C et A de mani` ere analogue. Pour quels r les droites AA , BB , CC sont-elles concourantes ? Quel est alors leur point d’intersection ?

3670.

Propos´ e par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie.

Soit n ≥ 2 un entier. Calculer
1 0 1 0 1 0

dxdydz x+y+z

3671

. Propos´ e par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie.

Soit M un point a ` l’int´ erieur d’un t´ etrah` edre ABCD . Est-il vrai, oui ou non, que [ACD ] [ABD ] [ABC ] 2 [BCD ] = = = = √ , AM 2 BM 2 CM 2 DM 2 3 si et seulement si le t´ etrah` edre est r´ egulier et que M est son centre de gravit´ e. On note ici l’aire de T par [T ].

3672.
Vietnam.

Propos´ e par Pham Van Thuan, Universit´ e de Science des Hano¨ ı, Hano¨ ı,

Soit x et y deux nombres r´ eels tels que x2 + y 2 = 1 Montrer que 1 1 1 3 + + ≥ 2 2 y 1+x 1+y 1 + xy 1 + x+ 2 Trouver quand cette in´ egalit´ e est valable.

390

3673.

Propos´ e par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie.
∞ n=2

Calculer le produit 1− 1 n2
(−1)n−1

. Propos´ e par Zhang Yun, High School attached to Xi An Jiao Tong / University, Xi An City, Shan Xi, Chine. Soit I le centre de la sph` ere inscrite dans un t´ etrah` edre ABCD et soit A1 , B1 , C1 , D1 les points respectivement sym´ etriques de I par rapport aux plans BCD , ACD , ABD , ABC . Les quatre droites AA1 , BB1 , CC1 , DD1 sontelles forc´ ement concourantes ?

3674

3675.

Propos´ e par Michel Bataille, Rouen, France.

Soit a, b et c les cˆ ot´ es d’un triangle et s son demi-p´ erim` etre. Soit respectivement r et R les rayons de ses cercles inscrit et circonscrit. Montrer que b(s − b) + c(s − c) 3R 6 ≤ ≤ . a(s − a) r cyclique .................................................................

3651.

Correction. Proposed by Hung Pham Kim, student, Stanford University, Palo Alto, CA, USA. Let a, b, and c be nonnegative real numbers such that a + b + c = 3. Prove that a2 b + b2 c + c2 a + abc + 4abc(3 − ab − bc − ca) ≤ 5 .

3664.
CA, USA. that

Proposed by Hung Pham Kim, student, Stanford University, Palo Alto,

Let a, b, and c be nonnegative real numbers such that a + b + c = 3. Prove

|(1 − a2 b)(1 − b2 c)(1 − c2 a)| ≤ 3|1 − abc| .

3665.

Proposed by Nguyen Thanh Binh, Hanoi, Vietnam.

Let the diagonals AC and BD of the cyclic quadrilateral ABCD intersect at M , and let the line joining M to the midpoint of BC meet AD at Q. Prove that M Q is perpendicular to AD if and only if the sides AD and BC are parallel (in which case ABCD is an isosceles trapezoid), or the diagonals are perpendicular (and we have Brahmagupta’s conﬁguration).

391

3666.

Proposed by Michel Bataille, Rouen, France.

Let R denote the set of all pairs of relatively prime integers and S = {(a, b) ∈ R : 5a + 42b ≡ 0 (mod 1789)} . Find an explicit bijection between S and R − S .

3667.

Proposed by Joe Howard, Portales, NM, USA.

Suppose bi > 0 for i = 1, 2, . . . , n; n ≥ 3; and
n i=1 n− 2 bi ≥ n i=1

Én
i=1

bi = 1. Prove

n−1 2

3668.

Proposed by Neven Juriˇ c, Zagreb, Croatia.

Suppose p, q and r are distinct prime numbers. How many positive integer 1 1 solutions has the equation x +y = 1 where y = pqr ? z

3669.

Proposed by Michel Bataille, Rouen, France.

Let ABC be a triangle inscribed in a circle Γ and let r be a real number − − → − − → − → − → − → (r = 0, 1). Points D , E , F are deﬁned by BD = r BC , CE = r CA, AF = − → r AB . The circle Γ meets again DA at D1 and the parallel to BC through D1 at B . Points C and A are constructed in a similar way. For which r are AA , BB , CC concurrent lines? What is then their point of concurrency?

3670.

Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.

Let n ≥ 2 be an integer. Calculate
1 0 1 0 1 0

dxdydz x+y+z

3671

. Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.

Let ABCD be a tetrahedron and let M be a point in its interior. Prove or disprove that [BCD ] AM 2 = [ACD ] BM 2 = [ABD ] CM 2 = [ABC ] DM 2 2 = √ , 3

if and only if the tetrahedron is regular and M is its centroid. Here [T ] denotes the area of T .

392

3672.
Vietnam.

Proposed by Pham Van Thuan, Hanoi University of Science, Hanoi,

Let x and y be real numbers such that x2 + y 2 = 1. Prove that 1 1+ x2 + 1 1+ y2 + 1 1 + xy ≥ 3 1+
x+y 2 2

When does this inequality occur?

3673.

Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.
∞ n=2

Calculate the product 1− 1 n2
(−1)n−1

3674

. Proposed by Zhang Yun, High School attached to Xi An Jiao Tong / University, Xi An City, Shan Xi, China.

Let I denote the centre of the inscribed sphere of a tetrahedron ABCD and let A1 , B1 , C1 , D1 denote their symmetric points of point I about planes BCD , ACD , ABD , ABC respectively. Must the four lines AA1 , BB1 , CC1 , DD1 be concurrent?

3675.

Proposed by Michel Bataille, Rouen, France.

Let a, b, and c be the sides of a triangle and let s be its semiperimeter. Let r and R denote its inradius and circumradius respectively. Prove that 6 ≤ b(s − b) + c(s − c) 3R ≤ . a ( s − a ) r cyclic

393

SOLUTIONS
No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems.

3542 . [2010 : 240, 242; 2011 : 244] Proposed by Cosmin Pohoat ¸a ˘, Tudor Vianu National College, Bucharest, Romania.
The mixtilinear incircles of a triangle ABC are the three circles each tangent to two sides and to the circumcircle internally. Let Γ be the circle tangent to each of these three circles internally. Prove that Γ is orthogonal to the circle passing through the incentre and the isodynamic points of the triangle ABC . [Ed.: Let ΓA be the circle passing through A and the intersection points of the internal and external angle bisectors at A with the line BC . The isodynamic points are the two points that ΓA , ΓB , and ΓC have in common.] Solution by Paul Yiu, Florida Atlantic University, Boca Raton, FL, USA.

A X I Y I B Z C O

1. The circle Γ has barycentric equation

(a + b + c)2 (a2 + b2 + c2 − 2ab − 2bc − 2ca)(a2 yz + b2 zx + c2 xy ) + 8abc(x + y + z )
cyclic

bc(b + c − a)x

= 0.

Proof. The mixtilinear incircles are deﬁned by the equations

394

(a + b + c)2 (a2 yz + b2 zx + c2 xy ) − (x + y + z )fa (x, y, z ) = 0,

(a + b + c)2 (a2 yz + b2 zx + c2 xy ) − (x + y + z )fb (x, y, z ) = 0,

(a + b + c)2 (a2 yz + b2 zx + c2 xy ) − (x + y + z )fc (x, y, z ) = 0,

where fa (x, y, z ) = 4b2 c2 x + c2 (c + a − b)2 y + b2 (a + b − c)2 z,

fc (x, y, z ) = b2 (b + c − a)2 x + a2 (c + a − b)2 y + 4a2 b2 z. Their radical center is the point deﬁned by fa (x, y, z ) = fb (x, y, z ) = fc (x, y, z ). Solving these equations we obtain the radical center in homogeneous barycentric coordinates

fb (x, y, z ) = c2 (b + c − a)2 x + 4c2 a2 y + a2 (a + b − c)2 z,

(a2 (b2 + c2 − a2 − 4bc) : b2 (c2 + a2 − b2 − 4ca) : c2 (a2 + b2 − c2 − 4ab)). This point divides OI in the ratio 2R : −r . The lines joining this radical center to the points of tangency with the circumcircle intersect the respective mixtilinear incircles again at the points
X = (a(a2 + 2a(b + c) − 3(b − c)2 ) : 2b2 (c + a − b) : 2c2 (a + b − c)), Y = (2a2 (b + c − a) : b(b2 + 2b(c + a) − 3(c − a)2 ) : 2c2 (a + b − c)), Z = (2a2 (b + c − a) : 2b2 (c + a − b) : c(c2 + 2c(a + b) − 3(a − b)2 )).

Γ is the circle containing these three points. Note: The center of Γ is the point (a2 (b2 + c2 − a2 +8bc) : b2 (c2 + a2 − b2 +8ca) : c2 (a2 + b2 − c2 +8ab)), which divides OI in the ratio 4R : r . 2. The line cyclic bc(b + c − a)x = 0 is the perpendicular bisector of II , where I is the inversive image of the incenter I in the circumcircle. Proof. The polar of I in the circumcircle is the line cyclic bc(b + c)x = 0. Replacing (x, y, z ) by 2(a + b + c)(x, y, z ) − (x + y + z )(a, b, c), we obtain the image of this polar under the homothety h I, 1 . This gives the 2 line in question. Since the pedal of I on its polar is the inversive image I , the line is the perpendicular bisector of II .
È È

395

3. One obtains the equation of the pencil of circles generated by a circle and a line by setting equal to zero a linear combination of the circle’s formula and the product of the line’s formula times that of the line at inﬁnity: x + y + z . (Circles are conics that pass through the conjugate imaginary points on the line at inﬁnity; all circles will contain those two points while the pencil generated by a circle and line will consist of all circles through the common pair of points of that circle and line—points which are possibly imaginary or coincident.) From the equation of Γ in (1), we conclude that it is a member in the pencil of circles generated by the circumcircle a2 yz + b2 zx + c2 xy = 0 and the perpendicular bisector of II’ with equation computed in (2). Now, any circle through I and I is orthogonal to the circumcircle. Since the isodynamic points are inverse in the circumcircle, the circle through I , I and one of them must also contain the other. In other words, the circle C through I and the isodynamic points contains I , and is orthogonal to every circle in the pencil generated by the circumcircle and the perpendicular bisector of II . Since Γ is a member of this pencil, it is orthogonal to circle C .
No other solutions were received.

3556.

[2010 : 315, 317] Proposed by Arkady Alt, San Jose, CA, USA.

For any acute triangle with side lengths a, b, and c, prove that (a + b + c) min{a, b, c} ≤ 2ab + 2bc + 2ca − a2 − b2 − c2 . I. Solution by Edmund Swylan, Riga, Latvia. Note that a + b + c = (−a + b + c) + (a − b + c) + (a + b − c), where each of the three terms on the right are positive. Now, −a2 + ab + ca = (−a + b + c)a, ab − b2 + bc = (a − b + c)b, ca + bc − c2 = (a + b − c)c.

II. Solution by Chip Curtis, Missouri Southern State University, Joplin, MO, USA. Taking a ≤ b ≤ c, the required inequality may be successively rewritten: 2a(b − a) + (c − b)(a + b − c) ≥ 0. ab + 2bc + ca − 2a2 − b2 − c2 ≥ 0,

The result follows, for any triangle, by adding these equations, replacing the ﬁnal factors on the right by min{a, b, c}, and using the ﬁrst identity above.

396

By the triangle inequality, both terms are nonnegative, proving the claim. III. Solution by Paolo Perfetti, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy We introduce the well-known change of variables a = x + z , b = x + y , −a c− b b− c ≥ 0, y = b+c ≥ 0 and z = a+2 ≥ 0. Taking c = y + z , or x = a+2 2 a ≤ b ≤ c gives y ≥ x, z and the required inequality becomes 2(x + y + z )(x + z ) ≤ 4(xy + yz + xz ), which is equivalent to (xy + zy ) ≥ x2 + z 2 , which holds because of the inequalities between x, y, z . The result follows.
ˇ ´ University of Sarajevo, Sarajevo, Bosnia Also solved by SEFKET ARSLANAGI C, and Herzegovina; GEORGE APOSTOLOPOULOS, Messolonghi, Greece; ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, France; PRITHWIJIT DE, Homi Bhabha Centre for Science Education, Mumbai, India; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; JOE HOWARD, Portales, NM, USA; KEE-WAI LAU, Hong Kong, China; ´ student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; CRISTINEL SALEM MALIKIC, MORTICI, Valahia University of Tˆ argovi¸ ste, Romania; ALBERT STADLER, Herrliberg, Switzerland; PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU, Com´ ane¸ sti, Romania; and the proposer. Most solvers observed that the result holds regardless of whether the triangle is acute. Arslanagi´ c, Geupel and Maliki´ c also noted the case of equality, a = b = c, but all appear to have missed two separate (degenerate) cases of equality, namely (i) c = 2a = 2b and (ii) a = 0, b = c.

3563.

[2010 : 316, 318] Proposed by Mikhail Kochetov and Sergey Sadov, Memorial University of Newfoundland, St. John’s, NL.

A square n × n array of lamps is controlled by an n × n switchboard. Flipping a switch in position (i, j ) changes the state of all lamps in row i and in column j . (a) Prove that for even n it is possible to turn oﬀ all the lamps no matter what the initial state of the array is. Demonstrate how to do it with the minimum number of switches. (b) Prove that for odd n it is possible to turn oﬀ all the lamps if and only if the initial state of the array has the following property: either the number of ON lamps in every row and every column is odd, or the number of ON lamps in every row and every column is even. If this property holds, provide an algorithm to turn oﬀ all the lamps. Solution by Steﬀen Weber, student, Martin-Luther-Universit¨ at, Halle, Germany. The order in which switches are ﬂipped does not aﬀect the ﬁnal state, and ﬂipping a switch twice has no eﬀect. A series of ﬂips, avoiding this redundancy, may be coded as an n × n (0, 1)-matrix in which “1” represents 2 a ﬂip in the corresponding position. It follows that there are at most 2n distinct transformations.

397

(a) If n is even, ﬂipping all switches which are in row i or column j changes only the state of lamp (i, j ). Doing this once for every ON lamp turns OFF 2 all lamps. It follows that there are at least 2n distinct transformations, and so n2 exactly 2 distinct transformations, in one-to-one correspondence with the coded matrices described above, which give the unique minimum series of ﬂips attaining each transformation. Eliminating redundant ﬂips from the above series of ﬂips to turn all lamps OFF we have that the unique minimum series of ﬂips is attained by ﬂipping each switch (i, j ) if and only if there are initially an odd number of ON lamps among the 2n − 1 positions in row i and column j . (b) If n is odd, ﬂipping any switch changes the parity of the number of ON lamps in every row and column. If all lamps are OFF this parity is 0 for every row and column, so as required, a solution is possible only if, in the initial state, the number of ON lamps in every row and column has the same parity. If this condition is met, follow the procedure described in (a) for n even to turn OFF all lamps in the initial (n − 1) × (n − 1) block. Because the parity property is preserved, either every lamp in the nth row and column is OFF (and we’re done), or all lamps in those positions are ON; in the latter case, ﬂipping switch (n, n) turns oﬀ these remaining lamps.
Also solved by SKIDMORE COLLEGE PROBLEM SOLVING GROUP, Skidmore College, Saratoga Springs, NY, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; GEORGE APOSTOLOPOULOS, Messolonghi, Greece; JOSEPH DiMURO, Biola University, La Mirada, CA, USA; ALBERT STADLER, Herrliberg, Switzerland; and the proposer. Some solvers observed that the solution described for odd n is also optimum. Some solvers explicitly derived relationships between the state matrix and the (coded) transformation, using arithmetic modulo 2 or representing both states and transformations (acting additively) as vectors in the space of n × n matrices over Z2 . The proposer notes that this problem belongs to a popular class of “switchboard problems” also known as “all ones” and “lights out” (the name of a commercial game). Problems of such type are found in many math competitions and popular math journals. The problem was inspired by [6]. However, optimality was not shown for even n and the solution for odd n was inelegant. Some other references are given below.

References
[1] Kvant, Problem M-665, v. 12, No 9 (1981), p. 26; Solution: v. 13, No 10 (1982), pp. 23–24. [2] Math. Intelligencer, Problem 88-8, v. 10, No. 3 (1988); Solution: v. 11, No. 2 (1989), pp. 31–32. [3] Amer. Math. Monthly, Problem 10197, v. 99 (1992), p. 162; Solution: v. 100 (1993), p. 806 . [4] M. Anderson, T. Feil, Turning lights out with linear algebra, Math. Magazine, v. 71, No. 4 (1998), pp. 300–303. [5] A. Shen, Lights out, Math Intelligencer, v. 22, No. 13 (2000), pp. 20–21. [6] P. Ara´ ujo, How to turn all the lights out, Elem. Math. v. 55 (2000), pp. 135–141.

398

3564. [2010 : 396, 398] Proposed by Pham Van Thuan, Hanoi University of Science, Hanoi, Vietnam.
Let a, b, c, d be positive real numbers. Prove that a3 + b3 + c3 + d3 + 32abcd a+b+c+d ≥ 3(abc + bcd + cda + dab) .

Solution by Joe Howard, Portales, NM, USA. This is Problem 8 of [1], left as an exercise. We will follow the solution to Problem 3, from the same article. Without loss of generality we can assume that d = min{a, b, c, d} = 1. Then, the inequality is: a3 + b3 + c3 + 1 + or equivalently (a + b + c + 1)4 + 32abc ≥ 3(a + b + c + 1)2 (ab + bc + ca + a + b + c) . Let p = a + b + c, q = ab + bc + ca and r = abc. Then p ≥ 3 and q ≥ 3. We need to prove The inequality 3q ≤ p2 is equivalent to ab + ac + bc ≤ a2 + b2 + c2 , which is easy to prove. Thus, we can ﬁnd some t ≥ 0 so that q= Now, by Theorem 1 in [1] we get 27 To complete the proof it suﬃces to show that (p + 1)4 − 3p(p + 1)2 − 3(p + 1)2 This simpliﬁes to (5p + 3)(p − 3)2 + t2 (27 + 27p2 − 42p − 64t) ≥ 0 . Using p ≥ 3 we get 14p2 ≥ 42p, while from Thus which proves (1), and hence completes the proof. 27p2 = 14p2 + 13p2 ≥ 42p + 78t ≥ 42p + 64t ,
p2 −t2 3

32abc a+b+c+1

≥ 3(abc + ab + ac + bc) ,

(p + 1)4 + 32r ≥ 3(p + 1)2 (p + q ) .

p2 − t2 3

r≥

p3 − 3pt2 − 2t3

p2 − t2 3

32 27

(p3 − 3pt2 − 2t3 ) ≥ 0 . (1)

p2 ≥ t2 + 9 ≥ 6t .

= q ≥ 3 we get

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; JOEL SCHLOSBERG, Bayside, NY, USA; ALBERT STADLER, Herrliberg, Switzerland; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer . There was one incorrect solution.

399

References
[1] P. V. Thuˆ a e V˜ ı, A useful inequality revisited, Crux, Vol 35(3), 2009, pp. 164-171 . n, Lˆ

3565.

[2010 : 396, 398] Proposed by Max Diaz, student, San Juan Bosco High School, Huancayo, Junin, Peru.

Find all positive integers n such that σ (τ (n)) = n, where τ (m) and σ (m) are, respectively, the number of positive divisors of the integer m and the sum of all the positive divisors of the integer m. Solution by Oliver Geupel, Br¨ uhl, NRW, Germany. The solutions are 1, 3, 4, and 12. Note that if m is a positive integer and √ m = ab where 1 ≤ a ≤ b ≤ m, √ then a ≤ m. Thus, for each d = 1, 2, . . . , m , there is at most one positive integer d such that dd = m. Hence, we have τ (m) ≤ 2
m . 2

√ √ m ≤ 2 m.

(1)
m , so 2

Furthermore, if d is a positive divisor of m with d < m, then clearly d ≤ d≤ Hence,
m 2

σ (m) ≤ m +

k=1

k ≤m+

1 2

m 2

+1 =

1 8

(m2 + 10m).

(2)

that

1 (m2 8

Now, if n is a solution to the given problem, then by (1), (2) and the fact + 10m) is an increasing function, we have √ n = σ (τ (n)) ≤ max σ (k)|1 ≤ k ≤ 2 n
k

≤

√ √ √ 1 (2 n)2 + 10(2 2) = (n + 5 n), 8 2 1

from which we easily deduce that n ≤ 25. Direct checking for n = 1, 2, . . . , 25 then reveal that only 1, 3, 4 and 12 satisfy the given condition and our proof is complete.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; HENRY RICARDO, Tappan, NY, USA; JOEL SCHLOSBERG, Bayside, NY, USA; ALBERT STADLER, Herrliberg, Switzerland; and the proposer.

400

3566.

[2010 : 396, 398] Proposed by an unknown proposer.

Given points A and C on a circle with centre O , choose B on the shorter arc AC . Let be the line tangent to the circle at B , and let P and Q be the points where intersects the bisectors of ∠AOB and ∠BOC , respectively. Prove that if E = AC ∩ OQ, then P E is perpendicular to OQ. Similar solutions by V´ aclav Koneˇ cn´ y, Big Rapids, MI, USA; and by Kee-Wai Lau, Hong Kong, China. We assume that the circle has unit radius, and set α = ∠AOP = ∠P OB and γ = ∠BOQ = ∠QOC . In triangle OEC we have ∠EOC = ∠QOC = γ and ∠OCE = ∠OCA = ∠OAC = 180◦ − 2(α + γ ) 2 = 90◦ − (α + γ ),

whence ∠OEC = 90◦ + α. By the sine law, OE = cos(α + γ ) cos α .

Moreover, from the right triangle OP B , OP = 1 ; cos α

= cos(α + γ ). But in triangle P OE , ∠P OE = α + γ ; therefore thus OE OP ∆P OE is a right triangle with hypotenuse OP and right angle at E . That is, P E is perpendicular to OQ, as desired.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece(2 solutions); CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; DAG JONSSON, Uppsala, ´ student, Sarajevo Sweden; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; SALEM MALIKIC, College, Sarajevo, Bosnia and Herzegovina; ALBERT STADLER, Herrliberg, Switzerland; and PETER Y. WOO, Biola University, La Mirada, CA, USA(2 solutions). All the submitted solutions were short and neat. Other nice methods came down to showing that P E is an altitude of triangle OP Q, or that the quadrilateral OAP E is cyclic with a right angle at A.

3567.

[2010 : 396, 398] Proposed by Albert Stadler, Herrliberg, Switzerland.

Prove that
∞ 0

e− x 1 − e− 2 x

x(1 − e−14x )

1 − e− 4 x

1 − e− 6 x

dx = ln 2 .

Solution by Oliver Geupel, Br¨ uhl, NRW, Germany. Let I denote the given integral.

401

With the substitution u = e−x , we easily obtain
1

I =−

(1 − u2 )(1 − u4 )(1 − u6 ) (1 − u14 ) ln u

du.

We apply the following known result [1]:
1 0

(1 − up )(1 − uq )(1 − ur )us−1 du (1 − ut ) ln u
´

= ln

Γ Γ

p+s Γ t p+q +s t

q +s s Γ r+ t t +s Γ q+r Γ t

p+q +r +s t p+r +s Γ s t t

where Γ denotes the gamma function and p, q , r , s, t are all positive. With p = 2, q = 4, r = 6, s = 1, and t = 14 we then obtain
´

I = ln Since it is known [2] that Γ Γ the result follows.
1 14 3 14

Γ Γ

1 14 3 14

Γ Γ

9 14 5 14

Γ Γ

11 14 13 14

Γ Γ

9 14 5 14

Γ Γ

11 14 13 14

= 2,

Also solved by MOHAMMED AASSILA, Strasbourg, France; MICHEL BATAILLE, Rouen, France; and the proposer.

References
[1] I. S. Gradshteyn, I. M. Ryzhik, Table of Integrals, Series, and Products, Academic Press, 5th edition, 1996; formula #32. [2] The American Mathematical Monthly, Problem 11426: 116(2009), p.365; solution in v. 117(2010), p. 842. Gamma Products, v.

3569 . [2010 : 397, 399] Proposed by Jian Liu, East China Jiaotang University, Nanchang City, China.
Let the point P lie inside the triangle ABC and let the point Q lie outside the triangle. Let w1 , w2 , w3 denote the lengths of the angle bisectors of ∠BP C , ∠CP A, ∠AP B , respectively. Does the inequality P A · QA + P B · QB + P C · QC ≥ 4(w1 w2 + w2 w3 + w3 w1 ) hold? [At http: // www. emis. de/ journals/ JIPAM/ article1162. html? sid= 1162 the proposer’s inequality is proved when Q lies inside the triangle.]

402

Comment by Oliver Geupel, Br¨ uhl, NRW, Germany. The cited article does not contain the promised inequality. Its main result is the weaker inequality P A · QA + P B · QB + P C · QC ≥ 4(r1 r2 + r2 r3 + r3 r1 ), (1)

where ri denotes the distances from P to the sides of the triangle, and both P and Q are interior points. We prove the following Lemma. For each point Q outside the triangle there exists a point Q on the boundary of the triangle such that QA > QA , QB > QB , and QC > QC . Proof. Drawing rays from the vertices of the triangle orthogonal to the sides partitions the plane outside the triangle into six regions: Three regions SA , SB , SC outwardly on the sides and three regions TA , TB , TC outwardly on the vertices. If Q lies in an “S ” region, deﬁne Q to be the orthogonal projection of Q onto the adjacent side of the triangle. If Q lies in an “T ” region deﬁne Q to be the adjacent vertex. Editor’s comment. Geupel’s lemma implies that inequality (1) holds for all points Q in the plane. The status of the required result for angle bisectors, however, remains in doubt until somebody produces either a correct reference or a valid proof. Of course, Geupel’s lemma shows that if the desired inequality holds when Q is an interior point, then it holds for arbitrary Q.
No solutions were received.

3570.

[2010 : 397, 399] Proposed by Arkady Alt, San Jose, CA, USA.

Let r , ra , rb , rc , and R be, respectively, the inradius, the exradii, and the circumradius of triangle ABC with side lengths a, b, c. Prove that
2 ra

2 ra

2 rb

2 rc

≥

4R + r 4R − r

Solution by Kee-Wai Lau, Hong Kong, China. Applying the Cauchy-Schwarz inequality to vectors
ra r r Ô , Ô 2 b 2, Ô c 2 2 b + rb a 2 + ra c 2 + rc

and

2, a 2 + ra

2 b2 + rb ,

2 c 2 + rc

we have
2 ra 2 a 2 + ra

2 rb

b2 +

2 rb

2 rc 2 c 2 + rc

≥

2 2 a2 + ra + b2 + rb + c 2 + rc

¡ 2

(ra + rb + rc )2
¡

¡ . (1)

403

Let s be the semiperimeter. We need the following known results. ra = 4R + r,
cyclic 2 ra = (4R + r )2 − 2s2 ,

(2) (3) (4)

cyclic

a2 = 2 s2 − r 2 − 4Rr .

Formulas (2) and (3) appear on p.61 (items 99, 103) and formula (4) on p.52 (item 5) [1]. With these formulas, we can get

a2 + ra + b2 + rb + c2 + rc

¡ 2

(ra + rb + rc )2
¡ 2

¡ = 2

(4R + r )2 (4R + r )2 − 2s2 + 2 (s2 − r 2 − 4Rr ) (4R + r )2 (4R + r )(4R − r ) = 4R + r 4R − r ,

= and this with (1) completes the solution.

Also solved by JOE HOWARD, Portales, NM, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; and the proposer. 2 + r 2 + r 2 = 16R2 − r 2 , Geupel and the proposer note that the identity a2 + b2 + c2 + ra c b used in the last step of the featured solution, is interesting on its own.

References
[1] D.S. Mitrinovi´ c et al., Recent Advances in Geometric Inequalities, Kluwer Academic Publishers, 1989.

3571.

[2010 : 397, 399] Proposed by Arkady Alt, San Jose, CA, USA.

Let n ≥ 1 be an integer. Among all increasing arithmetic progressions x1 , 2 2 x2 , . . . , xn such that x2 1 + x2 + · · · + xn = 1, ﬁnd the progression with the greatest common diﬀerence d. Solution by Oliver Geupel, Br¨ uhl, NRW, Germany. Since the case n = 1 is degenerate, let us assume n > 1. An arithmetic progression x1 , x2 , . . . , xn with common diﬀerence d > 0 has the property that 2 2 1 = x2 1 + x2 + · · · + xn if and only if 1=
n− 1 k=0

(x1 + kd)2 = nx2 1 + 2d

n− 1 k=0

x1 + d 2

n− 1 k=0

Equivalently, x1 is a real root of the following quadratic in x 1 = nx2 + (n − 1)nd · x + 1 6 (n − 1)n(2n − 1)d2 ,

404

which will happen if and only if n > 1 and d > 0 are such that its discriminant (n − 1)2 n2 d2 − 2 (n − 1)n2 (2n − 1)d2 + 4n 3

is non-negative and this is equivalent to saying (n − 1)n(n + 1)d2 ≤ 12. Consequently the greatest common diﬀerence is d= 12 (n − 1)n(n + 1) ,

and the ﬁrst term of the progression is the solution of the quadratic above with this value of d, namely 3(n − 1) x1 = − . n(n + 1)
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; ROY BARBARA, Lebanese University, Fanar, Lebanon; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; JOEL SCHLOSBERG, Bayside, NY, USA; ALBERT STADLER, Herrliberg, Switzerland; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. One incomplete solution was received.

3572. [2010 : 397, 399] Proposed by Jos´ e Luis D´ ıaz-Barrero, Universitat Polit` ecnica de Catalunya, Barcelona, Spain.
Let a, b, c be positive real numbers such that a + b + c = 1. Prove that ab
cyclic

c + ab

1 4
cyclic

√ ab

a+b

≥ 1.

Composite of similar solutions by Arkady Alt, San Jose, CA, USA; Salem Maliki´ c, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; and Albert Stadler, Herrliberg, Switzerland. Note ﬁrst that ab
cyclic

c + ab

=
cyclic

ab c(a + b + c) + ab 1 (a + b)(b + c)(c + a)

=
cyclic

ab (c + a)(c + b)

ab(a + b) .
cyclic

Hence the given inequality is equivalent to √ 4 ab(a + b) + (a + ab) ≥ 4(a + b)(b + c)(c + a),
cyclic cyclic

or (a +
cyclic

√

ab) ≥ 8abc .

405

By the AM-GM Inequality, we have √ √ √ √ (a + ab) = abc ( a + b)
cyclic cyclic

≥ 8 abc

√

√ √ √ √ √ √ a b b c c a = 8abc,
1 . 3

so our proof is complete. Clearly, equality holds if and only if a = b = c =

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; JOE HOWARD, Portales, NM, USA; PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; and the proposer.

3574.

[2010 : 398, 400, 548, 550] Proposed by Michel Bataille, Rouen, France.

Let x, y , and z be real numbers such that x + y + z = 0. Prove that cosh x ≤ cosh2
cyclic

cyclic

x−y 2

≤ 1+2

cosh x .
cyclic

Solution by Arkady Alt, San Jose, CA, USA. Let a := ex , b = ey , c = ez . Then a, b, c > 0 and abc = ex+y+z = 1. Let s := a + b + c, p := ab + ac + bc. Then cosh(x) =
cyc

1 2
cyc

(a + bc) =

s+p 2

Let’s observe that cosh x−y 2 = e
x− y 2

+e 2

y −x 2

a+b = √ = 2 ab Thus cosh2
cyc

2 √ (a + b) c 2

√ √ a b √ +√ a b .

(1)

x−y 2

1 4

(a + b)2 c =
cyc

1 4

a2 c + b2 c + 2 =
cyc

3 + sp . 4

Also cosh(x) =
cyc cyc

a + bc 2 1 8
cyc

=
cyc

a2 + 1 2a 2 + p2 + s2 − 2p − 2s 8 .

(a2 + 1) =

406

Thus, the inequality to prove becomes 1 2 or equivalently (s + p) ≤ 3 + sp 4 ≤1+ 2 + p2 + s2 − 2p − 2s 4 , (2)

2(s + p) ≤ 3 + sp ≤ 6 + p2 + s2 − 2(p + s) . √ √ 3 3 Observing that p ≥ 3 a2 b2 c2 = 3 and s ≥ 3 abc = 3 we obtain sp + 3 − 2(s + p) = (s − 3)(p − 3) + (s − 3) + (p − 3) ≥ 0 , which proves the left hand side of (3). To prove the RHS of (3) we note that 6+p2 + s2 − 2(p + s) − (3 + sp)
2

(3)

= (p − s) + (s − 3)(p − 3) + (s − 3) + (p − 3) ≥ 0 . This proves the RHS of (3), and thus completes the proof.

= 3 + p2 + s2 − 2p − 2s − sp = (p − s)2 + sp + 3 − 2(s + p)

(4)

Also solved by OLIVER GEUPEL, Br¨ uhl, NRW, Germany; ALBERT STADLER, Herrliberg, Switzerland; and the proposer.

3575.

[2010 : 398, 400] Proposed by Michel Bataille, Rouen, France.

Let ABC be a triangle with incentre I . Characterize the lines through I intersecting the sides AB and AC at D and E , respectively, such that DE = DB + EC and determine how many such lines there are in terms of ∠B and ∠C . Solution by Joel Schlosberg, Bayside, NY, USA. For each line DE through I intersecting sides AB and AC , ∠BID and ∠CIE are external to ∆BIC ; moreover, because ∠BID + ∠CIE = C 180◦ − ∠BIC = B + , each position of DE corresponds to a unique angle 2 C ◦ θ ∈ [− B , ] ⊂ [ − 45 , 45◦ ] such that 4 4 ∠BID = B 2 + 2θ A and ∠CIE = C 2 − 2θ.

π − (B + 2θ ) B/2 B/2

D
2θ B/2

I E
C/2

407

Because DE = DI + IE , the requirement that DE = DB + EC is equivalent to DI DB BI EC IE − = − . (1) BI BI CI CI CI By the Law of Sines applied to triangles DBI, BCI , and CEI , equation (1) is equivalent in turn to sin
B 2

sin(B + 2θ ) 2 sin(−θ ) cos 2 sin
B 2 B 2 B 2

− sin

B 2

+ 2θ

sin sin
C 2

C 2 B 2

= =

sin

C 2

sin(C − 2θ ) 2 sin(−θ ) cos 2 sin
C 2

− 2θ − sin
C 2

C 2

+ θ sin

+ θ cos

+ θ sin

B 2

− θ cos

C 2

−θ −θ

Clearly θ = 0 satisﬁes the condition; if θ = 0 then sin(−θ ) is nonzero and can be canceled, yielding B B + θ sin 2 2 B B B sin2 cos θ + sin cos sin θ 2 2 2 sin whence, tan θ = = = sin2 sin
C 2 C 2 C 2

C C − θ sin 2 2 C C C = sin2 cos θ − sin cos sin θ, 2 2 2 = sin

cos

+ sin

− sin2

B 2 B cos B 2 2

− 1 − 2 sin2

C + 1 − 2 sin2 B 2 2 C B B cos + 2 sin cos 2 sin C 2 2 2 2 B C −B sin 2 sin C + C − B 2 2 = tan C −B C +B 2 2 sin 2 cos 2

= .

− cos C + cos B sin C + sin B (2)

Since the tangent function is strictly increasing in the domain ä and bijective ç B C (−90◦ , 90◦ ), which contains both the interval − B , and the angle C − , 4 4 2 equation (2) is equivalent to θ = conditions imply that ∠B = ∠C, Our conclusion: • If ∠B = ∠C or
∠B ∠C C −B 2

for θ = 0 and θ ∈ − B , 4 B 2 +2 C−B 2

C 4

. These B 2

∠B ∠C

∈

1 2

,2 ,

and

∠BID =

=C−

desired property; it satisﬁes ∠BID = BC through I .

∈ /

1 ,2 2

, then there is a single line with the
B 2

and is therefore the parallel to

408
ä ç

• If ∠B = ∠C and

∠B ∠C

property; they satisfy ∠BID =

∈

1 ,2 2

, then there are two lines with the desired
B 2

and ∠BID = C −

B . 2

Also solved by the proposer. We also received one incorrect and one incomplete solution. The incomplete solution provided an appealing intuitive approach, as follows: One obvious line that satisﬁes the required equation is the line DE through I that is parallel to BC : in this case, because BI bisects ∠B , ∠DIB = ∠CBI = ∠IBD , whence DB = DI ; similarly EC = IE , so that DE = DB + EC , as desired. Another solution is obtained by reﬂecting this DE (namely the parallel to BC through I ) in the angle bisector AI to obtain E D , with E the image of D on AC and D the image of E on AB . Here D B+E C = = AB + AC − (AD + AE ) = AB + AC − (AE + AD ) DB + EC = DE = D E .

This second solution will not exist, however, if either D or E fall outside ∆ABC . E falls outside when C lies between E and A; that is, if ∠AE I < ∠ACI or, in terms of C B ∠B = ∠ADE = ∠AE D = ∠AE I , if ∠B < ∠2 . Similarly D falls outside if ∠C < ∠2 . The two lines coincide when ∠B = ∠C . In all other cases there will be at least two solutions. It remains to show that there cannot be more than two solutions. For this our correspondent makes claims about the monotonicity of the lengths of the relevant segments, ﬁrst as the point E moves along AC from its position where EI ||BC to the foot of the angle bisector BI , and then as it moves along AC to the vertex C ; he claimed that these claims are “easy to prove.” I suppose it should be comforting to know that the proof is easy; however, this editor has seen many very easy proofs that he was not clever enough to devise for himself, many of which were not only easy, but correct. Please, if the proof is easy but not routine, then supply the proof.

Crux Mathematicorum
with Mathematical Mayhem
Former Editors / Anciens R´ edacteurs: Bruce L.R. Shawyer, James E. Totten, V´ aclav Linek

Crux Mathematicorum
Founding Editors / R´ edacteurs-fondateurs: L´ eopold Sauv´ e & Frederick G.B. Maskell Former Editors / Anciens R´ edacteurs: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Crux Mathematicorum with Mathematical Mayhem is a problem-solving journal at the secondary and university undergraduate levels, published by the Canadian Mathematical Society eight times per year (February, March, April, May, September, October, November, and December). Its aim is primarily educational; it is not a research journal.

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Editor-in-Chief
Associate Editor Skoliad Editors Acting Mayhem Editor Mayhem Associate Editor Olympiad Editors Book Reviews Editor Articles Editor Problems Editors

Shawn Godin
Jeﬀ Hooper Lily Yen Mogens Hansen Shawn Godin Lynn Miller Robert Woodrow Nicolae Strungaru Amar Sodhi Robert Dawson Ian Aﬄeck Edward Barbeau Chris Fisher Chris Grandison Anna Kuczynska Cosmin Pohoata Nicolae Strungaru Edward Wang Johan Rudnick

Managing Editor

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Copyright
The Canadian Mathematical Society grants permission to individual readers of this publication to copy articles for their own personal use.

37: No 7
Published by:

November / Novembre 2011

Canadian Mathematical Society Soci´ et´ e math´ ematique du Canada 209 - 1725 St. Laurent Blvd. Ottawa, ON K1G 3V4, Canada Fax/T´ el´ ec. : 613 733 8994 c CANADIAN MATHEMATICAL SOCIETY 2011. ALL RIGHTS RESERVED. SYNOPSIS 409 Skoliad No. 136 Lily Yen and Mogens Hansen

- Swedish Junior High School Mathematics Contest, Final Round, 2010/2011 - Concours math´ ematique su´ edois, Niveau ´ ecole interm´ ediaire Ronde ﬁnale, 2010/2011 - Solutions to questions of the Niels Henrik Abel Mathematics Contest 2009–2010, 2nd Round 415 Mathematical Mayhem 415 Mayhem Problems: 418 Mayhem Solutions: Shawn Godin M507–M512 M470–M475 R.E. Woodrow and Nicolae Strungaru OC41–OC50

423 The Olympiad Corner: No. 297 423 Olympiad Corner Problems:

In this Corner are solutions from readers to some problems from - Olimpiada Nacional Escolar de Matematica 2009, Level 2 - S´ election OIM 2006 - 17th Japanese Mathematical Olympiad, First Round - 17th Japanese Mathematical Olympiad, Final Round - Croatian Mathematical Competition 2007 National Competition, 3rd Grade - Croatian Mathematical Competition 2007 National Competition, 4th Grade 447 Book Reviews Amar Sodhi

447 Magical Mathematics : The Mathematical Ideas That Animate Great Magic Tricks by Persi Diaconis and Ron Graham Reviewed by S. Swaminathan 448 Unsolved Crux Problem: 154

449 Recurring Crux Conﬁgurations 3 :

J. Chris Fisher

This new, occasionally appearing column, highlights situations that reappear in Crux problems. In this issue problem editor J. Chris Fisher examines triangles for which 2B = C + A. Enjoy! 454 Problems: 3676–3687 This month’s “free sample” is:

3687.

Proposed by Albert Stadler, Herrliberg, Switzerland.

Let n be a nonnegative integer. Prove that
∞

kn k!

k+1−

1 k!

∞

e−t tk+1 dt
1

=
k=0

S (n, k) k+2

k=0

where kn is taken to be 1 for k = n = 0 and S (n, k) are the Stirling numbers of the second kind that are deﬁned by the recursion S (n, m) = S (n−1, m−1)+mS (n−1, m), S (n, 0) = δ0,n , S (n, n) = 1 .

.................................................................

3687.

Propos´ e par Albert Stadler, Herrliberg, Suisse.

Soit n un entier non n´ egatif. Montrer que
∞

kn k!

k+1−

1 k!

∞

e−t tk+1 dt
1

=
k=0

S (n, k) k+2

k=0

o` u l’on pose kn = 1 pour k = n = 0 et o` u S (n, k) sont les nombres de Striling du second ordre, d´ eﬁnis par la r´ ecursion S (n, m) = S (n−1, m−1)+mS (n−1, m), S (n, 0) = δ0,n , S (n, n) = 1 .

459 Solutions: 3568, 3573, 3576–3580, 3582–3587

409

SKOLIAD

No. 136

Lily Yen and Mogens Hansen
Please send your solutions to problems in this Skoliad by October 1, 2012. A copy of CRUX with Mayhem will be sent to one pre-university reader who sends in solutions before the deadline. The decision of the editors is ﬁnal. Our contest this month is the Final Round of the Swedish Junior High School Mathematics Contest 2009/2010. Our thanks go to Paul Vaderlind, Stockholm University, Sweden for providing us with this contest and for permission to publish it. We also thank Rolland Gaudet, Universit´ e de Saint-Boniface, Winnipeg, MB, for translating the contest.

Swedish Junior High School Mathematics Contest Final Round, 2010/2011
3 hours allowed

1. The year 2010 is divisible by three consecutive primes. Find the last year before that with this property. 2.
Draw a line from the centre, O , of a circle with radius r to a point, P , outside the circle. Then choose two points, A and B , on the circle such that AB has length r and is parallel with OP . Find the area of the shaded region.
A B

3. 4.

Five distinct positive numbers are given. No matter which two of them you choose, one divides the other. The sum of the ﬁve numbers is a prime. Show that one of the ﬁve numbers is 1. A large cube consists of eight identical smaller cubes. The faces of each of the smaller cubes bear the numbers 3, 3, 4, 4, 5, and 5 such that opposite faces bear the same number. Assign to each face of the large cube the sum of the four visible numbers. Show that the numbers assigned to the faces of the large cube cannot be six consecutive integers.

5. 6.

The parallelogram ABCD has area 12. The point P is on the diagonal AC . The area of ABP is one third of the area of ABCD . Find the area of CDP .

Place ten numbers in the grid subject to the following rules: 1. For neighbours in the bottom row, the number on the right must be twice as large as the number on the left. 2. Other than in the bottom row, each number is the sum of the two numbers immediately below it. Find the smallest positive integer that you can place in the bottom left position such that the sum of all ten numbers is a square.

410

Concours math´ ematique su´ edois Niveau ´ ecole interm´ ediaire Ronde ﬁnale 2010/2011 Dur´ ee : 3 heures 1. En 2010, on constate que ce nombre est divisible par trois nombres premiers cons´ ecutifs. D´ eterminer la derni` ere ann´ ee avant ¸ ca, o` u cette propri´ et´ e tenait. 2. Tracer une ligne du centre, O, d’un cercle ayant rayon r
jusqu’` a un point, P , en dehors du cercle. Choisir alors deux points, A et B , sur le cercle et tels que AB est de longueur r et est parall` ele a ` OP . D´ eterminer la surface ombr´ ee.
A B

Cinq nombres positifs distincts vous sont donn´ es. Qu’importe lesquels deux vous choisissez, l’un d’eux divise l’autre. La somme des cinq nombres est un nombre premier. D´ emontrer que l’un des cinq nombres est 1.

4. Un

gros cube est form´ e de huit petits cubes identiques. Les faces des petits cubes portent les nombres 3, 3, 4, 4, 5 et 5, de fa¸ con a ` ce que les faces oppos´ ees portent le mˆ eme nombre. Assigner a ` chaque face du gros cube la somme de ses quatre nombres visibles. D´ emontrer que les nombres assign´ es aux faces du gros cube ne peuvent pas ˆ etre six entiers cons´ ecutifs.

5. Le parallelogramme ABCD a une surface de 12. Le point P se trouve sur la diagonale AC . La surface de ABP est le tiers de la surface de ABCD . D´ eterminer la surface de CDP . 6. Placer dix nombres sur la grille, sujet aux r` egles suivantes.
1. Pour des voisins dans la rang´ ee du bas, le nombre a ` droite doit ˆ etre deux fois celui a ` gauche. 2. Pour les rang´ ees autres que celle du bas, chaque nombre est la somme des deux nombres imm´ ediatement sous lui. D´ eterminer le plus petit entier positif qu’on peut placer en bas a ` l’extrˆ eme gauche, si la somme des dix nombres est un nombre carr´ e.

Next follow solutions to the Niels Henrik Abel Mathematics Contest, 2009– 2010, 2nd Round, given in Skoliad 130 at [2011 : 3–5].

411

Solution by Gesine Geupel, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany. The ﬁrst interesting numbers are 1005 and 1206. Since 1005 = 3 · 5 · 67 and 1206 = 2 · 32 · 67, their greatest common divisor is 3 · 67 = 201. Thus, d is a divisor of 201. On the other hand, if x is an interesting number whose last two digits form the number k, then the ﬁrst two digits of x form the number 2k, so x = 200k + k = 201k. Thus every interesting number is divisible by 201. Consequently, d = 201.
Also solved by DAVID GOU, student, Burnaby North Secondary School, Burnaby, BC; and NELSON TAM, student, John Knox Christian School, Burnaby, BC.

2. A calculator performs this operation: It multiplies by 2.1, then erases all digits to the right of the decimal point. For example, if you perform this operation on the number 5, the result is 10; if you begin with 11, the result is 23. Now, if you begin with the whole number k and perform the operation three times, the result is 201. Find k. ´ Solution by Rowena Ho, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC.
To undo the calculator’s operation, divide by 2.1 and round up. That is, 201/2.1 ≈ 95.7 → 96, 96/2.1 ≈ 45.7 → 46, and 46/2.1 ≈ 21.9 → 22. Then verify: 22 · 2.1 = 46.2 → 46, 46 · 2.1 = 96.6 → 96, and 96 · 2.1 ≈ 201.6 → 201. Thus k = 22 is a possible solution. If you begin with 23, the result after three steps is 210, and if you begin with 21, the result after three steps is 193. Thus k = 22 is the only integer solution.

´ Also solved by LENA CHOI, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; GESINE GEUPEL, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany; and KRISTIAN HANSEN, student, Burnaby North Secondary School, Burnaby, BC. Our solver’s claim that the inverse operation is to divide by 2¡ .1 and round up is not quite ¢ , 202 ≈ [95.72, 96.19) to 201. correct. The calculator sends any number in the interval 201 2.1 2.1 However, chasing such intervals is much more work than estimating a solution, as our solver does, verifying it, and checking uniqueness.

3. The pentagon ABCDE consists of a square, ACDE , with side
length 8, and an isosceles triangle, ABC , such that AB = BC . The area of the pentagon is 90. Find the area of BEC . Solution by Kristian Hansen, student, Burnaby North Secondary School, Burnaby, BC. Note that EB and EC slice the pentagon into three triangles. You can therefore ﬁnd the area of BEC by subtracting the areas of ABE and CDE from the area of the pentagon. Since ACDE is a square with side 8, ∠CDE = 90◦ , ·8 and the area of CDE is 82 = 32. Since ABC is isosceles with base 8, the distance to B from the line through A and E is 4. Therefore ABE has base |AE | = 8, ·4 height 4, and, thus, area 82 = 16. A

B C

E
4 B

A 8 E 8

C 8 D

412

The area of ABCDE is 90, so the area of 90 − 32 − 16 = 42.

BEC is

´ Also solved by LENA CHOI, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; GESINE GEUPEL, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany; ´ and ROWENA HO, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC.

4. In how many ways can one choose three diﬀerent integers between 0.5 and 13.5 such that the sum of the three numbers is divisible by 3? ´ Solution by Lena Choi, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC.
When adding integers and checking divisibility Rem. 1 2 0 (by 3), the only relevant property of the integers is the 1 2 3 remainder (after division by 3). The table lists the 4 5 6 7 8 9 remainders of the integers in question. 10 11 12 If all three integers leave remainder 0, so will the 13 sum. You can choose three of the four numbers 3, 6, 9, and 12 in four ways. (Choosing three out of four is the same as choosing the one to leave behind. Surely, you can choose one of four in four ways.) If all three numbers leave remainder 1, their sum will leave remainder 1 + 1 + 1 = 3 ≡ 0 (mod 3). You must now choose three of the ﬁve numbers 1, 4, 7, 10, and 13. You can choose the ﬁrst number in 5 ways, the second in 4 ways, and the third in 3 ways. Thus you can choose the three numbers in a speciﬁc order in 5 · 4 · 3 = 60 ways. Three numbers can be arranged in 6 ways, so you can choose three of ﬁve numbers in 60/6 = 10 ways without order. If all three numbers leave remainder 2, their sum will leave remainder 2 + 2 + 2 = 6 ≡ 0 (mod 3). You can choose three of the four numbers 2, 5, 8, and 11 in four ways. Finally, if you choose one number in each column in the table, the sum leaves remainder 1 + 2 + 0 = 3 ≡ 0 (mod 3). You have 5 choices in the ﬁrst column, 4 in the second, and 4 in the last. Thus you can choose the three numbers in 5 · 4 · 4 = 80 ways. All in all, you can choose three numbers whose sum leaves remainder 0 when divided by 3 in 4 + 10 + 4 + 80 = 98 ways.
Also solved by GESINE GEUPEL, student, Max Ernst Gymnasium, Br¨ uhl, NRW, ´ Germany; and ROWENA HO, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC.

Solution by the editors.

If a and b are positive integers such that a3 − b3 = 485, ﬁnd a3 + b3 .

Note that 485 = 5 · 97, so 485 can only be written as a product of two integers in two ways, 5 · 97 and 1 · 485. Now, 485 = a3 − b3 = (a − b)(a2 + ab + b2 ), and both a − b and a2 + ab + b2 are integers. Also a − b < a2 + ab + b2 since both a and b are positive integers. This leaves two cases: If a − b = 1 and a2 + ab + b2 = 485, then a = b + 1 and (b + 1)2 + (b + 1)b + b2 = 485, hence b2 + 2b + 1 + b2 + b + b2 = 485, so 3b2 + 3b − 484 = 0, thus 3(b2 + b) = 484. Since b is an integer and 484 is not a multiple of 3, this is not possible.

413

If a − b = 5 and a2 + ab + b2 = 97, then a = b + 5 and (b + 5)2 + (b + 5)b + b2 = 97, so b2 + 10b + 25 + b2 + 5b + b2 = 97, thus 3b2 + 15b − 72 = 0, hence 3(b + 8)(b − 3) = 0. Since b is positive, b = 3 and a = b + 5 = 8. Hence the only solution is that a = 8 and b = 3. Thus a3 + b3 = 83 +33 = 512 + 27 = 539.

6.
a

−2 2

If a and b are positive integers such that a3 + b3 = 2ab(a + b), ﬁnd b + a2 b−2 .

Solution by the editors. Note that a3 + b3 = (a + b)(a2 − ab + b2 ). Since a3 + b3 = 2ab(a + b) and a + b = 0, a2 − ab + b2 = 2ab, so a2 + b2 = 3ab. Now divide through by ab; 2 b b 2 b2 2 −2 +a = 3. Thus 32 = ( a +a ) = a +2 a ·b+a +2+ a−2 b2 , then a 2 = a b b b b2 b a −2 2 2 −2 2 and hence a b + a b = 3 − 2 = 7.

Let D be the midpoint of side AC in ABC . If ∠CAB = ∠CBD and the length of AB is 12, then ﬁnd the square of the length of BD . Solution by the editors.
C a D a A 12 x B b

Let a be |AD | (which equals |CD | since D is the midpoint), let b be |BC |, and let x be |BD |, as labeled in the diagram. Since ∠CAB = ∠CBD and ∠ACB = ∠BCD then ABC ∼ BDC . Therefore |AC | |BC | BC | |CD | a b b = | and | = | , so 2b = a and 12 = a , |BC | CD | |AB | BD | x 2 2 2 2 2 hence 2a = b and bx = 12a. Thus b x = 144a = 72 · 2a2 = 72b2 , so |BD |2 = x2 = 72.

8. If x, y, and z are whole numbers and xyz + xy +2yz + xz + x +2y +2z = 28 ﬁnd x + y + z .
Solution by the editors. Note: The question should have said that x, y , and z are positive integers. Noting that many of the terms in the given equation contain an x, one may try to factor out that x: x(yz + y + z + 1) + 2yz + 2y + 2z = 28. Noting that the x-free terms almost equal twice the expression in the brackets, one may try to reconstruct the expression in the brackets: x(yz +y +z +1)+2yz +2y +2z +2 = 30, so x(yz +y +z +1)+2(yz +y +z +1) = 30, so (x+2)(yz +y +z +1) = 30. Encouraged by this success, try the idea again: (x + 2) y (z + 1) + z + 1 = 30, so (x + 2)(y + 1)(z + 1) = 30 = 2 · 3 · 5. Since x, y , and z are all positive integers, x + 2 ≥ 3, y + 1 ≥ 2, and z + 1 ≥ 2. Therefore the only possible solutions are (x + 2, y + 1, z + 1) = (3, 2, 5), (x + 2, y + 1, z + 1) = (3, 5, 2), (x + 2, y + 1, z + 1) = (5, 2, 3), and (x + 2, y + 1, z + 1) = (5, 3, 2). That is, (x, y, z ) = (1, 1, 4), (x, y, z ) = (1, 4, 1), (x, y, z ) = (3, 1, 2), and (x, y, z ) = (3, 2, 1). In all four cases, x + y + z = 6.
If you allow x, y , and z to be whole numbers, you have to consider a few more cases for (x, y, z ), namely: (0, 0, 14), (0, 2, 4), (0, 4, 2), (0, 14, 0), (1, 0, 9), (1, 9, 0), (3, 0, 5), (3, 5, 0), (4, 0, 4), (4, 4, 0), (8, 0, 2), (8, 2, 0), (13, 0, 1), (13, 1, 0), and (28, 0, 0). This adds 8, 10, 14, and 28 as possible values of x + y + z .

414

Henrik’s math class needs to choose a committee consisting of two girls and two boys. If the committee can be chosen in 3630 ways, how many students are there in Henrik’s math class? Solution by the editors.

Say Henrik’s class has x girls and y boys. You can choose the ﬁrst girl for the committee in x ways and the second girl in x − 1 ways. Thus you can choose the two girls in a particular order in x(x − 1) ways. However, the order the two girls were chosen in is irrelevant, so the number of ways to choose the girls for the 1 x(x − 1). committee is 2 1 y (y − 1) ways. Thus you can choose Likewise, you can choose the boys in 2 1 1 the committee in 2 y (y − 1) · 2 x(x − 1) ways, so 1 4 so The left-hand side includes two pairs of consecutive integers. To match those on the right-hand side, one pair must be 11 and 2 · 5 = 10, while the other is 11 and 22 · 3 = 12. Thus x = 11 and y = 12 or the other way around. In either case, Henrik’s class has 23 students. x(x − 1)y (y − 1) = 14520 = 23 · 3 · 5 · 112 . x(x − 1)y (y − 1) = 3630,

10.

100!(1002 + 100 + 1). Find Solution by the editors.

Let S be 1!(12 + 1 + 1) + 2!(22 + 2 + 1) + 3!(32 + 3 + 1) + · · · +
S+1 . (As usual, k! = 1 · 2 · 3 · · · · · (k − 1) · k.) 101!

Each term in the sum has the form k!(k2 + k + 1) = k! (k2 + 2k + 1) − k = k! (k + 1)2 − k Therefore the sum is telescoping: S = = (2! · 2 − 1! · 1) + (3! · 3 − 2! · 2) + · · · + (100! · 100 − 99! · 99) + (101! · 101 − 100! · 100) 101! · 101 − 1! · 1 = 101! · 101 − 1 ,
S +1 101!

= k!(k + 1)2 − k! k = (k + 1)! (k + 1) − k! k

so S + 1 = 101! · 101, thus

= 101.

This issue’s prize of one copy of Crux Mathematicorum for the best ´ solutions goes to Rowena Ho, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC. We hope that the very low number of reader solutions this time was caused by the irregular production schedule that Crux Mathematicorum has suﬀered and the due date in September.

415

MATHEMATICAL MAYHEM
Mathematical Mayhem began in 1988 as a Mathematical Journal for and by High School and University Students. It continues, with the same emphasis, as an integral part of Crux Mathematicorum with Mathematical Mayhem. The interim Mayhem Editor is Shawn Godin (Cairine Wilson Secondary School, Orleans, ON). The Assistant Mayhem Editor is Lynn Miller (Cairine Wilson Secondary School, Orleans, ON). The other staﬀ members are Ann Arden (Osgoode Township District High School, Osgoode, ON), Nicole Diotte (Windsor, ON), Monika Khbeis (Our Lady of Mt. Carmel Secondary School, Mississauga, ON) and Daphne Shani (Bell High School, Nepean, ON).

Mayhem Problems
Please send your solutions to the problems in this edition by 15 October 2012. Solutions received after this date will only be considered if there is time before publication of the solutions. Each problem is given in English and French, the oﬃcial languages of Canada. In issues 1, 3, 5, and 7, English will precede French, and in issues 2, 4, 6, and 8, French will precede English. The editor thanks Rolland Gaudet, Universit´ e de Saint-Boniface, Winnipeg, MB, for translating the problems from English into French.

M507.

Proposed by the Mayhem Staﬀ.

M508.

Proposed by the Mayhem Staﬀ.

In 1770, Joseph Louis Lagrange proved that every non-negative integer can be expressed as the sum of the squares of four integers. For example 6 = 22 + 12 +12 +02 and 27 = 52 +12 +12 +02 = 42 +32 +12 +12 = 32 +32 +32 +02 (in the theorem it is acceptable to use 02 , or to use a square more than once). Notice that 27 had several diﬀerent representations. Show that there is a number, not greater than 1 000 000 that can be represented as a sum of four distinct non-negative integers in more than 100 ways. (Note that rearrangements are not considered diﬀerent, so 42 + 32 + 22 + 12 = 12 + 22 + 32 + 42 are the same representation of 30.)

416

M509.

Proposed by Titu Zvonaru, Com´ ane¸ sti, Romania.

Let ABC be a triangle with angles B and C acute. Let D be the foot of the altitude from vertex A. Let E be the point on AC such that DE ⊥ AC and let M be the midpoint of DE . Show that if AM ⊥ BE , then ABC is isosceles.

M510. Proposed by
and Herzegovina.

ˇ Sefket Arslanagi´ c, University of Sarajevo, Sarajevo, Bosnia

If a, b, c ∈ C such that |a| = |b| = |c| = r > 0 and a + b + c = 0, compute the value of the expression |ab + bc + ca| |a + b + c| in terms of r .

M511.
USA.

Proposed by Gili Rusak, student, Shaker High School, Latham, NY,

Pens come in boxes of 48 and 61. What is the smallest number of pens that can be bought in two ways if you must buy at least one box of each type?

M512.

Selected from a mathematics competition.

A class of 20 students was given a three question quiz. Let x represent the number of students that answered the ﬁrst question correctly. Similarly, let y and z represent the number of students that answered the second and the third questions correctly, respectively. If x ≥ y ≥ z and x + y + z ≥ 40, determine the smallest possible number of students who could have answered all three questions correctly in terms of x, y and z . .................................................................

M507.

´ Propos´ e par l’Equipe de Mayhem.

Un grillage 4 par 4 est form´ e de chevilles amovibles se situant a ` un centim` etre l’une de l’autre, tel qu’illustr´ e dans le sch´ ema. Des ´ elastiques sont attach´ es aux chevilles de fa¸ con a ` former des carr´ es ; deux carr´ es diﬀ´ erents de taille 2 par 2 sont illustr´ es dans le sch´ ema. Quel est le plus petit nombre de chevilles qui doivent ˆ etre enlev´ ees de fa¸ con a ` ce que pas un seul carr´ e puisse ˆ etre form´ e?

417

M508.

´ Propos´ e par l’Equipe de Mayhem.

En 1770, Joseph Louis Lagrange a d´ emontr´ e que tout entier non n´ egatif pouvait s’´ ecrire comme somme de carr´ es de quatre entiers. Par exemple, 6 = 22 + 12 + 12 + 02 et 27 = 52 + 12 + 12 + 02 = 42 + 32 + 12 + 12 . (Dans le th´ eor` eme de Lagrange il est permis d’utiliser 02 et de r´ ep´ eter un carr´ e.) Remarquer que 27 a plusieurs repr´ esentations diﬀ´ erentes. D´ emontrer qu’il y a un nombre inf´ erieur ou ´ egal a ` 1 000 000 qui peut ˆ etre repr´ esent´ e comme somme de carr´ es de quatre entiers non n´ egatifs distincts, de plus de 100 mani` eres. (Noter que les r´ earrangements ne sont pas consid´ er´ es distincts, c’est-` a-dire que 42 + 32 + 22 + 12 = 12 + 22 + 32 + 42 constituent la mˆ eme repr´ esentation de 30.)

M509.

Propos´ e par Titu Zvonaru, Com´ ane¸ sti, Roumanie.

Soit ABC un triangle avec angles aigus B et C . Soit D le pied de l’altitude a ` partir du sommet A. Soit E le point sur AC tel que DE ⊥ AC ; soit M le mipoint de DE . D´ emontrer que si AM ⊥ BE alors ABC est isoc` ele. ˇ Propos´ e par Sefket Arslanagi´ c, Universit´ e de Sarajevo, Sarajevo, Bosnie et Herz´ egovine. Si a, b, c ∈ C tels que |a| = |b| = |c| = r > 0 et a + b + c = 0, calculer la valeur de l’expression |ab + bc + ca| |a + b + c| en termes de r . ´ . E-U

M510.

M511.

Propos´ e par Gili Rusak, ´ etudiant, Shaker High School, Latham, NY,

Des plumes nous viennent en boˆ ıtes de 48 et de 61. Quel est le plus petit nombre de plumes qui peuvent ˆ etre achet´ ees de deux mani` eres diﬀ´ erentes si on doit acheter au moins une boˆ ıte de chaque type ?

M512.

S´ electionn´ ea ` partir de concours math´ ematiques.

Une classe de 20 ´ etudiants a subi un examen a ` trois questions. Soit x le nombre d’´ etudiants ayant r´ epondu correctement a ` la premi` ere question. De mˆ eme, soit y et z les nombres d’´ etudiants ayant r´ epondu correctement a ` la deuxi` eme puis a ` la troisi` eme question respectivement. Si x ≥ y ≥ z et x + y + z ≥ 40, d´ eterminer, en termes de x, y et z , le plus petit nombre possible d’´ etudiants ayant pu r´ epondre correctement aux trois questions.

418

Mayhem Solutions
M470.
Proposed by the Mayhem Staﬀ Vazz needs to buy desks and monitors for his new business. A desk costs $250 and a monitor costs $260. Determine all possible ways that he could spend exactly $10 000 on desks and monitors. Solution by Gili Rusak, student, Shaker High School, Latham, NY, USA. Let d and m represent the number of desks and monitors Vazz buys, respectively. Using the information from the problem we get 250d + 260m = 10 000 which reduces to 25d + 26m = 1000 or 26m = 1000 − 25d. Looking at this modulo 25 we get m ≡ 0 (mod 25). Since 25d + 26m = 1000, we must have 26m ≤ 1000, so m < 50. Therefore, m, being a non-negative integer, can only equal 0 or 25. When m = 0, d = 40 and when m = 25, d = 14. There are only two possibilities for Vazz to buy: 14 desks and 25 monitors or 40 desks and 0 monitors.
Also solved by FLORENCIO CANO VARGAS, Inca, Spain; JEREMY COOPER, student, Angelo State University, San Angelo, TX, USA; A. WIL EDIE, student, Missouri State University, Springﬁeld, MO, USA; MUHAMMAD HAFIZ FARIZI, student, SMPN 8, Yogyakarta, Indonesia; AFIFFAH NUUR MILA HUSNIANA, student, SMPN 8, Yogyakarta, Indonesia; MUHAMMAD ROIHAN MUNAJIH, ´ IES “Abastos”, Valencia, student, SMPN 8, Yogyakarta, Indonesia; RICARD PEIRO, Spain; CAO MINH QUANG, Nguyen Binh Khiem High School, Vinh Long, Vietnam; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; and NECULAI STANCIU, George Emil Palade Secondary School, Buz˘ au, Romania.

M471.

Proposed by the Mayhem Staﬀ

Square based pyramid ABCDE has a square base ABCD with side length 10. Its other four edges AE , BE , CE , and DE each have length 20. Determine the volume of the pyramid. Solution by Scott Brown, Auburn University, Montgomery, AL, USA. The volume of a square pyramid as E 2 shown in the diagram is V = b 3h . According to the information given, b = 10 and the edges EA = EB = EC = ED = 20. To ﬁnd h, we will ﬁrst ﬁnd the slant s h height s. Consider the triangle EBC and let s = EF , where F is the midpoint of side C BC . Now, triangle EF B is a right triangle, where EB = 20 and BF = 5. H A F Using the Pythagorean Theorem yields b 2 2 2 s = 20 B √− 5 = 375. So the slant height is s = 5 15. Now let H be the point where the height meets the square base of the

419

pyramid. Triangle EHF is a right triangle. Using the Pythagorean Theorem √ √ again yields h2 = (5 15)2 − 52 = 350. √ So the height is h = 5 14. 2 Thus the volume is V = b 3h = 5003 14 square units.
Also solved by FLORENCIO CANO VARGAS, Inca, Spain; MUHAMMAD HAFIZ FARIZI, student, SMPN 8, Yogyakarta, Indonesia; AFIFFAH NUUR MILA HUSNIANA, student, SMPN 8, Yogyakarta, Indonesia; MITEA MARIANA, No. 2 Secondary School, Cugir, Romania; MUHAMMAD ROIHAN MUNAJIH, ´ IES “Abastos”, Valencia, student, SMPN 8, Yogyakarta, Indonesia; RICARD PEIRO, Spain; CAO MINH QUANG, Nguyen Binh Khiem High School, Vinh Long, Vietnam; GILI RUSAK, student, Shaker High School, Latham, NY, USA; BRUNO SALGUEIRO FANEGO, Viveiro, Spain(two solutions); AND NECULAI STANCIU, George Emil Palade Secondary School, Buz˘ au, Romania. The problem could also be solved by using the Pythagorean Theorem to ﬁnd the length of the diagonal of the base, then using the Pythagorean Theorem a second time in a triangle such as EHB . About half the solutions used this method while the rest were similar to the featured solution.

M472.

Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania Suppose that x is a real number. Without using calculus, determine the maximum possible value of
x2 + 6x + 8 . x2 + 6x + 10 2x2 − 8x + 17 and the minimum possible value of x2 − 4x + 7

Solution by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain. We have 2x2 − 8x + 17 x2 − 4x + 7 = 2(x2 − 4x + 7) + 3 x2 − 4x + 7 3 x2 − 4x + 7 3 (x − 2)2 + 3

=2+ =2+

Hence maximizing

in turn, is equivalent to minimizing (x − 2)2 + 3. Since (x − 2)2 ≥ 0, the sum (x − 2)2 + 3 is a minimum when x − 2 = 0, i.e. when x = 2. Thus the maximum value of Similarly, we have
2x2 − 8x + 17 is 2 + x2 − 4x + 7
3 3

3 2x2 − 8x + 17 is equivalent to maximizing . This, x2 − 4x + 7 (x − 2)2 + 3

= 3.

x2 + 6x + 8 2 =1− . 2 x + 6x + 10 (x + 3)2 + 1 Hence minimizing
x2 + 6x + 8 2 is equivalent to maximizing . Thus x2 + 6x + 10 (x + 3)2 + 1

the minimum occurs when x = −3 and the minimum value is −1.

420

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; FLORENCIO CANO VARGAS, Inca, Spain; MITEA MARIANA, No. 2 Secondary School, Cugir, Romania; ´ IES “Abastos”, Valencia, Spain; CAO MINH QUANG, Nguyen Binh Khiem RICARD PEIRO, High School, Vinh Long, Vietnam; HENRY RICARDO, Tappan, NY, USA; GILI RUSAK, student, Shaker High School, Latham, NY, USA; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; and the proposer.

M473.

Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania Determine all pairs (a, b) of positive integers for which a2 + b2 − 2a + b = 5.

I. Solution by George Apostolopoulos, Messolonghi, Greece. Rearranging the equation we obtain a2 − 2a + b2 + b − 5 = 0 which we will treat as a quadratic equation in a with discriminant D = (−2)2 − 4 · 1(b2 + b − 5) = 4 − 4b2 − 4b + 20 = −4b2 − 4b + 24. In order for our quadratic equation to have real solutions we must have D ≥ 0, so −4b2 − 4b + 24 ≥ 0 ⇔ b2 + b − 6 ≤ 0 ⇔ (b − 2)(b + 3) ≤ 0 ⇔ −3 ≤ b ≤ 2. But b is a positive integer, so b = 1 or b = 2. For b = 1, we get a = 3 or a = −1. For b = 2, we get a = 1. Thus the solutions are the pairs (a, b) = (3, 1) and (a, b) = (1, 2). II. Solution by Samuel G´ omez Moreno, Universidad de Ja´ en, Ja´ en, Spain. First observe that (2(a − 1))2 + (2b + 1)2 = 4 (a − 1)2 + b + 1 2 1 4
2

= 4 a2 − 2a + 1 + b2 + b + = 4 (a − 2a + b + b) + =4 5+ 5 4 = 25.
2 2

5 4 (1)

Since there are only two possibilities of writing 25 as the sum of the squares of two non-negative integers, namely 25 = 02 + 52 and 25 = 32 + 42 , equation (1) gives us the solutions (a, b) = (3, 1) and (a, b) = (1, 2).

421

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; FLORENCIO CANO VARGAS, Inca, Spain; MITEA MARIANA, No. 2 Secondary School, ´ IES “Abastos”, Valencia, Spain; CAO MINH QUANG, Cugir, Romania; RICARD PEIRO, Nguyen Binh Khiem High School, Vinh Long, Vietnam; GILI RUSAK, student, Shaker High School, Latham, NY, USA; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; and the proposer. One incomplete solution was received.

M474.

Proposed by Dragoljub Miloˇ sevi´ c, Gornji Milanovac, Serbia

Let a, b and x be positive integers such that x2 − bx + a − 1 = 0. Prove that a2 − b2 is not a prime number. Solution by Cao Minh Quang, Nguyen Binh Khiem High School, Vinh Long, Vietnam. Assume that a2 − b2 is a prime number. Since a2 − b2 = (a − b)(a + b), and a, b > 0, hence a − b = 1. Therefore x2 − bx + b = 0, so (x2 − 1) − b(x − 1) = −1 which implies that (x − 1)(x + 1 − b) = −1. Since x is a positive integer, we must have x − 1 = 1, x + 1 − b = −1 or x = 2, b = 4 which implies that a = 5. This yields a2 − b2 = 9, which is not prime, a contradiction, and we are done.
´ Also solved by FLORENCIO CANO VARGAS, Inca, Spain; SAMUEL GOMEZ ´ IES “Abastos”, Valencia, MORENO, Universidad de Ja´ en, Ja´ en, Spain; RICARD PEIRO, Spain; GILI RUSAK, student, Shaker High School, Latham, NY, USA; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; NECULAI STANCIU, George Emil Palade Secondary School, Buz˘ au, Romania; and the proposer. The proposer and most of the solvers worked on the assumption that a prime is a positive integer. If we are working with integers, then we can consider numbers like −11 to be prime as well. In this case, 11 and −11 are called associates, and ±1 are called units. Then a prime number is only divisible by a unit or one of its associates. This idea is used in number theory when extending the idea of number system and hence extending the idea of what a prime is in this number system. Only Peir´ o considered the case where a2 − b2 < 0 which leads to a = 1, b = 2 and a2 − b2 = −3, a prime, which contradicts the problem.

M475.
ON

Proposed by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,

Let x denote the greatest integer not exceeding x. For example, 3.1 = 3 and −1.4 = −2. Let {x} denote the fractional part of the real number x, that is, {x} = x − x . For example, {3.1} = 0.1 and {−1.4} = 0.6. Show that there exist inﬁnitely many irrational numbers x such that x · {x} = x . Solution by Florencio Cano Vargas, Inca, Spain. First note that the only rational solution is x = 0 (this was problem M437 [2010 : 135,136; 2011 : 14]). Next note there there is no solution if x < 0. Indeed if x < 0, then x · {x} > x > x . Also note that if 0 < x < 1, then {x} = x so x · {x} = x2 > 0 = x . Clearly, x and {x} are either both rational or both irrational. Thus we will limit ourselves to the case x > 1, where x is irrational. We will solve the equation for {x}. Let x = k > 0, for some positive integer

422

k. Then x = k + {x} with 0 < {x} < 1. The given equation is then ({x} + k) · {x} = k, with solution given by {x} = k 2
æÖ

4 k

−1 ,

where we have discarded the “minus” solution since 0 < {x} < 1. We see that this solution falls in this range since
Ö

1+ and k 2
æÖ

4 k
é

−1>1−1 =0

4 k

−1 =

1 Ô 2 [ k + 4k − k ] 2 1 = [ (k + 2)2 − 4 − k] 2 1 < [ (k + 2)2 − k] = 1. 2

Therefore we end up with inﬁnitely many irrational solutions for {x}, and hence we get the inﬁnite family of solutions k x= 2 where k is any positive integer.
´ Also solved by SAMUEL GOMEZ MORENO, Universidad de Ja´ en, Ja´ en, Spain; ´ IES “Abastos”, Valencia, Spain; GILI RUSAK, student, Shaker High School, RICARD PEIRO, Latham, NY, USA; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; NECULAI STANCIU, George Emil Palade Secondary School, Buz˘ au, Romania; and the proposer.

æÖ

4 1+ +1 , k

423

THE OLYMPIAD CORNER
No. 297 R.E. Woodrow and Nicolae Strungaru
The problems from this issue come from the Indian IMO Selection Test, the Colombian Mathematical Olympiad, the Singapore Mathematical Olympiad, the Serbian Mathematical Olympiad, the Romanian National Olympiad, and the Finnish National Olympiad. Our thanks go to Adrian Tang for sharing the material with the editors.
The solutions to the problems are due to the editors by 1 November 2012. Each problem is given in English and French, the oﬃcial languages of Canada. In issues 1, 3, 5, and 7, English will precede French, and in issues 2, 4, 6, and 8, French will precede English. In the solutions’ section, the problem will be stated in the language of the primary featured solution. The editors thank Jean-Marc Terrier of the University of Montreal for translations of the problems.

OC41.

Let P be a point in the interior of a triangle ABC . Show that PA BC + PB AC + PC AB ≥ √ 3.

OC42. OC43.

Find the smallest n for which n! has at least 2010 diﬀerent divisors. Find all functions f : R → R satisfying f (x3 + y 3 ) = xf (x2 ) + yf (y 2 ) ; ∀x, y ∈ R .

OC44. In a scalene triangle ABC , we denote by α and β the interior angles at A and B . The bisectors of these angles meet the opposite sides of the triangle at points D and E respectively. Prove that the acute angle between the lines D β| and E does not exceed |α− . 3 OC45. OC46.
Let a1 , a2 , a3 , ..., a15 be prime numbers forming an arithmetic progression with common diﬀerence d > 0. If a1 > 15, prove that d > 30, 000.

Let p be a prime number, and let x, y, z be integers so that 0 < x < y < z < p. Suppose that x3 , y 3 and z 3 have the same remainders when divided by p. Prove that x2 + y 2 + z 2 is divisible by x + y + z .

424

OC47.

Let a, b be two distinct odd positive integers. Let an be the sequence deﬁned as a1 = a ; a2 = b ; an = the largest odd divisor of an−1 + an−2. Prove that there exists a natural number N so that, for all n ≥ N we have an = gcd(a, b).

π π The angles of a triangle ABC are π , 27 and 47 . The bisectors meet 7 the opposite sides at A , B and C . Prove that A B C is an isosceles triangle.

OC48.

OC49. Let N be a positive integer. How many non congruent triangles are there, whose vertices lie on the vertices of a regular 6N -gon? OC50.
Let n ≥ 2. If n divides 3n + 4n , prove that 7 divides n. un point int´ erieur d’un triangle ABC . Montrer que PA BC + PB AC + PC AB ≥ √ 3. .................................................................

OC41. Soit P

OC42. Trouver le plus petit n pour lequel n! poss` ede au moins 2010 diviseurs diﬀ´ erents. OC43. Trouver toutes les fonctions f : R → R satisfaisant
f (x3 + y 3 ) = xf (x2 ) + yf (y 2 ) ; ∀x, y ∈ R .

OC44. Dans un triangle scal` ene ABC , notons α et β les angles int´ erieurs en
A et B . Les bissectrices respectives de ces angles coupent les cˆ ot´ es oppos´ es du triangle aux points D et E . Montrer que l’angle aigu entre les droites D et E β| . n’exc` ede pas |α− 3

OC45. Soit a1, a2, a3, ..., a15 des nombres premiers formant une progression
arithm´ etique de raison d > 0. Si a1 > 15, montrer que d > 30, 000.

OC46. Soit p un nombre premier, et soit x, y, z trois entiers tels que 0 < x <
y < z < p. Supposons que x3 , y 3 et z 3 ont les mˆ emes restes lorsqu’on les divise par p. Montrer que x2 + y 2 + z 2 est divisible par x + y + z .

OC47. Soit a, b deux entiers positifs impairs distincts. Soit an la suite d´ eﬁnie
par a1 = a ; a2 = b ; an = le plus grand diviseur impair de an−1 + an−2. Montrer qu’il existe un nombre naturel N tel que, pour tous les n ≥ N , on a an = gcd(a, b). les cˆ ot´ es oppos´ es en A , B et C . Montrer que A B C est un triangle isoc` ele .
π π OC48. Les angles d’un triangle ABC sont π , 27 et 47 . Les bissectrices coupent 7

OC49. Soit N

un entier positif. Combien y a-t-il de triangles non congruents dont les sommets sont sur les sommets d’un 6N -gone r´ egulier ?

425

OC50. Soit n ≥ 2. Si n divise 3n + 4n , montrer que 7 divise n.

First the editor apologize to Paolo Perfetti, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; whose solutions were misﬁled. Correct solutions were received for: Indian Team Selection Test 2007, #2 and #7; Mediterranean Mathematics Competition 2007, #1 and #4; and Bulgarian Team First Selection Test, #1. Next we turn to the Olimpiada Nacional Escolar de Matematica 2009, Level 2, given at [2010: 373].

Let a, b, c and d be four integer numbers whose sum is 0. Let M = (bc − ad)(ac − bd)(ab − cd) .

Show that there is a whole number P such that P 2 = M . Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Henry Ricardo, Tappan, NY, USA; Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give Ricardo’s write-up. Replacing d by −a − b − c in each factor of M and factoring, we see that bc − ad = (a + b)(a + c), ac − bd = (a + b)(b + c), and ab − cd = (a + c)(b + c). Thus M = (bc − ad)(ac − bd)(ab − cd) = (a + b)2 (a + c)2 (b + c)2 = [(a + b)(a + c)(b + c)]2 = P 2 .

2. An equilateral triangle of side length 6 is divided into 36 small equilateral triangles of side length 1. The resulting chart is covered by m markers of type A and n markers of type B without doubling or leaving empty spaces. Markers of type A are formed by two equilateral triangles of side length 1 and markers of type B are formed from 3 small triangles, as shown in the ﬁgure. A B Determine all possible values of m.
Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA. Since each marker of type A covers two triangles and each marker of type B covers three triangles, we have 2m + 3n = 36, for nonnegative integers m and n. This implies that m ∈ {0, 3, 6, 9, 12, 15, 18}. We will show that m = 0, 3, 6, 9 are possible, while m = 12, 15, 18 are impossible. We orient the large triangle and number the smaller triangles as shown in the following diagrams.

426

1 2 5 10 17 26 27 18 11 6 3 4 8 9 15 16 24 25 35 36 26 17 27 10 18 5 11 2 6

1 3 4 8 9 15 16 24 25 35 36

7 13

12 20

14 22

12 20

14 22

19 29

21 31

23 33

19 29

21 31

23 33

m=0
1 2 5 10 17 26 27 18 11 6 3 4 8 9 15 16 24 25 35 36 26 17 27 10 18

m=3
1 2 5 11 6 3 4 8 9 15 16 24 25 35 36

7 13

12 20

14 22

12 20

14 22

19 29

21 31

23 33

19 29

21 31

23 33

m=6

m=9

The diagrams show that m = 0, 3, 6, 9 are possible. For m = 3, we replaced the triples (4, 8, 9) and (15, 16, 24) covered by markers of type B in the m = 0 case with the pairs (4, 8), (9, 15), (16, 24) covered by markers of type A. Similarly for m = 6, we make the further switch of (11, 10, 18) and (17, 27, 26) to (11, 10), (18, 17), and (27, 26); and for m = 9, we make the additional switch of (13, 14, 22) and (12, 20, 19) to (14, 22), (13, 12), and (20, 19). We note that of the 36 small triangles 21 have a vertex uppermost, while 15 have an edge uppermost. Each marker of type A always covers one triangle of each of these types, while each marker of type B covers two of one type and one of the other. If m = 12, then of the 12 triangles covered by markers of type B , nine have a vertex on top, and three have the edge on top, which is impossible since at least one third of the triangles covered by markers of type B must have an edge on top. Hence, m = 12. If m = 15, then all of the six triangles covered by markers of type B have a vertex on top. Hence, m = 15. Likewise, m = 18 since 18 markers of type A would require 18 triangles with an edge on top. Thus, as claimed, m ∈ {0, 3, 6, 9}.

√ For each positive integer n let d be the largest divisor of n with d ≤ n, and n deﬁne an = d − d. Show that in the sequence a1 , a2 , a3 . . ., each nonnegative integer k appears inﬁnitely often. Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA. We give the solution by Curtis.

427

List the primes 2 = p1 < p2 < p3 < · · · . Given a nonnegative integer k, for each positive integer j with pj > k we deﬁne nj (k) = pj (pj − k), and dj = pj − k. Then dj is the largest divisor of nj (k) with dj ≤ nj (k), and anj (k) =
nj (k) dj

− dj = k. Hence, k appears inﬁnitely often in the sequence {an }.

On a circle N ≥ 5 points are marked so that the N arcs formed have the same length. A coin is placed on each point, and Ricardo and Tom´ as play a game with the following rules: • They play alternately. • Ricardo starts. • A player may take a coin only if that coin forms an acute triangle with at least two other coins.

A player loses when he cannot take any coin during his turn. Does either player have a winning strategy? If so, what is it? Discussion and solution by Stan Wagon, Macalester College, St. Paul, MN, USA. The following solution is adapted from a solution in Spanish found by Witold Jarnicki at www.fileden.com/files/2008/5/31/1938837//CuartaFase2009.pdf. It is by Sergio Vera (the author of the problem). We use Alice and Bob for the two players, with Alice moving ﬁrst. Bob has a winning strategy for every N ≥ 4. When N is even, Bob can simply choose the point diametrically opposite to Alice. For if Alice chooses X there is an acute triangle Xab and neither a nor b can be −X and b cannot be −a. But then Bob has triangle (−X, −a, −b), all of which are available, by the symmetry that faced Alice. Thus Alice will always face a symmetric conﬁguration, and therefore Bob can always move after Alice does. Now assume N = 2m + 1. Call a conﬁguration C of 2k + 1 coins, balanced if for any point X ∈ C , the diameter through X splits the remaining 2k points into two equalized sets. Then the initial conﬁguration is balanced. To be precise: the diameter through X deﬁnes two semicircles. For C to be balanced, each semicircle has the point X and exactly k other points. The following easily proved fact is useful: Three points on a circle form an acute triangle if and only if for every diameter, the three points include one on one side of the diameter and one on the other. Two key facts: Fact 1: For a 5-coin balanced situation, the player to move loses. Fact 2: For k ≥ 3 and a 2k + 1-coin balanced situation C , any move on C can be followed by another move resulting in a 2k − 1-coin balanced situation. These two facts suﬃce, since starting from N Bob can keep things balanced, and so eventually the case N = 5 is reached and Alice loses.

428

Proof of Fact 1. If the ﬁrst player takes coin X , the second player takes any of the “neighbours” of X (moving around the circle). If the points are Pi in order, suppose the ﬁrst player takes P1 and second takes P2 . Then P2 P4 P5 is acute, for if not, there would be a diameter with P2 , P4 , and P5 on one side. But the diameter from P2 splits P4 and P5 , so any diameter with P2 , P4 , and P5 on one side would have to be on one side of diameter from P2 , impossible. Proof of Fact 2. If the ﬁrst player takes coin X , the second player takes the coin Y that is the farthest from X (thinking in terms of angular measure from the center): Let P and Q be the neighbours of X on the two sides. Then Y P Q is acute as in the proof of Fact 1, and the removal of X and Y leaves a balanced position. Note: One can summarize Bob’s strategy in all cases (except N = 5) by simply saying: If Alice chooses X , Bob chooses the point nearest −X . Alternate solution by Stephen Morris, Newbury, England; and Stan Wagon, Macalester College, St. Paul, MN, USA. Let Alice and Bob be the players who move ﬁrst or second, respectively. We let P1 to PN be the positions the coins occupy at the start, in counterclockwise order. If N is even, then Bob wins by always choosing the point diametrically opposite to Alice’s choice (as described in the other proof). Assume 5 ≤ N = 2m +1. Call two positions opposite if they are as close as is possible to being diametrically opposite; each position is opposite two positions (that is, P1 and either Pm+1 or Pm+2 ). We claim that a winning strategy for Bob is to take an immediately winning move when available and otherwise take the coin closest to being opposite to Alice’s last move. In what follows assume that Bob plays by this strategy. We assume the standard geometrical fact that a chord in a circle subtends an obtuse angle on its near side and an acute angle on its far side. Case 1. If a player has a legal move, then all available moves are legal. Proof. Let P be one of the remaining points, use the diameter from P to divide the points into two sets, and choose the points Q and R, respectively, in each set that is farthest from P . Then P QR is an acute triangle. The angles at Q and R are acute because these points straddle a diameter; the angle at P is acute because if not then Q and R, and therefore all the remaining coins, would be on the same side of the diameter parallel to QR, contradicting the fact that there is a legal move. Case 2. If there are four coins left and they do not lie on a semicircle then Bob has an immediately winning move. Proof. The diameter through one of the points divides the other three into a pair and a singleton. The singleton wins. Case 3. If a set of coin positions is contained in a semicircle, then there is a set of m contiguous vacancies. Proof. The diameter through any position partitions the rest into two sets of size m. A diameter not through a position can be rotated until it just touches

429

a position without aﬀecting the state. Case 4. At the start, and after each move by Bob, vacant positions can be paired so that they are opposite each other. Proof. At the start there are no vacancies. Suppose the assertion is true, Alice chooses a coin, and Bob has a legal choice. If Bob takes an opposite coin then it remains true. Suppose Bob cannot take an opposite coin. Suppose without loss of generality, that Alice takes P1 and Bob takes coin Pm+r+1 (a legal choice by Case 1). Positions m + 2, . . . , m + r must be vacant and be paired in the initial pairing. Since 1 is unavailable, m + 2 is paired with 2 and, similarly, these pairings must be (m + 2, 2), (m + 3, 3), . . . , (m + r, r ). Following Bob’s move there is a new pairing that includes (m + 2, 1), (m + 3, 2), . . . , (m + r + 1, r ) and is otherwise the same. Case 5. Alice never has an immediately winning move. This means that the given strategy wins for Bob. Proof. Suppose she did. If Bob had been faced with 4 coins and had a legal move, then he would have won by Case 2. Therefore, Alice was facing at least 5 coins. For Alice’s move to be winning, she must see a semicircle having only one coin in it. By Case 3, she is faced with a set of m contiguous positions containing one coin. No positions in this set are opposite so there are m − 1 vacancies paired with other positions in the pairing from Case 4. This leaves at most two other positions that can contain coins. But that only allows for three coins, contradiction.

We return to the ﬁles of solutions for the October 2010 number of the Corner and the S´ election OIM 2006 given at [2010: 374–376].

Dans le triangle ABC soit D le milieu du cˆ ot´ e BC et E la projection de C sur AD . On suppose que ∠ACE = ∠ABC . Montrer que le triangle ABC est soit isoc` ele, soit rectangle. Solved by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Amengual Covas.

We put CA = b, AB = c and denote by F the foot of the perpendicular from A to BC . We have ∠BAF = ∠EAC = ∠DAC (since both ∠BAF and ∠EAC are complementary to ∠ABC ). Thus the two lines AF and AD are equally inclined to the arms of ∠A. Since AD is a median, AF is the symmedian of ABC at A and so AF

D E

430

divides the side BC in the ratio of the square of the sides: BF FC = c2 b2 . (1)

Now, BF = c cos B and F C = b cos C . Hence, from equation (1), we get cos B cos C = c b = sin C sin B ,

by the law of sines, which is equivalent to sin B cos B − sin C cos C = 0 or, equivalently, sin 2B − sin 2C = 0 which we rewrite as 2 cos that is, cos(B + C ) sin(B − C ) = 0. Hence, either cos(B + C ) = 0 or sin(B − C ) = 0. Thus, if cos(B + C ) = 0, then B + C = 90◦ and ABC is right-angled at A. If sin(B − C ) = 0, then B = C and ABC is isosceles with b = c. 2B + 2C 2 sin 2B − 2C 2 = 0,

Soient 1 = d1 < d2 < . . . < dk = n les diviseurs positifs de n. D´ eterminer tous les n tels que 2 2n = d 2 5 + d6 − 1 . We claim that the only solution is n = 272. √ 2 2 Note ﬁrst that 2 n > d , and 2 n < 2 d , so that d < 2n and d6 > 5 5 6 √ n. These imply that n has at most 10, but at least 6 positive divisors. If r 1 r2 n = pr 1 p2 · · · p , where p1 < p2 < · · · < p are primes, then the number of divisors of n is (r1 + 1)(r2 + 1) · · · (r + 1) ≤ 10. Hence, ≤ 3.

Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA.

Case 1. Suppose n = pr , where p is prime. 1, p, p2 , p3 , . . . , pr . Thus, 5 ≤ r ≤ 9. Hence,

Then the divisors of n are

2pr = (p4 )2 + (p5 )2 − 1. This implies that p divides 1, a contradiction.
1 r2 Case 2. Suppose n = pr 1 p2 , where p1 < p2 are primes and r1 , r2 ≥ 1. If p1 2 and p2 are both odd, then 2n = d2 5 + d6 − 1 is odd, a contradiction. Hence, p1 = 2 .

431

Letting p = p2 , we have d5 = 2s1 ps2 and d6 = 2t1 pt2 for 0 ≤ si , ti ≤ ri , i = 1, 2. Thus the condition of the problem becomes 2r1 +1 pr2 = (2s1 ps2 )2 + (2t1 pt2 )2 − 1. If either s1 ≥ 1 and t1 ≥ 1 or s2 ≥ 1 and t2 ≥ 1, then a prime divides 1, a contradiction. Since (r1 + 1)(r2 + 1) ≤ 10, it follows that r1 , r2 ≤ 4. In the following table, showing the possible pairs (r1 , r2 ), the condition is satisﬁed by 2 primes p that satisfy d2 5 + d 6 − 1 − 2n = 0 .
(r1 , r2 ) (1, 2) (1, 3) (1, 4) (2, 1) (2, 1) (2, 2) (2, 2) (3, 1) (3, 1) (3, 1) (4, 1) (4, 1) (4, 1) (4, 1) n 2 p2 2 p3 2 p4 4p 4p 4 p2 4 p2 8p 8p 8p 16p 16p 16p 16p divisors of n in increasing order {1, 2, p, 2p, p2 , 2p2 } {1, 2, p, 2p, p2 , 2p2 , p3 , 2 p3 } Notes d5 p2 p2 p2 p≥5 p=3 p≥5 p=3 p ≥ 11 p = 5, 7 p=3 p ≥ 17 p = 11, 13 p = 5, 7 p=3 2p 2p 2p 2p p 8 2p 16 p 8 2p d6 2 p2 2 p2 2 p2 4p 4p 4p p2 2p 2p 8 p 16 2p 8
2 d2 5 + d6 −1 − 2 n

p None None None None None None None None None None 17 17 None None

5 p4 − 1 − 4 p2 5 p4 − 1 − 4 p3 5 p4 − 1 − 4 p4 20p2 − 1 − 8p 20p2 − 1 − 8p 20p2 − 1 − 8p2 4 p2 + p4 − 1 − 8 p2

{1, 2, p, 2p, p2 , 2 p2 , p3 , 2 p3 p4 , 2 p4 } {1, 2, 4, p, 2p, 4p} {1, 2, p, 4, 2p, 4p} {1, 2, 4, p, 2p, 4p, p2 , 2 p2 , 4 p2 } {1, 2, p, 4, 2p, p2 , 4p, 2p2 , 4p2 } {1, 2, 4, 8, p, 2p, 4p, 8p} {1, 2, 4, p, 8, 2p, 4p, 8p}

5p2 − 1 − 16p 64 + 4p2 −1 − 16p 4p2 + 64 −2 − 16p 256 + p2 −1 − 32p

{1, 2, p, 4, 2p, 8, 4p, 8p} {1, 2, 4, 8, 16, p, 2p, 4p, 8p, 16p}

{1, 2, 4, 8, p, 16, 2p, 4p, 8p, 16p}

{1, 2, 4, p, 8, 2p, 16, 4p, 8p, 16p}

p2 + 256 −1 − 32p 64 + 4p2 −1 − 32p 4p2 + 64 −1 − 32p

{1, 2, p, 4, 2p, 8, 4p, 16, 8p, 16p}

1 r2 r3 Case 3. Suppose n = pr 1 p2 p3 , with p1 < p2 < p3 primes and r1 , r2 , r3 ≥ 1. Then 2 · 2 · (r3 + 1) ≤ (r1 + 1)(r2 + 1)(r3 + 1) ≤ 10, implies that r3 ≤ 3 . 2 Hence, r3 = 1. Likewise, r1 = r2 = 1, so that n = p1 p2 p3 . As in case 2, p1 = 2. We have

Hence, the only solution in this case is p = 17, corresponding to n = 16 · 17 = 272. For this n, we have d5 = 16 and d6 = 17.

β2 2 β3 α2 2α3 4p2 p3 = 22α1 p2 p 3 + 2 2 β1 p 2 − 1, 2 p3 2

where, for i = 1, 2, 3, αi , βi ∈ {0, 1}, and either αi = 0 or βi = 0. Also, α1 and β1 cannot both be 0. We test the possibilities.

432

(α1 , β1 , α2 , β2 , α3 , β3 ) (0, 1, 0, 0, 0, 0) (0, 1, 0, 0, 0, 1) (0, 1, 0, 0, 1, 0) (0, 1, 0, 1, 0, 0) (0, 1, 0, 1, 0, 1) (0, 1, 0, 1, 1, 0) (0, 1, 1, 0, 0, 0) (0, 1, 1, 0, 0, 1) (0, 1, 1, 0, 1, 0) (1, 0, 0, 0, 0, 0) (1, 0, 0, 0, 0, 1) (1, 0, 0, 0, 1, 0) (1, 0, 0, 1, 0, 0) (1, 0, 0, 1, 0, 1) (1, 0, 0, 1, 1, 0) (1, 0, 1, 0, 0, 0) (1, 0, 1, 0, 0, 1) (1, 0, 1, 0, 1, 0)

α2 2α3 4p2 p3 = 22α1 p2 p3 + 2 β2 2β3 22 β 1 p2 p3 − 1 2 4p2 p3 = 1 + 4 − 1 = 4 2 4p2 p3 = 1 + 4p2 3 − 1 = 4p3 2 4p2 p3 = p2 3 + 4 − 1 = p3 + 3 4p2 p3 = 1 + 4p2 − 1 = 4 p2 2 2 2 − 1 = 4p2 p2 4p2 p = 1 + 4p2 p 2 3 2 3 2 4p2 p3 = p2 3 + 4p2 − 1 2 4p2 p3 = p2 2 + 4 − 1 = p2 + 3 2 2 4p2 p3 = p2 + 4p3 − 1 2 2 2 4p2 p3 = p2 2 p3 + 4 − 1 = p2 p3 + 3 4p2 p3 = 4 + 1 − 1 = 4 2 4p2 p3 = 4 + p2 3 − 1 = p3 + 3 2 4p2 p3 = 4p3 + 1 − 1 = 4p2 3 2 4p2 p3 = 4 + p2 2 − 1 = p2 + 3 2 2 2 4p2 p3 = 4 + p2 2 p3 − 1 = p2 p3 + 3 2 2 4p2 p3 = 4p3 + p2 − 1 2 4p2 p3 = 4p2 2 + 1 − 2 = 4p2 2 2 4p2 p3 = 4p2 + p3 − 1 2 2 2 4p2 p3 = 4p2 2 p3 + 1 − 1 = 4p2 p3

Implication p2 = p3 = 1 p2 = p3 p3 = 3 p2 = p3 p2 = p3 = 1 p2 = 3, p3 = 1

Possible? No No No No No No No No No No No No No No

p3 = 3 p2 = 3, p3 = 1 p2 p3 = 1 or 3 p3 = p2 p2 p3 = 1

Hence there are four possibilities to consider. We note that the divisors of n, in increasing order, are {1, 2, p2 , 2p2 , p3 , 2p3 , p2 p3 , 2p2 p3 } or {1, 2, p2 , p3 , 2p2 , 2p3 , p2 p3 , 2p2 p3 }. • Suppose (α1 , β1 , α2 , β2 , α3 , β3 ) = (0, 1, 0, 1, 1, 0). Then d5 = p3 and d6 = 2p2 . In neither of the above orders for the divisors of n does d6 = 2p2 . • Suppose (α1 , β1 , α2 , β2 , α3 , β3 ) = (0, 1, 1, 0, 0, 1). Then d5 = p2 and d6 = 2p3 . But p2 is the third smallest rather than the ﬁfth smallest divisor of n. • Suppose (α1 , β1 , α2 , β2 , α3 , β3 ) = (1, 0, 0, 1, 1, 0). Then d5 = 2p3 and d6 = p2 < d6 , a contradiction. • Suppose (α1 , β1 , α2 , β2 , α3 , β3 ) = (1, 0, 1, 0, 0, 1). Then d5 = 2p2 and d6 = p3 , which is also impossible since the sixth smallest divisor of n is 2p3 rather than p3 . Hence, the case of three distinct prime factors for n is impossible. Thus, the only solution is n = 272.

Trouver toutes les fonctions f : R → R telles que pour tout, x, y ∈ R on ait l’´ egalit´ e suivante f (f (x) − y 2 ) = f (x)2 − 2f (x)y 2 + f (f (y )) . (1)

433

Solved by Michel Bataille, Rouen, France. It is readily checked that the zero function x → 0 and the square function x → x2 are solutions. We show that there are no other solutions. To this aim, consider f , diﬀerent from the zero function, satisfying the given functional equation (1). Taking x = y = 0 in (1), we obtain f (0) = 0 and then with y = 0, (1) gives f (f (x)) = f (x)2 for all x ∈ R. It follows that (1) rewrites as f (f (x) − y 2 ) = f (x)2 − 2f (x)y 2 + f (y )2 . With x = 0 and x = y in turn, (2) yields f (−y 2 ) = f (y )2 and f (f (x) − x2 ) = 2f (x)(f (x) − x2 ). Using (3), we can rewrite (2) as f (f (x) − y 2 ) − (f (x) − y 2 )2 = f (−y 2 ) − (−y 2 )2 and then, taking the images under f and using (4), we are led to f (x)(f (x) − 2y 2 )(f (y )2 − y 4 ) = 0 (x, y ∈ R). (6) (5) (4) (3) (2)

Now, assume that for some real number f (a)2 = a4 . Then, a = 0 and from (3), we would have f (x) = 0 or f (x) = 2a2 for all x ∈ R. Taking b ∈ R such that f (b) = 0 (this is possible because f is not the zero function), we obtain f (b) = 2a2 and from (5) with x = b 4a4 − 4a2 y 2 = f (2a2 − y 2 ) − f (y )2 for all y ∈ R. Since the right-hand side can take at most 3 values (namely 0, 2a2 , −2a2 ) while the left-hand side can take inﬁnitely many values, we have reached a contradiction. We conclude that for all y , we have f (y )2 = y 4 . Recalling (1), f (−y 2 ) = y 4 , hence f (u) = u2 for all negative real numbers u. Also, since y 8 = f (−y 2 )2 = f (f (−y 2 )) = f (y 4 ), we see that f (u) = u2 when u is positive and ﬁnally, f is the square function.

Les trois z´ eros r´ eels du polynˆ ome P (x) = x3 − 2x2 − x + 1 sont a > b > c. Trouver la valeur de l’expression a2 b + b2 c + c2 a .

Solved by Arkady Alt, San Jose, CA, USA; Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give Zvonaru’s write-up. Since P (−1) = −1, P (0) = 1, P (1) = −1, P (2) = −1, P (3) = 7, it follows that a ∈ (2, 3), b ∈ (0, 1), c ∈ (−1, 0). We let α = a2 b + b2 c + c2 a, and β = ab2 + bc2 + ca2 .

434

Since c > −1 and 1 > b > 0, we have c > −1 ⇒ bc > −b > −1, hence bc + 1 > 0 or b2 c + b > 0. It follows that, because a > 1, α > a2 b + b2 c = a2 b + b2 c + b − b > a2 b − b = b(a2 − 1) > 0, hence α > 0. Since a, b, c are the roots of the equation P (x) = 0, we have a + b + c = 2, ab + bc + ca = −1, abc = −1. From these, we obtain −2 = (a + b + c)(ab + bc + ca) = a2 b + ab2 + b2 c + bc2 + c2 a + ca2 + 3abc, that is, α + β = 1. Furthermore, we have αβ = (a2 b + b2 c + c2 a)(ab2 + bc2 + ca2 ) = a3 b3 + b3 c3 + c3 a3 + 3a2 b2 c2 + abc(a3 + b3 + c3 ).
2

(1)

(2)

Since (x + y + z )3 = x3 + y 3 + z 3 + 3x2 y + 3xy 2 + 3y 2 z + 3yz 2 + 3x z + 3xz 2 + 6xyz , we deduce that 8 = (a + b + c)3 = a3 + b3 + c3 + 3α + 3β − 6 ⇒ a3 + b3 + c3 = −3α − 3β + 14 and −1 = (ab + bc + ca)3

(3)

= a3 b3 + b3 c3 + c3 a3 + 3a2 b3 c + 3a3 b2 c + 3ab3 c2 + 3a3 bc2 + 3ab2 c3 + 3a2 bc3 + 6a2 b2 c2

= a3 b3 + b3 c3 + c3 a3 + 3abc(α + β ) + 6a2 b2 c2 ⇒ a3 b3 + b3 c3 + c3 a3 = 3α + 3β − 7. Using (3) and (4), by (2) we obtain αβ = 3α + 3β − 7 + 3 + 3α + 3β − 14, that is αβ = −12. (5) (4)

435

Solving the system (1) and (5), α+β =1 αβ = −12, we obtain (α, β ) = (4, −3) and (α, β ) = (−3, 4). Since α > 0, we must have α = 4, hence a2 b + b2 c + c2 a = 4.
1 + 1 + 1 = 1. D´ emontrer Soient a, b, c des nombres r´ eels positifs avec a b c l’in´ egalit´ e suivante: √ √ Ô Ô Ô √ √ ab + c + bc + a + ca + b ≥ abc + a + b + c .

10.

Solved by Arkady Alt, San Jose, CA, USA; George Apostolopoulos, Messolonghi, Greece; Michel Bataille, Rouen, France; Paolo Perfetti, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; Henry Ricardo, Tappan, NY, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution by Ricardo. We have √ abc = √ abc 1 a + 1 b + 1 c
Ö Ö

bc a

ca b

ab c

Thus it is suﬃcient to prove that
Ô

cyclic

a + bc ≥
1 x

√ cyclic +
1 y

bc a

(1) √ x + yz ≥

For positive numbers x, y , z , with √ x + yz is equivalent to x x + yz ≥ x +

1 z

= 1, we see that

yz 1 1 √ + 2 yz = x + yz 1 − − x y z

√ + 2 yz,

√ or y + z ≥ 2 yz , which is true by the AGM inequality. Applying this result to each term of the left-hand cyclic sum in (1), our inequality is proved.

Next we turn to the Japanese Mathematical Olympiad, First Round, given at [2010 : 376–377].

Let ABCD be a convex quadrilateral with AB = 3, BC = 4, CD = 5, DA = 6 and ∠ABC = 90◦ . Find the area of ABCD .

436

Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Henry Ricardo, Tappan, NY, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the writeup of Zvonaru. By the pythagorean theorem we obtain AC = 5. It follows that the triangle ACD is isosceles. Let M be the midpoint of AD , hence CM is the altitude of triangle ACD . Thus we have AM = 3, and AC = 5, from which we deduce that M C = 4. It follows that D M A

B 3·4 2 + 6·4 2

C = 18.

area of ABCD = area of ABC + area of ACD =

Determine the tens place of 1112 of 1112 ).

(1213 th power of 11, not the 13th power

Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Norvald Midttun, Royal Norwegian Naval Academy, Sjøkrigsskolen, Bergen, Norway; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the write-up by Midttun. From the binomial theorem we ﬁnd
10

= (10 + 1) = 10 0

=
i=0

10 i

· 10i · 110−i
10

· 100 +

10 1
10

· 101 +

10 i

i=2

· 10i

= 1 + 100 + 100
i=2

10 · 10i−2 = 100N + 1, i

for some positive integer N . Similarly, we ﬁnd 1110n = (1110 )n = (100N + 1)n = 100N1 + 1 and 1213 = 10M + 2. Now we have (11)12
13

= 1110M +2 = 1110M · 112 +(100M1 +1)(100+21) = 100M2 +21.

So the tens digit is 2.

AB is a segment on a plane with length 7, and P is a point such that the distance between P and line AB is 3. Find the smallest possible value of AP × BP .

437

Let 1 be the line through A and B , and let 2 be a line parallel to 1 such that the distance between 1 and 2 is 3 units. Without loss of generality we can assume that P is on 2 . Let θ be the measure of the ∠AP B . The (AP )(P B ) sin θ , yet is also equal to area of the triangle AP B is equal to 1 2 1 1 21 1 (AB )(3) = 2 (7)(3) = 2 . Thus, 2 (AP )(P B ) sin θ = 21 and hence 2 2 (AP )(P B ) = 21 sin θ . (1)

Clearly, according to (1), the product AP · BP will be at minimum, when sin θ is at maximum, that is, when sin θ = 1, so θ = 90◦ . Note that there are two such positions of the point P along the line l2 which produce the smallest possible value of AP · BP . These two positions, P and P , are the intersection points of the line l2 and the semicircle with diameter AB = 7. Therefore, the smallest possible value of AP · BP is 21.

The tens digit of the 4-digit integer n is nonzero. If we take the ﬁrst 2 digits and the last 2 digits as two 2-digit integers, their product is a divisor of n. Determine all n with this property. Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Curtis.

Write n = 100x + y , where 10 ≤ x ≤ 99 and 10 ≤ y ≤ 99. Then xy divides 100x + y . This implies that x divides y . Write y = tx. The constraints x+y n x+tx t on x and y imply that 1 ≤ t ≤ 9. Note that xy = 100 = 100 = 100+ . xy tx2 tx In particular, t must divide 100. Thus, t ∈ {1, 2, 4, 5}. • If t = 1, then y = x, and 99.
n xy

101 , x

which is not an integer for 10 ≤ x ≤

n • If t = 2, then y = 2x, and xy = 102 = 51 . Since 99 ≥ y = 2x ≥ 20, 2x x we must have x = 17. This corresponds to n = 1734. n • If t = 4, then y = 4x, and xy = corresponding to n = 1352. n • If t = 5, then xy = appropriate interval. 105 5x 104 4x

26 . x

The only suitable x is x = 13,

21 . x

None of the divisors of 21 is in the

438

Hence, the only n’s with the desired property are n ∈ {1352, 1734}.

6. We have 15 cards numbered 1, 2, . . . , 15. How many ways are there to choose some (at least one) cards so that all numbers on these cards are greater than or equal to the number of cards chosen?
Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Norvald Midttun, Royal Norwegian Naval Academy, Sjøkrigsskolen, Bergen, Norway; and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON. We give Wang’s write-up. We consider the general problem for which there are n cards numbered 1, 2, . . . , n, respectively. Let f (n) denote the number of ways of choosing k (at least one) cards so that all numbers on these cards are at least k. We prove that f (n) = Fn+2 − 1 where Fn denotes the nth Fibonacci number deﬁned by F1 = F2 = 1 and Fn = Fn−1 + Fn−2 for n ≥ 3. Let the n cards be denoted by c1 , c2 , . . . , cn where the number on ci is i, 1 ≤ i ≤ n. Suppose we choose only k cards, k ≥ 1. Then clearly the given condition is satisﬁed if and only if all of these cards are chosen from ck , ck+1 , . . . , cn . . This is possible if and only if k ≤ n + 1 − k or k ≤ n+1 2 Hence the total number of possible selections is
n+1 2

f (n) =
k=1

n+1−k k

n 1

n−1 2

+ ··· +

n+1−

n+1 2

Now we recall the following well known fact which can be found, for example, on p. 87 of Basic Techniques of Combinatorial Theory by Daniel I.A. Cohen and can be proved by induction: n 0 + n−1 1 + n−2 2 + · · · = Fn+1 .

Replacing n by n + 1, we then have f (n) = n 1 + n−1 2 + · · · = Fn+2 − 1.

In particular, for the given problem, the answer is f (15) = F17 − 1. By straightforward computations, we ﬁnd F17 = 1597 so f (15) = 1596.

In how many ways can 100 be written as a sum of nonnegative powers of 3?

Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON. We give the write-up by Curtis. This is equivalent to ﬁnding the number of solutions in nonnegative integers to the equation 81x4 + 27x3 + 9x2 + 3x1 + x0 = 100.

439

Once x4 , x3 , x2 , and x1 are determined, x0 is determined. We must have x4 ∈ {0, 1}. Case 1. If x4 = 1, then x3 = 0, and x2 ∈ {0, 1, 2}. (a) If x2 = 2, then x1 = 0, yielding only one solution. (b) If x2 = 1, then x1 ∈ {0, 1, 2, 3}, yielding 4 solutions. (c) If x2 = 0, then x1 ∈ {0, 1, 2, 3, 4, 5, 6}, yielding 7 solutions. Case 2. If x4 = 0, then x3 ∈ {0, 1, 2, 3}. (a) If x3 = 3, then x2 ∈ {0, 1, 2}, yielding 12 solutions, as in the total for case 1. (b) If x3 = 2, then x2 ∈ {0, 1, 2, 3, 4, 5}, yielding (in reverse order) 1, 4, 7, 10, 13, and 16 solutions, for a total of 51. (c) If x3 = 1, then x2 ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8}, yielding (in reverse order) 1, 4, 7, 10, 13, 16, 19, 22, and 25 solutions, for a total of 117. (d) If x3 = 0, then x2 ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}, yielding (in reverse order) 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, and 34 solutions, for a total of 210. In case 1, we have 12 solutions; in case 2, we have 390, for a grand total of 402 solutions.

How many pairs of integers (a, b) satisfy a2 b2 = 4a5 + b3 ?

Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the write-up of Curtis, noting that the solution does not consider the possibility of negative values such as (−1, 2) and (2, −4). Any odd prime dividing a divides b3 and hence b; conversely, any odd prime dividing b divides 4a5 and hence a. Let p be an odd prime dividing a and b. Write a = pr u and b = ps v , with r, s ≥ 1, and gcd(p, u) = gcd(p, v ) = 1. Then p2r+2s u2 v 2 = 4p5r u5 + p3s v 3 . (1)

We claim that the largest exponent on p in equation (1) is 2r + 2s. Indeed, the smallest two exponents on p in equation (1) must be the same, since otherwise, dividing by the smallest power of p would give two terms still divisible by p while the third term would not be. • If 2r + 2s = 5r ≤ 3s, then r =
2 s, 3

so that

10 s 3

≤ 3s, a contradiction.

• If 2r + 2s = 3s ≤ 5r , then s = 2r , so that 6r ≤ 5r , a contradiction.

440

That proves the claim. Thus, 5r = 3s ≤ 2r + 2s, so that pm u 2 v 2 = 4 u 5 + v 3 ,
1 r= 3 r . Applying this argument where m = (2r + 2s) − 3s = 2r − s = 2r − 5 3 to each of the odd primes dividing a and b, we obtain k i=1
i pm · 22x+2y = 25x+2 + 23y . i

By the same reasoning as before, the two smallest exponents on 2 must be equal. Case 1. If x = 0, then
x i=1
i pm · 22 y = 4 + 23 y . i

If y ≥ 2, then all of the exponents on 2 are diﬀerent. Hence, y ∈ {0, 1}. (a) If y = 0, then
k
i pm = 5, i

i=1

so that the only odd prime dividing a and b is 5. Also, 1 = m = m1 = Therefore, r = 3 and s = 5 so that a = 53 and b = 55 . (b) If y = 1, then
k i=1
i pm · 4 = 12 i

1 r. 3

k
i pm = 3, i

i=1

so that the only odd prime dividing a and b is 3. We again obtain r = 3 and s = 5 so that a = 33 and b = 2 · 35 . Case 2. If x = 1, then
k i=1
i pm · 22+2y = 27 + 23y , i

and we must have y = 2. Thus,
k
i pm = 3. i

i=1

We obtain a = 2 · 33 and b = 22 · 35 .

441

Case 3. If x = 2, then
k i=1
i pm · 24+2y = 212 + 23y , i

and y = 4, so that
k
i pm = 2, i

i=1

a contradiction. Case 4. Now suppose x > 2. (a) If 2x + 2y = 5x + 2 ≤ 3y , then y = implying that x ≤ 2, a contradiction.
3 x 2 9 x 2

+ 1, so that 5x + 2 ≤

+ 3,

(b) If 2x + 2y = 3y ≤ 5x + 2, then y = 2x, and 6x ≤ 5x + 2, again yielding the contradiction x ≤ 2. Hence 5x + 2 = 3y ≤ 2x + 2y . Thus, y = Write x = 3z + 2. Then y = 5z + 4, and
k i=1
i pm · 216z+12 i

5 2 x+ 3 , 3

so that x ≡ 2 mod 3.

215z+12 + 215z+12

k i=1
i pm · · · 2z i

This implies that z = 1, and no odd primes divide a or b. Also, x = 5 and y = 9. Thus, a = 25 and b = 29 . In summary, (a, b) ∈ {(125, 3125), (27, 486), (54, 972), (32, 512)}.

10. A set of cards with positive integers on them is given, and the sum of these integers is 2007. For any integer k = 1, 2, . . . , 2006, there is only one way to choose some of these cards so that the sum of the numbers on them is k. How many such sets of cards are there?
Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA. To be able to obtain a sum of 1, there must be at least one card with the number 1. One possibility is that there are 2007 cards, all of which are 1’s. If there are exactly m cards with a 1, with m < 2007, then all of the sums from 1 through m and no others can be obtained with these m cards. Accordingly, the next highest card must be m + 1. Together with the m 1’s, each sum less than or equal to 2m + 1 can then be obtained. Therefore, the next card must be 2m + 2 = 2(m + 1). Continuing in this way, the subsequent cards must be 4(m + 1), 8(m + 1), 16(m + 1), and so on. The sum of the integers on the cards is then m + (m + 1)(1 + 2 + 4 + · · · + 2n ) = m + (m + 1)(2n+1 − 1).

442

If m < 2007, we must therefore have −1 + (m + 1)2n+1 = 2007 = 2008 = 23 · 251.

(m + 1)2 (m + 1)2n+1

n+1

Thus, n ∈ {0, 1, 2}. • If n = 0, then m = 1003. • If n = 1, then m = 501. • If n = 2, then m = 250. There are thus four suitable sets of cards: • 2007 cards, all of which are 1’s; • 1003 cards which are 1’s and a 1004; • 501 cards which are 1’s, a 502, and a 1004; • 250 cards which are 1’s, a 251, a 502, and a 1004.

Next we give readers’ solutions to problems of the 17th Japanese Mathematical Olympiad, Final Round, given at [2010: 378].

Let Γ be the circumcircle of triangle ABC . Let ΓA be the circle tangent to AB , AC and tangent internally to Γ, and let ΓB and ΓC be deﬁned similarly. Let ΓA , ΓB , ΓC be tangent to Γ at A , B , C , respectively. Prove that the lines AA , BB , CC are concurrent. Solved by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; and Ricardo Barroso Campos, University of Seville, Seville, Spain. We give the solution of Amengual Covas.

443

A E

A ΓA A∗

D bisector of ∠BAC

Let the excircle opposite A touch the side BC at A1 , and let B1 and C1 be deﬁned similarly. Let D , E and A∗ be the symmetric points of C , B and A1 respectively with respect to the bisector of ∠BAC . Then (a) the excircle opposite A touches DE at A∗ ; and (b) the lines AA1 and AA∗ are isogonal at A. √ We invert the diagram in the circle with center at A and radius AB · AC . The circumcircle Γ becomes the straight line through D and E ; the circle ΓA inverts into the excircle opposite A. Tangencies are preserved, so that A and A∗ are inverse points, implying that A, A and A∗ are collinear. That is, AA1 and AA are isogonal lines. Analogously, BB1 and BB are isogonal at B , and CC1 and CC are isogonal at C . Since AA1 , BB1 and CC1 concur (at the Nagel point of ABC ), the isogonal lines AA , BB and CC also concur, as desired.

Next, we look at the readers’ solutions to problems given in the November 2010 issue, the last issue that featured Olympiad problem sets and an invitation to submit solutions. We begin with the Croatian Mathematical Competition 2007, National Competition, given at [2010: 435–436].

444

3rd Grade

Let n be a positive integer such that n + 1 is divisible by 24.

(a) Prove that n has an even number of divisors (including 1 and n itself). (b) Prove that the sum of all divisors of n is divisible by 24. (Simpliﬁed from Putnam Competition 1969) Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Oliver Geupel, Br¨ uhl, NRW, Germany; David E. Manes, SUNY at Oneonta, Oneonta, NY, USA; Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution and comment by Wang. (a) We show that the conclusion actually holds under the weaker assumption αk 1 α2 that n + 1 is divisible by 4. Clearly n = 1. Let n = pα 1 p2 . . . pk be the prime powers decomposition of n where αi ’s are natural numbers and pi ’s are primes, i = 1, 2, . . . , k. Let τ (n) denote the number of (positive) Ék divisors of n. It is well known that τ (n) = i=1 (1 + αi ). Suppose τ (n) is odd. Then each 1 + αi is odd so αi must be even for all i = 1, 2, . . . , k. Thus, n is a perfect square. Let n = m2 where m is a natural number. Clearly n is odd so m is also odd. Hence m2 ≡ 1 (mod 4). Therefore, n + 1 = m2 + 1 ≡ 2 (mod 4) contradicting the fact that n + 1 ≡ 0 (mod 4). (b) This is exactly the same as problem B1 of the 1969 Putnam Math Competition. A full solution can be found on p. 63 of The William Lowell Putnam Math Competition, Problems and Solutions; edited by Gerald L. Alexanderson et al. The proof given there actually establishes the stronger fact that d + (n/d) is divisible by 24 for all divisors d of n. The conclusion then follows immediately from part (a).

In the triangle ABC , with ∠BAC = 120◦ , the bisectors of the angles ∠BAC , ∠ABC and ∠BCA intersect the opposite sides in the points D , E , and F , respectively. Prove that the circle with diameter EF passes through D . (British Mathematical Olympiad 2005) Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Oliver Geupel, Br¨ uhl, NRW, Germany; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solutions of Zvonaru.

First Solution. The bisector of ∠BAD meets the side BD at the point T . Since ∠T AC = ∠T AD + ∠DAC = 30◦ + 60◦ = 90◦ , B

A F E

445

we deduce that AC is the external bisector of ∠BAD . Because BE is the bisector of ∠ABD , it follows that DE is an external bisector of ∠ADB , that is DE is a bisector of ∠ADC . Similarly we obtain that DF is a bisector of ∠ADB ; it follows that 1 ∠EDF = 2 (∠ADC + ∠ADB ) = 90◦ , hence the circle with diameter EF passes through D . Second Solution. As usual we write a = BC , b = CA, c = AB . By the . Bisector’s Theorem we obtain DC = bab +c It is known that the bisector AD = b+c 2 = We have, again using the Bisector’s Theorem AD DC = c a = AE EC
2bc cos
A

bc . b+c

and, by the converse of the Bisector’s Theorem, it follows that DE is the bisector of ∠ADC . Similarly, DF is the bisector of ∠ADB , hence ∠EDF = 90◦ .

3. In triangle ABC the distances of vertex A from the centre of the circumscribed
circle and the orthocentre are equal. Determine the angle α = ∠BAC . proposal for IMO 1989) (USA Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Oliver Geupel, Br¨ uhl, NRW, Germany; and Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA. We give the solution by Curtis. Let a = BC , b = CA, c = AB , and let O and H be the circumcenter and orthocenter, respectively. Let C be the point at which line HC intersects side AB . By the extended Law of Sines, the circumradius is given by AO = b . 2 sin B

cos A . Thus, On the other hand, AC = ±b cos A, so that AH = ± bsin B

b b cos A =± . 2 sin B sin B Thus, cos A = ± 1 , so that α = 60◦ , 120◦ . 2 Next we give the solution of Geupel. Denote the circumcentre and the orthocentre by O and H , respectively. We prove that the condition AO = AH is equivalent to α = 60◦ , 120◦ . Let a, b, c, h, and o be the coordinates of A, B , C , H , and O in the plane of complex numbers. Without loss of generality let o = 0, c = 1, and b = eiϑ where 0 < ϑ ≤ π . We have h = a + b + c = a + b + 1. The condition AO = AH is therefore successively equivalent to |a| = |h − a|,

446

1 = | b + 1| , 1 = 1 + eiϑ 1 + e−iϑ , 1 cos ϑ = − , 2 ◦ ∠BOC = 120 , 240◦ , and ﬁnally, α= The proof is complete. 1 ∠BOC = 60◦ , 120◦ . 2

4th Grade

In a 5 × n table, where n is a positive integer, each 1 × 1 cell is painted either in red or in blue. Find the smallest possible n such that, for any painting of the table, one can always choose three rows and three columns for which the 9 cells in their intersection have all the same colour. Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution by Curtis. Each column has at least three reds or at least three blues. For k = 1, 2, . . . , n, form the quadruple (y1 (k), y2 (k), y3 (k), c(k)), where the yi (k) are positive integers such that 1 ≤ y1 (k) < y2 (k) < y3 (k) ≤ 5, (1)

And the cells ( ¡k, y1 (k)), (k, y2 (k)), (k, y3 (k)) each have colour c(k). There are at most 5 · 2 = 20 such quadruples. Thus, with n = 41, at least three 3 columns must have the same quadruple. Choose three such columns with their corresponding y -values. To see that no smaller n will work, assume n ≤ 40, and, in each column, colour either exactly three cells red or exactly three cells blue in such a way that each of the possible quadruples described above appears at most twice. Then no selection of three rows and three columns provides an intersection with all cells of the same colour.

That completes the Corner for this number.

447

BOOK REVIEWS
Amar Sodhi
Magical Mathematics: The Mathematical Ideas That Animate Great Magic Tricks by Persi Diaconis and Ron Graham, with a foreword by Martin Gardner Princeton University Press, 2012 ISBN 978-0-691-15164-9, 244 + xii pp., hardcover, US$29.95 Reviewed by S. Swaminathan, Dalhousie University, Halifax, N. S. Magic tricks are fascinating to young and old alike. Many simple magic tricks that work by themselves are based on mathematical principles. Consider the following trick: the magician hands four playing cards to a spectator, turns his back and gives instructions to the spectator to perform actions on the cards in diﬀerent ways such as turning a card face down, transferring cards from top to bottom either singly or collectively, and doing these actions any number of times in any order. When the spectator has ﬁnished performing the actions, he (she) is requested to perform one ﬁnal set of transfers and then asked to mention how many cards are facing the opposite way from the others. The answer turns out to be only one card that faces the opposite way. The magician tells correctly what that particular card is! This amazing trick is explained fully with rich illustrations in the ﬁrst chapter of the wonderful book under review. The authors are eminent mathematicians; Ron Graham of Bell Labs and UC, San Diego is an expert on combinatorial mathematics, and Persi Diaconis is a professor of statistics at Stanford University. Both of them are skilled performers of magic; Ron is a juggler and Persi is a skilled card magician. They have been mixing entertainment with mathematics for most of their lives besides teaching, publishing papers with deep results and conjecturing new results. The book consists of ten chapters with the following titles: Mathematics in the Air, In Cycles, Is This Stuﬀ Actually Good for Anything?, Universal Cycles, From Gilbreath Principle to the Mandelbrot Set, Neat Shuﬄes, The Oldest Mathematical Entertainment?, Magic in the Book of Changes, What Goes Up must Come Down, Stars of Mathematical Magic (and Some of the Best Tricks in the Book), Going Further, On Secrets, Notes, Index. Easy, step by step instructions are provided for each trick, explaining clearly how the eﬀect is set up and oﬀering tips on what to say and do during the performance. Each card trick introduces a new mathematical idea, and varying the tricks takes the readers to the very threshold of current mathematical knowledge. However, sophisticated math terminology is avoided. For example, the underlying theme of the trick mentioned in the ﬁrst paragraph of this review is that a large permutation group (’group’ in the mathematical sense) leaves an interesting set of invariants under the group ﬁxed while the cards are getting mixed, which contributes to the startling ﬁnal eﬀect. Other card tricks link to mathematical secrets of combinatorics, graph theory, number theory, topology, the Riemann

448

Hypothesis, and even Fermat’s Last Theorem. The book contains a wealth of conjuring lore, including some closely guarded secrets of legendary magicians. It discusses the mathematics of juggling, and shows how I Ching, an ancient Chinese fortune-telling book, connects to the history of probability and tricks both old and new. Stories underlying some tricks of eccentric and brilliant inventors of mathematical magic are discussed. Copious colourful illustrations and pictures are provided to illustrate the text. Readers are sure to enjoy this brilliant book. Their interest in magic will get kindled if it is not already there. They will get introduced to little-known mathematical theorems. The book will certainly become a classic.

Unsolved Crux Problem
As remarked in the problem section, no problem is ever closed. We always accept new solutions and generalizations to past problems. Recently, Chris Fisher published a list of unsolved problems from Crux [2010 : 545, 547]. Below is a sample of one of these unsolved problems: [1976 : 110, 159, 197, 225-226; 1977 : 20-22, 108-109, 191-193] Proposed by Kenneth S. Williams, Carleton University, Ottawa, ON. Let pn denote the nth prime, so that p1 = 2, p2 = 3, p3 = 5, p4 = 7, etc. Prove or disprove that the following method ﬁnds pn+1 given p1 , p2 , . . ., pn . In a row list the integers from 1 to pn − 1. Corresponding to each r an−1 an−1 a1 1 (1 ≤ r ≤ pn − 1) in this list, say r = pa in a 1 · · · pn−1 , put p2 · · · pn second row. Let be the smallest odd integer not appearing in the second row. The claim is that = pn+1 . Example. Given p1 = 2, p2 = 3, p3 = 5, p4 = 7, p5 = 11, p6 = 13. 1 ↓ 1 2 ↓ 3 3 ↓ 5 4 ↓ 9 5 ↓ 7 6 ↓ 15 7 ↓ 11 8 ↓ 27 9 ↓ 25 10 ↓ 21 11 ↓ 13 12 ↓ 45

154.

We observe that

= 17 = p7 .

449

RECURRING CRUX CONFIGURATIONS 3
J. Chris Fisher Triangles whose angles satisfy 2B = C + A
Because the angles of a triangle ABC sum to 180◦ , they are in arithmetic progression if and only if the intermediate angle measures 60◦ . I found nearly two dozen problems in CRUX with MAYHEM that deal with triangles having a 60◦ angle; there was considerable overlap, so for this month’s column instead of listing those problems, I will simply list and discuss the properties that readers have discovered, and provide references to where proofs can be found. As usual, A, B , and C will represent either the vertices of a triangle or the measure of its angles, depending on the context; a, b, and c represent either the opposite sides or their lengths; s = (a + b + c)/2 is the semiperimeter, while H, I, O , and G are the orthocentre, incentre, circumcentre, and centroid, respectively. We shall use 2B = C + A and B = 60◦ interchangeably. The ﬁrst eight properties came, in part, from Problem 724 [1982 : 78; 1983 : 92-94] (proposed by Hayo Ahlburg) and the comments found there. Property 1. ∠B = 60◦ if and only if sin(A − B ) = sin A − sin C . The simple proof of only if is on page 92; for the if part note that sin C = sin(A + B ) and expand that and sin(A − B ) to get an equation that reduces to 1 cos B = 2 .

Property 2 [1983 : 93] is just the cosine law. It forms the basis of an olympiad problem proposal of Murray Klamkin that was never used: One of the angles of a triangle is 60◦ if and only if the square of the side opposite that angle equals the sum of the cubes of the sides divided by the perimeter; that is, ∠B = 60◦ if and only if b2 = a3 + b3 + c3 a+b+c .

Property 2. ∠B = 60◦ if and only if a2 − b2 = c(a − c).

A more substantial use of Property 2 came in [3] to obtain a characterization of integer-sided triangles having an angle of 60◦ : Let p and q be (i) relatively prime integers with (ii) one of them odd, the other even, and (iii) p not a multiple of 3; use p and q to deﬁne the integers x = |p2 − 3q 2 | and y = 2pq . If x > y we set b = p2 + 3q 2 , a = x + y and c = 2y, x − y (there are two values of c for each a and b because the quadratic equation of Property 2 will have two integer zeros); if x < y we use the same b but set a = 2y and c = y ± x. Then in either case with either value of c, ABC is a triangle with ∠B = 60◦ and side lengths a > b > c that are relatively prime integers; conversely, for any such triangle there exists a pair of integers p and q that produce that triangle using the given recipe. For example, p = 1, q = 2 determines x = 11, y = 4 and triangles with sides 15, 13, 8 as well as 15, 13, 7; for p = 2, q = 1, the parameters are x = 1, y = 4, and

450

the resulting triangles have sides 8,7, 5, and 8, 7, 3. Similar results and further references can be found in [4]. Property 3. If ∠B = 60◦ then the points A, C, O, I, H lie on a circle that also contains the excentre Ib opposite vertex B ; its radius equals the circumradius of ∆ABC and its centre O is where the angle bisector BI again meets the circumcircle. The solution to Problem M1046 from the 1987 U.S.S.R. journal Kvant [1988 : 165; 1990 : 103] follows easily: If ∠B = 60◦ then one of the bisectors of the angle between the altitudes from A and C passes through O . One should take care with the converse of Property 3, which is the topic of problem 1521 [1990 : 74; 1991 : 126-127] (proposed by J.T. Groenman). If either A, I, O, C or A, H, I, C are concyclic, then it follows that ∠B = 60◦ and all ﬁve points are concyclic. Such is not the case, however, with A, H, O, C concyclic because then ∠B = 120◦ is also possible (which can be seen by changing the roles of the points B with H and O with O ). This property is clearly equivalent to the following result, which is the subject of Problem 998 [1984 : 319; 1986 : 65] (proposed by Andrew P. Guinand): If one angle of a triangle is either 60◦ or 120◦ , then the image of the orthocentre under inversion with respect to the circumcircle lies on the side (possibly extended) opposite that angle.
B
60◦

C C

O N

I H A

O A

Figure 1: The angles of ∆ABC satisfy 2B = C + A. Property 4. The circle of Property 3, with radius R and centre O , intersects the lines BA and BC at points A and C for which AA = CC = |c − a|. (Proof is on page [1983 : 93].) Property 5. In any triangle, if N is the centre of its nine-point circle (and, therefore, the midpoint of OH ), and P is the projection of the incentre I onto the Euler line OGN H , then P lies between G and H ; furthermore, P = N if and only if one angle of the triangle has measure 60◦ . This is Problem 260 [1977 : 155; 1978 : 58-60] (Proposed by W.J. Blundon). A variant of this property became Problem 5 on the 2007 Indian Team Selection

451

Test [2010 : 278; 2011 : 370-371]: For triangles that are not equilateral, the common tangent to the incircle and the nine-point circle is parallel to the Euler line if and only if the angles of the triangle are in arithmetic progression. This, of course, is because in all triangles that common tangent is perpendicular to IN . Property 6. If ∠B = 60◦ then N lies on the bisector of that angle; conversely, if the nine-point centre of a triangle lies on the interior bisector of ∠CBA, then the vertex B lies on the perpendicular bisector of AC or ∠B = 60◦ . This is Problem 2855 [2003 : 316; 2004 : 308-309] (Proposed by Antreas P. Hatzipolakis and Paul Yiu); the claim that BN bisects ∠B when ∠B = 60◦ is also proved as a part of Problem 724 [1983 : 94]. Property 7. If B is the intermediate angle of ∆ABC , then ∠B = 60◦ if and only if OI = IH , if and only if OIb = Ib H . The equivalence of ∠B = 60◦ and OI = IH is proved as part of Problem 260 (Property 5 above). It was proved yet again as part of Problem 1521 (see Property 3 above). This result also appeared as Problem 739 [1982 : 107; 1983 : 153-154, 210-211] (proposed by G.C. Giri), where there is another proof and references to textbooks where it appears as an exercise. There is also a reference to a stronger result [2]: If the angles of triangle ABC are labeled so that A ≤ B ≤ C then < 1, IO ∠B = 60◦ ⇒ HI = IO, HI ∠B < 60◦ ⇒ 1 < < 2. IO The proof of the result for OIb = Ib H can be found in [1]. √ Property 8. ∠B = 60◦ if and only if s = 3(R + r ). The proof is another part of the solution to Problem 260 [1978 : 58-60]. Property 9. ∠B = 60◦ if and only if the bisector BO of ∠B is perpendicular to the Euler line OH ; when these properties hold, then N is the common midpoint of BO and OH . (Recall that O was deﬁned in Property 3 to be where the angle bisector again intersects the circumcircle. Note that in any triangle, BH and OO are both perpendicular to AC .) The claims follow from the proof in Problem 1521 (see Property 3). Three more proofs can be found in [5]. An immediate consequence is Problem 3 of Round 2 of the 2006-2007 British Mathematical Olympiad [2010: 154; 2011: 165]: If the Euler line OH meets BA at P and BC at Q, then ∠B = 60◦ implies that OQ = HP . See [2011: 165] for an independent proof (although the problem and proof found there were unnecessarily restricted to acute-angled triangles). Another immediate consequence of Problem 1521 is Problem 1673 [1991 : 237; 1992 : 218-219] (proposed by D.J. Smeenk): ∠B > 60◦ ⇒ 0 < HI

452

Given ∆ABC let P be an arbitrary point of the line BA and Q be on BC , neither point coinciding with a vertex. If ∠B = 60◦ then the Euler lines of ∆ABC and ∆P BQ are parallel; moreover, if the two Euler lines coincide then the circumcircle P QR contains O . Jordi Dou added a proof that the sides P Q of those triangles whose Euler lines coincide with that of the given triangle are tangent to the parabola with focus O and directrix OH . Property 10. ∠B = 60◦ or 120◦ if and only if BH = BO . This is case (i) of Problem 1518 [1990 : 44; 1991 : 122] (proposed by K.R.S. Sastry). Compare Property 9; Problem 1232(b) in [5] says that BO ||OH if and only if ∠B = 120◦ . Property 11. ∠B = 60◦ or 120◦ if and only if its internal bisector divides an altitude in the ratio 1 : 2. This is Problem 2526 [2000 : 177; 2001 : 271-273] (proposed by K.R.S. Sastry). We devote the remainder of this compilation to properties that were discovered over the past 20 years by Toshio Seimiya. What a pity that we failed to invite him to write this article for us! For most of these properties we denote by D and E the points where the interior bisectors of angles A and C meet the opposite sides. Properties 12 through 16 are quite closely related.
B
60◦

D O C N P

Q I H A E

Figure 2: Seimiya’s properties 12, 13, 14, and 15. Property 12. If ∠B = 60◦ then the points D and E are two vertices of an equilateral triangle whose third vertex lies on AC and whose incentre is I . In other words, the bisectors of angles A and C meet the opposite sides at the centres of two circles with common radius DE that intersect on AC . This is Seimiya’s counterexample to the incorrect claim made by the proposer of Problem 1446(c) [1989 : 148; 1990 : 217-219] (namely, that the existence of this inscribed equilateral triangle implied that the original triangle ABC was necessarily equilateral). Property 13. Deﬁne P to be the point where the line perpendicular to DE

453

meets AC , and Q to be where it meets DE . Then IP = 2IQ if and only if ∠B = 60◦ . (Problem 2011 [1995 : 52; 1996 : 80]) Property 14. Deﬁne P to be the point where the bisector of ∠AIC meets AC , and Q to be where it meets DE . Then IP = 2IQ if and only if ∠B = 60◦ . (Problem 2939 [2004 : 229, 232; 2005 : 243-244]) Property 15. ∠ADE = 30◦ if and only if ∠B = 60◦ or ∠C = 120◦ . (Problem 2263 [1997 : 364; 1998 : 432-433]) Property 16. Call F the point where DE intersects AC . If ∆ABC has 1 (∠DAC − ∠ECA), then ∠B = 60◦ . (Problem BC > BA and ∠DF C = 2 2314 [1998 : 107; 1999 : 117]) Property 17. Given ∆ABC deﬁne P and Q to be points on the same side of AC as B , with P the point on BC for which P C = BA, and Q the point on BA for which QA = BC . Then ∠B = 60◦ if and only if O lies on P Q. (Problem 1692 [1991 : 301; 1992 : 284-285]) Property 18. An acute-angled triangle ABC is given, and equilateral triangles ABP and BCQ are drawn outwardly on the sides AB and BC . Suppose that AQ and CP meet BC and AB at R and T , respectively, and that AQ and CP intersect at S . If the area of the quadrilateral BRST is equal to the area of the triangle ASC , then ∠B = 60◦ . (Problem 2304 [1998 : 46; 1999 : 56-57]) In addition to his many problems, Semiya also wrote an article for CRUX with MAYHEM entitled “On Some Examples of Geometric Fallacies” [29:6 (October 2003) 393-396]. He began with a theorem and proposed two attempted converses, both of which came with very convincing arguments; he then pointed out the subtle but critical errors in the arguments, and provided counterexamples to show that those converses were indeed false. It is the theorem that is relevant here: Theorem. Let ABC be a triangle with ∠B = 60◦ . Let D be the point on BC produced beyond C such that CD = CA, and let E be the point on BA produced beyond A such that CA = AE . Then ∠DCA = 2∠AED , ∠CAE = 2∠EDC , and ∠EDA = 30◦ .

References
[1] M.N. Aref and William Wernick, Problems and Solutions in Euclidean Geometry, Dover, 1968, p.185. [2] Anders Bager, Solution to Problem E2282 (proposed by W.J. Blundon), American Mathematical Monthly, 47:79 (1972) 397-398. [3] Bob Burn, Triangles with a 60◦ angle and Sides of Integer Length. Mathematical Gazette, 87:508 (Mar. 2003) 148-153. [4] Russell A. Gordon, Properties of Eisenstein Triples Mathematics Magazine, 85:1 (Feb. 2012) 12-25. [5] J.T. Groenman and D.J. Smeenk, Problem 1232(a), Mathematics Magazine, 60:1 (Feb. 1987) 43-45. (Appeared in Feb. 1986.)

454

PROBLEMS
Solutions to problems in this issue should arrive no later than 1 November 2012. An asterisk ( ) after a number indicates that a problem was proposed without a solution. Each problem is given in English and French, the oﬃcial languages of Canada. In issues 1, 3, 5, and 7, English will precede French, and in issues 2, 4, 6, and 8, French will precede English. In the solutions’ section, the problem will be stated in the language of the primary featured solution. The editor thanks Jean-Marc Terrier of the University of Montreal for translations of the problems.

3676.

Proposed by Michel Bataille, Rouen, France.

Let a, b, and c be the sides of a triangle with semiperimeter s, inradius r and circumradius R. Let r and R be the inradius and circumradius of a triangle with sides a(s − a), b(s − b), and c(s − c). Prove that Rr ≥ R r.

3677.

Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.

Let n be a positive integer. Prove that
n− 1 k=1

(−1)k sinn (kπ/n) =

(1 + (−1)n )n 2n

· cos

nπ 2

3678.

Proposed by Michel Bataille, Rouen, France.

Let Γ1 , Γ2 be two intersecting circles and U one of their common points. Show that there exists inﬁnitely many pairs of lines passing through U and meeting Γ1 and Γ2 in four concyclic points. Give a construction of such pairs.

3679.
CA, USA. that

Proposed by Hung Pham Kim, student, Stanford University, Palo Alto,

Let a, b, and c be nonnegative real numbers such that a + b + c = 3. Prove (a2 b + c)(b2 c + a)(c2 a + b) ≤ 4(ab + bc + ca − abc) .

3680.

Proposed by Michel Bataille, Rouen, France.

In a system of axes (Ox, Oy, Oz ), let U (1, 1, 1), S (a, b, c) and H (ha , hb , hc ) where a, b, c are the sides of a triangle ABC and ha , hb , hc are the corresponding altitudes. Given that the lines OU and SH intersect at M 1 such that |HM | = 3 |HS |, ﬁnd the angles of ∆ABC .

455

3681.

Proposed by Nguyen Thanh Binh, Hanoi, Vietnam.

Let D, E , and F be the points where the incircle of ∆ABC touches the sides. Let Z be the Gergonne point (where AD, BE , and CF concur), and let M be the midpoint of BC . Deﬁne T to be the tangency point of the incircle with the circle through B and C that is tangent to it, and let the common tangent line at that point intersect AC at S . Prove that AB, SZ , and M E are concurrent.

3682.
Vietnam.

Proposed by Pham Van Thuan, Hanoi University of Science, Hanoi,

Let a, b, c, and d be nonnegative real numbers such that a2 + b2 + c2 + d2 = 1. Prove that 1 1 1 1 1 1 + + + + + ≤ 8. 1 − ab 1 − bc 1 − cd 1 − da 1 − bd 1 − ac

3683.

Proposed by Michel Bataille, Rouen, France.

Let n be an integer with n ≥ 2 and z a complex number with |z | ≤ 1. Prove that
n

kz n−k = 0 .
k=1

3684.

Proposed by Nguyen Thanh Binh, Hanoi, Vietnam.

Given two circles that are internally tangent at T , let the chord BC of the outer circle be tangent to the inner circle at D . Let the second tangents from B and C touch the inner circle at F and E respectively, and deﬁne J = EF ∩ DT and Z = BE ∩ CF . Prove that (a) JZ intersects BC at its midpoint, and (b) T D bisects ∠BT C . Comment. This result allows for a solution to a special case of the Problem of Apollonius: Construct a circle through two given points that is tangent to a given circle which, itself, is tangent to the line joining the given points.

3685.

Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.

Let f : [0, 1] → (0, ∞) be a bounded function which is continuous at 0. Find the value of
Õ
n

n→∞

lim

1 1

1 2

+ ··· +

1 n

Ǳn

456

3686.

Proposed by Michel Bataille, Rouen, France.
2

Let a, b, and c be real numbers such that abc = 1. Show that a− 1 a +b− 1 b +c− 1 c ≤ 2 a+ 1 a b+ 1 b c+ 1 c .

3687.

Proposed by Albert Stadler, Herrliberg, Switzerland.
∞ k=0

Let n be a nonnegative integer. Prove that kn k! k+1− 1 k!
∞ 1 n

e−t tk+1 dt =
k=0

S (n, k) k+2

where kn is taken to be 1 for k = n = 0 and S (n, k) are the Stirling numbers of the second kind that are deﬁned by the recursion S (n, m) = S (n − 1, m − 1) + mS (n − 1, m), S (n, 0) = δ0,n , S (n, n) = 1 . .................................................................

3676.

Propos´ e par Michel Bataille, Rouen, France.

Soit a, b et c les cˆ ot´ es d’un triangle de demi-p´ erim` etre s et dont les cercles inscrit et circonscrit sont de rayons respectifs r et R. Soit d’autre part les rayons r et R des cercles inscrit et circonscrit d’un triangle sont les cˆ ot´ es sont a(s − a), b(s − b)et c(s − c). Montrer que Rr ≥ R r.

3677.

Propos´ e par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie.
n− 1 k=1

Soit n un entier positif. Montrer que (−1)k sinn (kπ/n) = (1 + (−1)n )n 2n · cos nπ 2 .

3678.

Propos´ e par Michel Bataille, Rouen, France.

Soit U un des deux points d’intersection de deux cercles Γ1 et Γ2 . Montrer qu’il existe une inﬁnit´ e de paires de droites passant par U et coupant Γ1 et Γ2 en quatre points cocycliques. Donner une construction de telles paires.

3679.

Propos´ e par Pham Kim Hung, ´ etudiant, Universit´ e de Stanford, Palo ´ . Alto, CA, E-U

Soit a, b et c trois nombres r´ eels non n´ egatifs tels que a + b + c = 3. Montrer que (a2 b + c)(b2 c + a)(c2 a + b) ≤ 4(ab + bc + ca − abc) .

457

3680.

Propos´ e par Michel Bataille, Rouen, France.

Dans un syst` eme d’axes (Ox, Oy, Oz ), soit U (1, 1, 1), S (a, b, c) et H (ha , hb , hc ) o` u a, b, c sont les cˆ ot´ es d’un triangle ABC et ha , hb , hc en sont les hauteurs correspondantes. En supposant que les droites OU et SH se coupent en M de telle sorte que |HM | = 1 |HS |, trouver les angles de ∆ABC . 3

3681.

Propos´ e par Nguyen Thanh Binh, Hano¨ ı, Vietnam.

Soit D, E et F les points de contact du cercle inscrit du triangle ABC avec ses cˆ ot´ es. Soit Z le point de Gergonne (intersection de AD, BE et CF ), et soit M le point milieu de BC . Notons T le point de tangence du cercle inscrit avec le cercle qui lui est tangent et qui passe par B et C , et soit S le point d’intersection de la tangente commune en T avec AC . Montrer que AB, SZ et M E sont concourantes.

3682.
Vietnam.

Propos´ e par Pham Van Thuan, Universit´ e de Science des Hano¨ ı, Hano¨ ı,

Soit a, b, c et d quatre nombres r´ eels non n´ egatifs tels que a2 +b2 +c2 +d2 = 1. Montrer que 1 1 − ab + 1 1 − bc + 1 1 − cd + 1 1 − da + 1 1 − bd + 1 1 − ac ≤ 8.

3683.

Propos´ e par Michel Bataille, Rouen, France.

Soit n un entier avec n ≥ 2 et z un nombre complexe tel que |z | ≤ 1. Montrer que
n

kz n−k = 0 .
k=1

3684.

Propos´ e par Nguyen Thanh Binh, Hano¨ ı, Vietnam.

On donne deux cercles int´ erieurement tangents en T et une corde BC du cercle ext´ erieur tangente au cercle int´ erieur en D . Soit respectivement F et E les points de contact des secondes tangentes issues de B et C avec le cercle int´ erieur, et soit J = EF ∩ DT et Z = BE ∩ CF . Montrer que (a) JZ coupe BC en son point milieu, et (b) T D est la bissectrice de ∠BT C . Commentaire. Ce r´ esultat permet de r´ esoudre un cas sp´ ecial du probl` eme d’Apollonius : Construire un cercle passant par deux points donn´ es qui soit tangent a ` un cercle donn´ e qui, lui, est tangent a ` la droite joignant les deux points.

458

3685.

Propos´ e par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie.

Soit f : [0, 1] → (0, ∞) une fonction born´ ee, continue en 0. Trouver la valeur de Ǳn Õ Õ Õ 1 1 n n n f 1 + f + · · · + f 1 2 n lim . n→∞ n

3686.

Propos´ e par Michel Bataille, Rouen, France.

Soit a, b et c trois nombres r´ eels tels que abc = 1. Montrer que a− 1 1 1 +b− +c− a b c
2

≤ 2 a+

1 a

1 b

1 c

3687.

Propos´ e par Albert Stadler, Herrliberg, Suisse.

Soit n un entier non n´ egatif. Montrer que
∞ k=0

kn k!

k+1−

1 k!

∞ 1

e−t tk+1 dt =
k=0

S (n, k) k+2

o` u l’on pose kn = 1 pour k = n = 0 et o` u S (n, k) sont les nombres de Striling du second ordre, d´ eﬁnis par la r´ ecursion S (n, m) = S (n − 1, m − 1) + mS (n − 1, m), S (n, 0) = δ0,n , S (n, n) = 1 .

459

SOLUTIONS
A ﬁling error took place and as a result a few solvers were not acknowledged in the past few issues. The editors would like to recognize the following correct solutions: George Apostolopoulos, Messolonghi, Greece (3754); ˇ Sefket Arslanagi´ c, University of Sarajevo, Sarajevo, Bosnia and Herzegovina (3566, 3570, 3572, 3574); Michel Bataille, Rouen, France (3566, 3570, 3572); John Hawkins and David R. Stone, Georgia Southern University, Statesboro, GA, USA (3558); Oliver Geupel, Br¨ uhl, NRW, Germany (3564); Dragoljub Miloˇ sevi´ c, Gornji Milanovac, Serbia (3457); Paolo Perfetti, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy (3564); Henry Ricardo, Tappan, NY, USA (3558); Joel Schlosberg, Bayside, NY, USA (3556, 3563, 3566); Digby Smith, Mount Royal University, Calgary, AB (3571); and Titu Zvonaru, Com´ ane¸ sti, Romania (3572). If any other errors or omissions occur, please send an email to [email protected].

No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems.

3568.

[2010 : 396, 398] Proposed by Albert Stadler, Herrliberg, Switzerland.

Let n be a nonnegative integer and let ak be the coeﬃcient of z k in the McLaurin expansion of (z − 1)n ln(1 − z ). Prove that an = 1 + 1 2 + 1 3 + ··· + 1 n and ak = (n + 1) −1

k ¡ n+1

, k > n.

I. Solution by George Apostolopoulos, Messolonghi, Greece.
n

It is well known that (z − 1)n =
∞

i=0

(−1)n−i

n i z and i

ln(1 − z ) = −

i=1

zi , thus i
n

(z − 1)n ln(1 − z ) = From (1) we deduce that an =
n− 1 i=0

i=0

(−1)n−i

n i

−

∞ i=1

zi i

(1)

(−1)

n− i

n i

1 − n−i

n¡

=
j =1

(−1)

j +1

460
n¡
(−1)
i+1 i

Let g (n) = an = so g (n) =
i=1

i
i+1

. Recall that
n

n i

n−1 i
i−1

n i=1

(−1)

n−1¡
i

+
i=1

(−1)

i+1

n−1¡

n−1 i−1

(2)

Now, since

n i
n i=1

= 0 for i > n or i < 0, then

(−1)

i+1

n−1¡
i

i
n−1 i−1

n− 1 i=1

(−1)
n i
i+1

i+1

n−1¡
i

i
· 1 , hence n
i

= g (n − 1).

(3)

It is easy to verify that
n i=1

·
n

1 = i

(−1)

i+1

n−1¡
i−1

n ¡

=
i=1

(−1)
å

=−

1 n

n i=1

(−1)i
è

n i 1 n . (4)

=−

1 n

(1 − 1)n − (−1)0

n 0

1¡ 1 2 1 Thus (2), (3) and (4) yield g (n) = g (n − 1) + . Since g (1) = (−1) = 1, n 1 n n 1 1 we deduce that g (n) = , that is, an = . i i
i=1 i=1

From (1) we deduce that for k > n
n

ak =
i=0
n

(−1)n−i
n¡
i

n i

−

1 . k−i

Let f (n, k) =
i=0

(−1)

n−i

k−i

, then
n

f (n, k) =
i=0

(−1)

n− i

n−1¡
i

k−i

+
i=0 i

(−1)

n− i

n−1¡

=−

n− 1 i=0

(−1)

n− 1 − i

n − 1 ¡

k−i
n− 1 − j

i−1

k−i

n− 1 j =0

(−1)

n−1¡
j

k−1−j

= f (n − 1, k − 1) − f (n − 1, k). Now f (0, k) =
f (n − 1, k) = 1 1 = k ¡ for all positive integer values of k, and if k (0 + 1) 0+1

1 k ¡ holds for all k > n − 1, then f (n, k) = f (n − 1, k − 1) n n

−f (n − 1, k) for all k − 1 > n − 1, i.e. k > n. Thus

461

f (n, k) =

k−1¡ −
n

k¡ =
n

(n + 1)

k ¡ n+1

for k > n so ak =

−1 k ¡ for k > n. (n + 1) n+1
n ∞

II. Solution by Oliver Geupel, Br¨ uhl, NRW, Germany. Considering (z − 1)n ln(1 − z ) = have to prove that
n j =1

j =0

(−1)n−j

n j z j

−

j =1

1 j z , we j

(−1)j +1 n j j

=1+

1 2

1 3

+ ··· +

1 n

(1)

and

n j =0

(−1)j n k−n+j j

(n + 1)

k ¡ n+1

k > n.

(2)

For a proof of identity (1) we refer to [1]. We only prove (2). n n j (−1)j x = xk−n−1 (1 − x)n ; hence We have xk−n−1 j j =0
n j =0

k−n+j j
æ

(−1)j

=
0

xk−n−1 (1 − x)n dx.

Integration by parts for nonnegative m, n yields
1 0

(1 − x)n xm+1 x (1 − x) dx = m+1
m n

é1

+
0

n m+1

1 0

xm+1 (1 − x)n−1 dx

n m+1

1 0

xm+1 (1 − x)n−1 dx.

By repeated application of this formula, we obtain
1 0

xk−n−1 (1 − x)n dx = =

(k − n)(k − n + 1) · · · (k − 1) 1 = k ¡, (n + 1) n+1

(k − n)(k − n + 1) · · · (k − 1) n(n − 1) · · · 1

n(n − 1) · · · 1

1 0

xk−1 dx

1 k

which completes the proof.
Also solved by MICHEL BATAILLE, Rouen, France; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; JOEL SCHLOSBERG, Bayside, NY, USA; and the proposer.

462

References
[1] Loren C. Larson, Problem solving through problems, Springer, second printing, 1990, example 5.1.4, p.160.

3573.

[2010 : 397, 399] Proposed by A.A. Dzhumadil’daeva, Almaty Republic Physics and Mathematics School, Almaty, Kazakhstan.

Let (2n + 1)!! = 1 · 3 · · · (2n + 1) be the double factorial, so (for example) 7!! = 105. Make the convention that 0!! = (−1)!! = 1. Prove that for any nonnegative integer n, n (2i − 1)!!(2j − 1)!!(2k − 1)!! = (2n + 1)!! . i, j, k

i+j +k=n i,j,k≥0

Solution by Oliver Geupel, Br¨ uhl, NRW, Germany. From the identity (2m − 1)!! = 1 · 3 · · · (2m − 1) = we deduce the formal power series (1 − 4z )− 2 = =
m
1

(2m)! 2m m!

m! 2m 2m m

−1 2 (−4)m z m m (−4)m m! − 1 2 − 3 2 ··· − 2m − 1 2 zm . zm

=
m

2m (2m − 1)!! m! 2i 2j j

zm =
m

2m m

Hence, the number
i+j +k=n i,j,k≥0

2k k

i −3 2

is the coeﬃcient of z n in the series

(1 − 4z )− 2 = =

(−4)n z n − 3 2 − 5 2 ··· − 2 2n + 1 2 n+1 zn zn .

(−4)n
n

=
n

2n (2n + 1)!! n!

zn =
n

n + 1 2n + 2

463

We conclude n
i+j +k=n i,j,k≥0

i, j, k

(2i − 1)!!(2j − 1)!!(2k − 1)!! =
i+j +k=n i,j,k≥0

i, j, k i!j !k! 2i i 2j j

i ! 2i 2i i

j ! 2j 2j j

k ! 2k 2k k

2k k

i+j +k=n i,j,k≥0

(n + 1)! 2n + 2 2n+1 n+1

= (2n + 1)!!

which completes the proof.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; MICHEL BATAILLE, Rouen, France; JOEL SCHLOSBERG, Bayside, NY, USA; SKIDMORE COLLEGE PROBLEM SOLVING GROUP, Skidmore College, Saratoga Springs, NY, USA; ALBERT STADLER, Herrliberg, Switzerland; and the proposer.

3576.

[2010 : 459, 461] Proposed by Mehmet S ¸ ahin, Ankara, Turkey.

Let ABC be a triangle with interior points D , E , F such that ∠F AB = ∠EAC , ∠F BA = ∠DBC , ∠DCB = ∠ECA, AF = AE , BF = BD , and CD = CE . If R is the circumradius of ABC , r is the circumradius of EDF , and s is the semiperimeter of ABC , prove that the area of triangle EDF is Solution by John G. Heuver, Grande Prairie, AB. Let Ia , Ib , and Ic be the excentres of ∆ABC . The deﬁnition of points E and F implies that the lines AE and AF are symmetric by reﬂection in the internal bisector AIa of ∠BAC ; that is, AIa is the perpendicular bisector of EF . But AIa is also perpendicular to Ib Ic (which is the external bisector of ∠BAC ), whence EF ||Ib Ic . Analogous statements hold for F D and DE . We deduce ﬁrst that the sides of ∆DEF are parallel to the corresponding sides of ∆Ia Ib Ic , so that the two triangles are similar, and second that I is the circumcentre of ∆DEF . Furthermore, ∠EDF = ∠Ib Ia Ic = 1 (∠B + ∠C ), and, because I 2 is the circumcentre of ∆DEF , ∠EIF = 2∠EDF = ∠B + ∠C . Because the circumradius of ∆DEF is given to be r , similar reasoning for the angles at E
sr2 . 2R

464

and F allows us to deduce that Area(DEF ) = Area(EIF ) + Area(F ID ) + Area(DIE ) 1 = r 2 (sin(B + C ) + sin(C + A) + sin(A + B )) 2 1 1 a b c = r 2 (sin A + sin B + sin C ) = r 2 + + 2 2 2R 2R 2R r2 s = . 2 R
Also solved by ARKADY ALT, San Jose, CA, USA; MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; MICHEL BATAILLE, Rouen, France; PRITHWIJIT DE, Homi Bhabha Centre for Science Education, Mumbai, India; JOEL SCHLOSBERG, Bayside, NY, ¨ USA; MIHA¨ I STOENESCU, Bischwiller, France; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. Our featured solution proves only that if ∆DEF were to exist, its area would equal the predicted value. Bataille and Schlosberg both addressed the question of existence: our featured solution makes clear that reﬂection in AIa takes E to F , in BIb takes F to D , and in CIc takes D back to E . Because the product of three reﬂections is an opposite isometry, while I and E are points that are ﬁxed by this product, this isometry must be a reﬂection in IE . Deﬁne to be the line that makes a directed angle with CIc equal to the directed angle from AIa to BIb . We conclude that it is both necessary and suﬃcient that E be a point of inside ∆ABC diﬀerent from I , for which its reﬂections in AIa and CIc , namely F and D , also lie inside that triangle.

3577.

[2010 : 459,461]Proposed by Mehmet S ¸ ahin, Ankara, Turkey.

Let H be the orthocentre of the acute triangle ABC with A on the ray HA and such that A A = BC . Deﬁne B , C similarly. Prove that Area(A B C ) = 4 Area(ABC ) + Solution by Joel Schlosberg, Bayside, NY, USA. Editor’s comment. The statement of the problem is somewhat ﬂawed. What Schlosberg proves here is the theorem, Let H be the orthocentre of the arbitrary triangle ABC ; deﬁne A to be the unique point satisfying AA = BC that lies on the half-line which starts at A and extends along the line HA in the direction that misses the line BC . Deﬁne B , C similarly. Then Area(A B C ) = 2 2 +c2 4Area(ABC ) + a +b . 2 We have modiﬁed Schlosberg’s proof to make use of directed angles. Recall that ∠XY Z as a directed angle denotes that angle (whose measure ranges from 0◦ to 180◦ ) through which the line XY must be rotated about Y in the positive direction in order to coincide with Y Z . a2 + b2 + c2 2 .

465

Because corresponding sides of ∠BHC and ∠CAB are perpendicular, while ∠ABH and ∠HCA are complements of ∠CAB in right triangles, we have ∠BHC = ∠CAB = 90◦ − ∠ABH = 90◦ − ∠HCA. It follows that Area(HB C ) = 1 HB · HC sin ∠BHC 2 1 = (HB + b)(HC + c) sin ∠BHC 2 1 1 = HB · HC sin ∠BHC + cHB cos ∠ABH 2 2 1 1 + bHC cos ∠HCA + bc sin ∠CAB 2 2 1− 1− → − − → → − − → = Area(HBC ) + BA · BH + CA · CH + Area(ABC ). 2 2 1− 1− − → − − → → − − → CB · CH + AB · AH + Area(ABC ), 2 2 1− 1− → − − → − → − − → AC · AH + BC · BH + Area(ABC ). 2 2

Similarly, Area(HC A ) = Area(HCA) + and Area(HA B ) = Area(HAB ) + Consequently, Area(A B C ) = Area(HB C ) + Area(HC A ) + Area(HA B ) = Area(HBC ) + Area(HCA) + Area(HAB ) + 3Area(ABC ) 1− 1− − → − − → − − → → − − → − − → + BC · (BH + HC ) + CA · (CH + HA) 2 2 1− → − − → − − → + AB · (AH + HB ) 2 1− 1− 1− − → − − → → − → → − → = 4Area(ABC ) + BC · BC + CA · CA + AB · AB 2 2 2 a2 + b2 + c2 = 4Area(ABC ) + . 2
Also solved by ARKADY ALT, San Jose, CA, USA; MIGUEL AMENGUAL COVAS, ˇ ´ University of Sarajevo, Sarajevo, Cala Figuera, Mallorca, Spain; SEFKET ARSLANAGI C, Bosnia and Herzegovina; MICHEL BATAILLE, Rouen, France; PRITHWIJIT DE, Homi Bhabha Centre for Science Education, Mumbai, India; JOHN G. HEUVER, Grande Prairie, ¨ AB; ALBERT STADLER, Herrliberg, Switzerland; MIHA¨ I STOENESCU, Bischwiller, France; ERCOLE SUPPA, Teramo, Italy; PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU, Com´ ane¸ sti, Romania; and the proposer. The proposer’s submitted problem applied to all triangles, but since its statement was somewhat ambiguous the editor restricted it to acute triangles. Unfortunately, even the published statement of 3577 is ambiguous. We hope we got it right this time.

466

3578.
Romania.

[2010 : 459,462] Proposed by Ovidiu Furdui, Campia Turzii, Cluj,

Let a > 0 and b > 1 be real numbers and let f : [0, 1] → R be a continuous function. Find
n→∞

lim na/b

1 0

f (x) 1 + na xb

dx .

Solution by Mohammed Aassila, Strasbourg, France (expanded slightly by the editor). We show that the required limit is
πf (0) ¡. b sin π b
a

Let L denote the given limit. Using the substitution y = n b x, we have n
a b

1 0

f (x) 1 + na xb
Ê∞
0

dx =
0 1 dy . 1+yb
1

f (x) 1 + nbx

a

¡b · n dx =

a b

nb 0

f n− b y
a

1 + yb

dy.

Hence, L = f (0)

Let u = y b so y = u b and dy = L=
Ê∞

1 1 u b −1 du. b
1 b

Then we have (1)
Ê1

f (0) b

∞ 0

u du. u(1 + u)

Let Γ(x) = 0 e−t tx−1 dt, x > 0 and B (x, y ) = 0 tx−1 (1 − t)y−1 dt, x > 0, y > 0 denote the Gamma function and the Beta function, respectively. The following formulae are well known [1]: B (x, y ) = Let u = Γ(x)Γ(y ) Γ(x + y ) and Γ(x)Γ(1 − x) = π sin(πx) . (2)

from (1), (2) and the obvious fact that Γ(1) = 1, we have L= = f (0) b f (0) b ·
1 0

t 1 t . Then du = dt and u(1 + u) = . Hence, 1−t 1 − t2 (1 − t)2

t b −1 (1 − t)− b dt =
1 b

f (0) b ·

B π

1 b
π b

,1−

1 b

Γ(1)

Γ 1−

1 b

f (0) b

sin

which completes the proof.
´ Also solved by MICHEL BATAILLE, Rouen, France; MANUEL BENITO, OSCAR CIAURRI, EMILIO FERNANDEZ, and LUZ RONCAL, Logro˜ no, Spain; KEE-WAI LAU, Hong Kong, China; PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; ALBERT STADLER, Herrliberg, Switzerland; and the proposer. Ê ∞ dt There were three incomplete solutions all of which gave f (0) 0 1+ or its equivalent tb form, as the ﬁnal answer. Stadler remarked that it suﬃces to assume that f (x) is a bounded integrable function which is continuous from the right at 0.

467

References
[1] E.T. Whittaker and G.N.Watson, A course of Mathematical Analysis, Cambridge University Press, 1961.

3579.

[2010 : 459,462] Proposed by Peter Y. Woo, Biola University, La Mirada, CA, USA. Let α = x1 x2 x3
π and 13

= tan(4α) + 4 sin(α) = − tan(α) + 4 sin(3α) , = tan(α) + 4 sin(α) = − tan(4α) + 4 sin(3α) ,

= tan(6α) − 4 sin(6α) = tan(2α) + 4 sin(5α) .

Prove that the length x1 can be constructed with compass and straightedge and determine whether or not the same is true for x2 and x3 . Solution by Stan Wagon, Macalester College, St. Paul, MN, USA. Applying trigonometric expansion with the help of Mathematica, we ﬁnd that √ x1 = 13 − 2 13 which is clearly constructible since it contains only square √ roots. [Ed.: The proposer remarked that x1 = 13 − 2 13 can be proved from the solution to problem # 3305.] On the other hand, using trigonometric expansion and an algorithm to deduce the minimal polynomial satisﬁed by an algebraic number, we learn that each of x2 and x3 is a root of the irreducible polynomial x12 − 78x10 + 1963x8 − 20020x6 + 81991x4 − 138398x2 + 81133. Since it is a classical result that an algebraic number is constructible if and only if the degree of its minimal polynomial is a power of 2, we conclude that x2 and x3 are not constructible.
The proposer gave a partial answer by showing that x1 is constructible. No other solutions were received.

3580. [2010: 460, 462] Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.
Let k > 0 and m ≥ 0 be real numbers, and let {a} = a − a denote the fractional part of a. Calculate
1 0

1 1 − k x (1 − x)k

xm (1 − x)m dx .

Solution by Michel Bataille, Rouen, France.

468

Let I denote the integral. The substitution u = 1 − x yields
0

I =
1 1

1 u)k

=
0

(1 − uk 1 1 − k (1 − x)k x

−

um (1 − u)m (−du) xm (1 − x)m dx .

Thus 2I =

1 0

[{φ(x)} + {−φ(x)}] xm (1 − x)m dx ,

1 1 where φ(x) := (1− −x k . x)k Since φ(x) is continuous, strictly increasing on (0, 1), and limx→0+ φ(x) = −∞ ; limx→1− φ(x) = +∞, it follows that φ is a bijection from (0, 1) onto R. Thus, for each integer p ∈ Z, the equation φ(x) = p has exactly one solution ap ∈ (0, 1). It is easy to see that

0 {a} + {−a} = Thus 1

if a ∈ Z if a ∈ /Z

[{φ(x)} + {−φ(x)}] xm (1 − x)m = xm (1 − x)m outside the countable set {ap |p ∈ Z}. Hence I= 1
1 0 1 0

2 1 = 2

[{φ(x)} + {−φ(x)}] xm (1 − x)m dx xm (1 − x)m dx = (m!)2 (Γ(m + 1))2 = . 2Γ(2m + 2) 2(2m + 1)!

Also solved by PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; ALBERT STADLER, Herrliberg, Switzerland; and the proposer.

3582.

[2010 : 460, 462] Proposed by Panagiote Ligouras, Leonardo da Vinci High School, Noci, Italy.

Let Γ1 , Γ2 be circles of radius r with centres A, B (respectively), let {C , D } = Γ1 ∩ Γ2 , and suppose that ∠BCA = 90◦ . A line through C intersects Γ1 and Γ2 again at E and F , respectively. The circle Γ with centre O and radius R passes through points E and F . A second line passes through C , is perpendicular to the segment EF , and intersects the circle Γ in G and H . Prove that CH 2 + CG2 = 4(R2 − r 2 ).

469

Solution by Peter Y. Woo, Biola University, La Mirada, CA, USA. Let M be the midpoint of EC , N the midpoint of CF , a = AM , and b = BN . Because corresponding sides of ∆AM C and ∆CN B are perpendicular, while BC = AC = r , the triangles are congruent. In particular, a = CN = N F and b = M C = EM , so that EF = 2(b + a) and r 2 = a2 + b2 . Furthermore, if P is the midpoint of EF , then EP = b + a so that M P = a and P C = b − a. Next, let Q be the midpoint of GH and h = P O (= CQ). Then OQ = P C = b − a, whence QH 2 = R2 − (b − a)2 . Moreover, CH 2 + CG2 = (h + QH )2 + (h − QH )2 = 2 QH 2 + h2 . But R2 = OE 2 = h2 + (b + a)2 , hence h2 = R2 − (b + a)2 . Assembling the pieces, we conclude that CH 2 + CG2 = 2
ä ä

R2 − (b − a)2 + R2 − (b + a)2
ç

= 2 2R2 − 2 b2 + a2 as claimed.

= 4R2 − 4r 2 ,

Also solved by MICHEL BATAILLE, Rouen, France; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; JOEL SCHLOSBERG, Bayside, NY, USA; and the proposer.
F C r h A Q O O Q B N b E A a a r P M F C N b G h r B

G P r

b E

M a

Only Bataille observed that there arise two cases: C can lie between E and F (as in the ﬁrst diagram), or not between them (as in the second diagram). If we use directed line segments

470

along the lines EF and GH (so that XY = −Y X for points X and Y both on one of these lines or on a line parallel to one of them), then Woo’s notation has been modiﬁed by the editor so that his argument deals with both cases simultaneously. Otherwise, one must observe that when C is not between E and F (and therefore not between G and H ) and the diagram is labeled so that b ≥ a, then P E = b − a and P C = b + a, and the featured argument goes through without diﬃculty.

3583.

[2010: 460, 463] Proposed by Paolo Perfetti, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy. Let α and β be nonnegative real numbers and deﬁne
n

an pn

= =

n + ln(n + 1)
k=1

α + k + ln k β + (k + 1) + ln(k + 1)
n

, .

α + n + 1 + ln(n + 1)
k=1

α + k + ln k β + (k + 1) + ln(k + 1)
∞ È n=1

Find those nonnegative real numbers α and β for which determine the relation between α and β that ensures that
∞ n=1

an converges, and (α + 1)2 . 2

an − pn ln 1 +

1 n+1

= (α + 1)(α + 2 + ln 2) −

Solution by the proposer. We claim that the series converges if and only if β > α + 1. Let’s start by observing that α + n + 1 + ln(n + 1) β + n + 2 + ln(n + 2) and =1+ α − β − 1 + ln 1 −
1 n+2

β + 2 + n + ln(n + 2)

1 1 + ln 1 + n+1 n + 1 + ln(n + 2) =1+ . n + ln(n + 1) n + ln(n + 1)

Hence an+1 an = = α + n + 1 + ln(n + 1) n + 1 + ln(n + 2) β + n + 2 + ln(n + 2) 1+ n + ln(n + 1)
1 n+2

α − β − 1 + ln 1 −

β + 2 + n + ln(n + 2)
1 n+2

· + ·

1 + ln 1 +
1 n+1

1 n+1

n + ln(n + 1)

=1+ +

α − β − 1 + ln 1 −

1 + ln 1 +

β + 2 + n + ln(n + 2)
1 n+2

n + ln(n + 1)
1 n+1

α − β − 1 + ln 1 −

1 + ln 1 +

β + 2 + n + ln(n + 2)

n + ln(n + 1)

471

In particular an+1 an = − 1+ α−β n
1 n+2

α − β − 1 + ln 1 −

β + 2 + n + ln(n + 2) α − β − 1 + ln 1 −

−
1 n+2

α−β−1 n ·

1 + ln 1 +
1 n+1

1 n+1

n + ln(n + 1)

−

1 n

β + 2 + n + ln(n + 2) n + ln(n + 1) 1 1 1 1 − − = (α − β − 1) + β + 2 + n + ln(n + 2) n n + ln(n + 1) n + + α − β − 1 + ln 1 − ln 1 −
1 n+2 1 n+2

1 + ln 1 +

β + 2 + n + ln(n + 2)

1 + ln 1 +
1 n+1

1 n+1

n + ln(n + 1)

+ β + 2 + n + ln(n + 2) n + ln(n + 1) −β − 2 − ln(n + 2) ln(n + 1) = (α − β − 1) 2 − 2 n + βn + 2n + n ln(n + 2)) n + n ln(n + 1) + + =O α − β − 1 + ln 1 − ln 1 −
1 n+2 1 n+2

ln 1 +

β + 2 + n + ln(n + 2) +

1 + ln 1 +
1 n+1

1 n+1

n + ln(n + 1)

ln 1 +

β + 2 + n + ln(n + 2) ln(n) n2

n + ln(n + 1)

Thus, we get an+1 an =1+ α−β n +O ln(n) n2 . (1)

Convergence for β > α + 1. For all n ≥ 2 set bn = n ln1 2 (n) . Then n ln2 (n) bn+1 (n + 1) ln2 (n + 1) − n ln2 (n) = = 1 − bn (n + 1) ln2 (n + 1) (n + 1) ln2 (n + 1) =1− (n + 1) ln2 (n + 1) (n + 1) ln2 (n + 1) 1 (n ln(n + 1) − n ln(n)) (ln(n + 1) + ln(n)) − =1− (n + 1) (n + 1) ln2 (n + 1)
1 ln 1 + n (ln(n + 1) + ln(n)) 1 =1− − , (n + 1) (n + 1) ln2 (n + 1) n

(n + 1) ln2 (n + 1) − n ln2 (n + 1)

−

n ln2 (n + 1) − n ln2 (n)

472

so bn+1 bn =1− =1− − 1 n 1 n + + 1 n(n + 1) 1 n(n + 1)
1 n n

− −

ln 1 +

1 n n

(ln(n + 1) + ln(n))

(n + 1) ln2 (n + 1) 2 (n + 1) ln(n + 1)

ln 1 +

(n + 1) ln2 (n + 1) 1 1 ln(n) + ln(n + 1) =1− + − n n(n + 1) (n + 1) ln(n + 1) 1 n (1+ n ) ln (ln(n + 1) + ln(n)) e − (n + 1) ln2 (n + 1) 2 1 ln(n) +O =1− − n (n + 1) ln(n + 1) n2

(ln(n + 1) + ln(n)) − 2 ln(n + 1)

(2)

Since β > α + 1, by combining (1) and (2) we get that there exists some N so that for all n > N we have 0< Hence, for all n > N we have an+1 bn+1 It follows that the sequence < an bn . an+1 bn+1 ≤ . an bn

from above. Hence, there exists an M so that, for all n we have 0 < an < M bn . Since the series
È∞
n=2

an is eventually decreasing, and hence bounded bn

Comparison Test we get that Divergence for β ≤ α + 1. For all n ≥ 2 set cn = cn+1 cn =1− 1 n

1 is convergent by the integral test, by the n ln2 n È
n

an is convergent.

1 . Exactly as in (2) one can show that n ln2 (n)

−

1 (n + 1) ln(n + 1)

ln(n) n2

Thus, there exists some N so that, for all n > N we have an+1 an ≥ cn+1 cn > 0.

473

Hence, for all n > N we have an+1 cn+1 It follows that the sequence so that, for all n we have > an cn > 0.

an is eventually increasing. Hence, there exists an m cn

an > mcn . Moreover, m can be chosen as the smallest term of so that m > 0. È∞ 1 is divergent, by the Comparison Test we get Since the series n=2 n ln n È that n an is also divergent. This completes the ﬁrst part of the problem. For the second part, let’s observe ﬁrst that
n

an , and hence it can be chosen cn

an = (α + n + 1 + ln(n + 1))
k=1 n

α + k + ln(k) β + k + 1 + ln(k + 1)

− (α + 1) Deﬁne qn :=

k=1 n

α + k + ln(k) . β + k + 1 + ln(k + 1) α + k + ln(k)

k=1

β + k + 1 + ln(k + 1)

Then an = pn − (α + 1)qn . Since by the deﬁnition of qn we have 0 < qn ≤ an , it follows that È q n n are absolutely convergent. Let’s note that qn+1 qn In particular β (qn+1 − qn ) + (β − α)qn + (n + 2)qn+1 − qn (n + 1) = qn ln(n + 1) − qn+1 ln(n + 2) . Since
È È
n

pn and

α + n + 1 + ln(n + 1) β + n + 2 + ln(n + 2)

(3)

(4)

an is convergent, we are in the case β > α + 1. Thus qn+1 qn < β + n + ln(n + 1) β + n + 2 + ln(n + 2) < 1,

and hence qn is decreasing. We will employ the following well known lemma:

474
È

Lemma: Let qn be monotonic and positive. If lim nqn = 0 .
n

qn is convergent, then

Let’s denote

È
n

qn = U . By summing (4) from 1 to N we get
N

β (qN +1 − q1 ) + (β − α)

n=1

qn + (N + 2)qN +1 − 2q1 (n + 1) = q1 ln(2) − qN +1 ln(N + 2) .

Using the above Lemma and letting N → ∞ we get: −βq1 + (β − α)U − 2q1 = q1 ln(2) . Since q1 =
α+1 we get β + 2 + ln 2

U =

(β + 2 + ln(2))q1 α+1 = . β−α β−α

Now, let’s observe that pn+1 pn and hence β (pn+1 − pn ) + (β − α)pn + [(n + 2)pn+1 − (n + 1)pn ] − pn By summing we get
N N

α + n + 2 + ln(n + 2) β + n + 2 + ln(n + 2)

= [pn ln(n + 1) − pn+1 ln(n + 2)] + [pn ln(n + 2) − pn ln(n + 1)]

β (pN +1 − p1 ) + (β − α)

n=1

pn + [(N + 2)pN +1 − 2p1 ] −
N

pn
n=1

= [p1 ln(2) − pN +1 ln(N + 2)] + Again using the Lemma, and letting N → ∞ we get: −βp1 − 2p1 + (β − α) Since p1 = (α + 2 + ln(2))
∞ n=1

pn ln 1 +
n=1

1 n+1

pn −

∞ n=1

pn = p1 ln(2) +
n=1

pn ln 1 +

1 n+1

α+1 , β + 2 + ln(2)

475

we get (β − α)
∞ n=1

pn −

∞ n=1

pn −

pn ln 1 +
n=1

1 n+1

= (α + 2 + ln(2))(α + 1) ,

and hence
∞ n=1

pn =

1 β−α−1

pn ln 1 +
n=1

1 n+1

(α + 2 + ln(2))(α + 1) β−α−1

This shows that
∞ n=1

an − pn ln 1 +
∞

1 n+1
∞ n=1

= =

n=1

pn −

∞ n=1

(α + 1)qn −

pn ln 1 +
∞ n=1

1 n+1 pn ln 1 + 1 n+1 − (α + 1)2 β−α .

(α + 2 + ln(2))(α + 1) β−α−1

β−α−1

β−α−2

Thus, if β = α + 2, the second formula holds.
No other solution was received.

3584.

ˇ [2010: 460, 463] Proposed by Sefket Arslanagi´ c, University of Sarajevo, Sarajevo, Bosnia and Herzegovina. Let ABC be a triangle with inradius r , side lengths a, b, c and medians √
c a b 3 3 + 2 2 + 2 2 ≤ . 2 27r3 m2 mb mc mc ma a mb

ma , mb , mc . Prove that

Solution by Michel Bataille, Rouen, France. We ﬁrst observe that m2 a = = where p =
a+b+c 2

2b2 + 2c2 − a2 4 (b + c)2 − a2 4

≥

b2 + 2bc + c2 − a2 4

= p(p − a) ,

is the semiperimeter of the triangle ABC . Similarly we get
2 m2 b ≥ p(p − b) ; mc ≥ p(p − c) .

476

Thus, we get c
2 m2 a mb

a
2 m2 b mc

b
2 m2 a mc

≤

c p2 (p − a)(p − b)

− b)(p − c) b + 2 p (p − a)(p − c) c(p − c) + a(p − a) + b(p − b) = p2 (p − a)(p − b)(p − c)

a p2 (p

Let S denote the area of the triangle ABC . Since p(p − a)(p − b)(p − c) = S 2 = r 2 p2 and a(p − a) + b(p − b) + c(p − c) = 2[abc + (p − a)(p − b)(p − c)] p ,

to complete the proof we need to show that abc + (p − a)(p − b)(p − c) ≤ √ 3 p4 (1)

27r

By applying the AM-GM inequality to both terms on the left side we get abc + (p − a)(p − b)(p − c) ≤ a+b+c 3 + = Also r 2 p2 = p(p − a)(p − b)(p − c) ≤ p and hence Thus abc + (p − a)(p − b)(p − c) ≤ which proves (1).
Also solved by ARKADY ALT, San Jose, CA, USA; KEE-WAI LAU, Hong Kong, China; ˇ ´ Gornji Milanovac, Serbia; and the proposer. DRAGOLJUB MILO SEVI C,
3

(p − a) + (p − b) + (p − c) 3 = p3 3

8 p3 27

p3 27

(p − a)(p − b)(p − c) 3

≤

p4 27

√ 3 3r ≤ p . p3 p4 p4 = ≤ √ , 3 3p 9 3r

477

3585.

[2011 : 461,463] Proposed by Arkady Alt, San Jose, CA, USA.

Let Tn (x) be the Chebyshev polynomial of the ﬁrst kind deﬁned by the recurrence Tn+1 (x) = 2xTn (x) − Tn−1 (x) for n ≥ 1 and the initial conditions T0 (x) = 1 and T1 (x) = x. Find all positive integers n such that Tn (x) ≤ (2n−2 + 1)xn − 2n−2 xn−1 , x ∈ [1, ∞) . Solution by Albert Stadler, Herrliberg, Switzerland. The given recurrence deﬁning Tn (x) implies that it is a polynomial of degree n, whose leading coeﬃcient is 2n−1 for all n ≥ 1. If the required inequality is to hold for all x ≥ 1 at a particular positive integer n, then necessarily the leading coeﬃcient of Tn (x) must be at most 2n−2 + 1. Thus 2n−1 ≤ 2n−2 + 1, which implies that 2n−2 ≤ 1, which is true only when n = 1 or n = 2. 1 x− 2 , which clearly holds With n = 1, the inequality demands that x ≤ 3 2 for all x ≥ 1. However, with n = 2, the inequality demands that 2x2 − 1 ≤ 2x2 − x, which clearly fails for all x > 1. Thus n = 1 is the only positive integer with the required property.
Also solved by MICHEL BATAILLE, Rouen, France.

3586.

[2010 : 461,463] Proposed by Shai Covo, Kiryat-Ono, Israel.

For each positive integer n, an is the number of positive divisors of n of the form 4m + 1 minus the number of positive divisors of n of the form 4m + 3 (so a4 = 1, a5 = 2, and a6 = 0). Evaluate the sum Solution by the proposer. We show that
∞ ∞ È n=1

(−1)n+1

an . n

n=1

(−1)n+1

an n

π ln 2 4

(1)

To prove (1) we ﬁrst establish the following lemma: Lemma: Let xk = (−1)k+1 k and yk = (−1)k+1 (2k − 1), for each integer k ≥ 1. Then
n→∞

lim

1 1
|xi yj |≤n

x i yj

= √ 1 xi

∞ k=1

1 xk

∞ k=1

1 yk

π ln 2 4

Proof: For any ﬁxed n let m = 1 1 = x i yj 1
|xi yj |≤n n i=1

n . Then 1 yj
m n

=
i=1

bi +
i=m+1

|xi yj |≤n

where bi =

1 xi

478

For each ﬁxed i = 1, 2, · · · , m, let
|xi yj |≤n

1 π = + Ei . yj 4

Using the fact that

¬ ¬ ¬ ¬ k ¬π ¬ 1 1 j +1 ¬ − ¬≤ (−1) ¬4 ¬ 2j − 1 ¬ 2k + 1 ¬ j =1

we have |Ei | ≤ Thus,
¬ ¬ ¬ ¬ ¬
m i=1

1 2k + 1 π 4
m i=1

and (2k + 1)i > n.
¬ ¬ ¬ Ei ¬ m ¬ ¬< . ¬≤ ¬ ¬ ¬ xi xi n i=1

bi −

1¬ ¬

It follows that
n→∞

lim

bi =
i=1

π 4

∞ i=1

1 xi

π 4

ln 2.

Hence, it suﬃces to show that
n n→∞

lim

bi = 0.
i=m+1

(2)

We consider the sums |bm+1 + bm+2 |, |bm+3 + bm+4 |, · · · . (Note that 1 → 0 as n → ∞.) |bn | = n Now deﬁne w(i) = max |yj |
|xi yj |≤n

and consider an arbitrary but ﬁxed i ∈ {m + 1, m + 2, · · · } where i ≤ n − 1. If w(i) = w(i + 1), then
¬ ¬ ¬ ¬¬ ¬ ¬ ¬ ¬1 1 1 ¬ 1 ¬ ¬ ¬ ¬≤ |bi + bi+1 | = ¬ − . ¬ ¬ ¬ i+1 i(i + 1) ¬|xi yj |≤n yj ¬ i

(3)

If w(i) = w(i +1) then w(i +1) ≤ w(i) − 2 since clearly w(i +1) ≤ w(i) and both are odd integers. We now show that w(i + 1) ≥ w(i) − 2. Note ﬁrst that √ iw(i) ≤ n ⇒ (m + 1)w(i) ≤ n ⇒ n + 1 w(i) ≤ n √ √ ⇒ nw(i) < n ⇒ w(i) < n √ ⇒ w(i) ≤ n = m ⇒ w(i) ≤ i − 1 < 2(i + 1) (4)

479

so (i + 1)(w(i) − 2) = iw(i) + w(i) − 2(i + 1) < n. Returning now to the proof of (4) which clearly holds if w(i) = 1 since w(i + 1) ≥ 1. If w(i) ≥ 3, then w(i) − 2 > 0 and is odd. Since w(i) − 2 is of the form |yj | for some j and since w(i + 1) = max|xi+1 yj |≤n |yj |, we have (i+1)w(i+1) ≥ (i+1)|yj | = (i+1)(w(i)−2) from which w(i+1) ≥ w(i)−2 follows, establishing (4). Therefore, w(i + 1) = w(i) − 2. Using (3) we obtain |bi + bi+1 | ≤ 1 i(i + 1) 1 + 1 |xi w(i)| 1 = 1 + 1 iw(i) (5)

i(i + 1) 1 = + 1+ i(i + 1) (i + 1)w(i) i

If (i + 1)w(i) ≤ n, then |xi+1 |w(i) ≤ n would imply that |xi+1 ||yj | ≤ n for all yj such that |xi ||yj | ≤ n so w(i) < w(i + 1), a contradiction. Hence, (i + 1)w(i) > n. From (5) and (6) we obtain 1 i(i + 1) 2 n (6)

|bi + bi+1 | <

(7)

However, noting that w(m + 1) ≤ m so the number of indices i for which w(i + 1) = w(i) − 2 is bounded above by m. Since both
1 m 1 + + · · · and tend to 0 as (m + 1)(m + 2) (m + 3)(m + 4) n
n n→∞

n → ∞ we conclude that lim the proof.

bi = 0 establishing (2) and completing
i=m+1

Also solved by ALBERT STADLER, Herrliberg, Switzerland; who gave a 3-page proof based on the Dirichlet L-function associated with the non-trivial character (mod 4), and analytic continuation of some complex function deﬁned by Dirichlet series.

480

3587 . [2010 : 461,463] Proposed by Ignotus, Colegio Manablanca, Facatativ´ a, Colombia.
Deﬁne the prime graph of a set of positive integers as the graph obtained by letting the numbers be the vertices, two of which are joined by an edge if and only if their sum is prime. (a) Prove that given any tree T on n vertices, there is a set of positive integers whose prime graph is isomorphic to T . (b) For each positive integer n, determine t(n), the smallest number such that for any tree T on n vertices, there is a set of n positive integers each not greater than t(n) whose prime graph is isomorphic to T . No solutions have been received so this problem remains open.

Crux Mathematicorum
with Mathematical Mayhem
Former Editors / Anciens R´ edacteurs: Bruce L.R. Shawyer, James E. Totten, V´ aclav Linek

Crux Mathematicorum
Founding Editors / R´ edacteurs-fondateurs: L´ eopold Sauv´ e & Frederick G.B. Maskell Former Editors / Anciens R´ edacteurs: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

Information for Contributors
Readers are invited to send their solutions to any of the problems published in Crux Mathematicorum with Mathematical Mayhem. We also welcome your problem proposals and articles, as well as comments and suggestions. For the problems in Skoliad or Mathematical Mayhem, we are looking especially for solutions from pre-university students. On any correspondence, please give your name, school or other aﬃliation (if applicable), city, province or state, and country. Pre-university students should also include their grade in school. Please address your correspondence to the appropriate editor, as listed below. Correspondence may be sent by email or regular mail (but not by FAX). Material sent 1 × 11 or A4 paper. by regular mail should be typed or neatly hand-written on standard 8 2 For problem proposals and solutions, please sign each page, and use a new page when beginning a new problem. For email submissions, it would be appreciated if LaTeX could be used, but MSWord, WordPerfect, PDF ﬁles, and PostScript are also acceptable. Graphics ﬁles should be in epic format or encapsulated PostScript.

Articles should be carefully written, reasonably short, and expository in nature. Mayhem
articles should be at a level that could be understood by mathematically advanced pre-university students. When an article is accepted for publication, an author will be sent a “Consent to Publish and Transfer of Copyright” form, the return of which, duly completed, is a requirement for publication. Authors will be sent reprints in PDF format. CRUX articles should be sent to Robert Dawson, Department of Mathematics and Computing Science, Saint Mary’s University, Halifax, NS, Canada, B3H 3C3, or emailed to [email protected] For Mayhem articles, see Mathematical Mayhem below.

Skoliad: Contest proposals and solutions should be sent to Lily Yen and Mogens Hansen, 7255
Hewitt Street, Burnaby, BC, V5A 3M3, Canada, or emailed to [email protected] Mathematical Mayhem: Mayhem articles, proposals, and solutions should be sent to Shawn Godin, Cairine Wilson S.S, 975 Orleans Blvd, Orleans, ON, Canada, K1C 2Z5, or emailed to [email protected]

Olympiad Corner: All communication should be directed to Professor Nicolae Strungaru,
Department of Mathematics and Statistics, Grant MacEwan University, 10700 - 104 Avenue, Edmonton, AB, Canada, T5J 4S2, or emailed to [email protected]

CRUX Problems and Solutions: Problem proposals and solutions should be sent to
Shawn Godin, Cairine Wilson S.S, 975 Orleans Blvd, Orleans, ON, Canada, K1C 2Z5 or emailed to [email protected] Each problem proposal should be accompanied by a solution, or at least suﬃcient information to indicate that a solution is likely to be found. Please include any references or insights that might help the editor. Original problems are especially solicited. However, other interesting problems may be acceptable if they are not too well known and they are accompanied by appropriate references. Ordinarily, if the originator of a problem can be located, the originator’s permission should be obtained before the problem is submitted.

Acknowledgements
The support of the Ottawa Carleton District School Board is gratefully acknowledged.

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37: No 8
Published by:

December / D´ ecembre 2011

- Concours de L’Association math´ ematique de Qu´ ebec, 2011 Ordre secondaire - Mathematics Association of Quebec Contest, 2011 Secondary level - Solutions to questions of the Baden-W¨ urttemberg Mathematics Contest, 2010 488 Mathematical Mayhem Shawn Godin

488 Mayhem Year End Wrap Up 489 Mayhem Problems: 491 Mayhem Solutions: M513–M518 M476–M481 R.E. Woodrow and Nicolae Strungaru OC51–OC60

495 The Olympiad Corner: No. 298 495 Olympiad Corner Problems:

In this Corner are solutions from readers to some problems from - Croatian Mathematical Competition 2007, National Competition, 4th Grade - 51st National Mathematical Olympiad in Slovenia, Selection Examinations - Correspondence Mathematical Competition in Slovakia 2006/7 First Round, First Set - Latvian School Mathematical Olympiad, Grade 11 - Latvian Mathematical Olympiad, Grade 12 - Finnish National High School Mathematics Competition, Final Round - IX Olimpiada Matem´ atico de Centram´ erica y el Cariba, 2007 526 Book Reviews Amar Sodhi

526 Loving + Hating Mathematics: Challenging the Myths of Mathematical Life by Reuben Hersh and Vera John-Steiner

528 The Beauty of Fractals: Six Diﬀerent Views edited by Denny Gulick and Jon Scott Reviewed by Daryl Hepting 529 Recurring Crux Conﬁgurations 4 : J. Chris Fisher

This new, occasionally appearing column, highlights situations that reappear in Crux problems. In this issue problem editor J. Chris Fisher examines bicentric quadrilaterals. Enjoy! 535 That old root ﬂipping trick of Andrey Andreyevich Markov Gerhard J. Woeginger The author illustrates how a straightforward fact about the roots of certain quadratic equations can be used to solve a variety of questions. Examples include many problems from various mathematical Olympiads. 540 Problems: 3670, 3688–3700 This month’s “free sample” is:

3689.

Propos´ e par Ivaylo Kortezov, Institut de Math´ ematiques et Informatique, Acad´ emie des Sciences, Soﬁa, Bulgarie. Dans un groupe de n personnes, chacune poss` ede un livre diﬀ´ erent. Disons qu’une paire de personnes op` ere un ´ echange si elles s’´ echangent leur livre pr´ esentement en leur possession. Trouver le plus petit nombre possible d’´ echanges E (n), de sorte que chaque paire de personnes a proc´ ed´ e ` a au moins un ´ echange et que ﬁnalement chaque personne se retrouve avec son livre de d´ epart. .................................................................

3689.

Proposed by Ivaylo Kortezov, Institute of Mathematics and Informatics, Bulgarian Academy of Sciences, Soﬁa, Bulgaria. In a group of n people, each one has a diﬀerent book. We say that a pair of people performs a swap if they exchange the books they currently have. Find the least possible number E (n) of swaps such that each pair of people has performed at least one swap and at the end each person has the book he or she had at the start. 545 Solutions: 3589–3600 560 YEAR END FINALE 562 Index to Volume 37, 2011

481

SKOLIAD

No. 137

Lily Yen and Mogens Hansen
Please send your solutions to problems in this Skoliad by December 15, 2012. A copy of CRUX with Mayhem will be sent to one pre-university reader who sends in solutions before the deadline. The decision of the editors is ﬁnal. Our contest this month is the Mathematics Association of Quebec Contest, Secondary level, 2011. Our thanks go to Denis Lavigne, Royal Military College Saint-Jean, Quebec, for providing us with this contest and for permission to publish it.

Concours de l’Association math´ ematique du Qu´ ebec, 2011 Ordre secondaire
Dure : 3 heures

Otto aime tellement les palindromes (les nombres qui demeurent les mˆ emes lorsqu’on inverse l’ordre de leurs chiﬀres) qu’il a concoct´ e l’alpham´ etique suivant : AMQMA × 6 = LUCIE .

Trouver les valeurs des huit chiﬀres. (N. B. Un alpham´ etique est un petit casse-tˆ ete math´ ematique qui consiste en une ´ equation o` u les chiﬀres sont remplac´ es par des lettres. Le r´ esoudre consiste a ` trouver quelle lettre correspond a ` quel chiﬀre pour que l’´ equation soit vraie. Dans le probl` eme, le mˆ eme chiﬀre ne peut ˆ etre repr´ esent´ e par deux lettres diﬀ´ erentes et une lettre repr´ esente toujours le mˆ eme chiﬀre. Bien entendu, un nombre ne doit jamais commencer par z´ ero. Par exemple, l’alpham´ etique PAPA + PAPA = MAMAN a pour solution P = 7, A = 5, M = 1 et N = 0. Ainsi, en rempla¸ cant les lettres par les chiﬀres, on a bien 7575 + 7575 = 15150.)

2. Anik roule toujours a ` 108 km/h sur l’autoroute. En se rendant a ` un concours de
math´ ematique, elle d´ epasse un train qui longe l’autoroute et qui se dirige dans le mˆ eme sens qu’elle. Elle remarque qu’elle met exactement 77 secondes a ` le d´ epasser, i.e. franchir la distance qui s´ epare la queue du train et sa tˆ ete. Arriv´ ee a ` destination, elle r´ ealise qu’elle a oubli´ e sa calculatrice, alors elle rebrousse chemin. Elle recroise alors le train, qui roule a ` la mˆ eme vitesse, et cette fois, elle met exactement sept secondes a ` parcourir le train de la tˆ ete a ` la queue. Quelle est la longueur du train ? ` un coin de rue, le feu de circulation reste vert pendant 30 secondes et rouge 3. A pendant 30 secondes (on suppose le temps du feu jaune inclus a ` mˆ eme le temps du feu vert). Combien de temps perd-on, en moyenne, a ` attendre a ` ce coin de rue ? Justiﬁer.

4. Combien y a-t-il de nombres entiers entre 0 et 999 (inclusivement) dont l’´ ecriture d´ ecimale ne contient aucun 7 ? Quelle est la somme de ces nombres ?

482

5. Dans un cercle de rayon r, deux cordes AB et CD se coupent perpendiculairement en X . Montrer que
A

C X B

|XA|2 + |XB |2 + |XC |2 + |XD |2 = 4r 2 .

1733 = 5 177 717, 1923 = 7 077 888 et 13093 = D 2 242 946 629 sont trois exemples d’entiers N dont le cube compte le mˆ eme nombre de chiﬀres diﬀ´ erents que N lui-mˆ eme. Mais existe-t-il des entiers qui contiennent plus de chiﬀres diﬀ´ erents que leur cube ? Oui : le nombre 13 798 compte cinq chiﬀres diﬀ´ erents tandis que son cube, 2 626 929 525 592 n’en compte que quatre (2 ; 5 ; 6 et 9). On dira d’un tel nombre (quand il contient plus de chiﬀres diﬀ´ erents que son cube) qu’il est d´ eﬁcient. Montrer qu’il y a une inﬁnit´ e d’entiers d´ eﬁcients.

7. Sachant que le syst` eme d’´ equations
x=
Ô

11 − 2yz,

√ 12 − 2xz,

et z =

13 − 2xy

poss` ede des solutions r´ eelles, que vaut x + y + z ?

Mathematics Association of Quebec Contest, 2011 Secondary level
3 hours allowed

Otto likes palindromes (numbers that read the same forwards and backwards) so much that he has constructed this alphametic: AMQMA × 6 = LUCIE . Find the values of the eight digits. (Recall that an alphametic is a small mathematical puzzle consisting of an equation in which the digits have been replaced by letters. The task is to identify the value of each letter in such a way that the equation comes out true. Diﬀerent letters have diﬀerent values, diﬀerent digits are represented by diﬀerent letters, and no number begins with a zero. For example, the alphametic PAPA + PAPA = MAMAN has the solution P = 7, A = 5, M = 1, and N = 0, yielding 7575 + 7575 = 15150.) Anik is going 108 km/h on the highway. On her way to a math contest, she passes a train that travels beside the highway in the same direction as Anik. She notices that it takes her exactly 77 seconds to pass the train from the rear to the front. Upon arrival, she ﬁnds that she has forgotten her calculator and turns back. She again passes the train, which still travels at the same speed. This time it takes her seven seconds to pass from the front of the train to the rear. How long is the train?

483

3. At an intersection, the traﬃc light is red for 30 seconds and green for 30 seconds. (Ignore the yellow light.) How long do you have to wait, on average, at the intersection? Justify your answer. 4. How many integers from 0 to 999 (inclusive) do not contain the digit 7? What
is the sum of these numbers?

In a circle with radius r , the two chords AB and CD intersect at a right angle at X . Show that
A

C X B

|XA|2 + |XB |2 + |XC |2 + |XD |2 = 4r 2 .

The cubes 1733 = 5 177 717, 1923 = 7 077 888 and D 13093 = 2 242 946 629 are examples of a whole number N 3 3 that contains as many diﬀerent digits as its cube, N . If N contains fewer diﬀerent digits than N , then N is said to be deﬁcient. For example, 13 798 has ﬁve diﬀerent digits, while its cube, 2 626 929 525 592, has four (2, 5, 6, and 9), so 13 798 is deﬁcient. Show that there are inﬁnitely many deﬁcient whole numbers.

If x, y , and z are real numbers such that Ô √ x = 11 − 2yz, y = 12 − 2xz,

and z =

13 − 2xy

what is the value of x + y + z ?

Next follow solutions to the Baden-W¨ urttemberg Mathematics Contest, 2010, given in Skoliad 134 at [2011:259–260].

Sonja has nine cards on which the nine smallest two-digit prime numbers are printed. She wants to order these cards in such a way that neighbouring cards always diﬀer by a power of 2. In how many ways can Sonja order her cards? Solution by Elisa Kuan, student, Meadowridge School, Maple Ridge, BC. The ﬁrst nine two-digit primes are 11, 13, 17, 19, 23, 13 17 29, 31, 37, and 41. The relevant powers of 2 are 1, 2, 4, 11 8, 16, and 32. In the ﬁgure, those primes that diﬀer by a 19 power of 2 are connected. Since 41 is only connected to one 41 other prime, any arrangement of these primes must begin 23 (or end) 41, 37, 29. 37 If the arrangement then continues with 31, it must 29 31 go on as 41, 37, 29, 31 , 23, 19. At 19 you again have a choice: 41, 37, 29, 31, 23, 19 , 17, 13, 11 or 41, 37, 29, 31, 23, 19 , 11, 13, 17. Both of these arrangements work out (that is, they use all nine primes). If the arrangement instead has 13 following 29, you immediately have the choice: 11 or 17. In either case, you must go on to 19, so now you have 41, 37, 29, 13, 11, 19 or 41, 37, 29, 13, 17, 19. If you here go to 17 or 11 (whichever is

484

available), then 23 and 31 become stranded. If, on the other hand, you continue as 41, 37, 29, 13, 11/17, 19 , 23, 31, then 11 or 17 becomes stranded. Thus, the only possible arrangements are: 41, 37, 29, 31, 23, 19, 17, 13, 11; 41, 37, 29, 31, 23, 19, 11, 13, 17; and these reversed, so Sonja can arrange her cards in four ways.
´ Also solved by LENA CHOI, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; GESINE GEUPEL, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany; ´ RICHARD I. HESS, Rancho Palos Verdes, CA, USA; ROWENA HO, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; and SZERA PINTER, student, Moscrop Secondary School, Burnaby, BC.

2. A 50 cm by 30 cm by 28 cm box contains wooden blocks that all measure 10 cm by 9 cm by 7 cm. At most how many blocks can ﬁt in the box? Explain how to ﬁt that many blocks into the box.
Solution by Jay Chau, student, Burnaby Mountain Secondary School, Burnaby, BC.
10 10 10

9 9 9
10

7 7 7 7 9 9 9 9 7 7
10 10

The volume of the box is 50 · 30 · 28 cm3 = 42 000 cm3 , and the volume of each 000 block is 10 · 9 · 7 cm3 = 630 cm3 , so there is room for at most 42 = 200 = 630 3 2 66 3 = 66 blocks. The ﬁgure shows that ﬁtting 66 blocks is indeed possible.
Also solved by GESINE GEUPEL, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany. You may well wonder, how our solver found a way to squeeze all 66 blocks into the box. As a ﬁrst try, you might line the 10-side of the blocks against the 50-side of the box and the 7-side against the 28-side since both work out perfectly. However, this leaves room for only 30 = 3 layers, so only 5 · 4 · 3 = 60 blocks. The trouble is that the 9-side of the blocks does 9 not ﬁt perfectly against any side of the box. The 50-side has the most wiggle room, so ﬁtting as many 9’s as convenient could provide a better ﬁt. Trying out diﬀerent numbers of 9’s, you will ﬁnd that 4 · 9 + 2 · 7 = 50, which leads to our solver’s solution.

485

Five distinct positive numbers are given. Forming all possible sums of two of these numbers you obtain seven diﬀerent sums. Show that the sum of the ﬁve original numbers is divisible by 5. Solution by Gesine Geupel, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany. Suppose the ﬁve distinct numbers are a, b, c, d, and e, and that a < b < c < d < e. Then a +b < a+ c < a+ d < a +e < b +e < c+ e < d +e. This is already seven distinct sums, so the remaining three sums, b + c, b + d, and c + d must be in the seven sums listed above. Since a + d < b + d < b + e, it follows from the long inequality above that b + d = a + e. Since a + e = b + d < c + d < c + e, it similarly follows that c + d = b + e. Finally, since a + c < b + c < b + d = a + e, it follows that b + c = a + d. That is, b + d = a + e, c + d = b + e, and b + c = a + d. If you add the second and third of these equations and subtract the ﬁrst, you get that (c + d) + (b + c) − (b + d) = (b + e) + (a + d) − (a + e), so 2c = b + d. On the other hand, a + e = b + d, so a + e = 2c. Thus a + b + c + d + e = (a + e) + (b + d) + c = 2c + 2c + c = 5c, which clearly is divisible by 5.
´ Also solved by LENA CHOI, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; and ROWENA HO, student, ´ Ecole Dr. Charles Best Secondary School, Coquitlam, BC. As our solver clearly assumes (as does every other submitted solution), the ﬁve numbers should be integers. The problem should have made this clear; we apologise for the omission.

4. Three squares are arranged as in the ﬁgure. Show that the two shaded triangles
have the same area.

´ Solution by Lena Choi, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC. Let a and b denote the side lengths of the two outer squares, label the vertices as in the ﬁgure, and extend the top side of the right-hand square until it meets the vertical line through D .

486

F A a E B

C b

Since the sides of ABE are parallel to the the corresponding sides of DCF and |AB | = |DC |, it follows that ABE ∼ = DCF . Hence the distance from D to the base line is a + b. If you now use the horizontal side of each shaded triangle as the base, then the left-hand shaded triangle has base a and height b, while the right-hand shaded triangle has base b and height a. Therefore the area of either shaded triangle is 1 ab. 2
Also solved by GESINE GEUPEL, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; and ROWENA HO, student, ´ Ecole Dr. Charles Best Secondary School, Coquitlam, BC.

Triangle ABC is isosceles and ∠ACB = 90◦ . The point D is on the line AC beyond C , and the point E is on the line CB beyond B . Show that |CD | = |CE | if line BD is perpendicular to line AE .
D C

B E

´ Solution by Rowena Ho, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC. In the ﬁgure in the statement of the problem, BD does not seem perpendicular to AE , so move D farther out. Moreover, let F denote the intersection between BD and AE . Now similar triangles do the work: ∠EBF = ∠DBC and ∠BF E = 90◦ = ∠BCD , so BEF ∼ BDC . Also, BEF and AEC are both right-angled and share an angle, so they are also similar. Thus AEC ∼ BEF ∼ BDC , |CE | CD | so | = | . AC | |BC |
D

A F

B E

487

Also solved by LISA CHEN, student, Moscrop Secondary School, Burnaby, BC; LENA ´ CHOI, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; GESINE GEUPEL, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany; and JUSTINE HANSEN, student, Burnaby North Secondary School, Burnaby, BC.

Since |CD |.

ABC is given to be isosceles, |AC | = |BC |. Therefore, |CE | =

The product of three positive integers is three times as large as their sum. Find all such triples. Solution by Gesine Geupel, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany. Let a, b, and c denote the three positive integers. Since the order of the integers is not relevant, you may assume that a ≤ b ≤ c. The problem states b that abc = 3(a + b + c). Therefore, ab = 3( a +c + 1) ≤ 3(1 + 1 + 1) = 9, c a b because c ≤ 1 and c ≤ 1. Only a few pairs of integers, (a, b), such that 1 ≤ a ≤ b satisfy that ab ≤ 9, namely (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (2, 2), (2, 3), (2, 4), and (3, 3). Since abc = 3a +3b +3c, (ab − 3)c = 3a +3b, a+3b . For each possible pair, (a, b), you may now calculate c: so c = 3 ab−3
a 1 1 1 1 1 1 1 b 1 2 3 4 5 6 7 c −3 −9 impossible 15 9 7 6 but c is positive but c is positive a 1 1 2 2 2 3 b 8 9 2 3 4 3 c
27 5

5 12 5
18 5

but c is an integer but c ≥ b but c is an integer

but c ≥ b

Thus, the only solutions are (1, 4, 15), (1, 5, 9), (1, 6, 7), (2, 2, 12), (2, 3, 5), and (3, 3, 3).
´ Also solved by LENA CHOI, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; and ROWENA HO, student, ´ Ecole Dr. Charles Best Secondary School, Coquitlam, BC.

´ This issue’s prize for the best solutions goes to Lena Choi, student, Ecole Dr. Charles Best Secondary School, Coquitlam, BC. We hope that you will enjoy our featured contest, and we look forward to receiving your solutions at [email protected] or the postal address listed inside the back cover.

488

MATHEMATICAL MAYHEM
Mathematical Mayhem began in 1988 as a Mathematical Journal for and by High School and University Students. It continues, with the same emphasis, as an integral part of Crux Mathematicorum with Mathematical Mayhem. The interim Mayhem Editor is Shawn Godin (Cairine Wilson Secondary School, Orleans, ON). The Assistant Mayhem Editor is Lynn Miller (Cairine Wilson Secondary School, Orleans, ON). The other staﬀ members are Ann Arden (Osgoode Township District High School, Osgoode, ON), Nicole Diotte (Windsor, ON), Monika Khbeis (Our Lady of Mt. Carmel Secondary School, Mississauga, ON) and Daphne Shani (Bell High School, Nepean, ON).

Mayhem Year End Wrap Up
Shawn Godin
Hello Mayhem readers. Another “year” has ended and it marks the end of another chapter in the story of Mathematical Mayhem. Mayhem was created “by students, for students” in the fall of 1988 and was published 5 times a year for 8 years. After losing its funding, Mayhem joined Crux in 1997, volume 23. Now, with times changing, it is time for the journal to make yet another change. Mayhem will revert to a stand alone journal that appears 5 times a year and follows the school year (issues in September, November, January, March and May). The diﬀerence is that Mayhem will exist on-line. The plan is to expand the journal beyond just problems to include columns, articles, interactive material and possibly even videos. We are working hard toward the relaunch of Mayhem which should occur some time in early 2013. Keep your eye on the CMS web site and the Crux Facebook page for updates. At this point I need to thank the Mayhem staﬀ for their help preparing the material for each issue. The problems editors ANN ARDEN, NICOLE DIOTTE, MONIKA KHBEIS and DAPHNE SHANI who sift through all the submitted solutions and prepare the featured solutions, your work is very much appreciated. Also to my assistant editor LYNN MILLER, for your work behind the scenes and preparing solutions, I want to thank you from the bottom of my heart. You are always there when I need a little extra help. I wish all the best to our readers. I look forward to receiving your problem proposals and solutions. Keep your eye open for the new Mayhem. Shawn Godin

489

Mayhem Problems
Please send your solutions to the problems in this edition by 15 November 2012. Solutions received after this date will only be considered if there is time before publication of the solutions. Each problem is given in English and French, the oﬃcial languages of Canada. In issues 1, 3, 5, and 7, English will precede French, and in issues 2, 4, 6, and 8, French will precede English. The editor thanks Rolland Gaudet, Universit´ e de Saint-Boniface, Winnipeg, MB, for translating the problems from English into French.

M513.

´ Propos´ e par l’Equipe de Mayhem.

Un grillage triangulaire ´ equilat´ eral consiste de chevilles espac´ ees d’un centim` etre l’une de l’autre, tel qu’indiqu´ e au sch´ ema. Des bandes ´ elastiques sont plac´ ees autours des chevilles de fa¸ con a ` former des triangles ´ equilat´ eraux ; deux tels triangles ´ equilat´ eraux a ` deux centim` etres de cˆ ot´ e sont illustr´ es au sch´ ema. Combien de triangles ´ equilat´ eraux diﬀ´ erents sont possibles ?

M514.
AB .

Propos´ e par Dragoljub Miloˇ sevi´ c, Gornji Milanovac, Serbie.

Le nonagone ABCDEF GHI est r´ egulier. D´ emontrer que AE − AC = Propos´ e par Titu Zvonaru, Com´ ane¸ sti, Roumanie.

M515.

Sans utiliser les techniques du calcul diﬀ´ erentiel, d´ eterminer les valeurs minimales et maximales de 2x x2 + 2x + 2 o` u x est un nombre r´ eel.

M516.

Propos´ e par Syd Bulman-Fleming et Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON. D´ emontrer que pour tout k entier non nul il existe au moins quatre paires ordonn´ ees d’entiers, (x, y ), telles que y2 − 1 = k 2 − 1. x2 − 1 ˇ M517. Propos´ e par Sefket Arslanagi´ c, Universit´ e de Sarajevo, Sarajevo, Bosnie et Herz´ egovine. D´ eterminer toutes les solutions r´ eelles de l’´ equation Ô Ô √ 3 x + y + 2 8 − x + 6 − y = 14.

490

M518.

S´ electionn´ ea ` partir de concours math´ ematiques.

Un nombre de carr´ es unitaires sont plac´ es sur une ligne, tel qu’indiqu´ e au sch´ ema ci-bas. P Q

X Y O Soit O le coin inf´ erieur a ` gauche du premier carr´ e et soient P et Q les coins eme et 2012i` eme carr´ sup´ erieurs a ` droite des 2011i` es respectivement. Lorsque P et Q sont reli´ es a ` O , ils intersectent le cˆ ot´ e droit du premier carr´ e a ` X et Y respectivement. D´ eterminer la surface du triangle OXY . .................................................................

M513.

Proposed by the Mayhem Staﬀ.

An equilateral triangular grid is formed by removable pegs that are one centimetre apart as shown in the diagram. Elastic bands may be attached to pegs to form equilateral triangles, two diﬀerent equilateral triangles two centimetres on each side are shown in the diagram. How many diﬀerent equilateral triangles are possible?

M514. M515.

Proposed by Dragoljub Miloˇ sevi´ c, Gornji Milanovac, Serbia.

Nonagon ABCDEF GHI is regular. Prove that AE − AC = AB . Proposed by Titu Zvonaru, Com´ ane¸ sti, Romania. Without using calculus, determine the minimum and maximum values of 2x x2 + 2x + 2 where x is a real number.

M516. Proposed by Syd Bulman-Fleming and Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON.
Show that for any given nonzero integer k there exists at least four distinct ordered pairs (x, y ) of integers such that y2 − 1 = k 2 − 1.

x2 − 1

491

M517. Proposed by
and Herzegovina.

ˇ Sefket Arslanagi´ c, University of Sarajevo, Sarajevo, Bosnia

Find all real solutions of the equation Ô Ô √ 3 x + y + 2 8 − x + 6 − y = 14.

M518.

Selected from a mathematics competition.

A number of unit squares are placed in a line as shown in the diagram below. P Q

X Y O Let O be the bottom left corner of the ﬁrst square and let P and Q be the top right corners of the 2011th and 2012th squares respectively. When P and Q are connected to O they intersect the right side of the ﬁrst square at X and Y respectively. Determine the area of triangle OXY .

Mayhem Solutions
M476.
Proposed by the Mayhem Staﬀ Deﬁne s(n) to be the sum of the digits of the positive integer n. For example, s(2011) = 2 + 0 + 1 + 1 = 4. Determine the number of four-digit positive integers n with s(n) = 4. Solution by David E. Manes, SUNY at Oneonta, Oneonta, NY, USA. There are ﬁve ways to write 4 as the sum of positive integers; namely 4, 3 + 1, 2 + 2, 2 + 1 + 1, and 1 + 1 + 1 + 1. If s(n) = 4 + 0 + 0 + 0, then n = 4000 is the only such integer. If s(n) = 3 + 1, then there are six possible values for n; namely n = 3100, 3010, 3001, 1300, 1030, or 1003. If s(n) = 2 + 2, then the three possible values for n are n = 2200, 2020, or 2002. If s(n) = 2 + 1 + 1, then the nine possible values for n are n = 2011, 2101, 2110, 1021, 1012, 1102, 1120, 1210, or 1201. Finally, if s(n) = 1 + 1 + 1 + 1, then n = 1111 is the only such integer. Hence, there are 20 positive integers with s(n) = 4.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; GESINE ´ GEUPEL, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany; ALICIA GOMEZ ´ GOMEZ, Club Math´ ematique de l’Instituto de Ecuaci´ on Secundaria No. 1, Requena-Valencia,

492

´ IES “Abastos”, Valencia, Spain; JAIME PIQUERAS GARC´ Spain; RICARD PEIRO, IA, Club Math´ ematique de l’Instituto de Ecuaci´ on Secundaria No. 1, Requena-Valencia, Spain; and BRUNO SALGUEIRO FANEGO, Viveiro, Spain.

M477.

Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania

Let m be an integer parameter such that the equation x2 − mx + m +8 = 0 has one integer root. Determine the value of the parameter m. Solution by George Apostolopoulos, Messolonghi, Greece. If there is only one root then the discriminant must be zero, so (−m)2 − 4(m + 8) = 0 ⇔ ⇔ (m − 2)2 = 36 m − 2 = ±6

⇔

m = 8 or m = −4.

Also solved by DAVID E. MANES, SUNY at Oneonta, Oneonta, NY, USA. Two incorrect solutions were submitted.

M478.

Proposed by the Mayhem Staﬀ

Consider the set of points (x, y ) in the plane such that x2 + y 2 − 22x − 4y + 100 = 0 . is the largest. Determine the distance Let P be the point in this set for which x of P from the origin. Solution by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain. We recognize the given set of points as the circle with centre at (11, 2) and radius 5 and whose parametric equations are x = 11 + 5 cos θ, y = 2 + 5 sin θ Denote tan
θ 2

(1)
1 − tan2 1+
θ 2 tan2 θ 2

by t. Since sin θ =

2 tan 1+

(1) are equivalent to x= Hence,
y x

θ 2 tan2 θ 2

and cos θ =

, equations

6t2 + 16 1 + t2

, y=
d( y x) dt

2t2 + 10t + 2 1 + t2

are solutions of −3t + 2t + 8 = 0 or t = to a minimum, and t = 2 corresponds to a maximum. We ﬁnd x and y for t = 2 to be x = 8, y = 6, so that the distance of P from the origin is x2 + y 2 =
Ô

t2 +5t+1 , 3t2 +8 2

the derivative

5(−3t2 +2t+8) , and the critical values (3t2 +8)2 4 − 3 , 2. The value t = − 4 corresponds 3

82 + 62 = 10

493

´ Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; RICARD PEIRO, IES “Abastos”, Valencia, Spain; and PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy. Three incorrect solutions were submitted.

M479.

Proposed by Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania Let A = 1 · 2 · 3 · · · · · 2011 = 2011!. (a) Determine the largest positive integer n for which 3n divides exactly into A. (b) Determine the number of zeroes at the end of the base 10 representation of A. Solution by David E. Manes, SUNY at Oneonta, Oneonta, NY, USA. (a) The exponent n of the highest power of 3 that divides exactly into 2011! is given by n = 223, then
2011 33 2011 3k 2011 3k k=1 74, 2011 34 ∞ È

. One calculates

2011 3

= 670,

2011 32

= = 24, 2011 = 8, 2011 = 2, and if k ≥ 7, 35 36 = 0. Therefore, n = 670 + 223 + 74 + 24 + 8 + 2 = 1001.

(b) The number of zeroes with which the decimal representation of 2011! terminates is equal to the exponent, m, of the highest power of 10 that divides 2011!. Furthermore, m is also the exponent of the highest power of 5 that divides 2011!, that is, m =
∞ È

Hence, the base 10 representation of 2011! ends in 501 zeroes.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; GESINE ´ IES GEUPEL, student, Max Ernst Gymnasium, Br¨ uhl, NRW, Germany; RICARD PEIRO, “Abastos”, Valencia, Spain; and BRUNO SALGUEIRO FANEGO, Viveiro, Spain.

k=1

2011 5k

= 402 + 80 + 16 + 3 = 501.

M480.

Proposed by Dragoljub Miloˇ sevi´ c, Gornji Milanovac, Serbia

Let x, y , and k be positive numbers such that x2 + y 2 = k. Determine the minimum possible value of x6 + y 6 in terms of k. Solution by Bruno Salgueiro Fanego, Viveiro, Spain. Note that x6 + y 6 = (x2 )3 + (y 2 )3 = (x2 + y 2 )(x4 − x2 y 2 + y 4 ) = k((x + y 2 )2 − 3x2 y 2 ) = k(k2 − 3x2 y 2 ) attains its minimum possible value if and only if −3x2 y 2 attains its minimum value, or equivalently, if and only if x2 y 2 attains its maximum value. By the arithmetic mean-geometric mean inequality,
2

x2 y 2 ≤

k2 with equality if and only if x = y , that is x2 y 2 4 k attains its maximum value if and only if x = y = , so the minimum possible 2 k2 k3 = . value of x6 + y 6 is x6 + y 6 = k k2 − 3 4 4

x2 + y 2 2

494

Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; GEORGE APOSTOLOPOULOS, Messolonghi, Greece; DAVID E. MANES, SUNY at ´ IES “Abastos”, Valencia, Spain; PAOLO Oneonta, Oneonta, NY, USA; RICARD PEIRO, PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; and NECULAI STANCIU, George Emil Palade Secondary School, Buz˘ au, Romania;

M481.
ON

Proposed by Edward T.H. Wang, Wilfrid Laurier University, Waterloo,

Suppose that a, b, and x are real numbers with ab = 0 and a + b = 0. If and b.
sin4 x cos4 x 1 sin6 x cos6 x + = , determine the value of + in terms of a a b a+b a3 b3

Solution by Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania. From
cos4 x 1 sin4 x + = , it follows a b a+b

⇒

b2 sin4 x + a2 cos4 x + ab[(sin2 x + cos2 x)2 − 2 sin2 x cos2 x − 1] = 0 b2 sin4 x + a2 cos4 x + ab(1 − 2 sin2 x cos2 x − 1) = 0 b2 sin4 x + a2 cos4 x − 2ab sin2 x cos2 x = 0

b2 sin4 x + a2 cos4 x + ab(sin4 x + cos4 x − 1) = 0

=0 a b a+b b(a + b) sin4 x + a(a + b) cos4 x − ab = 0

sin4 x

cos4 x

−

(b sin2 x − a cos2 x)2 = 0.

Therefore we have sin4 x a and consequently, + cos4 x b = 1 a+b ⇒ (b sin2 x − a cos2 x)2 = 0,

sin2 x cos2 x 1 = = . Finally we obtain that a b a+b

sin6 x a3

cos6 x b3

= =

1 a+b 2 (a + b)3

+ .

1 a+b

Also solved by by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallorca, Spain; GEORGE APOSTOLOPOULOS, Messolonghi, Greece; DAVID E. MANES, SUNY at ´ IES “Abastos”, Valencia, Spain; and BRUNO Oneonta, Oneonta, NY, USA; RICARD PEIRO, SALGUEIRO FANEGO, Viveiro, Spain.

495

THE OLYMPIAD CORNER
No. 298 R.E. Woodrow and Nicolae Strungaru
The solutions to the problems are due to the editors by 1 December 2012. Each problem is given in English and French, the oﬃcial languages of Canada. In issues 1, 3, 5, and 7, English will precede French, and in issues 2, 4, 6, and 8, French will precede English. In the solutions’ section, the problem will be stated in the language of the primary featured solution. The editors thank Jean-Marc Terrier of the University of Montreal for translations of the problems.

OC51. D´ eterminer toutes les paires (a, b) d’entiers non n´ egatifs tels que ab + b divise a2b + 2b. Noter que, pour ce probl` eme, 00 = 1. OC52. Soit d, d
deux diviseurs de n avec d > d. Montrer que d >d+ d2 n .

OC53.Trouver tous les polynˆ omes P (x) ∈ R[x] tels que P (a) ∈ Z implique
a ∈ Z.

OC54.

On donne quatre points dans le plan tels que les cercles inscrits des quatre triangles form´ es par trois des quatre points sont ´ egaux. Montrer que les quatre triangles sont ´ egaux.

OC55. Soit d un entier positif. Montrer que, pour tout entier S , il existe un
entier n > 0 et une suite que S=
1 (1 1 , 2 , ..., n ,

o` u pour tout k,
3 (1

= 1 ou
n (1

= −1, tels

+ d)2 +

2 (1

+ 2d)2 +

+ 3d)2 + +

+ nd)2 .

OC56.

On suppose que f : N → N est une fonction telle que pour tout a, b ∈ N, l’expression af (a) + bf (b) + 2ab est un carr´ e parfait. Montrer que f (a) = a pour tout a ∈ N.

Soit ABC un triangle et A , B , C les points milieu respectifs de BC , CA, AB . Soit P et P deux points dans le plan tels que P A = P A , P B = P B , P C = P C . Montrer que tous les P P passent par un mˆ eme point.

OC57.

OC58. Trouver le plus petit n pour lequel il existe des polynˆ omes f1 (x), f2 (x), . . . , fn (x) ∈ Q[x] tels que
2 2 2 f1 (x) + f2 (x) + · · · + fn (x) = x2 + 7 .

496

OC59. Soit n un entier positif impair tel que φ(n) et φ(n + 1) sont des puissances de deux. Montrer que n + 1 est une puissance de deux ou n = 5. OC60. On ´ ecrit les nombres 1, 2, . . . , 20 au tableau noir. Une op´ eration consiste a ` choisir deux nombres a, b tels que b ≥ a + 2, eﬀacer a et b et les remplacer par a + 1 et b − 1. Trouver le nombre maximal d’op´ erations possibles.
.................................................................

OC51. Determine all pairs (a, b) of nonnegative integers so that ab + b divides a2b + 2b. Note, for this problem 00 = 1. OC52.
Let d, d be two divisors of n with d > d. Prove that d >d+ d2 n .

OC53.
a ∈ Z.

Find all the polynomials P (x) ∈ R[x] so that P (a) ∈ Z implies

OC54. Given four points in the plane so that the incircles of the four triangles formed by three of the four points are equal, prove that the four triangles are equal. OC55. Let d be a positive integer. Show that for every integer S there exists an integer n > 0 and a sequence 1 , 2 , ..., n , where for any k, k = 1 or k = −1, such that
S=
1 (1

+ d)2 +

2 (1

+ 2d)2 +

3 (1

+ 3d)2 + +

n (1

+ nd)2 .

OC56.

Suppose that f : N → N is a function so that for all a, b ∈ N the expression af (a) + bf (b) + 2ab is a perfect square. Prove that f (a) = a for all a ∈ N.

Let ABC be a triangle and A , B ,C be the midpoints of BC , CA, AB respectively. Let P and P be points in a plane such that P A = P A , P B = P B , P C = P C . Prove that all P P pass through a ﬁxed point.

OC57.

OC58. Find the smallest n for which there exists polynomials f1 (x), f2 (x), . . . , fn (x) ∈ Q[x] such that
2 2 2 f1 (x) + f2 (x) + · · · + fn (x) = x2 + 7 .

OC59. Let n be an odd positive integer such that both φ(n) and φ(n + 1) are powers of two. Prove n + 1 is a power of two or n = 5. OC60. On a blackboard we write the numbers 1, 2, ..., 20. A move consists of selecting two numbers a, b from the blackboard so that b ≥ a + 2, erasing a and b and writing instead a + 1 and b − 1. Find the maximum number of possible moves.

497

In this number of the Corner we will complete the ﬁles of solutions from the readers to problems given in the Corner, and also my time as editor of the Corner. Thanks to those who contributed solutions to problems discussed in 2011 numbers: Arkady Alt Miguel Amengual Covas George Apostolopoulos Mohammed Aassila Ricardo Barroso Campos Michel Bataille Chip Curtis Prithwijit De Jos´ e Luis D´ ıaz-Barrero Gesine Geupel Oliver Geupel Geoﬀrey A. Kandall Giulio Loddi David Manes Norvald Midttun Soohyun Park Paolo Perfetti Henry Ricardo Bruce Shawyer D.J. Smeenk Gheorge Ghita Stanciu Edward T.H.Wang Dexter Wei Konstantine Zelator Kaiming Zhao Titu Zvonaru

I’ve enjoyed my association with Crux Mathematicorum and the Corner since January 1988. I would like to express my thanks to all those over nearly a quarter century who have supported the Corner and Crux Mathematicorum by sending in problem sets, comments, and solutions. I would also like to express my thanks to Joanne Canape who has been transcribing my scribbles A into L TEX manuscripts over the last decades. With the next volume of Crux Mathematicorum there will be a new team and new features to draw your interest and support. R.E. Woodrow

First, we look at the readers’ solutions to problems from the Croatian Mathematical Competition 2007, National Competition, given at [2010: 435–436], that we started in the last issue.

4th Grade

Sequence (an )n≥0 is deﬁned recursively by a 0 = 3, a n = 2 + a 0 · a 1 · . . . · a n− 1 , n ≥ 1.

(a) Prove that any two terms of the sequence are relatively prime positive integers. (b) Determine a2007.

498

Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; David E. Manes, SUNY at Oneonta, Oneonta, NY, USA; Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solutions of Manes. For part (a), assume there exist two integers m > n > 0 such that gcd(am , an ) = d > 1. Then a0 a1 . . . am−1 = am − 2. Since the term an is one of the factors on the left side of this equation, it follows that d divides 2. But each an (n ≥ 0) is odd, therefore d = 2 is impossible. Hence, d = 1. For part (b), we begin by showing inductively that the terms an of the n sequence are given by the Fermat numbers 2(2 ) + 1. Note that a0 = 3 = 0 2(2 ) + 1. Assume that k ≥ 0 is a nonnegative integer and ak = 2 + a0 a1 · · · ak−1 = 2(2 or ak − 2 = a0 a1 · · · ak−1 = 2(2 Then ak+1 = (a0 a1 · · · ak−1 )ak + 2 = (2(2 = (2(2
k+1 k k k

)

+1 − 1.
k

)

− 1) + 2 = 2(2
n

k+1

)

− 1)(2(2 ) + 1) + 2

+ 1.

Therefore, by induction, an = 2(2

)

+ 1 for each n ≥ 0. Hence,
2007

a2007 = 2(2

)

+ 1.

In acute triangle ABC let A1 , B1 and C1 be the midpoints of sides BC , CA and AB , respectively. The radius of its circumscribed circle, with centre O , is 1. Prove that 1 1 1 + + ≥ 6. |OA1 | |OB1 | |OC1 |

Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution by Zvonaru. Since it is known that OA1 = R cos A, we have to prove that 1 cos A Let f : 0,
π 2

1 cos B

1 cos C
1 . cos x

≥ 6. We have

→ R be the function f (x) = f (x) =

sin x cos2 x

; = cos2 x + 2 sin2 x cos3 x > 0,

f (x) =

cos3 x − sin x(−2 cos x sin x) cos4 x

499

hence f is a convex function. Applying Jensen’s Inequality we get f (A) + f (B ) + f (C ) ≥ 3f that is 1 cos A + 1 cos B + 1 cos C ≥ A+B+C 3 3 cos
π 3

= 6. ABC is equilateral.

The equality holds if and only if A = B = C , that is

Next we move to readers’ solutions to problems of the 51st National Mathematical Olympiad in Slovenia, Selection Examinations for the IMO 2007, given at [2010: 436–437].

First Selection Examination, December 2006

Show that the inequality (1 + a2 )(1 + b2 ) ≥ a(1 + b2 ) + b(1 + a2 )

holds for any pair of real numbers a and b. Solved by George Apostolopoulos, Messolonghi, Greece; Chip Curtis, Missouri Southern State University, Joplin, MO, USA; David E. Manes, SUNY at Oneonta, Oneonta, NY, USA; Paolo Perfetti, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; Henry Ricardo, Tappan, NY, USA; Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Perfetti. We know that (x − y )2 ≥ 0 =⇒ x2 + y 2 ≥ 2|x||y | thus (1 + a2 )(1 + b2 ) = ≥ since |x| ≥ x. (1 + a2 )(1 + b2 )/2 + (1 + a2 )(1 + b2 )/2 |a|(1 + b2 ) + |b|(1 + a2 ) ≥ a(1 + b2 ) + b(1 + a2 )

2. Prove that any triangle can be decomposed into n isosceles triangles for every positive integer n ≥ 4.
Solved by Titu Zvonaru, Com´ ane¸ sti, Romania.

500

We will prove three claims. Claim 1. Any triangle can be decomposed into 4 isosceles triangles. Let ABC be a triangle. Suppose that ∠BAC is the A greatest angle of ABC . Denoting by D the projection of N M A onto BC , it follows that D lies between B and C (since the angles ∠ABC and ∠ACB are acute). C Let M , N be the midpoints of the sides AC and AB , B D respectively. Since in a right-angled triangle the median to the hypotenuse is half of the hypotenuse, it follows that the triangles BDN , N DA, DCM and DM A are isosceles. Claim 2. Any triangle can be decomposed into 5 isosceles triangles. (a) If ABC is not equilateral, then we may suppose A that AB < BC . We take a point D on the side BC such that AB = BD . We obtain an isosceles triangle ABD D B and for ADC we can apply claim 1. (b) Suppose that ABC is an equilateral triangle. A Let O be the circumcentre of ABC . The perpendicular through O to OC meets BC at M , and let N be O the midpoint of M C . ◦ Since ∠BOC = 120 , it follows that ∠BOM = 30◦ = ∠OBM . We deduce that the triangles OAB , B M N OAC , OBM , OM N and ON C are isosceles.

Claim 3. Any triangle can be decomposed into 6 A isosceles triangles. E Let I be the incentre of ABC . The incircle is F I tangent to the sides BC , CA, AB at the points D , E and F , respectively. We have the following 6 isosceles triangles: AF E , B D C IF E , BDF , IDF , CDE and IDE . Using Claim 1, we see that if we can decompose a triangle into k isosceles triangles, we can decompose this triangle into k + 3 isosceles triangles, and so on. Using Claims 1, 2 and 3 we deduce that any triangle can be decomposed into n ≥ 4 isosceles triangles. Comment. If ABC is an acute-angled triangle, then it can also be decomposed into 3 isosceles triangles (namely AOB , BOC , COA, where O is the circumcentre).

Let ABC be a triangle with |AC | < |BC | and denote its circumcircle by K. Let E be the midpoint of the arc AB that contains the point C and let D be a point on the segment BC , such that |BD | = |AC |. The line DE meets the circle K again in F . Prove that A, B , C and F are the vertices of an isosceles trapezoid.

501

Solved by George Apostolopoulos, Messolonghi, Greece; Oliver Geupel, Br¨ uhl, NRW, Germany; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Zelator. Since E is the midpoint of the arc AB [that contains C ], we E have C ϕ ϕ |BE | = |EA|. (1) θ θ ϕ D Also by hypothesis, B A |BD| = |AC |. And obviously, ∠EBC = ∠EAC = θ. (3) F (2) ϕ

From (1), (2), and (3), it follows that the triangles BDE and AEC are congruent. Therefore |ED| = |EC | and so triangle DEC is isosceles. ∠EDC = ϕ = ∠ECD. Furthermore, ∠EDC = ϕ = ∠BDF and ∠BF E = ϕ = ∠BCE. (6) From (4), (5), and (6); it follows that the triangle F BD is isosceles with |BD| = |BF |. (7) (5) (4)

Hence, from (7) and (2) ⇒ |BF | = |CA|, which further implies that ∠BCF = ∠CF A, which in turn implies that BC and F A are parallel. We have {|BF | = |CA| and BC F A} ⇒ BCAF is an isosceles trapezoid.

Second Selection Examination, February 2007

1. Every point in the plane with positive integer coordinates (x, y) such that x ≤ 19 and y ≤ 4 is colored green, red or blue. Prove that there exists a rectangle with sides parallel to the coordinate axes and with vertices of the same colour.
Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Zvonaru. We rephrase the problem as follows: “Each square of a 4 × 19 chessboard is coloured green, red or blue. Prove that there exists a rectangle whose four corner squares are all coloured the same.”

502

There are 19 columns each containing 4 squares. By the Pigeon-hole Principle, each column must contain some pair of squares of the same colour (we have 4 squares per column and only 3 colours). In order for a rectangle to have its four corners coloured the same colour, there must be two diﬀerent columns in which squares of the same colour are placed in the same two rows. ¡ ¡ ways to place a pair of green squares, 4 ways to place a There are 4 2 2 4¡ pair of red squares and 2 ways to place a pair of blue squares. Thus, there are ¡ 3· 4 = 3 · 6 = 18 ways altogether. Since there are 19 columns in the board, 2 there must be at least 2 diﬀerent columns in which a pair of squares of the same colour are placed in the same way. It follows that there exists a rectangle whose four corner squares are coloured with the same colour.
2 +1) ( t (t 2

Comment. In a similar way, we can solve the problem with a (t + 1) × + 1) chessboard and t colours.

2. The circles K1 and K2 of diﬀerent radii meet at A1 and A2 . Let t be the common tangent of the two circles, such that the distance from t to A1 is shorter than the distance from t to A2 . Let B1 and B2 be the points in which t touches K1 and K2 , respectively. Let K3 and K4 be the circles with radii |A1 B1 | and |A1 B2 | and the centre A1 . The circles K1 and K3 meet again at C1 , while the circles K2 and K4 meet again at C2 . Denote the intersection of the lines B1 C1 and B2 C2 by D and let E be the intersection of B1 C1 and K4 which lies on the same side of the line B2 C2 as C1 . Show that A1 D is perpendicular to EC2 .
Solved by Oliver Geupel, Br¨ uhl, NRW, Germany. Let O1 and O2 be the centres of K1 and K2 , respectively. Let F1 and G1 be the feet of the altitudes from B1 and A1 in triangle A1 B1 O1 , respectively. Let F2 and G2 be the feet of the altitudes from B2 and A1 in triangle A1 B2 O2 , respectively. By A1 G1 ⊥ B1 G1 and A2 G2 ⊥ B2 G2 , the quadrilateral B1 B2 G2 G1 is a rectangle. Hence, we see that B1 G1 = B2 G2 . Also, the triangles A1 B1 O1 and A1 B2 O2 are isosceles with O1 A1 = O1 B1 and O2 A1 = O2 B2 , which implies that A1 F1 = B1 G1 and A1 F2 = B2 G2 . Therefore, A1 F1 = A1 F2 , that is, the lines B1 E and B2 C2 have the same distance from the centre A1 of K4 . Hence, the two lines are symmetric with respect to the axis A1 D . Thus, the triangle DC2 E is isosceles with DC2 = DE and with the line A1 D as its axis of symmetry. Consequently, A1 D is perpendicular to EC2 .

503

K4
B2

K3
A1 G2

G1 O1

F1 A2

F2 O2

C1 E C2 D

Find a positive integer n such that n2 − 1 has exactly 10 positive divisors. Show that n2 − 4 cannot have exactly 10 positive divisors for any positive integer n. Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Oliver Geupel, Br¨ uhl, NRW, Germany; David E. Manes, SUNY at Oneonta, Oneonta, NY, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solutio by Manes. If n is even, then n2 − 1 = (n + 1)(n − 1) and gcd(n + 1, n − 1) = 1. Since τ (n), the number of positive divisors of n, is multiplicative, it follows that τ (n2 − 1) = τ (n + 1)τ (n − 1) = 10 = 5 · 2. Unique factorization implies one such factorization is τ (n + 1) = 2 and τ (n − 1) = 5. If n − 1 = 34 , then τ (n − 1) = 5. Therefore, n = 82 and τ (n + 1) = τ (83) = 2. Accordingly, if n = 82, then n2 −1 = 6723 = 34 · 83 and τ (n2 −1) = 10. The solution n = 82 is not unique since n = 2400 also satisﬁes the property that τ (n2 − 1) = 10. Note that if τ (m) = 10, then either m = p4 q or m = p9 for some distinct primes p and q . Thus, if τ (n2 − 4) = 10, then either n2 − 4 = p4 q or n2 − 4 = p9 . If n2 − 4 = (n + 2)(n − 2) = p9 , then unique factorization implies n + 2 = pr and n − 2 = ps where r > s and r + s = 9. Subtracting the two equations, it follows that pr − ps = 4, hence p = 2. Therefore, 2r−2 − 2s−2 = 1 implies r − 2 = 1 and s − 2 = 0, a contradiction. If n2 − 4 = (n +2)(n − 2) = p4 q , then either n +2 = pr and n − 2 = ps q or n + 2 = pr q and n − 2 = ps . If n + 2 = pr and n − 2 = ps q , then r + s = 4 and r > s. Subtracting the two equations, one obtains pr − ps q = 4, again implying that p = 2. Therefore, 2r − 2s q = 4 or 2r−2 = 2s−2 q + 1, whence r = 2 or s = 2, each of which is a contradiction since r + s = 4 and r > s. On the other hand, if n + 2 = pr q and n − 2 = ps , then r + s = 4. Solving for n in each equation, it follows that n = pr q − 2 = ps + 2 or pr q − ps = 4, whence p = 2. Therefore, 2r q − 2s = 4 so that 2r−2 q = 2s−2 + 1. But r + s = 4

504

implies (r − 2) + (s − 2) = 0 or r = s = 2. Hence, q = 2, a contradiction since p and q are distinct primes. Consequently, all of these contradictions imply that τ (n2 − 4) = 10 for any positive integer n.

Third Selection Examination, March 2007

Let x = 0.a1a2 a3 a4 . . . and y = 0.b1 b2 b3 b4 . . .

be the decimal representations of two positive real numbers. The equality bn = a2n holds for all positive integers n. Given that x is a rational number, show that y is rational, too. Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA. (i) If the decimal representation for x terminates, then so does that for y , and y is rational. (ii) Otherwise, the decimal representation for x eventually repeats with some smallest period t ∈ N. Thus, there is a positive integer k0 such that for k ≥ k0 , ak+t = ak . Hence, for j ∈ {0, 1, . . . , t − 1}, there exists rj ∈ N ∪ {0} such that for k ≥ k0 , ak = rj if k ≡ j mod t. Thus, bn = a2n = ri if 2n ≡ i mod t. (a) If gcd(2, t) = 1, that is, if t is odd, let α be the order of 2 mod t. Then 2α ≡ 1 mod t. Suppose 2n ≡ i mod t. Then 2n+α = 2n · 2α ≡ 2n ≡ i mod t, so that bn+α = a2n+α = ri = a2n = bn . Hence, α is a period for {bn }, so y is rational. (b) If t = 2s · u, with u odd, let α be the order of 2 mod u. Then 2α ≡ 1 mod u. For n ≥ s, 2s divides 2n+α − 2n , so that 2n+α = 2n · 2α ≡ 2n mod t. As before, this implies that y is rational.

Let ABCD be a trapezoid with AB parallel to CD and |AB | > |CD |. Let E and F be the points on segments AB and CD , respectively, such that DF | |AE | = | . Let K and L be two points on the segment EF such that |EB | |F C | ∠AKB = ∠DCB and ∠CLD = ∠CBA.

Show that K , L, B and C are concyclic. Solved by Oliver Geupel, Br¨ uhl, NRW, Germany. Let the perpendicular to BC at B intersect the perpendicular bisector of AB at O , and let the perpendicular to BC at C intersect the perpendicular bisector of CD at P . The line OP intersects AB and CD at points G and H , respectively, such that |BG| : |EG| = |CH | : |F H |. Hence, the lines BC , EF , and GH have a common intersection S . From ∠AKB = 180◦ − ∠ABC we see that the line BC is tangent to the circle (ABK ). Thus, O is the centre of (ABK ). Analogously, P is the centre of the circle (CDL). Hence, (ABK ) and (CDL) are homothetic with respect to the centre of homothety S .

505

If M is the second point of intersection of the line SK and the circle (ABK ), then (B, C ), (O, P ), (G, H ), (E, F ), and (M, L) are pairs of corresponding points under this homothety. Therefore, ∠KM B = ∠SM B = ∠SLC . Furthermore, ∠KM B = ∠KBC , because the line BC is tangent to the circle (ABKM ). Thus, ∠KBC = ∠SLC . If the point K is between S and L then ∠KBC = ∠KLC. (1)

F K

P G B

L E

Otherwise L is between S and K , and we conclude ∠KBC = 180 − ∠KLC.
◦

(2) M

Either of (1) and (2) implies that K , L, B , and C are concyclic.

Next, we look at solutions submitted to problems of the Correspondence Mathematical Competition in Slovakia 2006/7 First Round, First Set, given at [2010: 438–439].

There are some pigeons and some sparrows sitting on a fence. Five sparrows ﬂew away and there remained two pigeons for each sparrow. Then 25 pigeons ﬂew away and there remained three sparrows for each pigeon. Find the initial numbers of sparrows and pigeons. Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Oliver Geupel, Br¨ uhl, NRW, Germany; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Curtis. Let p and s be the initial numbers of pigeons and sparrows, respectively. Then p and s satisfy the system p = s−5 = 2(s − 5) 3(p − 25).

Substituting the ﬁrst equation into the second, solving for s, and then substituting back gives p = 30 s = 20.

3. We have eight cubes with digits 1, 2, 3, 4, 5, 6, 7, 9 (each cube has one digit written on one of its faces). In how many ways can we create four two-digit primes from the cubes?

506

Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Oliver Geupel, Br¨ uhl, NRW, Germany; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution by Curtis. We claim that there are four ways. We assume that the order in which the four primes occur is immaterial. Note ﬁrst that none of the four primes can end in 2, 4, 5, or 6. Hence these digits must be the ﬁrst digits of the four primes and 1, 3, 7, and 9 must be the second digits of the primes. The prime ending in 1 must be 41 or 61. The prime ending in 3 must be 23, 43, or 53. The prime ending in 7 must be 47 or 67. The prime ending in 9 must be 29 or 59. (i) Suppose that 41 is one of the primes. Then 47 is not one of the primes, so 67 must be. The other two primes can then be either 23 and 59 or 29 and 53. (ii) Suppose that 61 is one of the primes. Then 67 is not one of the primes, so 47 must be. As before, the other primes can then be either 23 and 59 or 29 and 53. Hence, there are four possible sets of primes: {41, 67, 23, 59}, {41, 67, 29, 53}, {61, 47, 23, 59}, {61, 47, 29, 53}. A nine-member committee was formed to select a chief of the KMS. There are three candidates for the chief. Each member of the committee orders the candidates and gives 3 points to the ﬁrst one, 2 points to the second one and 1 point to the last one. After summing the points of the candidates it turned out that no two candidates have the same number of points, hence the order of the candidates is clear. Someone noticed that if every member of the committee selected only one candidate, the resulting order of candidates would be reversed. How many points did the candidates get? Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the write-up of Zvonaru. Let C1 , C2 , C3 be the candidates, with the ﬁnal order C1 − C2 − C3 . We denote by s(C1 ), s(C2 ), s(C3 ) the sum of the points of the candidates C1 , C2 , C3 , respectively. For i = 1, 2, 3, we denote xi — the number of ﬁrst place received by Ci yi — the number of second place received by Ci zi — the number of third place received by Ci . We have s(Ci ) = 3xi + 2yi + zi , i = 1, 2, 3 with xi + yi + zi = 9, i = 1, 2, 3 and x1 + x2 + x3 = 9, y1 + y2 + y3 = 9, z1 + z2 + z3 = 9.

507

The sum of the points is 54. Since s(C1 ) > s(C2 ) > s(C3 ), we deduce that s(C3 ) < 18. Since x1 < x2 < x3 , it results that x3 ≥ 4; • if x3 ≥ 6, then s(C3 ) ≥ 18, a contradiction. • if x3 = 5, then y3 + z3 = 4 and s(C3 ) = 15 + 4 + y3 > 18, also a contradiction. It follows that x3 = 4, and from x1 + x2 = 5 and x1 < x2 we deduce that x1 = 2, x2 = 3. We have s(C1 ) = s(C2 ) = s(C3 ) = 6 + y1 + z1 + y1 = 13 + y1 9 + y2 + z2 + y2 = 15 + y2 12 + y3 + z3 + y3 = 17 + y3

Since s(C3 ) < 18 we get y3 = 0 and y1 + y2 = 9. Because s(C1 ) > s(C2 ) > s(C3 ) we obtain y1 > y2 + 2 and y2 > y3 + 2 = 2 . It is easy to see that the only possibility is y2 = 3, y1 = 6, hence s(C1 ) = 19, s(C2 ) = 18, n(C3 ) = 17.

5.
(a) Find all positive integers n such that both of the numbers 2n − 1 and 2n + 1 are primes. (b) Find all primes p such that both of the numbers 4p2 + 1 and 6p2 + 1 are primes. Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; David E. Manes, SUNY at Oneonta, Oneonta, NY, USA; Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the write-up of Wang. (a) The only such integer is n = 2. Let f (n) = 2n + 1 and g (n) = 2n − 1. Clearly g (1) = 1 which is not a prime. Next, f (2) = 5 and g (2) = 3 are both primes. Now suppose n ≥ 3. If n is odd, then n = 2k + 1 where k ≥ 1 so f (n) = 22k+1 + 1 = 2(4k ) + 1 ≡ 2(1) + 1 ≡ 0 (mod 3). Since f (n) > 3 and is divisible by 3, it is a composite. If n is even, then n = 2k where k ≥ 1 so g (n) = 22k − 1 = 4k − 1 ≡ 1 − 1 ≡ 0 (mod 3). Since g (n) > 3 and is divisible by 3, it is a composite. This completes the proof. (b) This is the same as Problem #3 of the Finnish Math Olympiad, 2006, (Final Round) the solution of which has appeared in Crux 36(7), 2010; p. 447.

Find all positive integers n such that n + 200 and n − 269 are cubes of integers.

508

Solved by George Apostolopoulos, Messolonghi, Greece; Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Oliver Geupel, Br¨ uhl, NRW, Germany; David E. Manes, SUNY at Oneonta, Oneonta, NY, USA; Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution by Geupel. We prove that the unique solution is n = 1997. Suppose that a positive integer n has the desired property. Let a3 = n + 200, b3 = n − 269, and d = a − b. We have d > 0 and 0 = a3 − (a − d)3 − 469 = 3da2 − 3d2 a + d3 − 469, a quadratic equation in a with the discriminant 9d4 − 12d(d3 − 469) = 3d(1876 − d3 ) ≥ 0. Hence 1 ≤ d ≤ 12. Moreover, the integer d is a divisor of a3 − b3 = 469 = 7 · 67; hence d ∈ {1, 7}. If d = 7, then the equation (1) becomes 0 = 21(a2 − 7a − 6), which has no integer solution, a contradiction. Consequently d = 1. Equation (1) becomes 0 = 3(a + 12)(a − 13); thus a = 13, b = 12, and n = 133 − 200 = 1997. [Ed.: Note that instead of (1) we note that a3 −b3 = (a−b)(a2 +ab+b2 ) = 469 = 7 × 67, which leads to four choices for a − b, from which the solution follows.] (1)

7. There were 33 children at a camp. Every child answered two questions: “How many other children have the same ﬁrst name as you?” and “How many other children at camp have the same family name as you?”. Among the answers each of the numbers from 0 to 10 occurred at least once. Show that there were at least two children with the same ﬁrst name and the same family name. (Mathematical Contests 1997–1998, 1.18 Russia, 29/95)
Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Oliver Geupel, Br¨ uhl, NRW, Germany. We give the solution by Geupel. If the number k occurs among the answers, then there is a name that occurs exactly k + 1 times. Hence, among the ﬁrst and family names of the 33 children, there are names N1 , N2 , . . . , N11 that occur exactly 1, 2, . . . , 11 times, respectively. Since 1 + 2 + · · · + 11 = 2 · 33, the names N1 , N2 , . . . , N11 cover all the ﬁrst and family names of the group. By symmetry there is no loss of generality in assuming that N11 is a ﬁrst name occurring 11 times. Then among the 11 children with ﬁrst name N11 at most 10 distinct family names occur. By the Pigeonhole Principle, two children among these 11 children have the same family name. The proof is complete.

Find all triples of integers x, y , z satisfying 2x + 3y = z 2 .

509

Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; David E. Manes, SUNY at Oneonta, Oneonta, NY, USA; Soohyun Park, Student, University of Toronto Schools, Toronto, ON; Neculai Stanciu, George Emil Palade Secondary School, Buz˘ au, Romania; Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON; and Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA. We give the solution of Wang. We show that the only solutions are (x, y, z ) = (0, 1, ±2), (3, 0, ±3) and (4, 2, ±5). Note ﬁrst that z = 0 since the left side is always positive. Thus, z 2 ≥ 1. If at least one of x and y is negative then the left side cannot be an integer, a contradiction. Hence, we may assume that x and y are nonnegative. We may also assume for the time being that z is positive. We ﬁrst establish a lemma. Lemma. If p is a prime, then the only solutions in nonnegative integers u and v to the equation pu + 1 = v 2 is (u, v ) = (3, 3) if p = 2, and (u, v ) = (1, 2) if p = 3. If p = 2, 3, then there are no solutions. Proof. Clearly u = 0 since 2 is not a square. Also, v = 0. We have pu = v 2 − 1 = (v − 1)(v + 1) so v − 1 = pa and v + 1 = pb for some integers a and b such that 0 < a < b with a + b = u. Then pb − pa = 2 or pa (pb−a − 1) = 2. If p = 2, then a = 0 and pb−a − 1 = 2 so pb = 3 which implies that p = 3 and b = 1. It follows that u = a + b = 1 and v = 2. If p = 2, then we have either (i) 2a = 1, 2b−a − 1 = 2; or (ii) 2a = 2, 2b−a − 1 = 1. In case (i) we get a = 0 and 2b = 3 which is impossible. In case (ii) we have a = 1 and 2b−1 − 1 = 1 so b = 2. It follows that u = a + b = 3 and v = 1 + 2 = 3 so (u, v ) = (3, 3) and the lemma is established. Back to the given problem, note that if x = 0, then the equation becomes 3y +1 = z 2 . Hence by the lemma above the only solution is (x, y, z ) = (0, 1, 2). Similarly, if y = 0, then the equation becomes 2x + 1 = z 2 , so by the lemma again, the only solution is (x, y, z ) = (3, 0, 3). Now we assume x ≥ 1 and y ≥ 1. If x = 1 then we have 2 + 3y = z 2 which is impossible since mod 3, 2 + 3y ≡ 2 while z 2 ≡ 0 or 1. Thus, x ≥ 2. Since 2x + 3y is odd so is z . Then z 2 ≡ 1 (mod 4). But mod 4, 2x ≡ 0 so 3y ≡ (−1)y ≡ 1. Thus, y is even. Let y = 2t where t ≥ 1. Then we have 2x = z 2 − 32t = (z − 3t )(z +3t ). Hence z − 3t = 2c and z + 3t = 2d where 0 ≤ c < d such that c + d = x. Subtracting the two equations above we get 2·3t = 2d −2c so 3t = 2c−1 (2d−c −1). Since 2·3t is even, 2c = 1 so c ≥ 1. Therefore, 2c−1 = 1 and 2d−c − 1 = 3. This yields c = 1 and 2d−1 − 1 = 3t . We now show that the only solution to this last equation is d = 3 and t = 1. Since t ≥ 1 we have 2d−1 ≥ 4 so d ≥ 3. Thus, 2d−1 ≡ 0 (mod 4) which implies that 2d−1 − 1 ≡ −1 (mod 4) ⇒ 3t ≡ (−1)t ≡ −1 (mod 4) so t is odd. Hence, we have 2d−1 = 3t + 1 = (3 + 1)(3t−1 − 3t−2 + · · · − 3 + 1). But 3t−1 − 3t−2 + · · · − 3 + 1, is the sum of an odd number of odd integers so it is odd. Since it divides 2 it must be 1. It then follows that 2d−1 = 4 so d = 3 which implies t = 1. Finally, y = 2t = 2, z = 3 + 2c = 5 from which x = 4 follows. This yields the third and last solution (4, 2, 5).

12.

We are given an acute triangle ABC with circumcentre O . Let T be the

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circumcentre of AOC . Let M be the midpoint of AC . The points D and E lie on the lines AB and CB respectively in such a way that the angles M DB and M EB are equal to the angle ABC . Prove that the lines BT and DE are perpendicular. Solved by Titu Zvonaru, Com´ ane¸ sti, Romania. As usual we write a = BC , b = CA, c = AB . Let R be the circumradius ABC , and let N be the midpoint of AB . We denote T A = R1 , and we suppose that a ≥ b ≥ c. Since M N BE and ∠M EB = D ∠N BE , the quadrilateral BEM N is an isosceles trapezoid with A of
T

MN =

a 2

ME = NB =

c 2

.
N M

Denoting by P the projection of N c cos B , onto BC , we deduce that BP = 2 O a hence BE = 2 + c cos B . Applying the C B E P Law of Sines we have a EC = − c cos B 2 = R(sin A − 2 sin C cos B ) = R[sin A − sin(C + B ) − sin(C − B )] = R[sin A − sin A + sin(B − C )] = R sin(B − C ). Similarly we obtain BD = c 2 + a cos B ; AD = R sin(A − B ).

Since ∠T CE = ∠T CO + ∠OCE = B + 90◦ − A = 90◦ − (A − B ) by the Law of Cosines we have T E 2 = T C 2 + EC 2 − 2T C · EC cos ∠T CE and similarly
2 T D 2 = R2 1 + AD − 2RR1 sin(A − B ) sin(B − C ),

2 = R2 1 + EC − 2RR1 sin(B − C ) sin(A − B )

hence T E 2 − T D 2 = EC 2 − AD 2 . On the other hand we deduce that EC 2 − AD 2 =(a − BE )2 − (BD − c)2 (1)

=a(a − a − 2c cos B ) − c(c − c − 2a cos B ) + BE 2 − BD 2 ,

=a(a − 2BE ) − c(c − 2BD ) + BE 2 − BD 2

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hence BE 2 − BD 2 = EC 2 − AD 2 . By (1) and (2) it results that the lines BT and DE are perpendicular. (2)

13. A line passing through the centroid T of the triangle ABC meets the side AB at P and the side CA at Q. Prove that
4 · P B · QC ≤ P A · QA. (R.B. Manfrino: Inequalities, 111/3.29, Spain 1998) Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the version of Zvonaru. Let M be the midpoint of BC . We denote a = BC , x = We will prove that xy = x + y If P Q BC , then x = y = = 2 TM and (1) is true. Suppose that P Q meets BC at the point S such that B lies between S and C . By Menelaus’ Theorem applied in ABC , we have BS SC that is SB SC
ay hence SB = . x−y AT
A

AQ AP ,y= . PB QC

(1)

CQ QA

AP PB

= −1,
S

P M C

y x

⇔

SB SB + a

y x

Now we apply Menelaus’ Theorem in SB T M P A · · =1⇔ SM T A P B ⇔

ABM to get · 1 ·x=1 2 2(x − y ) · x 2 =1

ay x− y ay +a x− y 2

x−y

a(2y + x − y )

hence xy = x + y . Using (1) and AM-GM Inequality we obtain √ xy = x + y ≥ 2 xy ⇒ (xy )2 ≥ 4xy ⇒ xy ≥ 4, that is PA PB · QA QC ≥ 4.

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14.

Integers x and y greater than 1 satisfy the relation 2x2 − 1 = y 15 .

(a) Prove that x is divisible by ﬁve. (b) Are there such integers x and y greater than 1 satisfying 2x2 − 1 = y 15 ? Could you ﬁnd all such numbers? (Russia 2004/05) Solved by David E. Manes, SUNY at Oneonta, Oneonta, NY, USA. (a) Note that if integers x, y > 1 satisfy 2x2 − 1 = y 15 then 2x2 − 1 = (y 3 )5 . Let z = y 3 > 1 and consider the Diophantine equation 2x2 − 1 = z 5 . Then z is odd and 2x2 = z 5 + 1 = (z + 1)(z 4 − z 3 + z 2 − z + 1). Let d = gcd(z + 1, z 4 − z 3 + z 2 − z + 1). Then d divides z + 1 and d | (z 4 − z 3 + z 2 − z + 1) = (z 3 − 2z 2 + 3z − 4)(z + 1) + 5. Hence, d divides 5 and so d = 1 or 5. If d = 5, then 5 | 2x2 whence 5 divides x and we are done. If d = 1, then z + 1 and z 4 − z 3 + z 2 − z + 1 are relatively prime and their product is twice a square. Since z 4 − z 3 + z 2 − z + 1 is odd, it must be a square. However, for y > 1, (2z 2 − z )2 < 4(z 4 − z 3 + z 2 − z + 1) < (2z 2 − z + 1)2 . Therefore, z 4 − z 3 + z 2 − z + 1 is not a square, a contradiction. Accordingly if 2x2 − 1 = y 15 and x, y > 1, then 5 divides x.

(b) Assume integers x, y > 1 exist so that 2x2 − 1 = y 15 . Then integers x, z = y 5 > 1 exist such that 2x2 − 1 = z 3 , By a result of Cohn, the only solution of this equation with x, y > 1 is x = 78, z = 23 (cf. Theorem 2, p. 27, J.H.E. Cohn, The Diophantine equations x3 = N y 2 ± 1, Quart. J. Math. Oxford, 42(1991), 27–30). Since 23 is not a ﬁfth power of any integer, it follows that the only solution of 2x2 − 1 = y 15 is x = y = 1.

Next we give readers’ solutions to the Latvian School Mathematical Olympiad, Grade 11, given at [2010: 440].

For a positive integer n:

(a) can the sums of digits of n and n + 2007 be equal? (b) can the sums of digits of n and n + 199 be equal?

513

Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; David E. Manes, SUNY at Oneonta, Oneonta, NY, USA; Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the write-up of Curtis. For a positive integer n, let d(n) denote the sum of the digits of n. Let m and n be positive integers. Write
r r

m=
k=0

ak · 10k

and

n=
k=0

bk · 10k ,

where the ak and bk are nonnegative integers. If r = 0 and a0 + b0 < 10, then d(m + n) = d[(a0 + b0 )] = a0 + b0 = d(m)+ d(n). If r = 0 and a0 + b0 ≥ 10, that is if a carry is required, then d(m + n) = d[1 · 10 + (a0 + b0 − 10)] = a0 + b0 − 9 = d(m) + d(n) − 9. The same occurs each time a carry is required, so d(m + n) = d(m) + d(n) minus 9 times the number of carries required. Hence, 9 | [d(m + n) − (d(m) + d(n))]. (a) If n = 3, then d(n) = 3 and d(n + 2007) = d(2010) = 3 = d(n), so the answer to the ﬁrst question is ‘yes’. (b) If d(n) = d(n + 199), then 9 | [d(n + 199) − d(n) − d(199)] implies that 9 divides d(199). But d(199) = 19, a contradiction. Hence, the answer to the second question is ‘no’.

2. Do there exist three quadratic trinomials such that each of them has at least one root, but the sum of any two quadratic trinomials doesn’t have any roots?
Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Zvonaru. The answer is YES. Let a, b, c be three real numbers such that a = b = c = a. (x − a)2 (x − c) (x − a)2 + (x − b)2 (x − b)
2 2

= 0 = 0 = 0 = 0 = 0 = 0

has the root a has the root b has the root c has no real roots has no real roots has no real roots.

(x − a)2 + (x − c)2 (x − b)2 + (x − c)2

In triangle ABC a point K lies on median AM and ∠BAC + ∠BKC = 180◦ . Prove that AB · KC = AC · KB .

Solved by George Apostolopoulos, Messolonghi, Greece; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Apostolopoulos.

514

We assume ∠BAC < 90◦ . On AM take M D = KM . Then KBDC is a parallelogram, since the diagonals bisect each other. So ∠BKC = ∠BDC and ∠BAC + ∠BDC = 180◦ hence ABDC is cyclic and ∠ABD + ∠ACD = 180◦ . Thus [ [ AB · BD ABD ] = , ACD ] AC · CD

K M

where [ ABD ], as usual, represents the area of ABC . Also [ ABD ] = [ ACD ] so AB · BD = AC · CD , but BD = KC and CD = KB namely AB · KC = AC · KB .

For a sequence of real numbers a1 , a2 , a3 , . . . we have a11 = 4, a22 = 2 and a33 = 1. In addition, the relation an+3 − an+2 an+3 + an+2 = an − an+1 an + an+1

holds for each n. Prove that: (a) ai = 0 for each i, (b) the sequence is periodic,
k k (c) ak 1 + a2 + · · · + a100 is a square of an integer for any arbitrary positive integer k.

= an+3 an − an+1 an+3 + an an+2 − an+1 an+2 ⇔ an an+2 = an+1 an+3 .

(1)

By the same relation we deduce that an = an+1 for each n. Using (1) we have ai ai+2 = ai+1 ai+3 = ai+2 ai+4 . Suppose that ai = 0; it results that ai+1 = 0 and from ai+1 ai+3 = 0 we get ai+3 = 0. Since ai+2 = ai+3 = ai+4 , we obtain a contradiction ai+2 ai+4 = 0. (b) By (1) we have an an+2 = an+1 an+3 ; an+1 an+3 = an+2 an+4 ,

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hence an an+2 = an+2 an+4 . Since an+2 = 0, it results that an = an+4 , that is the sequence is periodic. (c) We have a1 = a5 = a9 = . . . = a33 = 1 a2 = a6 = a10 = . . . = a22 = 2 a3 = a7 = a11 = 4 and from a1 a3 = a2 a4 we obtain a4 = 2. It follows that
k ak 1 + · · · + a100 k k k k k k = 25(ak 1 + a2 + a3 + a4 ) = 25(1 + 2 + 4 + 2 )

= 25((2k )2 + 2 · 2k + 1),

k k 2 hence ak 1 + · · · + a100 = [5(2 + 1)] .

Next we look at solutions to the Latvian Mathematical Olympiad Grade 12, given at [2010: 440–441].

What can be the values of nonnegative real numbers a and b, if it is known that equations x2 + a2 x + b3 = 0 and x2 + b2 x + a3 = 0 have a common real root? Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Paolo Perfetti, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We use the solution of Perfetti.

Proof. The solutions of the two equations are √ √ −a2 ± a4 − 4b3 −b2 ± b4 − 4a3 x1 , 2 = , x3 , 4 = 2 2 They are real if and only if {a4 ≥ 4b3 ∧ b4 ≥ 4a3 } ⇒ (ab)3 (ab) ≥ 16(ab)3 ⇒ {ab = 0 ∨ ab ≥ 16} Moreover we have {a ≥ 4b
4 3

Answer: a = b = 0, a = b ≥ 4.

´
4 3

∧ b ≥ 4a } =⇒

a ≥ (4b )
b4 4

3 1/4

∧ a≤

b4 4

1/3

The graphs of the functions (4b3 )1/4 and

show that a = b = 0 or

a, b ≥ 4. If a = b = 0 the roots are all equal to zero. If a = b ≥ 4 the two parabola coincide. If a = b = 4 they have one root with multiplicity two. If a = b > 4 the have two distinct roots.

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Thus let us suppose a = b and because of the symmetry with respect to the change (a, b) → (b, a) we may suppose b > a. Moreover we note that b4 − 4a3 > a4 − 4b3 so x3 is smaller than any other root and therefore we can have only x4 = x1 or x4 = x2 . First case x4 = x1 . −b2 + b4 − 4a3 = −a2 − a4 − 4b3 , or squaring 4a3 + 4b3 − 2a2 b2 =
Ô Ô Ô

b2 −a2 = b4 − 4a3
Ô

b4 − 4a3 + a4 − 4b3

a4 − 4b3

The left-hand side may be both negative and positive. For the values which make it positive we can square again and obtain 4(4a6 − 8a3 b3 − 4a5 b2 + 4b6 − 4b5 a2 + 4b7 + 4a7 ) = 0 By AGM we have (a7 + a7 + a7 + a7 + a7 + b7 + b7 )/7 ≥ a5 b2 , (a6 + b6 ) ≥ a3 b3 (1)

with equality in both cases if and only if a = b. If follows that (1) is never zero unless a = b. Second case. x4 = x2 . −b2 + b4 − 4a3 = −a2 + a4 − 4b3 , or squaring 4a3 + 4b3 − 2a2 b2 = −
Ô Ô Ô

b2 −a2 = b4 − 4a3
Ô

b4 − 4a3 − a4 − 4b3

a4 − 4b3

For the values which make the left hand side negative we can square again and obtain as above 4(4a6 − 8a3 b3 − 4a5 b2 + 4b6 − 4b5 a2 + 4b7 + 4a7 ) = 0 The consequences are the same. The conclusions are that if a = b = 0 the four solutions all coincide. If a = b ≥ 4 the solutions coincide pairwise. For any other values of (a, b) the solutions are all distinct.

Solve the system of equations sin2 x + cos2 y = y 2 sin2 y + cos2 x = x2

517

Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA, modiﬁed by the editor. From the two equations, we ﬁnd that sin2 x − sin2 y = y 2 − 1 = 1 − x2 = from which it follows that 2 sin2 x + x2 = 2 sin2 y + y 2 . (2) 1 2 (y 2 − x2 ),

The function 2 sin2 t + t2 is an even function of t that is strictly increasing for 0 ≤ t ≤ π . Since, from the original equations, we have that |x| and |y | do √2 , it follows that from (1) |x| = |y |. not exceed 2 < π 2 On the other hand, when this condition holds, we must have x2 = y 2 = 1, since x2 + y 2 = 2. So the equations are satisﬁed when (x, y ) ∈ {(1, 1), (1, −1), (−1, 1), (−1, −1)}.

4. Two circles w1 and w2 intersect in points A and B . Line t1 is drawn through point B with other intersection point with w1 being C and other intersection point with w2 being E . Line t2 is drawn through point B with other intersection point with w1 being D and other intersection point with w2 being F . Point B lies between C and E and between D and F . Midpoints of segments CE and DF are denoted by M and N . Prove that triangles ACD , AEF and AM N are similar.
Solved by Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give Zelator’s write-up.

A θ O1 O2 ω C D M ω θ ϕ B N γ θ γ ϕ E F θ

Looking at the ﬁgure and the cyclic quadrilaterals ACDB and ABEF , the following statements are clear: ∠CAD = ∠CBD = ∠EBF = ∠EAF = θ, ∠ABF = ∠AEF = ∠ACD = ϕ, ∠ACB = ∠ADB = ω, (1) (2) (3)

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∠BF A = ∠BEA = γ.

(4)

From (1) and (2) we have ∠CAD = ∠EAF = θ and ∠ACD = ∠AEF = ϕ which proves that the triangles ACD and AEF are similar (two-angle criterion). From (3) and (4), it follows that triangles ACE and ADF are similar (two-angle criterion) (also note that ∠CAE = θ + ∠DAE and ∠DAF = ∠DAE + θ ). The sides CE and DF lie across the congruent angles ∠CAE and ∠DAF . Thus, since N and M are the midpoints of CE and DF , and the triangles ACE and ADF are similar, it follows that ∠DAM = ∠CAN, ∠DAN + ∠N AM = ∠CAD + ∠DAN, ∠N AM = ∠CAD = θ by (1). And since ∠EBF = θ , it follows by (5) that ∠EBF = θ = ∠CAD (6) (5)

It follows from (6) that AN BM is a cyclic quadrilateral. Thus ∠AN M = ∠ABM = ϕ and by (2) we see that ∠AN M = ∠ACD = ϕ. (7)

Clearly (6) and (7) imply that the triangles ACD and AN M are similar (twoangle criterion). And since we have already shown that ACD and AEF are similar, we conclude that the triangles ACD , AEF , and AN M are similar.

5. The set of all positive integers has been split in several parts in such a way that
each integer belongs exactly to one part and each of the parts contains inﬁnitely many integers. Can this be done so that one part contains a multiple of any positive integer? Give the answer if (a) there are a ﬁnite number of parts, (b) there are an inﬁnite number of parts. Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Oliver Geupel, Br¨ uhl, NRW, Germany. We give the solution of Geupel. We prove that the answer is “Yes” in both cases. Let A1 and A2 denote the set of even numbers and the set of odd numbers, respectively. Then, A1 contains the double of each positive integer, which completes the proof for part (a). Let p1 = 2, p2 = 3, p3 = 5, . . . be the sequence of prime numbers. For k = 1, 2, 3, . . ., let Bk denote the set of positive integers which are divisible by pk but not divisible by any of the numbers p1 , p2 , . . . , pk−1 . Then, {B1 , B2 , B3 , . . .} is an inﬁnite partition of the set of positive integers and B1 is the set of even numbers, which completes the proof for part (b). While this was obvious, the following question is more interesting: Is it true that, for each partition of the set of positive integers, one part contains a multiple of each positive integer?

519

We prove that the answer is “Yes” for partitions into a ﬁnite number of parts. Suppose the contrary. Then there exists a partition {A1 , A2 , . . . , An } such that for each Ak there is a number ak with the property that no multiple of ak is contained in Ak . The number a = a1 a2 · · · an is in some Ak , say in An . By a = (a1 a2 · · · an−1 )an , it follows that a multiple of an is contained in An . This is a contradiction, which completes the proof. However, the answer is “No” if inﬁnitely many parts can occur. A counterexample is the partition {B1 , B2 , B3 , . . .} where Bk is the set of positive integers which have exactly k distinct prime divisors.

Next we turn to solutions from our readers to problems of the Finnish National High School Mathematics Competition, Final Round, given at [2010: 441–442].

1. Show that when a prime number is divided by 30, the remainder is either a prime number or 1. Is a similar claim true when the divisor is 60 or 90?
Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; David E. Manes, SUNY at Oneonta, Oneonta, NY, USA; Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution by Manes. Note that if n is a composite number less than 30, then gcd(30, n) = 1. Let p be a prime number and assume that the remainder r when p is divided by 30 is neither 1 nor a prime. Then p = 30m + r for some integers m and r with 0 < r < 30. Therefore, gcd(30, r ) = 1 implies there is a prime q such that q divides both of the integers 30 and r . Hence, q divides p and this implies q = p, a contradiction that proves the result. However, no such claim can be made when the divisor is 60 or 90 since the prime 229 divided by 60 or 90 leaves a remainder of 49.

Determine the number of real roots of the equation x8 − x7 + 2x6 − 2x5 + 3x4 − 3x3 + 4x2 − 4x + 5 = 0. 2

Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the write-up of Curtis. Let P (x) be the given polynomial. Then P (x) = = 1 2 1 2 [2x7 (x − 1) + 4x5 (x − 1) + 6x3 (x − 1) + 8x(x − 1) + 5] [2x(x − 1)(x6 + 2x4 + 3x2 + 4) + 5] > 0 for x > 1.

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Also, P (−x) = 1 2 (2x8 + 2x7 + 4x6 + 4x5 + 6x4 + 6x3 + 8x2 + 8x + 5).

Thus, P (−x) > 0 for x > 0, so that P (x) > 0 for x < 0. Hence, any real zero of P (x) lies in the interval [0, 1]. But, neither P (0) nor P (1) equals 0, and for < 2x(x − 1) < 0 and 0 < x6 + 2x4 + 3x2 + 4 < 10. This 0 < x < 1, − 1 2 implies that for 0 < x < 1, 0 < P (x) < 5. Thus, P (x) has no real zeros.

3. There are ﬁve points in the plane, no three of which are collinear. Show that some four of these points are the vertices of a convex quadrilateral.
Solved by Oliver Geupel, Br¨ uhl, NRW, Germany; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution by Geupel. The convex hull of the given points is a A convex n-gon where n ∈ {3, 4, 5}. If n = 4 then we are done. If n = 5 then any four of the ﬁve points are T E the vertices of a convex quadrilateral. D Finally suppose n = 3. The convex hull is S a triangle, say ABC , and the other two points, say D and E , are in the interior of that triangle. B C Because no three of the points A, B , C , D , and E are collinear, the line DE does not pass through a vertex of the triangle. Hence, the line DE meets two sides of the triangle, say AB and AC at inner points S and T , respectively. Assume DS < ES . Then ∠CBD < ∠CBA < 180◦ , ∠BDE < ∠SDE = 180◦ , ∠BCE < ∠BCA < 180◦ , ∠CED < ∠T ED = 180◦ .

Consequently, BCED is a convex quadrilateral. This completes the proof.

Show that there exists a polynomial P (x) with integer coeﬃcients such that the equation P (x) = 0 has no integer solutions but for each positive integer n there is an x ∈ Z such that n | P (x). Solved by David E. Manes, SUNY at Oneonta, Oneonta, NY, USA. Let P (x) = 6x2 + 5x + 1 = (3x + 1)(2x + 1). Then P (x) = 0 has no integer roots. Let n be an arbitrary positive integer and write n = 2r · s, where r is an integer ≥ 0 and s is odd. Since gcd(2r , 3) = 1 = gcd(s, 2), it follows that there exist positive integers u and v such that 3u ≡ −1 (mod 2r ) and 2v ≡ −1 (mod s) .

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Moreover, gcd(2r , s) = 1 so that the Chinese Remainder Theorem implies there is an integer m such that 3m ≡ −1 (mod 2r · s) and 2m ≡ −1 (mod 2r · s) .

Therefore, P (m) = (3m + 1)(2m + 1) = 0 mod n, whence n | P (m).

To complete this number of the Corner we turn to the solutions to problems of the IX Olimpiada Matem´ atico de Centram´ erica y el Cariba 2007, given at [2010: 442–443].

1. The OMCC is an annual mathematical competition.

The ninth olympiad takes place in the year 2007. Which positive integers n divide the year in which the nth olympiad takes place? Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the solution of Zelator. Since the OMCC is an annual event and the 9th olympiad takes place in 2007, it follows that the ﬁrst olympiad took place in the year 2007 − 8 = 1999. Thus the nth olympiad takes placed in the year 1999 + (n − 1) = 1998 + n. So if n | (1998 + n) we have 1998 + n = kn, for some positive integer k, so that 1998 = (k − 1) · n, whence n | 1998. So, the positive integers n which divide the year in which the nth olympiad takes place are precisely the positive divisors of 1998. Now, 1998 = 2(999) = 2 · 9 · (111) = 2 · 9 · 3 · 37 = 2 · 33 · 37. The number of divisors of 1998 is τ (1998) = (1 + 1) · (3 + 1) · (1 + 1) = 16. The sixteen divisors of 1998 are the positive integers: n = 1, 2, 2 · 3, 2 · 32 , 2 · 33 , 2 · 37, 3, 32 , 33 , 37, 3 · 37, 32 · 37, 33 · 37, 2 · 33 · 37, 2 · 32 · 37, 2 · 3 · 37. We list these sixteen positive integers in increasing order. n = 1, 2, 3, 6, 9, 18, 27, 37, 54, 74, 111, 222, 333, 666, 999, 1998.

2. Let ABC be a triangle, D and E points on the sides AC and AB , respectively,
such that the lines BD , CE and the angle bisector of angle A concur in an interior point P of the triangle. Prove that there is a circle tangent to the four sides of the quadrilateral ADP E if and only if AB = AC . Solved by Titu Zvonaru, Com´ ane¸ sti, Romania.

522

Let A be the point of intersection of BC with the angle bisector of angle A. (i) Suppose that AB = AC ; then A is the midpoint of BC . By Ceva’s Theorem we obtain AE EB hence DA AE = AB − AE AC − DA ⇔ AE · AC − AE · DA = AB · DA − AE · DA,
B

E P

BA AC

CD DA

that is AE = AD . It results that AEP and ADP are congruent (side-angle-side) hence AE + DP = AD + EP , which means that there is a circle tangent to the four sides of quadrilateral ADP E . (ii) Suppose that there is a circle tangent to the four sides of quadrilateral ADP E . Then the centre of this circle lies on the bisector of angle ∠EAD , that is on AP . We deduce that AP is the bisector of ∠EP D , and the triangles AEP and ADP are congruent (angle-side-angle), hence AE = AD . Because A lies on the bisector AP of ∠BAC , AB = BA . Inserting the last two equalities into Ceva’s AC A C theorem gives us =1 A C DA EB AB AB − AE ⇒ = ⇒ = AC DC AC AC − AE ⇒AB · AC − AB · AE = AB · AC − AC · AE, EB AB hence AB = AC . EA · AB · DC

3. Let S be a ﬁnite set of integers. For any two integers p, q ∈ S , with p = q, there are integers a, b, c in S , not necessarily distinct and with a = 0, such that the polynomial F (x) = ax2 + bx + c satisﬁes F (p) = F (q ) = 0. Determine the maximum number of elements set S can have.
Solved by Oliver Geupel, Br¨ uhl, NRW, Germany; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give Geupel’s write-up. We prove that max|S | = 3. A valid set with three elements is S = {−1, 0, 1}. We show by contradiction that |S | < 4. Suppose S is such that |S | ≥ 4. Then, there is a member p ∈ S such that |p| > 1. Let T = {−1, 0, 1, p}. If p > 1 then −a(p + 1) ∈ / T for each a ∈ T \ {0}. Hence, there are no a = 0, b, c ∈ T such that ax2 + bx + c = a(x − 1)(x − p) = ax2 − a(p + 1)x + ap.

523

If p < −1 then −a(p − 1) ∈ / T for each a ∈ T \ {0}. Hence, there are no a = 0, b, c ∈ T such that ax2 + bx + c = a(x + 1)(x − p) = ax2 − a(p − 1)x − ap. Therefore S = T . Thus, there are distinct members p, q ∈ S such that |p| > 1 and |q | > 1. Let p and q be the elements of S with the greatest absolute values. Then, apq ∈ / S for each a ∈ S \ {0}. Hence, there are no a = 0, b, c ∈ S such that ax2 + bx + c = a(x − p)(x − q ) = ax2 − a(p + q )x + apq, a contradiction. This completes the proof that |S | < 4.

The inhabitants of a certain island speak a language in which every word can be written with the following letters: a, b, c, d, e, f , g . A word is said to produce another one if the second word can be formed from the ﬁrst one applying any of the following rules as many times as needed: (i) Replace a letter by two letters according to one of the substitutions: a → bc, b → cd, c → de, d → ef, e → f g, f → ga, g → ab. (ii) If only one letter is between two letters that are the same, these two letters can be eliminated. Example: df d → f As another example, caf ed produces bf ed, since caf ed → cbcf ed → bf ed. Prove that every word on this island produces any other word. Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the write-up of Curtis. We note ﬁrst that each letter produces each letter as seen by the following progression: a → bc → cdc → d → ef → f gf → g → ab → bcb → c → de → ef e

→ f → ga → aba → b → cd → ded → e → f g → gag → a.

Suppose word M has m letters and word N has n letters and we wish to produce N from M . By the above, we can assume that each letter of M and N is ‘a’. If m = n, we are done. From a → bc → aa and aa → abc → cbc → b → a, we see that if m < n, we can expand using a → aa repeatedly: if m > n, we can condense using aa → a repeatedly. Hence we can produce N from M .

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5. Given two non-negative integers m and n, with m > n, we say that m ends in n if one can erase some consecutive digits from left of m to obtain n. For example, 329 ends only in 9 and in 29. Determine how many three-digit numbers end with the product of their digits.
Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; Konstantine Zelator, University of Pittsburgh, Pittsburgh, PA, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the write-up of Zvonaru. Let abc be the three-digit numbers, with a = 0. (i) If c = abc, then we have c = 0 or ab = 1. It results that the numbers ab0 (with a = 1, 2, . . . , 9 and b = 0, 1, 2, . . . , 9) and 11c (with c = 1, 2, . . . , 9) end in the product of their digits. In this case we have 90 + 9 = 99 numbers with the desired property. (ii) If bc = abc, then we have successively bc = abc ⇔ 10b + c = abc ⇔ b = ⇔ ab = Since ab is an integer, we obtain ac = 11, ac = 12, ac = 15, ac = 20, ab = 11 ab = 6 ab = 3 ab = 2 (1) (2) (3) (4) ac ac − 10 c ac − 10 10 .

⇔ ab = 1 +

ac − 10

The systems (1) and (4) have no solution. From (2) we deduce the following possibilities: (a = 2, b = 3, c = 6)(a = 3, b = 2, c = 4)(a = 6, b = 1, c = 2) and from (3) we deduce the solution (a = 3, b = 1, c = 5). It results that the numbers 236, 324, 612 and 315 end in the product of their digits. As a conclusion, there are 99 + 4 = 103 three-digit numbers which end in the product of their digits.

6. Let A and B be points on the circle Γ such that the lines P A and P B are tangent to Γ for an exterior point P . Let M be the midpoint of AB . The perpendicular bisector of AM intersects Γ at C which is interior to ABP , the line AC intersects the line P M at G, and the line P M intersects Γ at D , which is exterior to the triangle ABP . If BD is parallel to AC , prove that G is the point in which the medians of ABP concur.
Solved by Chip Curtis, Missouri Southern State University, Joplin, MO, USA; and Titu Zvonaru, Com´ ane¸ sti, Romania. We give the write-up of Zvonaru.

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A Γ C N P E G M O D

Let O be the centre of Γ, E be the intersection of P D with Γ and N be the midpoint of AM . a b We denote P E = a, P D = b; it results that OA = b− and P O = a+ . 2 2 2 By the power of the point P with respect to the circle Γ we obtain P A = 2ab ab, and by similitude in AP O we have AP 2 = P M · P O , hence P M = a +b
b− a ) 2ab and M D = b − a = b( . +b a+b By the Pythagorean theorem we deduce

AM 2 = AP 2 − P M 2 = ab −

, (a + b)2 (a + b)2 ab(b − a)2 b2 (b − a)2 (ab + b2 )(b − a)2 AD 2 = AM 2 + M D 2 = + = . 2 2 (a + b) (a + b) (a + b)2

4a2 b2

ab(b − a)2

Since AC BD , the cyclic quadrilateral ACBD is an isosceles trapezoid; because AD = BD = BC = 2AC , we have AB 2 = BD + AC 2
2 2 2

+ AD 2 −

BD − AC 2

⇔ 4AM = AD + AD · AC ⇔ 4AM 2 = ⇔ 3AD 2 2 = 3(b − a)2 · b(a + b) (a + b)2 4ab(b − a)2

(a + b)2 ⇔ 8a = 3a + 3b, hence

5a = 3b

(1)

Since AG BD and AM = M B we deduce that GM = M D . Using (1) we have 5a 5a ( − a) b(b − a) 5a GM = = 3 513 = , (2) a+b 12 +a 3 2ab 5a PM = = . (3) a+b 4 By (2) and (3) it follows that P M = 3GM ; since P M is a median in ABP , we deduce that G is the centroid of ABP . THE END

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BOOK REVIEWS
Amar Sodhi
Loving + Hating Mathematics: Challenging the Myths of Mathematical Life by Reuben Hersh and Vera John-Steiner Princeton University Press, 2011 ISBN: 978-0-691-14247-0 , Hardcover, 416 + x pp., US$24.95 Reviewed by Georg Gunther, Grenfell Campus (MUN), Brook, NL There is an anecdote about the brilliant, but notoriously absent-minded, mathematician Norbert Wiener. After moving to a new city he got lost on the way home from the university and stopped a little girl on the street to ask her if she knew where the Wieners lived. She took his hand and said “Yes Daddy, come with me.” Wiener was a child prodigy who received his Harvard PhD in mathematics at age 18. He exempliﬁed the stereotype of a mathematician: “someone who solved the Rubik’s cube at eight, took calculus at fourteen, and was tackling serious mathematics at sixteen. And he’s a guy”. [Haas and Henle, Notices of the AMS, September 2007]. How true to the real thing are these stereotypes? The book “Loving + Hating Mathematics: Challenging the Myths of Mathematical Life” challenges four commonly accepted myths about mathematicians. (1) mathematicians are diﬀerent from other people, lacking emotional complexity; (2) mathematics is a solitary pursuit; (3) mathematics is a ‘young man’s game’; and (4) mathematics is an eﬀective ﬁlter for higher education. This is not a book about mathematics; instead, it is about mathematicians and the kinds of people they are. The authors both have impeccable credentials; both have received prestigious awards for previous books. They are colleagues at the University of New Mexico, where they hold the rank of Professor Emeritus. In nine chapters, the authors lead the reader through a voyage that traces the lives of mathematicians from childhood and early years as students, through their productive years into maturity and old age. Each chapter begins with a simple question. For example, Chapter 1 asks: “How does a child ﬁrst begin to become a mathematician? Is it a predisposition or some special gift? What makes it possible, ﬁnally, to commit one’s life to this risky, forbidding pursuit”? This leads into the second chapter which explores the culture of the mathematical community. One common trait of mathematicians is their ability to focus, often for many years, upon one particular result. This can lead to obsessive and compulsive behaviors swinging in extreme situations into paranoia, sociopathy and other forms of madness. The British philosopher, A. N. Whitehead, described the pursuit of mathematics as a ‘divine madness of the human spirit’; in Chapter 3, the authors explore the boundaries that separate this ‘divine madness’ from more destructive forms of mental illness. Among the examples given is the tragic case of the logician Kurt G¨ odel, who in his later years suﬀered from the delusion that any food not

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prepared by his wife was poisoned. When she was hospitalized for a while he refused to eat anything at all: he subsequently died of malnutrition, weighing a mere 68 pounds. The next two chapters explore the importance of professional collaborations and supportive social networks. There is a subtle irony in this: of all forms of intellectual activity, doing mathematics is perhaps the one that most isolates its practitioners from the society that supports them; yet without fruitful collaboration, without colleagues who speak the same language, without a vital and vibrant community of peers, mathematics would wither and die. After a thought-provoking chapter that focuses on the complex issues of gender and age, the book concludes with two chapters dealing with various aspects of mathematical pedagogy. Chapter 8 looks in detail at two contrasting approaches to teaching at the post-secondary level. The ﬁrst of these was promoted by R .L. Moore who taught at the University of Texas for 49 years, producing 50 PhD students and 1678 doctoral descendants. His method was to recruit students early in their careers, so that he could exercise total draconian control over their mathematical education, to the extent that he forbade them to read or talk about mathematics outside of his classes. This pedagogical approach is compared to that of Clarence Stephens, who eventually became Chair of the Mathematics Department at SUNY (Potsdam), where he championed a student-centered approach to teaching that has been phenomenally successful. One measure of this success lies in the number of students who graduate with a major in mathematics; whereas nationally, this number lies at roughly 1% of all graduates, at Potsdam, it lies at approximately 20%. Finally, the authors consider the perplexing and troubling question of why so many students who graduate from high school do so hating mathematics and feeling hopelessly inadequate about their skill levels. The problem is magniﬁed when, in steadily increasing numbers, these young people enter college where they run into an impenetrable barrier in the form of a mathematics ﬁlter that makes no sense. Why, after all, should a student aiming for admission to medical school be required to study calculus? The book is well researched, and written in a style that is both informal and informative. At times, the prose is almost too episodic; in many instances, the reader is left looking for more depth and detail about some particular point or individual. However, this potential ﬂaw is mitigated by the presence of an extensive bibliography at the end of each chapter. Mathematicians will enjoy this book. They will be able to relate on a personal level to many of the issues that are raised; as well, they will be able to re-acquaint themselves on a more human level with many of the individuals whose names have been long familiar for their mathematical contributions. Nonmathematicians will ﬁnd this fascinating reading. By experiencing the human side of mathematics, they may come to appreciate in new ways a discipline that lives in their recollection as a seemingly impenetrable tangle of abstraction and complexity.

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The Beauty of Fractals: Six Diﬀerent Views edited by Denny Gulick and Jon Scott Mathematics Association of America, 2010 Electronic ISBN: 9780883859711, 95 + x pp., downloadable pdf, US$25 Reviewed by Daryl Hepting, University of Regina, Regina, SK The Beauty of Fractals: Six Diﬀerent Views comes out of a special session on chaos and fractals at the Annual General Meeting of the MAA. The editors, Denny Gulick and Jon Scott, together have organized a series of summer workshops and several contributed paper sessions at Joint Mathematics Meetings on these topics. They have also worked to advance the cause of visual thinking in various ways, not the least of which is this book on the beauty of fractals. I commend the editors and authors for assembling this volume. Being a graduate student when The Science of Fractal Images and The Algorithmic Beauty of Plants were ﬁrst published, I have many fond memories of the topics presented here. No matter how many times I approach these topics, I never tire of them. This book is sure to inspire many in their own explorations of fractals. The preface rightly mentions the books of Mandelbrot, Barnsley, and Devaney as important in making the mathematics of fractals accessible to a wide audience. It is, therefore, exciting to see a contribution from Robert Devaney in this volume. More interesting still, is the fact that his chapter deals with Tom Stoppard’s play Arcadia that oﬀers “teachers of both mathematics and the humanities to join forces in a unique and rewarding way.” Devaney and the other contributors do indeed provide a “taste of the breadth” of fractals; both in terms of topics and also of the disciplines involved. This small volume is certainly not exhaustive, but it provides many entry points. For those interested in the study of fractals, this book provides a good introduction and each chapter provides references that allow the reader to go further. There are many wonderful illustrations throughout the book. I believe that images like the ones found in this book are an important way to introduce the concepts of fractals and to build an appreciation of their beauty that comes from the underlying mathematics. My personal wish would have seen more time spent on the algorithms or software tools used to generate the images seen throughout the book so that interested readers could easily move to hands-on exploration with the aid of a computer. Anne Burns’ beautifully reproduced colour images in Chapter 1 (Mathscapes - Fractal Scenery) are included with little detail about how those ﬁnal images were pulled together. Without a clear path to recreating the images seen in the book, there is still much to recommend it. Each of the 6 chapters presents an interesting perspective, with solid mathematical foundations, that is valuable on its own.

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RECURRING CRUX CONFIGURATIONS 4
J. Chris Fisher Bicentric Quadrilaterals
A convex quadrilateral is called bicentric (or chord-tangent) if it has both a circumcircle (passing through all four vertices) and an incircle (tangent to all four sides). One sees the terminology inscribable (or inscriptible) for cyclic, and circumscribable (or circumscriptible) for containing an incircle, but in English these terms are often confused, as we explained before in CRUX with MAYHEM [1997: 530-531; 2011: 243-244]; it is therefore wise to avoid them when the context is not clear. It is standard practice to assume a cyclic polygon to be convex; thus, ABCD is cyclic if its vertices are arranged around a circle in the order in which they appear in the label. We begin with a brief look at L´ eo Sauv´ e’s essay, “On Circumscribable Quadrilaterals” [ 1976 : 63-67], which deals with sets of four lines that are tangent to the same circle. Four lines in general position intersect in six points and form what is called a complete quadrilateral; four of those points are vertices of a convex quadrangle ABCD , while the opposite sides of that quadrangle intersect at E = AB ∩ CD and F = AD ∩ BC . It is possible for a complete quadrilateral to have an incircle (as on the left of Figure 1) or an excircle (on the right). Both these conﬁgurations are composed of a pair of triangles ABF and AED having a common angle A and sides BF and ED opposite A that have a common point C . Howard Grossman [5] tied together earlier results of Pitot, Durrande, Urquhart, and others into two main theorems that were extended by Sokolowsky [1976 : 163-170]:
F D C F D A C B E

A B

Figure 1: Circles internally tangent (left) and externally tangent (right) to the four sides of a complete quadrilateral.

Theorem 1. If one of the following four properties holds, then they all hold and triangles ABF and AED have a common incircle. (a) AB − BC = AD − DC , (b) AE − EC = AF − F C , (c) DE − EB = BF − F D , (d) AB − BF + F A = AE − ED + DA. Theorem 2. If one of the following four properties holds, then they all hold and triangles ABF and AED have a common excircle.

530

(a) AB + BC = AD + DC , (b) AE + EC = AF + F C , (c) DE + EB = BF + F D , (d) ∆ABF and ∆AED have equal perimeters. Because bicentric quadrilaterals have an incircle by deﬁnition, Theorem 2 will be relevant in what follows only to Problem 777. Of course, a quadrangle requires more than the properties of Theorem 1 to be bicentric—the condition that ABCD be cyclic requires the further property that each pair of its opposite angles sums to 180◦ . A quite diﬀerent characterization of bicentric quadrilaterals came from Euler: if a circle of radius r is placed in the interior of a circle of radius R so that the distance x between their centres satisﬁes 1 (R − or, equivalently, x2 = R2 + r 2 − r 2 (4R2 + r 2 ), x)2 + 1 (R + x)2 = 1 r2

then every point of the outer circle is the vertex of a bicentric quadrilateral that is inscribed in the outer circle and circumscribed about the inner circle. As an immediate consequence of Euler’s formula we deduce that for any bicentric quadrilateral with circumradius R and inradius r , R2 ≥ 2r 2 , with equality if and only if the quadrilateral is a square. Useful formulas involving the semiperimeter s; consecutive sides a, b, c, d; area F ; and diagonals e, f of a bicentric quadrilateral can be found in standard references: s = a + c = b + d, F = √ abcd = rs, ac + bd = ef,

(ab + cd)(ac + bd)(ad + bc) 4F

Problem 777. [1982 : 246; 1984 : 20-22] (Proposed by O. Bottema) Let ABCD be a convex quadrilateral that is not a rectangle. If two of (i) ac + bd = ef , (ii) s equals √ the sum of two sides of the quadrilateral, or (iii) F = abcd = rs hold, then the third holds also. In other words, for a nonrectangular convex quadrilateral ABCD , two of the following three statements implies the third: (i) ABCD is cyclic. (ii) ABCD has either an incircle or an excircle. √ (iii) The area of ABCD is abcd.

531

Problem 1203. [1987 : 14; 1988 : 91-93] (Proposed by Milen N. Naydenov) Bicentric quadrilaterals satisfy √ √ a) 2 F ≤ s ≤ r + r 2 + 4R2 ; √ b) 6F ≤ ab + ac + ad + bc + bd + cd ≤ 4r 2 + 4R2 + 4r r 2 + 4R2 ; √ 2 c) 8sr 2 ≤ abc + abd + acd + bcd ≤ 2r r + r 2 + 4R2 ; ¡ d) 4F r 2 ≤ abcd ≤ 16 r 2 r 2 + 4R2 . 9 Equality holds on the left in each part and on the right in part (d) if and only if the bicentric quadrilateral is a square; it holds on the right of (a), (b), and (c) if and only if at least one pair of opposite angles of the quadrilateral are right angles. The featured proof by Murray Klamkin showed, among other things, that the left-hand inequalities in parts (a), (b), and (c) are immediate consequences of the AM-GM inequality, while the right-hand inequality of part (a) appeared earlier in [1]. Along the way he proved a result that is interesting in its own right, namely Ô ef = 2r r + r 2 + 4R2 . (1) Problem 1376. [1988 : 235; 1989 : 287-288] (Proposed by G.R. Veldkamp) The diagonals e and f of a bicentric quadrilateral satisfy ef 4r 2 − 4R2 ef = 1.

This equation appears (with a typo) on page 49 of [3]; it is an immediate consequence of formula (1). Problem 1983. [1994 : 259; 1995 : 257-258] (Proposed by K.R.S. Sastry) If the line through I parallel to the side AB of the bicentric quadrilateral ABCD s meets AD in A and BC in B , then A B = 2 . Problem 2027. [1995 : 90; 1996 : 94-95] (Proposed by D.J. Smeenk), and Problem 3338. [2008 : 239, 242; 2009 : 241-242] (Proposed by Toshio Seimiya) Let EF be the diameter of the circumcircle of bicentric quadrilateral ABCD that is perpendicular to the diagonal BD , labeled with E on the same side of BD as A. Let BD intersect EF at M and AC at X . Then AX XC = EM MF = AI 2 IC 2 .

The ﬁrst equality is Problem 2027, which (with a new proof) appeared again as Problem 3211 [2007 : 43, 46; 2008 : 61-62]; the second equality is Problem 3338. See also Problem 3598, whose solution appears later in this issue, for yet another result involving the quantities AI 2 , BI 2 , CI 2 , DI 2 . Problem 2194. [1996 : 362; 1997 : 530-532] (Proposed by Christopher J. Bradley) Find an integer-sided triangle ABC and a transversal LM N with N on side AB , M on side AC , and L on the extension of side BC such that BCM N

532

is an integer-sided bicentric quadrilateral, and also the segments M L, and CL are of integer length. Michael Lambrou began his featured solution to 2194 with a rational-sided triangle ABC and chose the transversal to be the line tangent to the incircle of ∆ABC for which ∠AM N = ∠B . Lambrou and Richard Hess each provided further examples in which ∠AM N = 90◦ , while B and L are reﬂections of M and A in the bisector of the exterior angle at N of ∆AM N . Problem 2203. [1997 : 46; 1998 : 112-113] (Proposed by Miguel Amengual Covas; reworded here) Given a cyclic quadrilateral P QRS , the tangents to its circumcircle at its vertices form a bicentric quadrilateral if and only if P R ⊥ QS .

K R

D N X C

I O O P

B L

Figure 2: Problems 2203 and 2209.

Two proofs were provided to this problem even though the result is readily found in the literature such as [2, Section 39, pages 188-191] where much more is provided, including the following consequence of Brianchon’s Theorem (called Newton’s theorem in [4, Sections 1274 and 1275, pp. 563-565]): If the sides of a quadrangle ABCD are tangent to an ellipse at the points P, Q, R, and S , then AC, BD, P R, and QS are concurrent (at X in Figure 2). Problem 2209. [1997 : 47; 1998 : 112-113] (Proposed by Miguel Amengual

533

Covas; reworded here) Denote by I the intersection point of the diagonals of a cyclic quadrilateral KLM N . The feet A, B, C, D of the perpendiculars from I onto the sides of KLM N are the vertices of a quadrangle that has an incircle with centre I ; moreover, if the diagonals KM and LN are perpendicular then ABCD is also cyclic (and therefore bicentric) and its circumcircle passes also through the midpoints of the sides of the given quadrilateral KLM N . Conversely, if ABCD is a bicentric quadrilateral, then the lines that are perpendicular at A, B, C , and D to the lines from the incentre I form the sides of a cyclic quadrilateral whose perpendicular diagonals meet at I . The closely related Problem 1836 [1993 : 113; 1994 : 84-85] (proposed by Jisho Kotani) deﬁned ABCD from the cyclic quadrilateral KLM N as in Problem 2209; the result there is that the sum of the areas of the lunes inside the circles with diameters KL, LM, M N, N K that lie outside the circumcircle equals the area of the quadrangle KLM N if and only if the diagonals KM and LN are perpendicular (and ABCD is therefore bicentric). Of course, when ABCD is bicentric the midpoints of the sides of KLM N are the vertices of a rectangle whose diagonals pass through the circumcentre O of ABCD . This result together with all of Problem 2209 forms part of Theorem 159 in Section 749 (pages 321-322) of [4], where further properties and references are provided. For example, • O is the midpoint between the incentre I = KM ∩ LN and the centre O of the circumcircle of KLM N ; • the line OI also contains the point X = AC ∩ BD = P R ∩ QS where the diagonals of ABCD and of P QRS meet; • it follows from the featured solution that the diagonal LIN passes through the intersection of AD with BC , while KIM passes through AB ∩ CD ; moreover, the line joining those two points is perpendicular to the line O OIX . (This last result was proved as part of Michel Bataille’s solution to Problem 3256 [2007 : 298, 300; 2008 : 312-313], proposed by V´ aclav Koneˇ cn´ y.) Returning to the quadrilateral ABCD , we have Theorem. A quadrilateral ABCD that has an incircle tangent to the sides AB, BC, CD , and DA at the points P, Q, R, and S , respectively, is cyclic if and only if AP × CR = BQ × DS .

The “if” part was Problem 2 on Peru’s Olimpiada Nacional Escolar de Matematica 2009, Level 3 [2010 : 374; 2011 : 381-383]; the converse is a simple exercise in trigonometry. Note how this theorem is related to Problem 2209: the product condition is equivalent to saying that the line P R divides the opposite sides AB and DC is the same ratio, namely AP : P B = DR : RC , while QS divides the opposite sides AD and BC in the same ratio. (Note that P R and QS bisect the angles at X formed by the diagonals AC and BD .) Constructions. We conclude with two constructions taken from the article “A Duality for Bicentric Quadrilaterals”, by Michel Bataille [2009 : 310-312]. The ﬁrst is based on the characterization given above in Problem 2203: A bicentric

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quadrilateral is uniquely determined by three of its sides tangent to a given circle. If the points of tangency are P, Q, and R, then the fourth side is the tangent to the circle at the point where it meets the perpendicular to P R through Q. (Bataille includes yet another proof of 2203.) The second construction can be thought of as a dual of the ﬁrst: A bicentric quadrilateral is uniquely determined by three of its vertices. It is a consequence of the following extension of Problems 2027 and 3338: Let ABCD be a convex quadrilateral inscribed in a circle with diagonals AC and BD meeting at X . Then ABCD is bicentric if and only if both (i) B and D are on the same side of the perpendicular bisector of AC , and (ii) BX = cos2 B BD . 2 Figure 3 shows how to construct D using K on the bisector BV of ∠ABC : K is the point where the perpendicular to BC through C meets BV , C is the point of BC on the perpendicular to BV at K , and D is the ﬁrst point of the circumcircle of ABC on the line through C parallel to AC .
B

K C

Figure 3: Construction of the bicentric quadrangle ABCD given the vertices A, B , and C .

References
[1] W.J. Blundon and R.H. Eddy, Solution to Problem 488 (proposed by Blundon and Eddy), Nieuw Archief voor Wiskunde, 26 (1978) 465-466. [2] Heinrich D¨ orrie, Triumph der Mathematik, W¨ urzburg, 1958. English translation: 100 Great Problems of Elementary Mathematics, Dover, 1965. [3] R.H. Eddy and G.C.W. Sabin, Inequalities in R, r, s for the Inscribed Circumscribed Quadrilateral, Nieuw Archief voor Wiskunde (4), 6:1-2 (1988) 47-56. [4] F. G.-M., Exercices de G´ eom´ etrie, 4i` eme ´ edition. 1907. [5] Howard Grossman, Urquhart’s Quadrilateral Theorem. The Mathematics Teacher, 66 (1973) 643-644.

535

That old root ﬂipping trick of Andrey Andreyevich Markov
Gerhard J. Woeginger A straightforward fact
All the mathematics in this article is based on the following straightforward fact. Fact. Let p and q be positive integers. If the equation x2 − px + q = 0 has a positive integer root, then also its second root is a positive integer. Why do we call this fact straightforward? Well, it easily follows from Vieta’s formulas x1 + x2 = p and x1 x2 = q for the roots x1 and x2 of a quadratic equation. As p and x1 are integers, also x2 = p − x1 is integer. And as q and x1 both are positive, also x2 = q/x1 is positive. So the fact can be proved in two lines and indeed is simple. Then what’s the reason for spending several pages on it? Well, the fact turns out to be surprisingly useful in the analysis of certain Diophantine equations. This article will illustrate this usefulness by a number of examples and also discuss some of its background.

Short history lesson
The Russian mathematician Andrey Andreyevich Markov (1856–1922) received his Master’s degree in 1880 from the university of St Petersburg. His supervisor was Pafnuty Lvovich Chebyshev (1821–1894), and the title of his thesis was “About binary quadratic forms with positive determinant”. Among many other contributions, the thesis contained as a side result the complete analysis of the following Diophantine equation: a2 + b2 + c2 = 3abc. (2) This equation is nowadays called the Markov equation. Let us take a closer look at it. For a solution triple (a, b, c) of (2) over the positive integers, we consider the following quadratic equation in x. x2 − (3bc)x + (b2 + c2 ) = 0. (3)

The quadratic equation (3) has two roots: one of them of course is a, and the other one is — at this place we apply our straightforward fact — the positive integer a = 3bc − a. Unless the two roots coincide, we have found a new solution triple (a , b, c) for equation (2). Symmetric observations for b and c yield the following: whenever (a, b, c) is a positive integer solution of (2), then also the triples (a , b, c), (a, b , c), and (a, b, c ) are positive integer solutions where a = 3bc − a, b = 3ac − b, c = 3ab − c.

536

The weight of triple (a, b, c) is deﬁned as the maximum of a, b, c. How does the weight of the new triples relate to the weight of the old triple? Let us assume for the moment that a, b, c are pairwise distinct and satisfy a > b > c. We then derive (a − b)(b − a ) = a2 − b2 + 3b2 c − 3abc = a2 − b2 + 3b2 c − (a2 + b2 + c2 )

= 3b2 c − 2b2 − c2 = 2b2 (c − 1) + c(b2 − c) > 0. This implies a < b < a, and analogous arguments lead to b > a and c > a. Hence by ﬂipping the largest coordinate from a into a we decrease the weight of the triple, and by ﬂipping one of the smaller coordinates we increase the weight of the triple. Now let us repeatedly ﬂip the largest coordinate, so that the weight of the resulting triples keeps decreasing. Since the weight cannot decrease below zero, we must eventually get stuck with a triple whose coordinates are not pairwise distinct anymore. By symmetry we may assume that the root ﬂipping process terminates with b = c. Then (2) becomes a2 = (3a − 2)b2 , which implies that b divides a. Thus a = kb, and substituting this into the equation and rewriting yields k(3b − k) = 2. Hence k must be a divisor of 2: if k = 1 then b = c = 1 and a = 1; if k = 2 then b = c = 1 and a = 2. Let us summarize our ﬁndings. For every positive integer solution (a, b, c) of the Markov equation (2) with pairwise distinct coordinates, root ﬂipping produces three neighbor solutions. One of these neighbors has smaller weight, and is called the predecessor of (a, b, c). The other two neighbors have larger weight, and are called the successors of (a, b, c). If we start in an arbitrary solution and follow the chain of predecessors, we eventually arrive in one of the special solutions (1, 1, 1) or (2, 1, 1). Vice versa, by starting in (1, 1, 1) and by repeately moving to successors, we can reach every possible solution of equation (2). Figure 1 lists the ﬁrst few solutions of the equation.

Three shiny examples
Next, we want to discuss three concrete problems from mathematical competitions where that old root ﬂipping trick of Andrey Andreyevich Markov serves as a crucial ¨ ¨ (13, 5, 1) (1, 1, 1) (2, 1, 1) (5, 2, 1) d d (29, 5, 2) ¨ ¨ r r (169, 29, 2) (433, 29, 5) r r (194, 13, 5) (34, 13, 1)

··· ··· ··· ···

Figure 1: Some solutions of the Markov equation.

537

tool. Our ﬁrst problem was posed as problem B4 on the 1978 William Lowell Putnam Mathematics Competition. Problem 1 Prove that for every positive integer N , the equation a2 + b2 + c2 + d2 = abc + abd + acd + bcd has a solution in integers a, b, c, d ≥ N . Consider an arbitrary positive integer solution (a, b, c, d), and assume without loss of generality that a with a ≤ b, c, d is the smallest coordinate. For applying Markov’s root ﬂipping trick, we introduce the quadratic equation x2 − (bc + bd + cd)x + (b2 + c2 + d2 − bcd) = 0. (5) (4)

One root of this equation (5) is a itself, and the other root is a = bc + bd + cd − a. Since we selected a as the smallest coordinate in the quadruple, the second root satisﬁes a ≥ 3a2 − a > a. Hence ﬂipping the smallest coordinate will automatically increase the value of that coordinate! The rest is easy. We start from the trivial solution (1, 1, 1, 1), and repeatedly ﬂip the smallest coordinate. Eventually this must bring all coordinates above any ﬁxed bound N . This fully settles the Putnam question. It is instructive to see the actual sequence of solutions. Starting from (1, 1, 1, 1) we get successively (2, 1, 1, 1), (2, 4, 1, 1), (2, 4, 13, 1), (2, 4, 13, 85). One can show — and we encourage the reader to do so — that from this point onwards the four coordinates are always pairwise distinct. Furthermore if the coordinates are ordered as a < b < c < d, then 2a ≤ b, 2b ≤ c, 2c ≤ d and 2d ≤ a . In the year 1988, Markov’s old root ﬂipping trick made its ﬁrst appearance at the International Mathematical Olympiad. Only eleven students managed to ﬁnd the solution to the following problem. Problem 2 Let a and b be two positive integers such that q = Show that q is a perfect square. a2 + b2 ab + 1 is integer.

We choose a pair (a, b) with the smallest possible sum a + b among all pairs of positive integers that yield this ratio q and hence satisfy a2 + b2 − abq − q = 0. (6)

Without loss of generality we assume a ≥ b. The next step is already routine for us: in order to apply Markov’s root ﬂipping trick, we introduce the quadratic equation x2 − bq · x + (b2 − q ) = 0. (7) The ﬁrst root of (7) is a, and the second root is a = bq − a. Clearly a is integer, but is it positive or negative? For settling this question we will distinguish three cases on the sign of b2 − q .

538

In the ﬁrst case we deal with b2 − q > 0. Then the root a = (b2 − q )/a is indeed positive. Furthermore a ≥ b implies a = (b2 − q )/a < (a2 − q )/a < a. The resulting contradiction a + b < a + b shows that this ﬁrst case cannot occur. In the second case we deal with b2 − q < 0. Then (6) yields a(a − bq ) = 2 q − b > 0, which implies a > bq . Since (6) also yields q = b2 + a(a − bq ) > a > bq , we get the contradiction 1 > b. Hence also the second case cannot occur. All in all, the only remaining possibility is the third case with b2 − q = 0. But then q = b2 indeed is a perfect square, and we have arrived at the desired conclusion. Our next problem is the central piece of a slightly more diﬃcult question from the 2007 International Mathematical Olympiad. Problem 3 Let a and b be positive integers such that 4ab − 1 divides (a − b)2 . Show that a = b. Once again Markov’s root ﬂipping trick applies. Suppose that for some integer q ≥ 0 there exists a pair (a, b) of positive integers with a2 − 2ab + b2 = (4ab − 1)q. (8)

Among all such pairs (a, b) we pick one that minimizes the value max{a, b}, and without loss of generality we furthermore assume a ≥ b. The quadratic equation x2 − (2b + 4bq ) · x + (b2 + q ) = 0. has a and a = (b2 + q )/a as positive integer roots. Since we considered a pair with smallest possible value max{a, b}, we conclude a ≥ a and hence b2 + q ≥ a2 . By plugging this into (8) we derive (a − b)2 = (4ab − 1)q ≥ q ≥ a2 − b2 , which implies b ≥ a. Since we started from the assumption a ≥ b, we arrive at the desired conclusion a = b. This implies q = 0 and completes the proof.

Homework exercises
We challenge the reader to settle the following ﬁve problems along the lines indicated above. Problem 4 Show that there are inﬁnitely many quadruples of positive integers (a, b, c, d) that satisfy the equation a2 + b2 + c2 + d2 = abcd. One possible approach to this problem ﬁrst guesses a solution with a = b = c = d and then repeatedly applies Markov’s root ﬂipping trick. Problem 5 For q ≥ 4, show that the following equation has no solution over the integers: a2 + b2 + c2 = q · abc.

539

Note that for q = 3 the given equation coincides with the Markov equation (2) that we have discussed in detail. For q ≥ 4 one can actually recycle the machinery from the q = 3 case: every solution triple with pairwise distinct coordinates generates another solution triple with strictly smaller weight. But this time the ﬁnal step and the conclusion of the argument are diﬀerent, as the chain of predecessors does never terminate! This leads to an inﬁnite descent argument and shows that there are no solutions. Problem 6 Determine all positive integers q for which the following equation has a solution (a, b) over the positive integers. (a) (b) (c) a2 + b2 + 1 = q · ab a2 + b2 + 6 = q · ab

a2 + b2 − 1 = q · ab

In parts (a) and (b) you will detect all possible values for q by setting a = b = 1. For showing that these are the only possible values, you should evoke the root ﬂipping trick. Part (c) is a trick question, and you may want to understand what’s going on for a = q and b = q 2 − 1. Problem 7 Determine all positive integers q for which the following equation has a solution (a, b) over the positive integers. (a) (b) a2 + b2 = q · (ab − 1) a2 + b2 + ab = q · (ab − 1)

Part (a) of this problem is from the 2002 USA team selection test for the International Mathematical Olympiad, and part (b) is from the 2004 Romanian team selection test for the International Mathematical Olympiad. It is not diﬃcult to guess that the answer to part (a) is q = 5, and that the answer to part (b) is q = 4 and q = 7. Both parts can be attacked by Markov’s trick, and in both parts the root ﬂipping process gets stuck as soon as a ≥ a ≥ b holds (where as usually a denotes the old root and a denotes the new root). All eﬀort then goes into understanding and characterizing these terminal situations. Problem 8 Show that there are inﬁnitely many pairs of positive integers (a, b) a+1 b+1 for which + is a positive integer. b a This problem is taken from the second round of the 2007 British Mathematical Olympiad. The pair (1, 1) yields the integer value 4, and then once again Markov’s root ﬂipping trick does the job. Gerhard J. Woeginger Department of Mathematics and Computer Science TU Eindhoven P.O. Box 513, NL-5600 MB Eindhoven The Netherlands [email protected]

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PROBLEMS
Toutes solutions aux probl` emes dans ce num´ ero doivent nous parvenir au plus tard le 1 decembre 2012. Une ´ etoile ( ) apr` es le num´ ero indique que le probl` eme a ´ et´ e soumis sans solution. Chaque probl` eme sera publi´ e dans les deux langues oﬃcielles du Canada (anglais et fran¸ cais). Dans les num´ eros 1, 3, 5 et 7, l’anglais pr´ ec´ edera le fran¸ cais, et dans les num´ eros 2, 4, 6 et 8, le fran¸ cais pr´ ec´ edera l’anglais. Dans la section des solutions, le probl` eme sera publi´ e dans la langue de la principale solution pr´ esent´ ee. La r´ edaction souhaite remercier Jean-Marc Terrier, de l’Universit´ e de Montr´ eal, d’avoir traduit les probl` emes.

3670.

Correction. Propos´ e par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie.
1 0 1 0 1 0

Calculer dxdydz x+y+z .

´ . 3688. Propos´ e par Arkady Alt, San Jos´ e, CA, E-U Soit Tn (x) le polynˆ ome de Chebychev de premi` ere esp` ece d´ eﬁni par la r´ ecurrence Tn+1 (x) = 2xTn (x) − Tn−1 (x) pour n ≥ 1 et les conditions initiales T0 (x) = 1 et T1 (x) = x. Trouver tous les entiers positifs n tels que Tn (x) ≥ (2n−2 + 1)xn − 2n−2 xn−1 , x ∈ [1, ∞) . [Ed. : Noter que le probl` eme 3585 a ´ et´ e originellement imprim´ e avec la fausse in´ egalit´ e.]

3689. Propos´ e par Ivaylo Kortezov, Institut de Math´ ematiques et Informatique, Acad´ emie des Sciences, Soﬁa, Bulgarie.
Dans un groupe de n personnes, chacune poss` ede un livre diﬀ´ erent. Disons qu’une paire de personnes op` ere un ´ echange si elles s’´ echangent leur livre pr´ esentement en leur possession. Trouver le plus petit nombre possible d’´ echanges E (n), de sorte que chaque paire de personnes a proc´ ed´ ea ` au moins un ´ echange et que ﬁnalement chaque personne se retrouve avec son livre de d´ epart.

3690.

Propos´ e par Michel Bataille, Rouen, France.

Soit a, b et c trois nombres r´ eels positifs distincts avec a + b + c = 1. Montrer que (5x2 − 6xy + 5y 2 )(a3 + b3 + c3 ) + 12(x2 − 3xy + y 2 )abc > (x − y )2 pour tous les nombres r´ eels x et y .

541

3691.

Propos´ e par Pham Kim Hung, ´ etudiant, Universit´ e de Stanford, Palo ´ . Alto, CA, E-U

Soit a, b et c trois nombres r´ eels non n´ egatifs tels que a + b + c = 3. Montrer que b2 c c2 a a2 b + + ≤ 1. 4 − bc 4 − ca 4 − ab

3692.

Propos´ e par Nguyen Thanh Binh, Hano¨ ı, Vietnam.

Etant donn´ e un point arbitraire M dans le plan d’un triangle ABC , on d´ eﬁnit D, E et F comme les deuxi` emes points o` u le cercle circonscrit coupe respectivement les droites AM, BM , et CM . Si O1 , O2 et O3 sont les centres respectifs des cercles BCM, CAM et ABM , montrer que DO1 , EO2 et F O3 sont concourants.

3693.

Propos´ e par Michel Bataille, Rouen, France.

Etant donn´ ek∈ 1 , 0 , soit {an }∞ eﬁnie par a0 = 2, a1 = 1 n=0 la suite d´ 4 et la r´ ecursion an+2 = an+1 + kan . Evaluer
∞ n=1

an . n2

3694.
Vietnam.

Propos´ e par Pham Van Thuan, Universit´ e de Science des Hano¨ ı, Hano¨ ı,

Soit x, y et z trois nombres r´ eels non n´ egatifs tels que x2 + y 2 + z 2 = 1. Montrer que 1− x+y 2
2

1−

y+z 2

1−

x+z 2

≥

√

3695.

Propos´ e par Michel Bataille, Rouen, France.

Soit a un nombre r´ eel positif. Trouver toutes les fonctions strictement monotones f : (0, ∞) → (0, ∞) telles que (x + a)f (x + y ) = af (yf (x)) pour tous les x, y positifs.

3696.

Propos´ e par Nguyen Thanh Binh, Hano¨ ı, Vietnam.

Supposons que le cercle inscrit du triangle ABC touche les cˆ ot´ es BC en D , CA en E , et AB en F . Construire avec la r` egle et le compas les trois cercles mutuellement tangents qui sont tangents int´ erieurement au cercle inscrit, un en D , un en E 4 et un en F .

542

3697.

Propos´ e par Michel Bataille, Rouen, France.

Pour un entier positif n, montrer que tan π 7
6n

+ tan

2π 7

+ tan

3π 7

est un entier et trouver la plus haute puissance de 7 qui divise cet entier.

3698.

Propos´ e par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie.

Trouver la valeur de
1 n→∞

lim

1 + nn xn2 dx .

3699.

Propos´ e par Mehmet S ¸ ahin, Ankara, Turquie.

Dans un triangle ABC , soit I le centre de son cercle inscrit, ρa , ρb et ρc les rayons respectifs des cercles inscrits. Montrer que 1 ρa + 1 ρb + 1 ρc ≥ 18 tan(75◦ ) a+b+c .

3700.

Propos´ e par Michel Bataille, Rouen, France.

Soit ABC un triangle et a = BC , b = CA, c = AB . Si l’on a que aP A2 + cP B 2 + bP C 2 = cP A2 + bP B 2 + aP C 2 = bP A2 + aP B 2 + cP C 2 pour un certain point P , montrer que ABC est ´ equilat´ eral. .................................................................

3670.

Correction. Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.
1 0 1 0 1 0

Calculate

dxdydz x+y+z

3688.

Proposed by Arkady Alt, San Jose, CA, USA.

543

3689. Proposed by Ivaylo Kortezov, Institute of Mathematics and Informatics, Bulgarian Academy of Sciences, Soﬁa, Bulgaria.
In a group of n people, each one has a diﬀerent book. We say that a pair of people performs a swap if they exchange the books they currently have. Find the least possible number E (n) of swaps such that each pair of people has performed at least one swap and at the end each person has the book he or she had at the start.

3690.

Proposed by Michel Bataille, Rouen, France.

Let a, b, and c be three distinct positive real numbers with a + b + c = 1. Show that (5x2 − 6xy + 5y 2 )(a3 + b3 + c3 ) + 12(x2 − 3xy + y 2 )abc > (x − y )2 for all real numbers x and y .

3691.
CA, USA. that

Proposed by Hung Pham Kim, student, Stanford University, Palo Alto,

Let a, b, and c be nonnegative real numbers such that a + b + c = 3. Prove a2 b 4 − bc + b2 c 4 − ca + c2 a 4 − ab ≤ 1.

3692.

Proposed by Nguyen Thanh Binh, Hanoi, Vietnam.

For an arbitrary point M in the plane of triangle ABC deﬁne D, E , and F to be the second points where the circumcircle meets the lines AM, BM , and CM , respectively. If O1 , O2 , and O3 are the respective centres of the circles BCM, CAM , and ABM , prove that DO1 , EO2 , and F O3 are concurrent.

3693.

Proposed by Michel Bataille, Rouen, France.

, 0 , let {an }∞ Given k ∈ 1 n=0 be the sequence deﬁned by a0 = 2, a1 = 1 4 and the recursion an+2 = an+1 + kan . Evaluate
∞ n=1

an . n2

3694.
Vietnam.

Proposed by Pham Van Thuan, Hanoi University of Science, Hanoi,

Let x, y , and z be nonnegative real numbers such that x2 + y 2 + z 2 = 1. Prove that 1− x+y 2
2

1−

y+z 2

1−

x+z 2

≥

√

544

3695.

Proposed by Michel Bataille, Rouen, France. Find all strictly monotone functions

Let a be a positive real number. f : (0, ∞) → (0, ∞) such that

(x + a)f (x + y ) = af (yf (x)) for all positive x,y .

3696.

Proposed by Nguyen Thanh Binh, Hanoi, Vietnam.

Let the incircle of triangle ABC touch the sides BC at D , CA at E , and AB at F . Construct by ruler and compass the three mutually tangent circles that are internally tangent to the incircle, one at D , one at E , and one at F .

3697.

Proposed by Michel Bataille, Rouen, France.

For positive integer n, prove that tan π 7
6n

+ tan

2π 7

+ tan

3π 7

is an integer and ﬁnd the highest power of 7 dividing this integer.

3698.

Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.

Find the value of
1 n→∞

lim

1 + nn xn2 dx .

3699.

Proposed by Mehmet S ¸ ahin, Ankara, Turkey.

Let ABC denote a triangle, I its incenter, and ρa , ρb , and ρc the inradii of IBC , ICA, and IAB , respectively. Prove that 1 1 1 18 tan(75◦ ) + + ≥ . ρa ρb ρc a+b+c

3700.

Proposed by Michel Bataille, Rouen, France.

Let ABC be a triangle and a = BC , b = CA, c = AB . Given that aP A2 + cP B 2 + bP C 2 = cP A2 + bP B 2 + aP C 2 = bP A2 + aP B 2 + cP C 2 for some point P , show that ABC is equilateral.

545

SOLUTIONS
No problem is ever permanently closed. The editor is always pleased to consider for publication new solutions or new insights on past problems.

3589.

[2010 : 548, 550] Proposed by V´ aclav Koneˇ cn´ y, Big Rapids, MI, USA.

Find all integers n > 6 for which there exists a convex n-gon with an interior point P such that P Ai = Ai Ai+1 for each i, where indices are taken modulo n. Solution by Roy Barbara, Lebanese University, Fanar, Lebanon. We will see how to construct a suitable n-gon for n ≥ 6; the accompanying ﬁgure shows the construction for n = 7. Lemma. For m ≥ 3 there exists a convex (m +1)-gon A0 A1 A2 · · · Am such that if P is the midpoint of A0 Am , then we have P Ai = Ai Ai+1 for 0 ≤ i ≤ m − 1; furthermore, the angles P Am Am−1 and A1 A0 P will both be at most 60◦ . Suppose the lemma has been proved and ﬁx n ≥ 6. Set m = n − 3 (so that m ≥ 3) and let A0 A1 A2 · · · Am be as in the lemma. Orient the polygon so that Am A0 is a horizontal line segment whose midpoint is P , and attach the bottom half of the regular hexagon Am Am+1 Am+2 A0 that has Am A0 as a main diagonal; speciﬁcally, let Am+1 and Am+2 be the vertices of equilateral triangles erected below segments Am P and P A0 (as in the ﬁgure). Note that P Ai = Ai Ai+1 for i = m, m + 1, and m + 2, whence the resulting convex (m + 3)-gon together with the point P satisfy the conditions P Ai = Ai Ai+1 , as required.

A3 Am

A1 A0 P

Am+1 O

Am+2

546

Proof of the Lemma. Draw a horizontal segment A0 B of length 2 and deﬁne P to be its midpoint. (At the end of the proof the ﬁgure will be adjusted so that Am = B .) Let O be a point below A0 B on its perpendicular bisector , “suﬃciently far” from P (the meaning of which will soon be speciﬁed). Let Γ denote the circle centred at O that passes through A0 and B , and let γ denote the open arc A0 B above the segment A0 B . Let A1 be the point of γ for which A0 A1 = 1 (= P A0 ). Note that as O tends to inﬁnity away from P along , the limit position of the arc γ is the segment A0 B , so that the limit position of A1 is P . Consequently, we may situate O suﬃciently far from P so that P A1 < Note further that a point X ∈ Γ is on γ if and only if P X < 1. (2) 1 2m . (1)

For 2 ≤ i ≤ m, we deﬁne Ai recursively to be the unique point of Γ counterclockwise from Ai for which Ai Ai−1 = P Ai−1 . Claim: A1 , A2 , · · · , Am all lie on γ . This holds for A1 by its deﬁnition; we prove the claim for the remaining points. For i = 2, · · · , m, the triangle inequality applied to P Ai−1 Ai implies that P Ai ≤ P Ai−1 + Ai−1 Ai . Since P Ai−1 = Ai−1 Ai (by our construction), we get P Ai ≤ 2 · P Ai−1 for all i. By induction we deduce that P Ai ≤ 2i−1 · P A1 Using (3) and (1) we obtain for i ≤ m, P Ai ≤ 2i−1 · P A1 ≤ 2m−1 · P A1 < 2m−1 1 2m−1 < 1. 2 ≤ i ≤ m. (3)

By virtue of (2) we conclude that the points Ai all lie on γ as claimed. To conclude the proof, it remains to adjust the position of O so that Am = B . If we let O move along toward P , the circle Γ as well as the points A1 , · · · , Am are subjected to a continuous motion. For some position of O ∈ the point Am will coincide with the point B : The limit of A3 is B when O reaches P (because then A0 A1 A2 A3 would form the top half of a regular hexagon); when m > 3, our construction established that Am lies on the arc γ between B and A3 , so that Am would necessarily reach B before O reaches P . The construction is complete for that position of O for which Am = B . Convexity is guaranteed because all m + 1 vertices of the upper portion of the polygon lie on the arc of a circle.
Also solved by OLIVER GEUPEL, Br¨ uhl, NRW, Germany; ALBERT STADLER, Herrliberg, Switzerland; EDMUND SWYLAN, Riga, Latvia; and the proposer. There was one incorrect submission. It would be nice if some reader could provide simple answers to two questions that arise naturally from this problem: It is clear that there are inﬁnitely many diﬀerent solution n-gons when n > 6; is the regular hexagon the only convex 6-gon that satisﬁes the conditions of the problem, namely, there exists an interior point P such that P Ai = Ai Ai+1 for all i? Are there any such n-gons for n < 6?

547

3590. [2010 : 548, 551] Proposed by G.W. Indika Amarasinghe, University of Kelaniya, Kelaniya, Sri Lanka.
Let ABP C be a quadrilateral such that BC bisects the segment AP and AP bisects ∠BAC . Let a = BC , b = AC , c = AB , p = BP , and q = P C . Prove that p2 q2 + = b+c. c b ˇ Composite of nearly identical solutions by Sefket Arslanagi´ c, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; and Salem Maliki´ c, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina.
C

Denote by S the intersection point of AP and BC , and by x the length of the equal segments AS = SP . Applying Stewart’s theorem to triangles P CB and ABC and their respective cevians P S and AS , we get

q S x p P

b x A c

p2 · SC + q 2 · BS = a(BS · SC + x2 ) = b2 · BS + c2 · SC. Because AS bisects ∠BAC , we have BS =

(1)

ac ab and SC = . b+c b+c

Inserting these values of BS and SC into (1), we deduce that p2 ab b+c + q 2 ac b+c = b2 ac b+c + c2 ab b+c .

Finally, multiply both sides of the last equality by

b+c to get the desired result. abc

Also solved by ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, France; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; RICHARD I. HESS, Rancho Palos Verdes, CA, USA; JOE HOWARD, Portales, NM, USA; D. KIPP JOHNSON, Valley Catholic School, Beaverton, OR, USA; DAG JONSSON, Uppsala, Sweden; KEE-WAI LAU, Hong Kong, China; MADHAV R. MODAK, formerly of Sir Parashurambhau College, Pune, India; SCOTT PAULEY, NATALYA WEIR, and ANDREW WELTER, students, Southeast Missouri State University, Cape Girardeau, MO, USA; BOB SERKEY, Tucson, AZ, USA; EDMUND SWYLAN, Riga, Latvia; HAOHAO WANG and YANPING XIA, Southeast Missouri State University, Cape Girardeau, MO, USA; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer.

3591.

[2010 : 548, 551] Proposed by Michel Bataille, Rouen, France.

Let E be an ellipse with centre O . At exactly four points P of E , the tangent to E makes a 45◦ angle with OP . What is the eccentricity of E ? Solution by V´ aclav Koneˇ cn´ y, Big Rapids, MI, USA.

548

is e = 1 − Because of the symmetry of E about the axes, there will be exactly one point P in each quadrant where the tangent makes a 45◦ angle with OP , so we will restrict our attention to the ﬁrst quadrant. If P has coordinates y1 (x1 , y1 ), OP will have slope m = x . Moreover, the tangent at P will have 1 yy1 xx1 equation a2 + b2 = 1, and its slope will be mt = − b2 x1 a2 y
1

We choose coordinates so that E has equation
b2 . a2

x2 a2

y2 b2

= 1; its eccentricity

= (e2 − 1)

1 m

If θ is the acute angle at P between OP and the tangent, then tan θ = 1 + mmt m − mt = 1 e2 m + (1 − e2 ) 1 m .

The condition θ = 45◦ yields m2 − e2 m + (1 − e2 ) = 0, whence Ô m =
e2 ±

only if e4 + 4e2 − 4 = 0; because the eccentricity of an ellipse must lie between 0 √ and 1, we conclude that e = 2( 2 − 1) (which is about .91). It is of interest √ 2 to note that m = e = 2 − 1, so that OP makes an angle of 22.5◦ with the 2 x-axis and, consequently, so does the tangent at P .
Also solved by CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; RICHARD I. HESS, Rancho Palos Verdes, CA, ´ USA; GERHARDT HINKLE, Student, Central High School, Springﬁeld, MO, USA; VACLAV ˇ Y, ´ Big Rapids, MI, USA(a second solution); MADHAV R. MODAK, formerly of Sir KONECN Parashurambhau College, Pune, India; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. Note that as a point P moves about any ellipse centred at O from one vertex to the next, the smaller angle that OP makes with the tangent at P decreases from 90◦ to some minimum value, and then increases back to 90◦ . When the ellipse is a circle (whose eccentricity is e = 0), that angle is constant at 90◦ ; as e increases to 1 the ellipse ﬂattens to a line segment, while that minimum angle decreases to zero. Our featured solution shows how to determine the invertible function that relates the eccentricity of an ellipse to the minimum angle between OP and the tangent at P .

e4 − 4(1 − e2 ) . This last equation has a unique solution if and 2

ˇ . [2010 : 549, 551] Proposed by Faruk Zejnulahi and Sefket Arslanagi´ c, University of Sarajevo, Sarajevo, Bosnia and Herzegovina. Let a, b, and c be nonnegative real numbers such that a + b + c = 3. Prove or disprove that 19 20 ≤ 1 1+a+ b2 + 1 1+b+ c2 + 1 1+c+ a2 ≤ 27 20 .

3592

Solution by Oliver Geupel, Br¨ uhl, NRW, Germany. We show that both inequalities are false.

549

Let f (a, b, c) = Then 3 3 f 0, , 2 2 = 4 4 2 13 9 16 38 19 + + < + + = = 13 19 5 40 40 40 40 20 1 1+a+ b2 + 1 1+b+ c2 + 1 1 + c + a2 .

which invalidates the left inequality. A counterexample for the right inequality is given by f (0, 0.1, 2.9) = 1 1.01 + 1 9.51 + 1 3. 9 > 0.99+0.1051+0.2563 = 1.3514 > 27 20 .

Also solved by RICHARD I. HESS, Rancho Palos Verdes, CA, USA; D. KIPP JOHNSON, Valley Catholic School, Beaverton, OR, USA; PAOLO PERFETTI, Dipartimento di Matematica, Universit` a degli studi di Tor Vergata Roma, Rome, Italy; C. R. PRANESACHAR, Department of Mathematics, Indian Institute of Science, Bangalore, India; and STAN WAGON, Macalester College, St. Paul, MN, USA. 3 Johnson set a = 2 − x, b = 3 + x and c = 0 with − 3 ≤x≤ 3 and considered the 2 2 2 resulting function f (x). Using a TI84+ graphing calculator he found the minimum and maximum values of f (x) to be f (−0.0815361) ≈ 0.91692 < 19 , and f (1.441528) ≈ 20 , respectively. Using the Maximize and Minimize commands of the symbolic 1.3532 > 27 20 algebra, Mathematica, as he did in the past, Wagon found that these values obtained by Johnson are in fact the true extrema of the function f (a, b, c), and furthermore, they are the two real roots of the following polynomial of degree 8: 1753490160x8 − 7242829056x7 + 11338524468x6 − 8267726008x5 + 2802538373x4 − 513184720x3 + 132643611x2 − 8858430x + 2068173. In an 8-page solution, Pranesachar, with the help of MAPLE, obtained the same polynomial and actually gave a detailed and complete proof.

3593. [2010 : 549, 551] Proposed by Daryl Tingley, University of New Brunswick,
Fredericton, NB. Show that for all nonnegative integers n the rightmost nonzero digit of (4 · 5n )! is 4. Furthermore, show that if n ≥ k ≥ 0, then the string of k + 1 consecutive digits with this digit 4 at the right is independent of n. Solution by the proposer, modiﬁed and expanded by the editor. We ﬁrst deﬁne Pn =
∞ i=1

t such that 5t | (4 · 5n )! is given by
ë

(4 · 5n )! . By a well known formula, the largest integer 105n −1

E5 ((4 · 5n )!) =

4 · 5n 5i

= 4(5n−1 + 5n−2 + · · · + 5 + 1) = 5n − 1.
n

Since clearly E5 ((4 · 5n )!) ≤ E2 ((4 · 5n )!), 105 −1 | (4 · 5n )! so Pn is an integer. Note that the second part of the problem can now be restated as follows: show that if n ≥ k ≥ 0 then Pn ≡ Pk mod 10k+1 .

550

We also note in passing that E2 ((4 · 5 )!) = ≥
n ∞

4 · 5n 2i

∞ i=1

i=1 ∞ ê i=1

4·
∞ i=1

5 2

n− i

2i+2 · 5n−i = 2 3

22 · 5n ·

2 5

= 4 (5n )

2 3

= 2 · 5n +
n

· 5n > 2 · 5n + 1

so 22·5 +1 |Pn and in particular, 2k+1 |Pn if 0 ≤ k ≤ n. This fact will be used later on in the proof. and let S = {1, 2, · · · , 4 · 5n }. Fix k where Next deﬁne Rn = 55n −1 k + 1 ≤ n. Let A = {m ∈ S : 5 | m} and B = {m ∈ S : 5 m}. Then (4 · 5n )! = An Bn where An = m and Bn = m. For each m ∈ A factor out one copy of 5 to obtain
m∈A m∈B

(4 · 5n )!

An = 54·5

n−1

4 · 5n− 1 .

(1)

On the other hand, for each m ∈ B , we have m = q · 5k+1 + r where 0 ≤ q ≤ 4 · 5n−k−1 − 1 and 1 ≤ r < 5k+1 such that (r, 5) = 1. Hence, Bn =
4·5n−k−1 −1 q =0 5k+1

r =1, (r,5)=1

q · 5k+1 + r

(2)

Now recall the following generalization of Wilson’s Theorem by Gauss: [Ed.: cf. e.g., Elementary Number Theory and its Applications by Kenneth H. Rosen; 4th ed., page 204.] Theorem. For any integer m ≥ 1, k ≡ −1 (mod m) if m = 4, pa , 2pa 1 (mod m) otherwise

1≤k≤m, (k,m)=1

where p is an odd prime and a is any positive integer. By the theorem above we have
5k+1

r =1, (r,5)=1

q · 5k+1 + r ≡ −1 mod 5k+1

so (2) becomes Bn ≡
4·5n−k−1 −1 q =0

(−1) = (−1)4·5

n−k−1

= 1 mod 5k+1 .

(3)

551

From (1) and (3) we have An Bn 54 · 5 Rn = 5n −1 ≡ 5
n−1

55 n −1

4 · 5n− 1 !

4 · 5n− 1 ! = = Rn−1 55n−1 −1

mod 5k+1 .

k + 1 ≤ n we have, since 2, 5 φ that 25
n

Simple induction yields Rn ≡ Rk
k+1

mod 5k+1

= 1, by Euler’s Theorem on totient function

for k ≤ n. Furthermore, for

−1

= =

1 2 1 2

· 24 · 5

n−1

· 25 )

n−1

1 2

24 · 5
n−1

5n−k−1

· 25

n−1

2φ(5

k+1

5n−k−1

· 25

≡
−1

1 2

(1) · 25

n−1

= 25

n−1

−1

mod 5k+1 .

Inductively, we then have 25 25
k

−1

≡ 25

mod 5k+1 for k ≤ n. So,
k

−1

Pn ≡ 25
k

−1

Pn = Rn ≡ Rk = 25

−1

mod 5k+1

Also solved by ALBERT STADLER, Herrliberg, Switzerland. A partial solution for k ≤ 1 was submitted by OLIVER GEUPEL, Br¨ uhl, NRW, Germany.

mod 10k+1 . Finally, P0 = 4! ≡ 4 (mod 10) so the rightmost nonzero digit of (4 · 5n )! is 4 for all n ≥ 0, and our proof is complete.

for k ≤ n. Since 25 −1 , 5k+1 = 1, it follows that Pn ≡ Pk mod 5k+1 . Since both Pn and Pk are divisible by 2k+1 as mentioned and proved above, we have Pn ≡ Pk mod 2k+1 . Since 5k+1 , 2k+1 = 1, it follows that Pn ≡ Pk

3594.

[2010 : 549, 551] Proposed by Michel Bataille, Rouen, France.

Let x, y , z be three indeterminates and A = (y − z )(y + x)(x + z ), B = (z − x)(z + y )(y + x), C = (x − y )(x + z )(z + y ). Find all polynomials P , Q, R ∈ C[x, y, z ] such that x2 P + y 2 Q + z 2 R xP + yQ + zR Solution by the proposer. Let N = x2 A + y 2 B + z 2 C , D = xA + yB + zC and F = N . Since D N and D both vanish when x = y or y = z or z = x, they are both divisible by (x − y )(y − z )(z − x). Thus, F is not in its lowest form and we simplify it. = x2 A + y 2 B + z 2 C xA + yB + zC .

552

Denoting by S the polynomial xy + yz + zx, we have D = x(y − z )(x2 + S ) + y (z − x)(y 2 + S ) + z (x − y )(z 2 + S )

= x3 (y − z ) + y 3 (z − x) + z 3 (x − y ) = (z − y )(y − x)(x − z )(x + y + z ) and similarly N = x4 (y − x) + y 4 (z − x) + z 4 (x − y )

= x3 (y − z ) + y 3 (z − x) + z 3 (x − y ) + S · [x(y − z ) + y (z − x) + z (x − y )]

= (z − y )(y − x)(x − z )(x + y + z )2 .

= (z − y )(y − x)(x − z )(x2 + y 2 + z 2 + S ) + S · (z − y )(y − x)(x − z )

+ S · [x2 (y − z ) + y 2 (z − x) + z 2 (x − y )]

[Ed.: By induction, we see that for every n ≥ 2, xn (y − z )+ y n (z − È integer n a b c x) + z (x − y ) = (z − y )(y − x)(x − z ) x y z , where the sum is extended to all triples (a, b, c) of nonnegative integers for which a + b + c = n − 2.] From these results, we deduce that F = x + y + z . The problem reduces to ﬁnding P , Q, R such that x2 P + y 2 Q + z 2 R = (x + y + z )(xP + yQ + zR), that is, xy (P + Q) + yz (Q + R) + zx(R + P ) = 0. (1) Thus, z must divide P + Q and we can set P + Q = 2zW . Similarly Q + R = 2xU and R + P = 2yV for some polynomials U , V , W . Substituting in (1), we see that U + V + W = 0. Solving for P , Q, R we easily ﬁnd P = −xU + yV + zW, Q = xU − yV + zW, R = xU + yV − zW.

We can select polynomials K1 , K2 , K3 (in many ways) such that U = K2 − K3 , V = K3 − K1 , W = K1 − K2 . We obtain P = (z − y )K1 − (z + x)K2 + (x + y )K3 , Q = (y + z )K1 + (x − z )K2 − (x + y )K3 ,

R = −(y + z )K1 + (z + x)K2 + (y − x)K3 .

Conversely, if K1 , K2 , K3 are arbitrary polynomials in x, y , z , then (1) is satisﬁed and we have the general form of the solution. Note that the obvious solutions (A, B, C ) is obtained with K1 = −x2 , K2 = −y 2 , K3 = −z 2 .

553

There were no other solutions. However, Albert Stadler, Herrliberg, Switzerland noted that (x2 P + y 2 Q + z 2 R)(xA + yB + zC ) − (xP + yQ + zR)(x2 A + y 2 B + z 2 C ) =(x − y )(y − z )(z − x)(x + y + z )[x(y + z )P + y (z + x)Q + z (x + y )R]. The expression in square brackets must vanish and Stadler isolated some properties of the polynomials P , Q, R.

3595.
that

[2010 : 549, 551-552] Proposed by Bill Sands, University of Calgary, Calgary, AB. Let a, b, n be positive integers satisfying a < b and n < a + b, and so exactly
1 of the integers a2 , a2 + 1, a2 + 2, . . . , b2 are squares. n

(1)

Do the following: (a) Prove that also exactly
1 of the consecutive integers n

(n − a)2 , (n − a)2 + 1, (n − a)2 + 2, . . . , b2 are squares. (b) Exactly the integers (n − 1)2 = n2 − 2n + 1, n2 − 2n + 2, . . . , n2 are squares. Thus, for every integer n ≥ 3, the values a = 1, b = n and a = n − 1, b = n always satisfy (1). For which integers n ≥ 3 are these the only solutions of (1)? Solution by Kathleen E. Lewis, University of the Gambia, Brikama, Gambia. (a) If a < n − a, the list of integers a2 , a2 + 1, . . . , (n − a)2 − 1 contains (n − a)2 − a2 = n2 − 2an integers, of which (n − a) − a = n − 2a, or total, are squares. Therefore if then it follows that
1 of the list a2 , a2 + 1, . . . , b2 consists of squares, n 1 of the n 1 1 of the integers 1, 2, . . . , n2 are squares, and also exactly of n n

b2 must consist of squares. In the same way, if n − a < a, the sequence (n − a)2 , (n − a)2 + 1, . . . , a2 − 1 contains a2 − (n − a)2 = 2an − n2 numbers of which
1 of them, are squares, so once again, the list n 1 (n − a)2 , (n − a)2 + 1, . . . , b2 must have of its elements being squares. n

1 of the diﬀerence of the lists, (n − a)2 , (n − a)2 + 1, . . . , n

a − (n − a) = 2a − n, which is

squares, so the proportion of squares is

(b) The list a2 , a2 +1, . . . , b2 contains b2 − a2 +1 numbers, of which b − a +1 are
b−a+1 1 . Setting this equal to gives b2 − a 2 + 1 n

b2 − a2 +1 = (b − a +1)n, which can be rearranged to (b − a)(b + a − n) = n − 1. Setting one of the factors on the left equal to 1 and the other equal to n − 1 yields the trivial solutions (a, b) = (1, n) and (a, b) = (n − 1, n). If n − 1 is prime, these are the only solutions. Also, in the case that n − 1 is a power of 2, they are

554

the only solutions. This is because the only nontrivial factorization of a power of 2 consists of two even numbers. But if n − 1 is even, n is odd and b − a and b + a − n have opposite parity. Suppose that n − 1 is odd and not a prime. Then it can be written as a nontrivial product of two odd integers. Since b − a and b + a − n have the same parity, they can be set equal to these integers to get integer values of a and b. On the other hand, if n − 1 is even but not a power of 2, it can be written as the product of an even and an odd integer, both exceeding 1. Since n is odd, b − a and b + a − n have opposite parity and we can once again get a nontrivial solution for a and b.
Also solved by OLIVER GEUPEL, Br¨ uhl, NRW, Germany; JOEL SCHLOSBERG, Bayside, NY, USA; and the proposer.

3596. [2010 : 549, 552] Proposed by Paolo Perfetti, Dipartimento di Matematica,
Universit` a degli studi di Tor Vergata Roma, Rome, Italy. Let x, y and z be positive real numbers. Prove that x(y + z )
cyclic

(x + 2y +

2z )2

≤

(x + y )(x + y + 2z )
cyclic

(3x + 3y + 4z )2

ˇ Composite of similar solutions by Sefket Arslanagi´ c, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; and D. Kipp Johnson, Valley Catholic School, Beaverton, OR, USA, expanded slightly by the editor. Since both summations are homogeneous we may assume that x + y + z = 1. Then the given inequality is equivalent, in succession, to (y + z )(y + z + 2x)
cyclic

(3y + 3z +

4x)2

−

x(y + z ) (x + 2y + 2z )2 ≥0
¡

≥0

cyclic

(1 − x)(1 + x) x(1 − x) − (3 + x)2 (2 − x)2

(1 − x) (1 + x)(2 − x)2 − x(3 + x)2
cyclic

(3 + x)2 (2 − x)2 − x)2 + +

≥0 ≥ 0 as (3x − 1) = 0

(1 − x)(4 − 9x − 9x2 )
cyclic

(3 +

x)2 (2 −

27(3x − 1) 250 ≥0

cyclic

(3 +

9x3 − 13x + 4 x)2 (2

x)2

27(3x − 1) 250

(3x − 1)(250(3x2 + x − 4) + 27(3 + x)2 (2 − x)2 )
cyclic

250(3 + x)2 (2 − x)2

≥0

555

(3x − 1)(27x4 + 54x3 + 453x2 − 74x − 28)
cyclic

250(3 + x)2 (2 − x)2

≥0

(3x − 1)2 (9x3 + 21x2 + 158x + 28)
cyclic

250(3 + x)2 (2 − x)2

≥0

The last inequality above is clearly true. Clearly the equality holds if and only if x = y = z .
Also solved by CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Br¨ uhl, NRW, Germany; KEE-WAI LAU, Hong Kong, China; JOEL SCHLOSBERG, Bayside, NY, USA; and the proposer.

3597.

[2010 : 550, 552] Proposed by Johan Gunardi, student, SMPK 4 BPK PENABUR, Jakarta, Indonesia. One hundred students take an exam consisting of 50 true or false questions. Prove that there exist three students whose answers coincide for at least 13 questions. Solution by D. Kipp Johnson, Valley Catholic School, Beaverton, OR, USA.

Suppose that for each question i, 1 ≤ i ≤ 50, ai students answer ’TRUE’ and 100 − ai students answer ’FALSE’. The number of sets of three students whose answers agree for question i is then ai 3
¡

100 − ai 3

= 0 when n < k. with the usual convention that n k Now for any integer x, 1 ≤ x ≤ 100, x 100 − x + 3 3 = 49(x2 − 100x + 3300) = 49((x − 50)2 + 800) ≥ 39200

Thus the average number of coinciding answers among all sets of three students is
50 i=1

ai 3

100 − ai 3 ≥

100 3

(50)(39200) ≈ 12.12 161700

However, the actual number of coinciding answers for any set of three students is an integer, so at least one set of three students must have coinciding answers for at least 13 questions.
Also solved by OLIVER GEUPEL, Br¨ uhl, NRW, Germany; and the proposer.

556

[2010 : 550, 552] Proposed by Zhang Yun, High School attached to Xi / An Jiao Tong University, Xi An City, Shan Xi, China.

3598.

The quadrilateral ABCD has both a circumscribed circle and an inscribed circle, the latter with centre I . Put a = AB , b = BC , c = CD , and d = DA. Prove that IB 2 IC 2 ID 2 IA2 + + + = 2. ab bc cd da I. Solution by Salem Maliki´ c, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina. Let ∠DAB = α and ∠ABC = β . Since ABCD is cyclic, ∠BCD = 180◦ − α and ∠CDA = 180◦ − β . Also, since I is the centre of the inscribed circle, AI bisects ∠BAD , so ∠BAI = ∠DAI = α . Similar results hold for 2 angles at other vertices. Using The Law of Sines, we have IB 2 ab IA
2

= =

IB a IA d

· ·

IB b IA a

= =

α 2 β sin α+ 2 cos β 2 β cos α− 2

sin

· ·

α 2 α cos β− 2 sin β 2 β sin α+ 2

cos

Summing the last two equalities, we have IB 2 ab + IA2 da = sin
α 2

cos

sin

α 2 α+β 2

+ sin · cos +

β cos β 2 2 α−β 2

sin α + sin β sin α + sin β

= 1.

Since for similar reasons we have holds.

IB 2 ab

IA2 da

= 1, then the desired proposition

II. Solution by Oliver Geupel, Br¨ uhl, NRW, Germany. Let E , F , G, and H denote the H v D feet of the perpendiculars from the point A s I to the lines AB , BC , CD , and DA, v Es respectively. Since I is the centre of r G the inscribed circle, then EI = HI , r t AE = AH , BE = BF , CF = I CG, and DG = DH . Let these lengths be denoted r , s, t, u, and v , B u O respectively. Since the quadrilateral is t ◦ cyclic, we have ∠HDI = 90 − F ∠EBI = ∠BIE . Hence, the right u triangles BEI , and IHD are similar, C which yields r = v . Analogously, we t r r u have s = r . Therefore r 2 = su = tv.

557

Since IB 2 · c + IC 2 · a = (r 2 + t2 )(u + v ) + (r 2 + u2 )(s + t)

= t2 (u + v ) + u2 (s + t) + r 2 (s + t + u + v ) = t2 (u + v ) + u2 (s + t) + stv + stu + tuv + suv = t2 (u + v ) + u2 (s + t) + st(u + v ) + uv (s + t) = (u + v )t(s + t) + (s + t)u(u + v ) = (s + t)(t + u)(u + v ) = abc,

then

IC 2 IB 2 + = 1. ab bc ID 2 cd + IA2 da = 1,

Analogously,

hence the proposition.
ˇ Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Greece; SEFKET ´ University of Sarajevo, Sarajevo, Bosnia and Herzegovina; ROY BARBARA, ARSLANAGI C, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, France; PRITHWIJIT DE, Homi Bhabha Centre for Science Education, Mumbai, India; KEE-WAI LAU, Hong Kong, China; MADHAV R. MODAK, formerly of Sir Parashurambhau College, Pune, India; JOEL SCHLOSBERG, Bayside, NY, USA; EDMUND SWYLAN, Riga, Latvia; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer.

3599

. [2010 : 550, 552] Proposed by Cristinel Mortici, Valahia University of Tˆ argovi¸ ste, Romania. Let m and n be positive integers such that 2m − 3n ≥ n. Prove that 2m − 3n ≥ m .

Solution by Chip Curtis, Missouri Southern State University, Joplin, MO, USA. The claim is false, as can be seen by taking (m, n) = (2, 1). The inequality 2m − 3n ≥ n becomes 1 ≥ 1, whereas the inequality 2m − 3n ≥ m becomes 1 ≥ 2. The following modiﬁcation of the problem does have a solution. Suppose m and n are positive integers such that 2m−1 − 3n ≥ n. Then 2m − 3n ≥ m + n. (2) (1)

558

Set Then

For the proof, ﬁx n, and suppose (1) holds. Then m ≥ 1 + log2 (3n + n). fn (m) = 2m − m − 3n . fn (m) = 2m · ln 2 − 1,

which is greater than 0 for m ≥ 1. Hence fn (m) is increasing on [1, ∞). Therefore, fn (m) ≥ fn (1 + log2 (3n + n)) = 21+log2 (3
n

+n)−[1+log2 (3n +n)]−3n

= (3n + 2n − 1) − log2 (3n + n). For t > 0, deﬁne g (t) = t − log2 t. Since g (t) = 1 −
1 , t ln 2

= 2(3n + n) − 1 − log2 (3n + n) − 3n

1 we have g (t) > 0 if and only if t > ln . In particular, g (t) > 0 for t ≥ 2. 2 Hence, g (t) is increasing on [2, ∞), so g (t) ≥ g (2) = 1 for t ≥ 2. Also, g (1) = 1. Thus, g (k) ≥ 1 for all positive integers k. We now have

fn (m) ≥ g (3n + n) + (n − 1) ≥ n. This implies that 2m − 3n ≥ m + n, as claimed.
OLIVER GEUPEL, Br¨ uhl, NRW, Germany also pointed out the claim was false, citing the same counterexample as Curtis.

3600.
Romania.

[2010 : 550, 552] Proposed by Ovidiu Furdui, Campia Turzii, Cluj,

Let k ≥ 1 be a nonnegative integer. Prove that
∞ n1 ,n2 ,...,nk =1

1 (n1 + n2 + · · · + nk )!

= (−1)k−1

k− 1 j =0

(−1)j j!

− 1

Solution by Michel Bataille, Rouen, France.

559

For positive integer m, the cardinality of the set of all k −tuples of positive −1¡ integers (n1 , n2 , . . . , nk ) with n1 + n2 + · · · + nk = m is m , this number k− 1 being 0 if k > m. Let S denote the left side. Since all terms in the sum are nonnegative, S= that is S= 1 (k − 1)!
∞ m=1 ∞ m=1 n1 +n2 +···+nk =m

1 (n1 + n2 + · · · + nk )!

∞ m=1

m−1 k−1

1 m!

(m − 1)(m − 2) . . . (m − k + 1)

1 1 = f (k−1) (1), m! (k − 1)!

where, for positive x, f (x) = By Leibniz’ formula, f
(k−1)

ex x

−

1 x

∞ m=1

xm−1 m!

(x) =

k− 1 j =0

k−1 x e · (−1)j j !x−(j +1) − (−1)k−1 (k − 1)!x−k . j

Hence 1 S= (k − 1)! =e
k− 1 j =0 k− 1 j =0

k−1 e · (−1)j j ! − (−1)k−1 (k − 1)! j − (−1)k−1
¿

(k − 1 − j )!
¾
k− 1 j =0 k− 1 j =0

(−1)j

= (−1)k−1 e
¾

(k − 1 − j )!
¿

(−1)j −k+1

−1

= (−1)k−1 e

(−1)j j!

−1 .

Also solved by DON KRUG, Northern Kentucky University, Highland Heights, KY, USA; and the proposer. The proposer showed that the sum was equal to Sk−1 (1)/(k − 1)!, where
∞

Sk−1 (x) =
j =k

xj = (j − k)!

tk−1 et dt.
0

Integration by parts yields the recurrence relation Sk−1 (1) = e − (k − 1)Sk−2 (1), from which the desired result follows. Krug followed a similar strategy.

560

YEAR END FINALE
We have made it! It has been an interesting year here at CRUX with MAYHEM. There were delays getting me appointed, changes in personnel, technical diﬃculties, and more delays renewing the contract between the Ottawa Carleton District School Board, my employer, and the CMS. At the same time we made some changes to CRUX with MAYHEM and developed a plan for the journal in the future. The journal saw some changes this year, with the introduction of a new Olympiad Corner format, and there will be more changes to come. We have worked out a plan for CRUX with MAYHEM that we will be implementing in volume 39. As a result, volume 38 will be streamlined as we prepare for the changes that will take place the following year. In volume 38 we will drop from 8 to 6 issues. Also, Mathematical Mayhem and Skoliad will be separating from Crux Mathematicorum. The solutions to all problems from these sections that have appeared in print in CRUX with MAYHEM will appear in volume 38, but there will be no new problems. Mathematical Mayhem (with Skoliad) will carry on separately and be expanded. To ﬁll the void, a new problems column will be created so there will be more problems for you to solve. The solutions to all problems from the regular problems section that have appeared in volume 37 will appear in volume 38 and we will continue to publish 100 problems per volume. Also the solutions to the new Olympiad Corner problems will start to appear. As Vazz pointed out at the end of the last volume, he had cleared the backlog of articles, which left us with few for this year. You may notice that we ended 8 pages short, which was due, in part, to the shortage of articles. To help us get caught up, we are planning not to publish any articles in the next volume. The hope is that articles will return in volume 39 on a more regular basis. We will work though volume 38 as quickly as possible, and let you know when we have a ﬁrm idea of what Crux Mathematicorum will look like for volume 39 and onward. Our hope is that the publication schedule will be back on track by the end of volume 39. I would like to thank our readers for their patience in this time of change and delays. The kind words and encouragement that you have sent to me are greatly appreciated. I will continue to work hard to make Crux Mathematicorum the best it can be. I am not alone putting the journal together, there are a number of people that I must ´ thank. First, I must thank VACLAV (VAZZ) LINEK. Since I started this position, Vazz was always there to help me learn the ropes. Through numerous emails and a few phone calls he helped me quickly ﬁgure out all that I needed to do. Even though the number of times I contact him has greatly dropped oﬀ, when I need him, he is always there. All the best to you in the future Vazz and thanks. Next, I must thank the editorial staﬀ. I thank my associate editor JEFF HOOPER, whose detailed proofreading picks up many of the things I have missed. I thank CHRIS FISHER for his work as problems editor and for all his other contributions like his history of CRUX with MAYHEM and his current column Recurring Crux Conﬁgurations. His dedication and expertise are greatly appreciated. I thank EDWARD WANG for his work as problems editor as well as his continued contribution of nice problems and solutions to the MAYHEM section. Our readers greatly beneﬁt from his contributions. I thank NICOLAE STRUNGARU for his dual role as problem editor and new editor to the Olympiad Corner. His willingness to take over the column, while continuing his old role until it could be ﬁlled, has helped to ﬁll the void left when Robert stepped down last year. I also thank CHRIS GRANDISON for his precise work as a problems editor. This year IAN AFFLECK, ROBERT CRAIGEN, and COSMIN POHOATA have left the editorial board. I thank them for their time with us, their work has been greatly appreciated. I want to thank EDWARD BARBEAU, RICHARD GUY and BILL SANDS. When I took over, I approached these gentlemen for some “unoﬃcial” help. They were a great help to me proofreading and working my way through some of the backlog of problems from our ﬁles. I am thrilled that Ed has agreed to join the board as a problems editor and that Bill has agreed to become the Editor at Large. I also want to welcome ANNA KUCZYNSKA to the board as problems editor and thank her for the great job she has done so far. I must thank JOANNE CANAPE and ROBERT WOODROW at the University of Calgary, for continuing to send those last sets of solutions for the Olympiad Corner. Your dedication

561

to this column and CRUX with MAYHEM for over 20 years is greatly appreciated and will be sorely missed. I thank JEAN-MARC TERRIER and ROLLAND GAUDET for providing French translations. The translations are always done quickly and occasionally there are comments on how to improve the English part as well! I thank ROBERT DAWSON for taking over as the articles editor and for providing us with such nice articles this year. I thank AMAR SODHI for keeping me in good supply of interesting book reviews (I have purchased and read several of these books already). I thank LILY YEN and MOGENS LEMVIG HANSEN for their work as Skoliad Editors and the great job they do. I thank the MAYHEM staﬀ ANN ARDEN, NICOLE DIOTTE, MONIKA KHBEIS and DAPHNE SHANI for their help preparing solutions. I especially thank my assistant editor LYNN MILLER for her dedication and extra help behind the scenes. I thank IAN VANDERBURGH for his time doing The Problem of the Month, it was a pleasure working with you and your column will be missed. I thank the staﬀ at the CMS head oﬃce in Ottawa. In particular I thank JOHAN RUDNICK, DENISE CHARRON, and STEVE LA ROCQUE. Their behind the scenes support of me at CRUX with MAYHEM is greatly appreciated. I thank TAMI EHRLICH and the staﬀ at Thistle Printing for taking the ﬁles that I send them and turning them into the journal you have in your hands. I must also thank the OTTAWA DISTRICT SCHOOL BOARD for partnering with the CMS to allow me to become the Editor-in-Chief. In particular I would like to thank director JENNIFER ADAMS, superintendent PINO BUFFONE, Human Resources Oﬃcer JENNIFER BALDELLI and principal KEVIN GILMORE for their support and encouragement. I look forward to working with all of you through the next n volumes. I also want to thank my wife JULIE and my sons SAMUEL and SIMON for their support during this past year. You guys are my rock (and roll!). Finally, I sincerely thank all of the readers of CRUX. Without your problems, solutions, articles and comments there is no CRUX. I look forward to your contributions, letters and email for the next volume. Shawn Godin

Crux Mathematicorum
with Mathematical Mayhem
Former Editors / Anciens R´ edacteurs: Bruce L.R. Shawyer, James E. Totten, V´ aclav Linek

Crux Mathematicorum
Founding Editors / R´ edacteurs-fondateurs: L´ eopold Sauv´ e & Frederick G.B. Maskell Former Editors / Anciens R´ edacteurs: G.W. Sands, R.E. Woodrow, Bruce L.R. Shawyer

562

INDEX TO VOLUME 37, 2011
Skoliad Lily Yen and Mogens Lemvig Hansen February No. 130 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 March No. 131 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 April No. 132 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 May No. 133 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 September No. 134 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 October No. 135 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337 November No. 136 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 December No. 137 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481 Mathematical Mayhem Shawn Godin February ........................................................... 9 March . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 April . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 May . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 September . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266 October . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 November . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415 December . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 488 Mayhem Problems February M470–M475 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 March M476–M481 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 April M482–M487 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 May M488–M494 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 September M495–M500 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 October M501–M506 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 November M507–M512 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415 December M513–M518 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489 Mayhem Solutions February M432–M437 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 March M438, M439, M441, M443, M444 . . . . . . . . . . . . . . . . . . . . . . . . 78 April M440, M442, M445–M450 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 May M451–M456 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 September M457–M462 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 October M463–M469 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 November M470–M475 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 418 December M476–M481 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 491 Problem of the Month Ian VanderBurgh February . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 March . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 April . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 May . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206 September . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 Miscellaneous Mayhem Editorial Shawn Godin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266 Mayhem Year End Wrap Up Shawn Godin . . . . . . . . . . . . . . . . . . . . . . . . . 488

563

The Olympiad Corner R.E. Woodrow and Nicolae Strungaru February No. 291 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 March No. 292 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 April No. 293 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 May No. 294 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 September No. 295 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275 October No. 296 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352 November No. 297 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423 December No. 298 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496 Book Reviews Amar Sodhi Alex’s Adventures in Numberland, by Alex Bellos Reviewed by Bruce Shawyer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 The Calculus of Friendship, by Steven Strogatz Reviewed by Georg Gunther . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 Pythagoras’ Revenge: A Mathematical Mystery, by Arturo Sangalli Reviewed by Mark Taylor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168 Icons of Mathematics: An Exploration of Twenty Key Images, by Claudi Alsina and Roger B. Nelsen Reviewed by Edward J. Barbeau . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232 Lobachevski Revisited, by Seth Braver Reviewed by J. Chris Fisher . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301 Charming Proofs: A Journey Into Elegant Mathematics, by Claudi Alsina and Roger B. Nelsen Reviewed by R. P. Gallant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384 Magical Mathematics : The Mathematical Ideas That Animate Great Magic Tricks, by Persi Diaconis and Ron Graham Reviewed by S. Swaminathan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447 Loving + Hating Mathematics: Challenging the Myths of Mathematical Life, by Reuben Hersh and Vera John-Steiner Reviewed by Georg Gunther . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 526 The Beauty of Fractals: Six Diﬀerent Views, edited by Denny Gulick and Jon Scott Reviewed by Daryl Hepting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 528 Recurring Crux Conﬁgurations J. Chris Fisher
304 385 449 304

Triangles for which 2b2 = c2 + a2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Triangles for which 2b = c + a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Triangles for which 2B = C + A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Bicentric Quadrilaterals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Crux Articles

Robert Dawson

Summations according to Gauss, by Gerhard J. Woeginger . . . . . . . . . . . . . . . . . 308 A nest of Euler Inequalities, by Luo Qi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312 That old root ﬂipping trick of Andrey Andreyevich Markov Gerhard J. Woeginger . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 535

564

Problems
February March April May September October November December 3601–3612 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 3613–3625 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 3626–3637 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170 3638–3650 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234 3650, 3651–3663 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318 3651, 3664–3675 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388 3676–3687 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454 3670, 3688–3700 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 540 3501–3513 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 3515–3520, 3523–3525 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 3521, 3522, 3527–3538 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175 3526, 3539–3550 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240 3224, 3551–3555, 3557–3562 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323 3542, 3556, 3563–3567, 3569–3572, 3574, 3575 . . . . . . . . . . . . . . . . . 393 3576–3587 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454 3589–3600 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 545

Solutions
February March April May September October November December

Miscellaneous
New Editor-in-Chief for CRUX with MAYHEM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Editorial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Crux Chronology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 Editorial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 Editorial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 Editorial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257 Unsolved Crux Problems 342 and 1754 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303 Unsolved Crux Problem 154 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 448 Year End Finale . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 560

565

Proposers and solvers appearing in the SOLUTIONS section in 2011:

Proposers
Anonymous Proposer 3525, 3566 Yakub N. Aliyev 3505, 3518 Arkady Alt 3556, 3570, 3571, 3585, 3688 G.W. Indika Amarasinghe 3590 George Apostolopoulos 3603, 3628, 3644 ˇ Sefket Arslanagi´ c 3584, 3592, Vahagn Aslanyan 3555, 3562 Roy Barbara 3640 Ricardo Barroso Campos 3520 Michel Bataille 3514, 3529, 3532, 3545, 3546, 3574, 3575, 3591, 3594, 3604, 3608, 3617, 3623, 3631, 3634, 3638, 3642, 3648, 3650, 3656, 3659, 3666, 3669, 3675, 3676, 3678, 3680, 3683, 3686, 3693, 3695, 3697, 3700 Mih´ aly Bencze 3534, 3561 K.S. Bhanu 3531 J´ anos Bodn´ ar 3516 N. Javier Buitrago Aza 3552 Cao Minh Quang 3526, 3533 Shai Covo 3586 Max Diaz 3565 Jos´ e Luis D´ ıaz-Barrero 3502, 3515, 3539, 3547, 3605, 3627, 3641, 3645 A.A. Dzhumadil’daeva 3573 Juan Jos´ e Egozcue 3645 J. Chris Fisher 3224 Ovidiu Furdui 3512, 3530, 3550, 3551, 3578, 3580, 3612, 3618, 3624, 3633, 3637, 3646, 3652, 3655, 3670, 3671, 3673, 3677, 3685, 3698, Dinu Ovidiu Gabriel 3616 Samuel G´ omez Moreno 3536 Johan Gunardi 3558, 3597 John G. Heuver 3620 Joe Howard 3667 Hung Pham Kim 3508, 3509, 3527, 3549, 3619, 3639, 3651, 3664, 3679, 3691 Ignotus 3587 Walther Janous 3535 Neven Juriˇ c 3614, 3668 Hiroshi Kinoshita 3528 Mikhail Kochetov 3563 V´ aclav Koneˇ cn´ y 3517, 3589, 3606 Ivaylo Kortezov 3689 Panagiote Ligouras 3582, 3609, 3632, 3647, Jian Liu 3569 Thanos Magkos 3559, 3626, 3657 Dorin M˘ arghidanu 3521, 3522 Marian Marinescu 3537 George Miliakos 3607 Dragoljub Miloˇ sevi´ c 3660 Cristinel Mortici 3599 Nguyen Duy Khanh 3519 Nguyen Thanh Binh 3665, 3681, 3684, 3692, 3696 Victor Oxman 3538 Pedro Henrique O. Pantoja 3506, 3663 Paolo Perfetti 3557, 3583, 3596 Pham Huu Duc 3507, 3554 Pham Van Thuan 3511, 3548, 3560, 3564, 3602, 3615, 3625, 3636, 3643, 3649, 3654, 3672, 3682, 3694 Cosmin Pohoat ¸a ˘ 3510, 3542 Pantelimon George Popescu 3539 Mariia Rozhkova 3504 Josep Rubi´ o-Masseg´ u 3515 Sergey Sadov 3563 Mehmet S ¸ ahin 3543, 3544, 3576, 3577, 3635, 3699 Bill Sands 3595, 3601 Hassan A. ShahAli 3501, 3513 Bruce Shawyer 3503 Slavko Simic 3523 D.J. Smeenk 3540, 3541 Albert Stadler 3567, 3568, 3687 Neculai Stanciu 3611, 3613 Daryl Tingley 3593 George Tsapakidis 3622 Harley Weston 3224 Peter Y. Woo 3579, 3610, 3653, Paul Yiu 3661 Katsuhiro Yokota 3528 Zhang Yun 3598, 3674 Faruk Zejnulahi 3592 Titu Zvonaru 3524, 3621

3553, 3629, 3662, 3690,

3572,

3600, 3658,

3630,

Featured Solvers — Individuals
Anonymous Solver 3572, 3574 Mohammed Aassila 3503, 3520, 3578 Arkady Alt 3543, 3544, 3549 George Apostolopoulos 3511, 3517, 3524, 3525, 3527, 3536, 3544, 3561, 3568 ˇ Sefket Arslanagi´ c 3528, 3590, 3596 Dionne Bailey 3553 Roy Barbara 3517, 3589 Michel Bataille 3502, 3504a, 3517, 3520, 3522, 3529, 3534, 3538, 3543, 3546, 3552, 3580, 3584, 3594, 3600 Elsie Campbell 3553 Cao Minh Quang 3533 Emmanuel Lance Christopher 3544 Shai Covo 3586 Chip Curtis 3556, 3599 Charles R. Diminnie 3553 Richard Eden 3521, 3532, 3547 Oliver Geupel 3513, 3517, 3518, 3519, 3526, 3528, 3529, 3530, 3544, 3545, 3550, 3555, 3565, 3567, 3569, 3571, 3573, 3592, 3598 Johan Gunardi 3510 John Hawkins 3555 John G. Heuver 3520, 3540, 3576 Joe Howard 3507, 3554, 3559, 3564 Hung Pham Kim 3508 D. Kipp Johnson 3596, 3597 Hiroshi Kinoshita 3528 V´ aclav Koneˇ cn´ y 3517, 3566, 3591 Kee-Wai Lau 3566, 3570 Kathleen E. Lewis 3595 Salem Maliki´ c 3515, 3528, 3572, 3590, 3598 Gr´ egoire Nicollier 3224 Ricard Peir´ o 3520 Paolo Perfetti 3583 Pham Van Thuan 3548 Henry Ricardo 3539 Mariia Rozhkova 3504b Joel Schlosberg 3531, 3544, 3551, 3560, 3562, 3575, 3577 Harry Sedinger 3501 Shailesh Shirali 3516 Slavko Simic 3523 D.J. Smeenk 3541 Albert Stadler 3512, 3514, 3535, 3537, 3548, 3557, 3559, 3572, 3585 David Stone 3555 Edmund Swylan 3556 Daryl Tingley 3593 Stan Wagon 3579 Steﬀen Weber 3563 Peter Y. Woo 3538, 3544, 3582 Paul Yiu 3542 Katsuhiro Yokota 3528 Li Zhou 3510 Titu Zvonaru 3502, 3544

Featured Solvers — Groups
Skidmore College Problem Solving Group 3558

566

Other Solvers — Individuals
Mohammed Aassila 3520, 3567 Yakub N. Aliyev 3505, 3518 Arkady Alt 3515, 3521, 3525, 3526, 3528, 3532, 3533, 3547, 3553, 3556, 3560, 3571, 3576, 3577, 3584 G.W. Indika Amarasinghe 3590 Miguel Amengual Covas 3528, 3529, 3576, 3577 George Apostolopoulos 3501, 3502, 3504a, 3505, 3506, 3507, 3510, 3511, 3512, 3514, 3515, 3516, 3520, 3521, 3526, 3528, 3529, 3530, 3531, 3532, 3535, 3537, 3539, 3540, 3545, 3547, 3548, 3549, 3553, 3554, 3555, 3556, 3558, 3559, 3560, 3562, 3563, 3564, 3565, 3566, 3571, 3572, 3573, 3598 Vahagn Aslanyan 3555, 3562 Victor Arnaiz 3531, 3536 ˇ Sefket Arslanagi´ c 3501, 3504a, 3505, 3511, 3515, 3525, 3526, 3532, 3533, 3539, 3540, 3547, 3549, 3553, 3554, 3556, 3559, 3560, 3561, 3566, 3570, 3572, 3574, 3577, 3584, 3598 Dionne T. Bailey 3533, 3539, 3560 Muriel Baker 3539 Roy Barbara 3501, 3502, 3506, 3514, 3520, 3525, 3539, 3540, 3555, 3556, 3558, 3559, 3560, 3561, 3571, 3590, 3598 Ricardo Barroso Campos 3520 Michel Bataille 3501, 3503, 3505, 3510, 3512, 3514, 3515, 3516, 3521, 3523, 3524, 3525, 3526, 3528, 3532, 3536, 3539, 3540, 3544, 3545, 3547, 3548, 3553, 3555, 3556, 3560, 3566, 3567, 3568, 3570, 3572, 3573, 3574, 3575, 3576, 3577, 3578, 3582, 3585, 3590, 3591, 3598 Brian D. Beasley 3502, 3506, 3539 Mih´ aly Bencze 3561 Manuel Benito 3578 K.S. Bhanu 3531 Mihaela Blanariu 3528, 3532 Paul Bracken 3512, 3539, 3550 Scott Brown 3504a N. Javier Buitrago Aza 3552 Elsie M. Campbell 3533, 3539, 3560 Cao Minh Quang 3526 Pedro A. Castillejo 3531, 3536 ´ Oscar Ciaurri 3578 Chip Curtis 3501, 3505, 3506, 3510, 3514, 3524, 3525, 3526, 3529, 3532, 3536, 3539, 3540, 3544, 3547, 3558, 3559, 3560, 3565, 3566, 3568, 3571, 3572, 3582, 3590, 3591, 3596 Paul Deiermann 3552, 3560 M.N. Deshpande 3531 Jos´ e Luis D´ ıaz-Barrero 3502, 3515, 3539, 3547, 3572 Max Diaz 3565 Charles R. Diminnie 3502, 3525, 3533, 3539, 3547, 3560 Joseph DiMuro 3501, 3506, 3555, 3558, 3560, 3562, 3563 Marian Dinc˘ a 3527, 3532 A.A. Dzhumadil’daeva 3573 Richard B. Eden 3505, 3528, 3534, 3540, 3548 Keith Ekblaw 3531 Oleh Faynshteyn 3515, 3532, 3540, 3543, 3544, 3548, 3553 Emilio Fernandez 3578 Ovidiu Furdui 3512, 3530, 3550, 3551, 3578, 3580, 3600 Ian June L. Garces 3501 Francisco Javier Garc´ ıa Capit´ an 3520 Oliver Geupel 3501, 3502, 3503, 3504a, 3505, 3506, 3507, 3510, 3511, 3512, 3514, 3515, 3516, 3520, 3521, 3525, 3531, 3532, 3533, 3534, 3535, 3536, 3539, 3540, 3541, 3543, 3546, 3547, 3551, 3552, 3553, 3554, 3556, 3557, 3558, 3559, 3560, 3561, 3562, 3563, 3564, 3566, 3570, 3572, 3574, 3589, 3590, 3591, 3593, 3595, 3596, 3597, 3599 Samuel G´ omez Moreno 3536 Johan Gunardi 3501, 3503, 3505, 3506, 3515, 3558, 3597 John Hawkins 3551, 3558, 3560 Richard I. Hess 3502, 3506, 3558, 3560, 3590, 3591, 3592 John G. Heuver 3505, 3528, 3577 Gerhardt Hinkle 3591 Joe Howard 3504a, 3512, 3556, 3570, 3572, 3590 Peter Hurthig 3506 D. Kipp Johnson 3590, 3592 Dag Jonsson 3566 Michael Josephy 3501, 3506 Neven Juriˇ c 3551, 3560 Geoﬀrey A. Kandall 3503 Hung Pham Kim 3527, 3549 V´ aclav Koneˇ cn´ y 3503, 3520, 3528, 3529, 3532, 3540, 3589, 3591 Mikhail Kochetov 3563 Anastasios Kotronis 3512 Don Krug 3600 Kee-Wai Lau 3525, 3527, 3528, 3548, 3556, 3558, 3559, 3560, 3570, 3578, 3584, 3590, 3596, 3598 R. Laumen 3506 Kathleen E. Lewis 3501, 3531, 3558 Panagiote Ligouras 3582 Htet Naing Lin 3539 Thanos Magkos 3504a Salem Maliki´ c 3539, 3547, 3553, 3554, 3555, 3556, 3559, 3561, 3566 David E. Manes 3539 Dorin M˘ arghidanu 3521, 3522 Marian Marinescu 3537 James Mayer 3539 Norvald Midttun 3558, 3560 Dragoljub Miloˇ sevi´ c 3557, 3584 M.R. Modak 3590, 3591, 3598 Cristinel Mortici 3539, 3556, 3560 Victor Oxman 3538 Pedro Henrique O. Pantoja 3506, 3526 Scott Pauley 3590 Paolo Perfetti 3507, 3511, 3514, 3515, 3521, 3525, 3526, 3527, 3530, 3533, 3535, 3537, 3548, 3549, 3550, 3551, 3554, 3557, 3564, 3572, 3578, 3580, 3592, 3596 Pham Huu Duc 3507, 3554 Pham Van Thuan 3511, 3560, 3564 Cosmin Pohoat ¸a ˘ 3510 John Postl 3539 C.R. Pranesachar 3592 Hannah Prest 3539 Prithwijit De 3505, 3514, 3525, 3528, 3532, 3556, 3576, 3577, 3598 James Reid 3539 Henry Ricardo 3558, 3565 Luz Roncal 3578 Mariia Rozhkova 3504a Josep Rubi´ o-Masseg´ u 3515 Sergey Sadov 3563 Mehmet S ¸ ahin 3543, 3555, 3576, 3577 Bill Sands 3595 Joel Schlosberg 3501, 3502, 3505, 3510, 3528, 3532, 3536, 3552, 3555, 3556, 3558, 3561, 3563, 3564, 3565, 3566, 3568, 3571, 3573, 3576, 3582, 3595, 3596, 3598 Bob Serkey 3590 Hassan A. ShahAli 3501, 3513 Bruce Shawyer 3503 Shailesh Shirali 3520 D.J. Smeenk 3520 Digby Smith 3501, 3539, 3558, 3560, 3571 Cullan Springstead 3539 Albert Stadler 3502, 3506, 3515, 3517, 3518, 3520, 3521, 3522, 3525, 3526, 3528, 3529, 3530, 3531, 3532, 3534, 3539, 3540, 3547, 3550, 3551, 3552, 3553, 3554, 3555, 3556, 3561, 3562, 3563, 3564, 3565, 3566, 3567, 3568, 3571, 3573, 3574, 3577, 3578, 3580, 3586, 3589, 3593 Miha¨ ı Sto¨ enescu 3505, 3576, 3577 David R. Stone 3551, 3558, 3560 Ercole Suppa 3577 Edmund Swylan 3501, 3502, 3503, 3506, 3558, 3560, 3589, 3590, 3598 Daniel Tsai 3525 Stan Wagon 3527, 3548, 3549, 3592 Haohao Wang 3539, 3549, 3590 Elizabeth Wamser 3539 Natalya Weir 3590 Andrew Welter 3590 Brent Wessel 3539 Daniel Winger 3539 Jerzy Wojdylo 3539, 3549 Peter Y. Woo 3501, 3503, 3505, 3510, 3511, 3514, 3515, 3516, 3520, 3521, 3525, 3526, 3527, 3528, 3529, 3532, 3533, 3537, 3539, 3540, 3541, 3547, 3548, 3551, 3553, 3556, 3558, 3559, 3560, 3561, 3564, 3566, 3570, 3576, 3577, 3590, 3591, 3598 Yanping Xia 3590 Paul Yiu 3542 Zhang Yun 3598 Konstantine Zelator 3505, 3525 Li Zhou 3506 Titu Zvonaru 3503, 3510, 3511, 3524, 3528, 3539, 3540, 3556, 3572

Other Solvers — Groups
Missouri State University Problem Solving Group 3501, 3503, 3524 Skidmore College Problem Solving Group 3560, 3561, 3563, 3573

Crux Mathematicorum 37

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