65Contributor Pro�les:Arkady AltArkady Alt was born in Odessa, Ukraine,by the Bla k Sea. His interest in mathemati sbegan when he was in middle s hool. At �rst,his mother had to onstantly demand persis-ten e from him and �ght his negligen e whenhe was solving math problems, but sin e hewas very interested in astronomy and physi she qui kly realized that to really understandthose subje ts he had to know mathemati s.His professional areer initially began inapplied mathemati s, fo using on engineeringand e onomi s, but in 1981 he swit hed to anedu ational areer path and began tea hing atHigh S hool No. 100 in Odessa, where he re-ated a spe ialized math ourse that exposed students to advan ed topi s atthe Olympiad and University level. This evolved into a spe ial two-yearprogram whi h united Mathemati s and Computer S ien e into one ourse,unique in the ountry at the time. Many graduates from this program sub-sequently parti ipated in and won national level Mathemati s ompetitions,with some going into a ademia and beyond.When the old war ended hemoved his family to Israel, where he taught hildren and adults ranging from high s hool to university and Olympiadlevel. Several of his students ompleted their high s hool math requirementsearly and enrolled in university prior to full high s hool graduation. Two ofthem won gold and silver medals at the IMO in 1995 { 1997.At university Arkady Alt was drawn to Topology and Algebra. Soon aftergraduating he dis overed Serge Lang's Algebra. Finally group theory, num-ber theory, topology, and linear algebra were all yoked together and he wasinspired to publish on those topi s in onjun tion with on rete problems.At his suggestion many students have read this book from over to over.After his immigration to Israel he began working in the �eld of inequal-ities, where he saw amazing opportunities to apply a host of te hniques. Hedelights in �nding elegant solutions free of heavy ma hinery so he an sharethe problems with his students.He now lives in San Jose, California, with his wife and two adult hil-dren, who were also raised to love mathemati s. He still tea hes and is urrently working on a book on inequalities that will ontain the methods hehas developed over the years. He is an a tive problem solver and proposer inmany urrent mathemati s publi ations, where he often �nds new problemsfor himself and his students.His other interests in lude philosophy, history, musi , and reading.

66EDITORIALV�a lav Linek

This issue of CRUX with MAYHEM marks the return of ContributorPro�les, and there will be more oming up this year.Sin e we are living in an ele troni age and the possibilities for pro ess-ing and sending information are virtually limitless, some feedba k is in orderfor CRUX with MAYHEM ontributors who are savvy in this area.For Contributor Pro�les, bla k and white photos of the ontributor aregood starting materials. If an ele troni �le is sent, then a high-resolution(at least 600 dpi) bla k and white en apsulated posts ript �le (EPS) with abounding box is preferred. We an also a ept olour JPG �les, but these areharder to work with as the �nal result will be bla k and white.We ontinue to a ept all �le formats, but some are easier to work withthan others. Plain text (TXT) �les, TEX, and LATEX �les are preferred to DOCX�les, as the latter leave \ghost hara ters" when they are opied into TEX orLATEX �les. The optimal ombination is a LATEX �le with the PDF output. Ingeneral DVI and PS �les are less preferable, sin e these will be onverted toPDF �les in any ase and PS �les take up a onsiderable amount of spa e. Ifyou are sending a bat h of �les you may onsider ompressing them into aZIP �le to save spa e and keep them all together in your email.Regarding email etiquette, please hoose an informative subje t head-ing that gives a good idea of what the email is about, rather than (say) hitting\reply" and generating a dis onne ted subje t heading. For example, puttingthe number(s) of the CRUX with MAYHEM problem(s) you have just solvedinto the subje t immediately alerts us as to what your email is about. If pos-sible, please use plain Roman hara ters in your emails, sin e (for example)Greek, Cyrilli , or Chinese hara ters are often onverted into question markswhen they are re eived. Please type at least a few words in your email ratherthan leaving the body blank and letting us guess what is intended.Choosing a good �le name also helps, su h as (again) putting the num-ber(s) of the CRUX with MAYHEM problem(s) you have just solved into the�le name. If you solve big bat hes of problems from a single issue then youmay want to in lude the volume number, issue number, month, and year intothe �le name of a zipped �le (for example: CRUX v35n4 May2009 solns.ZIP).We of ourse still love to re eive your letters and exquisite handwrittenmathemati s on all sizes, shapes, and olours of paper!Finally, we urge all problem solvers and proposers to start ea h so-lution or proposal on a new page, with the full name and aÆliation of the ontributor. V �a lav Linek

67SKOLIAD No. 123Lily Yen and Mogens HansenPlease send your solutions to problems in this Skoliad by 1 August, 2010. A opy of CRUXwill be sent to one pre-university reader who sends in solutionsbefore the deadline. The de ision of the editors is �nal.

This month's Skoliad ontest is the City Competition of the CroatianMathemati al So iety, 2009, Se ondary Level, Grade 1. Our thanks go to�Zeljko Hanj�s, University of Zagreb, Croatia, for providing us with this ontestand for permission to publish it.La r �eda tion souhaite remer ier Rolland Gaudet, de Coll �ege universi-taire de Saint-Bonifa e, Winnipeg, MB, d'avoir traduit e on ours.Comp�etition 2009 de la So i �et �e math �ematique roateNiveau se ondaire, premi �ere ann �ee1. R �eduire la fra tion

a4 − 2a3 − 2a2 + 2a + 1

(a + 1)(a + 2).

2. Si on �e rit le hi�re 3 �a gau he d'un entier �a deux positions d �e imales,alors on obtient, bien sur, un entier �a trois positions d �e imales. Si le doublede l'entier �a trois positions �egale 27 fois l'entier �a deux positions, quel �etaitl'entier �a deux positions, au d �epart ?3. D �eterminer le plus grand entier n tel que 3�n − 5

3

�−2(4n+1) > 6n+5.4. D �eterminer le nombre de diviseurs de 288.

5. Dans la �gure, ABCDEF est unhexagone r �egulier tandis que EFGHIest un pentagone r �egulier. D �eterminerl'angle ∠GAF . ............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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6. Dans un trap �eze ABCD, l'angle �a B est re tangle, et la diagonale BDest perpendi ulaire au ot �e AD. Le ot �e BC est de longueur 5, tandis que ladiagonale BD est de longueur 13. D �eterminer la surfa e du trap �eze ABCD.

687. Lors de la fete de Th �er �ese, le premier oup de sonnette annon� a la premi �ereinvit �ee. Ensuite, �a haque oup de sonnette, le nombre d'invit �ees qui sepr �esentaient augmentait de deux. Si la sonnette a retentit n fois, ombiend'invit �ees se sont-elles pr �esent �ees �a la fete ?8. D �eterminer tous les entiers positifs n tels que n2 − 440 est le arr �e d'unentier.

City Competition of the Croatian Mathemati alSo iety, 2009Se ondary Level, Grade 11. Redu e the fra tiona4 − 2a3 − 2a2 + 2a + 1

(a + 1)(a + 2).

2. If you write the digit 3 on the left side of a two-digit number, you ob-tain, of ourse, a three-digit number. If twi e the three-digit number equals27 times the two-digit number, what is the original two-digit number?3. Find the largest integer n su h that 3

�n − 5

3

�− 2(4n + 1) > 6n + 5.4. Find the number of divisors of 288.

5. In the �gure, ABCDEF is a regu-lar hexagon while EFGHI is a regularpentagon. Determine the angle ∠GAF ..........................................................................................................................

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6. In a trapezoid ABCD, the angle at B is a right angle, and the diagonalBD is perpendi ular to the leg AD. The length of the leg BC is 5, and thelength of the diagonal BD is 13. Find the area of the trapezoid ABCD.7. At Tihana's birthday party, the �rst guest arrived the �rst time the bellrang. Ea h time the bell rang thereafter the number of guests arriving wastwo more than the number that had arrived the previous time the bell rang.If the bell rang n times, how many guests attended the party?8. Determine all positive integers n su h that n2 − 440 is the square of aninteger.

69Next follow solutions to the Calgary Mathemati al Asso iation JuniorHigh S hool Mathemati s Contest, Part B, 2009, that appeared previously inSkoliad 117 [2009 : 194{196℄.1. 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?Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto,ON. Let d denote the distan e to the park (in kilometres). Then the ost(in ents) of using the �rst taxi ompany is 1000 + 50d while the ost ofusing the se ond ompany is 400 + 80d. Sin e these two osts are equal,

1000 + 50d = 400 + 80d, so 600 = 30d, and, thus, d = 20.Also solved by GESINE GEUPEL, student, Max Ernst Gymnasium, Br �uhl, NRW, Germany.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.Solution by Gesine Geupel, student, Max Ernst Gymnasium, Br �uhl, NRW,Germany.The table shows a possible hoi e of guests.Winner First guest Se ond guestAli e Bob CarolBob Carol DavidCarol David EvaDavid Eva Ali eEva Ali e BobAlso solved by JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON.3. 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.

70(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.Solution by Gesine Geupel, student, Max Ernst Gymnasium, Br �uhl, NRW,Germany.(a) The table shows a possible set of s ores where two students have ahigher average than Adrian.First Test Se ond Test AverageAdrian 7 5 6.0Bob 6 9 7.5Carol 10 4 7.0(b) To get a better average than Adrian a student must s ore better thanAdrian on at least one test. Sin e only one student s ored better than Adrianon ea h of two tests, at most two students an have a higher average thanAdrian.Also solved by JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON.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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Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto,ON. The diagonal of the re tangle is √62 + 82 = 10 by the PythagoreanTheorem. But this diagonal is also a diameter of the inner ir le, and thediameter of the inner ir le equals the length of the side of the square.Thus the square has side length 10. By the Pythagorean Theorem again,the length of the diagonal of the square is √

102 + 102 =√

200 = 10√

2.Sin e the diagonal of the square is also a diameter of the bigger ir le, theradius of the bigger ir le is 5√

2. It follows that the area of the bigger ir leis π�5√

2�2

= 50π.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.

71Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto,ON. Sin e (A/100)B = 400, it follows that AB = 40000 = 26 · 54. Sin eA does not ontain the digit zero, it must have fa tors of only 2 or only 5.Therefore, the possible two-digit values for A are 25, 16, 32, and 64.Solving the equation for B yields that B = 40000/A, so (A, B) is oneof (25, 1600), (16, 2500), (32, 1250), and (64, 625). Sin e B has three digits,the only solution is (A, B) = (64, 625).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.Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto,ON. The re tangle with sides 1

4and 5

4has perimeter 3. This solves the �rstpart. The re tangle with sides 2 +

√2 and 2 −

√2 has perimeter 8 and area�

2 +√

2��

2 −√

2�

= 22 −�√

2�2

= 4 − 2 = 2 .This solves the last two parts.This issue's prize of one opy of Crux Mathemati orum for the best so-lutions goes to JixuanWang, student, DonMills Collegiate Institute, Toronto,ON. The urrent editors of this se tion inherited the name Skoliad, and wehave wondered about its origins. The best we have been able to re onstru tis this: Olympiad is a derivative of Mount Olympus; the Skoliad se tion fea-tures problems that are more a essible than Math Olympiad problems andis therefore named after a less exalted pla e, namely the town Skolos (orS olus) as mentioned in The Iliad, se ond song, line 497.The paragraph above prompted the proofreaders of this issue of Skoliadto inform us that the name originated with Ri hard K. Guy, who found Skolosas the name of a mountain on a map of Gree e at the University of Calgarylibrary.We soli it information on the exa t lo ation of Skolos (whi h seems tobe an open problem in Greek ar haeology) as well as reader solutions to thefeatured ontest.

72MATHEMATICAL 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 (Our Lady of Mt. Carmel Se ondaryS hool, Mississauga, ON) and Eri Robert (Leo Hayes High S hool, Freder-i ton, NB).

Mayhem ProblemsVeuillez nous transmettre vos solutions aux probl �emes du pr �esent num�eroavant le 15 juin 2010. 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.M426. Propos �e par l' �Equipe de Mayhem.D �eterminer le nombre d'entiers positifs plus petits ou �egaux �a 1 000 000et qui soient divisibles par tous les entiers 2, 3, 4, 5, 6, 7, 8, 9 et 10.M427. Propos �e par l' �Equipe de Mayhem.Dans un demi- er le de diam�etreAB, on dessine un triangle �equilat �eralABC. Trouver l'aire de la r �egion omprise �a l'int �erieur du triangle mais �al'ext �erieur du demi- er le.M428. Proposed by Ne ulai Stan iu, �E ole se ondaire George Emil Palade,Buz�au, Roumanie.Trouver tous les entiers x pour lesquels on a

(4 − x)4−x + (3 − x)3−x + 20 = 4x + 3x .M429. Propos �e par Samuel G �omez Moreno, Universit �e de Ja �en, Ja �en,Espagne.Trouver tous les triplets (a, b, c) d'entiers positifs, ave a(bc) =�ab�c.

73M430. Propos �e par Edward T.H. Wang, Universit �e Wilfrid Laurier,Waterloo, ON.Soit pn le ne nombre premier. Montrer que pn > 3n pour tous lesn ≥ 12.M431. Propos �e par Shailesh Shirali, �E ole Rishi Valley, Inde.Dans un triangle a utangle ABC, soit D le pied de la perpendi ulaireabaiss �ee de A sur BC, E le pied de la perpendi ulaire abaiss �ee de D surAC. Soit F un point sur le segment DE tel que DF

F E=

cot C

cot B. Montrer que

AF et BE sont perpendi ulaires..................................................................M426. Proposed by the Mayhem Sta�.Determine the number of positive integers less than or equal to 1 000 000that are divisible by all of the integers 2, 3, 4, 5, 6, 7, 8, 9, and 10.M427. Proposed by the Mayhem Sta�.A semi ir le has diameter AB. Equilateral triangle ABC is drawn onthe same side of AB as the semi ir le. Determine the area of the region thatlies inside the triangle and outside the semi ir le.M428. Proposed by Ne ulai Stan iu, George Emil Palade Se ondaryS hool, Buz�au, Romania.Determine all integers x for whi h(4 − x)4−x + (3 − x)3−x + 20 = 4x + 3x .M429. Proposed by Samuel G �omez Moreno, Universidad de Ja �en, Ja �en,Spain.Determine all triples (a, b, c) of positive integers with a(bc) =

�ab�c.M430. Proposed by Edward T.H. Wang, Wilfrid Laurier University,Waterloo, ON.Let pn be the nth prime number. Prove that pn > 3n for all n ≥ 12.M431. Proposed by Shailesh Shirali, Rishi Valley S hool, India.In a ute triangle ABC, the foot of the perpendi ular from A to BCis D, and the foot of the perpendi ular from D to AC is E. Point F islo ated on line segment DE su h that DF

F E=

cot C

cot B. Prove that AF and BEare perpendi ular.

74Mayhem SolutionsM394. 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.Solution by all the solvers below indi ated by a star.We write the numbers a, b, c, d, and e as n − 2, n − 1, n, n + 1, and

n + 2, respe tively. Then1

2

�c2 + e2

�− 1

2

�a2 + c2

�=

1

2

�e2 − a2

�=

1

2

�(n + 2)2 − (n − 2)2

�=

1

2(2n · 4) = 4n = 4c ,as required.Solved by ∗EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosnia andHerzegovina; ∗JACLYN CHANG, student, Western Canada High S hool, Calgary, AB; ∗RICHARDI. HESS, Ran ho Palos Verdes, CA, USA; ∗HUGO LUYO S�ANCHEZ, Ponti� ia UniversidadCat �oli a del Peru, Lima, Peru; ∗RICARD PEIR �O, IES \Abastos", Valen ia, Spain; ∗BRUNOSALGUEIRO FANEGO, Viveiro, Spain; ∗JOS �E JAIME SAN JUAN CASTELLANOS, student, Uni-versidad te hnol �ogi a de la Mixte a, Oaxa a, Mexi o; ∗JIXUAN WANG, student, Don MillsCollegiate Institute, Toronto, ON; ∗GUSNADI WIYOGA, student, SMPN 8, Yogyakarta,Indonesia; ∗OSCAR XIA, student, St. George's S hool, Van ouver, BC; and KONSTANTINEZELATOR, University of Pittsburgh, Pittsburgh, PA, USA.M395. Proposed by the MayhemSta�.The quadrilateral ABCD is su hthat ea h of its sides is tangent to agiven ir le, as shown. If AB = AD,prove that BC = CD.

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Solution by Ja lyn Chang, student, Western Canada High S hool, Calgary,AB. In the diagram, AB = AD, andAB, BC, CD, DA are tangent to the ir le at E, F , G, H, respe tively.Be ause of the theorem that saysthat the two tangents to a ir le froma given exterior point have the samelength, then AE = AH, BE = BF ,CF = CG, and DG = DH.

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B

C

D

EF

GH

q

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q

75Also, sin e AB = AD, then AH +DH = AE +BE. But AH = AE,so DH = BE. But DG = DH and BE = BF , so DG = BF . Sin e

CF = CG, then BC = BF + CF = DG + CG = CD, as required.Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo,Bosnia and Herzegovina; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; HUGO LUYOS�ANCHEZ, Ponti� ia Universidad Cat �oli a del Peru, Lima, Peru; RICARD PEIR �O, IES\Abastos", Valen ia, Spain; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; JIXUAN WANG,student, Don Mills Collegiate Institute, Toronto, ON; GUSNADI WIYOGA, student, SMPN 8,Yogyakarta, Indonesia; and KONSTANTINE ZELATOR, University of Pittsburgh, Pittsburgh, PA,USA. There was one in omplete solution submitted.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.Solution by Gusnadi Wiyoga, student, SMPN 8, Yogyakarta, Indonesia.We know that AD = CB, AB = CD, and BD = DB. Hen e △ABDis ongruent to △BCD. This means that the two in ir les have equal radii.Next, we �nd the radius, r, of these ir les by �nding the radius of thein ir le of △ABD (see the �rst �gure below). Conne t O1 to ea h of A, B,and D. Also, let points P , Q, and R on AB, BD, and DA, respe tively, bethe points of tangen y of the ir le to the sides of the triangle; onne t O1 toP , Q, and R.

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A B

CD

O1

O2

P

RX

Y

Sin e AD = 6 and AB = 8, then by the Pythagorean Theorem, wehaveBD =

pAD2 + AB2 =

p62 + 82 = 10 .Also, the area of △ABD is 1

2(AD)(AB) =

1

2(8)(6) = 24. But this areaalso equals the sum of the areas of △AO1B, △BO1D, and △DO1A. Sin e

O1P , O1Q, and O1R are perpendi ular to AB, BD, and DA, respe tively,then these areas equal 1

2(AB)(O1P ), 1

2(BD)(O1Q), and 1

2(DA)(O1R),respe tively. Therefore, 1

2(8r) +

1

2(10r) +

1

2(6r) = 24, or 12r = 24, andso r = 2.

76Lastly, onstru t re tangle O1XO2Y with sides parallel to the sides ofthe original re tangle (see the se ond �gure on the pre eding page). Notethat O1X = 8−2r, sin e O1 is r units from AD and O2 is r units from BC.Thus, O1X = 4. Similarly, XO2 = 6 − 2r = 2. Therefore,

O1O2 =p

42 + 22 =√

20 = 2√

5 .Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; JACLYN CHANG,student, Western Canada High S hool, Calgary, AB; HUGO LUYO S�ANCHEZ, Ponti� iaUniversidad Cat �oli a del Peru, Lima, Peru; RICARD PEIR �O, IES \Abastos", Valen ia, Spain;BRUNO SALGUEIRO FANEGO, Viveiro, Spain; and JIXUAN WANG, student, Don MillsCollegiate Institute, Toronto, ON. There were two in omplete solutions submitted.M397. Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.Determine all pairs (x, y) of integers su h thatx4 − x + 1 = y2 .Solution by Edin Ajanovi , student, First Bosniak High S hool, Sarajevo,Bosnia and Herzegovina.We onsider four ases: x ≤ −1, x = 0, x = 1, and x ≥ 2.If x ≤ −1, then x < 1, so (x2)2 = x4 < x4 − x + 1. Also, x < 0 and

2x + 1 ≤ −1 < 0, so x(2x + 1) > 0, whi h yields 2x2 > −x andx4 − x + 1 < x4 + 2x2 + 1 =

�x2 + 1

�2 .Therefore, (x2)2 < x4 − x + 1 < (x2 + 1)2. Sin e x4 − x + 1 is stri tlybetween two onse utive perfe t squares, then it annot be a perfe t squareitself, so it annot equal y2 in this ase.If x = 0, then the equation be omes y2 = 1, so y = ±1. This yieldsthe solutions (x, y) = (0, 1) and (0, −1).If x = 1, then the equation be omes y2 = 1, so y = ±1. This yieldsthe solutions (x, y) = (1, 1) and (1, −1).If x ≥ 2, then x > 1 so x4 − x +1 < x4 = (x2)2. Also, x(2x − 1) > 0so −x > −2x2, whi h yieldsx4 − x + 1 > x4 − 2x2 + 1 =

�x2 − 1

�2 .Therefore, (x2 −1)2 < x4 −x+1 < (x2)2. Sin e x4 −x+1 is again stri tlybetween two onse utive perfe t squares, then it annot be a perfe t squareitself, so it annot equal y2 in this ase.This overs all possible ases. Therefore, the solutions are (0, 1), (0, −1),(1, 1), and (1, −1).Also solved by RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; RICARD PEIR �O, IES\Abastos", Valen ia, Spain; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON;and JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON. There were threein omplete solutions submitted.

77M398. 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 equations

x + y + z = a ,xy + yz + zx = −1 ,

xyz = −a .Solution by Gusnadi Wiyoga, student, SMPN 8, Yogyakarta, Indonesia.(a) Sin e r, s, and t are the roots of w3 − bw2 + cw −d = 0, then we havew3 − bw2 + cw − d = (w − r)(w − s)(w − t) ;w3 − bw2 + cw − d = w3 − (r + s + t)w2 + (rs + st + rt)w − rst .Sin e these ubi s are equal for any value of w, the orresponding oeÆ ientsare equal, so we have b = r + s + t, c = rs + st + rt, and d = rst.(b) Suppose that x, y, and z are the roots of the equation

m3 − (x + y + z)m2 + (xy + yz + xz)m − xyz = 0 .From the given information, this means that x, y, and z are the roots ofm3 − am2 − m + a = 0, whi h an be rewritten as

m2(m − a) − (m − a) = 0 ;(m2 − 1)(m − a) = 0 ;

(m − 1)(m + 1)(m − a) = 0 .Therefore, the roots are m = 1, m = −1, and m = a.Therefore, the possible values of x, y, and z are 1, −1, and a. In orderto satisfy the given equations, x, y, and z need to take all three of thesevalues, in some order. Therefore, the possible triples (x, y, z) are (1, −1, a),(−1, 1, a), (a, 1, −1), (a, −1, 1), (1, a, −1), and (−1, a, 1).Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosniaand Herzegovina; GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; RICHARD I. HESS,Ran ho Palos Verdes, CA, USA;HUGO LUYO S�ANCHEZ, Ponti� ia Universidad Cat �oli a del Peru,Lima, Peru; RICARD PEIR �O, IES \Abastos", Valen ia, Spain; BRUNO SALGUEIRO FANEGO,Viveiro, Spain; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; JIXUANWANG, student, Don Mills Collegiate Institute, Toronto, ON; and KONSTANTINE ZELATOR,University of Pittsburgh, Pittsburgh, PA, USA. There were two in omplete solutions submitted.Zelator noted that some of these solutions are redundant when a = 1 or a = −1.M399. Proposed by Ne ulai Stan iu, George Emil Palade Se ondaryS hool, Buz�au, Romania.Determine all triples (a, b, c) of positive integers for whi h 3ab − 1

abc + 1isa positive integer.

78Solution by Os ar Xia, student, St. George's S hool, Van ouver, BC.Suppose that 3ab − 1

abc + 1= n, where n is a positive integer. Then we have

3ab − 1 = nabc + n, or 3ab − nabc = n + 1, or ab =n + 1

3 − nc.Sin e a, b, c, and n are positive integers, then 3−nc is a positive integer(sin e ab is positive) so (n, c) = (1, 1), (2, 1), or (1, 2).If (n, c) = (1, 1), then ab = 1 so (a, b) = (1, 1).If (n, c) = (2, 1), then ab = 3 so (a, b) = (3, 1) or (1, 3).if (n, c) = (1, 2), then ab = 2 so (a, b) = (2, 1) or (1, 2).Therefore, the �ve triples are (a, b, c) = (1, 1, 1), (3, 1, 1), (1, 3, 1),

(2, 1, 2), (1, 2, 2). (We an he k that ea h triple satis�es the requirements.)Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo,Bosnia and Herzegovina; GEOFFREY A. KANDALL, Hamden, CT, USA; RICARD PEIR �O, IES\Abastos", Valen ia, Spain; BRUNO SALGUEIRO FANEGO, Viveiro, Spain; EDWARDT.H. WANG, Wilfrid Laurier University, Waterloo, ON; and KONSTANTINE ZELATOR,University of Pittsburgh, Pittsburgh, PA, USA. There were two in orre t solutions and onein omplete solution submitted.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.Solution by Bruno Salgueiro Fanego, Viveiro, Spain, modi�ed by the editor.Suppose that an + bn = cn for some positive integer n ≥ 2. Sin e thenumbers a, b, and c are positive, then a < c and b < c.Suppose that k is a positive integer with 1 ≤ k < n. To show thatak, bk, and ck are the side lengths of a triangle, we need to prove threeinequalities, namely we need to prove that ak + bk > ck, that ak + ck > bk,and that bk + ck > ak.Sin e a < c, then bk + ck > ak. Sin e b < c, then ak + ck > bk. Itremains to prove that ak + bk > ck.Sin e 0 < a < c and 0 < b < c and k − n < 0, then ak−n > ck−n > 0and bk−n > ck−n > 0. Therefore,

ck = ck−ncn

= ck−n(an + bn)

= ck−nan + ck−nbn

< ak−nan + bk−nbn

= ak + bk ,as required.Therefore, ak, bk, and ck are the side lengths of a triangle.There were two in orre t solutions submitted.

79Problem of the Month

Ian VanderBurghSometimes seemingly unrelated problems have surprising onne tions.

Problem 1 (2001 Fermat Contest) Inthe diagram, square ABCD has sidelength 2, with M the midpoint ofBC and N the midpoint of CD.What is the area of the shaded regionBMND?

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A B

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M

NProblem 2 (2001 Fermat Contest) A sealed bottle, whi h ontains water, hasbeen onstru ted by atta hing a ylinder of radius 1 m to a ylinder of radius3 m, as shown in Figure A. When the bottle is right side up, the height of thewater inside is 20 m, as shown in the ross-se tion of the bottle in Figure B.When the bottle is upside down, the height of the liquid is 28 m, as shownin Figure C. What is the total height, in m, of the bottle?

Figure A

20 cm

Figure B

28 cm

Figure C

height ofliquid

These two problems have a more obvious onne tion and a less obvious onne tion. The super� ial onne tion is pretty lear { they ome from thesame ontest in the same year! But there is a more subtle onne tion.Solution to Problem 1 We al ulate the area of the region BMND by al- ulating the area of the entire square ABCD and subtra ting the areas oftriangles BAD and MCN .Sin e the side length of square ABCD is 2, then the area of squareABCD is 2 · 2 = 4.Triangle BAD is right-angled at A and has BA = AD = 2, so hasarea 1

2· BA · AD =

1

2· 2 · 2 = 2.Triangle MCN is right-angled at C. Sin e M and N are the midpointsof BC and CD, then MC = CN = 1. Therefore, the area of triangleMCNis 1

2· MC · CN =

1

2· 1 · 1 =

1

2.Thus, the area of BMND is 4 − 2 − 1

2=

3

2.

80It's hard to ome up with mu h of a onne tion between these problemswithout a tually trying the se ond problem. Let's look at what to me seemslike the most natural way of solving the se ond problem.Solution 1 to Problem 2 Let the height of the larger ylinder be H m and theheight of the smaller ylinder be h m. We want to determine the value of

H + h. We do this by al ulating the total volume of water in the bottle intwo di�erent ways, using Figure B and Figure C.In Figure B, the water entirely �lls the larger ylinder and partially �llsthe smaller ylinder. Put another way, we have H m of water in the large ylinder and 20 m of water in the bottle, so (20 − H) m of water in thesmall ylinder. The total volume of water in the bottle, in ubi entimetres,is thus π · 32 · H + π · 12 · (20 − H) = 9πH + π(20 − H) = 20π + 8πH.Similarly, in Figure C, we have h m of water in the small ylinder and28 m of water in the bottle, so we have (28 − h) m of water in the large ylinder. The total volume of water in the bottle, in ubi entimetres, isthus π · 12 · h + π · 32 · (28 − h) = πh + 9π(28 − h) = 252π − 8πh.Equating these expressions for the amount of water in the bottle yields20π + 8πH = 252π − 8πh, hen e 8πh + 8πH = 232π and h + H = 29.Therefore, the total height of the bottle is 29 m.This solution was not all that diÆ ult, ex ept for a bit of thinking and abit of algebra. We note that we didn't a tually al ulate h or H (nor ould wehave). This method of solving Problem 2 is the most natural in some sense,but it does not give us any hint as to the onne tion with the �rst problem.So what is this elusive onne tion? In the �rst problem, we found thearea of the shaded region by al ulating the \missing area" and subtra tingfrom the total area of the �gure. We did not even onsider trying to �ndthe dimensions of the trapezoid (yes, BMND is a trapezoid) and �nding thearea this way, sin e this method would be a fair bit more involved.Can we apply this type of thinking to Problem 2? Yes, we an, bymakinga key observation. In Solution 1, we used the fa t that the volume of waterwas the same in both on�gurations.Solution 2 to Problem 2 Let x m be the total height of the bottle. Then theheight of the olumn of air in Figure B is x−20 and the height of the olumnof air in Figure C is x − 28. Noti e that the olumn of air in Figure B is a ylinder of radius 1 and in Figure C it is a ylinder of radius 3.Sin e the volume of water is the same in Figure B as in Figure C, thenthe volume of air in ea h on�guration is also the same. The volume of air inFigure B is π ·12 ·(x−20) and the volume of air in Figure C is π ·32 ·(x−28).Therefore, π(x − 20) = 9π(x − 28), or 9(28) − 20 = 9x − x andso 8x = 252 − 20 = 232, when e x = 29. Hen e, the total height of the ylinder is 29 m.This type of thinking is often really useful in mathemati s. Often, we an solve a problem (or solve it more easily) by fo ussing on what is missing,rather than what is there.

81THE OLYMPIAD CORNERNo. 284

R.E. WoodrowWe start this issue with the problems of the four Team Sele tion Testsfor BMO 2007 and IMO 2007 of the Republi of Moldova. My thanks go toBill Sands, Canadian Team Leader to the IMO in Vietnam, for olle ting themfor our use. REPUBLIC OF MOLDOVATeam Sele tion Tests for BMO 2007 and IMO 2007First Test | 5 Mar h 20071. In triangle ABC the points M , N , and P are the midpoints of the sidesBC, AC, and AB, respe tively. The lines AM , BN , and CP interse t the ir um ir le of ABC at A1, B1, and C1, respe tively. Prove that the areaof the triangle ABC does not ex eed the sum of the areas of the trianglesBA1C, AB1C, and AC1B.2. Let p be a prime number, p 6= 2, and m1, m2, . . . , mp onse utivepositive integers, and σ a permutation of the set A = {1, 2, . . . , p}. ProvethatA ontains two distin t numbers k, l su h that p | (mkmσ(k)−mlmσ(l)).3. Inside the triangle ABC there is a point T su h that

∠ATB = ∠BTC = ∠CTA = 120◦ .Prove that the Euler lines of the triangles ATB, BTC, ATC are on urrent.4. Let P = A1A2 . . . An be a onvex polygon. For any point M in theinterior, let Bi be the point where AiM interse ts the perimeter. We saythat P is balan ed if for some su h M the points B1, B2, . . . , Bn are interiorto distin t sides of P . Determine all n for whi h there exists a balan edpolygon with n sides. Se ond Test | 23 Mar h 20075. Determine the smallest positive integers m and k su h that(a) there exist 2m + 1 onse utive positive integers whose ubes sum to aperfe t ube;(b) there exist 2k + 1 onse utive positive integers whose squares sum toa perfe t square.

826. Let I be the in entre of triangle ABC and let R be its ir umradius.Prove that AI + BI + CI ≤ 3R.7. Let U , V be two points inside the angle BAC su h that ∠BAU =∠CAV . Let X1, X2 be the proje tions of U onto the angle sides AC, AB;and let Y1, Y2 be the proje tions of V onto the angle sides AC, AB. Let thelines X2Y1 and X1Y2 interese t at W . Prove that U , V , W are ollinear.8. The onvex hull of �ve points in the plane has area S. Prove that threeof these points form a triangle of area not greater than �5 −

√5

10

�S.Third Test | 24 Mar h 20079. Let a1, a2, . . . , an (n ≥ 2) be real numbers in the interval [0, 1]. Let

S = a31 + a3

2 + · · · + a3n. Prove that

nXi=1

ai

2n + 1 + S − a3i

≤ 1

3.

10. Find all polynomials f with integer oeÆ ients, su h that f(p) is a primefor every prime p.11. Let ABC be a triangle with a = BC, b = AC, c = AB, inradius r,and ir umradius R. Let rA, rB, and rC be the radii of the ex ir les of thetriangle ABC. Prove thata2

�2

rA

− r

rBrC

�+ b2

�2

rB

− r

rCrA

�+ c2

�2

rC

− r

rBrA

�= 4(R + 3r) .

12. Consider n distin t points in the plane, n ≥ 3, arranged su h that thenumber r(n) of segments of length l is maximized. Prove that r(n) ≤ n2

3.Fourth Test | 25 Mar h 200713. Prove that the plane annot be overed by the inner regions of �nitelymany parabolas.14. Let b1, b2, . . . , bn (n ≥ 1) be nonnegative real numbers at least one ofwhi h is positive. Prove that P (X) = Xn − b1Xn−1 − · · · − bn−1X − bnhas a single positive root p, whi h is simple, and that the absolute value ofea h root of P (X) is not greater than p.15. A ir le is tangent to the sides AB and AC of the triangle ABC and toits ir um ir le at P , Q, and R respe tively. Prove that if PQ ∩ AR = {S},then ∠SBA = ∠SCA.

8316. Prove that there are in�nitely many primes p for whi h there exists apositive integer n su h that p divides n! + 1 and n does not divide p − 1.

Next we give sele ted problems of the Thai Mathemati al OlympiadExaminations 2006. Thanks again go to Bill Sands, Canadian team Leader tothe IMO in Vietnam, for olle ting them for us.THAI MATHEMATICAL OLYMPIADEXAMINATIONS 2006Sele ted Problems1. Suppose f : R → R is a fun tion satisfyingf(x2 + x + 3) + 2f(x2 − 3x + 5) = 6x2 − 10x + 17for all real x. Find f(85).2. Evaluate

8000Xk=84

�k

84

��8084 − k

84

� .3. Find all integers n su h that n2 + 59n + 881 is a perfe t square.4. Find the least positive integer n su h that

√3zn+1 − zn − 1 = 0has a omplex root z with |z| = 1.5. Let pk denote the kth prime number. Find the remainder when

2550Xk=2

pp4

k−1kis divided by 2550.

6. Find all fun tions f : R → R su h that2005Xi=1

f(xi + xi+1) + f

2006Xi=1

xi

!≤

2006Xi=1

f(2xi)for all real numbers x1, x2, . . . , x2006.7. A triangle has perimeter 2s, inradius r, and the distan es from its in entreto the verti es are sa, sb, and sc. Prove that3

4+

r

sa

+r

sb

+r

sc

≤ s2

12r2.

848. Let N be the set of positive integers. Is there a bije tion f : N → N withthe three properties below?(a) f(n + 2006) = f(n) + 2006 for all n ∈ N;(b) f(f(n)) = n + 2 for n = 1, 2, 3, . . . , 2004;( ) f(2549) > 2550.9. Find all prime numbers p su h that 2p−1 − 1

pis a perfe t square.10. In a s hool yard 229 boys and 271 girls are divided into 10 groups of 50students ea h, with the students in ea h group numbered from 1 to 50. Fourstudents are sele ted from two groups su h that two pairs of students haveidenti al numbers and the number of girls is odd. Show that the number ofways to sele t four students in this manner is odd.11. Find all fun tions f : R → R su h that

f(x + cos(2007y)) = f(x) + 2007 cos(f(y))for all real numbers x and y.Next we give the problems of the TurkishMathemati al Olympiad 2006.Thanks again go to Bill Sands, Canadian Team Leader to the IMO in Vietnam,for providing these for our use.14th TURKISH MATHEMATICAL OLYMPIAD 2006De ember 16{17, 20061. Let E and F be points on the side CD of a onvex quadrilateral ABCDsatisfying 0 < DE = FC < CD. The ir um ir les of triangles ADE and

ACF interse t at K 6= A, and the ir um ir les of triangles BDE and BCFinterse t at L 6= B. Prove that A, B, K, and L are on y li .2. Find the largest real number t su h that, in any s hool with 2006 studentsand 14 tea hers where every student is a quainted with at least one tea her,a student and a tea her an be found su h that they are a quainted and theratio of the number of students who are a quainted with the tea her to thenumber of tea hers who are a quainted with the student is at least t.3. Find all positive integers n for whi h every oeÆ ient of the polynomialPn(x) = (x2 +x+1)n − (x2 +x)n − (x2 +1)n − (x+1)n +x2n +xn +1is divisible by 7.

854. Let a1, a2, . . . , an (n ≥ 2) be positive real numbers satisfying the relationt = a1 + a1 + · · · + an = a2

1 + a22 + · · · + a2

n. Prove thatXi6=j

ai

aj

≥ (n − 1)2t

t − 1.

5. Let A1, B1, C1 be the feet of the altitudes from verti es A, B, C ina ute triangle ABC, respe tively, and let OA, OB, OC be the in entres ofthe triangles AB1C1, BC1A1, CA1B1, respe tively. Let TA, TB, TC bethe points of tangen y of the in ir le of ABC to the sides BC, CA, AB,respe tively. Show TAOCTBOATCOB is a regular hexagon.6. Prove that there exists no triangle whose side lengths, area, and angles(measured in degrees) are all rational numbers.As a �nal set for this number we give the Turkish Team Sele tion Testfor the IMO 2007. Thanks again go to Bill Sands, Canadian Team Leader tothe IMO in Vietnam, for olle ting them for us.TURKISH TEAM SELECTION TEST FOR IMO 2007Mar h 24{25, 20071. An airline ompany is planning to run two-way ights between some ofthe six ities A, B, C, D, E, and F . Determine the number of ways these ights an be arranged so that it is possible to travel between any two ofthese six ities using only the ights of this ompany.2. Let A and B be distin t points on a ir le Γ. For a variable point P on Γdistin t from A and B, �nd the lo us of the point M su h that PM is theopposite ray to the angle bise tor of ∠APB and MP = AP + PB.3. Let a, b, c be positive real numbers su h that a + b + c = 1. Prove that

1

ab + 2c2 + 2c+

1

bc + 2a2 + 2a+

1

ca + 2b2 + 2b≥ 1

ab + bc + ca.

4. The a ute triangle ABC is similar to the triangle A1B1C1 whose ver-ti es B1, C1, A1 lie on the rays AC, BA, CB, respe tively. Prove that theortho entre of A1B1C1 is the ir um entre of ABC.5. Determine all positive odd integers n for whi h there exist odd integersx1, x2, . . . , xn su h that

x21 + x2

2 + · · · + x2n = n4 .

866. In how many ways an the numbers 1 and −1 be assigned to the unitsquares of a 2007×2007 hessboard so that the absolute value of the sum ofthe numbers in any square made up from the unit squares of the hessboarddoes not ex eed 1?

Next we return to solutions from our readers to problems proposed butnot used at the 47th International Mathemati al Olympiad 2006 in Sloveniagiven at [2008: 461{464℄.G2. Let ABCDE be a onvex pen-tagon su h that

∠BAC = ∠CAD = ∠DAE ;∠ABC = ∠ACD = ∠ADE .The diagonalsBD andCE interse t at

P . Prove that the line AP bise ts theside CD. ..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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A

B

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D

E

P

Solved by Mi hel Bataille, Rouen, Fran e; and Titu Zvonaru, Com�ane�sti,Romania. We give Bataille's version.Let S denote the dire t similar-ity with entre A transforming B intoC. From the hypotheses, we haveS(C) = D and S(D) = E. LetU be the point of interse tion of theline segments AC and BD. Sin eS(AC) = AD and S(BD) = CE,the image V of U under S is the pointof interse tion of AD and CE. It fol-lows that ...........................................................................................................................................................................

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B

C

D

E

PU

V

W

UA

UC=

V A

V D. (1)Now, if AP meets CD at W , we have from Ceva's theorem

AU

UC· DV

V A· CW

WD= 1

and using (1), it follows that CW

W D= 1. This means that W is the midpointof CD, so the proof is omplete.

87G3. A point D is hosen on the sideAC of a triangle ABC with

∠ACB < ∠BAC < 90◦in su h a way that BD = BA. Thein ir le of ABC is tangent to AB andAC at points K and L, respe tively.Let J be the in entre of triangleBCD.Prove that the line KL interse ts theline segment AJ at its midpoint. .......................................................................................................................................................................................................................................................................................................................................................................................................................................

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Solved by Mi hel Bataille, Rouen, Fran e; Oliver Geupel, Br �uhl,NRW, Germany; and Titu Zvonaru, Com�ane�sti, Romania. We give Geupel'swrite-up.It is a basi fa t that, if the in ir le of △PQR is tangent to the sidePQ at the point T , then 2PT = PQ + PR − QR. (See, for example,H.S.M. Coxeter and S.L. Greitzer, Geometry Revisited, The Mathemati alAsso iation of Ameri a, 1967, Theorem 1.4.1, page 11.)Let L′ be the point on the side AC su h that JL′‖LK. Denote thepoint of interse tion of AJ and KL by M . Let the in ir le of △BCD meetCD at point T . Be ause J is on the internal bise tor of ∠BDC, we havethat ∠JDL′ = 90◦ − 1

2∠ADB = ∠DL′J ; hen e DL′ = 2DT . Using thebasi fa t above, we obtain

DL′ = 2DT = BD + CD − BC = AB + CD − BCand2AL = AB + AC − BC .Consequently,

AL′ = AC − CD + DL′

= AC − CD + (AB + CD − BC) = 2AL .We on lude that AJ

AM=

AL′

AL= 2, whi h ompletes the proof.

G4. In triangle ABC, let J be the entre of the ex ir le tangent to sideBC at A1 and to the extensions ofsides AC and AB at B1 and C1, re-spe tively. Suppose that the linesA1B1 and AB are perpendi ular andinterse t at D. Let E be the footof the perpendi ular from C1 to lineDJ. Determine the angles ∠BEA1and ∠AEB1.

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q q q q

qq

q

q q

A D B C1

E

J

A1

C

B1

88Solution by Titu Zvonaru, Com�ane�sti, Romania.As usual we write a = BC, b = CA, c = AB, and s =

1

2(a + b + c).It is known that the following hold: BA1 = BC1 = s− c, AB1 = AC1 = s,and CA1 = CB1 = s − b. Sin e B1D ⊥ AB, we have

AD = s cos A , DB = (s − c) cos B .By Menelaus' theorem we obtainB1C

B1A· DA

DB· A1B

A1C= 1 ⇐⇒ s − b

s· s cos A

(s − c) cos B· s − c

s − b= 1

⇐⇒ cos A = cos B ,hen e the given triangle ABC is isos eles with ∠A = ∠B.We denote by h the altitude from C to the line AB, and by [ABC] thearea of △ABC. We now have the following al ulations and dedu tions:JC1 =

[ABC]

s − a=

ch

c= h ; h2 = a2 − c2

4.

DC1 = DB + BC1 = (s − c) cos B + (s − c)

= (s − c)�1 +

c

2a

�=�a − c

2

� �a +

c

2

�· 1

a=

h2

a.

DJ2 = h2 +h4

a2=⇒ DJ =

hα

a, where α =

pa2 + h2 .

DE =DC2

1

DJ=

h3

aα; EJ =

JC21

DJ=

ah

α;

DE ·EJ ·DJ =h3

aα· ah

α· hα

a=

h5

aα.By Stewart's theorem we obtain the following four dedu tions:

A1D2 · EJ − A1E2 · DJ + A1J2 · DE = EJ · DJ · DE

=⇒ A1E2 · hα

a= (s − c)2 · h2

a2· ah

α+ h2 · h3

aα− h5

aα

=⇒ A1E2 =(s − c)2 · h2

a2 + h2(1)

BD2 · EJ − BE2 · EJ + BJ2 · DE = EJ · DJ · DE

=⇒ BE2 · hα

a= (s − c)2 · c2

4a2· ah

α+ [(s − c)2 + h2] · h3

aα− h5

aα

=⇒ BE2 =(s − c)2 · a2

a2 + h2(2)

89AD2 · EJ − AE2 · DJ + AJ2 · DE = DE · EJ · DJ

=⇒ AE2 · hα

a= s2 · c2

4a2· ah

α+ (s2 + h2) · h3

aα− h5

aα

=⇒ AE2 =s2a2

a2 + h2(3)

B1D2 · EJ − B1E2 · DJ + B1J2 · DE = DE · EJ · DJ

=⇒ B1E2 · hα

a= s2 · h2

a2· ah

α+ h2 · h3

aα− h5

aα

=⇒ B1E2 =s2h2

a2 + h2. (4)It follows from (1)-(4) that

A1E2 + BE2 = BA21 and AE2 + B1E2 = AB2

1 ,so by the onverse of the Pythagorean Theorem, ∠BEA1 = ∠AEB1 = 90◦.G5. Cir lesω1 and ω2 with entresO1and O2 are externally tangent at pointD and internally tangent to a ir le ωat points E and F , respe tively. Linet is the ommon tangent of ω1 and ω2at D. Let AB be the diameter of ωperpendi ular to t, so that A, E, andO1 are on the same side of t. Provethat the lines AO1, BO2, EF , and tare on urrent. q qq

q q

q

q

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ω1

ω2

tSolved by Mi hel Bataille, Rouen, Fran e; Oliver Geupel, Br �uhl, NRW,Germany; and Titu Zvonaru, Com�ane�sti, Romania. We �rst give the solu-tion of Geupel.Let O be the entre of ω. The triangles O1ED and OEB are isos eleswith O1E‖OE and O1D‖OB; hen e ED‖EB; whi h means that points B,D, andE are ollinear. Similarly,A, D, and F are ollinear. Denote byG theinterse tion of AO1 with BO2. We apply Pappus' theorem to the ollinearpoints A, B, O, and the ollinear points O2, O1, D, thus obtaining that thepoints E = BD∩OO1, F = AD∩OO2, and G = AO1 ∩BO2 are ollinear.It therefore remains to prove that G lies on the line t.The line O1D interse ts ω1 and ω2 again at points H and I, respe -tively. The homothety with entre E that maps O1 to O also maps ω1 toω and thus H to A. Therefore, H lies on the line AE. Similarly, I lies onBF . Denote by C the interse tion of AE and BF . Sin e AF ⊥ BC and

90BE ⊥ AC it follows that D = AF ∩ BE is the ortho entre of △ABC andt is the third altitude of △ABC, whi h shows that C lies on t.Let J denote the interse tion of AB and t. Be ause the triangles HICand ABC are homotheti ,

DO1

DO2

=DH

DI=

JA

JB.Consequently, the point G at the interse tion of the lines AO1 and BO2 alsolies on the line t, whi h ompletes the proof.Next we give the solution of Bataille.Sin e AB is a diameter of ω, we have AF ⊥ FB and AE ⊥ EB. Itfollows that the point H of interse tion of AE and BF is the ortho entre of

△ADB. This said, we shall make use of the inversion I with pole D su hthat I(ω) = ω. We learly have I(E) = B and I(F ) = A so that the line EFis transformed into the ir um ir le Γ of △ADB. We denote by U the pointof interse tion of tand EF , W the entre of Γ, O themidpoint of AB,and U ′ = I(U)Note that U ′ is ont and Γ with U ,U ′ on either sideof D on t and that−−→DH = 2

−−→WO.Let tB bethe tangent to ωat B and O′

1 bethe re e tion ofD in tB. FromI(O1) = O′

1 (be- ause I(ω1) = tB),we dedu e that theimage of the lineAO1 under I is the ir le (DFO′

1).Let V be the en-tre of this ir le.Clearly, V is ontB and also onthe perpendi ularbise tor of DF .

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A B

D

E

F

H

I

O

O1O′

1

O2

U

U ′

VW

t

r r

r

r

r

r

r

r

r rr

r

r

rr

Sin e the latter is parallel to BH and passes through the midpoint I of HD(note that the ir le with diameter DH passes through E and F ), it follows

91that HIV B is a parallelogram. Thus −−→

BV =−→HI =

−−→OW and V is on theperpendi ular to DU ′ through W , that is, on the perpendi ular bise tor of

DU ′. As a result, the ir le (DFO′1) passes through U ′ and its inverse, andthe line AO1 passes through U . Similarly, the line BO2 passes through Uand we are done.G6. In a triangle ABC, let Ma, Mb,

Mc be the respe tive mid-points ofthe sides BC, CA, AB and let Ta,Tb, Tc be the mid-points of the ar sBC, CA, AB of the ir um ir le ofABC not ontaining A, B, C, respe -tively. For ea h i ∈ {a,b,c}, let ωibe the ir le with diameter MiTi. Letpi be the ommon external tangent toωj , ωk su h that {i,j,k} = {a,b,c}and su h that ωi lies on one side ofpi while ωj , ωk lie on the other side.Prove that the lines pa, pb, pc form atriangle similar to ABC and �nd theratio of similitude.

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......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................r r

r

r rr

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r

r

A

B C

Mb

Ma

Mc

Tc

Tb

Ta

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ωa

ωb

ωc

Solution by Oliver Geupel, Br �uhl, NRW, Germany.Let a = BC, b = CA, c = AB, and 2s = a + b + c. Let F = [ABC]be the area of △ABC with ir umradius R, ir um entre O, and in entre I.Lemma. Lines pa and MbMc are parallel, and their distan e is F (s − a)

2sa.Proof. For ea h i ∈ {a, b, c}, let Oi, ri be the entre and the radius of

ωi. The perpendi ular from Ob to BC meets MbMc and ωb at Nb and Pb,respe tively, with Nc, Pc de�ned similarly. Then rb = ObTb =R − OMb

2=

R(1 − cos B)

2and ObNb = ObMb cos C = rb cos C; hen e

NbPb = rb − ObNb =R(1 − cos B)(1 − cos C)

2

=abc

2F· (s − a)(s − c)

ac· (s − a)(s − b)

ab=

F (s − a)

2sa.

Similarly, NcPc =F (s − a)

2sa. We on lude that pa = PbPc.Corollary. The triangles △(pa, pb, pc) and MaMbMc are homotheti with entre I and ratio 2. Thus, △(pa, pb, pc) ∼ △ABC with ratio 4.Proof. Denoting by −→

P the position ve tor of point P , we have−→I =

a−→A + b

−→B + c

−→C

a + b + c=

(s − a)−→Ma + (s − b)

−→Mb + (s − c)

−→Mc

s.

92Thus, d(I, MbMc) =

2[IMbMc]

MbMc=

4(s − a)[MaMbMc]

sa= 2d(pa, MbMc),and by entirely similar al ulations we have d(I, McMa) = 2d(pb, McMa)and d(I, MaMb) = 2d(pc, MaMb), where d denotes the Eu lidean distan e.G8. Points A1, B1, C1 are on the sides BC, CA, AB of a triangle ABC, re-spe tively. The ir um ir les of triangles AB1C1, BC1A1, CA1B1 interse tthe ir um ir le of triangle ABC again at points A2, B2, C2, respe tively(that is, A2 6= A, B2 6= B, C2 6= C). Points A3, B3, C3 are symmet-ri to A1, B1, C1 with respe t to the midpoints of the sides BC, CA, ABrespe tively. Prove that the triangles A2B2C2 and A3B3C3 are similar.

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...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................r r

r

r

r

r

r

r

r

r

A

B C

B2

C2

B1

A2

C1

A1

Solution by Oliver Geupel, Br �uhl, NRW, Germany.Let us onsider the situation in the plane of omplex numbers, and leta, b, c, . . . denote the oordinates of the points A, B, C, . . . . It is well-known (see Titu Andrees u, Dorin Andri a, Complex numbers fromA to... Z,Birkh�auser, Boston, 2006, page 68) that triangles PQR and STU are similar(with the same orientation) if and only if p − r

q − r=

s − u

t − u.First we re ognize ins ribed angles on the ir um ir les of △ABC and

△AB1C1. We have∠A2BC1 = ∠A2BA = ∠A2CA = ∠A2CB1and∠A2C1B = 180◦ − ∠A2C1A

= 180◦ − ∠A2B1A = ∠A2B1C .Therefore, the (likewise) oriented triangles A2BC1 and A2CB1 are similar,whi h implies thata2 − c1

b − c1

=a2 − b1

c − b1

.

93We obtain a2 =

bb1 − cc1

b + b1 − c − c1

. Similarly,b2 =

ccaa1

c + c1 − a − a1

, and c2 =aa1 − bb1

a + a1 − b − b1

.Sin e A3 is symmetri to A1 with respe t to the midpoint of BC, we havea3 − c = b − a1, and hen e a3 = b + c − a1. Similarly b3 = c + a − b1 andc3 = a + b − c1.By the hara terization given above, it suÆ es to prove that

(a2 − c2)(b3 − c3) = (b2 − c2)(a3 − c3) .But this an easily be veri�ed by employing the relations above and learingdenominators. This ompletes the proof.N1. Given x ∈ (0, 1) let y ∈ (0, 1) be the number whose nth digit after thede imal point is the (2n)th digit after the de imal point of x. Prove that if xis a rational number, then y is a rational number.Solution by Oliver Geupel, Br �uhl, NRW, Germany.For ea h positive integer n, let xn be the nth digit of x after the de i-mal point. Be ause x is rational, there exists a positive integer N su h thatthe sequen e (xn)n≥N is periodi with some period q. By the PigeonholePrin iple, there exist positive integers m and n su h that 2m > 2n ≥ N and2m ≡ 2n (mod q). Then the sequen e of the least nonnegative remainders(2k mod q)k≥n is periodi with a period not greater than m − n. Conse-quently, the sequen e (yk)k≥n = (x2k)k≥n is periodi ; hen e the de imalexpansion of the number y is eventually periodi and therefore y is rational.N2. For ea h positive integer n let

f(n) =1

n

�jn

1

k+jn

2

k+ · · · +

jn

n

k� ,where ⌊x⌋ is the greatest integer not ex eeding x.(a) Prove that f(n + 1) > f(n) for in�nitely many n.(b) Prove that f(n + 1) < f(n) for in�nitely many n.Solved by Oliver Geupel, Br �uhl, NRW, Germany.Let d(n) denote the number of positive divisors of the integer n. Ob-serve that for ea h positive integer k, there are exa tly jn

k

k numbers in theset {1, 2, . . . , n} whi h are divisible by k. Therefore,jn

1

k+jn

2

k+ · · · +

jn

n

k= d(1) + d(2) + · · · + d(n) .

94This implies that f(n) is the arithmeti mean of the numbers of divisors ofthe �rst n positive integers.Be ause the fun tion d is unbounded, there are in�nitely many numbersn, su h that d(n + 1) > max{d(k) : 1 ≤ k ≤ n}. By rewriting this as

f(n + 1) =d(1) + d(2) + · · · + d(n + 1)

n + 1

>d(1) + d(2) + · · · + d(n)

n= f(n)we have ompleted part (a).For part (b), on the one hand sin e f(6) =

7

3> 2 and d(k) ≥ 2 forea h k > 1, it follows that f(n) > 2 whenever n > 5. On the other hand,for ea h prime number n + 1 we have d(n + 1) = 2. Consequently, for anyprime number n + 1 ≥ 7 we have f(n + 1) < f(n), and we have ompletedpart (b).N4. Let a and b be relatively prime integers with 1 < b < a. De�ne theweight of an integer c, denoted by w(c), to be the minimum possible valueof |x| + |y| taken over all pairs of integers x and y su h that

ax + by = c .An integer c is alled a lo al hampion if w(c) ≥ max{w(c ± a), w(c ± b)}.Find all lo al hampions and determine their number.Solution by Oliver Geupel, Br �uhl, NRW, Germany.We prove a series of lemmas.Lemma 1. Let x, y, and c be integers su h that y ≥ 0 andax + by = c . (1)Then the ondition

w(c) = |x| + |y| (2)is equivalent to|x + b| + |y − a| ≥ |x| + y . (3)Proof: If (2) holds, then (3) follows immediately from the de�nition of theweight. Conversely, assume (3). Note that y < a. We are to show that forea h integer n,

|x + nb| + |y − na| ≥ |x| + y . (4)If n > 0, then (4) follows from|x + nb| + |y − na|

= |x + nb| + na − y ≥ |x + nb| + (n − 1)b + a − y

= |x + nb| + |(1 − n)b| + |y − a| ≥ |x + b| + |y − a| ≥ |x| + y .

95If n ≤ 0, then we derive (4) as follows:

|x + nb| + |y − na| = |x + nb| − na + y

≥ |x + nb| + | − nb| + y ≥ |x| + y .Lemma 2. Let d and x be integers su h that −b + 1 ≤ d ≤ b. Then the onditions|x + b| ≥ |x| + d (5)and

2x + b ≥ d (6)are equivalent. Also the onditions |x + b| ≤ |x| + d and 2x + b ≤ d areequivalent.Proof: If x ≤ −b, then both (5) and (6) are false. If −b < x < 0, then|x + b| = x + b and |x| = −x; hen e (5) is equivalent to x + b ≥ −x + d;that is to (6). Finally, if x ≥ 0, then both (5) and (6) are satis�ed. The proofof the se ond laim is similar.Lemma 3. Let x, y, and c be integers satisfying (1). Then, (2) is satis�ed ifand only if one of the following holds:(a) −

�a + b

2

�≤ y < −

�a − b

2

� and x ≤ y +a + b

2;(b) −

�a − b

2

�≤ y ≤ a − b

2;( ) a − b

2< y ≤ a + b

2and y ≤ x +

a + b

2.Proof: We an assume without loss of generality that y > 0, sin e we answit h from x, y, c to −x, −y, −c, whenever y < 0. Now, (2) is equivalentto (3) by Lemma 1. Again y < a holds; hen e (3) is equivalent to

|x + b| ≥ |x| + 2y − a . (7)If y ≤ a − b

2, then 2y − a ≤ −b and (7) is true. Next, if a − b

2< y ≤ a + b

2,then by Lemma 2 inequality (7) is equivalent to 2x + b ≥ 2y − a. Finally,for y >

a + b

2no solution exists.Lemma 4. Let x, y, c be integers satisfying (1), (2). Let a − b

2< y ≤ a + b

2.Then c is a lo al hampion if and only if x < 0 and

x +

�a + b

2

�= y . (8)Proof: If c is a lo al hampion, then by Lemma 3

y ≤ x +a + b

2< x + 1 +

a + b

2.

96Again by Lemma 3, we obtain w(c + a) = |x + 1| + y. Be ause c is a lo al hampion, it follows that |x + 1| ≤ |x|; hen e x < 0. If y ≤ x − 1 +

a + b

2,then we obtain from Lemma 3 that w(c − a) = |x − 1| + y = w(c) + 1,whi h ontradi ts the hypothesis that c is a lo al hampion. Therefore,

x +a + b

2− 1 < y ≤ x +

a + b

2,that is (8).Conversely, it remains to prove that c is indeed a lo al hampion. Weshow that w(c ± a) ≤ w(c) and w(c ± b) ≤ w(c).First, w(c + a) = |x + 1| + y ≤ |x| + y = w(c) holds.Se ond, we have c − a = a(x + b − 1) + b(y − a), as well as

−�

a + b

2

�< y − a < −

�a − b

2

�and x + b − 1 ≤ y − a +

a + b

2. By Lemma 3, it follows w(c − a) =

|x+ b−1|+a−y. From x+a + b − 2

2≤ y, we obtain 2x+ b−1 ≤ 2y −a.The se ond part of Lemma 2 leads to |s + b − 1| ≤ |x| + 2y − a; thus

w(c − a) ≤ |x + b − 1| + a − y ≤ |x| + y = w(c) .Third, c + b = a(x + b) + b(y + 1 − a); hen ew(c + b) ≤ b + a − y − 1 = a + b − 1 −

�a + b

2

�≤�

a + b

2

�= w(c) .Fourth, c − b = ax + b(y − 1); thus

w(c−b) ≤ |x|+y−1 = w(c)−1 .Lemma 5. Let x, y, and c be integers satisfying inequalities (1) and (2). If−�

a + b

2

�≤ y <

a − b

2, then c is a lo al hampion if and only if x > 0 and

x = y+j

a + b

2

k. Moreover, there are no lo al hampions c with |y| ≤ a − b

2.Proof: The �rst part follows from Lemma 4 by repla ing x, y by −x, −y. Let

|y| ≤ a − b

2. We on lude by Lemma 3 that w(c − a) = |x − 1| + |y| and

w(c + a) = |x + 1| + |y|. Clearly, one of the numbers |x − 1| and |x + 1| isgreater than |x|, whi h ompletes the proof of the se ond part.Corollary 6. There are b − 1 lo al hampions if ab is odd and 2(b − 1) lo al hampions if ab is even.Proof: We an des ribe the lo al hampions expli itly. If ab is odd, then byLemma 4 and Lemma 5 the set of lo al hampions is�(a + b)(2 − b)

2+ n(a + b) : 0 ≤ n ≤ b − 2

� .

97If ab is even, then Lemma 4 and Lemma 5 yield the disjoint sets

M =

�2a − ab + b − b2

2+ n(a + b) : 0 ≤ n ≤ b − 2

� ,N =

�2a − ab + b − b2

2+ n(a + b) + b : 0 ≤ n ≤ b − 2

� ,respe tively. The set M is an arithmeti progression with di�eren e a + b,and the set N is the same progression with o�set b.Next we turn to solutions from our readers to problems of the SwedishMathemati al Contest 2005/2006 given at [2008 : 464{465℄.1. Find all solutions in integers x and y of the equation

(x + y2)(x2 + y) = (x + y)3 .Solved by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain; Mi helBataille, Rouen, Fran e; and Titu Zvonaru, Com�ane�sti, Romania. We givethe solution of Amengual Covas.The equation is equivalent to xy + x2y2 = 3x2y + 3xy2, whi h an berewritten as xy(1 + xy − 3x − 3y) = 0. We observe that (x, y) = (0, m),(m, 0), where m is an integer, are solutions.Now suppose that xy 6= 0. Then 1 + xy − 3x − 3y = 0. Sin e no pairs(x, y) with x = 3 an satisfy this last equation, we rewrite it in the formy =

3x − 1

x − 3, or y = 3 +

8

x − 3.It follows that y is an integer if and only if x − 3 divides 8. Thus,

x − 3 ∈ {±1, ±2, ±4, ±8}, so that x ∈ {−5, −1, 1, 2, 4, 5, 7, 11}.We on lude that the omplete solution set for (x, y) is{(0, m), (m, 0) : m ∈ Z}[

{(2, −5), (−5, 2), (−1, 1), (1, −1), (4, 11), (11, 4), (5, 7), (7, 5)} .2. A queue in front of a ounter onsists of 12 persons. The ounter is then losed be ause of a te hni al problem and the 12 people are redire ted toanother one. In how many di�erent ways an the new queue be formed ifea h person maintains the same position as before, or is one step loser tothe front, or is one step farther from the front?Solution by Oliver Geupel, Br �uhl, NRW, Germany.Let Qn be the number of distin t permutations π of {1, 2, . . . , n} su hthat for 1 ≤ k ≤ n we have k − 1 ≤ π(k) ≤ k + 1. For n ≥ 3, if π hasthe desired property, then we have π(n) ∈ {n − 1, n}. First onsider the

98 ase π(n) = n − 1. Then π(n − 1) = n, and the numbers 1, 2, . . . , n − 2 an be arranged in Qn−2 ways. Se ond, if π(n) = n, then the numbers1, 2, . . . , n − 1 an be arranged in Qn−1 ways. Hen e, for n ≥ 3, we havethe re ursion Qn = Qn−2 + Qn−1 with Q1 = 1 and Q2 = 2.Therefore, Qn is the (n + 1)st Fibona i number Fn+1 (spe i� ally, thesolution of our exer ise is Q12 = F13 = 233).3. In the triangle ABC the angle bise tor from A interse ts the side BCin the point D and the angle bise tor from C interse ts the side AB in thepoint E. The angle at B is greater than 60◦. Prove that AE + CD < AC.Solved by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain; GeorgeApostolopoulos, Messolonghi, Gree e; Mi hel Bataille, Rouen, Fran e; andTitu Zvonaru, Com�ane�sti, Romania. We give the version of Apostolopoulos.We have cos B <

1

2. By the Law ofCosines, b2 = a2 + c2 − 2ac · cos B,so cos B =

a2 + c2 − b2

2ac<

1

2. Then

a2 + c2 < ac + b2 ,bc + a2 + c2 + ab < ab + ac + b2 + bc ,

c(b + c) + a(a + b) < (a + b)(b + c) ,bc

a + b+

ab

b + c< b .

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

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

A

B CD

E

q

q

aa

From the Bise tor Theorem, we have AE =bc

a + band CD =

ab

b + c, hen ethe on lusion AE + CD < AC.Comment by Zvonaru. This problem is part ( ) of Problem 3 of the ontestTrenti �eme Olympiad Math �ematique Belge, [2008 : 80℄.4. The polynomial f(x) is of degree four. The zeroes of f are real and forman arithmeti progression, that is, the zeroes are a, a+d, a+2d, and a+3dwhere a and d are real numbers. Prove that the three zeroes of f ′(x) alsoform an arithmeti progression.Solved by Mi hel Bataille, Rouen, Fran e; and Titu Zvonaru, Com�ane�sti,Romania. We give Bataille's version.First, note that the zeroes of f ′(x) are also real numbers. This is obvi-ous if d = 0 and follows from Rolle's theorem otherwise.Let xk = a + kd, (k = 0, 1, 2, 3) and set λ =

x0 + x3

2=

x1 + x2

2.Then,

f(x) = m3Y

k=0

(x − xk) = m�x2 − 2λx + p

� �x2 − 2λx + q

� ,

99where m is a nonzero real number, p = x0x3, and q = x1x2. It readilyfollows that

f ′(x) = 4m(x − λ)�x2 − 2λx +

p + q

2

� .The zeroes of f ′(x) are λ and real numbers y1, y2 su h that y1 +y2 = 2λ. Itfollows that y1, λ, y2 is an arithmeti progression, and the proof is omplete.5. Ea h square on a 2005×2005 hessboard is painted either bla k or white.This is done in su h a way that ea h 2 × 2 \sub- hessboard" ontains an oddnumber of bla k squares. Show that the number of bla k squares among thefour orner squares is even. In how many di�erent ways an the hessboardbe painted so that the above ondition is satis�ed?Solution by Oliver Geupel, Br �uhl, NRW, Germany.Consider more generally integers m ≥ 2, n ≥ 2, and an m × n boardrepresented by a matrix �s11 · · · s1n

· · · · · · · · ·sm1 · · · smn

�where

sij =

§1 if square (i, j) is painted bla k,0 if square (i, j) is painted white.Call the board odd if for ea h 2 × 2 sub-board

Bij =

�sij si,j+1

si+1,j si+1,j+1

�the sum of its entries bij = sij + si,j+1 + si+1,j + si+1,j+1 is odd, where1 ≤ i ≤ m − 1 and 1 ≤ j ≤ n − 1.Proposition 1. For ea h odd m × n board, if 2 | (m − 1)(n − 1), then thesum of its orner entries is even.Proof: By hypothesis, the number

m−1Xi=1

n−1Xj=1

bij = 4

m−1Xi=2

n−1Xj=2

sij + 2

m−1Xi=2

(si1 + sin)

+ 2

n−1Xj=2

(s1j + smj) + (s11 + s1n + sm1 + smn)

is even. Hen e, the sum s11 + s1n + sm1 + smn is also even.Proposition 2. For m ≥ 2 and n ≥ 2, the number of odd m × n boardsis 2m+n−1.

100Proof: By indu tion on n.For n = 2, there are eight ways to paint the topmost 2 × 2 sub-board.The rows 3, 4, . . . , m an be su essively painted, where we have two hoi esfor ea h row. We obtain 8 · 2m−2 = 2m+1 di�erent painted boards.In an m × n board the leftmost n − 1 olumns form an m × (n − 1)board: �

s11 · · · s1,n−1 s1n... ... ...sm1 · · · sm,n−1 smn

� ,whi h an be painted in 2m+n−2 ways by the indu tion hypothesis. We nowhave two hoi es for s1n and s2n. Ea h further entry sin in the last olumnis uniquely determined by the elements si−1,n−1, si−1,n, and si,n−1. Wetherefore obtain 2m+n−2 · 2 = 2m+n−1 painted boards.Remark: For m = n = 2005 we obtain 24009 painted boards.

That brings us to the start of the problems given in numbers of theCorner for 2009, and solutions from our readers to problems of the GermanMathemati al Olympiad, Final Round, Grades 12{13, Muni h, April 29{May2, 2006 given at [2009 : 21℄.1. Determine all positive integers n for whi h the numberzn = 101 · · · 101| {z }

2n+1 digitsis a prime.Solved byMatthew Babbitt, home-s hooled student, Fort Edward, NY, USA;Alex Song, student, Elizabeth Ziegler Publi S hool, Waterloo, ON; EdwardT.H. Wang, Wilfrid Laurier University, Waterloo, ON; and KonstantineZelator, University of Pittsburgh, Pittsburgh, PA, USA. We give the write-up of Babbit, modi�ed by the editor.We havezn =

nXi=0

100i =100n+1 − 1

100 − 1

=102n+2 − 1

99=

�10n+1 + 1

� �10n+1 − 1

�99

.It is obvious that 9 | (10n+1 − 1) for all positive integers n. Note that 99divides (10n+1 + 1)(10n+1 − 1) be ause zn is an integer, and 11 divides 99,hen e 11 divides (10n+1 + 1)(10n+1 − 1).

101When 11 does not divide 10n+1 − 1, then it divides 10n+1 + 1, whi his always greater than 11. Also, 9 divides 10n+1 − 1, whi h is always greaterthan 9. Therefore, zn is not prime in this ase.When 11 does divide 10n+1 − 1, then we have that 10n+1 − 1 is amultiple of 99. Sin e 10n+1 + 1 is always greater than 1, zn an only beprime when 10n+1 − 1 = 99, or n = 1.Therefore, zn is not prime for n > 1, and z1 = 101 is prime.5. Let x be a nonzero real number satisfying the equation ax2 + bx+ c = 0.Furthermore, let a, b, and c be integers satisfying |a| + |b| + |c| > 1. Provethat

|x| ≥ 1

|a| + |b| + |c| − 1.Solved by Mi hel Bataille, Rouen, Fran e; and Konstantine Zelator, Univer-sity of Pittsburgh, Pittsburgh, PA, USA. We give the solution of Bataille.Sin e a, b, and c are integers, |a| + |b| + |c| ≥ 2. We rewrite theinequality as �

|a| + |b| + |c|�|x| ≥ 1 + |x| , (1)whi h ertainly holds if |x| ≥ 1, sin e then

(|a| + |b| + |c|)|x| ≥ 2|x| ≥ 1 + |x| .Therefore, we suppose that 0 < |x| < 1 in what follows. Now, |x| > |x|2,so we have |a| · |x|+ |b| · |x| ≥ |a| · |x|2 + |b| · |x| ≥ |ax2 + bx| = | − c| = |c|and it follows that(|a| + |b| + |c|)|x| ≥ |c|(1 + |x|) .Thus, (1) ertainly holds if |c| ≥ 1, that is, if c 6= 0.Finally, if c = 0, then ax + b = 0 (sin e x is nonzero) and b 6= 0 sin eotherwise a = 0 as well and |a| + |b| + |c| > 1 would not be true. Theleft-hand side of (1) then be omes (|a| + |b|)|x| and

(|a| + |b|)|x| = |ax| + |b|·|x|= | − b| + |b|·|x|= |b|(1 + |x|) ≥ 1 + |x| ,sin e b is a nonzero integer. Thus, (1) ontinues to hold in that ase and theproof is omplete.6. Let a ir le through B and C of a triangle ABC interse t AB and AC in

Y and Z, respe tively. Let P be the interse tion of BZ and CY , and let Xbe the interse tion of AP and BC. Let M be the point that is distin t fromX and on the interse tion of the ir um ir le of the triangle XY Z with BC.Prove that M is the midpoint of BC.

102Solution by Oliver Geupel, Br �uhl, NRW, Germany.We redu e the geometri problem to an algebrai one, whi h an besolved by dire t omputation. Let a, b, c, x, y, z be the oordinates of A, B,C, X, Y , Z in the omplex plane, respe tively. Without loss of generalitylet b = −1 and c = 1. It suÆ es to prove that 0, x, y, z are on y li . Itis well known that distin t non ollinear points p1, p2, p3, p4 are on y li if (p3 − p2)(p1 − p4)

p1 − p2)(p3 − p4)is a real number (see Titu Andrees u, Dorin Andri a,Complex numbers fromA to...Z, Birkh�auser, Boston, 2006, page 68). There-fore, it suÆ es to show that

(y − x)z

x(y − z)∈ R . (1)Let AY = λ · AB; that is y = a − λ(a + 1), where λ ∈ R. Thefa t that B, C, Y , Z are on y li implies that △ABC ∼ △AZY , hen e

AZ =AB2

AC2· AY

AB·AC; that is, z = a− |a + 1|2

|a − 1|2 λ(a−1). By Ceva's theoremwe see that −−→BX =

−−→BY−−→AY

·−→AZ−→CZ

· −−→XC, that is x + 1 =

(y + 1)(z − a)

(y − a)(z − 1)· (1 − x),and thus

x =|a + 1|2 − |a − 1|2

|a + 1|2 + |a − 1|2 − 2λ|a + 1|2.Substituting these expressions into the fra tion in (1) and learing real fa torsyields the expression�

2λ2|a + 1|2(a + 1) − λ(|a + 1|2(3a + 1) + |a − 1|2(a + 1))

+ |a + 1|2(a − 1) + |a − 1|2(a + 1)�

·�λ|a + 1|2(a − 1) − |a − 1|2a

�·�

a − 1

a + 1

� .We rewrite this last expression as f3(a)λ3+f2(a)λ2+f1(a)λ+f0(a), wheref3(a) = 2|a + 1|4|a − 1|2 ,f2(a) = |a + 1|2|a − 1|2(|a − 1|2 − 5|a|2 − a − a − 1) ,f1(a) = |a − 1|2(5|a|2 − 2|a| − a2 − a2 − 2a − 2a − 1 + |a + 1|2|a − 1|2) ,f0(a) = −2|a|2|a + 1|2|a − 1|2 .Ea h of the four oeÆ ients above is a real number, and so is λ. This om-pletes the proof of the relation (1).

That ompletes the Corner for this issue. Send me your ni e solutionsand generalizations.

103BOOK REVIEWSAmar SodhiThe Mathemati s of the Heavens and the Earth: The Early Historyof TrigonometryBy Glen Van Brummelen, Prin eton University Press, 2009ISBN13: 978-0-691-12973-0, hard over, 329+xvii pages, US$39.50Reviewed byMenolly Lysne, Simon Fraser University, Burnaby, BCIn his book The Mathemati s of the Heavens and the Earth, Glen VanBrummelen takes on the task of giving a histori al treatment of trigonometryfrom its earliest roots until the beginnings of its modern development. Hetakes the reader on a voyage through time and spa e, moving from An ientEgypt, Babylon and Gree e, to Alexandrian Gree e through medieval Indiaand Islam, ending in 1550 in Renaissan e Europe. Van Brummelen begins byasking the question \What is trigonometry?" His answer is twofold, involv-ing �nding lengths given angles and �nding angles given lengths. With thisde�nition in mind, the author begins with the early so ieties that ould doone or the other of these su h as the An ient Egyptians, who la ked angularmeasure but were able to make omputations for ar hite tural purposes, andthe Babylonians, who had angular measure. Then there were so ieties that ould do both, su h as the Alexandrian Greeks who produ ed hord tables sothat a person ould �nd the hord for a given angle. This allowed the Greeksto determine shadow lengths, hen e permitting them to move from anglesto lengths. This journey through the development of trigonometry is boundtogether by an investigation into the transmission of knowledge from oneregion to another and the ontroversy that surrounds this transmission. Su h ontroversy is in India where there are disputes as to whether there was atransmission of knowledge from Gree e or whether the knowledge was de-veloped in India. After ea h text, Van Brummelen provides an explanation,whi h allows the sometimes elusive meaning to be ome apparent.This book would be a wonderful addition to the library of any student,espe ially the graduate student resear hing the early history of mathemati s.Van Brummelen provides an ex ellent bibliography and also points out themany ontroversies that took pla e over the interpretation of events. Whilethe author does show one side of the arguments in the text, through thoroughfootnotes the reader is given the opportunity to investigate other interpre-tations if so in lined. The book often provides only a histori al sket h, butat any point where the reader might ask, \Where an I learn more aboutthis?" the reader is dire ted to a variety of primary and se ondary sour esfrom annotated translated texts to biographies of key �gures.This book may not be ideal for the high s hool tea her or studentfor a variety of reasons. While The Mathemati s of the Heavens and theEarth provides ex ellent histori al ba kground, whi h ould avour a lass ontrigonometry, a large portion of the book is devoted to areas whi h, while

104important in the histori al ontext, are rarely addressed in the modern set-ting. These in lude a thorough depi tion of hord and sine tables and their onstru tions, as well as the development and uses of spheri al trigonometry.These are both highly important, espe ially in the ontext of astronomy, butneither is investigated in the high s hool setting, making this book diÆ ultto in orporate into the lassroom.The book itself is well written and provides a solid histori al a ountthough a small failing an be found in its layout. The diagram for an exampleis often found on the page after the orresponding written des ription. Thismakes following many of the examples somewhat frustrating. However, themathemati s required to follow the examples are easily at the high s hoolor early university level. Aside from this small in onvenien e, in pra ti allyevery other sense the book ows well and is easy to read.Overall, this book is an ex ellent resour e for a university student andan interesting read for anyone with a solid grasp of high s hool mathemati sand an interest in history.Homage to a Pied Puzzler and Mathemati al Wizardry for a GardnerEdited by Ed Pegg, Jr., Alan S hoen, and Tom RodgersPublished by A K Peters, Ltd., 2009ISBN 978-1-56881-315-8, hard over, 285+xxii pages, US$49.00; andISBN 978-1-56881-447-6, hard over, 262+xvii pages, US$49.00 (resp.)Reviewed by David Ehrens, Ma alester College, St. Paul, MN, USAThese books are ompiled from the seventh \Gathering for Gardner",also known as G4G7. The presentations, and thus the arti les, are on any-thing related to the types of re reational mathemati s made popular byMartin Gardner in his de ades of arti les for S ienti� Ameri an. This leadsto a huge variety of topi s, with only the additional theme of the numberseven helping to link them all. For example, a map olouring on a torus re-quires at most seven olours, and a number of people onne ted that sevento this year's theme and brought in related ideas. One of the olour plates inthese well-illustrated books depi ts a map (and its dual) that require seven olours; the map is knitted whilst the dual is ro heted!While there may not be strong onne tions between the arti les, thevariety of topi s does mean that anyone who enjoys puzzles and games will�nd many arti les to suit their parti ular interests. Pure math, paper folding,and even the reation and marketing of puzzles are dealt with by a variety ofpeople, tantalizing us with what's possible. Anyone who enjoys the works ofGardner will �nd a dire tion to follow in these olle tions.Puzzle history is well represented, with the 14-15 sliding blo k puzzleand the invention laims of Sam Loyd examined. That same puzzle is ex-panded upon, with a variety of moves suggested instead of the traditionalsliding. I grew up with the small plasti versions of the game instead of theoriginal blo ks in a box, so the idea of other movements, like jumping or

105swit hing, was new to me. Other Loyd puzzles, like the Two Ovals-to-Tableproblem, are also dis ussed.Many arti les explore the jump needed to turn something that is math-emati ally possible into a physi al reality. While the mathemati ian may beintrigued by a game that takes ten thousand moves to �nish, this would notmake for a popular toy. In Homage to a Pied Puzzler, Adrian Fisher, the reator of Navigati (whi h �rst appeared in the Daily Mail in 2006) des ribeswhat it takes to reate a puzzle for a newspaper audien e.When pondering the audien e for this book, I was �rst stru k by thevariety of topi s. This makes it a �ne book for the puzzler who likes thetopi s that Gardner brought to the forefront. I was also intrigued by theidea that a talented or interested high s hool student ould reate physi alexamples of these mathemati al ideas. Many of the paper folding arti lesseemed to be perfe t for these students to do as proje ts for lass. The gamesand ard tri ks ould also lead to �ne demonstrations. The arti les in thesebooks ould be put to use in the lassroom to popularize mathemati s just asGardner's arti les popularized mathemati s for earlier generations.Lessons in Play: An Introdu tion to Combinatorial Game TheoryBy Mi hael H. Albert, Ri hard J. Nowakowski, and David WolfePublished by A K Peters, Ltd., 2007ISBN 1-56881-277-9, hard over, 288+xi pages, US$69.00Reviewed by Sarah K.M. Aldous, South Island S hool, Hong KongImagine you are playing a game of Nim with a lassmate, olleague orfriend: you start with three heaps of pennies. On your turn you an hooseone heap and remove any number of pennies. The winner will be the lastperson to be able to play, and the afternoon o�ee will be bought by theloser. You s our your brain. Surely there is a way to des ribe the optimalstrategy, guaranteeing you a free o�ee and muÆn?Combinatorial game theory explores the strategies behind two-playergames su h as Nim in whi h there are no han e elements (for example, di erolling or oin ipping). It is a ri h area of very interesting mathemati s thatis easily a essible to students and instru tors. This book is a omprehensiveintrodu tory textbook designed to be used for a �rst ourse in ombinatorialgame theory at university level. The authors are three respe ted games re-sear hers who aimed to reate a book that an be used by a student alone,by an instru tor new to game theory, or by an experien ed game theorist.There have been two other standard texts used for ombinatorial gametheory ourses, OnNumbers and Games (Conway, 1976, se ond edition 2001)and Winning Ways for Your Mathemati al Plays (Berlekamp, Conway, andGuy; 1982, se ond edition 2001). Naturally, Lessons in Play draws heavily onthe ideas and material presented in these two books. However, the ex ellentorganisation of Lessons in Play lends itself mu h better to its purpose as a ourse textbook. In parti ular, the text is easy to follow and the se tions are

106very well laid out; a motivated student would be able to follow independentlythe lear explanations. Students reading on their own are also en ouragedto pro eed with a pen il in hand, trying out ideas through short exer isesinterspersed in the text. Instru tors new to tea hing game theory wouldbe helped by notes before ea h hapter giving ideas about how to stru turetheir le tures. There are also preparation questions before ea h hapter that ould be given to students as an assignment before ea h se tion is overed in lass. Instru tors will also appre iate support from a website, from whi h asolutions manual is available.The book has many strengths, the foremost of whi h is the lear, logi alorganisation. This will make the text more useful to beginning students andtea hers and easier to follow than other books. Furthermore, an appendixlists ea h ombinatorial game mentioned in the book, giving its set of rulesand an example. Another strength is the referen e to CGSuite, a pie e ofsoftware that performs algebrai manipulation of games. An appendix in thebook gives details about (freely) obtaining the software and some examplesof how to use it. There is a link from the textbook website to the CGSuitepage. Also helpful are the o asional dis ussions in the early hapters of howresear h into games is done. This will en ourage readers to try some resear hfor themselves. Indeed, there are lots of open problems being urrently re-sear hed in game theory, and some of these are mentioned throughout thebook and in the hapter entitled \Further Dire tions".The text ontains a few errors, and orre tions are listed on the book'swebsite. At times the readability ould be improved by adding examplesor having a de�nition appear earlier, su h as the de�nition of a \simpler"number. These are very infrequent, however, and they do not detra t mu hfrom the overall quality of the book. Perhaps the sets of rules ould be im-proved by adding referen es to the parts of the text that mention them andalso to literature so readers ould �nd further analysis of the games thatinterest them.Overall, Lessons in Play is a wel ome addition to the sto k of ombina-torial game theory textbooks. Students and instru tors will �nd it thoroughand easy to use. Reading the book and working through the exer ises mayhelp you win your morning o�ee games, and will ertainly open up areas offurther resear h and reading.

107PROBLEMSToutes solutions aux probl �emes dans e num�ero doivent nous parvenir au plustard le 1er septembre 2010. 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.

3505. Corre tion. Propos �e par Yakub N. Aliyev, Universit �e de Qafqaz,Khyrdalan, Azerba��djan.Les er les Γ1 et Γ2 ont meme entre O, et Γ1 est �a l'int �erieur de Γ2.Soit A 6= O un point �a l'int �erieur de Γ1 ; un rayon non parall �ele �a AO etissu de A oupe respe tivement Γ1 et Γ2 aux points B et C. Soit D le pointd'interse tion des tangentes aux points B et C des er les orrespondants.Soit E un point sur la droite BC tel que DE soit perpendi ulaire �a BC.Montrer que AB = EC si et seulement si OA est perpendi ulaire �a BC.3514. Propos �e par Mi hel Bataille, Rouen, Fran e.Soit m un nombre r �eel positif et a, b et c des nombres r �eels tels quea(a − b) + b(b − c) + c(c − a) = m .Quel est le domaine de ab(a − b) + bc(b − c) + ca(c − a) ?3515. Propos �e par Jos �e Luis D��az-Barrero et Josep Rubi �o-Masseg �u,Universit �e Polyte hnique de Catalogne, Bar elone, Espagne.Soit x, y et z trois nombres r �eels positifs. Montrer que�{x}2

y+

[x]2

z

�+

�{y}2

z+

[y]2

x

�+

�{z}2

x+

[z]2

y

�≥ x2 + y2 + z2

x + y + z,o �u ⌊a⌋ est le plus grand entier n'ex �edant pas a, et {a} = a − ⌊a⌋.3516. Propos �e par J�anos Bodn�ar, Budapest, Hongrie.Soit P et Q deux points int �erieurs du triangle ABC. Soit AA′, BB′ et

CC′ trois �eviennes passant par P . La droite par A′ parall �ele �a AQ ouperespe tivement les droites BQ et CQ aux points L et M ′. La droite par B′parall �ele �a BQ oupe respe tivement les droites CQ et AQ aux points M etN ′. La droite par C′ parall �ele �a CQ oupe respe tivement les droites AQ etBQ aux points N et L′.Est- e vrai que les triangles LMN et L′M ′N ′ ont meme aire ?

1083517. Propos �e par V�a lav Kone� n�y, Big Rapids, MI, �E-U.Y a-t-il un triangle s al �ene ABC pour lequel il existe un point P dansle plan de ABC tel que, pour haque droite ℓ par P , la somme des arr �esdes distan es de A, B et C �a ℓ est onstante ?3518. Propos �e par Yakub N. Aliyev, Universit �e de Qafqaz, Khyrdalan,Azerba��djan.Montrer que si n et i sont des entiers ave 0 ≤ i ≤ n, alors

0 < 1 − Ci

22i+1−�10(2i+1)n ·

p102n − 1

©<

1

102n,o �u {x} d �esigne la partie fra tionnaire du nombre r �eel x et Ci le ie nombrede Catalan, Ci =

1

i + 1

�2ii

�, i ≥ 0.3519. Propos �e par Nguyen Duy Khanh, �etudiant, Universit �e des S ien esde Hanoi, Hanoi, Vietnam.Deux triangles ABC et A′B′C′ ont les aires respe tives S et S′. Soitwa, wb et wc les longueurs des bisse tri es internes des angles de ABC aux ot �es BC, AC, AB, respe tivement ; on d �e�nit w′

a, w′b et w′

c de mani �ereanalogue. Montrer si oui ou non, on a waw′a + wbw

′b + wcw

′c ≥ 3

√3SS′.3520. Propos �e par Ri ardo Barroso Campos, Universit �e de S �eville, S �eville,Espagne.Construire un triangle ABC tel que la droite passant pas les entresdes er les ins rit et ir ons rit soit parall �ele au ot �e AB.3521. Propos �e par Dorin M�arghidanu, Colegiul Nat�ional \A.I. Cuza",Corabia, Roumanie.Soit x1, x2, . . . , xn des nombres r �eels dans l'intervalle [e, ∞) et pour haque indi e k soit ek =

x1 + x2 + · · · + xk

xk. Montrer que

xe1

1 + xe2

2 + · · · + xen

n ≥ nx1 + (n − 1)x2 + · · · + 2xn−1 + xn .3522. Propos �e par Dorin M�arghidanu, Colegiul Nat�ional \A.I. Cuza", Co-rabia, Roumanie.Si a, b, c et d sont des nombres r �eels positifs tels que abcd = 1, montrerque�1 +

a

b

�cd�1 +

b

c

�da �1 +

c

d

�ab�1 +

d

a

�bc

≥ 2

�16

a2 + b2 + c2 + d2

� .

1093523. Propos �e par Slavko Simi , Institut de Math �ematiques SANU, Bel-grade, Serbie.Soit f : R → R une fon tion ontinument di� �erentiable. R �esoudrel' �equation fon tionnelle

f(x) + f(y) − 2f�

x + y

2

�= p(x − y)

�f ′(x) − f ′(y)

� ,o �u p est un param�etre r �eel ind �ependant de x, y.3524. Propos �e par Titu Zvonaru, Com�ane�sti, Roumanie.Soit a1, a2, . . . , an+1 des nombres r �eels positifs satisfaisant la ondi-tion an+1 = min{a1, a2, . . . , an+1}. Montrer quean+1

1 + an+12 + · · · + an+1

n+1 − (n + 1)a1a2 · · · an+1

≥ (n + 1)an+1

�(a1 − an+1)

n + (a2 − an+1)n + · · · + (an − an+1)

n

− n(a1 − an+1)(a2 − an+1) · · · (an − an+1)� .3525. Propos �e par un proposeur anonyme.Soit 0 ≤ a, b ≤ 1. Montrer que a + b

1 + ab≤�√

a +√

b −√

ab�2.3526. Propos �e par Cao Minh Quang, Coll �ege Nguyen Binh Khiem, VinhLong, Vietnam.Soit a, b et c trois nombres r �eels positifs. Montrer queX y lique aÈ

a2 + 2(b + c)2≥ 1 ..................................................................3505. Corre tion. Proposed by Yakub N. Aliyev, Qafqaz University,Khyrdalan, Azerbaijan.The ir les Γ1 and Γ2 have a ommon entre O, and Γ1 lies inside Γ2.The point A 6= O lies inside Γ1; a ray not parallel to AO that starts at Ainterse ts Γ1 and Γ2 at the points B and C, respe tively. Let tangents to orresponding ir les at the points B and C interse t at the point D. Let Ebe a point on the line BC su h that DE is perpendi ular to BC. Prove that

AB = EC if and only if OA is perpendi ular to BC.3514. Proposed by Mi hel Bataille, Rouen, Fran e.Let m be a positive real number and a, b, c real numbers su h thata(a − b) + b(b − c) + c(c − a) = m .What is the range of ab(a − b) + bc(b − c) + ca(c − a)?

1103515. Proposed by Jos �e Luis D��az-Barrero and Josep Rubi �o-Masseg �u,Universitat Polit �e ni a de Catalunya, Bar elona, Spain.Let x, y, and z be positive real numbers. Prove that�

{x}2

y+

[x]2

z

�+

�{y}2

z+

[y]2

x

�+

�{z}2

x+

[z]2

y

�≥ x2 + y2 + z2

x + y + z,where ⌊a⌋ is the greatest integer not ex eeding a, and {a} = a − ⌊a⌋.3516. Proposed by J�anos Bodn�ar, Budapest, Hungary, Budapest, Hun-gary. Let P and Q be interior points of triangle ABC. Let AA′, BB′, and

CC′ be three on urrent evians through P . The line through A′ parallel toAQ interse ts the lines BQ and CQ at points L and M ′, respe tively. Theline through B′ parallel to BQ interse ts the lines CQ and AQ at points Mand N ′, respe tively. The line through C′ parallel to CQ interse ts the linesAQ and BQ at points N and L′, respe tively.Is it true that triangles LMN and L′M ′N ′ have the same area?3517. Proposed by V�a lav Kone� n�y, Big Rapids, MI, USA.Is there a s alene triangle ABC for whi h there exists a point P in theplane of ABC su h that, for ea h line ℓ through P , the sum of the squaresof the distan es of A, B, and C to ℓ is onstant?3518. Proposed by Yakub N. Aliyev, Qafqaz University, Khyrdalan,Azerbaijan.Prove that if n and i are integers with 0 ≤ i ≤ n, then

0 < 1 − Ci

22i+1−�10(2i+1)n ·

p102n − 1

©<

1

102n,where {x} denotes the fra tional part of the real number x and Ci is the ithCatalan number, Ci =

1

i + 1

�2ii

�, i ≥ 0.3519. Proposed by Nguyen Duy Khanh, student, Hanoi University ofS ien e, Hanoi, Vietnam.Two triangles ABC and A′B′C′ have areas S and S′, respe tively. Letwa, wb, wc be the lengths of the internal angle bise tors of ABC to thesides BC, AC, AB, respe tively, and de�ne w′

a, w′b, w′

c similarly. Prove ordisprove that waw′a + wbw

′b + wcw

′c ≥ 3

√3SS′.3520. Proposed by Ri ardo Barroso Campos, University of Seville, Seville,Spain.Constru t a triangle ABC su h that the line through the in entre andthe ir um entre is parallel to the side AB.

1113521. Proposed by Dorin M�arghidanu, Colegiul Nat�ional \A.I. Cuza",Corabia, Romania.Let x1, x2, . . . , xn be real numbers in the interval [e, ∞) and for ea hindex k let ek =

x1 + x2 + · · · + xk

xk. Prove that

xe1

1 + xe2

2 + · · · + xen

n ≥ nx1 + (n − 1)x2 + · · · + 2xn−1 + xn .3522. Proposed by Dorin M�arghidanu, Colegiul Nat�ional \A.I. Cuza",Corabia, Romania.If a, b, c, and d are positive real numbers satisfying abcd = 1, provethat�1 +

a

b

�cd�1 +

b

c

�da �1 +

c

d

�ab�1 +

d

a

�bc

≥ 2

�16

a2 + b2 + c2 + d2

� .3523. Proposed by Slavko Simi , Mathemati al Institute SANU, Belgrade,Serbia.Let f : R → R be a ontinuously di�erentiable fun tion. Solve thefun tional equation

f(x) + f(y) − 2f

�x + y

2

�= p(x − y)

�f ′(x) − f ′(y)

� ,where p is a real parameter independent of x, y.3524. Proposed by Titu Zvonaru, Com�ane�sti, Romania.Let a1, a2, . . . , an+1 be positive real numbers satisfying the onditionan+1 = min{a1, a2, . . . , an+1}. Prove thatan+1

1 + an+12 + · · · + an+1

n+1 − (n + 1)a1a2 · · · an+1

≥ (n + 1)an+1

�(a1 − an+1)

n + (a2 − an+1)n + · · · + (an − an+1)

n

− n(a1 − an+1)(a2 − an+1) · · · (an − an+1)� .3525. Proposed by an anonymous proposer.Let 0 ≤ a, b ≤ 1. Prove that a + b

1 + ab≤�√

a +√

b −√

ab�2.3526. Proposed by Cao Minh Quang, Nguyen Binh Khiem High S hool,Vinh Long, Vietnam.Let a, b, and c be positive real numbers. Prove thatX y li aÈ

a2 + 2(b + c)2≥ 1 .

112SOLUTIONSAu 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.We a knowledge a orre t solution to problem 3354 by Paolo Perfetti,Dipartimento di Matemati a, Universit �a degli studi di Tor Vergata Roma,Rome, Italy. Due to swit hed LATEX ommands, we instead in orre tly red-ited Gottfried Perz, Pestalozzigymnasium, Graz, Austria. We further a knowl-edge a orre t solution to problem 3412 by Salem Maliki� , student, SarajevoCollege, Sarajevo, Bosnia and Herzegovina. Our apologies for these errors.

3415. [2009 : 108, 111℄ Proposed by Cezar Lupu, student, University ofBu harest, Bu harest, Romania.Let a, b, and c be positive real numbers su h that abc = 1. Prove that3√

a +3√

b + 3√

c ≤ 3

È3(3 + a + b + c + ab + bc + ca) .Comment: �Sefket Arslanagi � , University of Sarajevo, Sarajevo, Bosnia andHerzegovina informs us that this problem has appeared as problem O98 inthe journal Mathemati al Re e tions, Vol. 5 (2008), pp. 38-39, by the sameproposer. Three solutions are given there using a variety of te hniques.Solved by ARKADY ALT, San Jose, CA, USA; GEORGE APOSTOLOPOULOS, Messolonghi,Gree e; MICHEL BATAILLE, Rouen, Fran e; CAO MINH QUANG, Nguyen Binh Khiem HighS hool, Vinh Long, Vietnam; CHIP CURTIS, Missouri Southern State University, Joplin, MO,USA; OLIVER GEUPEL, Br �uhl, NRW, Germany; JOE HOWARD, Portales, NM, USA; DUNGNGUYEN MANH, High S hool of HUS, Hanoi, Vietnam; MADHAV R. MODAK, formerly ofSir Parashurambhau College, Pune, India; ALBERT STADLER, Herrliberg, Switzerland; PETERY. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU, Com�ane�sti, Romania; and theproposer.

3416. [2009 : 108, 111℄ Proposed by Mi hel Bataille, Rouen, Fran e.Let the sequen e (an) be de�ned by a0 = 6 and the re ursionan+1 =

1

13

�8an

È3a2

n + 13 − 6a2n − 13

�for n ≥ 0. Prove that ea h an is a positive integer, and that a2

n − an+1 isdivisible by 13 for ea h n ≥ 0.Solution by Roy Barbara, Lebanese University, Fanar, Lebanon.Setf(x) =

2

13x�4p

3x2 + 13 − 3x�

− 1 ;

113by hypothesis, an+1 = f(an). Let

ω = 2 +√

3 , ω = 2 −√

3 ;α =

1

2

�4 +

√3� , α =

1

2

�4 −

√3� .As an aid to al ulating, we observe that ωω = 1, αα =13

4, and α−α =

√3.Part 1. De�ne in reasing sequen es of positive integers (un) and (vn) by

u0 = 4 , u1 = 11 , un+2 = 4un+1 − un ;v0 = 1 , v1 = 6 , vn+2 = 4vn+1 − vn .Solving for un and vn, we obtain

un = αωn + α ωn ,vn =

1√

3

�αωn − α ωn

� . (1)From (1) we easily dedu e that 3v2

n + 13 = u2n, so that un =

È3v2

n + 13;hen e, f(vn) =2

13vn(4un − 3vn) − 1. Using (1) again we have

4un − 3vn = α�4 −

√3�ωn + α

�4 +

√3�ωn

= 2αα�ωn + ωn

�=

13

2

�ωn + ωn

� .Hen e,f(vn) =

2

13· 1

√3

�αωn − α ωn

�· 13

2(ωn + ωn) − 1

=1

√3

�ωn + ωn

��αωn − α ωn

�− 1

=1

√3

�αω2n − α ω2n

� . (2)For n ≥ 0 de�ne the positive integer bn = v2n. Using (2) and (1) we havef(bn) = f(v2n) =

1√3

�αω2n+1 − α ω2n+1

�= v2n+1 = bn+1. Finally, from

a0 = b0 = 6, an+1 = f(an), and bn+1 = f(bn), indu tion yields thatan = bn for n ≥ 0, and the result follows.Part 2. For n ≥ 0 set θn =

1

12

�ω2n+1

+ ω2n+1 − 2�. Sin e θn is the solutionof the re ursion θn+1 = 12θ2

n + 4θn with initial ondition θ0 = 1, it is apositive integer. As an = v2n, then from (1) we havean =

1√

3

�αω2n − α ω2n

� .

114It follows that a2

n − an+1 =13

12(ω2n+1

+ ω2n+1 − 2) = 13θn, when e 13divides a2n − an+1, as desired.Also solved by ARKADY ALT, San Jose, CA, USA; BRIAN D. BEASLEY, PresbyterianCollege, Clinton, SC, USA; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA;PAUL DEIERMANN, Southeast Missouri State University, Cape Girardeau, MO, USA; OLIVERGEUPEL, Br �uhl, NRW, Germany; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; MADHAVR. MODAK, formerly of Sir Parashurambhau College, Pune, India; JOEL SCHLOSBERG,Bayside, NY, USA; ALBERT STADLER, Herrliberg, Switzerland; and the proposer. There wasone in omplete solution submitted.Many solvers began with the general Pell equation B2 − 3A2 = 13. With un and vnas de�ned in the featured solution above, we see that the pair B = un , A = vn , provides anin�nite family of solutions to this Pell equation. Deiermann began instead with the equation

B2 − dA2 = N, where d > 0 is a nonsquare integer and N > 0 is an integer for whi h thepair B = B0, A = A0 is the minimal positive solution for an equivalen e lass of solutions,while the pair B1 and A1 is the next largest solution. He proved that if (an) is de�ned bya0 = A1 and

an+1 =1

N

�2

ÈdA2

0 + N · an

pda2

n + N − 2A0da2n − A0N

�for n ≥ 0, then ea h an is a positive integer and a2

n − A0an+1 is divisible by N for ea hn ≥ 0. (In our problem N = 13, d = 3, A0 = 1, B0 = 4, A1 = 6, and B1 = 11.) Hessproved a similar generalization.3417. [2009 : 109, 111℄ Proposed by Mi hel Bataille, Rouen, Fran e.Let Sp(n) = 1p + 2p + · · · + np. Prove that

∞Xn=1

�S−1(n)

�2S1(n)

=1

2

∞Xn=1

S−1(n)

S3(n)

!+ 2

∞Xn=1

S−1(n)

S1(n)

! .Solution by Joel S hlosberg, Bayside, NY, USA.Let Hn = 1−1 + 2−1 + · · · + n−1 be the nth harmoni number. ThenHn = Hn−1 +

1

n. As usual let ζ(k) =

∞Pn=1

1

nkbe the Riemann zeta fun tion.A known identity due to Euler (see Jonathan Sondow and Eri W. Weisstein\Harmoni Number." http://mathworld.wolfram.com/HarmonicNumber.html)states that

∞Xn=1

Hn

n2= 2ζ(3) . (1)It is well known that Hn is asymptoti to ln n and that ln n = o(nk) for anypositive k. Therefore,

limn→∞

Hn

n + 1= 0 (2)and

limn→∞

H2n

n + 1= 0 . (3)

115By (2),

∞Xn=1

Hn

n(n + 1)= lim

N→∞

NXn=1

Hn

n(n + 1)= lim

N→∞

NXn=1

Hn

�1

n− 1

n + 1

�= lim

N→∞

H1 · 1

1+

NXn=2

1

n(Hn − Hn−1) − HN

N + 1

!= lim

N→∞

1

12+

NXn=2

1

n2

!− lim

N→∞

HN

N + 1

=∞X

n=1

1

n2= ζ(2) . (4)

By (1),∞X

n=1

Hn

(n + 1)2=

∞Xn=2

Hn−1

n2=

∞Xn=2

Hn − 1n

n2=

∞Xn=1

Hn − 1n

n2

=∞X

n=1

Hn

n2−

∞Xn=1

1

n3= 2ζ(3) − ζ(3) = ζ(3) . (5)

By (1), (4), and (5),∞X

n=1

Hn

n2(n + 1)2=

∞Xn=1

Hn

�1

n2+

1

(n + 1)2− 2

n(n + 1)

�=

∞Xn=1

Hn

n2+

∞Xn=1

Hn

(n + 1)2− 2

∞Xn=1

Hn

n(n + 1)

= 2ζ(3) + ζ(3) − 2ζ(2) = 3ζ(3) − 2ζ(2) . (6)By (1) and (3),∞X

n=1

H2n

n(n + 1)

= limN→∞

NXn=1

H2n

n(n + 1)= lim

N→∞

NXn=1

H2n

�1

n− 1

n + 1

�= lim

N→∞

H2

1 · 1

1+

NXn=2

1

n(H2

n − H2n−1) − H2

N

N + 1

!= lim

N→∞

1 +

NXn=2

1

n(Hn + Hn−1)(Hn − Hn−1)

!− lim

N→∞

H2N

N + 1

= limN→∞

1 +

NXn=2

1

n

�2Hn − 1

n

��1

n

�!

116= lim

N→∞

2 + 2

NXn=2

Hn

n2

!− lim

N→∞

1 +

NXn=2

1

n3

!= 2

∞Xn=1

Hn

n2−

∞Xn=1

1

n3= 2 · 2ζ(3) − ζ(3) = 3ζ(3) . (7)Note that S−1(n) = Hn. By the well-known formulas for the sums ofthe �rst n positive integers and the �rst n ubes,

S1(n) =n(n + 1)

2and S3(n) =

n2(n + 1)2

4. (8)By (4), (6), and (8),

1

2

∞Xn=1

S−1(n)

S3(n)

!+ 2

∞Xn=1

S−1(n)

S1(n)

!= 2

∞Xn=1

Hn

n2(n + 1)2+ 4

∞Xn=1

Hn

n(n + 1)

= 2(3ζ(3) − 2ζ(2)) + 4ζ(2) = 6ζ(3) . (9)By (7),∞X

n=1

(S−1(n))2

S1(n)= 2

∞Xn=1

H2n

n(n + 1)= 6ζ(3) ,

1

2

∞Xn=1

S−1(n)

S3(n)

!+ 2

∞Xn=1

S−1(n)

S1(n)

!= 6ζ(3) ,hen e the two sides are equal with a ommon value of 6ζ(3).Also solved by ARKADY ALT, San Jose, CA, USA; CHARLES R. DIMINNIE, Angelo StateUniversity, San Angelo, TX, USA; OLIVER GEUPEL, Br �uhl, NRW, Germany; WALTHER JANOUS,Ursulinengymnasium, Innsbru k, Austria; ALBERT STADLER, Herrliberg, Switzerland; and theproposer.

3418. [2009 : 109, 111℄ Proposed by Pantelimon George Popes u,Bu harest, Romania and Jos �e Luis D��az-Barrero, Universitat Polit �e ni a deCatalunya, Bar elona, Spain.Let I(φ) be the set of all antiderivatives of a ontinuous fun tion φ.(a) Determine the ontinuous fun tion f : Ip ⊂ R → R\{0} su h thatf(0) = 1 and f−p ∈ I(f), where p is an odd natural number and theinterval Ip ontains zero and is maximal for the given properties of f.(b) Prove that p = q if and only if Ip = Iq.

117Solution by Mi hel Bataille, Rouen, Fran e.(a) Let I be an interval ontaining 0 and let f : I → R\{0} be su h thatf(0) = 1 and f−p ∈ I(f). Then f−p is di�erentiable on the interval I and(f−p)′(x) = f(x) for all x ∈ I.Sin e p is an odd natural number, the fun tion h(x) = xp is a bije tionfromR toR. Moreover, its inverse is di�erentiable ex ept at 0. The omposi-tion h◦f−1(x) = f−p(x) is di�erentiable, thus the fun tion f−1(x) =

1

f(x)is di�erentiable, and so is the fun tion f. Note also that f(x) > 0 for allx ∈ I (sin e f is ontinuous, does not vanish, and f(0) > 0).Now let p = 2m − 1, where m is a positive integer. Sin e the fun tionf is di�erentiable,

f(x) = (f−p)′(x) = −pf−p−1(x)f ′(x) = −(2m − 1)f ′(x)

f(x)2m,so that f satis�es

(2m − 1)f ′(x) +�f(x)

�2m+1= 0 (x ∈ I) .An easy al ulation shows that the fun tion g(x) = f(x)−m satis�es

g′(x)g(x) =m

2m − 1.

Integrating and substituting g(0) = 1, it follows that �g(x)�2

=2mx

2m − 1+1.Thus, 2mx

2m − 1+ 1 > 0 and g(x) =

�2mx

2m − 1+ 1

�1/2 for x ∈ I. We on ludethatI ⊂ Ip =

�−p

p + 1, ∞

� and f(x) =

�(p + 1)x

p+ 1

�− 1p+1 . (1)Conversely, it is easy to he k that the fun tion f de�ned by (1) on Ip satis�es

(f−p)′(x) =

"�(p + 1)x

p+ 1

� p

p+1

#′

= f(x)

for all x ∈ Ip.(b) p = q is equivalent to pp+1

= qq+1

, hen e it is equivalent to Ip = Iq.Also solved by CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA;FRANCISCO JAVIER GARC�IA CAPIT �AN, IES �Alvarez Cubero, Priego de C �ordoba, Spain;OLIVER GEUPEL, Br �uhl, NRW, Germany; ALBERT STADLER, Herrliberg, Switzerland; and theproposers.The di�erentiability of the fun tion f was not addressed by any of the other solvers.

1183419. [2009 : 109, 111℄ Proposed by Vo Quo Ba Can, Can Tho Universityof Medi ine and Pharma y, Can Tho, Vietnam.Let a, b, and c be positive real numbers.(a) Prove that P y li ra2 + 4bc

b2 + c2≥ 2 +

√2.

(b) Prove that P y li ra2 + bc

b2 + c2≥ 2 +

13√

2.

Solution to (a) by Oliver Geupel, Br �uhl, NRW, Germany; and the proposer.We will show that the inequality holds even if one of a, b, c is zero.Without loss of generality, assume that a ≥ b ≥ c ≥ 0.Case 1. If 4b3 ≥ a2c, then we havea2 + 4bc

b2 + c2− a2

b2=

c(4b3 − a2c)

b2(b2 + c2)≥ 0 ,

b2 + 4ca

c2 + a2− b2

a2=

c(4a3 − b2c)

a2(c2 + a2)≥ 0 ,and

c2 + 4ab

a2 + b2≥ 4ab

a2 + b2.The result now follows after several appli ations of the AM{GM Inequality:X y li Êa2 + 4bc

b2 + c2≥ a

b+

b

a+ 2

Êab

a2 + b2

=

�1 − 1

√2

�a2 + b2

ab+

a2 + b2

√2ab

+ 2

Êab

a2 + b2

!≥ 2

�1 − 1

√2

�+ 2

4

Ê2(a2 + b2)

ab

≥ 2

�1 − 1

√2

�+ 2

√2 = 2 +

√2 .

Case 2. If 4b3 ≤ a2c, then a ≥ 2b and we havea2 + 4bc

b2 + c2− a2 + 4b2

2b2=

(b − c)[a2(b + c) − 4b2(b − c)]

2b2(b2 + c2)≥ 0 .Hen e, X y li Êa2 + 4bc

b2 + c2≥Ê

a2 + 4b2

2b2+

b

a+ 2

Êab

a2 + b2.

119Setting x =

a

b≥ 2, we need to prove that

f(x) =

Êx2

2+ 2 +

1

x+ 2

rx

x2 + 1≥ 2 +

√2 .Sin e x ≥ 2, we have

x(x2 + 1) − (x + 1)2 = x3

�1 − 1

x− 1

x2− 1

x3

�≥ x3

�1 − 1

2− 1

4− 1

8

�=

1

8x3 > 0 ;hen e Èx(x2 + 1) > x + 1, and then

f ′(x) =x

2

Éx2

2+ 2

− 1

x2− x2 − 1

(x2 + 1)È

x(x2 + 1)

>1

2

É1

2+

2

x2

− 1

x2− x2 − 1

(x2 + 1)(x + 1)

≥ 1

2− 1

x2− x − 1

x2 + 1=

x2 − 4

4x2+

(x − 2)2 + 1

4(x2 + 1)> 0 .Thus, f(x) is an in reasing fun tion on [2, ∞), and therefore,

f(x) ≥ f(2) =5

2+ 2

r2

5> 2 +

√2 .There was one in orre t solution submitted.Geupel redits the proposer and his publi ation [1℄ at the MathLinks website for hissolution to (a). The proposer submitted a solution to part (b) as well; however, his argu-ment relies on a graphi al omputational pa kage result, whi h has not been veri�ed otherwise.Geupel remarked that the author has published a solution to part (b) in [3℄. Geupel also men-tioned a similar inequality,X y li Éa2 + bc

b2 + c2≥É

2(a2 + b2 + c2)

ab + bc + ca+

1√

2,proved by the proposer in [2℄.Referen es[1℄ Vo Quo Ba Can, www.mathlinks.ro/viewtopic.php?t=197674,download �le toanhocmuonmaumain.pdf (in Vietnamese), Problem 1.46, pages 39f.[2℄ Vo Quo Ba Can, One more inequality,

http://www.mathlinks.ro/viewtopic.php?t=186338[3℄ Vo Quo Ba Can, Some new results,http://www.mathlinks.ro/viewtopic.php?t=186334&start=25

1203420. [2009 : 109, 112℄ Proposed by Mih�aly Ben ze, Brasov, Romania.Prove that

nYk=1

�(k + 1)2

k(k + 2)

�k+1

< n + 1 <nY

k=1

�k2 + k + 1

k(k + 1)

�k+1.Similar solutions by Cao Minh Quang, Nguyen Binh Khiem High S hool,Vinh Long, Vietnam and Oliver Geupel, Br �uhl, NRW, Germany.We re ognize that the produ t on the left is a teles oping produ t, thus

nYk=1

�(k + 1)2

k(k + 2)

�k+1

=2(n + 1)n+2

(n + 2)n+1=

2(n + 1)�1 + 1

n+1

�n+1.

By the Bernoulli Inequality, we have�1 +

1

n + 1

�n+1

> 1 + (n + 1)1

n + 1= 2 ,whi h proves the left inequality.Again by the Bernoulli Inequality, we have�

k2 + k + 1

k(k + 1)

�k+1

=

�1 +

1

k(k + 1)

�k+1

>

�1 +

1

k

�=

k + 1

k.Hen e,

nYk=1

�k2 + k + 1

k(k + 1)

�k+1

>nY

k=1

k + 1

k= n + 1 .Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia andHerzegovina; ARKADY ALT, San Jose, CA, USA; ROY BARBARA, Lebanese University, Fanar,Lebanon; MICHEL BATAILLE, Rouen, Fran e; PAUL BRACKEN, University of Texas, Edinburg,TX, USA; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; CHARLESR. DIMINNIE, Angelo State University, San Angelo, TX, USA; RICHARD I. HESS, Ran hoPalos Verdes, CA, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; SALEMMALIKI �C, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; MADHAV R. MODAK,formerly of Sir Parashurambhau College, Pune, India; JOEL SCHLOSBERG, Bayside, NY, USA;ALBERT STADLER, Herrliberg, Switzerland; EDMUND SWYLAN, Riga, Latvia; PETER Y. WOO,Biola University, LaMirada, CA, USA; TITU ZVONARU, Com�ane�sti, Romania; and the proposer.

3421. [2009 : 109, 112℄ Proposed by CaoMinh Quang, Nguyen Binh KhiemHigh S hool, Vinh Long, Vietnam.Let a, b, and c be positive real numbers su h that abc ≤ 1. Prove thata

b2 + b+

b

c2 + c+

c

a2 + a≥ 3

2.

121Solution by Arkady Alt, San Jose, CA, USA.After the substitution (a, b, c) =

�1

x,1

y,1

z

�, the inequality be omesequivalent toy2

x(y + 1)+

z2

y(z + 1)+

x2

z(x + 1)≥ 3

2,where x, y, and z are positive real numbers with xyz ≥ 1.Sin e by the Cau hy-S hwartz Inequality we have�

(xy + x) + (yz + y) + (zx + z)� � y2

x(y + 1)+

z2

y(z + 1)+

x2

z(x + 1)

�≥ (x + y + z)2 ,it suÆ es to prove that

(x + y + z)2

xy + yz + zx + x + y + z≥ 3

2,or

2(x + y + z)2 ≥ 3(xy + yz + zx) + 3(x + y + z) .However, the last inequality follows immediately from(x + y + z)2 ≥ 3(xy + yz + zx) ,and

(x + y + z)2 = (x + y + z)(x + y + z)

≥ 3 3√

xyz(x + y + z)

≥ 3(x + y + z) .Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia andHerzegovina; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVERGEUPEL, Br �uhl, NRW, Germany; DUNG NGUYEN MANH, High S hool of HUS, Hanoi,Vietnam; SALEM MALIKI �C, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina;PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. There was onein orre t solution and two in omplete solutions submitted.3422. [2009 : 110, 112℄ Proposed by CaoMinh Quang, Nguyen Binh KhiemHigh S hool, Vinh Long, Vietnam.Let a, b, and c be positive real numbers su h that a + b + c ≤ 1. Provethat

a

a3 + a2 + 1+

b

b3 + b2 + 1+

c

c3 + c2 + 1≤ 27

31.

122Solution by Dung Nguyen Manh, High S hool of HUS, Hanoi, Vietnam.Set a =

x

3, b =

y

3, and c =

z

3. Then x, y, and z are positive realnumbers satisfying x + y + z ≤ 3, and the inequality is equivalent toX y li x

x3 + 3x2 + 27≤ 3

31.

By the AM-GM Inequality, we havex3 + 1 + 1 ≥ 3x , and 3

�x2 + 1

�≥ 6x .Therefore, it suÆ es for us to prove thatX y li x

9x + 22≤ 3

31⇐⇒ 3 −

X y li 22

9x + 22≤ 27

31.

By the Cau hy{S hwarz Inequality, � P y li 1

9x + 22

�� P y li 9x + 22�

≥ 9, so itfollows that X y li 22

9x + 22≥ 198

9(x + y + z) + 66

≥ 198

27 + 66

=66

31,whi h settles the inequality on the right of the double impli ation above.Equality holds if and only if x = y = z = 1, or a = b = c =

1

3.Also solved by ARKADY ALT, San Jose, CA, USA; GEORGE APOSTOLOPOULOS,Messolonghi, Gree e; �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia andHerzegovina; ROY BARBARA, Lebanese University, Fanar, Lebanon; PAUL BRACKEN, Uni-versity of Texas, Edinburg, TX, USA; CHIP CURTIS, Missouri Southern State University, Joplin,MO, USA; OLIVER GEUPEL, Br �uhl, NRW, Germany; RICHARD I. HESS, Ran ho Palos Verdes,CA, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; KEE-WAI LAU, HongKong, China; SALEM MALIKI �C, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina;ALBERT STADLER, Herrliberg, Switzerland; PETER Y. WOO, Biola University, La Mirada, CA,USA; TITU ZVONARU, Com�ane�sti, Romania; and the proposer.Janous de�ned the numbers xI = 1.100203739 . . . and xM = 0.657298106 . . ., therespe tive (unique) real roots of 3x4 + 3x3 + x2 −6x −3 and 2x3 + x2 −1, and he showedthat if a, b, and c are positive reals with a + b + c = K and 0 < K/3 ≤ xI , thenX y li a

a3 + a2 + 1≤

8><>: 27K

K3 + 3K2 + 27, if 0 < K/3 ≤ xM ,

3xM

x3M + x2

M + 1, if xM < K/3 ≤ xI .

1233423. [2009 : 110, 112℄ Proposed by CaoMinh Quang, Nguyen Binh KhiemHigh S hool, Vinh Long, Vietnam.Let n ≥ 2 be an integer and x1, x2, . . . , xn positive real numbers su hthat x1 + x2 + · · · + xn = 2n. Prove that

nXj=1

�nX

i=1i6=j

xjÈx3

i + 1

Ǒ≥ 2n(n − 1)

3.

Solution by Arkady Alt, San Jose, CA, USA.Let xi = 2ti, 1 ≤ i ≤ n. Then the given inequality is equivalent tonX

j=1

�nX

i=1i6=j

tjÈ1 + 8t3i

Ǒ≥ n(n − 1)

3, (1)

where ea h ti is positive and t1 + t2 + · · · + tn = n.Sin e È1 + 8t3i ≤ 1 + 2t2i is equivalent to 1 + 8t3i ≤ 1 + 4t2i + 4t4i , or2ti ≤ 1 + t2i , whi h is learly true, we then have

nXj=1

�nX

i=1i6=j

tjÈ1 + 8t3i

Ǒ≥

nXj=1

�nX

i=1i6=j

tj

1 + 2t2i

Ǒ .Hen e, to establish (1), it suÆ es to show that

nXj=1

�nX

i=1i6=j

tj

1 + 2t2i

Ǒ≥ n(n − 1)

3. (2)

Now,nX

j=1

�nX

i=1i6=j

tj

1 + 2t2i

Ǒ=

nXj=1

− tj

1 + 2t2j+

nXi=1

tj

1 + 2t2i

!=

nXi=1

�nX

j=1

tj

1 + 2t2i

�−

nXj=1

tj

1 + 2t2j

=nX

i=1

n

1 + 2t2i−

nXj=1

tj

1 + 2t2j

=nX

i=1

n − ti

1 + 2t2i.

124Thus, (2) be omes

nXi=1

n − ti

1 + 2t2i≥ n(n − 1)

3. (3)To prove (3) we show that the inequality below holds for all positivereal numbers x and all positive integers n, with equality if and only if x = 1:

n − x

1 + 2x2≥�

7n − 4

9

�−�

4n − 1

9

�x . (4)Note that (4) is equivalent, in su ession, to

9(n − x) ≥�(7n − 4) − (4n − 1)x

� �1 + 2x2

� ,0 ≤ (8n − 2)x3 − (14n − 8)x2 + (4n − 10)x + (2n + 4) ,0 ≤

�(4n − 1)x + (n + 2)

�(x − 1)2 ,and learly the last inequality is true.Using (4) and the fa t that t1 + t2 + · · · + tn = n, we then have

nXi=1

n − ti

1 + 2t2i≥

nXi=1

�7n − 4

9−�

4n − 1

9

�ti

�=

(7n − 4)n

9− (4n − 1)n

9=

n(3n − 3)

9=

n(n − 1)

3,establishing (3) and ompleting our proof.Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; ALBERT STADLER,Herrliberg, Switzerland; PETER Y. WOO, Biola University, La Mirada, CA, USA; and theproposer.

3424. Corre tion. [2009 : 110, 112; 233, 235℄ Proposed by YakubN. Aliyev,Qafqaz University, Khyrdalan, 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.Solution by Oliver Geupel, Br �uhl, NRW, Germany.If 0 ≤ i ≤ m, then σ−1(i) = 2i; and if m + 1 ≤ i ≤ 2m, thenσ−1(i) = 2(i − m) − 1 ≡ 2i (mod 2m + 1). Therefore, we have

σ−1(i) ≡ 2i (mod 2m + 1) (1)for ea h i ∈ {0, 1, . . . , 2m}.

125By Euler's theorem, we have

2φ(2m+1) ≡ 1 (mod 2m + 1) , (2)where φ denotes Euler's totient fun tion.Using (1) and (2), we then haveσ−φ(2m+1)(i) ≡ 2φ(2m+1)i ≡ i (mod 2m + 1) ,and it follows that σφ(2m+1)(i) = i for ea h i.Hen e, σφ(2m+1)+1 = σ, and sin e 2 ≤ φ(2m + 1) + 1 ≤ 2m + 1, anappropriate hoi e for k would be k = φ(2m + 1) + 1.Also solved by MICHEL BATAILLE, Rouen, Fran e; EDMUND SWYLAN, Riga, Latvia;and the proposer.

3425. [2009 : 110, 112℄ Proposed by Slavko Simi , Mathemati al InstituteSANU, Belgrade, Serbia.For real x 6= −1, let f(x) =ex

x + 1. Prove that if f(x) = f(y) for some

x 6= y, then �√x + 1 −

py + 1

�2≥ ln f(y) .Solution by Albert Stadler, Herrliberg, Switzerland.The fun tion f(x) is negative and de reasing on (−∞, −1), positiveand de reasing on (−1, 0], and positive and in reasing on [0, ∞). So, if

f(x) = f(y) for some x 6= y, then we an assume that −1 < x ≤ 0 ≤ y.Upon setting u = x + 1 and v = y + 1, the problem statement be omes:Let g(u) =eu

u, u > 0. Prove that if g(u) = g(v) for some u 6= v, then

(√

u −√

v )2 ≥ v − 1 − ln v . (1)Thus, g(1) = e is a lo al minimum value, and 0 < u ≤ 1 ≤ v with u 6= v.Case 1. Here we assume that v ≥ 2. Then 0 < u < 0.5, sin e g(2) > 3.6and g(0.5) < 3.3. From g(u) = g(v) we obtain 1

u≤ eu

u≤ ev

v≤ e0.5

u, andhen e ve−v ≤ u ≤ ve0.5−v. We laim that

ve−v + 1 + ln v ≥ 2ve14− v

2 . (2)holds for v ≥ 2. Indeed, the inequality (2) holds for v = 2 and alsod

dv

�ve−v + 1 + ln v − 2ve

14− v

2

�= −(v − 1)e−v +

1

v− (v − 2)e

14− v

2

=e−v

v

�v(v − 2)e

14+ v

2 + ev − v(v − 1)�

≥ 0 ,

126sin e ev ≥ v(v − 1) for v > 0; inequality (2) follows from these two fa ts.We on lude that u + 1 + ln v ≥ ve−v + 1 + ln v ≥ 2ve

14− v

2 ≥ 2√

uv,whi h is equivalent to (1). This on ludes Case 1.Case 2. Here we assume that 1 ≤ v ≤ 2. We will need two lemmas in orderto settle this ase.Lemma 1 If 0 ≤ t ≤ 0.8, then1 −

√1 − t ≥

rt2 + 0.5t − ln(t2 + 0.5t + 1)

t2 + 0.5t + 1. (3)

Proof: Both sides of (3) are positive, so we obtain an equivalent inequalityby squaring and rearranging terms:1 − t − 2

√1 − t ≥

r−1 − ln(t2 + 0.5t + 1)

t2 + 0.5t + 1,so we want to show that

Ψ(t) = 2 ln(t2 + 0.5t + 1) − 2t3 + t2 − t + 4 − (4t2 + 2t + 4)√

1 − t ≥ 0 .We have Ψ(0) = 0 andΨ′(t)

t=

(−12t3 + 2t2 + 12t − 11)√

1 − t + 5(2t − 1)(2t2 + t + 2)

(2t2 + t + 2)√

1 − t,so we are left to prove that the numerator is nonnegative for 0 < t ≤ 0.8.For 0 ≤ t ≤ 0.6 we have 11 − 12t3 − 2t2 − 12t and √

1 − t ≥ 1 − 23t,so that

(−12t3 + 2t2 + 12t − 11)√

1 − t + 5(2t − 1)(2t2 + t + 2)

≥ (−12t3 + 2t2 + 12t − 11)(1 − 2

3t) + 5(2t − 1)(2t2 + t + 2)

=1

3(24t4 + 28t3 + 18t2 − 13t + 3) ≥ 1

3(28t2 − 13t + 3) > 0 ,and the last inequality holds for t > 0.For 0.6 ≤ t ≤ 0.8 we have���(−12t3 + 2t2 + 12t − 11)

√1 − t + 5(2t − 1)(2t2 + t + 2)

���≥ 5(2t − 1)(2t2 + t + 2 −

��12t3 + 2t2 + 12t − 11��

≥§

3.32 − 2.49 > 0 , if 0.6 ≤ t ≤ 0.7 ,7.63 − 6.1 > 0 , if 0.7 ≤ t ≤ 0.8 .This on ludes the proof of Lemma 1.

127Let t =

v − u

v. Then 0 < t < 1, and we dedu e from eu

u=

ev

vthat

v − u = ln�

v

u

�, or equivalentlyt = −1

vln(1 − t) =

1

v

∞Xk=1

tk

k. (4)Dividing by t we obtain

v − 1 =∞X

k=2

tk−1

k. (5)

The fun tion ϕ(x) =∞P

k=2

xk−1

ksatis�es ϕ(0) = 1, is stri tly in reasing on

(0, 1), and ϕ(x) → ∞ as x → 1. Thus, for ea h v ≥ 1, there is a uniquet = t(v) that satis�es (4) and t(v) stri tly de reases to 0 as v tends to 1. Wenow estimate t(v) from below.Lemma 2 If 1 ≤ v ≤ 2, then t =

v − u

vsatis�es t2 +

t

2+ 1 ≥ v, and thislatter inequality holds if and only if t ≥ −1

4+

1

4

È1 + 16(v − 1).Proof: Sin e g(0.4) > g(2), we have 0 < t(2) < 0.8. Then we obtain that

t =v − u

v<

2 − 0.4

2= 0.8. We dedu e from (4) and (5) that

v − 1 =∞X

k=2

tk−1

k

=t

2+ t2

19Xk=3

tk−3

k+ t18

∞Xk=20

tk−19

k

≤ t

2+ t2

19Xk=3

tk−3

k− t18 ln(1 − t)

=t

2+ t2

19Xk=3

tk−3

k+ t19v ≤ t

2+ t2 ,

sin e 19Pk=3

0.8k−3

k+ 2(0.8)17 < 1. This yields t2 +

t

2+ 1 ≥ v. By straightfor-ward al ulations using the quadrati formula, we see that t2+

t

2−(v−1) ≥ 0is equivalent to t ≥ −0.5 +

p0.25 + 4(v − 1)

2= −1

4+

1

4

È1 + 16(v − 1).Now φ(x) =

x − 1 − log x

xis monotoni ally in reasing for x > 1. Then

t2 + 0.5t − ln(t2 + 0.5t + 1)

t2 + 0.5t + 1≥ v − 1 − log v

v,

128be ause t2 +

t

2+ 1 ≥ v by Lemma 2. Finally, by using Lemma 1, we obtain

√v −

√u =

√v

�1 −

rv − u

v

�=

√v(1 −

√1 − t)

≥√

v

rt2 + 0.5t − ln(t2 + 0.5t + 1)

t2 + 0.5t + 1

≥√

v

Év − 1 − log v

v

=È

v − 1 − log v ,and squaring yields inequality (1).This on ludes Case 2, and all is proved.One in omplete solution was re eived whi h used the method of Lagrange multipliersbut without beforehand establishing the existen e of extrema.The proposer ommented that he dis overed the result by \some other onsiderations"and that he sought an elementary solution to the problem.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