WholeIssue 35 4 - Canadian Mathematical Society · plus egal a 10. Adrien r emar que que, pour le...

193EDITORIALV�a lav Linek

It has ome to our attention that some problems appearing in CRUXwith MAYHEM have been submitted to other pla es; in parti ular someCRUX with MAYHEM problems have appeared on ertain problem-solvingwebsites. While the trend of online problem solving is well established andquite popular, it is nevertheless still the ase that CRUX with MAYHEMis a traditional print journal. This means that submissions to CRUX withMAYHEM should not be submitted elsewhere and should not have alreadyappeared elsewhere (although some rare ex eptions may be permitted forolder problems that may not be that well known today). This in ludes prob-lem proposals, solutions, and arti les: if you submit them to CRUX withMAYHEM, then we ask that you do not submit them anywhere else!I would like to point out that material that is submitted to CRUX withMAYHEM re eives the attention of the Editorial Board, whereasmany onlinesour es do not provide this kind of attention. For instan e, problems andsolutions 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, wehave a ba klog of problem proposals to pro ess be ause only 100 four-digitproblems are published per year, whereas we re eive onsiderably more than100 problem proposals a year. If the wait is too long, then (if we have notalready used your problem) you an inform us that you would like to submityour problem elsewhere and we will then remove it from the queue. Thesame applies for other materials.If it omes to our attention that some dupli ation has o urred, thenwe 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 L �eo Sauv �e in 1975, and itdid 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 notexist 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 JimTotten spe ial issue slated for September. I thank you, the readers, for yourpatien e with the delays of the last eight months. With time available forpreparation this summer (unlike last summer) ir umstan es are favourablefor autumn to go a ording to s hedule.

194SKOLIAD No. 117

Lily Yen and Mogens HansenPlease 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 solu-tions before the deadline. The de ision of the editors is �nal.Our ontest this month is the Calgary Mathemati al Asso iation JuniorHigh S hool Mathemati s Contest, Part B, 2008. Our thanks go to JoanneCanape of the University of Calgary, Calgary, Alberta, for providing us withthis 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 math �ematique de CalgaryNiveau pr �ese ondaireRonde �nale, partie B, 20081. Ri hard a besoin de se rendre en taxi de sa maison �a un par pas tr �esloin. Deux ompagnies de taxi o�rent leurs servi es. La premi �ere fa ture ses lients �a un taux �xe de 10,00$, auquel s'ajoute un taux variable de 0,50$pour haque kilom�etre du trajet, tandis que la deuxi �eme a un taux �xe de4,00$ et un taux variable de 0,80$ le kilom�etre. Ri hard onstate que le outest le meme, quelle que soit la ompagnie hoisie. Quelle est la distan e enkilom�etres de sa maison au par ?2. Un poste de radio lan e un on ours o �u ha un des gagnants pourra as-sister �a deux mat hs des Canadiens de Montr �eal, 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 di� �erent. La han e a fait que les amis Ali e, Bertrand,Carole, David et Evelyne ont tous �et �e d �e lar �es gagnants au on ours. Mon-trer 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 unmat h ensemble.3. Deux tests sont administr �es �a un groupe d' �etudiants. Chaque �etudiantre� oit un s ore pour ha un des deux tests, e s ore �etant un entier non n �egatifau 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 sons ore ; il en est de meme pour le deuxi �eme test. Apr �es que le professeur aaÆ h �e 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.

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 oremoyenplus �elev �e que elui d'Adrien ? Expliquer lairement pourquoi votre r �eponseest orre te.4. On ommen e par tra er un re tangle detaille 6 m par 8 m. Ensuite, on tra e le er le ir ons rit de e re tangle, puis un arr �e ir ons rit autour du er le. En�n, ontra e le er le ir ons rit du arr �e tout juste onstruit. D �eterminer la surfa e de e der-nier er le en m2. .......................................................................................................

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5. Soit A un entier positif �a deux hi�res d �e imaux, ne ontenant au un z �eroet soit B un entier positif �a trois hi�res d �e imaux. Si A% de B donne 400,d �eterminer toutes les valeurs possibles de A et B.6. D �eterminer un re tangle ayant les deux propri �et �es suivantes : (i) sonp �erim �etre est un entier impair ; et (ii) au un de ses ot �es n'est entier.D �eterminermaintenant un re tangle ayant les deux propri �et �es suivantes :(i) sa surfa e est un entier pair ; et (ii) au un de ses ot �es n'est entier.En�n, d �eterminer un quadrilat �ere, pas n �e essairement re tangulaire,ayant les trois propri �et �es suivantes : (i) son p �erim �etre est un entier positif ;(ii) sa surfa e est un entier positif ; et (iii) au un de ses ot �es n'est entier.Calgary Mathemati al Asso iationJunior High S hool Mathemati s ContestFinal Round, Part B, 20081. Ri hard needs to go from his house to the park by taking a taxi. Thereare 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 om-pany 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 htaxi ompany he hooses. What is the distan e in km from his house to thepark?2. A radio station runs a ontest in whi h ea h winner will get to attendtwo Calgary Flames playo� games and to take one guest to ea h game. Thewinner 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 behis or her guests, so that ea h pair of the �ve friends gets to go to at leastone playo� game together.

1963. A lass was given two tests. In ea h test ea h student was given a non-negative integer s ore with a maximum possible s ore of 10. Adrian noti edthat in ea h test, only one student s ored higher than he did and nobody gotthe same s ore as he did. But then the tea her posted the averages of thetwo s ores for ea h student, and now there was more than one student withan average s ore higher than Adrian.(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 youranswer is orre t.4. A re tangle with dimensions 6 m by 8 mis drawn. A ir le is drawn ir ums ribingthis re tangle. A square is drawn ir um-s ribing this ir le. A se ond ir le is drawnthat ir ums ribes this square. What is thearea in m2 of the bigger ir le? .......................................................................................................

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5. If A is a two-digit positive integer that does not ontain zero as a digit, Bis a three-digit positive integer, and A% of B is 400, �nd all possible valuesof A and B.6. Find a re tangle with the following two properties: (i) its perimeter is anodd integer; and (ii) none of its sides is an integer.Next, �nd a re tangle with the following two properties: (i) its area isan even integer; and (ii) none of its sides is an integer.Finally, �nd a quadrilateral (not ne essarily a re tangle) with the fol-lowing three properties: (i) its perimeter is a positive integer; (ii) its area isa positive integer; and (iii) none of its sides is an integer.We now give solutions to the sele ted questions of the 2007 ChristopherNewport University Mathemati s Contest in Skoliad 111 [2008 : 257-259℄.1. Find the midpoint of the domain of the fun tion f(x) =

√

4 −√

2x + 5.(A) 1

4(B) 3

2(C) 2

3(D) −2

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

2x ≥ −5, hen e x ≥ −52.

197The �rst inequality is the same as 4 ≥

√2x + 5, so 16 ≥ 2x+5. Thus,

11 ≥ 2x and 112

≥ x. Hen e the domain of f is the interval [−52, 11

2

]. Themidpoint of this interval is 12

(

−52

+ 112

)

= 32, so the answer is (B).Also solved by JIXUANWANG, student, DonMills Collegiate Institute, Toronto, ON; andLUYUN ZHONG-QIAO, Columbia International College, Hamilton, ON.2. The sum a+ b, the produ t ab, and the di�eren e a2 − b2 for two positivenumbers a and b is the same nonzero number. What is b?(A) 2 (B) 1 +

√5

2(C) √

5 (D) 3 −√

5

3Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto,ON. If ab = a + b and ab = a2 − b2, then1 =

ab

ab=

a2 − b2

a + b=

(a − b)(a + b)

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 ±√

5

2. Sin e b isgiven 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. Findthe oeÆ ient of x in f(x).(A) 2 (B) 3 (C) 1 (D) −2Solution 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 = f(3) = 9a + 3b + c

9 = f(−3) = 9a − 3b + cSubtra ting the se ond equation from the �rst yields 6 = 6b, so b = 1, andthe 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 thenumbers falling uppermost is 7 or 11 if it is known that one of the numbersis a 5? (A) 2

9(B) 7

36(C) 1

9(D) 4

11

198Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto,ON. One an roll at least one 5 in eleven ways:First die 1 2 3 4 5 6 5 5 5 5 5Se ond die 5 5 5 5 5 5 1 2 3 4 6Sum 6 7 8 9 10 11 6 7 8 9 11Of these the sum is 7 or 11 in four ases, so the probability is 4

11and theanswer is (D).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 Ahappens given that B did happen. In the ase of rollingthe di e, A is \the sum is 7 or 11" while B is \at least one 5". The probability of rolling at leastone 5 with two di e is 1

6· 56

+ 56

· 16

+ 16

· 16or 11

36. When rolling two di e, one may get a sum of

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) = 436

.The formula above now yields P(A|B) =4/36

11/36= 4

11, as before.5. The number √24 +

√572 an be written in the form √

a +√

b, where aand b are whole numbers and b > a. What is the value b − a?(A) 4 (B) −2 (C) 2 (D) 3Solution by an unknown solver.Note that √572 =

√22 · 11 · 13 = 2

√11

√13. Thus,

√

24 +√

572 =

√

11 + 13 + 2√

11√

13

=

√√11

2+ 2

√11

√13 +

√13

2

=

√

(√

11 +√

13)2

=√

11 +√

13 .Hen e b − a = 13 − 11 = 2, and the answer is (C).Also solved by JIXUANWANG, student, DonMills Collegiate Institute, Toronto, ON; andLUYUN ZHONG-QIAO, Columbia International College, Hamilton, ON.6. Let x =1

2 + 13+ 1

2+ 13+···

be the indi ated ontinued fra tion. Whi h one ofthe following is equal to x?(A) √15 + 1

2(B) √

2 + 1

3(C) −3 +

√15

2(D) −

√15 − 3

2

199Solution by an unknown solver.Note that

x =1

2 +1

3 +1

2 +1

3 + · · ·

=1

2 +1

3 + x

.Therefore

x =1

2(3 + x) + 1

3 + x

=3 + x

7 + 2x.

Hen e 7x + 2x2 = 3 + x, so 2x2 + 6x − 3 = 0, and the quadrati formulayieldsx =

−6 ±√

60

4=

−6 ± 2√

15

4=

−3 ±√

15

2.Sin e x is learly positive, x =

−3 +√

15

2and the answer is (C).Also solved by JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON.7. If f

(√

1 + x

1 − x

)

= 5x, �nd f(2).(A) −15 (B) 15√−1 (C) 3 (D) −4Solution by an unknown solver.If √1 + x

1 − x= 2, then 1 + x

1 − x= 4, so 1 + x = 4 − 4x, when e x = 3

5.Substituting x = 3

5into the given equation now yields f(2) = 5 · 3

5= 3, andthe answer is (C).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 inthe transition of Skoliad editors. If you submit solutions on paper, or s ansof paper solutions, we request that you write your name and aÆliation onea h sheet.This issue's prize of one opy of Crux with Mayhem for the best solu-tions goes to Jixuan Wang, student, Don Mills Collegiate Institute, Toronto,ON. We would very mu h appre iate re eiving more solutions from our pre-university readers. Solutions to just some of the problems are also wel ome.

200MATHEMATICAL MAYHEMMathemati al Mayhem began in 1988 as a Mathemati al Journal for and byHigh 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). Theother sta� members are Monika Khbeis (As ension of Our Lord Se ondaryS hool, Mississauga) and Eri Robert (Leo Hayes High S hool, Frederi ton).

Mayhem ProblemsVeuillez nous transmettre vos solutions aux probl �emes du pr �esent num�eroavant le 31 aout 2009. Les solutions re� ues apr �es ette date ne seront prises en ompteque s'il nous reste du temps avant la publi ation des solutions.Chaque probl �eme sera publi �e dans les deux langues oÆ ielles du Canada(anglais et fran� ais). Dans les num�eros 1, 3, 5 et 7, l'anglais pr �e �edera le fran� ais,et dans les num�eros 2, 4, 6 et 8, le fran� ais pr �e �edera l'anglais.La r �eda tion souhaite remer ier Jean-Mar Terrier, de l'Universit �e deMontr �eal, d'avoir traduit les probl �emes.M394. Propos �e par l' �Equipe de Mayhem.Les nombres a, b, c, d et e sont inq entiers su essifs. Montrer que ladi� �eren e entre la moyenne des arr �es de c et e et elle des arr �es de a et cest �egale �a quatre fois c.M395. Propos �e par l' �Equipe deMayhem.Le quadrilat �ere ABCD est telque ha un de ses ot �es est tangent �aun er le donn �e, omme dans la �gure i- ontre. Si AB = AD, montrer queBC = CD.

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B

C

D

M396. Propos �e par l' �Equipe de Mayhem.Dans un re tangle ABCD de ot �es AB = 8 et BC = 6, on ins ritrespe tivement deux er les de entre O1 et O2 dans les triangles ABD etBCD. Trouver la distan e entre O1 et O2.M397. Propos �e par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie.Trouver toutes les paires (x, y) d'entiers tels que

x4 − x + 1 = y2 .

201M398. Propos �e par l' �Equipe de Mayhem.(a) Soit r, s et t les ra ines de l' �equation ubique w3 − bw2 + cw − d = 0.D �eterminer b, c et d en termes de r, s et t.(b) On suppose que a est un nombre r �eel. D �eterminer toutes les solutionsdu syst �eme d' �equations

x + y + z = a ,xy + yz + zx = −1 ,

xyz = −a .M399. Propos �e par Ne ulai Stan iu, �E ole Te hnique Sup �erieure de SaintMu eni Sava, Ber a, Roumanie.Trouver tous les triplets (a, b, c) d'entiers positifs tels que 3ab − 1

abc + 1soitun entier positif.M400. Propos �e par Mih�aly Ben ze, Brasov, Roumanie.Supposons que a, b et c sont trois nombres r �eels positifs. Supposons deplus que an + bn = cn pour un ertain entier positif n ave n ≥ 2. Montrerque si k est un entier positif ave 1 ≤ k < n, alors ak, bk et ck sont leslongueurs des ot �es d'un triangle..................................................................M394. Proposed by the Mayhem Sta�.The numbers a, b, c, d, and e are �ve onse utive integers, in that order.Prove that the di�eren e between the average of the squares of c and e andthe average of the squares of a and c is equal to four times c.

M395. Proposed by the MayhemSta�.The quadrilateral ABCD is su h thatea h of its sides is tangent to a given ir le, as shown. If AB = AD, provethat BC = CD..............................................................................................................................

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

B

C

D

M396. Proposed by the Mayhem Sta�.The re tangle ABCD has side lengths AB = 8 and BC = 6. Cir leswith entres O1 and O2 are ins ribed in triangles ABD and BCD. Deter-mine the distan e between O1 and O2.

202M397. Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.Determine all pairs (x, y) of integers su h that

x4 − x + 1 = y2 .M398. Proposed by the Mayhem Sta�.(a) The ubi equation w3 − bw2 + cw − d = 0 has roots r, s, and t.Determine b, c, and d in terms of r, s, and t.(b) Suppose that a is a real number. Determine all solutions to the systemof equationsx + y + z = a ,

xy + yz + zx = −1 ,xyz = −a .M399. Proposed by Ne ulai Stan iu, Saint Mu eni Sava Te hnologi alHigh S hool, Ber a, Romania.Determine all triples (a, b, c) of positive integers for whi h 3ab − 1

abc + 1isa positive integer.M400. Proposed by Mih�aly Ben ze, Brasov, Romania.Suppose that a, b, and c are positive real numbers. In addition, supposethat an + bn = cn for some positive integer n with n ≥ 2. Prove that if k isa positive integer with 1 ≤ k < n, then ak, bk, and ck are the side lengthsof a triangle.

Mayhem SolutionsM357. Proposed by the Mayhem Sta�.Determine all real numbers x that satisfy 32x+2 + 3 = 3x + 3x+3.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 form32(3x)2 + 3 = 3x + 333x. Substituting k = 3x, we obtain the equivalentquadrati 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 = 1

9.Substituting k = 3x, we have 3x = 3 or 3x = 1

9, so x = 1 or x = −2.

203Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosniaand Herzegovina; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; CAO MINH QUANG,Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; JACLYN CHANG, student, WesternCanada High S hool, Calgary, AB; LUIS DE SOUSA, student, IST-UTL, Lisbon, Portugal; JOS �EHERN�ANDEZ SANTIAGO, student, Universidad Te nol �ogi a de la Mixte a, Oaxa a, Mexi o;GIANNIS G. KALOGERAKIS, Canea, Crete, Gree e; RICARD PEIR �O, IES \Abastos", Valen ia,Spain; KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; MRIDUL SINGH,student, Kendriya Vidyalaya S hool, Shillong, India; ALEX SONG, student, Elizabeth ZieglerPubli S hool, Waterloo, ON; GEORGE TSAPAKIDIS, Agrinio, Gree e; EDWARD T.H. WANG,Wilfrid Laurier University, Waterloo, ON; JIXUAN WANG, student, Don Mills CollegiateInstitute, Toronto, ON; and CARLIN WHITE, student, California State University, Fresno, CA,USA. There were two in orre t solutions submitted.M358. Proposed by Ne ulai Stan iu, Saint Mu eni Sava Te hnologi alHigh S hool, Ber a, Romania.How many integers in the list 1, 2008, 20082, . . . , 20082009 are simul-taneously perfe t squares and perfe t ubes?Solution by Luis De Sousa, student, IST-UTL, Lisbon, Portugal, modi�ed bythe editor.The integer 2008 has prime fa torization 23 · 251. Thus, 2008k hasprime fa torization (23 · 251

)k= 23k · 251k. For a positive integer greaterthan 1 to be a perfe t square, its prime fa torization must in lude only evenexponents; for a positive integer greater than 1 to be a perfe t ube, its primefa torization must in lude only exponents divisible by 3.The exponents 3k and k are both even if and only if k is even. Theexponents 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 oneswith even exponents. Also, the powers of 2008 whi h are perfe t ubes arethose with exponents that are multiples of 3. Hen e, the powers of 2008whi h are perfe t squares and perfe t ubes are those whose exponents areboth multiples of 2 and of 3. In other words, they are those whose exponentsare 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 ofthe integers 20081, 20082, . . . , 20082009 are simultaneously perfe t squaresand perfe t ubes. The integer 1 is also in the list and is both a perfe t squareand a perfe t ube. Therefore, there are 334 + 1 = 335 integers in the listthat are simultaneously perfe t squares and perfe t ubes.Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosniaand Herzegovina; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; JACLYN CHANG,student, Western Canada High S hool, Calgary, AB; CHANTHOEUN CHAP and JUSTINHENDERSHOTT, students, California State University, Fresno, CA; USA; JOS �E HERN�ANDEZSANTIAGO, student, Universidad Te nol �ogi a de la Mixte a, Oaxa a, Mexi o; RICHARDI. HESS, Ran ho Palos Verdes, CA, USA; RICARD PEIR �O, IES \Abastos", Valen ia, Spain;KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; MRIDUL SINGH, stu-dent, 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.

204M359. Proposed by the Mayhem Sta�.A trapezoid DEFG is ir um-s ribed about a ir le of radius 2, asshown in the diagram. The side DEhas length 3 and there are right anglesat E and F . Determine the area of thetrapezoid.

E

FG

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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 ofthe 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 tan-gents to the same ir le from the samepoint have equal lengths. Let GZ =x.Then GY = x by equal tangents.Sin e OX and OW are perpen-di ular to DE and EF , respe tively,and the trapezoid has a right angle atE, then OXEW is a square.

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Sin e the radius of the ir le is 2, then OW = 2, so XE = 2. Sin eDE = 3, then DX = DE − XE = 1. By equal tangents, DY = 1.Now DEFA 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 toEF . Thus, AZ = DX = 1.We know thatDA = XO+OZ = 2+2 = 4, GD = GY +Y D = x+1,and GA = GZ − AZ = x − 1. By the Pythagorean Theorem,

GD2 = GA2 + AD2 ;(x + 1)2 = (x − 1)2 + 42 ;

x2 + 2x + 1 = x2 − 2x + 1 + 16 ;hen e 4x = 16 and x = 4. Therefore, GF = GZ + ZF = 4 + 2 = 6, and sothe area of the trapezoid is 12(GF + DE)(DA) or 1

2(6 + 3)(4) = 18.Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosniaand Herzegovina; SERDAR ALTUNTAS, student, University of Karlsruhe, Karlsruhe, Germany;MIGUEL AMENGUAL COVAS, Cala Figuera, Mallor a, Spain; SCOTT BROWN, Auburn Uni-versity, 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 Universityat 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,Gree e; RICARD PEIR �O, IES \Abastos", Valen ia, Spain; KUNAL SINGH, student, KendriyaVidyalaya 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 onein omplete solution submitted.

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

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 theequation in the form3x = 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 shouldbe 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 whendivided by 3, so is not divisible by 3. This is a ontradi tion, so there are nopositive integer solutions.Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosniaand Herzegovina; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam;LUIS DE SOUSA, student, IST-UTL, Lisbon, Portugal; RICARD PEIR �O, IES \Abastos",Valen ia, Spain; ALEX SONG, student, Elizabeth Ziegler Publi S hool, Waterloo, ON; GEORGETSAPAKIDIS, Agrinio, Gree e; and JIXUAN WANG, student, Don Mills Collegiate Institute,Toronto, ON. There were two in omplete solutions submitted.M361. Proposed by George Tsapakidis, Agrinio, Gree e.Let a, b, and c be positive real numbers. Prove thatab(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

a2b + ab2 + b2c + bc2 + c2a + ca2

6

≥ 6

√

(

a2b)(

ab2)(

b2c)(

bc2)(

c2a)(

ca2)

=6√

a6b6c6 = abc .It follows thata2b + ab2 + b2c + bc2 + c2a + ca2 ≥ 6abc .

206Rearranging the last expression, we �nd that

a2b + ab2 − abc + b2c + bc2 − abc + c2a + ca2 − abc ≥ 3abc ,whi h is equivalent toab(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, Bosniaand Herzegovina; MIGUEL AMENGUAL COVAS, Cala Figuera, Mallor a, Spain; GEORGEAPOSTOLOPOULOS, Messolonghi, Gree e; CAO MINH QUANG, Nguyen Binh Khiem HighS hool, Vinh Long, Vietnam; COURTIS G. CHRYSSOSTOMOS, Larissa, Gree e; LUIS DE SOUSA,student, IST-UTL, Lisbon, Portugal; JOS �E LUIS D�IAZ-BARRERO, Universitat Polit �e ni a deCatalunya, 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 S�ANCHEZ, Ponti� ia Universidad Cat �oli a del Peru, Lima, Peru; MISSOURISTATE UNIVERSITY PROBLEM SOLVING GROUP, Spring�eld, MO, USA; RICARD PEIR �O, IES\Abastos", Valen ia, Spain; KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong,India; ALEX SONG, student, Elizabeth Ziegler Publi S hool, Waterloo, ON; EDWARDT.H. WANG, Wilfrid Laurier University, Waterloo, ON; and JIXUAN WANG, student, Don MillsCollegiate Institute, Toronto, ON.M362. Proposed by Mih�aly Ben ze, Brasov, Romania.Suppose that x1, x2, . . . , xn is a sequen e of integers su h that∣

∣|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 andxk+1, xk+2, . . . , xn are not positive (if this is not the ase, then we anrearrange 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|) .

207It follows that |xk+1| + |xk+2| + · · · + |xn| = 1. Sin e the sum onsists ofnonnegative 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, wesee 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 STATEUNIVERSITY 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 MonthIan VanderBurgh

Contest problems involving omplex numbers don't appear very often...Problem (2007 Ameri an Invitational Mathemati s Challenge A) The omplexnumber z is equal to 9 + bi, where b is a positive real number and i2 = −1.Given that the imaginary parts of z2 and z3 are equal, �nd b....and when they do, they're often easier than they look. The one pie eof 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!

208Solution Sin e z = 9 + bi, then

z2 = (9 + bi)2 = (9 + bi)(9 + bi) = 81 + 18bi + b2i2

= 81 + 18bi − b2 =(

81 − b2)

+ (18b)i ,whi h givesz3 = (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 z2 is 18b and the imaginary part of z3 is 243b − b3. Ifthese 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 theMonth 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 tryto answer the question for a general omplex number z = a + bi. I did thisfor two reasons. First, I was going to use it to verify my answer. Se ond, Iwas urious how the \15" appeared.For z = a + bi,z2 = (a + bi)2 = (a + bi)(a + bi) = a2 + 2abi + b2i2

= a2 + 2abi − b2 =(

a2 − b2)

+ (2ab)iandz3 = (a + bi)3 = (a + bi)2(a + bi)

=[(

a2 − b2)

+ (2ab)i]

(a + bi)

=(

a3 − ab2)

+(

a2b − b3)

i +(

2a2b)

i +(

2ab2)

i2

=(

a3 − ab2)

+(

a2b − b3)

i +(

2a2b)

i −(

2ab2)

=(

a3 − 3ab2)

+(

3a2b − b3)

i .In this general ase, the imaginary part of z2 is 2ab and the imaginary partof z3 is 3a2b − b3. If these are equal, we have2ab = 3a2b − b3 ,

b3 − 3a2b + 2ab = 0 ,b[

b2 −(

3a2 − 2a)]

= 0 .Thus, b = 0 or b = ±√

3a2 − 2a. However b > 0, so b =√

3a2 − 2a.Can you see why 3a2 − 2a is not negative if a is an integer? In the originalquestion, a = 9, so that b =√

3(92) − 2(9) =√

225 = 15. This addresses

209my �rst on ern. It also gives somewhat of an answer to the se ond one aswell { the 15 doesn't appear in any really obvious way. (In other words, weshouldn'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 otherquestions we ould ask. One question that immediately ame to my mindwas whether or not it was surprising that we got an integer answer for b witha = 9. (Given that this problem appears on the AIME, in whi h every answeris an integer, it shouldn't be that surprising.) Put a bit di�erently, though,for a random positive integer a, is b =

√3a2 − 2a always an integer or is

a = 9 somewhat spe ial? If a = 1, we get b =√

3(12) − 2(1) =√

1 = 1,whi h is an integer; if a = 10, we get b =√

3(102) − 2(10) =√

280, whi his not an integer.Next question: For what positive integer values of a is b =√

3a2 − 2aa positive integer?This is a fair bit harder, but is worth looking at. We'll take what seemsinitially like a ba kwards approa h by starting with a positive integer b andthen 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 = b2and 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 ofb. When we do this, we obtaina =

−(−2) ±√

(−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 givepositive integer values for a. (A tually, we're really interested in the otherdire 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?So we fo us on the ase a =

1 +√

1 + 3b2

3. For a to be an integer, whatneeds to be true? Certainly, √

1 + 3b2 needs to be a positive integer, thoughthis is a tually not quite enough sin e √1 + 3b2 needs to be an integer andwe want 1 +

√1 + 3b2 to be divisible by 3.Let's fo us on the �rst issue. Suppose that √

1 + 3b2 = m for somepositive integer m. In this ase, 1 + 3b2 = m2 or m2 − 3b2 = 1. Do youre ognize this equation? This is an example of a Pell Equation.Oh dear { this keeps getting more ompli ated. Let's summarize for ase ond to see where we are. We are trying to �nd positive integers a for whi hb =

√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 wehave 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 thes ope of this olumn. If you've never seen how to �nd the positive integersolutions to m2 − 3b2 = 1, here's a qui k summary:

210Step 1 Find the smallest pair of positive integers that is a solution. You ando 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 smallestpositive 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 anin�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 enter-prising ould try to write down some kind of losed form solution. (As anadditional hallenge, try using the relation between (M, B) and (m, b) tosolve for m and b in terms of M and B. If you an do this, then you an usea larger solution to generate smaller ones.)This method gives us a list of values of b that we might try. At the veryleast, these values of b make √

1 + 3b2 an integer, though they might notmake a an integer be ause of the extra ondition. Let's try the �rst few:• If b = 1, then a =

1 +√

1 + 3b2

3=

1 +p

1 + 3(12)

3= 1, whi h is aninteger.

• If b = 4, then a =1 +

√1 + 3b2

3=

1 +p

1 + 3(42)

3=

8

3, whi h is notan integer.

• If b = 15, then we've already seen that a = 9.• If b = 56, then a =

1 +√

1 + 3b2

3=

1 +p

1 + 3(562)

3=

98

3, whi h isnot an integer.

• If b = 209, then a =1 +

√1 + 3b2

3=

1 +p

1 + 3(2092)

3= 121, whi his an integer.At a qui k glan e, it appears that every other value of b in this sequen egives 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 in-volving omplex numbers. We then talked about some of the stru ture of theproblem and tried to do some investigation into what was happening \behindthe s enes". We've found two more values of a that would give us integervalues 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 ledus 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 ofz2 and z4 might be equal.

211THE OLYMPIAD CORNERNo. 278R.E. WoodrowWe start this number with problems of the Hungarian Mathemati alOlympiad 2005-2006, National Olympiad, Grades 11-12, Se ond Round andFinal Round. Thanks go to Robert Morewood, Canadian Team Leader to the47th IMO in Slovenia, for olle ting them for our use.Hungarian Mathemati al Olympiad 2005{2006National Olympiad, Grades 11-12Se ond Round1. Find the positive values of x that satisfy

x(2 sin x−cos 2x) <1

x.

2. For f(x) = ax2 − bx + c we know that 0 < |a| < 1, f(a) = −b, andf(b) = −a. Prove that |c| < 3.3. The onvex quadrilateral ABCD satis�es ∠ABC = ∠ACD. The mid-points of BC and AD are E and F (respe tively), O = AC ∩ BD, andG 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.4. 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. Final Round1. De�ne the fun tion t(n) on the nonnegative integers by t(0) = t(1) = 0,t(2) = 1, and for n > 2 let t(n) be the smallest positive integer whi h doesnot divide n. Let T (n) = t

(

t(

t(n))). Find the value of S if

S = T (1) + T (2) + T (3) + · · · + T (2006) .2. Let A and B be two verti es of a tree with 2006 edges. We move alongthe edges starting from A and would like to get to B without ever turningba k. At any vertex we hoose the next edge among all possible edges (ex eptthe one on whi h we arrived) with equal probability.Over all possible trees and hoi es of verti es A and B, �nd the mini-mum probability of getting from A to B.

2123. A unit ir le k with entre K and a line e are given in the plane. Theperpendi ular from K to e interse ts e in point O and KO = 2. Let H bethe 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∠APB = α for any ir le in H with diameter AB on e. Determine α andthe lo ation of P .

Next we give the �rst round of the Hungarian Mathemati al Olympiad2005{2006 for Spe ialized Mathemati al Classes. Thanks again go to RobertMorewood, Canadian Team Leader to the 47th IMO in Slovenia, for olle tingit for our use.Hungarian Mathemati al Olympiad 2005{2006Spe ialized Mathemati al Classes, First Round1. 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 reversingits digits yields the same number, for example, 12321 is a palindrome.)2. We have �nitely many (but at least two) numbers of the form 1

2kwhosesum is at most 1. Prove that they an be divided into two groups su h thatthe sum of the numbers in ea h group is at most 1

2.3. 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 betweena red point and a blue point. How many pairs of blue and red points a hievethis minimum distan e?4. A tetrahedron has at least four edges ea h at most 1 unit in length. De-termine the maximum possible volume of the tetrahedron.5. Let k be a ir le with entre O and let AB be a hord of k whose midpoint,M , is distin t from O. The ray from O through M meets k at R. Let P bea point on the minor ar AR of k, let PM meet k again at Q, and let ABmeet QR at S. Whi h segment is longer, RS or PM?

The next problem set is the 2005 K�urs h �ak Competition (a Hungarian ompetition). Thanks again go to Robert Morewood, Team Leader to the47th IMO in Slovenia, for olle ting it for our use.

2132005 K�urs h �ak Competition1. Let N > 1 and assume that the sum of the nonnegative real numbers

a1, a2, . . . , aN is at most 500. Prove that there exists an integer k ≥ 1 andthere exist integers 1 = n0 < n1 < · · · < nk = N su h thatk∑

i=1

niani−1< 2005 .

2. Ann and Bob are playing tennis. The winner of a mat h is the player whois the �rst to win at least four games, being at least two games ahead of hisor her opponent. Ann wins a game with probability p ≤ 12independentlyof the out ome of the previous games. Prove that Ann wins the mat h withprobability 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 alsoa 10×11 re tangle onsisting of 55 dominos that exa tly overs the previouslevel. 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 ofsome domino of the tower. What is the minimum number of levels a rigidtower may have?Next we give the problems of the 8th Hong Kong (China) Mathemati alOlympiad 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.8th Hong Kong (China) Mathemati al Olympiad1. On a planet there are 3 · 2005! aliens and 2005 languages. Ea h pair ofaliens ommuni ate with ea h other in exa tly one language. Show that thereare 3 aliens who ommuni ate with ea h other in one ommon language.2. Suppose that there are 4n line segments of unit length inside a ir le ofradius n. Given a straight line ℓ, prove that there exists a straight line ℓ′ thatis either parallel to or perpendi ular to ℓ and su h that ℓ′ interse ts at leasttwo of the given line segments.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)

+ 18.4. Show that there exist in�nitely many squarefree positive integers n thatdivide 2005n − 1. (An integer is squarefree if it has no fa tor of the form d2for an integer d > 1.)

214Next we give the two Hong Kong Team Sele tion Tests for the Inter-national Mathemati al Olympiad 2006. Thanks go to Robert Morewood,Canadian Team Leader to the 47th IMO in Slovenia, for olle ting them forthe Corner. Hong Kong Team Sele tion Test 11. Find the integer solutions of the equation 7(x + y) = 3

(

x2 − xy + y2).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), and( ) f(x + 1, y + 1) = f

(

x, f(x + 1, y)).Find f(3, 2005) and f(4, 2005).3. In triangle ABC, the altitude, angle bise tor, and median from C divide

∠C into four equal angles. Find ∠B.4. Let x, y, and z be positive real numbers su h that x + y + z = 1. Fora positive integer n, let Sn = xn + yn + zn. Also, let P = S2S2005 andQ = S3S2004.(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 formedby any three of them is less than 1. Show that all of the points lie inside oron the boundary of a triangle with area less than 4.6. Find 22006 positive integers su h that(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 21. 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 arethe verti es of a y li quadrilateral.

2152. Let ABCD be a y li quadrilateral with BC = CD. The diagonals ACand 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 Ware on y li if and only if AB = AD.3. Points A and B lie in a plane and ℓ is a line in that plane passing throughA but not through B. The point C moves from A toward in�nity along ahalf-line of ℓ. The in ir le of △ABC tou hes BC at D and AC at E. Showthat the line DE passes though a �xed point.4. Let △ABC have ir umradius R. Let AB = c, BC = a, CA = b, andlet k1 and k2 be the ir les with diameters CA and CB, respe tively. Let kbe 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

4R.

Next we give the 20th Nordi Mathemati al Contest written on Mar h30, 2006. Thanks go to Robert Morewood, Canadian Team Leader to the 47thIMO in Slovenia, for obtaining it for the Corner.20th Nordi Mathemati al ContestMar h 30, 20061. Let B and C be points on two given rays from the same point A, su hthat AB + AC is onstant. Prove that there exists a point D distin t frompoint A su h that the ir um ir les of the triangles ABC pass through D forall hoi es of B and C subje t to the given onstraint.2. The real numbers x, y, and z are not all equal and satisfyx +

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 there ursion an+1 = a5

n + 487 for all n ≥ 0. Determine all values of m forwhi h the sequen e ontains as many square numbers as possible.4. The squares of a 100 × 100 hessboard are oloured with 100 di�erent olours. Ea h square is painted with one olour only and ea h olour is usedexa tly 100 times. Show that there exists a row or a olumn on the hessboardsu h that at least 10 di�erent olours are used to paint its squares.

216As three sets of problems for your problem solving pleasure over thebreak 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 inSlovenia, for obtaining them for our use. We remark that the odd-numberedquestions of the 10th form orrespond to the odd-numbered questions on the9th form, hen e they are omitted.XXXII Russian Mathemati al Olympiad 2005{2006Final Round, 9th formFirst Day

1. 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 thatthese segments form a losed broken line that does not interse t itself andthat is symmetri with respe t to one of the diagonals. Prove that the lengthof the broken line is at most 200 units.2. Show that there exist four integers a, b, c, and d whose absolute valuesare greater than 1 000 000 and whi h satisfy1

a+

1

b+

1

c+

1

d=

1

abcd.

3. On a ir le 2006 points are given. Peter olours ea h of these points withone of 17 olours. After that Mi hael onne ts these points by segments sothat two endpoints of ea h segment have the same olour and the segmentsdo 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 hin-der him. Find the greatest number of segments Mi hael an draw regardlessof Peter's olouring.4. A ir le ω tou hes the ir um ir le of a triangle ABC at A, interse ts sideAB at K, and interse ts side BC. A tangent CL to ω, with L on ω, is su hthat the segment KL interse ts side BC at T . Prove that the length of BTequals the length of the tangent from B to ω.9th formSe ond Day5. 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.

2176. The points P , Q, and R lie on the sides AB, BC, and CA of a triangleABC su h that AP = CQ and the quadrilateral RPBQ is y li . The tan-gents to the ir um ir le of triangle ABC at A and C meet the respe tivelines RP and RQ at X and Y . Prove that RX = RY .7. 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 twoadja ent squares if there is a ut between them. A player loses if he or shere onne ts the board (thus allowing the board to be lifted by a orner withoutit falling apart). Who has a winning strategy, the �rst player or the se ondplayer?8. A quadrati polynomial f(x) = x2 + ax + b is given. Suppose that theequation f

(

f(x))

= 0 has four distin t real roots and that the sum of two ofthem is equal to −1. Prove that b ≤ −14.10th formFirst Day2. Assume that the sum of the ubes of three onse utive positive integersis a ube of some positive integer. Prove that the middle number of thesethree numbers is divisible by 4.4. Let ABC be an isos eles triangle with AB = AC. Let ω be a ir letou hing the sides AB and AC and meeting the side BC at K and L. Thesegment AK meets ω for the se ond time at M . The points P and Q arethe re e tions of K with respe t to B and C, respe tively. Prove that the ir um ir le of △PMQ is tangent to ω.10th formSe ond Day6. Let K and L be points lying on the ar s AB and BC of the ir um ir leof △ABC, respe tively, so that the lines KL and AC are parallel. Provethat the in entres of △ABK and △CBL are equidistant from the midpointof the ar ABC.8. A 3000 × 3000 square is divided into dominos (that is, into 2 × 1 re t-angles). Prove that one an paint the dominos with three olours su h thatea h olour is used equally often and ea h pie e shares a side with no morethan two pie es of the same olour.11th formFirst Day1. Prove that sin

√x <

√sin x whenever 0 < x < π

2.

2182. Assume that the sum and the produ t of some pure periodi de imalfra tions is a pure de imal fra tion with period T . Prove that ea h of theoriginal fra tions has a period not ex eeding T .3. Consider a 49×69 grid. All the 50·70 gridpoints are marked. Two playerstake turns onne ting pairs of gridpoints by a segment so that no gridpointis the endpoint of two segments. The segments are drawn until there areno free gridpoints. The �rst player wins if, after this pro ess, he an orientea h of the segments so that the sum of all ve tors obtained is equal to thenull ve tor; otherwise the se ond player wins. Whi h player has a winningstrategy?4. The angle bise tors BB1 and CC1 of △ABC (with B1 on AC and C1on AB) meet at I. The line B1C1 meets the ir um ir le of △ABC at Mand N . Prove that the ir umradius of △MIN is twi e the ir umradius of△ABC. 11th formSe ond Day5. The sequen es of positive numbers {xn} and {yn} satisfy

xn+2 = xn + x2n+1 ,

yn+2 = y2n + yn+1 ,for all positive integers n. Prove that if x1, x2, y1, and y2 are ea h greaterthan 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, andtou hes AB, BC, CA at D, E, F , respe tively. The points A′, B′, C′ lie onthe 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. Supposethat SI is an altitude of the pyramid. Prove that S′ is equidistant from A′,B′, and C′.7. The polynomial (x + 1)n − 1 is divisible by a polynomial

P (x) = xk + ck−1xk−1 + ck−2xk−2 + · · · + c1x + c0of even degree k su h that c0, c1, . . . , ck−1 are odd integers. Prove that n isdivisible by k + 1.8. A group of pioneers has arrived at summer amp. Ea h pioneer has at least50 and at most 100 friends among the others. Prove that one an distribute�eld aps that ome in 1331 di�erent olors one ea h to the pioneers so thatthe friends of ea h pioneer have aps of at least 20 di�erent olors.

219Now we return to the �le of solutions from our readers to problems inthe May 2008 number of the Corner and the XIX Olimpiada Iberoameri anade Matemati as, given at [2008 : 214℄.2. Let A be a �xed exterior point with respe t to a given ir le with entre

O and radius r. Let M be a point on the ir le and let N be diametri allyopposite to M with respe t to O. Find the lo us of the entres of the ir lespassing through A, M , and N , as the point M is varied on the ir le.Solved by Mi hel Bataille, Rouen, Fran e; and Titu Zvonaru, Com�ane�sti,Romania. We give Bataille's write up.Let U be the ir um- entre of △AMN . Sin eO is the midpoint of MN ,UO is orthogonal to MNand UM2 = UO2 + r2.But UM = UA, so wehave UA2 − UO2 = r2.It follows that U is on aline ℓ perpendi ular to AO.More pre isely, if M0N0 isthe diameter perpendi ulartoAO andU0 is the ir um- entre of △AM0N0, thenℓ is the perpendi ular toAO through U0. Note thatU0 6= O.

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Conversely, Let U be any point on ℓ (so that UA2 −UO2 = r2) and letMN be the diameter perpendi ular toUO. Then UO2+r2 = UM2 = UN2so that UM = UN = UA and U is the ir um entre of △AMN (note thatA, M , and N are not ollinear be ause U0 6= O).3. 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) = 1and k = a + b.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 bep1, p2, . . . , pr. Note that ea h pj is odd and does not divide both k − 1 andk + 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 takeb = k − a. Then, ab ≡ aj(k − aj) (mod pj) with aj 6≡ 0 (mod pj) andk − aj is 1 or −1. Hen e, ab 6≡ 0 (mod pj) for ea h j. It follows thatgcd(a, n) = gcd(b, n) = 1 and k = a + b.

220We 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 −1and b = 1 satisfy the requirements.Otherwise, we denote by p1, p2, . . . , pr the odd prime divisors of nand let aj ∈ {k − 1, k + 1} be su h that pj does not divide aj. The ChineseRemainder Theorem provides an integer a su h that a ≡ aj (mod pj) forea h j and a ≡ 1 (mod 2). Setting b = k − a (note that b is odd), it isreadily 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 twodigits, and su h that 100a + b and 201a + b are perfe t squares with fourdigits ea h.Solution by Titu Zvonaru, Com�ane�sti, RomaniaLet 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 tingthe two equations yields

101a = y2 − x2 = (y − x)(y + x) .Be ause 101 is prime, y − x < 101, and y + x < 2 · 101, it follows thaty + 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 bhas 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 nowdedu e that x = 42, y = 59, a = 17, and b = 64 is the unique solution tothe problem.5. In a s alene triangle ABC, the interior bise tors of the angles A, B, andC meet the opposite sides at points A′, B′, and C′ respe tively. Let A′′ bethe interse tion of BC with the perpendi ular bise tor of AA′, let B′′ be theinterse tion of AC with the perpendi ular bise tor of BB′, and let C′′ bethe interse tion of AB with the perpendi ular bise tor of CC′. Prove thatA′′, B′′, and C′′ are ollinear.Solution by Titu Zvonaru, Com�ane�sti, RomaniaLet a = BC, b = CA,and c = AB. Let M bethe midpoint of AA′. By theBise tor Theorem we haveA′C =

ab

b + c, BA′ =

ac

b + c,and (with α =

A

2) it is knownthat AA′ =

2bc cos α

b + c. Sin e

∠MA′A′′ = α + C, in the ...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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A

A′A′′B C

M

α

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

221triangle A′′A′M we obtain

A′A′′ =MA′

cos(α + B)=

AA′

2 cos(α + C)=

bc cos α

(b + c) cos(α + C).Using the Law of Sines we obtain:

A′′B

A′′C=

A′A′′ − A′B

A′A′′ + A′C=

bc cos α − ac cos(α + C)

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

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

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· sin B − 2 sin α cos(α + C)

sin C + 2 sin α cos(α + C)

=c

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

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

=c

b· sin B − sin(A + C) + sin C

sin C + sin(A + C) − sin C

=c

b· sin B − sin B + sin C

sin C + sin B − sin C=

c

b· sin C

sin B,

hen e A′′B

A′′C=

c2

b2.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 SwedishMathemati al Contest 2004/2005 Quali� ation Round given at [2008 : 215℄.1. The ities A, B, C, D, and E are onne ted by straight roads (more thantwo ities may lie on the same road). The distan e from A to B, and fromC 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 thedistan e from C to E.Solved by Titu Zvonaru, Com�ane�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).

222We have

AC2 = 52 = 32 + 42

= CD2 + AD2 ,hen e, by the onverse of thePythagorean Theorem, CD isperpendi ular to AD. Bythe Pythagorean Theorem, wenow obtain CE = 5.

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A B

C

D E

1

3

3

4

5

. . . . . . . . . . . . . . . . . . . . . . ......................................................................s s

s

s s

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, obtainingthe sums a + b, a + c, . . . , c + d, but she forgets to write down one of thepossible sums. The �ve sums she obtains are 7, 11, 12, 18, and 23. Whi hsum did Linda forget? What are the positive integers a, b, c, d ?Solved by Jean-DavidHoule, student, M Gill University, Montreal, QC; JohnGrant M Loughlin, University of New Brunswi k, Frederi ton, NB; and TituZvonaru, Com�ane�sti, Romania. We give Houle's solution, modi�ed by theeditor.Without loss of generality, let the unknown sum be c + d. Then weknow that 3(a + b) + 2(c + d) = 7 + 11 + 12 + 18 + 23 = 71. Sin e 71 isodd, 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 iseither 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, itfollows 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 anumber from {c, d}, so we may take c = a + 1. Then a + c = 2a + 1 iseither 7 or 23, so a = 3 or a = 11. However, a+ b = 11, hen e a 6= 11. Wededu e that (a, b, c, d) = (3, 8, 4, 15), whi h yields all of the required sums.3. Determine the greatest and the least value ofmn

(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; andTitu Zvonaru, Com�ane�sti, Romania. We give Zvonaru's write up.By the AM{GM Inequality we have mn

(m + n)2≤ mn

4mn=

1

4, hen e thegreatest value is 1

4, a hieved when m = n.

223Let t =

m

n. Sin e 1 ≤ m, n ≤ 2004, then 1

2004≤ t ≤ 2004. We have

mn

(m + n)2≥ 2004

20052

⇐⇒ t

(t + 1)2≥ 2004

(2004 + 1)2

⇐⇒ 2004t2 + 4008t + 2004 − 20042t − 4008t − t ≤ 0

⇐⇒ 2004t2 − t − 20042t + 2004 ≤ 0

⇐⇒ t(2004t − 1) − 2004(2004t − 1) ≤ 0

⇐⇒ (t − 2004)(2004t − 1) ≤ 0 ,and the last inequality is true sin e 1

2004≤ t ≤ 2004. Hen e the least valueis 2004

20052, a hieved when (m, n) = (2004, 1) or (m, n) = (1, 2004).4. Let k and n be integers with 1 < k < n. If a set of n real numbers hasthe property that the mean value of any k of them is an integer, show thatall n numbers are integers.Solved by Jean-David Houle, student, M Gill University, Montreal, QC; andTitu Zvonaru, Com�ane�sti, Romania. We give Zvonaru's write up.Let A = {a1, a2, . . . , an} be a set of real numbers satisfying thehypotheses. There are integers z2, z2, . . . , zk+1 su h that

a1 + a2 + · · · + ak = kzk+1

a1 + a2 + · · · + ak−1 + ak+1 = kzk... ...a1 + a2 + a4 + · · · + ak + ak+1 = kz3

a1 + a3 + a4 + · · · + ak + ak+1 = kz2Adding these equations we obtainka1 + (k − 1)(a2 + a3 + · · · + ak+1) = k(z2 + z3 + · · · + zk+1) .Now, there is an integer z1 su h that a2 + a3 + · · · + ak+1 = kz1, hen e

a1 = z2 + z3 + · · · + zk+1 − (k − 1)z1 and a1 is an integer.It follows that a1, a2, . . . , an are integers.6. 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 theother n points yellow. Show that there are n segments, ea h with one blueendpoint and one yellow endpoint, su h that ea h of the 2n points is anendpoint of one of the n segments and none of the segments have a point in ommon.

224Solved by Oliver Geupel, Br �uhl, NRW, Germany; and Titu Zvonaru, Com�ane�sti,Romania. We give Geupel's solution.There are �nitely many (namely n!) bije tive orresponden es betweenthe 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 min-imum value. We will prove that ea h su h Γ has the desired property.Assume the ontrary. Then, thereare two blue points, say A and B, andtwo yellow points, say X and Y , su hthat the segments AX and BY aredrawn and interse t in a point, say S.We onstru t another orresponden e Γ′from Γ by removing the segments AXand BY and adding the segments AYand BX [Ed.: note that by the hypothe-ses AY and BX ontain no olouredpoints ex ept their endpoints℄. The or-responden e Γ′ is also bije tive. By thetriangle inequality,

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A (blue) B (blue)S

X (yellow)Y (yellow)

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.

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 alContest 2004/2005 Final Round, given at [2008 : 216℄.1. 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?Solved by George Apostolopoulos, Messolonghi, Gree e; and Titu Zvonaru,Com�ane�sti, Romania. We give Apostolopoulos' version.Sin e ∠O1AO2 = 90◦,half the area of the lens-shaped region of interse tionequals the area of a quar-ter ir le minus the area of

△AO2B. Hen e, the requiredarea is2(

1

4πR2 − 1

2R2)

=π − 2

2·R2.

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A

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O1O2

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

2253. The fun tion f satis�es f(x) + xf(1 − x) = x2 for all real numbers x.Determine the fun tion f.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, Com�ane�sti, Romania. We give Wang's write up.The only su h fun tion is f(x) = −x + 2 +

2x − 2

x2 − 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 havexf(1 − x) + x(1 − x)f(x) = x(1 − x)2 . (2)Subtra ting the identity from (2) yields (x − x2 − 1

)

f(x) = x − 3x2 + x3.Hen e,f(x) =

−x3 + 3x2 − x

x2 − x + 1= −x + 2 +

2x − 2

x2 − x + 1. (3)Conversely, if f(x) is given by (3), then

f(1 − x) = −(1 − x) + 2 +2(1 − x) − 2

(1 − x)2 − (1 − x) + 1

= x + 1 − 2x

x2 − x + 1;

f(x) + xf(x − 1) = −x + 2 + x2 + x − 2x2 − 2x + 2

x2 − x + 1

= x2 + 2 − 2 = x2 ,and the proof is omplete.4. If tan v = 2v and 0 < v < π2, then does it follow that sin v < 20

21?Solved by Mi hel Bataille, Rouen, Fran e; and Pavlos Maragoudakis, Pireas,Gree e. We give Bataille's write up.Yes, it does. To prove this, we introdu e the fun tion φ(x) = tan x−2xfor x ∈

(

0,π

2

), whose derivative is φ′(x) = sec2 x−2. Note that φ(0) = 0,that φ de reases on (0,π

4

) and in reases on (π

4,

π

2

), and that φ(x) → +∞as x → π

2

−. Therefore, φ has a unique root v ∈(

0,π

2

) and φ(x) < 0 forx ∈ (0, v) and φ(x) > 0 for x ∈

(

v,π

2

).This said, we havesin2 v = 1 − 1

1 + tan2 v= 1 − 1

1 + 4v2.

226It follows that sin v <

20

21is equivalent to v <

10√41

, whi h in turn is equiv-alent to φ

(

10√41

)

> 0.Now, 10√41

>5π

12, so that tan

(

10√41

)

> tan(

5π

12

)

= 2 +√

3. Also,20√41

<10

3< 2 +

√3, so we on lude that φ

(

10√41

)

> 0, as desired.5. A square of integer side n, where n ≥ 2, is divided into n2 squares ofside 1. Next, n−1 straight lines are drawn so that the interior of ea h of thesmall squares (the boundary not being in luded in the interior) is interse tedby at least one of the straight lines.(a) Give an example whi h shows that this an be a hieved for some n ≥ 2.(b) Show that among the n − 1 straight lines there are two lines whi hinterse t in the interior of the square of side n.Solution by Oliver Geupel, Br �uhl, NRW, Germany.We use Cartesian oordinates with (0, 0), (n, 0), (n, n), and (0, n) be-ing the orners of the big square Sn. For 0 ≤ i, j < n, let S(i, j) denote theunit square whose southwest orner is the point (i, j).

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Figure BFor 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 (12, 3) and (5

2, 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 squaresS(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 thelines 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 forn = N .

227Assume that m lines ℓ1, ℓ2, . . . , ℓm are drawn so that (a) and (b) holdfor 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 theS(k, k), where p+ q ≥ N −1. Therefore, P ontains a square of side p, andQ ontains a square of side q. By the indu tion hypothesis, at least p andq of the lines ℓk interse t P and Q, respe tively. (This remains true if p orq 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 tingQ. 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 Competition2004-2005 given at [2008 : 216-217℄.1. (a) A positive integer m is triangular if m = 1 + 2 + · · · + n for someinteger 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 integerlengths. The height of the pyramid is also an integer. Show that the volumeof the pyramid is an even integer.Solution by Titu Zvonaru, Com�ane�sti, Romania.(a) We have 1 + 2 + · · · + n =n(n + 1)

2, so all triangular numbersare of this form. Suppose that 8m + 1 is an odd perfe t square. Then theequation n(n + 1)

2= m has a solution if and only if n2 + n − 2m = 0 hasa positive integer root. However, sin e 8m + 1 is an odd perfe t square,

n =−1 +

√1 + 8m

2is a positive integer root of the last equation; hen e mis triangular.(b) Sin e the base of the pyramid is a right-angled triangle with sides ofinteger 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 thepyramid is h, then its volume is V = 13k2mn(m − n)(m + n)h.It suÆ es to show that V ′ = mn(

m2 − n2)

≡ 0 (mod 6) in order toshow that V is even.If m or n is even, then mn is even, and if both m and n are odd, thenm2 − n2 is even. Hen e, V ′ is even.Similarly, if m or n is divisible by 3, then mn is divisible by 3, and ifboth m and n are not divisible by 3, then m2 − n2 is divisible by 3 (sin ex2 ≡ 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.

2282. (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 foundtwo �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. Showthat for any nine points in A, there are at least two points among them su hthat the midpoint of the segment joining the two is also a point in A.Solution by Titu Zvonaru, Com�ane�sti, Romania.(a) Divide the aquarium into 8 ubes ea h 1 metre on a side. There are atleast two �sh in one of the ubes by the Pigeonhole Prin iple, and they areno 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. Themidpoint of the segment joining P1 to P2 is M

(

x1 + y1

2,

x2 + y2

2,

x3 + y3

2

)and M is in the set A if and only if the parities of orresponding oordinatesof P1 and P2 agree.For any nine points in A, by the Pigeonhole Prin iple, at least �vepoints 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 thesethree 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) whosemidpoint is in A.3. (a) Let △ABC be isos eles with AB = AC, and let D be the midpointof BC. The points P and Q lie respe tively on the segments AD and ABsu h that PQ = PC and Q 6= B. Show that ∠PQC = 12∠BAC.(b) Let ABCD be a rhombus with ∠BAD = 60◦. Let F , G, and H bepoints on the segments AD, CA, and DC respe tively su h that DFGH isa 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, Com�ane�sti,Romania. We give the solution of Amengual Covas.(a) Sin e PB = PC, we have PB = PC = PQ. If the base angles inisos eles triangles PQB, PBC, and PCQ are x, y, and z, respe tively, thenin△BCQwe have (x+y)+(y+z)+(z+x) = 180◦, hen e z = 90◦−(x+y).Thus, ∠PQC = z = 90◦ − (x + y) = 90◦ − ∠ABD = ∠BAD = 1

2∠BAC.(b) Sin e GH‖AD, we have ∠CGH = ∠CAD. However, ABCD is arhombus, so ∠CAD = ∠ACD. Therefore, ∠CGH = ∠ACD = ∠GCH,and hen e CH = GH = FD. Sin e △ABD and △BCD are equilat-eral, we also have BC = BD and ∠BCH = ∠BDF . Thus, △BCHand △BDF are ongruent with (i) BH = BF , and (ii) ∠CHB = ∠DFB.By (i), △BFH is isos eles; by (ii) DFBH is a y li quadrilateral, so that

∠FBH = 180◦ − ∠HDF = 180◦ − (60◦ + 60◦) = 60◦. We on lude that△BHF is equilateral, as desired.

229

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Figure B

4. (a) Let a, b, and c be positive real numbers. Prove that(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,Com�ane�sti, Romania. We use Bataille's solution.(a) Equivalently, we show that L = (a + b)2(a + c)2 − 4abc(a + b + c) isnonnegative. We have that (a + b)2(a + c)2 =(

a2 + (ab + bc + ca))2 andalso (ab + bc + ca)2 = a2b2 + b2c2 + c2a2 + 2abc(a + b + c). As a result,

L = a2(

a2 + b2 + c2)

+ b2c2 + 2a2(ab + bc + ca) − 2abc(a + b + c)

= a2(a + b + c)2 − 2abc(a + b + c) + b2c2

=(

a(a + b + c) − bc)2 .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 thehypothesis, we obtain s

(

a2 + b2 + c2)

≥ s(ab + bc + ca) > s2.Sin e s2 = (a + b + c)2 =(

a2 + b2 + c2)

+ (2ab + 2bc + 2ca), addinga 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 solu-tions and generalizations.

230BOOK REVIEWS

Amar SodhiThe M�obius Strip: Dr. August M�obius's Marvelous Band in Mathemati s,Games, Literature, Art, Te hnology, and CosmologyBy Cli�ord Pi kover, Thunder's Mouth Press, New York, 2006ISBN 978-0-7394-7542-3, soft over, 244+xxiii pages, US$15.95Reviewed by Edward J. Barbeau, University of Toronto, Toronto, ONQuoth mother of four year old Pete:\You must not ross M�obius 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 thisbook! Cli�ord A. Pi kover is employed at the IBM Thomas J. WatsonResear h Center in Yorktown, New York. Although his do torate from Yaleis in mole ular biology and bio hemistry, where he studied X-ray s atteringand 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 ludingmathemati s), but also some s ien e � tion. He is the author of over 200papers and the holder of 30 patents \mostly on erned with novel featuresfor 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 sampledand returned to frequently. One idea leads to another, and the text is pun tu-ated 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 toAugust Ferdinand M�obius (1790-1868) himself, who studied astronomy un-der Gauss and later dire ted the Observatory in Leipzig in 1848. We learnabout his times, his distinguished an estry (in luding Martin Luther on hismother's side) and progeny, and see a pi ture of his skull, reprodu ed froma book by his grandson, a neurologist who made a phrenologi al study ofmathemati ians' heads to see if there was a onne tion between bumps andmathemati al ability.The book opens with an introdu tion to the M�obius Strip and its singu-lar 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 be-gin to explore topology, higher dimensions, turning spheres and doughnutsinside out, the Klein Bottle and the Alexander Horned Sphere. However,the name M�obius is familiar to mathemati ians in another ontext, throughhis number theoreti fun tion. This provides the opportunity for a detour toMertens' onje ture, series involving π, and the Riemann zeta fun tion.

231Then we enter the world of the imagination, of literature, art, osmol-ogy, 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 ulationson multiple universes and arti� ial life. In all, this is a ri h and rewardingjourney.

The Contest Problem Book IX: Ameri an Mathemati s Competitions(AMC 12) 2001-2007Compiled and edited by David Wells and J. Douglas Faires,Mathemati al Asso iation of Ameri a, 2008ISBN 978-0-88385-826-4, soft over, 214+xiv pages, US$49.95Reviewed by John Grant M Loughlin, University of New Brunswi k,Frederi ton, NBAnother olle tion of ontest problems has been prepared by the MAA.The high quality of this olle tion of problems, based on the standard grade12 urri ulum, is not surprising as MAA problem books generally feature ahealthy range of problems organized in a \resour eful" manner. This book ispotentially valuable for tea hers working with students in high s hool math ontest venues. The book onsists of 13 ontest papers with 25 questionsea h. The detailed solutions to these 325 problems are a pra ti al asset toany 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 theheading \AMC 12 Problems". All subsequent years (2002-2007) feature twosets 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 topermit two separate o�erings of the same ontest. AMC 12A and AMC 12Bare o�ered just two weeks apart from one another for di�erent students.Hen e, the problem setters are required to ensure that the diÆ ulty leveland the balan e of topi s be onsistent within these two papers while notproviding an unfair advantage to those who write the 12B ontest two weeksafter the 12A ontest has been ompleted. It is noted that al ulus is notrequired to solve any of the problems, as al ulus is not a standard grade 12topi .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 Faireso�er 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 tedhere, a knowledging the tensions between on ise problem statements anddetail, among other issues. The informative prefa e is one of three featuresthat strike me as being ex eptional about this resour e. Familiarity withother publi ations in this series has reinfor ed my appre iation and expe ta-tion for a se ond noteworthy feature: the "Problem DiÆ ulty" summary forea 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

232the 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 hasbeen valuable to me as a mathemati ian working in a Fa ulty of Edu ationwith tea hers at di�erent levels. Finally, the third outstanding feature isthe Index of Problems that identi�es areas, su h as Algebra, with many sub-headings (e.g., AM{GM Inequality; parametri equations of lines; sum andprodu t of roots) ea h ontaining a listing of the spe i� problem numbersthat apply to the topi . In the ase of Geometry, the subtopi s are further ategorized under six broad areas: ir les, oordinates, polygons, quadrilat-erals, solids, and triangles. Any tea her looking for problems on a parti ulartopi an work from here { always a good strategy to onsider.While applauding the merits of the book, it is noted here that its ostseems 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 ProblemsBy Edward Barbeau, Canadian Mathemati al So iety, 2009ISBN 978-0-919558-21-2, soft over, 59+ii pages, US$15.00Reviewed by Amar Sodhi, Sir Wilfred Grenfell College, Corner Brook, NLThe CAUT Bulletin is a monthly newspaper published by the CanadianAsso iation of University Tea hers. This periodi al ontains arti les of in-terest to both tenured and non-tenured fa ulty and, by virtue of having alarge se tion devoted to job listings, is invaluable to anyone who is sear hingfor an a ademi position at a Canadian university. It is therefore a pleasantsurprise to �nd that a paper whi h es hews both a rossword puzzle and aSudoku nevertheless in ludes a mathemati al brainteaser.The \Homework!" is set by Edward Barbeau, a mathemati ian wellknown (among other things) for his work with the Canadian Mathemati alOlympiad. However, it is his ability to onvey mathemati s to a general au-dien e whi h omes to the forefront in The CAUT Problems. This olle tion offorty-two puzzles, whi h have no doubt intrigued many a s holar of the hu-manities or so ial s ien es, are bound to please a person who enjoys ta klinga problem whi h requires thought, but no more than basi skills in algebra.Undoubtedly The CAUT Problems is a valuable resour e for tea herswho 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 littlemagi in the lassroom. As with other volumes in the ATOM series, this bookis 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 bookwhi h a student an enjoy sharing with grandparents during a summer va a-tion. I should give a sample problem here, but my teenage daughter has runo� with the book again!

233PROBLEMSToutes solutions aux probl �emes dans e num�ero doivent nous parvenir au plustard le 1er novembre 2009. Une �etoile (⋆) apr �es le num�ero indique que le probl �emea �et �e soumis sans solution.Chaque probl �eme sera publi �e dans les deux langues oÆ ielles du Canada(anglais et fran� ais). Dans les num�eros 1, 3, 5 et 7, l'anglais pr �e �edera le fran� ais,et dans les num�eros 2, 4, 6 et 8, le fran� ais pr �e �edera l'anglais. Dans la se tion dessolutions, le probl �eme sera publi �e dans la langue de la prin ipale solution pr �esent �ee.La r �eda tion souhaite remer ier Jean-Mar Terrier, de l'Universit �e deMontr �eal, d'avoir traduit les probl �emes.

3419. Corre tion. Propos �e par Vo Quo Ba Can, Universit �e de M�ede ineet Pharma ie de Can Tho, Can Tho, Vietnam.Soit a, b et c trois nombres r �eels positifs.(a) Montrer que ∑ y lique√a2 + 4bc

b2 + c2≥ 2 +

√2.

(b) Montrer que ∑ y lique 3

√

a2 + bc

b2 + c2≥ 2 +

13√

2.

3424. Corre tion. Propos �e par Yakub N. Aliyev, Universit �e d'Etat de Ba-kou, Bakou, Azerba��djan.Pour un entier positif m, soit σ la permutation de {0, 1, 2, . . . , 2m}d �e�nie par σ(2i) = i pour i = 0, 1, 2, . . . , m et σ(2i − 1) = m + i pouri = 1, 2, . . . , m. Montrer qu'il existe un entier positif k tel que σk = σ et1 < k ≤ 2m + 1.3439. Propos �e par D.J. Smeenk, Zaltbommel, Pays-Bas.On suppose que dans le triangle ABC la hauteur ha = AD est �egale �aa = BC. Soit H l'ortho entre, M le point milieu de BC et E le point milieude AD. Montrer que HM = HE. Quand la r �e iproque est-elle valide ?3440. Propos �e par Hidetoshi Fukugawa, Kani, Gifu, Japon.Sur une table, on a N pi �e es de monnaie, toutes de meme dimension.Ces N pi �e es de monnaie peuvent etre arrang �ees aussi bien en un arr �e qu'enun triangle �equilat �eral. Trouver N .3441⋆. Propos �e par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie.Soit ABCD un quadrilat �ere onvexe et P un point dans l'int �erieur deABCD tel que PA =

AB√2, PB =

BC√2, PC =

CD√2, et PD =

DA√2. Con�r-mer ou in�rmer que ABCD est un arr �e.

2343442. Propos �e par Iyoung Mi helle Jung, �etudiante, Coll �ege de Langues�Etrang �eres Hanyoung, S �eoul, Cor �ee du Sud et Sung Soo Kim, Universit �eHanyang, S �eoul, Cor �ee du Sud.Soit C un one ir ulaire droit et D un disque de rayon �x �e situ �e dans labase du one C. Montrer que l'aire A de la partie du one situ �ee dire tementau-dessus de D est ind �ependante de la position du disque D.3443. Propos �e par Cao Minh Quang, Coll �ege Nguyen Binh Khiem, VinhLong, Vietnam.Soit a, b et c trois nombres r �eels positifs tels que a+b+c = 3. Montrerque

∑ y lique a2(b + 1)

a + b + ab≥ 2 .

3444. Propos �e par Cao Minh Quang, Coll �ege Nguyen Binh Khiem, VinhLong, Vietnam.Soit a, b et c trois nombres r �eels positifs tels que a+b+c = 1. Montrerque∑ y lique ab

3a2 + 2b + 3≤ 1

12.

3445. Propos �e par �Sefket Arslanagi � , Universit �e de Sarajevo, Sarajevo,Bosnie et Herz �egovine, �a la m�emoire de Murray S. Klamkin.Soit a, b et c trois nombres r �eels non n �egatifs tels que ab+bc+ac = 1.Montrer que(a) ∑ y li a

1 + bc≥ 3

√3

4; (b) ∑ y li a2

1 + a≥

√3

√3 + 1

.3446. Proposed by Mih�aly Ben ze, Brasov, Roumanie.Pour tout entier positif n montrer que⌊

√

n2 − n + 1 +√

n2 + n + 1⌋

+⌊

√

n2 + n +√

n2 + 3n + 2⌋

=⌊

√

4n2 + 3⌋

+⌊

√

4n2 + 8n + 3⌋ ,o �u ⌊x⌋ d �esigne le plus grand entier ne d �epassant pas x.3447. Propos �e par Mih�aly Ben ze, Brasov, Roumanie.Soit n un nombre entier positif. Montrer que

2

n!(n + 2)!<

n∏

k=1

(

k+1

√

k + 1

k− 1

)

<1

(n + 1)(n!)2.

2353448. Propos �e par Jos �e Luis D��az-Barrero and Miquel Grau-S �an hez, Uni-versit �e 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

a2Fn + b2Fn+1 + c2Fn+2 ≥ 4S

(

n+2∑

k=1

F 2k − F 2

n+1

)1/2

pour tout triangle ABC dont a, b, c sont les longueurs respe tives des ot �eset S sa surfa e.3449. Propos �e par un proposeur anonyme.Soit ABCD un arr �e de ot �e unit �e, M et N les milieux respe tifs deAB et CD. Y a-t-il un point P sur MN tel que les longueurs AP et PCsont toutes les deux des nombres rationnels ?3450. Propos �e par Dragoljub Milo�sevi� , Gornji Milanova , Serbie.Soit ABC un triangle ave r omme rayon de son er le ins rit, ra, rbet rc les rayons respe tifs de ses er les exins rits, et ha, hb, hc ses hauteursrespe tives. Montrer que

ha + 2ra

r + ra

+hb + 2rb

r + rb

+hc + 2rc

r + rc

≥ 27

4.

.................................................................3419. Corre tion. Proposed by Vo Quo Ba Can, Can Tho University ofMedi ine and Pharma y, Can Tho, Vietnam.Let a, b, and c be positive real numbers.(a) Prove that ∑ y li √a2 + 4bc

b2 + c2≥ 2 +

√2.

(b) Prove that ∑ y li 3

√

a2 + bc

b2 + c2≥ 2 +

13√

2.

3424. 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 forea h i = 1, 2, . . . , m. Prove that there exists a positive integer k su h thatσk = σ and 1 < k ≤ 2m + 1.

2363439. 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 midpointof AD. Prove that HM = HE. When does the onverse hold?3440. Proposed by Hidetoshi Fukugawa, Kani, Gifu, Japan.There are N oins on a table all of the same size. These N oins anbe arranged in a square and they an also be arranged into an equilateraltriangle. Find N .3441⋆. Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.LetABCD be a onvex quadrilateral and letP be a point in the interiorof ABCD su h that PA =

AB√2, PB =

BC√2, PC =

CD√2, and PD =

DA√2.Prove or disprove that ABCD is a square.3442. Proposed by Iyoung Mi helle Jung, student, Hanyoung ForeignLanguage High S hool, Seoul, South Korea and Sung Soo Kim, Hanyang Uni-versity, Seoul, South Korea.Let C be a right ir ular one and let D be a disk of �xed radius lyingwithin the base of the one C. Prove that if A is the area of that part ofthe one lying dire tly above D, then A is independent of the position of thedisk D.3443. 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. Provethat

∑ y li a2(b + 1)

a + b + ab≥ 2 .

3444. 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. Provethat∑ y li ab

3a2 + 2b + 3≤ 1

12.

3445. 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(a) ∑ y li a

1 + bc≥ 3

√3

4; (b) ∑ y li a2

1 + a≥

√3

√3 + 1

.

2373446. Proposed by Mih�aly Ben ze, Brasov, Romania.For any positive integer n prove that⌊

√

n2 − n + 1 +√

n2 + n + 1⌋

+⌊

√

n2 + n +√

n2 + 3n + 2⌋

=⌊

√

4n2 + 3⌋

+⌊

√

4n2 + 8n + 3⌋ ,where ⌊x⌋ denotes the greatest integer not ex eeding x.3447. Proposed by Mih�aly Ben ze, Brasov, Romania.Let n be a positive integer. Prove that

2

n!(n + 2)!<

n∏

k=1

(

k+1

√

k + 1

k− 1

)

<1

(n + 1)(n!)2.

3448. Proposed by Jos �e Luis D��az-Barrero and Miquel Grau-S �an hez,Universitat Polit �e ni a de Catalunya, Bar elona, Spain.Let Fn be the nth Fibona i number, that is, F0 = 0, F1 = 1, andFn = Fn−1 + Fn−2 for n ≥ 2. Prove that

a2Fn + b2Fn+1 + c2Fn+2 ≥ 4S

(

n+2∑

k=1

F 2k − F 2

n+1

)1/2

holds for any triangle ABC, where a, b, c, and S are the side lengths andarea of the triangle, respe tively.3449. Proposed by an anonymous proposer.Let ABCD be a unit square, M the midpoint of AB, and N the mid-point of CD. Is there a point P on MN su h that the lengths of AP andPC are both rational numbers?3450. Proposed by Dragoljub Milo�sevi� , Gornji Milanova , Serbia.Let △ABC have inradius r, exradii ra, rb, rc, and altitudes ha, hb, hc.Prove that

ha + 2ra

r + ra

+hb + 2rb

r + rb

+hc + 2rc

r + rc

≥ 27

4.

238SOLUTIONSAu un probl �eme n'est immuable. L' �editeur est toujours heureux d'en-visager la publi ation de nouvelles solutions ou de nouvelles perspe tivesportant sur des probl �emes ant �erieurs.

3330. [2008 : 171, 174℄ Proposed by Arkady Alt, San Jose, CA, USA.Letn be a natural number, let r be a real number, and let a1, a2, . . . , anbe positive real numbers satisfying n∏

k=1

ak = rn; prove thatn∑

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, Br �uhl, NRW, Germany, solutionto part (d) by the proposer.(a) The statement is not orre t in the stri t sense, be ause for ea h r > 0the inequality is satis�ed by a1 = a2 = r (and similarly for part (d)). Weprove instead that for r > 0, the inequality

1

(1 + a)3+

1

(1 + b)3≥ 2

(1 + r)3, (1)holds for all positive real numbers a and b satisfying ab = r2 if and only if

r ≥ 1

3.If ab ≥ 1

9, then the inequality (1) follows from the result given in CRUXwith Mayhem, problem 3319 (solution at [2009 : 121-122℄).Conversely, suppose that r <

1

3. Let f : [0, ∞) → R be given by

f(x) =1

(1 + x)3+

x3

(

x + r2)3 .

We have f ′′(r) =6(3r − 1)

r(1 + r)5< 0 and f ′′ is ontinuous. It follows that thereexists an x0 > 0 su h that f(x0) < f(r) =

2

(1 + r)3. We on lude that

239a = x0 and b =

r2

aviolate the inequality (1). This ompletes the proof ofpart (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 =

√a1a2. Then x ≥ r, and bypart (a) we have

1

(1 + a1)3+

1

(1 + a2)3≥ 2

(1 + x)3.It therefore suÆ es to show that

2

(1 + x)3+

x6

(

x2 + r3)3 ≥ 3

(1 + r)3.Clearing denominators and rearranging terms in this last inequality, we �ndthat it is equivalent to

(x − r)27∑

k=0

pk(r)xk ≥ 0 ,where

p0(r) = r7(

2r3 + 6r2 + 6r − 1)

p1(r) = r6(

4r3 + 12r2 + 3r − 2)

p2(r) = r4(

6r4 + 15r3 + 18r2 + 15r − 3)

p3(r) = r3(

5r4 + 18r3 + 33r2 + 5r − 6)

p4(r) = r(

4r5 + 21r4 + 27r3 + 13r2 + 9r − 3)

p5(r) = r(

3r4 + 15r3 + 21r2 + 21r − 3)

− 6

p6(r) = 2r4 + 9r3 + 15r2 + 5r − 6

p7(r) = r3 + 3r2 + 3r − 2It 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, thepolynomials 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 lossof generality, let a4 ≤ r and put x = 3√

a1a2a3. Then x ≥ r, and by part (b)we have1

(1 + a1)3+

1

(1 + a2)3+

1

(1 + a3)3≥ 3

(1 + x)3.It therefore suÆ es to show that

3

(1 + x)3+

x9

(

x3 + r4)3 ≥ 4

(1 + r)3.

240Clearing denominators and rearranging terms in this last inequality, we �ndthat it is equivalent to

(x − r)210∑

k=0

qk(r)xk ≥ 0 ,

whereq0(r) = r10

(

3r3 + 9r2 + 9r − 1)

q1(r) = r9(

6r3 + 18r2 + 6r − 2)

q2(r) = r8(

9r3 + 15r2 + 3r − 3)

q3(r) = r6(

8r4 + 21r3 + 27r2 + 23r − 3)

q4(r) = r5(

7r4 + 27r3 + 51r2 + 13r − 6)

q5(r) = r4(

6r4 + 33r3 + 39r2 + 3r − 9)

q6(r) = r2(

5r5 + 27r4 + 36r3 + 20r2 + 15r − 3)

q7(r) = r(

4r5 + 21r4 + 33r3 + 37r2 + 3r − 6)

q8(r) = r(

3r4 + 15r3 + 30r2 + 18r − 9)

− 9

q9(r) = 2r4 + 9r3 + 15r2 + 3r − 9

q10(r) = r3 + 3r2 + 3r − 3It 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, thepolynomials 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, . . . , ansubje t to the given onstraint. Let x be a positive real number, let ai = xfor i = 1, 2, . . . , n − 1, and let an =rn

xn−1. Then

n − 1

(1 + x)3+

x3n−3

(

xn−1 + rn)3 ≥ n

(1 + r)3holds for all x > 0. Taking the limit as x → ∞ yields n

(1 + r)3≤ 1, hen e

r ≥ 3√

n − 1.Conversely, suppose that r ≥ 3√

n − 1. We will prove by Mathemati alIndu tion that if n ≥ 4, r ≥ max{

13√

4, 3√

n − 1}, and a1, a2, . . . , an satisfythe given onstraint, then n

∑

k=1

1

(1 + ak)3≥ n

(1 + r)3.Note that the statement is true for n = 4 by part ( ).Now suppose that the statement is true for some n ≥ 4 and that

r ≥ max{

13√

4, 3√

n + 1 − 1}

= 3√

n + 1 − 1, and let a1, a2, . . . , an+1be positive real numbers su h that a1a2 · · · an+1 = rn+1. By symmetry,

241we may assume that a1 ≥ a2 ≥ · · · ≥ an+1. Let x = n

√a1a2 · · · an, then

x ≥ an+1 =rn+1

xnand xn+1 ≥ rn+1, so that x ≥ r ≥ 3

√n + 1 − 1 > 1

3√4.By indu tion, we have n

∑

k=1

1

(1 + ak)3≥ n

(1 + x)3, hen e

n∑

k=1

1

(1 + ak)3≥ n

(1 + x)3+

x3n

(

xn + rn+1)3 .

Let h(x) be the fun tion of x on the right side of the above inequality forx ≥ r. After some (tedious) al ulations we �nd that

h′(x) =3n(

xn+1 − rn+1)

P (x)

(1 + x)4(

xn + rn+1)4 ;

P (x) = 6x2nrn+1 + 4x2n+1rn+1 + 4xnr2n+2

+ x2n+2rn+1 + xn+1r2n+2 + r3n+3 − x3n−1 .Now P (r) = r3n−1(r + 1)3(3r − 1) > 0, sin e r > 13√

4>

1

3, 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,the fun tion Cn

xn+

Cn−1

xn−1+ · · · + C1

x+ C0 is de reasing, and P (x)

x3n−1is of thisform.℄ 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,min

x∈[r,x0]h(x) = h(r) =

n

(1 + r)3+

r3n

(

rn + rn+1)3 =

n + 1

(1 + r)3and for any x ∈ [x0, ∞) we haveh(x) > lim

x→∞h(x) = 1 ≥ n + 1

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

Therefore, the minimum value of h(x) on [r, ∞) is h(r) =n + 1

(1 + r)3, whi h ompletes the indu tion step and the proof.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 valuesof r for whi h parts (b) and ( ) hold.

3338. [2008 : 239, 242℄ Proposed by Toshio Seimiya, Kawasaki, Japan.A onvex y li quadrilateral ABCD has an in ir le with entre I. LetP be the interse tion of AC and BD. Prove that AP : CP = AI2 : CI2.

242Solution 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

CP=

1 + cos θ

1 − cos θ= cot2

θ

2. (1)On the other hand, if r denotes the radius of the in ir le, then

AI =r

sin θ2

and CI =r

sin(

90◦ − θ2

) =r

cos θ2

,so that

AI2

CI2=

cos2 θ2

sin2 θ2

= cot2θ

2. (2)The desired result follows from (1) and (2).Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia andHerzegovina; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam;OLIVER GEUPEL, Br �uhl, NRW, Germany; JOHN G. HEUVER, Grande Prairie, AB; TAICHIMAEKAWA, Takatsuki City, Osaka, Japan; SALEM MALIKI �C, student, Sarajevo College,Sarajevo, Bosnia and Herzegovina; MADHAV R. MODAK, formerly of Sir ParashurambhauCollege, Pune, India; SON HONG TA, Hanoi, Vietnam; PETER Y. WOO, Biola University, LaMirada, CA, USA; and the proposer.

3339. [2008 : 239, 242℄ Proposed by Toshio Seimiya, Kawasaki, Japan.Let Γ1 and Γ2 be two noninterse ting ir les ea h lying in the exteriorof 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 Γ2at 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 NA and NB with Γ1and Γ2, respe tively, di�erent from A and B. Prove that CP , DQ, and MNare on urrent.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 lesthat have ommon external tangents, whether or not they interse t. Let C′and D′ be the points where CP and DQ meet MN . We are to prove thatC′ = D′. Be ause of the symmetry about the line of entres of the two ir les, AMNC and BMND are isos eles trapezoids; thus, AC||MN and

243MN ||BD and ∠AMN = ∠MNC = ∠C′NC. Moreover, be ause NC istangent to Γ1 while CP is a hord, we have

∠C′CN = ∠PCN = ∠CAP = ∠CAN = ∠ANM ,when e △CNC′ ∼ △NMA. Consequently, CN

NM=

NC ′

MA, or

NC′ =CN · MA

NM.By the analogous argument the triangles DND′ and NMB are similar, sothat

ND′ =DN · MB

NM.Be ause CN = DN and MA = MB, it follows that NC′ = ND′; further-more, sin e C′ and D′ lie on the same side of N on the line MN , the points

C′ and D′ oin ide, as desired.II. Solution by Daniel Reisz, Auxerre, Fran e.La droiteMN est l'axe radi al|lieu des points ayantmeme puissan e|des deux er les Γ1 et Γ2. Il suÆt don de montrer que le point K in-terse tion des droites CP et DQ a meme puissan e par rapport aux deux er les pour pouvoir on lure que K se situe aussi sur la droite MN . DeNP · NA = NQ · NB, 'est �a dire de NP

NQ=

NB

NA, on d �eduit la similitudedes triangles NPQ et NBA, et don l' �egalit �e d'angles

∠NPQ = ∠NBA = ∠BDQ = ∠BDK .Montrons maintenant que les triangles KPQ et KDC sont semblables. Ilsont d �ej �a un angle ommun en K. Par ailleurs : (i) ∠BDC et ∠DCA sontsuppl �ementaires ( �a ause de AC||BD), (ii) ∠DCA et ∠CPA sont suppl �e-mentaires (CN est la tangente �a Γ1 en C, CA est une orde) et (iii) ∠CPA =∠KPN .Don , ∠BDC = ∠KPN . Or on sait d �ej �a que ∠NPQ = ∠NBA = ∠BDQ,don par soustra tion, ∠KPQ = ∠KDC d'o �u la similitude de nos deuxtriangles, e qui permet d' �e rire KP

KQ=

KD

KC; 'est �a dire

KP · KC = KD · KQ .Le point K a don meme puissan e par rapport aux deux er les et se trouvedon sur la droite MN .Also solved by MICHEL BATAILLE, Rouen, Fran e; FRANCISCO BELLOT ROSADO,I.B. Emilio Ferrari, Valladolid, Spain; SALEM MALIKI �C, student, Sarajevo College, Sarajevo,Bosnia and Herzegovina; PETER Y. WOO, Biola University, La Mirada, CA, USA; and theproposer.

2443340. [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 ondpoint D. Suppose that AB2 + AC2 = 2AD2. Prove that the angle ofinterse tion of AD and BC is 45◦.Similar solutions by D.J. Smeenk, Zaltbommel, the Netherlands and PeterY. Woo, Biola University, La Mirada, CA, USA.Let α = ∠CAB, β = ∠ABC,γ = ∠BCA, and let R be the ir umradius. Using the Lawof Sines for trianglesABC andABD, we have

AB

sin γ=

AC

sin β

=AD

sin(

α2

+ β) = 2R ,so that we an rewrite the on-dition AB2+AC2 = 2AD2 as

sin2 γ+sin2 β = 2 sin2(

α

2+ β

) .

..

..

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A

BC

D

E

βα2

α2 α

2

γ

This yields su essivelycos 2β + cos 2γ = 2 cos(α + 2β) ,cos 2β + cos 2γ = −2 cos(β − γ) ,

cos(β + γ) cos(β − γ) = − cos(β − γ) ,cos(β − γ)

[

cos(β + γ) + 1]

= 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 = −π

2. It follows that

α + 2β =π

2, sin e α + β + γ = π, and therefore, ∠AEC =

α

2+ β =

π

4, as laimed.Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallor a, Spain; GEORGEAPOSTOLOPOULOS, Messolonghi, Gree e; �SEFKET ARSLANAGI �C, University of Sarajevo,Sarajevo, Bosnia and Herzegovina; ROY BARBARA, Lebanese University, Fanar,Lebanon; MICHEL BATAILLE, Rouen, Fran e; IAN JUNE L. GARCES, Ateneo de ManilaUniversity, Quezon City, The Philippines; OLIVER GEUPEL, Br �uhl, NRW, Germany; KEE-WAILAU, Hong Kong, China; TAICHI MAEKAWA, Takatsuki City, Osaka, Japan; SALEM MALIKI �C,student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; MADHAV R. MODAK, formerlyof Sir Parashurambhau College, Pune, India; MICHAEL PARMENTER, Memorial Universityof Newfoundland, St. John's, NL; BOB SERKEY, Leonia, NJ, USA; SON HONG TA, Hanoi,Vietnam; TITU ZVONARU, Com�ane�sti, Romania; and the proposer. There were two in orre tsolutions submitted.

2453341. [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 esfrom the in entre of △ABC to the verti es A, B, and C, respe tively.Solution by George Apostolopoulos, Messolonghi, Gree e.Let s be the semiperimeter of triangle ABC. We haveRa =

s − a

cos A2

=s − a√

s(s−a)bc

=

√bc

√s − a

√s

=√

bc

√

1 − a

s.

Similarly,Rb =

√ca

√

1 − b

sand Rc =

√ab

√

1 − c

s.Using the Cau hy{S hwarz Inequality, we obtain

(Ra + Rb + Rc)2 =

(

√bc

√

1 − a

s+

√ca

√

1 − b

s+

√ab

√

1 − c

s

)2

≤ (bc + ca + ab)

(

1 − a

s+ 1 − b

s+ 1 − c

s

)

= (ab + bc + ca)

(

3 − 2s

s

)

= ab + bc + ca ,or √3(Ra + Rb + Rc) ≤

√

3(ab + bc + ca) .It suÆ es to show that√

3(ab + bc + ca) ≤ a + b + c .The last inequality is equivalent to a2 + b2 + c2 ≥ ab + bc + ca, whi h iswell known and easy to prove. This ompletes the solution.Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia andHerzegovina; 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,Br �uhl, NRW, Germany; JOE HOWARD, Portales, NM, USA; WALTHER JANOUS, Ursulinen-gymnasium, Innsbru k, Austria; THANOS MAGKOS, 3rd High S hool of Kozani, Kozani, Gree e;SALEM MALIKI �C, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; MADHAVR. MODAK, formerly of Sir Parashurambhau College, Pune, India; PETER Y. WOO, BiolaUniversity, La Mirada, CA, USA; TITU ZVONARU, Com�ane�sti, Romania; and the proposer.

2463342. [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

2∑ y li sin A

2sin

B

2≤ 1 +

r

R.

Similar solutions by CaoMinh Quang, Nguyen Binh KhiemHigh S hool, VinhLong, Vietnam; Oliver Geupel, Br �uhl, NRW, Germany; and Dung NguyenManh, High S hool of HUS, Hanoi, Vietnam.It is well known that cos A + cos B + cos C = 1 +r

R. Thus, it suÆ esto show that

2∑ y li sin A

2sin

B

2≤

∑ y li cos A .For positive x we have 2 ≤ x +

1

x. Taking x =

cos(A/2)

cos(B/2)and multiplying by

sinA

2sin

B

2, we obtain2 sin

A

2sin

B

2≤ 1

2

(

sin A tanB

2+ sin B tan

A

2

) ,hen e,2∑ y li sin A

2sin

B

2≤ 1

2

∑ y li tan A

2(sin B + sin C) .By sum to produ t formulas, we have

1

2

∑ y li tan A

2(sin B + sin C)

=∑ y li sin A

2

cos A2

sinB + C

2cos

B − C

2

=∑ y li cos B + C

2cos

B − C

2=

∑ y li cos A ,as desired.Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; �SEFKETARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; MICHEL BATAILLE,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 MALIKI �C,student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; JUAN-BOSCO ROMEROM�ARQUEZ, Universidad de Valladolid, Valladolid, Spain; XAVIER ROS, student, UniversitatPolit �e ni a de Catalunya, Bar elona, Spain; and the proposer. There was one in ompletesolution submitted.

2473343. [2008 : 240, 242℄ Proposed by Stan Wagon, Ma alester College,St. Paul, MN, USA.If the fa torials are deleted in theMa laurin series for sin x, one obtainsthe series for arctan x. Suppose instead that one alternates fa torials in theseries. Does the resulting series have a losed form? That is, an one �nd anelementary expression for the fun tion whose Ma laurin series is

x − x3

3+

x5

5!− x7

7+

x9

9!− x11

11+ · · · ?Essentially the same solution by Mi hel Bataille, Rouen, Fran e; Ri hardI. Hess, Ran ho Palos Verdes, CA, USA; V�a lav Kone� n�y, Big Rapids, MI,USA; Ralph Lozano, student, Missouri State University, Missouri, USA;Xavier Ros, student, Universitat Polit �e ni a de Catalunya, Bar elona, Spain;Digby Smith, Mount Royal College, Calgary, AB; and Peter Y. Woo, BiolaUniversity, La Mirada, CA, USA.Let f(x) denote the given series. The following Ma laurin series areall well known:

sin x = x − x3

3!+

x5

5!− x7

7!+ · · · , − ∞ < x < ∞ ; (1)

sinh x = x +x3

3!+

x5

5!+

x7

7!+ · · · , − ∞ < x < ∞ ; (2)

tan−1 x = x − x3

3+

x5

5− x7

7+ · · · , − 1 < x < 1 ; (3)

tanh−1 x = x +x3

3+

x5

5+

x7

7+ · · · , − 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

1

2

(

tan−1 x − tanh−1 x)

= −(

x3

3+

x7

7+

x11

11+ · · ·

) . (6)From (5) and (6) we on lude thatf(x) =

1

2

(

sin x + sinh x + tan−1 x − tanh−1 x) , − 1 < x < 1 .Also solved by CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA;OLIVER GEUPEL, Br �uhl, 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, SUNYOswego, Oswego, NY, USA; and the proposer.With the ex eption of the proposer, all of these solvers gave the same answer; namely

12

[

sin x+ sinh x+ tan−1 x− 12

ln( 1+x1−x

)] whi h is, of ourse, equivalent to the answer givenabove.

2483344. [2008 : 240, 242℄ Proposed by Hung Pham Kim, student, StanfordUniversity, Palo Alto, CA, USA.Let n be a positive integer, n ≥ 4, and let a1, a2, . . . , an be positivereal numbers su h that a1 + a2 + · · · + an = n. Prove that

1

a1

+1

a2

+ · · · +1

an

− n ≥ 3

n

(

a21 + a2

2 + · · · + a2n − n

) .Solution by Oliver Geupel, Br �uhl, NRW, Germany.We will use the following theorem.Theorem (Right Convex Fun tion Theorem) Let f(u) be a fun tion on an in-terval 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, thenf(x1) + f(x2) + · · · + f(xn) ≥ nf

(

x1 + x2 + · · · + xn

n

)

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

n≥ s.The desired inequality may be rewritten as

f(−a1) + f(−a2) + · · · + f(−an) ≥ nf

(

−a1 + a2 + · · · + an

n

) ,where f(u) = −3u2 − n

ufor −n < u < 0. Sin e f ′′(u) = −2n

u3− 6 > 0 for

u ≥ −1, the fun tion f is onvex for u ≥ s = −1. It suÆ es to prove thatf(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)y+ (n − 1)

(

−3y2 − n

y

)

≥ n(n − 3)for −1 ≤ y < 0. After learing denominators and rearranging terms, itbe omes n(n − 1)(y + 1)2[

(3(n − 1)y2 + 3ny + n]

≥ 0, or equivalentlyn(n − 1)(y + 1)2

[

(

y +n

2(n − 1)

)2

+n(n − 4)

12(n − 1)2

]

≥ 0 ,whi h is learly true.Equality holds if and only if a1 = a2 = · · · = an = 1.Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; CHIP CURTIS,Missouri Southern State University, Joplin, MO, USA; WALTHER JANOUS, Ursulinengymnas-ium, Innsbru k, Austria; SALEM MALIKI �C, student, Sarajevo College, Sarajevo, Bosnia andHerzegovina; DUNG NGUYEN MANH, High S hool of HUS, Hanoi, Vietnam; PETER Y. WOO,Biola University, La Mirada, CA, USA; and the proposerFor further appli ations of the Right Convex Fun tion Theorem, the interested reader isreferred to the arti le "The Proof of Three Open Inequalities" by V. C�rtoaje, [2008 : 231-238℄.

2493345. [2008 : 240, 243℄ Proposed by Hung Pham Kim, student, StanfordUniversity, Palo Alto, CA, USA.Let a, b, c, and d be positive real numbers su h that a + b+ c + d = 4.Prove that

a

1 + b2c+

b

1 + c2d+

c

1 + d2a+

d

1 + a2b≥ 2 .Solution by Dung Nguyen Manh, High S hool of HUS, Hanoi, Vietnam.By the AM{GM Inequality, we have

a

1 + b2c= a − ab2c

1 + b2c≥ a − ab2c

2b√

c= a − ab

√c

2

= a − b√

a · ac

2≥ a − b(a + ac)

4.Taking the y li sum a ross the inequality derived above and using the re-lation a + b + c + d = 4, we obtain

∑ y li a

1 + b2c≥ 4 − 1

4

∑ y li ab − 1

4

∑ y li abc . (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)

4+

(c + d)2(a + b)

4

=(a + b)(c + d)

4(a + b + c + d)

= (a + b)(c + d) ≤ (a + b + c + d)2

4= 4 . (3)From inequalities (1), (2), and (3) we then have

∑ y li a

1 + b2c≥ 4 − 1

4(4 + 4) = 2 .Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; �SEFKETARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; MICHEL BATAILLE,Rouen, Fran e; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam;JOE HOWARD, Portales, NM, USA; SALEM MALIKI �C, student, Sarajevo College, Sarajevo,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.

2503346. [2008 : 240, 243℄ Proposed by Bin Zhao, student, YunYuanHuaZhongUniversity of Te hnology and S ien e, Wuhan, Hubei, China.Given triangle ABC, prove that

π∑ y li 1

A≥

∑ y li sin A

2

∑ y li csc A

2

.Solution by Mi hel Bataille, Rouen, Fran e.Note that π = A + B + C and set x =

A

2, y =

B

2, and z =

C

2(so that

x, y, z ∈(

0, π2

)), the inequality is then rewritten as(x+y+z)

(

1

x+

1

y+

1

z

)

≥ (sin x+sin y+sin z)

(

1

sin x+

1

sin y+

1

sin z

)

or(

x

y− sin x

sin y+

y

x− sin y

sin x

)

+

(

y

z− sin y

sin z+

z

y− sin z

sin y

)

+

(

z

x− sin z

sin x+

x

z− sin x

sin z

)

≥ 0 ,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) =

sin u

uand g(u) = u sin u.Now, f is de reasing on (0,

π

2

) and g is in reasing on (0,π

2

). Thus,(

f(u) − f(v))(

g(v) − g(u))

≥ 0 whenever u, v ∈(

0,π

2

) and so the left sideof (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.3347. [2008 : 241, 243℄ Proposed by Mih�aly Ben ze, Brasov, Romania.Let A1A2A3A4 be a onvex quadrilateral. Let Bi be a point on AiAi+1for i ∈ {1, 2, 3, 4}, where the subs ripts are taken modulo 4, su h that

B1A1

B1A2

=B3A4

B3A3

=A1A4

A2A3

and B2A2

B2A3

=B4A1

B4A4

=A1A2

A3A4

.Prove that B1B3 ⊥ B2B4 if and only if A1A2A3A4 is a y li quadrilateral.

251Solution by Oliver Geupel, Br �uhl, NRW, Germany.If A1A2 6 ‖ A3A4, then let C = A1A2 ∩ A3A4 and v be the bise tor of∠A1CA4; otherwise, if A1A2 ‖ A3A4 de�ne v to be the midline between thetwo parallel lines. Similarly, if A1A4 6 ‖ A2A3, then let D = A2A3 ∩ A1A4and w be the bise tor of ∠A3DA4; otherwise, if A1A4 ‖ A2A3 de�ne w tobe their midline. The desired result follows readily from the following twolemmas:Lemma 1. The quadrilateral A1A2A3A4 is y li if and only if v ⊥ w.Lemma 2. It is the ase that B2B4 ‖ v and B1B3 ‖ w.Proof of Lemma 1. Assume that the �gure has been labeled so that C andA2A3 are on opposite sides of line A1A4 while D and A1A2 are on opposite

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A4

C

DE

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v

w qq

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A1

A2

A3

A4

C

DGw

B3B1 H I

qq

qq..................................................................................................................................................................................................................................................

sides of A3A4; denote the angles at the verti es Ai by αi, and de�neE = v ∩ w , F = v ∩ A1A4 , and G = w ∩ A3A4 .We will �rst prove that ∠(v, A1A4) = 90◦ +

α1 − α4

2. If A1A2 ‖ A3A4 then

α4 = 180◦ − α1; therefore ∠EFA4 = α1 = 90◦ +α1 − α4

2, as laimed. If

A1A2 6 ‖ A3A4, note that ∠CA1A4 = 180◦ −α1, ∠CA4F = 180◦ −α4, and∠A1CE = ∠A4CF =

α1 + α4

2− 90◦. Hen e,

∠A4FE = ∠A4CF + ∠CA4F =

(

α1 + α4

2− 90◦

)

+ (180◦ − α4)

= 90◦ +α1 − α4

2,

whi h is the desired angle between v and A1A4. Similarly, we have that∠A4GE = 90◦ +

α3 − α4

2.

252Inspe ting the angles in quadrilateral EGA4F , we see that

∠(v, w) = ∠FEG

= 360◦ −(

90◦ +α1 − α4

2

)

− α4 −(

90◦ +α3 − α4

2

)

= 180◦ − α1 + α3

2.Hen e, ∠(v, w) = 90◦ is equivalent to α1 + α3 = 180◦ whi h, in turn, isequivalent to A1A2A3A4 being y li .Proof of Lemma 2. If A1A4 ‖ A2A3 then, be ause parallels ut proportionalsegments from transversals, B1B3 ‖ A1A4 ‖ w. It remains to onsider the ase A1A4 6 ‖ A2A3. Let H and I be the points where the line B1B3 meets

A1A4 and A2A3, respe tively. We now apply the hypothesis and Menelaus'theorem to triangles A1A2D and A3A4D and transversal B1B3 to obtainA1A4 · IA2 · HD

A2A3 · ID · HA1

=B1A1 · IA2 · HD

B1A2 · ID · HA1

= 1 =B3A4 · IA3 · HD

B3A3 · ID · HA4

=A1A4 · IA3 · HD

A2A3 · ID · HA4

. (1)By omparing the �rst and last fra tions in (1), we see thatIA3

HA4

=IA2

HA1

=IA3 + A3A2

HA4 + A4A1

;when e, A3A2

A4A1

=IA3

HA4

. We substitute this last fra tion into the �nal fra tionappearing in (1) and dedu e that HD

ID= 1. That is, HD = ID. It followsthat

∠GDH =∠A2DH

2=

∠DIH + ∠DHI

2= ∠DHI .Consequently, w = GD ‖ HI = B1B3. Analogously, v ‖ B2B4.Also solved by MICHEL BATAILLE, Rouen, Fran e; KHANH BAO NGUYEN, High S hoolfor Gifted Students, Hanoi University of Edu ation, Hanoi, Vietnam; and PETER Y. WOO, BiolaUniversity, 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 appearedmore 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 theassumption that the given quadrilateral is y li , the diagonals of the quadrilateral divide itsinterior into two pairs of similar triangles, and the lines B1B3 and B2B4 are the bise tors ofthe angles formed by those diagonals. It seems as if the onverse is not so easily proved.

2533348. [2008 : 241, 243℄ Proposed by Mih�aly Ben ze, Brasov, Romania.Let ABC be an a ute-angled triangle. Let A1, B1, and C1 be pointson the sides BC, CA, and AB, respe tively, su h that the angles ∠AC1B1,∠BC1A1, and ∠ACB are all equal. Let M , N , and P be the ir um entresof △AC1B1, △BA1C1, and △CB1A1, respe tively. Prove that AM , BN ,and CP are on urrent if and only if AA1, BB1, and CC1 are altitudes of△ABC.Solution by Peter Y. Woo, Biola University, La Mirada, CA, USA, revised bythe editor.Denote the angles of △ABC by ∠A, ∠B, ∠C, and let AA′, BB′, andCC′ be the altitudes. The given angle assumptions imply that the trianglesA1BC1 and AB1C1 are both similar to △ABC. Be ause the angle at the ir um entre N that is subtended by the side BC1 of △A1BC1 is twi e theangle at A1 (whi h equals ∠A), we dedu e that ∠NBC1 = 90◦ −∠A for anyposition of C1 between B and A. Moreover, sin e ∠B′BA = 90◦ − ∠A, weknow that N must lie on the altitude BB′. Similarly, M must lie on the alti-tude CC′. Be ause the lines AM and BN therefore meet at the ortho entreH 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 CPpasses through H if and only if C1 = C′ (the foot of the altitude from C toAB).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, ∠A1B1C = ∠B and,as we argued earlier, ∠A1CP = 90◦ − ∠B so that P is on the altitudeCC1 = CH, as desired.We now prove that onversely, when C1 6= C′, then H is not on theline CP . When C1 is taken between B and C′ then (be ause by our angleassumption C1B1 ‖ C′B′) B1 must lie beyond B′ on the hal ine from A inthe dire tion of C. Similarly A1 lies beyond A′ on the hal ine from C in thedire tion of B. It follows that ∠A1B1C > ∠A′B′C; when e

∠A1CP = 90◦ − ∠A1B1C < 90◦ − ∠A′B′C = ∠A′CH ,and we on lude thatH is not on the lineCP . A similar argument establishesthat 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;OLIVER GEUPEL, Br �uhl, NRW, Germany; SALEM MALIKI �C, student, Sarajevo College,Sarajevo, Bosnia and Herzegovina; JOEL SCHLOSBERG, Bayside, NY, USA; and the proposer.3349. [2008 : 241, 243℄ Proposed by Mih�aly Ben ze, Brasov, Romania.Let a, b, and c be positive real numbers. Show that6∏ y li a3 + 1

a2 + 1≥ max

∑ y li a(1 + bc)(a2 + 1)

a3 + 1, ∑ y li ab(1 + c)(a2b2 + 1)

a3b3 + 1

.

254Solution 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≥√

t2 − t + 1 ≥ 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 + 1≥

√t2 − t + 1 of part (a) redu es to t(t − 1)2 ≥ 0, whi h is true.Similarly, the inequality √

t2 − t + 1 ≥ 4

√

t4 + 1

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

(

t2 + t + 1), whi h is true.We now pro eed to prove that 6

∏ y li a3 + 1

a2 + 1≥

∑ y li a(1 + bc)(a2 + 1)

a3 + 1.By part (a) of the Lemma, we have

(

b3 + 1)(

c3 + 1)

(

b2 + 1)(

c2 + 1) ≥ 4

√

(

b4 + 1)(

c4 + 1)

4≥

√

b2c2 + 1

2≥ bc + 1

2and(

a3 + 1

a2 + 1

)2

≥

√

a4 + 1

2≥ a .Hen e,

2∏ y li a3 + 1

a2 + 1≥ a(1 + bc)(a2 + 1)

a3 + 1.Adding a ross the y li permutations of the last inequality gives the desiredresult.We next pro eed to prove that 6

∏ y li a3 + 1

a2 + 1≥∑ y li ab(1 + c)

`

a2b2 + 1´

a3b3 + 1.By parts (a) and (b) of the Lemma, we have

(

a3 + 1)(

b3 + 1)

(

a2 + 1)(

b2 + 1) ≥ ab + 1

2≥ ab

(

a2b2 + 1)

a3b3 + 1and also c3 + 1

c2 + 1≥ 1 + c

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

2∏ y li a3 + 1

a2 + 1≥ ab(1 + c)(a2b2 + 1)

a3b3 + 1.

255Adding a ross the y li permutations of this last inequality gives the se onddesired 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. Provethat

yz

1 + x+

zx

1 + y+

xy

1 + z≤ 1

4.I. Similar solutions by George Apostolopoulos, Messolonghi, Gree e; CaoMinh Quang, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; KhanhBao Nguyen, High S hool for Gifted Students, Hanoi University of Edu a-tion, Hanoi, Vietnam; Babis Stergiou, Chalkida, Gree e; SonHong Ta, Hanoi,Vietnam; and Titu Zvonaru, Com�ane�sti, Romania.For positive real numbers a, b, and c we have that (b + c)2 ≥ 4bc,hen e a

b + c≤ a

4

(

1

c+

1

b

). Thus,∑ y li yz

1 + x=

∑ y li yz

(x + y) + (z + x)≤

∑ y li yz

4

(

1

x + y+

1

z + x

)

=∑ y li xy + zx

4(y + z)=

∑ y li x

4=

1

4.

Equality holds if and only if x = y = z = 13.II. Solution by Arkady Alt, San Jose, CA, USA, ondensed by the editor.Let e1 = x + y + z, e2 = xy + yz + zx, e3 = xyz, and S =

∑′ xy

kz + 1(∑′ denotes a y li sum over x, y, z). We will prove that if k, x, y, z arepositive real numbers and e1 = 1, then S ≤ max{

1

4, 1

k + 3

}.Let k ∈ (0, 1]. Sin e S = e3

∑′(

1

z− k

kz + 1

)

= e2 − ke3

∑′ 1

kz + 1and∑′ 1

kz + 1≥ 9

(∑′(kz + 1)

)−1=

9

k + 3, it follows that S ≤ e2 − 9ke3

k + 3.It therefore suÆ es to prove that e2 − 9ke3

k + 3≤ 1

k + 3, whi h is equivalent to

(k + 3)e2 − 9ke3 ≤ 1 . (1)The inequality (1) follows from the two inequalitiese2 ≥ 9e3 , (2)

4e2 ≤ 1 + 9e3 , (3)

256sin e 1− (k+3)e2 +9ke3 = (1−4e2 +9e3)+(1−k)(e2 −9e3) and k ≤ 1.Now, (2) follows from 3 3

√e3 ≤ e1 = 1 and 3 3

√

e23 ≤ e2, and these follow fromthe AM{GM Inequality, while (3) follows by rewriting the S hur Inequalitymodulo the relation e1 = 1; that is, one rewrites ∑′

x(x − y)(x − z) ≥ 0as 2e31 − 6e1e2 − e2

1 + 2e2 + 9e3 ≥ 0 and puts e1 = 1. This ompletes theproof of the inequality for k ∈ (0, 1].Note that if k = 1, then we obtain the original inequality to be proved,while if k > 1 then S = S(k) < S(1) ≤ 1

4= max

{

1

4, 1

k + 3

}.Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia andHerzegovina; MICHEL BATAILLE, Rouen, Fran e; PAUL BRACKEN, University of Texas, Edin-burg, TX, USA; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; REBECCAEVERDING and JENNIFER PAJDA, students, Southeast Missouri State University, Cape Gi-rardeau, MO, USA; IAN JUNE L. GARCES, Ateneo de Manila University, Quezon City, ThePhilippines; OLIVER GEUPEL, Br �uhl, NRW, Germany; JOE HOWARD, Portales, NM, USA;WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; WEI-DONG, Weihai Vo ationalCollege, Weihai, Shandong Provin e, China; NGUYEN THANH LIEM, Tran Hung Dao HighS 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. ofEle 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; STANWAGON, Ma alesterCollege, St. Paul, MN, USA; PETER Y. WOO, Biola University, La Mirada, CA, USA; and theproposer.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 atedthat sin e x + y + z = 1, our problem appears as problem 35 (solved on pp. 48-49) in thebook Old and New Inequalities by T. Andrees u, V. C�rtoaje, G. Dospines u, and M. Las u; GILPublishing House.Crux Mathemati orumwith Mathemati al MayhemFormer Editors / An iens R �eda teurs: Bru e L.R. Shawyer, James E. TottenCrux Mathemati orumFounding Editors / R �eda teurs-fondateurs: L �eopold Sauv �e & Frederi k G.B. MaskellFormer Editors / An iens R �eda teurs: G.W. Sands, R.E. Woodrow, Bru e L.R. ShawyerMathemati al MayhemFounding Editors / R �eda teurs-fondateurs: Patri k Surry & Ravi VakilFormer Editors / An iens R �eda teurs: Philip Jong, Je� Higham, J.P. Grossman,Andre Chang, Naoki Sato, Cyrus Hsia, Shawn Godin, Je� Hooper

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