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193 EDITORIAL Va lav Linek It has ome to our attention that some problems appearing in CRUX have been submitted to o...

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193

EDITORIAL Va lav Linek

It has ome to our attention that some problems appearing in CRUX have been submitted to other pla es; in parti ular some CRUX with MAYHEM problems have appeared on ertain problem-solving websites. While the trend of online problem solving is well established and quite popular, it is nevertheless still the ase that CRUX with MAYHEM is a traditional print journal. This means that submissions to CRUX with MAYHEM should not be submitted elsewhere and should not have already appeared elsewhere (although some rare ex eptions may be permitted for older problems that may not be that well known today). This in ludes problem proposals, solutions, and arti les: if you submit them to CRUX with MAYHEM, then we ask that you do not submit them anywhere else! I would like to point out that material that is submitted to CRUX with MAYHEM re eives the attention of the Editorial Board, whereas many online sour es do not provide this kind of attention. For instan e, problems and solutions featured in CRUX with MAYHEM are edited and typeset with are, and we do our best to lter the problem proposals we re eive. Sometimes it takes a while to pro ess your material. For example, we have a ba klog of problem proposals to pro ess be ause only 100 four-digit problems are published per year, whereas we re eive onsiderably more than 100 problem proposals a year. If the wait is too long, then (if we have not already used your problem) you an inform us that you would like to submit your problem elsewhere and we will then remove it from the queue. The same applies for other materials. If it omes to our attention that some dupli ation has o

urred, then we may not publish the dupli ated material. Having said this, it is obvious that our journal has hanged with time. Indeed, \Crux" was originally founded as Eureka by Leo Sauve in 1975, and it did not in lude book reviews, ontributor pro les, or Mayhem at that time, and it was ertainly not posted on the internet be ause the internet did not exist then! Moving forward, it is inevitable that CRUX with MAYHEM will

hange yet again. Finally, I am happy to report that work has already started on the Jim Totten spe ial issue slated for September. I thank you, the readers, for your patien e with the delays of the last eight months. With time available for preparation this summer (unlike last summer) ir umstan es are favourable for autumn to go a

ording to s hedule. with MAYHEM

194

SKOLIAD No. 117 Lily Yen and Mogens Hansen

Please send your solutions to problems in this Skoliad by 1 November, 2009. A opy of Crux will be sent to one pre-university reader who sends in solutions before the deadline. The de ision of the editors is nal. Our ontest this month is the Calgary Mathemati al Asso iation Junior High S hool Mathemati s Contest, Part B, 2008. Our thanks go to Joanne Canape of the University of Calgary, Calgary, Alberta, for providing us with this ontest and for permission to publish it. We also thank Rolland Gaudet, University College of St. Bonifa e, Winnipeg, MB for translating this ontest. Con ours de l'Asso iation mathematique de Calgary Niveau prese ondaire

Ronde nale, partie B, 2008 . Ri hard a besoin de se rendre en taxi de sa maison a un par pas tres loin. Deux ompagnies de taxi orent leurs servi es. La premiere fa ture ses

lients a un taux xe de 10,00$, auquel s'ajoute un taux variable de 0,50$ pour haque kilometre du trajet, tandis que la deuxieme a un taux xe de 4,00$ et un taux variable de 0,80$ le kilometre. Ri hard onstate que le out ^ est le m^eme, quelle que soit la ompagnie hoisie. Quelle est la distan e en kilometres de sa maison au par ? 1

. Un poste de radio lan e un on ours ou ha un des gagnants pourra assister a deux mat hs des Canadiens de Montreal, puis d'y amener un ami a ha un de es deux mat hs, soit un ami aux deux mat hs, soit deux amis,

ha un a un mat h dierent. La han e a fait que les amis Ali e, Bertrand, Carole, David et Evelyne ont tous et e de lar es gagnants au on ours. Montrer omment ha un de es gagnants peut hoisir ses amis parmi le groupe, de fa on a e que haque paire d'amis parmi les inq assiste a au moins un mat h ensemble.

2

. Deux tests sont administres a un groupe d'etudiants. Chaque etudiant re oit un s ore pour ha un des deux tests, e s ore etant un entier non negatif au plus egal a 10. Adrien remarque que, pour le premier test, seulement un etudiant a re u un s ore plus elev e que lui et que personne n'a egal e son s ore ; il en est de m^eme pour le deuxieme test. Apres que le professeur a aÆ he les s ores moyens des deux tests, Adrien onstate qu'il y a plus qu'un etudiant ave un s ore moyen plus elev e que le sien. 3

195 (a) Donner un exemple on ret, in luant tous les s ores, pour montrer que

e i est possible. (b) Quel est le plus grand nombre d'etudiants pouvant avoir un s ore moyen plus elev e que elui d'Adrien ? Expliquer lairement pourquoi votre reponse est orre te. . On ommen e par tra er un re tangle de taille 6 m par 8 m. Ensuite, on tra e le

er le ir ons rit de e re tangle, puis un

arre ir ons rit autour du er le. En n, on tra e le er le ir ons rit du arre tout juste

onstruit. Determiner la surfa e de e dernier er le en m2 . 4

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

. Soit A un entier positif a deux hires de imaux, ne ontenant au un zero et soit B un entier positif a trois hires de imaux. Si A% de B donne 400, determiner toutes les valeurs possibles de A et B . 5

. Determiner un re tangle ayant les deux propriet es suivantes : (i) son perim etre est un entier impair ; et (ii) au un de ses ot ^ es n'est entier. Determiner maintenant un re tangle ayant les deux propriet es suivantes : (i) sa surfa e est un entier pair ; et (ii) au un de ses ot ^ es n'est entier. En n, determiner un quadrilatere, pas ne essairement re tangulaire, ayant les trois propriet es suivantes : (i) son perim etre est un entier positif ; (ii) sa surfa e est un entier positif ; et (iii) au un de ses ot ^ es n'est entier. 6

Calgary Mathemati al Asso iation Junior High S hool Mathemati s Contest Final Round, Part B, 2008

. Ri hard needs to go from his house to the park by taking a taxi. There are two taxi ompanies available. The rst taxi ompany harges an initial

ost of $10.00, plus $0.50 for ea h kilometre travelled. The se ond taxi ompany harges an initial ost of $4.00, plus $0.80 for ea h kilometre travelled. Ri hard realises that the ost to go to the park is the same regardless of whi h taxi ompany he hooses. What is the distan e in km from his house to the park? 1

2. A radio station runs a ontest in whi h ea h winner will get to attend two Calgary Flames playo games and to take one guest to ea h game. The winner does not have to take the same guest to the two games. Lu kily, ve s hool friends Ali e, Bob, Carol, David, and Eva are all winners of this

ontest. Show how ea h winner an hoose two others from this group to be his or her guests, so that ea h pair of the ve friends gets to go to at least one playo game together.

196 . A lass was given two tests. In ea h test ea h student was given a nonnegative integer s ore with a maximum possible s ore of 10. Adrian noti ed that in ea h test, only one student s ored higher than he did and nobody got the same s ore as he did. But then the tea her posted the averages of the two s ores for ea h student, and now there was more than one student with an average s ore higher than Adrian. 3

(a) Give an example (using exa t s ores) to show that this ould happen. (b) What is the largest possible number of students whose average s ore

ould be higher than Adrian's average s ore? Explain learly why your answer is orre t. . A re tangle with dimensions 6 m by 8 m is drawn. A ir le is drawn ir ums ribing this re tangle. A square is drawn ir ums ribing this ir le. A se ond ir le is drawn that ir ums ribes this square. What is the area in m2 of the bigger ir le?

4

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

. If A is a two-digit positive integer that does not ontain zero as a digit, B is a three-digit positive integer, and A% of B is 400, nd all possible values of A and B . 5

6. Find a re tangle with the following two properties: (i) its perimeter is an odd integer; and (ii) none of its sides is an integer. Next, nd a re tangle with the following two properties: (i) its area is an even integer; and (ii) none of its sides is an integer. Finally, nd a quadrilateral (not ne essarily a re tangle) with the following three properties: (i) its perimeter is a positive integer; (ii) its area is a positive integer; and (iii) none of its sides is an integer.

We now give solutions to the sele ted questions of the 2007 Christopher Newport University Mathemati s Contest in Skoliad 111 [2008 : 257-259℄.

1. Find the midpoint of the domain of the fun tion f (x) = (A)

1 4

(B)

3 2

(C)

2 3

(D)

p √ 4 − 2x + 5.

−2 5

Solution by an unknown solver. The √ domain of the fun tion f is the set of real numbers, x, su h that both 4 − 2x + 5 ≥ 0 and 2x + 5 ≥ 0. The latter inequality is the same as 2x ≥ −5, hen e x ≥ − 25 .

197 √

The rst inequality is the same as 4 ≥ 2x + 5, so 16 ≥2x + 5.Thus, ≥ x. Hen e the domain of f is the interval − 25 , 11 11 ≥ 2x and 11 . The 2 2 1 5 11 3 midpoint of this interval is 2 − 2 + 2 = 2 , so the answer is (B).

Also solved by JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON; and LUYUN ZHONG-QIAO, Columbia International College, Hamilton, ON.

2. The sum a + b, the produ t ab, and the dieren e a2 − b2 for two positive numbers a and b is the same nonzero number. What is b? (A) 2

(B)

√ 1+ 5 2

(C)

√ 5

(D)

√ 3− 5 3

Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON. If ab = a + b and ab = a2 − b2 , then 1 =

ab a2 − b2 (a − b)(a + b) = = = a− b, ab a+b a+b

hen e a = b + 1. Sin e ab = a + b, it follows that (b + 1)b =√(b + 1) + b, so b2 − b − 1 = 0. The quadrati formula yields that b = 1 ±2 5 . Sin e b is given to be positive, b =

√ 1+ 5 , 2

and the answer is (B).

Also solved by LUYUN ZHONG-QIAO, Columbia International College, Hamilton, ON.

3. Let f (x) be a quadrati polynomial with f (3) = 15 and f (−3) = 9. Find the oeÆ ient of x in f (x). (A) 2

(B) 3

(C) 1

(D) −2

Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON. Sin e f (x) is a quadrati polynomial, f (x) = ax2 + bx + c for some

oeÆ ients a, b, and c. Now 15 = 9 =

f (3) = 9a + 3b + c f (−3) = 9a − 3b + c

Subtra ting the se ond equation from the rst yields 6 = 6b, so b = 1, and the answer is (C). Also solved by LUYUN ZHONG-QIAO, Columbia International College, Hamilton, ON. One orre t solution was submitted by an unknown solver.

4. A pair of fair di e is ast. What is the probability that the sum of the

numbers falling uppermost is 7 or 11 if it is known that one of the numbers is a 5? (A)

2 9

(B)

7 36

(C)

1 9

(D)

4 11

198 Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON. One an roll at least one 5 in eleven ways: First die Se ond die Sum Of these the sum is answer is (D).

7

1 5 6

2 5 7

or

11

3 5 8

4 5 9

5 5 10

6 5 11

5 1 6

5 2 7

5 3 8

5 4 9

5 6 11

in four ases, so the probability is

4 11

and the

One may also use the formula for onditional probability, P (A|B) =

P (A and B) P (B)

,

where P (A|B) is the probability that A happens given that B did happen. In the ase of rolling the di e, A is \the sum is 7 or 11" while B is \at least one 5". The probability of rolling at least one 5 with two di e is 16 · 56 + 56 · 61 + 16 · 16 or 11 . When rolling two di e, one may get a sum of 36 4 7 or 11 and at least one 5 in four ways, 2 + 5, 5 + 2, 5 + 6, and 6 + 5, so P (A and B) = 36 . 4/36 4 The formula above now yields P (A|B) = 11/36 = 11 , as before.

5. The number

p √ 24 + 572

√

an be written in the form a + and b are whole numbers and b > a. What is the value b − a? (B) −2

(A) 4

(C) 2

√ b,

where a

(D) 3

Solution by an unknown solver. √ √ √ √ Note that 572 = 22 · 11 · 13 = 2 11 13. Thus, q √ 24 + 572

= = = =

q √ √ 11 + 13 + 2 11 13 q √ 2 √ √ √ 2 11 + 2 11 13 + 13 q √ √ 2 11 + 13 √ √ 11 + 13 .

Hen e b − a = 13 − 11 = 2, and the answer is (C).

Also solved by JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON; and LUYUN ZHONG-QIAO, Columbia International College, Hamilton, ON.

6. Let x =

1 2+

be the indi ated ontinued fra tion. Whi h one of

1 3+

1

2+

1 3+···

the following is equal to x? (A)

√

15 + 1 2

(B)

√

2+1 3

(C)

−3 + 2

√

15

(D)

−

√

15 − 3 2

199 Solution by an unknown solver. Note that 1

x = 2+

2+

1

3+ 2+

Therefore

1

=

1

.

1 3+x

1 3 + ···

1 3+x = . 2(3 + x) + 1 7 + 2x 3+x 7x + 2x2 = 3 + x, so 2x2 + 6x − 3 = 0, and the x =

Hen e yields

−6 ± x = 4

√ 60

√ √ −6 ± 2 15 −3 ± 15 = = 4 2

Sin e x is learly positive, x =

7. If

quadrati formula

−3 + 2

√

15

.

and the answer is (C).

Also solved by JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON. r 1+x f = 5x, nd f (2). 1−x

(A) −15

√

(B) 15

−1

(C) 3

(D) −4

Solution by an unknown solver. If

r

1+x = 2, 1−x

then

1+x = 4, 1−x

so 1 + x

= 4 − 4x,

when e x

Substituting x = 35 into the given equation now yields f (2) = 5 · 35 the answer is (C).

=

3 5

.

= 3, and

Also solved by JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON; and LUYUN ZHONG-QIAO, Columbia International College, Hamilton, ON.

As you will have noti ed, the identity of one of our solvers was lost in the transition of Skoliad editors. If you submit solutions on paper, or s ans of paper solutions, we request that you write your name and aÆliation on ea h sheet. This issue's prize of one opy of Crux with Mayhem for the best solutions goes to Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON. We would very mu h appre iate re eiving more solutions from our preuniversity readers. Solutions to just some of the problems are also wel ome.

200

MATHEMATICAL MAYHEM Mathemati al Mayhem began in 1988 as a Mathemati al Journal for and by High S hool and University Students. It ontinues, with the same emphasis, as an integral part of Crux Mathemati orum with Mathemati al Mayhem. The Mayhem Editor is Ian VanderBurgh (University of Waterloo). The other sta members are Monika Khbeis (As ension of Our Lord Se ondary S hool, Mississauga) and Eri Robert (Leo Hayes High S hool, Frederi ton).

Mayhem Problems Veuillez nous transmettre vos solutions aux problemes du present numero avant le 31 aout^ 2009. Les solutions re ues apres ette date ne seront prises en ompte que s'il nous reste du temps avant la publi ation des solutions. Chaque probleme sera publie dans les deux langues oÆ ielles du Canada (anglais et fran ais). Dans les numeros 1, 3, 5 et 7, l'anglais pre edera le fran ais, et dans les numeros 2, 4, 6 et 8, le fran ais pre edera l'anglais. La reda tion souhaite remer ier Jean-Mar Terrier, de l'Universite de Montreal, d'avoir traduit les problemes. . Propose par l'Equipe de Mayhem. Les nombres a, b, c, d et e sont inq entiers su

essifs. Montrer que la dieren e entre la moyenne des arres de c et e et elle des arres de a et c est egale a quatre fois c. M394

. Propose par l'Equipe de Mayhem. Le quadrilatere ABCD est tel que ha un de ses ot ^ es est tangent a un er le donne, omme dans la gure

i- ontre. Si AB = AD, montrer que BC = CD .

D

M395

C

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

A

B

. Propose par l'Equipe de Mayhem. Dans un re tangle ABCD de ot ^ es AB = 8 et BC = 6, on ins rit respe tivement deux er les de entre O1 et O2 dans les triangles ABD et BCD . Trouver la distan e entre O1 et O2 . M396

. Propose par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie. Trouver toutes les paires (x, y) d'entiers tels que

M397

x4 − x + 1 = y 2 .

201 . Propose par l'Equipe de Mayhem.

M398

(a) Soit r, s et t les ra ines de l'equation

ubique w3 − bw2 + cw − d = 0. Determiner b, c et d en termes de r , s et t. (b) On suppose que a est un nombre reel. Determiner toutes les solutions du systeme d'equations x+y+z xy + yz + zx xyz

= = =

a, −1 , −a .

M399. Propose par Ne ulai Stan iu, E ole Te hnique Superieure de Saint Mu eni Sava, Ber a, Roumanie. Trouver tous les triplets (a, b, c) d'entiers positifs tels que un entier positif.

3ab − 1 abc + 1

soit

. Propose par Mihaly Ben ze, Brasov, Roumanie. Supposons que a, b et c sont trois nombres reels positifs. Supposons de plus que an + bn = cn pour un ertain entier positif n ave n ≥ 2. Montrer que si k est un entier positif ave 1 ≤ k < n, alors ak , bk et ck sont les longueurs des ot ^ es d'un triangle.

M400

................................................................. . Proposed by the Mayhem Sta. The numbers a, b, c, d, and e are ve onse utive integers, in that order. Prove that the dieren e between the average of the squares of c and e and the average of the squares of a and c is equal to four times c.

M394

D

. Proposed by the Mayhem

M395

Sta. The quadrilateral ABCD is su h that ea h of its sides is tangent to a given

ir le, as shown. If AB = AD, prove that BC = CD.

C

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

A

B

. Proposed by the Mayhem Sta. The re tangle ABCD has side lengths AB = 8 and BC = 6. Cir les with entres O1 and O2 are ins ribed in triangles ABD and BCD. Determine the distan e between O1 and O2 .

M396

202 . Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania. Determine all pairs (x, y) of integers su h that

M397

x4 − x + 1 = y 2 .

. Proposed by the Mayhem Sta.

M398

(a) The ubi equation w3 − bw2 + cw − d = Determine b, c, and d in terms of r, s, and t.

0

has roots r, s, and t.

(b) Suppose that a is a real number. Determine all solutions to the system of equations x+y+z xy + yz + zx xyz

= = =

a, −1 , −a .

M399. Proposed by Ne ulai Stan iu, Saint Mu eni Sava Te hnologi al High S hool, Ber a, Romania.

Determine all triples (a, b, c) of positive integers for whi h a positive integer.

3ab − 1 abc + 1

is

. Proposed by Mihaly Ben ze, Brasov, Romania. Suppose that a, b, and c are positive real numbers. In addition, suppose that an + bn = cn for some positive integer n with n ≥ 2. Prove that if k is a positive integer with 1 ≤ k < n, then ak , bk , and ck are the side lengths of a triangle. M400

Mayhem Solutions . Proposed by the Mayhem Sta. Determine all real numbers x that satisfy 32x+2 + 3 = 3x + 3x+3 .

M357

Solution by Shamil Asgarli, student, Burnaby South Se ondary S hool, Burnaby, BC. The equation 32x+2 + 3 = 3x + 3x+3 an be written in the form 2 x 2 3 (3 ) + 3 = 3x + 33 3x . Substituting k = 3x , we obtain the equivalent quadrati equations 9k2 + 3 = k + 27k and 9k2 − 28k + 3 = 0. Continuing to solve for k by fa toring, we obtain (k − 3)(9k − 1) = 0. Hen e, k = 3 or k = 19 . Substituting k = 3x , we have 3x = 3 or 3x = 19 , so x = 1 or x = −2.

203 Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; JACLYN CHANG, student, Western Canada High S hool, Calgary, AB; LUIS DE SOUSA, student, IST-UTL, Lisbon, Portugal; JOSE HERNANDEZ SANTIAGO, student, Universidad Te nologi a de la Mixte a, Oaxa a, Mexi o; IES \Abastos", Valen ia, GIANNIS G. KALOGERAKIS, Canea, Crete, Gree e; RICARD PEIRO, Spain; KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; MRIDUL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; ALEX SONG, student, Elizabeth Ziegler Publi S hool, Waterloo, ON; GEORGE TSAPAKIDIS, Agrinio, Gree e; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON; and CARLIN WHITE, student, California State University, Fresno, CA, USA. There were two in orre t solutions submitted.

. Proposed by Ne ulai Stan iu, Saint Mu eni Sava Te hnologi al High S hool, Ber a, Romania. How many integers in the list 1, 2008, 20082 , . . . , 20082009 are simultaneously perfe t squares and perfe t ubes? M358

Solution by Luis De Sousa, student, IST-UTL, Lisbon, Portugal, modi ed by the editor. The integer 2008 has prime fa torization 23 · 251. Thus, 2008k has k 3 prime fa torization 2 · 251 = 23k · 251k . For a positive integer greater than 1 to be a perfe t square, its prime fa torization must in lude only even exponents; for a positive integer greater than 1 to be a perfe t ube, its prime fa torization must in lude only exponents divisible by 3. The exponents 3k and k are both even if and only if k is even. The exponents 3k and k are both multiples of 3 if and only if k is a multiple of 3. Therefore, the powers of 2008 whi h are perfe t squares are the ones with even exponents. Also, the powers of 2008 whi h are perfe t ubes are those with exponents that are multiples of 3. Hen e, the powers of 2008 whi h are perfe t squares and perfe t ubes are those whose exponents are both multiples of 2 and of 3. In other words, they are those whose exponents are multiples of 6. The largest multiple of 6 less than 2009 is 2004 = 6 · 334, so there are 334 multiples of 6 in the list 1, 2, 3, . . . , 2008, 2009. This tells us that 334 of the integers 20081 , 20082 , . . . , 20082009 are simultaneously perfe t squares and perfe t ubes. The integer 1 is also in the list and is both a perfe t square and a perfe t ube. Therefore, there are 334 + 1 = 335 integers in the list that are simultaneously perfe t squares and perfe t ubes. Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; JACLYN CHANG, student, Western Canada High S hool, Calgary, AB; CHANTHOEUN CHAP and JUSTIN HENDERSHOTT, students, California State University, Fresno, CA; USA; JOSE HERNANDEZ SANTIAGO, student, Universidad Te nologi a de la Mixte a, Oaxa a, Mexi o; RICHARD IES \Abastos", Valen ia, Spain; I. HESS, Ran ho Palos Verdes, CA, USA; RICARD PEIRO, KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; MRIDUL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; ALEX SONG, student, Elizabeth Ziegler Publi S hool, Waterloo, ON; GEORGE TSAPAKIDIS, Agrinio, Gree e; and JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON.

204 . Proposed by the Mayhem Sta. A trapezoid DEF G is ir ums ribed about a ir le of radius 2, as shown in the diagram. The side DE has length 3 and there are right angles at E and F . Determine the area of the trapezoid. M359

D................................................................... E

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

G

F

Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON. Let O be the entre of the ir le. Let A be the point of interse tion of the perpendi ular from D to GF . Let the points of tangen y of EF , DE , GD , and GF to the ir le be W , X , Y , and Z , respe tively. We twi e use the fa t that tanD Xq E gents to the same ir le from the same Y q point have equal lengths. Let GZ = x. qO qW Then GY = x by equal tangents. Sin e OX and OW are perpendi ular to DE and EF , respe tively, q G F and the trapezoid has a right angle at A Z E , then OXEW is a square. Sin e the radius of the ir le is 2, then OW = 2, so XE = 2. Sin e DE = 3, then DX = DE − XE = 1. By equal tangents, DY = 1. Now DEF A is also a re tangle sin e it has three right angles (at E , F , and A) hen e the fourth angle is also a right angle. Also, XZ is parallel to EF . Thus, AZ = DX = 1. We know that DA = XO+OZ = 2+2 = 4, GD = GY +Y D = x+1, and GA = GZ − AZ = x − 1. By the Pythagorean Theorem, ...................................................................... ........ .................. . ..... ..... ............. .... .. ..... ... ... ..... ... .... . . . ...... ... ..... . . . ..... .... ..... ....... ... .... .. ..... .... ... ... ... ... ..... . .. . . . .. .. .... ...... . . . . . . ... .. . ... .... . . . . . . . .... .. .. .. .... .... ... ... ... ....... ... ...... .... ...... .... ....... . .................................................................................................................................................

GD 2 = (x + 1)2 = 2 x + 2x + 1 =

GA2 + AD 2 ; (x − 1)2 + 42 ; x2 − 2x + 1 + 16 ;

hen e 4x = 16 and x = 4. Therefore, GF = GZ + ZF = 4 + 2 = 6, and so the area of the trapezoid is 12 (GF + DE)(DA) or 12 (6 + 3)(4) = 18. Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; SERDAR ALTUNTAS, student, University of Karlsruhe, Karlsruhe, Germany; MIGUEL AMENGUAL COVAS, Cala Figuera, Mallor a, Spain; SCOTT BROWN, Auburn University, Montgomery, AL, USA; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; JACLYN CHANG, student, Western Canada High S hool, Calgary, AB; COURTIS G. CHRYSSOSTOMOS, Larissa, Gree e; JORDAN CRIST, student, Auburn University at Montgomery, Montgomery, AL, USA; LUIS DE SOUSA, student, IST-UTL, Lisbon, Portugal; JOSH GUZMAN and AMANDA PERCHES, students, California State University, Fresno, CA; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; GIANNIS G. KALOGERAKIS, Canea, Crete, IES \Abastos", Valen ia, Spain; KUNAL SINGH, student, Kendriya Gree e; RICARD PEIRO, Vidyalaya S hool, Shillong, India; ALEX SONG, student, Elizabeth Ziegler Publi S hool, Waterloo, ON; and GEORGE TSAPAKIDIS, Agrinio, Gree e. There was one in orre t and one in omplete solution submitted.

205 . Proposed by Ne ulai Stan iu, Saint Mu eni Sava Te hnologi al High S hool, Ber a, Romania. Determine all positive integers x that satisfy M360

3x = x3 + 3x2 + 2x + 1 .

Solution by Shamil Asgarli, student, Burnaby South Se ondary S hool, Burnaby, BC. We will show that there are no su h positive integers. Suppose that there was a positive integer solution x. We rewrite the equation in the form 3x = x x2 + 3x + 2 + 1 = x(x + 1)(x + 2) + 1 .

Sin e x > 0, the left side is divisible by 3. Therefore, the right side should be divisible by 3 as well. But x(x + 1)(x + 2) is divisible by 3, sin e it is the produ t of three

onse utive integers. Thus x(x + 1)(x + 2) + 1 gives a remainder of 1 when divided by 3, so is not divisible by 3. This is a ontradi tion, so there are no positive integer solutions. Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; IES \Abastos", LUIS DE SOUSA, student, IST-UTL, Lisbon, Portugal; RICARD PEIRO, Valen ia, Spain; ALEX SONG, student, Elizabeth Ziegler Publi S hool, Waterloo, ON; GEORGE TSAPAKIDIS, Agrinio, Gree e; and JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON. There were two in omplete solutions submitted.

. Proposed by George Tsapakidis, Agrinio, Gree e. Let a, b, and c be positive real numbers. Prove that

M361

ab(a + b − c) + bc(b + c − a) + ca(c + a − b) ≥ 3abc .

Solution by Shamil Asgarli, student, Burnaby South Se ondary S hool, Burnaby, BC. First, by the Arithmeti Mean{Geometri Mean Inequality, we nd that a2 b + ab2 + b2 c + bc2 + c2 a + ca2

≥ =

6 q 6

a2 b ab2 b2 c bc2 c2 a ca2

√ 6 a6 b6 c6 = abc .

It follows that a2 b + ab2 + b2 c + bc2 + c2 a + ca2 ≥ 6abc .

206 Rearranging the last expression, we nd that a2 b + ab2 − abc + b2 c + bc2 − abc + c2 a + ca2 − abc ≥ 3abc ,

whi h is equivalent to

ab(a + b − c) + bc(b + c − a) + ca(c + a − b) ≥ 3abc ,

as required.

Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia and Herzegovina; MIGUEL AMENGUAL COVAS, Cala Figuera, Mallor a, Spain; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; COURTIS G. CHRYSSOSTOMOS, Larissa, Gree e; LUIS DE SOUSA, student, IST-UTL, Lisbon, Portugal; JOSE LUIS D I AZ-BARRERO, Universitat Polite ni a de Catalunya, Bar elona, Spain; ANA GUTIERREZ and MARYLOV INSISIENGMAY, students, California State University, Fresno, CA, USA; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; HUGO LUYO SANCHEZ, Ponti ia Universidad Catoli a del Peru, Lima, Peru; MISSOURI IES STATE UNIVERSITY PROBLEM SOLVING GROUP, Spring eld, MO, USA; RICARD PEIRO, \Abastos", Valen ia, Spain; KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; ALEX SONG, student, Elizabeth Ziegler Publi S hool, Waterloo, ON; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; and JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON.

. Proposed by Mihaly Ben ze, Brasov, Romania. Suppose that x1 , x2 , . . . , xn is a sequen e of integers su h that

M362

|x1 | + |x2 | + · · · + |xn | − |x1 + x2 + · · · + xn | = 2 .

Prove that at least one of x1 , x2 , . . . , xn equals 1 or −1.

Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON. Without loss of generality, assume that x1 , x2 , . . . , xk are positive and xk+1 , xk+2 , . . . , xn are not positive (if this is not the ase, then we an rearrange the numbers to make it so). There are now two ases to onsider. Case 1 We have x1 + x2 + x3 + · · · + xn ≥ 0. Then |x1 + x2 + x3 + · · · + xn | = x1 + x2 + · · · + xn = x1 + x2 + · · · + xk − (−xk+1 ) − (−xk+2 ) − · · · − (−xn ) = |x1 | + |x2 | + · · · + |xk | − |xk+1 | + |xk+2 | + · · · + |xn | .

From this, we have 2

= |x1 | + |x2 | + · · · + |xn | − |x1 + x2 + · · · + xn | = |x1 | + |x2 | + · · · + |xn | − |x1 | + |x2 | + · · · + |xk | − |xk+1 | + |xk+2 | + · · · + |xn | = 2 |xk+1 | + |xk+2 | + · · · + |xn | = 2 |xk+1 | + |xk+2 | + · · · + |xn | .

207 It follows that |xk+1 | + |xk+2 | + · · · + |xn | = 1. Sin e the sum onsists of nonnegative integers, one term in the sum is equal to 1 (and the rest are 0), hen e one of the (nonpositive) numbers xk+1 , xk+2 , . . . , xn equals −1. Case 2 We have x1 + x2 + x3 + · · · + xn ≤ 0. In a similar way to Case 1, we see that |x1 + x2 + x3 + · · · + xn | = −(x1 + x2 + x3 + · · · + xn ) = −x1 − x2 − · · · − xk + (−xk+1 ) + (−xk+2 ) + · · · + (−xn ) = |xk+1 | + |xk+2 | + · · · + |xn | − |x1 | + |x2 | + · · · + |xk | .

Therefore, 2

= |x1 | + |x2 | + · · · + |xn | − |x1 + x2 + · · · + xn | = |x1 | + |x2 | + · · · + |xn | − |xk+1 | + |xk+2 | + · · · + |xn | − |x1 | + |x2 | + · · · + |xk | = 2 |x1 | + |x2 | + · · · + |xk | = 2 |x1 | + |x2 | + · · · + |xk | .

Therefore, |x1 | + |x2 | + · · · + |xk | = 1. The sum onsists of positive integers, so it has exa tly one term, whi h is equal to 1. Hen e, k = 1 and x1 = 1. By the two ases, one of x1 , x2 , . . . , xn is equal to 1 or −1.

Also solved by LUIS DE SOUSA, student, IST-UTL, Lisbon, Portugal; MISSOURI STATE UNIVERSITY PROBLEM SOLVING GROUP, Spring eld, MO, USA; ALEX SONG, student, Elizabeth Ziegler Publi S hool, Waterloo, ON; and GEORGE TSAPAKIDIS, Agrinio, Gree e. There was one in omplete solution submitted. Case 2 an be avoided by repla ing x1 , x2 , . . . , xn with −x1 , −x2 , . . . , −xn .

Problem of the Month Ian VanderBurgh

Contest problems involving omplex numbers don't appear very often... (2007 Ameri an Invitational Mathemati s Challenge A) The omplex number z is equal to 9 + bi, where b is a positive real number and i2 = −1. Given that the imaginary parts of z 2 and z 3 are equal, nd b. Problem

...and when they do, they're often easier than they look. The one pie e of information that we need to remember is that the imaginary part of a

omplex number is the oeÆ ient of i. As the famous slogan (almost) says, just al ulate it!

208 Solution

Sin e z = 9 + bi, then z2

whi h gives z3

= (9 + bi)2 = (9 + bi)(9 + bi) = 81 + 18bi + b2 i2 = 81 + 18bi − b2 = 81 − b2 + (18b)i , = = = = =

(9 + bi)3 = (9 + bi)2 (9 + bi) 81 − b2 + (18b)i (9 + bi) 729 − 9b2 + 81b − b3 i + (162b)i + 18b2 i2 729 − 9b2 + 81b − b3 i + (162b)i − 18b2 729 − 27b2 + 243b − b3 i .

The imaginary part of z 2 is 18b and the imaginary part of z 3 is 243b − b3 . If these are equal, then 243b − b3 = 18b, or b3 − 225b = 0, or b(b2 − 225) = 0, or b(b − 15)(b + 15) = 0. Sin e b > 0, then b = 15. At this point, you're all probably relieved. Finally, a Problem of the Month olumn with a short solution. But you an't get rid of me that easily! I was a tually a bit surprised by the answer b = 15, so de ided to try to answer the question for a general omplex number z = a + bi. I did this for two reasons. First, I was going to use it to verify my answer. Se ond, I was urious how the \15" appeared. For z = a + bi, z2

and z3

= (a + bi)2 = (a + bi)(a + bi) = a2 + 2abi + b2 i2 = a2 + 2abi − b2 = a2 − b2 + (2ab)i = (a + bi)3 = (a + bi)2 (a + bi) 2 = a − b2 + (2ab)i (a + bi) = a3 − ab2 + a2 b − b3 i + 2a2 b i + 2ab2 i2 = a3 − ab2 + a2 b − b3 i + 2a2 b i − 2ab2 = a3 − 3ab2 + 3a2 b − b3 i .

In this general ase, the imaginary part of z 2 is 2ab and the imaginary part of z 3 is 3a2 b − b3 . If these are equal, we have

2ab = 3a2 b − b3 , b3 − 3a2 b + 2ab = 0 , 2 b b − 3a2 − 2a = 0. √ √ Thus, b = 0 or b = ± 3a2 − 2a. However b > 0, so b = 3a2 − 2a. Can you see why 3a2 − 2a is p not negative if a is√an integer? In the original question, a = 9, so that b = 3(92 ) − 2(9) = 225 = 15. This addresses

209 my rst on ern. It also gives somewhat of an answer to the se ond one as well { the 15 doesn't appear in any really obvious way. (In other words, we shouldn't have been able to guess \15" by inspe tion right o the bat.) But wait, there's more! It's often interesting to think about what other questions we ould ask. One question that immediately ame to my mind was whether or not it was surprising that we got an integer answer for b with a = 9. (Given that this problem appears on the AIME, in whi h every answer is an integer, it shouldn't be that surprising.) Put a bit dierently, though, √ for a random positive integer a, is b = 3a2 −p2a always an integer √ or is a = 9 somewhat spe ial? If a = 1, we getpb = 3(12 ) − 2(1) = 1 = 1, √ whi h is an integer; if a = 10, we get b = 3(102 ) − 2(10) = 280, whi h is not an integer. √ Next question: For what positive integer values of a is b = 3a2 − 2a a positive integer? This is a fair bit harder, but is worth looking at. We'll take what seems initially like a ba kwards approa h by starting with a positive integer b and then trying to nd the orresponding values of a that are a tually integers. √ Suppose that b = 3a2 − 2a is a positive integer. Then 3a2 − 2a = b2 and so 3a2 − 2a − b2 = 0. While it may seem a bit unnatural at this point, a good approa h is to apply the quadrati formula to solve for a in terms of b. When we do this, we obtain a=

−(−2) ±

p (−2)2 − 4(3)(−b2 ) 2(3)

=

2±

√ 4 + 12b2 6

=

1±

√

1 + 3b2 3

.

Remember that we're trying to determine the positive integers b that give positive integer values for a. (A tually, we're really interested in the other dire tion, but this is the approa h that we're taking.) Sin e we want a to be positive, whi h sign do we take? We want the \+". Can you see why? √ 2 So we fo us on the ase a = 1 + 13 + 3b . For a to be an integer, what √ needs to be true? Certainly, 1 + 3b2 needs to be a positive integer, though √ this is a tually√not quite enough sin e 1 + 3b2 needs to be an integer and we want 1 + 1 + 3b2 to be divisible by 3. √ Let's fo us on the rst issue. Suppose that 1 + 3b2 = m for some positive integer m. In this ase, 1 + 3b2 = m2 or m2 − 3b2 = 1. Do you re ognize this equation? This is an example of a Pell Equation. Oh dear { this keeps getting more ompli ated. Let's summarize for a se ond √ to see where we are. We are trying to nd positive integers a for whi h b = 3a2 − 2a is a positive integer. We've now onverted this problem into nding positive integer solutions to the Pell Equation m2 − 3b2 = 1. If we have a solution (m, b) to this equation, we take the value of b and it is a

andidate to give a value of a that works. Finding all of the solutions to su h a Pell Equation is a bit beyond the s ope of this olumn. If you've never seen how to nd the positive integer solutions to m2 − 3b2 = 1, here's a qui k summary:

210 Find the smallest pair of positive integers that is a solution. You an do this by starting with b = 1 and in reasing b until you nd a solution. Here, we're lu ky and b = 1 gives m = 2, so (m, b) = (2, 1) is the smallest positive integer solution. Step 2 Starting with a solution (m, b), we an get another solution (M, B) by al ulating (M, B) = (2m + 3b, m + 2b). By this method we an get an in nite number of positive integer solutions to the equation m2 − 3b2 = 1. The pair (2, 1) leads to (7, 4) whi h leads to (26, 15) whi h leads to (97, 56) whi h yields (362, 209), and so forth. Those of you who are more enterprising ould try to write down some kind of losed form solution. (As an additional hallenge, try using the relation between (M, B) and (m, b) to solve for m and b in terms of M and B . If you an do this, then you an use a larger solution to generate smaller ones.) This method gives us a list√of values of b that we might try. At the very least, these values of b make 1 + 3b2 an integer, though they might not make a an integer be ause of the extra ondition. Let's try the rst few: Step 1

•

• • •

If b = 1, then a integer.

1+

=

√

√

1+ 1 + 3b2 = 3

1 + 3(12 ) = 1, 3

p

2)

1 + 3(4 If b = 4, then a = 3 an integer. If b = 15, then we've already seen that a = 9. 1+

If b = 56, then a not an integer.

=

1+

1+ 1 + 3b2 = 3

p

√

1+ 1 + 3b2 = 3

√

=

8 , 3

whi h is an

whi h is not

p 1 + 3(562 ) 98 = , 3 3

1+

p

1 + 3(2092 )

whi h is

= 121, whi h If b = 209, then a = 1 + 13 + 3b = 3 is an integer. At a qui k glan e, it appears that every other value of b in this sequen e gives an integer value of a, but we have by no means a tually proven this. Certainly, every value of b that we get out will give us an integer value for √ 1 + 3b2 , but this does not guarantee us integer values for a. Let's wrap up what we've done. We started out with a problem involving omplex numbers. We then talked about some of the stru ture of the problem and tried to do some investigation into what was happening \behind the s enes". We've found two more values of a that would give us integer values of b. It looks as if this method might lead to many more su h values, though we haven't proven this. So we started with a problem on omplex numbers that looked s ary, but wasn't a tually that ompli ated. This not-so- ompli ated problem led us down some pretty unexpe ted paths. There's lots more investigating that

ould be done here too, for instan e looking at when the imaginary parts of z 2 and z 4 might be equal. •

2

211

THE OLYMPIAD CORNER No. 278

R.E. Woodrow

We start this number with problems of the Hungarian Mathemati al Olympiad 2005-2006, National Olympiad, Grades 11-12, Se ond Round and Final Round. Thanks go to Robert Morewood, Canadian Team Leader to the 47th IMO in Slovenia, for olle ting them for our use. Hungarian Mathemati al Olympiad 2005{2006 National Olympiad, Grades 11-12

Se ond Round 1

. Find the positive values of x that satisfy x(2 sin x−cos 2x) <

1 x

. For f (x) = ax2 − bx + c we know that 0 f (b) = −a. Prove that |c| < 3. 2

.

< |a| < 1, f (a) = −b,

and

. The onvex quadrilateral ABCD satis es ∠ABC = ∠ACD. The midpoints of BC and AD are E and F (respe tively), O = AC ∩ BD, and G and H are the perpendi ular proje tions of O on the lines AB and CD (respe tively). Prove that the lines EF and GH are perpendi ular. 3

. Let a, b, c, d, and n be integers su h that n|ab, n|(bc + ad), and n|bd. Prove that n|bc and n|ad.

4

Final Round . De ne the fun tion t(n) on the nonnegative integers by t(0) = t(1) = 0, and for n > 2 let t(n) bethe smallest positive integer whi h does not divide n. Let T (n) = t t t(n) . Find the value of S if

1

t(2) = 1,

S = T (1) + T (2) + T (3) + · · · + T (2006) .

. Let A and B be two verti es of a tree with 2006 edges. We move along the edges starting from A and would like to get to B without ever turning ba k. At any vertex we hoose the next edge among all possible edges (ex ept the one on whi h we arrived) with equal probability. Over all possible trees and hoi es of verti es A and B , nd the minimum probability of getting from A to B . 2

212 . A unit ir le k with entre K and a line e are given in the plane. The perpendi ular from K to e interse ts e in point O and KO = 2. Let H be the set of all ir les entred on e and externally tangent to K . Prove that there is a point P in the plane and an angle α > 0 su h that ∠AP B = α for any ir le in H with diameter AB on e. Determine α and the lo ation of P . 3

Next we give the rst round of the Hungarian Mathemati al Olympiad 2005{2006 for Spe ialized Mathemati al Classes. Thanks again go to Robert Morewood, Canadian Team Leader to the 47th IMO in Slovenia, for olle ting it for our use. Hungarian Mathemati al Olympiad 2005{2006

Spe ialized Mathemati al Classes, First Round

1. Is it true that there are in nitely many palindromes in the arithmeti progression 7k + 3, k = 0, 1, 2, . . . ? (A number is a palindrome if reversing its digits yields the same number, for example, 12321 is a palindrome.)

. We have nitely many (but at least two) numbers of the form 21k whose sum is at most 1. Prove that they an be divided into two groups su h that the sum of the numbers in ea h group is at most 12 .

2

. The interval [0, 1] is divided by 999 red points into 1000 equal parts and by 1110 blue points into 1111 equal parts. Find the minimum distan e between 3

a red point and a blue point. How many pairs of blue and red points a hieve this minimum distan e? 4. A tetrahedron has at least four edges ea h at most 1 unit in length. Determine the maximum possible volume of the tetrahedron.

. Let k be a ir le with entre O and let AB be a hord of k whose midpoint, is distin t from O. The ray from O through M meets k at R. Let P be a point on the minor ar AR of k, let P M meet k again at Q, and let AB meet QR at S . Whi h segment is longer, RS or P M ?

5

M,

The next problem set is the 2005 Kurs h ak Competition (a Hungarian

ompetition). Thanks again go to Robert Morewood, Team Leader to the 47th IMO in Slovenia, for olle ting it for our use.

213 2005 Kurs h ak Competition

. Let N > 1 a1 , a2 , . . . , aN

and assume that the sum of the nonnegative real numbers is at most 500. Prove that there exists an integer k ≥ 1 and there exist integers 1 = n0 < n1 < · · · < nk = N su h that

1

k X

ni ani−1 < 2005 .

i=1

2. Ann and Bob are playing tennis. The winner of a mat h is the player who is the rst to win at least four games, being at least two games ahead of his or her opponent. Ann wins a game with probability p ≤ 21 independently of the out ome of the previous games. Prove that Ann wins the mat h with probability at most 2p2 . 3. A tower is built using dominos of size 2×1. The rst level of the tower is a 10 × 11 re tangle onsisting of 55 dominos, and ea h subsequent level is also a 10 × 11 re tangle onsisting of 55 dominos that exa tly overs the previous

level. The tower thus built is alled rigid if above ea h internal point of the rst 10 × 11 re tangle whi h is not a gridpoint, there is an internal point of some domino of the tower. What is the minimum number of levels a rigid tower may have?

Next we give the problems of the 8th Hong Kong (China) Mathemati al Olympiad written on De ember 3, 2005. Thanks go to Robert Morewood, Canadian Team Leader to the 47th IMO in Slovenia, for olle ting them for us. th

8

Hong Kong (China) Mathemati al Olympiad

. On a planet there are 3 · 2005! aliens and 2005 languages. Ea h pair of aliens ommuni ate with ea h other in exa tly one language. Show that there are 3 aliens who ommuni ate with ea h other in one ommon language.

1

. Suppose that there are 4n line segments of unit length inside a ir le of radius n. Given a straight line ℓ, prove that there exists a straight line ℓ′ that is either parallel to or perpendi ular to ℓ and su h that ℓ′ interse ts at least two of the given line segments.

2

3. Let a, b, c, and d be positive real numbers su h that a + b + c + d = 1. Prove that 6 a3 + b3 + c3 + d3 ≥ a2 + b2 + c2 + d2 + 18 .

. Show that there exist in nitely many squarefree positive integers n that divide 2005n − 1. (An integer is squarefree if it has no fa tor of the form d2 for an integer d > 1.) 4

214 Next we give the two Hong Kong Team Sele tion Tests for the International Mathemati al Olympiad 2006. Thanks go to Robert Morewood, Canadian Team Leader to the 47th IMO in Slovenia, for olle ting them for the Corner. Hong Kong Team Sele tion Test 1

. Find the integer solutions of the equation 7(x + y) = 3

1

2

x2 − xy + y 2

.

. The fun tion f (x, y), de ned for nonnegative integers x and y, satis es

(a)

f (0, y) = y + 1,

(b)

f (x + 1, 0) = f (x, 1),

( )

f (x + 1, y + 1) = f x, f (x + 1, y) .

and

Find f (3, 2005) and f (4, 2005).

. In triangle ABC , the altitude, angle bise tor, and median from C divide into four equal angles. Find ∠B .

3

∠C

. Let x, y, and z be positive real numbers su h that x + y + z = 1. For a positive integer n, let Sn = xn + yn + z n . Also, let P = S2 S2005 and Q = S3 S2004 .

4

(a) Find the smallest possible value of Q. (b) If x, y, and z are distin t, determine whi h of P or Q is the larger. 5. Finitely many points lie in a plane su h that the area of the triangle formed by any three of them is less than 1. Show that all of the points lie inside or on the boundary of a triangle with area less than 4.

. Find 22006 positive integers su h that

6

(a) ea h positive integer has 22005 digits; (b) ea h digit of ea h positive integer is a 7 or an 8; ( ) any two positive integers have at most half of their digits in ommon. Hong Kong Team Sele tion Test 2

. Let ABCD be a y li quadrilateral. Show that the ortho entres of △ABC , △BCD , △CDA, and △DAB are the verti es of a quadrilateral

ongruent to ABCD and show that the entroids of the same triangles are

1

the verti es of a y li quadrilateral.

215 2. Let ABCD be a y li quadrilateral with BC = CD . The diagonals AC and BD interse t at E . Let X , Y , Z , and W be the in entres of △ABE , △ADE , △ABC , and △ADC , respe tively. Show that X , Y , Z , and W are on y li if and only if AB = AD.

. Points A and B lie in a plane and ℓ is a line in that plane passing through but not through B . The point C moves from A toward in nity along a half-line of ℓ. The in ir le of △ABC tou hes BC at D and AC at E . Show that the line DE passes though a xed point.

3

A

4. Let △ABC have ir umradius R. Let AB = c, BC = a, CA = b, and let k1 and k2 be the ir les with diameters CA and CB , respe tively. Let k be the ir le of radius r whi h is tangent to k1 , k2 , and the line AB .

(a) Express r in terms of a, b, c, and R. (b) Find ∠C if r =

1 R 4

.

Next we give the 20th Nordi Mathemati al Contest written on Mar h 30, 2006. Thanks go to Robert Morewood, Canadian Team Leader to the 47th IMO in Slovenia, for obtaining it for the Corner. th

20

Nordi Mathemati al Contest

Mar h 30, 2006

1. Let B and C be points on two given rays from the same point A, su h that AB + AC is onstant. Prove that there exists a point D distin t from point A su h that the ir um ir les of the triangles ABC pass through D for all hoi es of B and C subje t to the given onstraint.

. The real numbers x, y, and z are not all equal and satisfy

2

x+

1 y

= y+

1 z

= z+

1 x

= k.

Determine all possible values of k. 3. The sequen e {an } of positive integers is de ned by a0 = m and the re ursion an+1 = a5n + 487 for all n ≥ 0. Determine all values of m for whi h the sequen e ontains as many square numbers as possible.

. The squares of a 100 × 100 hessboard are oloured with 100 dierent

olours. Ea h square is painted with one olour only and ea h olour is used exa tly 100 times. Show that there exists a row or a olumn on the hessboard su h that at least 10 dierent olours are used to paint its squares.

4

216 As three sets of problems for your problem solving pleasure over the break we give the questions of the 9th , the 10th , and the 11th form of the fth ( nal) round of the 2005{2006, XXXII Russian Mathemati al Olympiad. Thanks go to Robert Morewood, Canadian Team Leader to the 47th IMO in Slovenia, for obtaining them for our use. We remark that the odd-numbered questions of the 10th form orrespond to the odd-numbered questions on the 9th form, hen e they are omitted. XXXII Russian Mathemati al Olympiad 2005{2006 th

Final Round, 9 form First Day

. A square board 15 × 15 is divided into 152 unit squares. Some pairs of

entres of neighbouring (along a side) ells are onne ted by segments so that these segments form a losed broken line that does not interse t itself and that is symmetri with respe t to one of the diagonals. Prove that the length of the broken line is at most 200 units.

1

. Show that there exist four integers a, b, c, and d whose absolute values are greater than 1 000 000 and whi h satisfy

2

1 1 1 1 1 + + + = a b c d abcd

.

. On a ir le 2006 points are given. Peter olours ea h of these points with one of 17 olours. After that Mi hael onne ts these points by segments so that two endpoints of ea h segment have the same olour and the segments do not interse t (in parti ular, the segments do not have ommon endpoints). Mi hael wants to draw as many segments as possible, but Peter tries to hinder him. Find the greatest number of segments Mi hael an draw regardless of Peter's olouring. 3

. A ir le ω tou hes the ir um ir le of a triangle ABC at A, interse ts side at K , and interse ts side BC . A tangent CL to ω, with L on ω, is su h that the segment KL interse ts side BC at T . Prove that the length of BT equals the length of the tangent from B to ω.

4

AB

9th form Se ond Day 5. Let a1 , a2 , . . . , a10 be positive integers su h that a1 < a2 < · · · < a10 . Let bk be the greatest divisor of ak su h that bk < ak . If b1 > b2 > · · · > b10 , prove that a10 > 500.

217 . The points P , Q, and R lie on the sides AB , BC , and CA of a triangle su h that AP = CQ and the quadrilateral RP BQ is y li . The tangents to the ir um ir le of triangle ABC at A and C meet the respe tive lines RP and RQ at X and Y . Prove that RX = RY . 6

ABC

. A 100×100 square board is ut into dominos (that is, into 2×1 re tangles). Two players play a game. At ea h turn, a player may glue together any two adja ent squares if there is a ut between them. A player loses if he or she re onne ts the board (thus allowing the board to be lifted by a orner without it falling apart). Who has a winning strategy, the rst player or the se ond player? 7

. A quadrati polynomial f (x) = x2 + ax + b is given. Suppose that the equation f f (x) = 0 has four distin t real roots and that the sum of two of them is equal to −1. Prove that b ≤ − 41 .

8

10th form First Day 2. Assume that the sum of the ubes of three onse utive positive integers is a ube of some positive integer. Prove that the middle number of these three numbers is divisible by 4.

. Let ABC be an isos eles triangle with AB = AC . Let ω be a ir le tou hing the sides AB and AC and meeting the side BC at K and L. The segment AK meets ω for the se ond time at M . The points P and Q are the re e tions of K with respe t to B and C , respe tively. Prove that the

ir um ir le of △P M Q is tangent to ω.

4

10th form Se ond Day 6. Let K and L be points lying on the ar s AB and BC of the ir um ir le of △ABC , respe tively, so that the lines KL and AC are parallel. Prove that the in entres of △ABK and △CBL are equidistant from the midpoint of the ar ABC . 8. A 3000 × 3000 square is divided into dominos (that is, into 2 × 1 re tangles). Prove that one an paint the dominos with three olours su h that ea h olour is used equally often and ea h pie e shares a side with no more than two pie es of the same olour.

11th form

First Day 1

. Prove that sin

√ √ x < sin x whenever 0 < x <

π 2

.

218 2. Assume that the sum and the produ t of some pure periodi de imal fra tions is a pure de imal fra tion with period T . Prove that ea h of the original fra tions has a period not ex eeding T .

. Consider a 49×69 grid. All the 50·70 gridpoints are marked. Two players take turns onne ting pairs of gridpoints by a segment so that no gridpoint is the endpoint of two segments. The segments are drawn until there are no free gridpoints. The rst player wins if, after this pro ess, he an orient ea h of the segments so that the sum of all ve tors obtained is equal to the null ve tor; otherwise the se ond player wins. Whi h player has a winning strategy?

3

4. The angle bise tors BB1 and CC1 of △ABC (with B1 on AC and C1 on AB ) meet at I . The line B1 C1 meets the ir um ir le of △ABC at M and N . Prove that the ir umradius of △M IN is twi e the ir umradius of △ABC .

11th form Se ond Day

. The sequen es of positive numbers {xn } and {yn } satisfy

5

xn+2

= xn + x2n+1 ,

yn+2

2 = yn + yn+1 ,

for all positive integers n. Prove that if x1 , x2 , y1 , and y2 are ea h greater than 1, then xn > yn for some positive integer n. 6. Let SABC be a tetrahedron. The in ir le of △ABC has in entre I , and tou hes AB , BC , CA at D, E , F , respe tively. The points A′ , B ′ , C ′ lie on the edges SA, SB , SC , respe tively, so that AA′ = AD, BB ′ = BE , and CC ′ = CF . Let SS ′ be a diameter of the ir umsphere of SABC . Suppose that SI is an altitude of the pyramid. Prove that S ′ is equidistant from A′ , B ′ , and C ′ .

. The polynomial (x + 1)n − 1 is divisible by a polynomial

7

P (x) = xk + ck−1 xk−1 + ck−2 xk−2 + · · · + c1 x + c0

of even degree k su h that c0 , c1 , . . . , ck−1 are odd integers. Prove that n is divisible by k + 1. . A group of pioneers has arrived at summer amp. Ea h pioneer has at least and at most 100 friends among the others. Prove that one an distribute eld aps that ome in 1331 dierent olors one ea h to the pioneers so that the friends of ea h pioneer have aps of at least 20 dierent olors. 8

50

219 Now we return to the le of solutions from our readers to problems in the May 2008 number of the Corner and the XIX Olimpiada Iberoameri ana de Matemati as, given at [2008 : 214℄. . Let A be a xed exterior point with respe t to a given ir le with entre and radius r. Let M be a point on the ir le and let N be diametri ally opposite to M with respe t to O. Find the lo us of the entres of the ir les passing through A, M , and N , as the point M is varied on the ir le.

2

O

Solved by Mi hel Bataille, Rouen, Fran e; and Titu Zvonaru, Comane sti, Romania. We give Bataille's write up. Let U be the ir um entre of △AM N . Sin e O is the midpoint of M N , U U O is orthogonal to M N M0 and U M 2 = U O2 + r2 . But U M = U A, so we N have U A2 − U O2 = r2 . It follows that U is on a line ℓ perpendi ular to AO. A U0 O More pre isely, if M0 N0 is M the diameter perpendi ular to AO and U0 is the ir um entre of △AM0 N0 , then ℓ ℓ is the perpendi ular to N0 AO through U0 . Note that U0 6= O . Conversely, Let U be any point on ℓ (so that U A2 − U O2 = r2 ) and let M N be the diameter perpendi ular to U O . Then U O 2 +r 2 = U M 2 = U N 2 so that U M = U N = U A and U is the ir um entre of △AM N (note that A, M , and N are not ollinear be ause U0 6= O ). ..... ............... ..

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

. Let n and k be positive integers su h that n is odd or n and k are even. Prove there exist two integers a and b su h that gcd(a, n) = gcd(b, n) = 1 and k = a + b.

3

Solution by Mi hel Bataille, Rouen, Fran e. If n = 1, we take a = k − 1 and b = 1; so we suppose that n > 1. We rst onsider the ase where n is odd. Let the prime divisors of n be p1 , p2 , . . . , pr . Note that ea h pj is odd and does not divide both k − 1 and k + 1 (otherwise pj divides (k + 1) − (k − 1) = 2). Let aj ∈ {k − 1, k + 1} be su h that pj does not divide aj . By the Chinese Remainder Theorem, there exists an integer a su h that a ≡ aj (mod pj ) for ea h j . We take b = k − a. Then, ab ≡ aj (k − aj ) (mod pj ) with aj 6≡ 0 (mod pj ) and k − aj is 1 or −1. Hen e, ab 6≡ 0 (mod pj ) for ea h j . It follows that gcd(a, n) = gcd(b, n) = 1 and k = a + b.

220 We onsider next the ase where both n and k are even. If n = 2m for some positive integer m, then (sin e k is even) a = k − 1 and b = 1 satisfy the requirements. Otherwise, we denote by p1 , p2 , . . . , pr the odd prime divisors of n and let aj ∈ {k − 1, k + 1} be su h that pj does not divide aj . The Chinese Remainder Theorem provides an integer a su h that a ≡ aj (mod pj ) for ea h j and a ≡ 1 (mod 2). Setting b = k − a (note that b is odd), it is readily seen that gcd(a, n) = gcd(b, n) = 1 and k = a + b. 4. Find all pairs of positive integers (a, b) su h that a and b ea h have two digits, and su h that 100a + b and 201a + b are perfe t squares with four digits ea h.

Solution by Titu Zvonaru, Comane sti, Romania Let x, y be positive integers su h that 100a+b = x2 and 201a+b = y2 . Sin e x2 and y2 have four digits ea h, we have 32 ≤ x, y ≤ 99. Subtra ting the two equations yields 101a = y 2 − x2 = (y − x)(y + x) .

Be ause 101 is prime, y − x < 101, and y + x < 2 · 101, it follows that y + x = 101 and y − x = a. Therefore, y = 101 − x, a = 101 − 2x, and b = x2 + 200x − 10100. We√have x2 + 200x − 10100 > 9 sin e b has two digits, hen e x > −100 + 20109 > 41. On √ the other hand, sin e x2 + 200x − 10100 ≤ 99, we have x ≤ −100 + 20199 < 43. We now dedu e that x = 42, y = 59, a = 17, and b = 64 is the unique solution to the problem.

. In a s alene triangle ABC , the interior bise tors of the angles A, B , and meet the opposite sides at points A′ , B ′ , and C ′ respe tively. Let A′′ be the interse tion of BC with the perpendi ular bise tor of AA′ , let B ′′ be the interse tion of AC with the perpendi ular bise tor of BB ′ , and let C ′′ be the interse tion of AB with the perpendi ular bise tor of CC ′ . Prove that A′′ , B ′′ , and C ′′ are ollinear. 5

C

Solution by Titu Zvonaru, Comane sti, Romania A Let a = BC , b = CA, and c = AB . Let M be the midpoint of AA′ . By the Bise tor Theorem we have ab A′ C = , BA′ = b ac , b+c +c and (with α = A ) it is known 2 2bc cos that AA′ = b + c α . Sin e ∠M A′ A′′ = α + C ,

in the

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

A′′

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

α

B

M

A′

C

221 triangle A′′ A′ M we obtain A′ A′′ =

M A′ AA′ bc cos α = = cos(α + B) 2 cos(α + C) (b + c) cos(α + C)

.

Using the Law of Sines we obtain: A′′ B A′′ C

A′ A′′ − A′ B bc cos α − ac cos(α + C) = ′ ′′ ′ AA +AC bc cos α + ab cos(α + C)

=

c

=

b

·

sin B cos α − sin A cos(α + C) sin C cos α + sin A cos(α + C)

c sin B cos α − 2 sin α cos α cos(α + C) · b sin C cos α + 2 sin α cos α cos(α + C)

=

.

Sin e cos α 6= 0, we obtain A′′ B A′′ C

= = = =

′′

c b c b c b c b

· · · ·

sin B − 2 sin α cos(α + C) sin C + 2 sin α cos(α + C)

sin B − sin(α + α + C) − sin(α − α − C) sin C + sin(α + α + C) + sin(α − α − C)

sin B − sin(A + C) + sin C sin C + sin(A + C) − sin C

sin B − sin B + sin C sin C + sin B − sin C

=

c b

·

sin C sin B

,

2

B c hen e A = 2. A′′ C b By the onverse of Menelaus' theorem, it follows that A′′ , B ′′ , and C ′′ are ollinear.

Next we look at solutions from our les to problems of the Swedish Mathemati al Contest 2004/2005 Quali ation Round given at [2008 : 215℄. . The ities A, B , C , D, and E are onne ted by straight roads (more than two ities may lie on the same road). The distan e from A to B , and from C to D , is 3 km. The distan e from B to D is 1 km, from A to C it is 5 km, from D to E it is 4 km, and nally, from A to E it is 8 km. Determine the distan e from C to E . 1

Solved by Titu Zvonaru, Comane sti, Romania. Sin e AB + BD + DE = 3 + 1 + 4 = 8 = AE , it follows that the

ities A, B , D, and E are ollinear (on the same road).

222 We have = =

52 = 32 + 42 CD 2 + AD 2 ,

A E B D ..s......................3.....................s...............s........................4...........................s ...... 1 .. .. .............. ... ... ................

AC 2

... ...... ...... .. 3 ...... . 5 ...... ...... ... ...... .. ...s

hen e, by the onverse of the Pythagorean Theorem, CD is perpendi ular to AD. By the Pythagorean Theorem, we now obtain CE = 5.

C

2. Linda writes the four positive integers a, b, c, and d on a pie e of paper. Sin e she is amused by arithmeti , she adds the numbers in pairs, obtaining the sums a + b, a + c, . . . , c + d, but she forgets to write down one of the possible sums. The ve sums she obtains are 7, 11, 12, 18, and 23. Whi h sum did Linda forget? What are the positive integers a, b, c, d ?

Solved by Jean-David Houle, student, M Gill University, Montreal, QC; John Grant M Loughlin, University of New Brunswi k, Frederi ton, NB; and Titu Zvonaru, Comane sti, Romania. We give Houle's solution, modi ed by the editor. Without loss of generality, let the unknown sum be c + d. Then we know that 3(a + b) + 2(c + d) = 7 + 11 + 12 + 18 + 23 = 71. Sin e 71 is odd, a + b is also odd, hen e a + b is 7, 11, or 23 and then c + d is 25, 19, or 1, respe tively. Obviously c + d = 1 is impossible, so a + b + c + d is either 7 + 25 = 32 or 11 + 19 = 30. Sin e two numbers in {11, 12, 18, 23} must also add up to a + b + c + d but no two from this set add up to 32, it follows that a + b = 11 and that Linda forgot the sum c + d = 19. Now, a + b = 11 and 12 is the sum of a number from {a, b} and a number from {c, d}, so we may take c = a + 1. Then a + c = 2a + 1 is either 7 or 23, so a = 3 or a = 11. However, a + b = 11, hen e a 6= 11. We dedu e that (a, b, c, d) = (3, 8, 4, 15), whi h yields all of the required sums. . Determine the greatest and the least value of

3

mn (m + n)2

,

if m and n are positive integers, ea h not greater than 2004. Solved by George Apostolopoulos, Messolonghi, Gree e; Jean-David Houle, student, M Gill University, Montreal, QC; Pavlos Maragoudakis, Pireas, Gree e; Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON; and Titu Zvonaru, Comane sti, Romania. We give Zvonaru's write up. By the AM{GM Inequality we have

mn mn 1 ≤ = , (m + n)2 4mn 4

greatest value is 14 , a hieved when m = n.

hen e the

223 Let t =

m . n

Sin e 1 ≤ m, n ≤ 2004, then mn n)2

≥

1 ≤ t ≤ 2004. 2004

We have

2004

⇐⇒

(m + 20052 t 2004 ≥ 2 (t + 1) (2004 + 1)2

⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒

2004t2 + 4008t + 2004 − 20042 t − 4008t − t ≤ 0 2004t2 − t − 20042 t + 2004 ≤ 0 t(2004t − 1) − 2004(2004t − 1) ≤ 0 (t − 2004)(2004t − 1) ≤ 0 ,

1 and the last inequality is true sin e 2004 ≤ t ≤ 2004. Hen e the least value 2004 is 20052 , a hieved when (m, n) = (2004, 1) or (m, n) = (1, 2004).

. Let k and n be integers with 1 < k < n. If a set of n real numbers has the property that the mean value of any k of them is an integer, show that all n numbers are integers. 4

Solved by Jean-David Houle, student, M Gill University, Montreal, QC; and Titu Zvonaru, Comane sti, Romania. We give Zvonaru's write up. Let A = {a1 , a2 , . . . , an } be a set of real numbers satisfying the hypotheses. There are integers z2 , z2 , . . . , zk+1 su h that a1 + a2 + · · · + ak a1 + a2 + · · · + ak−1 + ak+1

.. .

a1 + a2 + a4 + · · · + ak + ak+1

a1 + a3 + a4 + · · · + ak + ak+1

= kzk+1 = kzk

.. . = kz3 = kz2

Adding these equations we obtain

ka1 + (k − 1)(a2 + a3 + · · · + ak+1 ) = k(z2 + z3 + · · · + zk+1 ) .

Now, there is an integer z1 su h that a2 + a3 + · · · + ak+1 a1 = z2 + z3 + · · · + zk+1 − (k − 1)z1 and a1 is an integer. It follows that a1 , a2 , . . . , an are integers.

= kz1 ,

hen e

. Let 2n (where n ≥ 1) points lie in the plane so that no straight line

ontains more than two of them. Paint n of the points blue and paint the other n points yellow. Show that there are n segments, ea h with one blue endpoint and one yellow endpoint, su h that ea h of the 2n points is an endpoint of one of the n segments and none of the segments have a point in

ommon. 6

224 Solved by Oliver Geupel, Bruhl, NRW, Germany; and Titu Zvonaru, Comane sti, Romania. We give Geupel's solution. There are nitely many (namely n!) bije tive orresponden es between the blue and the yellow points. Hen e, there is at least one orresponden e, Γ, where the sum, s(Γ), of the Eu lidean lengths of its n segments has minimum value. We will prove that ea h su h Γ has the desired property. Assume the ontrary. Then, there are two blue points, say A and B , and B (blue) two yellow points, say X and Y , su h .r ........ A (blue) . . . . . . . . . . . . that the segments AX and BY are . ..r... ..... ..... ........ ..... drawn and interse t in a point, say S . ........ . . ..... . . . . . ... ..... ..... ........ We onstru t another orresponden e Γ′ ..... ............. . . . . . .... ... from Γ by removing the segments AX ........ ..... ........ S .......... . . . . . and BY and adding the segments AY . . ..... ...... ..... ........ ..... and BX [Ed.: note that by the hypothe........ . .....r . . . . . ..r ses AY and BX ontain no oloured X (yellow) points ex ept their endpoints℄. The or- Y (yellow) ′ responden e Γ is also bije tive. By the triangle inequality, . ... .. .. ... .. .. .. .. .. . ... ... .. .. ... .. .. .. .. .. . ... .. .. .. .. . .

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

s(Γ′ ) = =

s(Γ) + AY + BX − AX − BY s(Γ) + (AY − AS − Y S) + (BX − SX − SB) < s(Γ) ,

ontradi ting the fa t that s(Γ) has minimum value. The proof is omplete. Next we look at solutions to problems of the Swedish Mathemati al Contest 2004/2005 Final Round, given at [2008 : 216℄. . Two ir les in the plane of the same radius R interse t at a right angle. How large is the area of the region whi h lies inside both ir les?

1

Solved by George Apostolopoulos, Messolonghi, Gree e; and Titu Zvonaru, Comane sti, Romania. We give Apostolopoulos' version. Sin e ∠O1 AO2 = 90◦ , A half the area of the lensshaped region of interse tion R R equals the area of a quarter ir le minus the area of O2 O1 △AO2 B . Hen e, the required area is R R 2

1 1 πR2 − R2 4 2

=

π−2 2 ·R . 2

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

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

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

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

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

B

[Ed.: Similarly, if the ir les interse t at θ radians, then the area between them is (θ − sin θ)R2 .℄

225 . The fun tion f satis es f (x) + xf (1 − x) Determine the fun tion f . 3

for all real numbers x.

= x2

Solved by George Apostolopoulos, Messolonghi, Gree e; Mi hel Bataille, Rouen, Fran e; Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON; and Titu Zvonaru, Comane sti, Romania. We give Wang's write up. −2 The only su h fun tion is f (x) = −x + 2 + x22x . −x+1 Repla ing x by 1 − x in the identity for f , we obtain

f (1 − x) + (1 − x)f (x) = (1 − x)2 .

(1)

Multiplying ea h side of (1) by x, we have Subtra ting the Hen e,

xf (1 − x) + x(1 − x)f (x) = x(1 − x)2 . (2) identity from (2) yields x − x2 − 1 f (x) = x − 3x2 + x3 .

f (x) =

−x3 + 3x2 − x x2

−x+1

2x − 2

= −x + 2 +

x2

Conversely, if f (x) is given by (3), then f (1 − x)

f (x) + xf (x − 1)

−(1 − x) + 2 +

=

x+1−

=

−x + 2 + x2 + x −

=

x2 + 2 − 2 = x2 ,

and the proof is omplete.

2x

.

(3)

2(1 − x) − 2 (1 − x)2 − (1 − x) + 1

=

x2

−x+1

−x+1

;

2x2 − 2x + 2 x2 − x + 1

. If tan v = 2v and 0 < v < π2 , then does it follow that sin v <

4

20 21

?

Solved by Mi hel Bataille, Rouen, Fran e; and Pavlos Maragoudakis, Pireas, Gree e. We give Bataille's write up. Yes, To prove this, we introdu e the fun tion φ(x) = tan x−2x it does. π for x ∈ 0, 2 , whose derivative is φ′ (x) = sec2 x − 2. Note that φ(0) = 0, that φ de reases on as

π− x → . 2

π 0, 4

and in reases on

π π , 4 2

Therefore, φ has a unique root v π x ∈ (0, v) and φ(x) > 0 for x ∈ v, . 2 This said, we have sin2 v = 1 −

∈

, and that φ(x) → +∞

0,

π 2

and φ(x)

1 1 = 1− 2 1 + tan v 1 + 4v 2

.

< 0

for

226 It follows that sin v < alent to φ

Now,

20 21

10 > 0. √ 41 10 5π , > √ 12 41

√ 20 10 < < 2 + 3, √ 3 41

is equivalent to v <

so that tan

10 √ 41

so we on lude that

10 √ , 41

whi h in turn is equiv-

√ 5π > tan = 2 + 3. 12 10 φ √ > 0, as desired.

Also,

41

. A square of integer side n, where n ≥ 2, is divided into n2 squares of side 1. Next, n − 1 straight lines are drawn so that the interior of ea h of the small squares (the boundary not being in luded in the interior) is interse ted by at least one of the straight lines. (a) Give an example whi h shows that this an be a hieved for some n ≥ 2. 5

(b) Show that among the n − 1 straight lines there are two lines whi h interse t in the interior of the square of side n.

Solution by Oliver Geupel, Bruhl, NRW, Germany. We use Cartesian oordinates with (0, 0), (n, 0), (n, n), and (0, n) being the orners of the big square Sn . For 0 ≤ i, j < n, let S(i, j) denote the unit square whose southwest orner is the point (i, j). ...................................................................................................................................... ... ... ... ... ... ..... .... .... .... ... .. ... .. ... ... ... ... ... . . . . . ... . . ... . . .... .... . . ..... . . . . .... . . . .... ..... .... . . . . . .... ...... ..... . . ... ................................................................................................................................................... . ... ... ..... ... ........ ... ... .... ................ ... ... ....... .. ... ... ... . .... .... ................. ..... .... .... ... .......... ..... .... .... . . . . . . . . ... . .. . .. ........................................................................................................................................................ . . . . . . . . ... . . ... . .... .. ..... .... ............. ..... .... .............. .... ........... .... .. ... ... ..... ... .... .... .... .... .... ... ... ... ... .. . ...................................................................................................................................

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

Figure A

Figure B

For part (a) we give an example with n = 3. Let g be the line through 0, 12 and 3, 52 , and let h be the line through 21 , 3 and 25 , 0 . The line g interse ts (see Figure A) the interiors of S(0, 0), S(0, 1), S(1, 1), S(2, 1), and S(2, 2), while the line h interse ts the interiors of the remaining squares S(1, 0), S(2, 0), S(0, 2), and S(1, 2). For part (b), it is suÆ es to prove the following statement for all n ≥ 1:

If m straight lines are drawn so that (a) the interior of ea h S(i, j) is interse ted by at least one of the lines, and (b) no two of the lines interse t in the interior of Sn , then m ≥ n.

The proof is by Mathemati al Indu tion. The laim is trivial for n = 1. Assume that it is valid for ea h integer n with 1 ≤ n < N . We prove it for n = N.

227 Assume that m lines ℓ1 , ℓ2 , . . . , ℓm are drawn so that (a) and (b) hold for SN . We sele t the appropriate latti e point of SN as the origin of a

oordinate system so that the in lination φ of ℓ1 satis es −90◦ ≤ φ < 0. The line ℓ1 interse ts the interior of at most one of the diagonal squares S(k, k), where 0 ≤ k < N , thus dividing SN in a southwest segment P

ontaining p of the S(k, k), and a northeast segment Q ontaining q of the S(k, k), where p + q ≥ N − 1. Therefore, P ontains a square of side p, and Q ontains a square of side q . By the indu tion hypothesis, at least p and q of the lines ℓk interse t P and Q, respe tively. (This remains true if p or q is zero.) By (b), ℓ1 does not interse t any other line in the interior of SN . Therefore, the p lines interse ting P are distin t from the q lines interse ting Q. We on lude that m ≥ 1 + p + q ≥ N , whi h ompletes the indu tion. Now we turn to solutions to the problems of the Abel Competition 2004-2005 given at [2008 : 216-217℄. . (a) A positive integer m is triangular if m = 1 + 2 + · · · + n for some integer n > 0. Show that m is triangular if 8m + 1 is a perfe t square. (b) The base of a pyramid is a right-angled triangle with sides of integer lengths. The height of the pyramid is also an integer. Show that the volume of the pyramid is an even integer.

1

Solution by Titu Zvonaru, Comane sti, Romania. (a) We have 1 + 2 + · · · + n = n(n2+ 1) , so all triangular numbers are of this form. Suppose that 8m + 1 is an odd perfe t square. Then the equation n(n2+ 1) = m has a solution if and only if n2 + n − 2m = 0 has a positive integer root. However, sin e 8m + 1 is an odd perfe t square, √ −1 + 1 + 8m n= is a positive integer root of the last equation; hen e m 2 is triangular. (b) Sin e the base of the pyramid is a right-angled triangle withsides of integer lengths, the legs of this triangle are k · 2mn and k m2 − n2 , where m, n, and k are positive integers and m > n > 0. If the height of the pyramid is h, then its volume is V = 13 k2 mn(m − n)(m + n)h. ′ 2 2 It suÆ es to show that V = mn m − n ≡ 0 (mod 6) in order to show that V is even. If m or n is even, then mn is even, and if both m and n are odd, then m2 − n2 is even. Hen e, V ′ is even. Similarly, if m or n is divisible by 3, then mn is divisible by 3, and if both m and n are not divisible by 3, then m2 − n2 is divisible by 3 (sin e x2 ≡ 1 (mod 3) for x 6≡ 0 (mod 3)). Hen e, V ′ is divisible by 3. It follows that the volume of the pyramid is an even integer.

228 . (a) There are nine small sh in a ubi aquarium with a side of 2 metres, entirely lled with water. Show that at any given√time there an be found two sh whose distan e apart is not greater than 3 metres. (b) Let A be the set of all points in spa e with integer oordinates. Show that for any nine points in A, there are at least two points among them su h that the midpoint of the segment joining the two is also a point in A. 2

Solution by Titu Zvonaru, Comane sti, Romania. (a) Divide the aquarium into 8 ubes ea h 1 metre on a side. There are at least two sh in one of the ubes by the Pigeonhole Prin iple, and they are √ no further apart than the diagonal of that ube, whi h is 3 metres. (b) Let P1 (x1 , y1 , z1 ) and P2 (x2 , y2 , z2 ) be two points in the set A. The y1 x2 + y2 x3 + y3 midpoint of the segment joining P1 to P2 is M x1 + , , 2 2 2 and M is in the set A if and only if the parities of orresponding oordinates of P1 and P2 agree. For any nine points in A, by the Pigeonhole Prin iple, at least ve points agree in the parity of the rst oordinate. Among these ve points, at least three agree in the parity of the se ond oordinate, and among these three points, at least two points agree in the parity of the third oordinate. Thus, we obtain at least two points P1 (x1 , y1 , z1 ) and P2 (x2 , y2 , z2 ) whose midpoint is in A. 3. (a) Let △ABC be isos eles with AB = AC , and let D be the midpoint of BC . The points P and Q lie respe tively on the segments AD and AB su h that P Q = P C and Q 6= B . Show that ∠P QC = 12 ∠BAC . (b) Let ABCD be a rhombus with ∠BAD = 60◦ . Let F , G, and H be points on the segments AD, CA, and DC respe tively su h that DF GH is a parallelogram. Show that △BHF is equilateral.

Solved by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain; D.J. Smeenk, Zaltbommel, the Netherlands; and Titu Zvonaru, Comane sti, Romania. We give the solution of Amengual Covas. (a) Sin e P B = P C , we have P B = P C = P Q. If the base angles in isos eles triangles P QB , P BC , and P CQ are x, y, and z , respe tively, then in △BCQ we have (x+y)+(y+z)+(z+x) = 180◦ , hen e z = 90◦ −(x+y). Thus, ∠P QC = z = 90◦ − (x + y) = 90◦ − ∠ABD = ∠BAD = 21 ∠BAC . (b) Sin e GHkAD, we have ∠CGH = ∠CAD. However, ABCD is a rhombus, so ∠CAD = ∠ACD. Therefore, ∠CGH = ∠ACD = ∠GCH , and hen e CH = GH = F D. Sin e △ABD and △BCD are equilateral, we also have BC = BD and ∠BCH = ∠BDF . Thus, △BCH and △BDF are ongruent with (i) BH = BF , and (ii) ∠CHB = ∠DF B . By (i), △BF H is isos eles; by (ii) DF BH is a y li quadrilateral, so that ∠F BH = 180◦ − ∠HDF = 180◦ − (60◦ + 60◦ ) = 60◦ . We on lude that △BHF is equilateral, as desired.

229 A

A .... ...... .. ... .. .. .. .. .. ..... .... . ... ... ... .. .. .. .. ..... .... .. ... ... .. ... . .. ... .... ... ... ... ... ... .. .... . . ... .. . . ... .... .. . . ... .. . . .. .... .. ... . ... ... ... . ... ......... . ... . .... .. ............ ... . ... .. ... ...... ... . ... ..... .... ... .. . ... ..... ... .. .. . . . ... ....... ... .. . . . . ... ... ...... .. . . . ... ... .... ....... .. ... . . . .... ... . . . ... . . . .... ... ... .. . ... . . .... ... ... .. .. . . . . ... .. .... ... .. . . . . . .... ....... ... .. . . . .... ..... ... .. . . . . . . . . . .... ... ...... .... ............... . . . . .. . . . . . . . . . .......... .. .... ... ...... . . . . .. . . . . . . . . . . . . . .... .... .......... . . ...... . . . . . . . . . . . . . . . . . . . . . .......... ... .... ... ....................... .................. . . . . . . . . .......... ..... ... .... .. .......................... . . . . . . ................... .. .......... .. .. ... .. ............................................................................................................................................................................................................

Figure A

Figure B

G

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

B

x z

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

D

z y

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

B

60

60

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

P

y

F

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

Q

x

.... ......... ... ... ... .. ... ... .. .... .... . . . ... ... ... ... ... ... ... ... ... ... .. ... .. .... . . . ... .. ....... . . . .. .................. ..... . . . ... ... ........ .. .. . . . . . . ... ..... .. .. . . . . ... . . . . . .. .. ... ....... .. .. ...... ..... ..... ... ... . . . . . . . . ... .. .. .... . . . ... . . . . . . ... . ...... .. ............ ... . .... . .. .......... . ........ . ....... ◦..... . . .. ... ... ............. . . . . . ....................................................................................................................................................... ... ............ . .... ..... ... ........... ◦..... ... ... ......... ... . ... ......... .. ... ......... .... ... ... ... ......... ... . . . ... .......... ... . .. ... ... . ..... ..... .................. .... ..... ... ............ .. ... ... ........ ... . . . . ... . ..... ... ... ... ... ... ... .... ... ... ... .. . ... . ..... ... ... ... .. ... ... .... ... ... ... ... ... ... ... ... ... ... ......... ....

C

D

H

C

. (a) Let a, b, and c be positive real numbers. Prove that

4

p (a + b)(a + c) ≥ 2 abc(a + b + c) .

(b) Let a, b, and c be real numbers su h that ab + bc + ca > a + b + c > 0. Prove that a + b + c ≥ 3.

Solved by George Apostolopoulos, Messolonghi, Gree e; Mi hel Bataille, Rouen, Fran e; Pavlos Maragoudakis, Pireas, Gree e; and Titu Zvonaru, Comane sti, Romania. We use Bataille's solution. (a) Equivalently, we show that L = (a + b)2 (a + c)2 − 4abc(a + b + c) is nonnegative. We have that (a + b)2 (a + c)2 = a2 + (ab + bc + ca) 2 and also (ab + bc + ca)2 = a2 b2 + b2 c2 + c2 a2 + 2abc(a + b + c). As a result, L

= = =

a2 a2 + b2 + c2 + b2 c2 + 2a2 (ab + bc + ca) − 2abc(a + b + c) a2 (a + b + c)2 − 2abc(a + b + c) + b2 c2 2 a(a + b + c) − bc .

Thus, L ≥ 0 and the inequality is proved. (b) Let s = a + b + c. We are given that ab + bc + ca > s, and s is positive, hen e we obtain s(2ab + 2bc + 2ca) > 2s2 . From the known inequality a2 + b2 + c2 ≥ ab + bc + ca and the hypothesis, we obtain s a2 + b2 + c2 ≥ s(ab + bc + ca) > s2 . Sin e s2 = (a + b + c)2 = a2 + b2 + c2 + (2ab + 2bc + 2ca), adding a ross the two inequalities we obtained yields s3 = s · s2 > 2s2 + s2 = 3s2 , and s > 3 follows immediately. That ompletes the Corner for the May Crux. Send me your ni e solutions and generalizations.

230

BOOK REVIEWS Amar Sodhi

The Mobius Strip: Dr. August Mobius's Marvelous Band in Mathemati s, Games, Literature, Art, Te hnology, and Cosmology By Cliord Pi kover, Thunder's Mouth Press, New York, 2006 ISBN 978-0-7394-7542-3, soft over, 244+xxiii pages, US$15.95 Reviewed by Edward J. Barbeau, University of Toronto, Toronto, ON Quoth mother of four year old Pete: \You must not ross Mobius Street." But an easy walk, On e round the blo k, Allowed him to manage the feat. [A winning limeri k in a ontest sponsored by the author.℄ What a wealth of onne tions and divergent thinking there is in this book! Cliord A. Pi kover is employed at the IBM Thomas J. Watson Resear h Center in Yorktown, New York. Although his do torate from Yale is in mole ular biology and bio hemistry, where he studied X-ray s attering and protein stru ture, his interest and knowledge is e le ti . His over thirty ve books in lude not only a number of popularizations of s ien e (in luding mathemati s), but also some s ien e tion. He is the author of over 200 papers and the holder of 30 patents \mostly on erned with novel features for omputers". His website http://www.pickover.com is worth a visit. This book is a ri h feast that ould either be read through or sampled and returned to frequently. One idea leads to another, and the text is pun tuated with puzzles, games, literary quotes, vignettes, diagrams of inventions, and pi tures of works of art. Of ourse, the author has to introdu e us to August Ferdinand Mobius (1790-1868) himself, who studied astronomy under Gauss and later dire ted the Observatory in Leipzig in 1848. We learn about his times, his distinguished an estry (in luding Martin Luther on his mother's side) and progeny, and see a pi ture of his skull, reprodu ed from a book by his grandson, a neurologist who made a phrenologi al study of mathemati ians' heads to see if there was a onne tion between bumps and mathemati al ability. The book opens with an introdu tion to the Mobius Strip and its singular properties, whi h leads to a dis ussion of knots and their lassi ation. A

hapter deals with various inventions inspired by the strip, and then we begin to explore topology, higher dimensions, turning spheres and doughnuts inside out, the Klein Bottle and the Alexander Horned Sphere. However, the name Mobius is familiar to mathemati ians in another ontext, through his number theoreti fun tion. This provides the opportunity for a detour to Mertens' onje ture, series involving π, and the Riemann zeta fun tion.

231 Then we enter the world of the imagination, of literature, art, osmology, and philosophy. No summary an do justi e to this part of the book, but the reader may be hallenged and fas inated by the author's spe ulations on multiple universes and arti ial life. In all, this is a ri h and rewarding journey. The Contest Problem Book IX: Ameri an Mathemati s Competitions (AMC 12) 2001-2007 Compiled and edited by David Wells and J. Douglas Faires, Mathemati al Asso iation of Ameri a, 2008 ISBN 978-0-88385-826-4, soft over, 214+xiv pages, US$49.95 Reviewed by John Grant M Loughlin, University of New Brunswi k, Frederi ton, NB Another olle tion of ontest problems has been prepared by the MAA. The high quality of this olle tion of problems, based on the standard grade 12 urri ulum, is not surprising as MAA problem books generally feature a healthy range of problems organized in a \resour eful" manner. This book is potentially valuable for tea hers working with students in high s hool math

ontest venues. The book onsists of 13 ontest papers with 25 questions ea h. The detailed solutions to these 325 problems are a pra ti al asset to any tea her or individual wishing to work with su h a olle tion. The opening set of problems is the 2001 ontest that appears under the heading \AMC 12 Problems". All subsequent years (2002-2007) feature two sets of problems, AMC 12A and AMC 12B. Ea h problem set is a stand-alone

ontest of 25 multiple hoi e problems. The format was hanged in 2002 to permit two separate oerings of the same ontest. AMC 12A and AMC 12B are oered just two weeks apart from one another for dierent students. Hen e, the problem setters are required to ensure that the diÆ ulty level and the balan e of topi s be onsistent within these two papers while not providing an unfair advantage to those who write the 12B ontest two weeks after the 12A ontest has been ompleted. It is noted that al ulus is not required to solve any of the problems, as al ulus is not a standard grade 12 topi . The prefa e is an honest and ontextually valuable re e tion on the

hallenges, in luding pitfalls, of putting together a ontest. Wells and Faires oer helpful insight to people who have never tried reating su h papers. My own experien es in various settings are onsistent with those re e ted here, a knowledging the tensions between on ise problem statements and detail, among other issues. The informative prefa e is one of three features that strike me as being ex eptional about this resour e. Familiarity with other publi ations in this series has reinfor ed my appre iation and expe tation for a se ond noteworthy feature: the "Problem DiÆ ulty" summary for ea h paper. Here the authors present the distribution of student responses (% A, B, C, D, E, and Omitted) for ea h problem while learly identifying

232 the orre t response. These tables telling the popularity of some wrong

hoi es and/or the low level of orre t responses inform me as to mis on eptions or troublesome areas of mathemati s. Insight into su h matters has been valuable to me as a mathemati ian working in a Fa ulty of Edu ation with tea hers at dierent levels. Finally, the third outstanding feature is the Index of Problems that identi es areas, su h as Algebra, with many subheadings (e.g., AM{GM Inequality; parametri equations of lines; sum and produ t of roots) ea h ontaining a listing of the spe i problem numbers that apply to the topi . In the ase of Geometry, the subtopi s are further

ategorized under six broad areas: ir les, oordinates, polygons, quadrilaterals, solids, and triangles. Any tea her looking for problems on a parti ular topi an work from here { always a good strategy to onsider. While applauding the merits of the book, it is noted here that its ost seems to be the lone disin entive to a quiring it as an individual resour e. The book's value would be enhan ed as a shared resour e through a library, math department, or ommon browsing area. A Taste of Mathemati s, Volume IX, The CAUT Problems By Edward Barbeau, Canadian Mathemati al So iety, 2009 ISBN 978-0-919558-21-2, soft over, 59+ii pages, US$15.00 Reviewed by Amar Sodhi, Sir Wilfred Grenfell College, Corner Brook, NL The CAUT Bulletin is a monthly newspaper published by the Canadian Asso iation of University Tea hers. This periodi al ontains arti les of interest to both tenured and non-tenured fa ulty and, by virtue of having a large se tion devoted to job listings, is invaluable to anyone who is sear hing for an a ademi position at a Canadian university. It is therefore a pleasant surprise to nd that a paper whi h es hews both a rossword puzzle and a Sudoku nevertheless in ludes a mathemati al brainteaser. The \Homework!" is set by Edward Barbeau, a mathemati ian well known (among other things) for his work with the Canadian Mathemati al Olympiad. However, it is his ability to onvey mathemati s to a general audien e whi h omes to the forefront in The CAUT Problems. This olle tion of forty-two puzzles, whi h have no doubt intrigued many a s holar of the humanities or so ial s ien es, are bound to please a person who enjoys ta kling a problem whi h requires thought, but no more than basi skills in algebra. Undoubtedly The CAUT Problems is a valuable resour e for tea hers who are sear hing for nifty posers to give to their keenest math students. The bonus hapter on ard tri ks may even allow a tea her to perform a little magi in the lassroom. As with other volumes in the ATOM series, this book is well suited as a prize in a high s hool mathemati s event. However, unlike the other volumes in the ATOM series, this is a book whi h a student an enjoy sharing with grandparents during a summer va ation. I should give a sample problem here, but my teenage daughter has run o with the book again!

233

PROBLEMS Toutes solutions aux problemes dans e numero doivent nous parvenir au plus tard le 1er novembre 2009. Une e toile (⋆) apres le numero indique que le probleme a e te soumis sans solution. Chaque probleme sera publie dans les deux langues oÆ ielles du Canada (anglais et fran ais). Dans les numeros 1, 3, 5 et 7, l'anglais pre edera le fran ais, et dans les numeros 2, 4, 6 et 8, le fran ais pre edera l'anglais. Dans la se tion des solutions, le probleme sera publie dans la langue de la prin ipale solution presentee. La reda tion souhaite remer ier Jean-Mar Terrier, de l'Universite de Montreal, d'avoir traduit les problemes. . Corre tion. Propose par Vo Quo Ba Can, Universite de Mede ine et Pharma ie de Can Tho, Can Tho, Vietnam. Soit a, b et c trois nombres reels positifs. 3419

(a) Montrer que (b) Montrer que

P

y lique

P

y lique

r

√ a2 + 4bc ≥ 2 + 2. b2 + c2

r

a2 + bc 1 ≥ 2+ √ . 3 b2 + c2 2

3

. Corre tion. Propose par Yakub N. Aliyev, Universite d'Etat de Bakou, Bakou, Azerbadjan. Pour un entier positif m, soit σ la permutation de {0, 1, 2, . . . , 2m} de nie par σ(2i) = i pour i = 0, 1, 2, . . . , m et σ(2i − 1) = m + i pour i = 1, 2, . . . , m. Montrer qu'il existe un entier positif k tel que σ k = σ et 1 < k ≤ 2m + 1. 3424

. Propose par D.J. Smeenk, Zaltbommel, Pays-Bas. On suppose que dans le triangle ABC la hauteur ha = AD est egale a a = BC . Soit H l'ortho entre, M le point milieu de BC et E le point milieu de AD. Montrer que HM = HE . Quand la re iproque est-elle valide ? 3439

. Propose par Hidetoshi Fukugawa, Kani, Gifu, Japon. Sur une table, on a N pie es de monnaie, toutes de m^eme dimension. Ces N pie es de monnaie peuvent e^ tre arrangees aussi bien en un arre qu'en un triangle equilat eral. Trouver N . 3440

⋆. Propose par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie.

3441

Soit ABCD un quadrilatere onvexe et P un point dans l'interieur de AB BC CD DA ABCD tel que P A = √ , P B = √ , P C = √ , et P D = √ . Con r2 2 2 2 mer ou in rmer que ABCD est un arre.

234 3442. Propose par Iyoung Mi helle Jung, etudiante, College de Langues Etrang eres Hanyoung, Seoul, Coree du Sud et Sung Soo Kim, Universite Hanyang, Seoul, Coree du Sud. Soit C un one ^ ir ulaire droit et D un disque de rayon xe situe dans la base du one ^ C . Montrer que l'aire A de la partie du one ^ situee dire tement de la position du disque D. au-dessus de D est independante

. Propose par Cao Minh Quang, College Nguyen Binh Khiem, Vinh Long, Vietnam. Soit a, b et c trois nombres reels positifs tels que a+b+c = 3. Montrer que X a2 (b + 1) ≥ 2. 3443

y lique

a + b + ab

. Propose par Cao Minh Quang, College Nguyen Binh Khiem, Vinh Long, Vietnam. Soit a, b et c trois nombres reels positifs tels que a+b+c = 1. Montrer que X ab 1 ≤ . 2 3444

y lique

3a + 2b + 3

12

. Propose par Sefket Arslanagi , Universite de Sarajevo, Sarajevo, Bosnie et Herzegovine, a la memoire de Murray S. Klamkin. Soit a, b et c trois nombres reels non negatifs tels que ab + bc + ac = 1. Montrer que 3445

(a)

X

y li

a

≥

1 + bc

√ 3 3 4

;

(b)

X

y li

√ 3

a2 1+a

≥ √ 3+1

.

. Proposed by Mihaly Ben ze, Brasov, Roumanie. Pour tout entier positif n montrer que

3446

jp k jp k p p n2 − n + 1 + n2 + n + 1 + n2 + n + n2 + 3n + 2 =

jp k jp k 4n2 + 3 + 4n2 + 8n + 3

,

le plus grand entier ne depassant pas x. ou ⌊x⌋ designe

. Propose par Mihaly Ben ze, Brasov, Roumanie. Soit n un nombre entier positif. Montrer que

3447

2 n!(n + 2)!

<

n Y

k=1

r

k+1

k+1 k

!

−1

<

1 (n + 1)(n!)2

.

235 . Propose par Jose Luis Daz-Barrero and Miquel Grau-San hez, Universite Polyte hnique de Catalogne, Bar elone, Espagne. Soit Fn le ne nombre de Fibona

i, -a-d, F0 = 0, F1 = 1 et Fn = Fn−1 + Fn−2 pour n ≥ 2. Montrer qu'on a 3448

2

2

2

a Fn + b Fn+1 + c Fn+2 ≥ 4S

n+2 X k=1

Fk2

−

2 Fn+1

!1/2

pour tout triangle ABC dont a, b, c sont les longueurs respe tives des ot ^ es et S sa surfa e. 3449. Propose par un proposeur anonyme. Soit ABCD un arre de ot ^ e unite, M et N les milieux respe tifs de AB et CD . Y a-t-il un point P sur M N tel que les longueurs AP et P C sont toutes les deux des nombres rationnels ? 3450. Propose par Dragoljub Milosevi , Gornji Milanova , Serbie. Soit ABC un triangle ave r omme rayon de son er le ins rit, ra , rb et rc les rayons respe tifs de ses er les exins rits, et ha , hb , hc ses hauteurs respe tives. Montrer que hb + 2rb hc + 2rc 27 ha + 2ra + + ≥ r + ra r + rb r + rc 4

.

................................................................. . Corre tion. Proposed by Vo Quo Ba Can, Can Tho University of Medi ine and Pharma y, Can Tho, Vietnam. Let a, b, and c be positive real numbers. 3419

(a) Prove that (b) Prove that

P

y li

P

y li

r

√ a2 + 4bc ≥ 2 + 2. b2 + c2

r

a2 + bc 1 ≥ 2+ √ . 3 b2 + c2 2

3

. Corre tion. Proposed by Yakub N. Aliyev, Baku State University, Baku, Azerbaijan. For a positive integer m, let σ be the permutation of {0, 1, 2, . . . , 2m} de ned by σ(2i) = i for ea h i = 0, 1, 2, . . . , m and σ(2i − 1) = m + i for ea h i = 1, 2, . . . , m. Prove that there exists a positive integer k su h that σ k = σ and 1 < k ≤ 2m + 1. 3424

236 . Proposed by D.J. Smeenk, Zaltbommel, the Netherlands. In triangle ABC suppose that the altitude ha = AD equals a = BC . Let the ortho entre be H , M be the midpoint of BC , and E be the midpoint of AD. Prove that HM = HE . When does the onverse hold? 3439

. Proposed by Hidetoshi Fukugawa, Kani, Gifu, Japan. There are N oins on a table all of the same size. These N oins an be arranged in a square and they an also be arranged into an equilateral triangle. Find N . 3440

⋆.

Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania. Let ABCD be a onvex quadrilateral and let P be a point in the interior BC CD DA of ABCD su h that P A = AB √ , P B = √ , P C = √ , and P D = √ . 2 2 2 2 Prove or disprove that ABCD is a square. 3441

. Proposed by Iyoung Mi helle Jung, student, Hanyoung Foreign Language High S hool, Seoul, South Korea and Sung Soo Kim, Hanyang University, Seoul, South Korea. Let C be a right ir ular one and let D be a disk of xed radius lying within the base of the one C . Prove that if A is the area of that part of the one lying dire tly above D, then A is independent of the position of the disk D. 3442

. Proposed by Cao Minh Quang, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam. Let a, b, and c be positive real numbers su h that a + b + c = 3. Prove that X a2 (b + 1) ≥ 2. a + b + ab

3443

y li

. Proposed by Cao Minh Quang, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam. Let a, b, and c be positive real numbers su h that a + b + c = 1. Prove that X 1 ab ≤ . 3a2 + 2b + 3

3444

12

y li

. Proposed by Sefket Arslanagi , University of Sarajevo, Sarajevo, Bosnia and Herzegovina, in memory of Murray S. Klamkin. Let a, b, and c be nonnegative real numbers su h that ab + bc + ac = 1. Prove that 3445

(a)

X

y li

a 1 + bc

≥

√ 3 3 4

;

(b)

X

y li

a2 1+a

√ 3

≥ √ 3+1

.

237 . Proposed by Mihaly Ben ze, Brasov, Romania. For any positive integer n prove that

3446

jp k jp k p p n2 − n + 1 + n2 + n + 1 + n2 + n + n2 + 3n + 2 =

jp k jp k 4n2 + 3 + 4n2 + 8n + 3

,

where ⌊x⌋ denotes the greatest integer not ex eeding x.

. Proposed by Mihaly Ben ze, Brasov, Romania. Let n be a positive integer. Prove that

3447

2 n!(n + 2)!

<

n Y

r

k+1

k=1

k+1 k

!

−1

<

1 (n + 1)(n!)2

.

. Proposed by Jose Luis Daz-Barrero and Miquel Grau-San hez, Universitat Polite ni a de Catalunya, Bar elona, Spain. Let Fn be the nth Fibona

i number, that is, F0 = 0, F1 = 1, and Fn = Fn−1 + Fn−2 for n ≥ 2. Prove that 3448

2

2

2

a Fn + b Fn+1 + c Fn+2 ≥ 4S

n+2 X

Fk2

k=1

−

2 Fn+1

!1/2

holds for any triangle ABC , where a, b, c, and S are the side lengths and area of the triangle, respe tively. . Proposed by an anonymous proposer. Let ABCD be a unit square, M the midpoint of AB , and N the midpoint of CD. Is there a point P on M N su h that the lengths of AP and P C are both rational numbers?

3449

. Proposed by Dragoljub Milosevi , Gornji Milanova , Serbia. Let △ABC have inradius r, exradii ra , rb , rc , and altitudes ha , hb , hc . Prove that ha + 2ra hb + 2rb hc + 2rc 27 + + ≥ . 3450

r + ra

r + rb

r + rc

4

238

SOLUTIONS Au un probleme n'est immuable. L'editeur est toujours heureux d'envisager la publi ation de nouvelles solutions ou de nouvelles perspe tives portant sur des problemes anterieurs. . [2008 : 171, 174℄ Proposed by Arkady Alt, San Jose, CA, USA. Let n be a natural number, let r be a real number, and let a1 , a2 , . . . , an n Q be positive real numbers satisfying ak = r n ; prove that 3330

k=1

n X

k=1

1

(1 + ak

)3

≥

n

(1 + r)3

,

(a) for n = 2 if and only if r ≥ 13 ;

(b) for n = 3 if r ≥

1 √ 3 4

;

( ) for n = 4 if r ≥

1 √ 3 4

;

(d) for n ≥ 5 if and only if r ≥

√ 3 n − 1.

Solution to parts (a)-( ) by Oliver Geupel, Bruhl, NRW, Germany, solution to part (d) by the proposer. (a) The statement is not orre t in the stri t sense, be ause for ea h r > 0 the inequality is satis ed by a1 = a2 = r (and similarly for part (d)). We prove instead that for r > 0, the inequality 1 1 2 + ≥ 3 3 (1 + a) (1 + b) (1 + r)3

,

(1)

holds for all positive real numbers a and b satisfying ab = r2 if and only if 1 r≥ . 3 If ab ≥ 91 , then the inequality (1) follows from the result given in CRUX with Mayhem, problem 3319 (solution at [2009 : 121-122℄). Conversely, suppose that r < 13 . Let f : [0, ∞) → R be given by f (x) =

1 (1 + x)3

We have f ′′ (r) =

6(3r − 1) <0 r(1 + r)5

exists an

su h that

x0 > 0

+

x3 x + r2

3

.

and f ′′ is ontinuous. It follows that there

f (x0 ) < f (r) =

2 . (1 + r)3

We on lude that

239 2

and b = ra violate the inequality (1). This ompletes the proof of part (a). Equality holds if and only if a = b = r. (b) We prove the result under the less restri tive

ondition r ≥ 0.47. With√ out loss of generality, let a3 ≤ r and put x = a1 a2 . Then x ≥ r, and by part (a) we have a = x0

1 1 2 + ≥ 3 3 (1 + a1 ) (1 + a2 ) (1 + x)3

.

It therefore suÆ es to show that 2 (1 +

x)3

+

x6 x2 + r 3

3 ≥

3 (1 + r)3

.

Clearing denominators and rearranging terms in this last inequality, we nd that it is equivalent to (x − r)2

where p0 (r) p1 (r) p2 (r) p3 (r) p4 (r) p5 (r) p6 (r) p7 (r)

7 X

k=0

pk (r)xk ≥ 0 ,

= r 7 2r 3 + 6r 2 + 6r − 1

= r 6 4r 3 + 12r 2 + 3r − 2

= r 4 6r 4 + 15r 3 + 18r 2 + 15r − 3 = r 3 5r 4 + 18r 3 + 33r 2 + 5r − 6

= r 4r 5 + 21r 4 + 27r 3 + 13r 2 + 9r − 3 = r 3r 4 + 15r 3 + 21r 2 + 21r − 3 − 6 = 2r 4 + 9r 3 + 15r 2 + 5r − 6 = r 3 + 3r 2 + 3r − 2

It suÆ es to prove that pk (r) > 0 for r ≥ 0.47 and 0 ≤ k ≤ 7. Using a al ulator, we verify that pk (0.47) > 0 for ea h k. Moreover, the polynomials pk (r) are in reasing fun tions for real arguments r ≥ 0.47. This

ompletes the proof. Equality holds if and only if a1 = a2 = a3 = r. ( ) We prove the result under the weaker

ondition r ≥ 0.59. Without loss √ of generality, let a4 ≤ r and put x = a1 a2 a3 . Then x ≥ r, and by part (b) we have 3

1 (1 + a1 )3

+

1 (1 + a2 )3

+

1 (1 + a3 )3

≥

3 (1 + x)3

It therefore suÆ es to show that 3 x9 4 + 3 ≥ 3 (1 + x) (1 + r)3 x3 + r 4

.

.

240 Clearing denominators and rearranging terms in this last inequality, we nd that it is equivalent to (x − r)2

where q0 (r)

=

q1 (r)

=

q2 (r)

=

q3 (r)

=

q4 (r)

=

q5 (r)

=

q6 (r)

=

q7 (r)

=

q8 (r)

=

q9 (r) q10 (r)

= =

10 X

k=0

qk (r)xk ≥ 0 ,

r 10 3r 3 + 9r 2 + 9r − 1 r 9 6r 3 + 18r 2 + 6r − 2 r 8 9r 3 + 15r 2 + 3r − 3

r 6 8r 4 + 21r 3 + 27r 2 + 23r − 3 r 5 7r 4 + 27r 3 + 51r 2 + 13r − 6 r 4 6r 4 + 33r 3 + 39r 2 + 3r − 9

r 2 5r 5 + 27r 4 + 36r 3 + 20r 2 + 15r − 3 r 4r 5 + 21r 4 + 33r 3 + 37r 2 + 3r − 6 r 3r 4 + 15r 3 + 30r 2 + 18r − 9 − 9

2r 4 + 9r 3 + 15r 2 + 3r − 9 r 3 + 3r 2 + 3r − 3

It suÆ es to prove that qk (r) > 0 for r ≥ 0.59 and 0 ≤ k ≤ 10. Using a al ulator, we verify that qk (0.59) > 0 for ea h k. Moreover, the polynomials qk (r) are in reasing fun tions for real arguments r ≥ 0.59. This

ompletes the proof. Equality holds if and only if a1 = a2 = a3 = a4 = r. (d) Suppose that r is su h that the inequality holds for all a1 , a2 , . . . , an subje t to the given onstraint. Let x bena positive real number, let ai = x for i = 1, 2, . . . , n − 1, and let an = xrn−1 . Then n−1

(1 + x)3

+

x3n−3 xn−1 + r n

3 ≥

n (1 + r)3

holds for all x > 0. Taking the limit as x → ∞ yields (1 +n r)3 ≤ 1, hen e √ r ≥ n − 1. √ Conversely, suppose that r ≥ − 1. Weowill prove by Mathemati al n n√ 1 √ Indu tion that if n ≥ 4, r ≥ max 4 , n − 1 , and a1 , a2 , . . . , an satisfy 3

3

3

3

n P

1 n the given onstraint, then ≥ . 3 (1 + a ) (1 + r)3 k k=1 Note that the statement is true for n = 4 by part ( ). Now nsuppose that theostatement is true for some n ≥ 4 and that √ √ r ≥ max √14 , n + 1 − 1 = n + 1 − 1, and let a1 , a2 , . . . , an+1 be positive real numbers su h that a1 a2 · · · an+1 = rn+1 . By symmetry, 3

3

3

241 √ n

we may assume that a1 ≥ a2 ≥ · · · ≥ an+1 . Let x = a1 a2 · · · an , then √ r n+1 x ≥ an+1 = and xn+1 ≥ rn+1 , so that x ≥ r ≥ n + 1 − 1 > √14 . xn 3

3

By indu tion, we have n X

k=1

Let

n P

k=1

1 n , ≥ (1 + ak )3 (1 + x)3

hen e

n x3n 1 ≥ + 3 3 (1 + ak ) (1 + x)3 xn + r n+1

h(x) be the fun tion of x on the right side of the x ≥ r . After some (tedious) al ulations we nd that 3n xn+1 − r n+1 P (x) ′ h (x) = 4 ; (1 + x)4 xn + r n+1 P (x)

.

above inequality for

= 6x2n r n+1 + 4x2n+1 r n+1 + 4xn r 2n+2 + x2n+2 r n+1 + xn+1 r 2n+2 + r 3n+3 − x3n−1 .

Now P (r) = r3n−1 (r + 1)3 (3r − 1) > 0, sin e r > √14 > 13 , and by degree

onsiderations P (x) → −∞ as x → ∞, hen e P (x) has exa tly one root x0 ∈ [0, ∞). [Ed.: note that for positive x and positive C1 , C2 , . . . , Cn , P (x) Cn−1 C n the fun tion C + n−1 + · · · + 1 + C0 is de reasing, and 3n−1 is of this xn x x x form.℄ So, P (x) > 0 for x ∈ [r, x0 ) and P (x) < 0 for x ∈ (x0 , ∞). Hen e, h(x) is in reasing on [r, x0 ) and de reasing on (x0 , ∞). Thus, 3

min h(x) = h(r) =

x∈[r,x0 ]

n (1 +

r)3

+

and for any x ∈ [x0 , ∞) we have h(x) >

lim h(x) = 1 ≥

x→∞

r 3n r n + r n+1

3 =

n+1 (1 + r)3

n+1 = h(r) . (1 + r)3

Therefore, the minimum value of h(x) on [r, ∞) is h(r)

ompletes the indu tion step and the proof.

=

n+1 , (1 + r)3

whi h

Also solved by the proposer (parts (a)-( )). There was one in omplete solution submitted. The proposer leaves Crux readers with the problem of determining the minimum values of r for whi h parts (b) and ( ) hold.

. [2008 : 239, 242℄ Proposed by Toshio Seimiya, Kawasaki, Japan. A onvex y li quadrilateral ABCD has an in ir le with entre I . Let be the interse tion of AC and BD. Prove that AP : CP = AI 2 : CI 2 .

3338

P

242 Solution by Mi hel Bataille, Rouen, Fran e. It has been shown (Crux problem 2027, [1995 : 90; 1996 : 94, 95℄) that AP CP

=

1 − cos ∠BCD 1 − cos ∠BAD

.

Set θ = ∠BAD; then ∠BCD = π − θ, be ause the quadrilateral ABCD is

y li . Thus AP 1 + cos θ θ = = cot2 . (1) CP

1 − cos θ

2

On the other hand, if r denotes the radius of the in ir le, then AI =

so that

r sin

θ 2

and

CI =

r sin

90◦

−

cos2 θ2 AI 2 2 θ = 2 θ = cot 2 CI 2 sin 2

.

θ 2

=

r cos

θ 2

,

(2)

The desired result follows from (1) and (2). University of Sarajevo, Sarajevo, Bosnia and Also solved by SEFKET ARSLANAGIC, Herzegovina; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; OLIVER GEUPEL, Bruhl, NRW, Germany; JOHN G. HEUVER, Grande Prairie, AB; TAICHI student, Sarajevo College, MAEKAWA, Takatsuki City, Osaka, Japan; SALEM MALIKIC, Sarajevo, Bosnia and Herzegovina; MADHAV R. MODAK, formerly of Sir Parashurambhau College, Pune, India; SON HONG TA, Hanoi, Vietnam; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer.

. [2008 : 239, 242℄ Proposed by Toshio Seimiya, Kawasaki, Japan. Let Γ1 and Γ2 be two noninterse ting ir les ea h lying in the exterior of the other. Let ℓ1 and ℓ2 be the ommon external tangents to Γ1 and Γ2 . Let ℓ1 meet Γ1 and Γ2 at A and B , respe tively, and let ℓ2 meet Γ1 and Γ2 at C and D, respe tively. Let M and N be the midpoints of AB and CD, respe tively, and let P and Q be the interse tions of N A and N B with Γ1 and Γ2 , respe tively, dierent from A and B . Prove that CP , DQ, and M N are on urrent. 3339

I. Composite of similar solutions by George Apostolopoulos, Messolonghi, Gree e and by John G. Heuver, Grande Prairie, AB. We shall see that the result holds, more generally, for any two ir les that have ommon external tangents, whether or not they interse t. Let C ′ and D′ be the points where CP and DQ meet M N . We are to prove that C ′ = D ′ . Be ause of the symmetry about the line of entres of the two

ir les, AM N C and BM N D are isos eles trapezoids; thus, AC||M N and

243 and ∠AM N = ∠M N C = ∠C ′ N C . Moreover, be ause N C is tangent to Γ1 while CP is a hord, we have M N ||BD

∠C ′ CN = ∠P CN = ∠CAP = ∠CAN = ∠AN M

when e △CN C ′ ∼ △N M A. Consequently, N C′ =

CN N C′ = , NM MA

CN · M A NM

,

or

.

By the analogous argument the triangles DN D′ and N M B are similar, so that DN · M B N D′ = . NM

Be ause CN = DN and M A = M B , it follows that N C ′ = N D′ ; furthermore, sin e C ′ and D′ lie on the same side of N on the line M N , the points C ′ and D ′ oin ide, as desired. II. Solution by Daniel Reisz, Auxerre, Fran e. La droite M N est l'axe radi al|lieu des points ayant m^eme puissan e| des deux er les Γ1 et Γ2 . Il suÆt don de montrer que le point K interse tion des droites CP et DQ a m^eme puissan e par rapport aux deux

er les pour pouvoir on lure que K se situe aussi sur la droite M N . De NP NB N P · N A = N Q · N B , 'est a dire de = , on deduit la similitude NQ NA des triangles N P Q et N BA, et don l'egalit e d'angles ∠N P Q = ∠N BA = ∠BDQ = ∠BDK .

Montrons maintenant que les triangles KP Q et KDC sont semblables. Ils ont dej a un angle ommun en K . Par ailleurs : (i) ∠BDC et ∠DCA sont supplementaires (a ause de AC||BD), (ii) ∠DCA et ∠CP A sont supple mentaires (CN est la tangente a Γ1 en C , CA est une orde) et (iii) ∠CP A = ∠KP N . Don , ∠BDC = ∠KP N . Or on sait dej a que ∠N P Q = ∠N BA = ∠BDQ, don par soustra tion, ∠KP Q = ∠KDC d'ou la similitude de nos deux KP KD triangles, e qui permet d'e rire = ; 'est a dire KQ KC KP · KC = KD · KQ .

Le point K a don m^eme puissan e par rapport aux deux er les et se trouve don sur la droite M N . Also solved by MICHEL BATAILLE, Rouen, Fran e; FRANCISCO BELLOT ROSADO, student, Sarajevo College, Sarajevo, I.B. Emilio Ferrari, Valladolid, Spain; SALEM MALIKIC, Bosnia and Herzegovina; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer.

244 . [2008 : 239, 242℄ Proposed by Toshio Seimiya, Kawasaki, Japan. The bise tor of ∠BAC interse ts the ir um ir le of △ABC at a se ond point D. Suppose that AB 2 + AC 2 = 2AD2 . Prove that the angle of interse tion of AD and BC is 45◦ . 3340

Similar solutions by D.J. Smeenk, Zaltbommel, the Netherlands and Peter Y. Woo, Biola University, La Mirada, CA, USA. Let α = ∠CAB , β = ∠ABC , γ = ∠BCA, and let R be the

ir umradius. Using the Law of Sines for triangles ABC and ABD , we have

=

AC AB = sin γ sin β AD = 2R , +β sin α 2

so that we an rewrite the ondition AB 2 +AC 2 = 2AD2 as sin2 γ+sin2 β = 2 sin2

α +β 2

This yields su

essively

...... ............................. ........................................ ........... ............. .......... ......... ......... ........ . . . . . . . ....... ....... ...... ....... ..... . . . . ..... ... . . . .... .. . . .... . .. .... . . . .... ... . . ... .. . ... . ... ... . ... .. . ... .. . .. .. .. . . .. ... .. .. ... .. ... ... .. .... .. .. .. ... ... ... ... . ... ....... . ... . . . . . . . . . . . . .. ....... ........... . . . . . . . . . . . . . .. .. .... ........ . . . . . . . .. . . . . . . . . . . .... ... .... .. .. ............ α .. .... ... .... ............ .. .... . ............ .. ............ 2......... α ..... ..... .. ............ . . . . . . . .. . . . . . .. .. ... ........ ... 2 . . . . . . . . . . . . . . . . . . ..... ... .. .. .... ............ ... .. . .... ............ .. ... .... ... ............ .... .. .. ............ ... .... ........ ............ . . . . .... . . . . . . . . . . . . . ............................................................................................................................................................................................................................ ............... .... .... α .... .......... .... ... .... .......... .... .... .... ......... 2 .... .... . ...... .... .... . . . ...... ................. . . . . . .. . ........ ..... .... ......... ................. ........ .... .. .......... ......... .... ............ ................. ......... ................... ......... ........ ............................. ..................................

.

cos 2β + cos 2γ cos 2β + cos 2γ cos(β + γ) cos(β − γ) cos(β − γ) cos(β + γ) + 1

γ

β

B

A

E

C

D = = = =

2 cos(α + 2β) , −2 cos(β − γ) , − cos(β − γ) , 0.

Sin e cos(β + γ) 6= −1, it follows that cos(β − γ) = 0. We an assume, without loss of generality, that β ≤ γ . Then β − γ = − π2 . It follows that π α π α + 2β = , sin e α + β + γ = π , and therefore, ∠AEC = + β = , as 2 2 4

laimed. Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallor a, Spain; GEORGE University of Sarajevo, APOSTOLOPOULOS, Messolonghi, Gree e; SEFKET ARSLANAGIC, Sarajevo, Bosnia and Herzegovina; ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, Fran e; IAN JUNE L. GARCES, Ateneo de Manila University, Quezon City, The Philippines; OLIVER GEUPEL, Bruhl, NRW, Germany; KEE-WAI LAU, Hong Kong, China; TAICHI MAEKAWA, Takatsuki City, Osaka, Japan; SALEM MALIKIC, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; MADHAV R. MODAK, formerly of Sir Parashurambhau College, Pune, India; MICHAEL PARMENTER, Memorial University of Newfoundland, St. John's, NL; BOB SERKEY, Leonia, NJ, USA; SON HONG TA, Hanoi, Vietnam; TITU ZVONARU, Comane sti, Romania; and the proposer. There were two in orre t solutions submitted.

245 . [2008 : 240, 242℄ Proposed by Arkady Alt, San Jose, CA, USA. For any triangle ABC with sides of lengths a, b, and c, prove that √ 3(Ra + Rb + Rc ) ≤ a + b + c, where Ra , Rb , and Rc are the distan es from the in entre of △ABC to the verti es A, B , and C , respe tively. 3341

Solution by George Apostolopoulos, Messolonghi, Gree e. Let s be the semiperimeter of triangle ABC . We have Ra =

s−a

cos

A 2

Similarly, Rb

s−a = q

s(s−a) bc

r √ b = ca 1 − s

√ √ r √ a bc s − a = bc 1 − = √ s s

and

Rc

√ = ab

r c 1− s

.

.

Using the Cau hy{S hwarz Inequality, we obtain 2

(Ra + Rb + Rc )

= ≤ =

or

!2 r r r √ √ a b c 1 − + ca 1 − + ab 1 − s s s a b c (bc + ca + ab) 1 − + 1 − + 1 − s s s 2s (ab + bc + ca) 3 − = ab + bc + ca , s √ bc

p √ 3(Ra + Rb + Rc ) ≤ 3(ab + bc + ca) .

It suÆ es to show that

p 3(ab + bc + ca) ≤ a + b + c .

The last inequality is equivalent to a2 + b2 + c2 ≥ ab + bc + ca, whi h is well known and easy to prove. This ompletes the solution. University of Sarajevo, Sarajevo, Bosnia and Also solved by SEFKET ARSLANAGIC, Herzegovina; ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, Fran e; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Bruhl, NRW, Germany; JOE HOWARD, Portales, NM, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; THANOS MAGKOS, 3rd High S hool of Kozani, Kozani, Gree e; student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; MADHAV SALEM MALIKIC, R. MODAK, formerly of Sir Parashurambhau College, Pune, India; PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU, Comane sti, Romania; and the proposer.

246 . [2008 : 240, 242℄ Proposed by Arkady Alt, San Jose, CA, USA. Let r and R be the inradius and ir umradius of △ABC , respe tively. Prove that X A B r 2 sin . sin ≤ 1+

3342

2

y li

2

R

Similar solutions by Cao Minh Quang, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; Oliver Geupel, Bruhl, NRW, Germany; and Dung Nguyen Manh, High S hool of HUS, Hanoi, Vietnam. r It is well known that cos A + cos B + cos C = 1 + R . Thus, it suÆ es to show that X X A B 2 sin sin ≤ cos A . 2

y li

2

y li

For positive x we have 2 ≤ x + x1 . Taking x = sin

A B sin , 2 2

cos(A/2) cos(B/2)

and multiplying by

we obtain

A 1 B 2 sin sin ≤ 2 2 2

hen e, 2

X

sin

y li

B A sin A tan + sin B tan 2 2

,

A A 1 X B sin tan (sin B + sin C) . ≤ 2 2 2 y li 2

By sum to produ t formulas, we have

1 X A tan (sin B + sin C) 2 y li 2 =

X sin

y li

=

X

y li

as desired.

cos

cos

A 2 A 2

sin

B+C 2

cos

B−C 2

X B+C B−C cos = cos A , 2 2

y li

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; SEFKET University of Sarajevo, Sarajevo, Bosnia and Herzegovina; MICHEL BATAILLE, ARSLANAGIC, Rouen, Fran e; SCOTT BROWN, Auburn University, Montgomery, AL, USA; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; JOE HOWARD, Portales, NM, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; KEE-WAI LAU, Hong Kong, China; THANOS MAGKOS, 3rd High S hool of Kozani, Kozani, Gree e; SALEM MALIKIC, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; JUAN-BOSCO ROMERO MARQUEZ, Universidad de Valladolid, Valladolid, Spain; XAVIER ROS, student, Universitat Polite ni a de Catalunya, Bar elona, Spain; and the proposer. There was one in omplete solution submitted.

247 3343. [2008 : 240, 242℄ Proposed by Stan Wagon, Ma alester College, St. Paul, MN, USA. If the fa torials are deleted in the Ma laurin series for sin x, one obtains the series for arctan x. Suppose instead that one alternates fa torials in the series. Does the resulting series have a losed form? That is, an one nd an elementary expression for the fun tion whose Ma laurin series is

x−

x5 x7 x9 x11 x3 + − + − + ··· 3 5! 7 9! 11

?

Essentially the same solution by Mi hel Bataille, Rouen, Fran e; Ri hard I. Hess, Ran ho Palos Verdes, CA, USA; Va lav Kone ny, Big Rapids, MI, USA; Ralph Lozano, student, Missouri State University, Missouri, USA; Xavier Ros, student, Universitat Polite ni a de Catalunya, Bar elona, Spain; Digby Smith, Mount Royal College, Calgary, AB; and Peter Y. Woo, Biola University, La Mirada, CA, USA. Let f (x) denote the given series. The following Ma laurin series are all well known: sin x = sinh x = tan−1 x = tanh−1 x =

x3 x5 + 3! 5! x3 x5 x+ + 3! 5! x5 x3 + x− 3 5 x3 x5 x+ + 3 5 x−

x7 + ··· 7! x7 + + ··· 7! x7 − + ··· 7 x7 + + ··· 7 −

,

− ∞ < x < ∞;

(1)

,

− ∞ < x < ∞;

(2)

,

− 1 < x < 1;

(3)

,

− 1 < x < 1.

(4)

From (1) and (2) we have 1 2

(sin x + sinh x) = x +

x5 5!

+

x9 9!

+ ···

.

(5)

From (3) and (4) we have 3 1 x x7 x11 −1 −1 tan x − tanh x = − + + + ··· 2 3 7 11

.

(6)

From (5) and (6) we on lude that

1 sin x + sinh x + tan−1 x − tanh−1 x , − 1 < x < 1. 2 Also solved by CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; OLIVER GEUPEL, Bruhl, NRW, Germany; DOUGLASS L. GRANT, Cape Breton University, Sydney, NS; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; JOHN KLASSEN, Champlain Regional College { St. Lawren e, Ste. Foy, QC; KATHLEEN E. LEWIS, SUNY Oswego, Oswego, NY, USA; and the proposer. With the ex eption of the proposer,all of these solvers gave the same answer; namely 1+x 1 sin x + sinh x + tan−1 x − 1 ln( 1−x ) whi h is, of ourse, equivalent to the answer given 2 2 above. f (x) =

248 . [2008 : 240, 242℄ Proposed by Hung Pham Kim, student, Stanford University, Palo Alto, CA, USA. Let n be a positive integer, n ≥ 4, and let a1 , a2 , . . . , an be positive real numbers su h that a1 + a2 + · · · + an = n. Prove that 3344

1 1 1 3 2 + + ··· + −n ≥ a1 + a22 + · · · + a2n − n . a1 a2 an n

Solution by Oliver Geupel, Bruhl, NRW, Germany. We will use the following theorem.

(Right Convex Fun tion Theorem) Let f (u) be a fun tion on an interval I ⊂ R, whi h is onvex for u ≥ s, s ∈ I . If f (x)+(n−1)f (y) ≥ nf (s) for all x, y ∈ I su h that x ≤ s ≤ y and x + (n − 1)y = ns, then

Theorem

f (x1 ) + f (x2 ) + · · · + f (xn ) ≥ nf

for all x1 , x2 , . . . , xn ∈ I su h that

x1 + x2 + · · · + xn n

x1 + x2 + · · · + xn ≥ s. n

The desired inequality may be rewritten as a1 + a2 + · · · + an f (−a1 ) + f (−a2 ) + · · · + f (−an ) ≥ nf − n

,

where f (u) = −3u2 − nu for −n < u < 0. Sin e f ′′ (u) = − 2n − 6 > 0 for u3 u ≥ −1, the fun tion f is onvex for u ≥ s = −1. It suÆ es to prove that f (x) + (n − 1)f (y) ≥ nf (−1) (1) whenever −n < x ≤ −1 ≤ y < 0 and x + (n − 1)y = −n. Substituting x = −n − (n − 1)y in (1), we obtain the inequality −3 n + (n − 1)y

2

+

n n + (n − 1) −3y 2 − ≥ n(n − 3) n + (n − 1)y y

for −1 ≤ y < 0. After learing denominators and rearranging terms, it be omes n(n − 1)(y + 1)2 (3(n − 1)y2 + 3ny + n ≥ 0, or equivalently 2

n(n − 1)(y + 1)

"

n y+ 2(n − 1)

2

n(n − 4) + 12(n − 1)2

whi h is learly true. Equality holds if and only if a1 = a2 = · · · = an = 1.

#

≥ 0,

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; WALTHER JANOUS, Ursulinengymnas student, Sarajevo College, Sarajevo, Bosnia and ium, Innsbru k, Austria; SALEM MALIKIC, Herzegovina; DUNG NGUYEN MANH, High S hool of HUS, Hanoi, Vietnam; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer For further appli ations of the Right Convex Fun tion Theorem, the interested reader is referred to the arti le "The Proof of Three Open Inequalities" by V. C^rtoaje, [2008 : 231-238℄.

249 . [2008 : 240, 243℄ Proposed by Hung Pham Kim, student, Stanford University, Palo Alto, CA, USA. Let a, b, c, and d be positive real numbers su h that a + b + c + d = 4. Prove that 3345

a 1+

b2 c

+

b 1+

c2 d

+

c 1+

d2 a

+

d 1 + a2 b

≥ 2.

Solution by Dung Nguyen Manh, High S hool of HUS, Hanoi, Vietnam. By the AM{GM Inequality, we have a 1 + b2 c

√ ab2 c ab2 c ab c ≥ a − = a − √ 1 + b2 c 2b c 2 √ b a · ac b(a + ac) = a− ≥ a− . 2 4 = a−

Taking the y li sum a ross the inequality derived above and using the relation a + b + c + d = 4, we obtain X

y li

a 1 X 1 X ≥ 4 − ab − abc . 1 + b2 c 4 y li 4 y li

(1)

Applying the AM{GM Inequality again, we have ab + bc + cd + da = (a + c)(b + d) ≤

(a + b + c + d)2 = 4 4

(2)

and abc + bcd + cda + dab = ab(c + d) + cd(a + b) (a + b)2 (c + d) (c + d)2 (a + b) + 4 4 (a + b)(c + d) = (a + b + c + d) 4 (a + b + c + d)2 = (a + b)(c + d) ≤ = 4. 4 ≤

(3)

From inequalities (1), (2), and (3) we then have X

y li

a 1 ≥ 4 − (4 + 4) = 2 . 2 1+b c 4

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; SEFKET University of Sarajevo, Sarajevo, Bosnia and Herzegovina; MICHEL BATAILLE, ARSLANAGIC, Rouen, Fran e; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; student, Sarajevo College, Sarajevo, JOE HOWARD, Portales, NM, USA; SALEM MALIKIC, Bosnia and Herzegovina; TRAN THANH NAM, Tomsk Polyte hni University, Tomsk, Russia; and the proposer. A solution was re eived that appears to be orre t, but due to its length and

omplexity ould not be veri ed in the available time.

250 . [2008 : 240, 243℄ Proposed by Bin Zhao, student, YunYuan HuaZhong University of Te hnology and S ien e, Wuhan, Hubei, China. Given triangle ABC , prove that 3346

   X 1 X X A A π ≥  sin   csc  2 2 A

y li

y li

y li

.

Solution by Mi hel Bataille, Rouen, Fran e.

Note that π = A + B + C and set x = A ,y = 2 x, y , z ∈ 0, π2 ), the inequality is then rewritten as (x+y+z)

or

1

x

+

1

y

+

1

z

≥ (sin x+sin y+sin z)

B , 2

and z =

1 sin x

+

C 2

1 sin y

(so that

+

1 sin z

x sin x y sin y − + − y sin y x sin x y sin y z sin z z sin z x sin x + − + − + − + − ≥ 0, z sin z y sin y x sin x z sin z

that is,

f (x) − f (y) g(y) − g(x) +

sin x sin y f (y) − f (z) g(z) − g(y) sin y sin z

+

f (z) − f (x) g(x) − g(z) sin z sin x

≥ 0,

where the fun tions f and g are de ned by f (u) = sinu u and g(u) = u sin u. Now, f is de reasing on 0, π2 and g is in reasing on 0, π2 . Thus, π f (u) − f (v) g(v) − g(u) ≥ 0 whenever u, v ∈ 0, and so the left side 2 of (1) is the sum of three nonnegative terms. The inequality (1) follows.

Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer.

. [2008 : 241, 243℄ Proposed by Mihaly Ben ze, Brasov, Romania. Let A1 A2 A3 A4 be a onvex quadrilateral. Let Bi be a point on Ai Ai+1 for i ∈ {1, 2, 3, 4}, where the subs ripts are taken modulo 4, su h that 3347

B1 A1 B3 A4 A1 A4 = = B1 A2 B3 A3 A2 A3

and

B2 A2 B4 A1 A1 A2 = = B2 A3 B4 A4 A3 A4

.

Prove that B1 B3 ⊥ B2 B4 if and only if A1 A2 A3 A4 is a y li quadrilateral.

251 Solution by Oliver Geupel, Bruhl, NRW, Germany. If A1 A2 6 k A3 A4 , then let C = A1 A2 ∩ A3 A4 and v be the bise tor of ∠A1 CA4 ; otherwise, if A1 A2 k A3 A4 de ne v to be the midline between the two parallel lines. Similarly, if A1 A4 6 k A2 A3 , then let D = A2 A3 ∩ A1 A4 and w be the bise tor of ∠A3 DA4 ; otherwise, if A1 A4 k A2 A3 de ne w to

be their midline. The desired result follows readily from the following two lemmas: . The quadrilateral A1 A2 A3 A4 is y li if and only if v ⊥ w.

Lemma 1

. It is the ase that B2 B4 k v and B1 B3 k w.

Lemma 2

Proof of Lemma 1. Assume that the gure has been labeled so that C and A2 A3 are on opposite sides of line A1 A4 while D and A1 A2 are on opposite C

C

.... ....... ...... .. .. .. .. ..... .... . .. .. .. .. .. .. .. .. .. .. ... .. .. .... .... ... . .. . ... .... .. .. ... .. .. .. ... .. .. ... .. ... .. .... . .. . 1 .. ... .. . ... ..................... ..... . .. ................. . . . 4 . ... ................ ... .................. ..... ... .. ...................... . ... . ................ .. .. .... ................ . . .. . ................ .. ... .. ................ .. . .. .. ................................................................................................................................................................................................................................................ .. ... .. . ................. .. .. ... ................. . . . . . .. . . . . . . . . . . . . . . . .. .. . . ................. .... .. .................. . . . . . . .. . .. . . . . . . . . . . . . .. ................... .. .................. ... ................. ..... 3 .. ..................... ... ................. ...

... ..... .. .. .. .. .. .... . ... .. .. .. .. .. .. .. .. .. ... .. . .. . .. . ... ... . .. . . .. ... .. . ... . 1 .. .. . .. ....................... .. . ................ .. 4 . . ................ .... ................. .. . .. ....................... .. . ................ .. 1 .... .............. .. ..................................................................................................................................................................................... ......................................................................3 . ..... .. ..... ........................................................................................................................................................................ .. .. ................. .. ... ................. . .. . . . . . . . .. . . . . . . . . . . .. . . ................... .. .................. .. ................. .. ................. .. ................. . . . . . . .. . . . . . . . . . . . . 3 .. ..................... .................

◦◦

A

A

F

w

E

A

A

G

B

D

qq

w

v

I

q

q qq

G

D

A

A

A2

H

B

A2

sides of A3 A4 ; denote the angles at the verti es Ai by αi , and de ne E = v ∩ w , F = v ∩ A1 A4 ,

and

G = w ∩ A3 A4 .

α4 We will rst prove that ∠(v, A1 A4 ) = 90◦ + α1 − . If A1 A2 k A3 A4 then 2 α − α4 α4 = 180◦ − α1 ; therefore ∠EF A4 = α1 = 90◦ + 1 , as laimed. If 2 ◦ A1 A2 6 k A3 A4 , note that ∠CA1 A4 = 180 − α1 , ∠CA4 F = 180◦ − α4 , and α + α4 − 90◦ . Hen e, ∠A1 CE = ∠A4 CF = 1 2

∠A4 F E

=

∠A4 CF + ∠CA4 F =

=

90◦ +

α1 − α4 2

,

whi h is the desired angle between α − α4 ∠A4 GE = 90◦ + 3 . 2

v

α1 + α4

and

2

A1 A4 .

− 90◦

+ (180◦ − α4 )

Similarly, we have that

252 Inspe ting the angles in quadrilateral EGA4 F , we see that ∠(v, w) = ∠F EG α1 − α4 α3 − α4 = 360◦ − 90◦ + − α4 − 90◦ + 2 2 α + α 1 3 = 180◦ − . 2

Hen e, ∠(v, w) = 90◦ is equivalent to α1 + α3 = 180◦ whi h, in turn, is equivalent to A1 A2 A3 A4 being y li . Proof of Lemma 2. If A1 A4 k A2 A3 then, be ause parallels ut proportional segments from transversals, B1 B3 k A1 A4 k w. It remains to onsider the

ase A1 A4 6 k A2 A3 . Let H and I be the points where the line B1 B3 meets A1 A4 and A2 A3 , respe tively. We now apply the hypothesis and Menelaus' theorem to triangles A1 A2 D and A3 A4 D and transversal B1 B3 to obtain A1 A4 · IA2 · HD A2 A3 · ID · HA1 = 1 =

B1 A1 · IA2 · HD

=

B1 A2 · ID · HA1

B3 A4 · IA3 · HD B3 A3 · ID · HA4

=

A1 A4 · IA3 · HD A2 A3 · ID · HA4

.

(1)

By omparing the rst and last fra tions in (1), we see that IA3 IA2 IA3 + A3 A2 = = HA4 HA1 HA4 + A4 A1 3 A2 when e, A A A 4

1

=

IA3 . H A4

We substitute this last fra tion into the nal fra tion

appearing in (1) and dedu e that that ∠GDH =

;

∠A2 DH 2

Consequently, w = GD k HI

HD = 1. ID

=

That is, HD

∠DIH + ∠DHI

= B1 B3 .

2

= ID .

It follows

= ∠DHI .

Analogously, v k B2 B4 .

Also solved by MICHEL BATAILLE, Rouen, Fran e; KHANH BAO NGUYEN, High S hool for Gifted Students, Hanoi University of Edu ation, Hanoi, Vietnam; and PETER Y. WOO, Biola University, La Mirada, CA, USA. There was one in omplete submission. Lemma 1 is readily found as a problem in textbooks and on ontests. It has appeared more than on e in CRUX; see [1980 : 226-230℄ for an alternative proof, several referen es, and a related result. Geupel found the \if" part of our problem on the Mathlinks website, http://www.mathlinks.ro/viewtopic.php?t=200396. The proof there is very easy: Under the assumption that the given quadrilateral is y li , the diagonals of the quadrilateral divide its interior into two pairs of similar triangles, and the lines B1 B3 and B2 B4 are the bise tors of the angles formed by those diagonals. It seems as if the onverse is not so easily proved.

253 . [2008 : 241, 243℄ Proposed by Mihaly Ben ze, Brasov, Romania. Let ABC be an a ute-angled triangle. Let A1 , B1 , and C1 be points on the sides BC , CA, and AB , respe tively, su h that the angles ∠AC1 B1 , ∠BC1 A1 , and ∠ACB are all equal. Let M , N , and P be the ir um entres of △AC1 B1 , △BA1 C1 , and △CB1 A1 , respe tively. Prove that AM , BN , and CP are on urrent if and only if AA1 , BB1 , and CC1 are altitudes of △ABC . 3348

Solution by Peter Y. Woo, Biola University, La Mirada, CA, USA, revised by the editor. Denote the angles of △ABC by ∠A, ∠B , ∠C , and let AA′ , BB ′ , and ′ CC be the altitudes. The given angle assumptions imply that the triangles A1 BC1 and AB1 C1 are both similar to △ABC . Be ause the angle at the

ir um entre N that is subtended by the side BC1 of △A1 BC1 is twi e the angle at A1 (whi h equals ∠A), we dedu e that ∠N BC1 = 90◦ − ∠A for any position of C1 between B and A. Moreover, sin e ∠B ′ BA = 90◦ − ∠A, we know that N must lie on the altitude BB ′ . Similarly, M must lie on the altitude CC ′ . Be ause the lines AM and BN therefore meet at the ortho entre H of △ABC , when the three lines AM, BN , and CP are on urrent, their

ommon point must be H . Our problem is thus redu ed to proving that CP passes through H if and only if C1 = C ′ (the foot of the altitude from C to AB ). When C1 = C ′ , it is a standard result that ∠BC ′ A′ = ∠AC ′ B ′ = ∠C , when e A1 = A′ and B1 = B ′ . As a onsequen e, ∠A1 B1 C = ∠B and, as we argued earlier, ∠A1 CP = 90◦ − ∠B so that P is on the altitude CC1 = CH , as desired. We now prove that onversely, when C1 6= C ′ , then H is not on the line CP . When C1 is taken between B and C ′ then (be ause by our angle assumption C1 B1 k C ′ B ′ ) B1 must lie beyond B ′ on the hal ine from A in the dire tion of C . Similarly A1 lies beyond A′ on the hal ine from C in the dire tion of B . It follows that ∠A1 B1 C > ∠A′ B ′ C ; when e ∠A1 CP = 90◦ − ∠A1 B1 C < 90◦ − ∠A′ B ′ C = ∠A′ CH , and we on lude that H is not on the line CP . A similar argument establishes that H is not on the line CP when C1 is taken between C ′ and A. Also solved by CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; student, Sarajevo College, OLIVER GEUPEL, Bruhl, NRW, Germany; SALEM MALIKIC, Sarajevo, Bosnia and Herzegovina; JOEL SCHLOSBERG, Bayside, NY, USA; and the proposer.

. [2008 : 241, 243℄ Proposed by Mihaly Ben ze, Brasov, Romania. Let a, b, and c be positive real numbers. Show that

3349

6

Y a3 + 1

y li

a2 + 1

≥ max

 X 

y li

2

a(1 + bc)(a + 1) a3 + 1

,

 X ab(1 + c)(a2 b2 + 1)  a 3 b3 + 1 

y li

.

254 Solution by Tran Thanh Nam, Tomsk Polyte hni University, Tomsk, Russia. We rst prove a lemma. Lemma

If t is a positive real number, then

(a)

t3 + 1 t2 + 1

p ≥ t2 − t + 1 ≥

s 4

t4 + 1 2

,

(b)

t+1 2

≥

t t2 + 1 t3 + 1

.

Proof: After squaring, olle ting like terms, and fa toring, the inequality √ t3 + 1 ≥ t2 − t + 1 of part (a) redu es to t(t − 1)2 ≥ 0, whi h is true. t2 + 1 √ t2 − t + 1 ≥

r

4

t +1 Similarly, the inequality redu es to (t − 1)4 ≥ 0, 2 whi h is true. Colle ting the fra tions in inequality (b) on one side and fa toring redu es it to (t − 1)2 t2 + t + 1 , whi h is true.

We now pro eed to prove that 6

By part (a) of the Lemma, we have

and

4

Q a3 + 1

y li

a2 + 1

≥

P a(1 + bc)(a2 + 1)

y li

a3 + 1

.

s s b3 + 1 c3 + 1 b4 + 1 c4 + 1 b2 c2 + 1 bc + 1 4 ≥ ≥ ≥ b2 + 1 c2 + 1 4 2 2

Hen e, 2

a3 + 1 a2 + 1

2

Y a3 + 1

y li

a2 + 1

≥

≥

s

a4 + 1 2

≥ a.

a(1 + bc)(a2 + 1) a3 + 1

.

Adding a ross the y li permutations of the last inequality gives the desired result. ` ´ Q a3 + 1 P ab(1 + c) a2 b2 + 1 We next pro eed to prove that 6 ≥ . a2 + 1 a 3 b3 + 1

y li

By parts (a) and (b) of the Lemma, we have

and also

y li

a3 + 1 b3 + 1 ab + 1 ab a2 b2 + 1 ≥ ≥ a2 + 1 b2 + 1 2 a3 b3 + 1

c3 + 1 1+c ≥ c2 + 1 2

2

(as it is equivalent to (c + 1)(c − 1)2 ≥ 0). Hen e,

Y a3 + 1

y li

a2 + 1

≥

ab(1 + c)(a2 b2 + 1) a3 b3 + 1

.

255 Adding a ross the y li permutations of this last inequality gives the se ond desired result. The overall inequality follows from the two results we have obtained. Also solved by Arkady Alt, San Jose, CA, USA; and the proposer.

3350. [2008 : 241, 243℄ Proposed by Panos E. Tsaoussoglou, Athens, Gree e. Let x, y, and z be positive real numbers su h that x + y + z = 1. Prove that yz zx xy 1 + + ≤ .

1+x

1+y

1+z

4

I. Similar solutions by George Apostolopoulos, Messolonghi, Gree e; Cao Minh Quang, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; Khanh Bao Nguyen, High S hool for Gifted Students, Hanoi University of Edu ation, Hanoi, Vietnam; Babis Stergiou, Chalkida, Gree e; Son Hong Ta, Hanoi, Vietnam; and Titu Zvonaru, Comane sti, Romania. For positive real numbers a, b, and c we have that (b + c)2 ≥ 4bc, a 1 1 a hen e b + c ≤ 4 c + b . Thus, X

y li

yz

1+x

=

X

y li

=

yz

(x + y) + (z + x)

≤

X yz

y li

X x X xy + zx 1 = = 4(y + z) 4 4

y li

y li

4

1

x+y

+

1

z+x

.

Equality holds if and only if x = y = z = 13 . II. Solution by Arkady Alt, San Jose, CA, USA, ondensed by the editor. P Let e1 = x + y + z , e2 = xy + yz + zx, e3 = xyz , and S = ′ kzxy +1 P ( ′ denotes a y li sum over x, y, z ). We will prove that if k , x , y , z are o n 1 1 positive real numbers and e1 = 1, then S ≤ max 4 , k + 3 . Let k

and

P′

P 1 P k 1 = e3 ′ − = e2 − ke3 ′ z kz + 1 kz + 1 −1 P′ 9 ≥9 (kz + 1) = , it follows that S ≤ e2 − 9ke3 .

∈ (0, 1].

1 kz + 1

Sin e S

It therefore suÆ es to prove that

k+3 9ke3 1 e2 − ≤ , k+3 k+3

(k + 3)e2 − 9ke3 ≤ 1 .

k+3

whi h is equivalent to (1)

The inequality (1) follows from the two inequalities e2 4e2

≥ 9e3 , ≤ 1 + 9e3 ,

(2) (3)

256 sin e 1 − (k + 3)e2 + 9ke3 = (1 − 4e2 + 9e3p ) + (1 − k)(e2 − 9e3 ) and k ≤ 1. √ Now, (2) follows from 3 e3 ≤ e1 = 1 and 3 e23 ≤ e2 , and these follow from the AM{GM Inequality, while (3) follows by rewriting P the S hur Inequality modulo the relation e1 = 1; that is, one rewrites ′ x(x − y)(x − z) ≥ 0 as 2e31 − 6e1 e2 − e21 + 2e2 + 9e3 ≥ 0 and puts e1 = 1. This ompletes the proof of the inequality for k ∈ (0, 1]. Note that if k = 1, then we obtain the originaln inequalityo to be proved, 1 while if k > 1 then S = S(k) < S(1) ≤ 14 = max 14 , k + . 3 3

3

University of Sarajevo, Sarajevo, Bosnia and Also solved by SEFKET ARSLANAGIC, Herzegovina; MICHEL BATAILLE, Rouen, Fran e; PAUL BRACKEN, University of Texas, Edinburg, TX, USA; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; REBECCA EVERDING and JENNIFER PAJDA, students, Southeast Missouri State University, Cape Girardeau, MO, USA; IAN JUNE L. GARCES, Ateneo de Manila University, Quezon City, The Philippines; OLIVER GEUPEL, Bruhl, NRW, Germany; JOE HOWARD, Portales, NM, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; WEI-DONG, Weihai Vo ational College, Weihai, Shandong Provin e, China; NGUYEN THANH LIEM, Tran Hung Dao High S hool, Phan Thiet, Vietnam; THANOS MAGKOS, 3rd High S hool of Kozani, Kozani, Gree e; DUNG NGUYEN MANH, High S hool of HUS, Hanoi, Vietnam; D.P. MEHENDALE (Dept. of Ele troni s) and M.R. MODAK, (formerly of Dept. Mathemati s), S. P. College, Pune, India; TRAN THANH NAM, Tomsk Polyte hni University, Tomsk, Russia; STAN WAGON, Ma alester College, St. Paul, MN, USA; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. The following solvers submitted multiple solutions: Alt (5 solutions), Apostolopoulos (3 solutions) and Cao (2 solutions). Salem Maliki , student, Sarajevo College, Sarajevo, Bosnia and Herzegovina indi ated that sin e x + y + z = 1, our problem appears as problem 35 (solved on pp. 48-49) in the book Old and New Inequalities by T. Andrees u, V. C^rtoaje, G. Dospines u, and M. Las u; GIL Publishing House.

Crux Mathemati orum

with Mathemati al Mayhem Former Editors / An iens Reda teurs: Bru e L.R. Shawyer, James E. Totten

Crux Mathemati orum

Founding Editors / Reda teurs-fondateurs: Leopold Sauve & Frederi k G.B. Maskell Former Editors / An iens Reda teurs: G.W. Sands, R.E. Woodrow, Bru e L.R. Shawyer

Mathemati al Mayhem Founding Editors / Reda teurs-fondateurs: Patri k Surry & Ravi Vakil Former Editors / An iens Reda teurs: Philip Jong, Je Higham, J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia, Shawn Godin, Je Hooper

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