193�EDITORIALV�a lav Linek

Vous avez probablement d �ej �a re� u, �a travers les orrespondan es, latriste nouvelle que notre o �editeur bien aim �e, Jim Totten, est mort le 9 mars2008. Une n �e rologie suit et �edtiorial.En 2009, on onsa ra un num�ero sp �e ial �a Jim. Les propositions deprobl �emes onsa r �es �a Jim que nous re evons avant le 1er janvier 2009 se-ront onsid �er �ees pour un ensemble sp �e ial de probl �emes �a appara�tre dansle num�ero sp �e ial.Les le teurs peuvent �egalement envoyer des arti les onsa r �es �a Jimpour le num�ero sp �e ial avant le 1er d �e embre 2008. Les le teurs qui veulent ontribuer des projets sp �e iaux devraient nous onsulter aussitot que pos-sible, au as o �u une attention parti uli �ere serait requise pour a omplir leprojet et pour �eviter la dupli ation (d'autres peuvent avoir des projets sem-blables �a soumettre).Vue le grand r �eseau des amis math �emati iens de Jim partout dans lemonde, il est fort possible qu'un seul num�ero sp �e ial n'est pas suÆsant pour ontenir tous les propositions et les projets. Dans e as nous distribueronsle mat �eriel suppl �ementaire dans les num�eros suivants.Je vous remer ie, le teurs, de votre appui au ours de ette p �eriodediÆ ile. Je suis maintenu �a ot par vos mots aimables, et votre en ourage-ment me donne la motivation pour ontinuer �a faire mon travail. Au plaisirde ommuniquer ave vous dans le pro hain futur..................................................................Some of you already know the sad news through orresponden e, thatour herished Co-Editor, Jim Totten, passed away on Mar h 9, 2008. Anobituary follows this editorial.In 2009 a spe ial issue will be dedi ated to Jim. Problem proposals were eive before January 1, 2009 and dedi ated to Jim will be onsidered for aspe ial set of problems to appear in the spe ial issue.Readers may also send arti les dedi ated to Jim for the spe ial issuebefore De ember 1, 2008. Readers wishing to ontribute spe ial proje tsshould onsult with us as soon as possible, in ase spe ial attention is re-quired to omplete the proje t and to avoid dupli ation (others may havesimilar proje ts in mind).Given Jim's extensive network ofmathemati al friends all over the world,it is quite possible there will be morematerial than one spe ial issue an hold.In that ase we will distribute the extra material in subsequent issues.I thank you, the readers, for your support during this diÆ ult period.I am buoyed by your kind words, and your en ouragement keeps me going.I very mu h look forward to orresponding with all of you in the future.

194IN MEMORIAMJames Edward Totten, 1947{2008

With the sudden death of Jim (James Edward Totten) on Mar h 9, 2008,theMathemati s Community lost someone who was dedi ated to mathemat-i s edu ation and to mathemati al outrea h. Jim was born August 9, 1947 inSaskatoon and raised in Regina. After obtaining his Ba helor's degree atthe University of Saskat hewan, Jim then earned a Master's and PhD degreein Mathemati s from the University of Waterloo, after whi h he joined thefa ulty at Saint Mary's University in Halifax. One of us (Robert) �rst got toknow Jim when he and Jim shared an oÆ e at the University of Saskat hewanwhile Jim visited there during 1978-1979. That was a long old winter, butJim's a tive interest and enthusiasm for mathemati s and the tea hing ofmathemati s made the year a memorable one. The next year Jim took upa position at Cariboo College, where he remained as it evolved into theUniversity College of the Cariboo and then eventually into Thompson RiversUniversity, retiring as Professor Emeritus in 2007.During his years in Kamloops, Jim was a mainstay of the CaribooContest, an annual event whi h brought students in to the ollege and whi hfeatured a keynote speaker, often drawn from Jim's list of mathemati alfriends. This on e in luded an invitation to Robert, whi h featured a talkon publi key en ryption mostly memorable for the failure of te hnology ata key moment, mu h to Jim's amusement.Jim be ame a member of the CMS in 1981, and joined the editorialboard of Crux in 1994. When Bru e was looking for someone to su eed himas Editor-in-Chief, there was no doubt in his mind whom to approa h. Bru espent a week in Kamloops staying at Jim's home and working with Jim andBru e Crofoot to smooth the transition. Jim's attention to detail and arewas appre iated by all, parti ularly those ontributing opy that was arefully he ked, as Robert gladly on�rms from his ontinued asso iation with Jimthrough the Olympiad Corner, an asso iation whi h ontinued to the end.Jim loved his Oldtimers' ho key and was an avid golfer. When not play-ing ho key or golf, he was a tive with the Kamloops Outdoors Club. Jim wasnever just a parti ipant, always an a tive volunteer.Jim is survived by his loving wife of 40 years, Lynne, his son Dean,daughter-in-law Christie and granddaughter Mikayla of Se helt, his fatherWilf Totten of Edmonton, sister Judy Totten of Regina, sister Josie Laing(Neil) of Onoway, sister-in-law Marilyn Totten of Red Deer, sister-in-lawConstan e Ladell (David Dahl) of Kamloops, many ousins, family friendsand mathemati al olleagues. All miss his warmth and love.To honour the memory of Jim, the Thompson Rivers UniversityFoundation is now a epting ontributions for the Jim Totten S holarship.As remembered by two of his olleagues from Crux,Bru e Shawyer and Robert Woodrow

195SKOLIAD No. 110

Robert BilinskiPlease send your solutions to the problems in this edition by 1 O tober,2008. A opy of MATHEMATICAL MAYHEM Vol. 4 will be presented toone pre-university reader who sends in solutions before the deadline. Thede ision of the editor is �nal.Nos questions proviennent e mois- i du Con ours de Comt �e, So i �et �eMath �ematique de Croatie 2007 (niveau se ondaire). Nous remer ions Hanj�s�Zeljko, Universit �e de Zagreb, pour la opie du on ours.

Con ours de Comt �eSo i �et �e Math �ematique de Croatie(niveau se ondaire, 1er grade) 9 mars 20071. Trouver toutes les solutions enti �eres de l' �equation x2 + 112 = y2.2. Dans un er le ave entre S et rayon r = 2, deux rayons SA et SB sonttra �es. L'angle entre eux est 45◦. Soit K l'interse tion de la ligne AB et laperpendi ulaire de AS passant par S. Soit L le pied de la hauteur du sommetB dans le triangle ABS. D �eterminer l'aire du trap �eze SKBL.3. Soient a, b et c trois nombres r �eels non-nuls donn �es. Trouver x, y et z si

ay + bx

xy=

bz + cy

yz=

cx + az

zx=

4a2 + 4b2 + 4c2

x2 + y2 + z2.

4. Soient a et b deux nombres r �eels positifs tels que a > b et ab = 1. Prouvezl'in �equationa − b

a2 + b2≤

√2

4.D �eterminer a + b quand l' �egalit �e tient.5. Le ratio des longueurs des deux ot �es d'un re tangle est 12 : 5. Les dia-gonales divisent le re tangle en quatre triangles. Les er les ins rits de deuxtriangles qui ont un ot �e ommun sont tra �es. Soient r1 et r2 les rayons.Trouver le ratio r1 : r2.

196County CompetitionThe Croatian Mathemati al So iety(se ondary level, 1st grade) Mar h 9, 20071. Find all integer solutions to the equation x2 + 112 = y2.2. In a ir le with entre S and radius r = 2, two radii SA and SB aredrawn. The angle between them is 45◦. Let K be the interse tion of the line

AB and the perpendi ular to line AS through point S. Let L be the footof the altitude from vertex B in ∆ABS. Determine the area of trapezoidSKBL.3. Let a, b, and c be given nonzero real numbers. Find x, y, and z if

ay + bx

xy=

bz + cy

yz=

cx + az

zx=

4a2 + 4b2 + 4c2

x2 + y2 + z2.

4. Let a and b be positive real numbers su h that a > b and ab = 1. Provethe inequalitya − b

a2 + b2≤

√2

4.Determine a + b if equality holds.5. The ratio between the lengths of two sides of a re tangle is 12 : 5. Thediagonals divide the re tangle into four triangles. Cir les are ins ribed in twoof them having a ommon side. Let r1 and r2 be their radii. Find the ratio

r1 : r2.Next we give solutions to the 23rd W.J. Blundon Mathemati s Contestgiven at [2007 : 321-322℄.23rd W.J. Blundon Mathemati s ContestSponsored by the Canadian Mathemati al So ietyin ooperation withThe Department of Mathemati s and Statisti sMemorial University of NewfoundlandFebruary 22, 20061. If loga x = logb y, show that ea h is also equal to logab xy.Solution by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto,ON. Let N = loga x = logb y. Then aN = x and bN = y or (ab)N = xy.This means that N is also equal to logab xy.

1972. In how many ways an 20 dollars be hanged into dimes and quarters,with at least one of ea h oin used?Solution by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto,ON. An amount of 20 dollars is equal to 2000 ents. The last digit being 0,the number of quarters used is even, sin e the other denomination gives usmultiples of 10. The number of nonzero even multiples of 25 whi h are lessthan 2000 is 39. Therefore there are 39 ways that 20 dollars an be hangedinto dimes and quarters, with at least one of ea h used.3. If one of the women at a party leaves, then 20% of the people remainingat the party are women. If, instead, another woman arrives at the party,then 25% of the people at the party are women. How many men are at theparty?Solution by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto,ON. Fra tions 20% and 25% are equal to 1

5and 1

4, respe tively. Let x bethe number of women at the party and let y be the number of people at theparty. A ording to the problem, we have

x − 1 =1

5(y − 1) ,

x + 1 =1

4(y + 1) .These equations have the solution (x, y) = (7, 31), and therefore the numberof men at the party is y − x = 24.4. Find two fa tors of 248 − 1 between 60 and 70.Solution by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto,ON. Repeatedly using the formula x2 − y2 = (x − y)(x + y), we have

248 − 1 = (224 + 1)(212 + 1)(26 + 1)(26 − 1) .Sin e 26 = 64, two fa tors of 248 − 1 between 60 and 70 are 63 and 65.5. The yearly hanges in the population ensus of a town for four onse utiveyears are, respe tively, 25% in rease, 25% in rease, 25% de rease, and 25%de rease. Find the net per ent hange to the nearest per ent over the fouryears.

198Solution by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto,ON. Let A be the number of people in the town originally. The numberof people after four years is A(1.25)(1.25)(0.75)(0.75) = 0.87890625A.Therefore, the net per ent hange to the nearest per ent over four years is a12% de rease.6. If x + y = 5 and xy = 1, �nd x3 + y3.Solution by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto,ON. The steps in the solution are

(x + y)3 = 53 ,x3 + 3x2y + 3xy2 + y3 = 125 ,

x3 + y3 = 125 − 3xy(x + y) ,x3 + y3 = 125 − 3(1)(5) ,x3 + y3 = 110 .7. The point (4, 1) is on the line that passes through the point (4, 1) and isperpendi ular to the line y = 2x + 1. Find the area of the triangle formedby the line y = 2x + 1, the given perpendi ular line, and the x-axis.Solution by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto,ON. Let y = mx + b be the equation of the unknown line. Sin e it isperpendi ular to the line y = 2x + 1 and it passes through the point (4, 1),its equation is y = −1

2x + 3. From the equations, we get the oordinates ofall the verti es and the dimensions of the triangle. Its base is 6.5 units andits height is 2.6 units. Therefore, its area is 8.45 square units.

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r

r

r

r(6, 0)

(4, 1)

(− 1

2, 0)

(a, 2a + 1) = (4

5, 13

5)

y = 2x + 1

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

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

8. An arbitrary point is sele ted inside an equilateral triangle. From thispoint perpendi ulars are dropped to ea h side of the triangle. Show thatthe sum of the lengths of these perpendi ulars is equal to the length of thealtitude of the triangle.

199OÆ ial Solution, modi�ed by the editor.Let s be the length of the equal sides, let h bethe altitude, and let x, y, and z be the threedistan es from the interior point to the sidesof the triangle. Then,Area of triangle =

1

2sx +

1

2sy +

1

2sz

=1

2s(x + y + z) .

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....But we also have Area of triangle =1

2sh .Can elling 1

2s from ea h expression for the area, we have x + y + z = h.Also solved by RUIQI YU, student, Stephen Lea o k Collegiate Institute, Toronto, ON9. Find all positive integer triples (x, y, z) satisfying the equations

x2 + y − z = 100 and x + y2 − z = 124 .Solution by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto,ON. Subtra t the �rst equation from the se ond and then fa tor the result toobtain (y − x)(x + y − 1) = 24. Trying out all pairs of fa tors of 24, namely(1, 24), (2, 12), (3, 8), and (4, 6) for ea h of the two expressions leads to twopositive integer solutions for x and y, namely (x, y) = (3, 6) or (12, 13).However, the �rst solution for x and y leads to a negative z, and must beex luded. Hen e (x, y, z) = (12, 13, 57).10. How many roots are there of the equation sin x = 1

100x?Solution by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto,ON. If x is measured in degrees, then we have 3 roots, sin e in this asethe line y = x

100only interse ts one period of y = sin x. If x is measuredin radians, then the line has a y-value bigger than 1 for x bigger than 100.From x = 0 to x = 100, the line y = x

100 uts through approximately 15.91periods of y = sin x. So we see that the line uts the urve 32 times when x isnot negative. By symmetry, it uts the urve 32 times when x is not positive.But we ount (0, 0) twi e, hen e the line interse ts the urve 32+32−1 = 63times. Hen e there are 63 roots when x is measured in radians.

That brings us to the end of another issue. This month's winner ofa past Volume of Mayhem is Ruiqi Yu. Congratulations Ruiqi! Continuesending in your ontests and solutions.

200MATHEMATICAL MAYHEMMathemati al Mayhem began in 1988 as a Mathemati al Journal for and byHigh S hool and University Students. It ontinues, with the same emphasis,as an integral part of Crux Mathemati orum with Mathemati al Mayhem.The Mayhem Editor is Ian VanderBurgh (University of Waterloo). Theother sta� members are Monika Khbeis (As ension of Our Lord Se ondaryS hool, Mississauga), Eri Robert (LeoHayesHigh S hool, Frederi ton), LarryRi e (University of Waterloo), and Ron Lan aster (University of Toronto).

Mayhem ProblemsVeuillez nous transmettre vos solutions aux probl �emes du pr �esent num�eroavant le 15 juillet 2008. Les solutions re� ues apr �es ette date ne seront prises en ompte que 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.M344. Propos �e par l' �Equipe de Mayhem.On onstruit le tableau arr �e

1 2 34 5 67 8 9form�e des nombres de 1 �a 9 �e rits en ordre roissant, ligne apr �es ligne. Lasomme des entiers sur les deux diagonales est 15. Si l'on onstruit un tableau arr �e semblable ave les entiers de 1 �a 10 000, quelle sera la somme desnombres sur ha une des diagonales ?M345. Propos �e par John Grant M Loughlin, Universit �e du Nouveau-Brunswi k, Frederi ton, NB.L'aire d'un triangle iso �ele ABC est q

√15. Si AB = 2BC, exprimerle p �erim �etre du triangle ABC en fon tion de q.M346. Propos �e par l' �Equipe de Mayhem.D �eterminer le nombre de hi�res que omporte l'entier 280 sans l'aided'une al ulatri e.

201M347. Propos �e par l' �Equipe de Mayhem.Quatre nombres entiers positifs a, b, c et d sont tels que

(a + b + c)d = 420 ,(a + c + d)b = 403 ,(a + b + d)c = 363 ,(b + c + d)a = 228 .Trouver es quatre nombres.M348. Propos �e par l' �Equipe de Mayhem.Le p �erim �etre d'un se teur de er le est 12 (le p �erim �etre in lut les deuxrayons et l'ar ). D �eterminer le rayon du er le qui maximise l'aire de e se -teur.M349. Propos �e par l' �Equipe de Mayhem.(a) Trouver toutes les paires ordonn �ees d'entiers (x, y) ave 1

x+

1

y=

1

5.(b) Combien y a-t-il de paires ordonn �ees d'entiers (x, y) ave

1

x+

1

y=

1

1200?

.................................................................M344. Proposed by the Mayhem sta�.Consider the square array1 2 34 5 67 8 9formed by listing the numbers 1 to 9 in order in onse utive rows. The sumof the integers on ea h diagonal is 15. If a similar array is onstru ted usingthe integers 1 to 10 000, what is the sum of the numbers on ea h diagonal?M345. Proposed by John GrantM Loughlin, University of New Brunswi k,Frederi ton, NB.The area of isos eles △ABC is q

√15. Given that AB = 2BC, expressthe perimeter of △ABC in terms of q.M346. Proposed by the Mayhem Sta�.Without using a al ulator, �nd the number of digits in the integer 280.

202M347. Proposed by the Mayhem Sta�.Four positive integers a, b, c, and d are su h that

(a + b + c)d = 420 ,(a + c + d)b = 403 ,(a + b + d)c = 363 ,(b + c + d)a = 228 .Find the four integers.M348. Proposed by the Mayhem Sta�.The perimeter of a se tor of a ir le is 12 (the perimeter in ludes thetwo radii and the ar ). Determine the radius of the ir le that maximizes thearea of the se tor.M349. Proposed by the Mayhem Sta�.(a) Find all ordered pairs of integers (x, y) with 1

x+

1

y=

1

5.(b) How many ordered pairs of integers (x, y) are there with

1

x+

1

y=

1

1200?

Mayhem SolutionsM294. Proposed by Bru e Shawyer, MemorialUniversity of Newfoundland, St. John's, NL.Nine ir les of radius 1/2 are externally tan-gent to a ir le of radius 1 and are tangent to oneanother, as shown.Determine the distan e between the entresof the �rst and last of the ir les of radius 1/2.

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Solution by Cao Minh Quang, Nguyen Binh Khiem High S hool, Vinh Long,Vietnam, adapted by the editor.Label the entre of the �rst ir le of radius 1

2with A, the entre of these ond ir le with B, the entre of the last (ninth) ir le with C, and the entre of the ir le of radius 1 with O. We need to determine the length of

AC. We know that AO, BO, and CO are ea h equal to 1 + 1

2= 3

2, be ausethe smaller ir les are tangent to the larger ir le, and AB = 1

2+ 1

2= 1,be ause the smaller ir les with entres A and B are tangent.

203Let ∠AOB = α. Then ∠AOC = 360◦ − 8α, be ause there are 8 pairsof ir les determining an angle of α at O. Applying the Law of Cosines to

△AOC and using cos θ = cos(360◦ − θ), we haveAC2 = AO2 + CO2 − 2AO · CO cos ∠AOC

=

(3

2

)2+

(3

2

)2− 2

(3

2

)(3

2

)cos(360◦ − 8α)

=9

2(1 − cos 8α) .To �nd cos 8α, we �rst �nd cos α using the Law of Cosines in △AOB:

AB2 = AO2 + BO2 − 2AO · BO cos ∠AOB ,12 =

(3

2

)2+

(3

2

)2− 2

(3

2

)(3

2

)cos α ,

and solving for cos α gives cos α = 7

9. Therefore,

cos 2α = 2 cos2 α − 1 = 2

(7

9

)2− 1 =

17

81,

cos 4α = 2 cos2 2α − 1 = 2

(17

81

)2− 1 = −5983

6561,

cos 8α = cos2 4α − 1 = 2

(5983

6561

)2− 1 =

28545857

43046721,

and hen eAC =

√9

2(1 − cos 8α) =

√9

2

(1 − 28545857

43046721

)=

1904√

2

2187.

Also solved by HASAN DENKER, Istanbul, Turkey; ANGELA DREI, Riolo Terme, Italy;RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; R. LAUMEN, Deurne, Belgium; MISSOURISTATE UNIVERSITY PROBLEM SOLVING GROUP, Spring�eld, MO, USA; RICARD PEIR �O, IES\Abastos", Valen ia, Spain; KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong,India; and J. SUCK, Essen, Germany. There was 1 in orre t solution submitted.M295. Proposed by Bru e Shawyer, Memorial University of Newfound-land, St. John's, NL.Square ABCD is ins ribed in one-eighth ofa ir le of radius 1 so that there is one vertex onea h radius and two verti es on the ar .Determine the exa t area of the square in theform a + b√

c

d, where a, b, c, and d are integers.

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204Solved independently by Hasan Denker, Istanbul, Turkey; Salem Maliki� ,student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; R. Laumen,Deurne, Belgium; and Ri ard Peir �o, IES \Abastos", Valen ia, Spain.Let N and M represent the midpoints ofAD and BC, respe tively. In right triangleOAN , we have AN = 1

2BC and ∠AON = π

8.Therefore, 1

2BC

ON= tan π

8. We al ulate tan π

8.Sin e

1 = tanπ

4=

2 tan π8

1 − tan2 π8

, ...............................................................................................................................................................................................................................

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A

B

C

DO

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we let x = tan π8and we simplify to obtain x2 + 2x − 1 = 0. Now x ispositive, so by the quadrati formula tan π

8= x =

−2 +√

8

2=

√2 − 1.Thus 1

2BC

ON=

√2 − 1 and ON =

BC

2√

2 − 2, and by the PythagoreanTheorem in triangle BOM we have

OB2 = BM2 + OM2 = BM2 + (ON + NM)2 .Sin e OB = 1, NM = BC, BM = 1

2BC, and ON =

BC

2√

2 − 2, we obtain

1 =1

4BC2 +

(BC

2√

2 − 2+ BC

)2

=1

4BC2 +

(2√

2 − 1

2√

2 − 2

)2

BC2

=1

4BC2 +

((2

√2 − 1)(2

√2 + 2)

4

)2

BC2

=1

4BC2 +

(11 + 6

√2

4

)BC2 =

(6 + 3

√2

2

)BC2 ,

hen e the area is BC2 =2

6 + 3√

2, whi h upon rationalizing is 2 −

√2

3.Also solved by SAMUEL G�OMEZMORENO, Universidad de Ja �en, Ja �en, Spain; RICHARDI. HESS, Ran ho Palos Verdes, CA, USA; CAOMINH QUANG, Nguyen Binh Khiem High S hool,Vinh Long, Vietnam; KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; andJ. SUCK, Essen, Germany.M296. Proposed by Daniel Tsai, student, Taipei Ameri an S hool, Taipei,Taiwan.Let n be a positive integer. In the Cartesian plane, onsider the points

ak = (k, n) and bk = (k, 0) for k = 1, 2, . . . , n. We onne t ea h pairak, bk by a straight (verti al) line segment. Then we draw an arbitrary �nitenumber of horizontal line segments, ea h onne ting two adja ent verti alline segments, su h that no one point on any verti al segment is the end-point of two horizontal segments.

205Let A = {a1, a2, . . . , an} and B = {b1, b2, . . . , bn}. De�ne a mapfrom A to B as follows: starting from ai, travel down the segment untilyou meet the end-point of a horizontal segment, go to the other end-pointof that segment, and ontinue on down the new verti al line, repeating thisuntil there are no more horizontal segments to meet, �nally ending at bj forsome j. Show that no two points of A map to the same point of B.Solution by Ri hard I. Hess, Ran ho Palos Verdes, CA, USA.The pro ess des ribed is uniquely reversible, so that starting from bjgets you to ai if going from ai got you to bj. Therefore, when going in reverse,there is no possibility of splitting paths from bj to ai, and therefore no twopoints in A an map to the same point in B.Also solved by JACLYN CHANG, student, Western Canada High S hool, Calgary, AB; andthe proposer.M297. Proposed by John GrantM Loughlin, University of New Brunswi k,Frederi ton, NB.Numbers su h as 34543 and 713317 whose digits an be reversed with-out hanging the number are alled palindromes. Show that all four-digitpalindromes are multiples of 11.Solved independently by Angela Drei, Riolo Terme, Italy; Samuel G �omezMoreno, Universidad de Ja �en, Ja �en, Spain; Ri hard I. Hess, Ran ho PalosVerdes, CA, USA; R. Laumen, Deurne, Belgium; Salem Maliki� , student,Sarajevo College, Sarajevo, Bosnia and Herzegovina; and Kunal Singh, stu-dent, Kendriya Vidyalaya S hool, Shillong, IndiaEvery four-digit palindrome an be written in the form abba, where aand b are integers, su h that 1 ≤ a ≤ 9 and 0 ≤ b ≤ 9. We an then on ludethat

abba = 1000a + 100b + 10b + a = 1001a + 110b = 11(91a + 10b) .Thus, every four-digit palindrome is a multiple of 11.Also solved by JACLYN CHANG, student, Western Canada High S hool, Calgary, AB; andDANIEL TSAI, student, Taipei Ameri an S hool, Taipei, Taiwan.M298. Proposed by John GrantM Loughlin, University of New Brunswi k,Frederi ton, NB.(a) Given that a number is a four-digit palindrome, what is the probabilitythat the number is a multiple of 99?(b) Given that a four-digit number is a multiple of 99, what is the proba-bility that the number is a palindrome?

206Solution by Samuel G �omez Moreno, Universidad de Ja �en, Ja �en, Spain.(a) A four-digit palindrome has the form abba, where a and b are integerswith 1 ≤ a ≤ 9 and 0 ≤ b ≤ 9. There are therefore 9 · 10 = 90four-digit palindromes. By Problem M297, we know that all four-digitpalindromes are multiples of 11. For a four-digit palindrome, being amultiple of 99 is therefore equivalent to being a multiple of 9.A positive integer is a multiple of 9 if and only if the sum of its digits isalso a multiple of 9, so the four-digit palindromes whi h are multiplesof 99 are pre isely those for whi h 2(a + b) is a multiple of 9, thatis, 2(a + b) equals 18 or 36. A omplete list of these palindromes is

1881, 2772, 3663, 4554, 5445, 6336, 7227, 8118, 9009, and 9999. Thusthe probability is 10

90= 1

9.(b) There are a total of 91 four-digit multiples of 99, namely the multi-ples from 99(11) = 1089 to 99(101) = 9999. By part (a), we knowthat among these there are ten four-digit palindromes. Therefore therequired probability is 10

91.Also solved by ANGELA DREI, Riolo Terme, Italy; RICHARD I. HESS, Ran ho PalosVerdes, CA, USA; R. LAUMEN, Deurne, Belgium; and DANIEL TSAI, student, Taipei Ameri anS hool, Taipei, Taiwan. There were 2 in orre t solutions submitted.M299. Proposed by Titu Zvonaru, Com�ane�sti, Romania.Let a, b, and c be positive real numbers with ab + bc + ca = 3. Provethat

ab

c2 + 1+

bc

a2 + 1+

ca

b2 + 1≥ 3

2.Solution by Cao Minh Quang, Nguyen Binh Khiem High S hool, Vinh Long,Vietnam.Using the Arithmeti Mean-Geometri Mean Inequality, we have

(abc)2/3 = 3√

(ab)(bc)(ac) ≤ ab + bc + ca

3= 1 ,hen e abc ≤ 1. The desired inequality is equivalent to

(ab − ab

c2 + 1

)+

(bc − bc

a2 + 1

)+

(ca − ca

b2 + 1

)≤ 3

2,whi h is the same as

abc2

c2 + 1+

bca2

a2 + 1+

cab2

b2 + 1≤ 3

2.We have (x − 1)2 ≥ 0, hen e x2 + 1 ≥ 2x, and hen e 1

x2 + 1≤ 1

2xwhen xis positive. Thus,

abc2

c2 + 1+

bca2

a2 + 1+

cab2

b2 + 1≤ abc2

2c+

bca2

2a+

cab2

2b=

3

2abc ≤ 3

2.

207Equality holds if and only if a = b = c = 1

3.Also solved by ARKADY ALT, San Jose, CA, USA (two solutions); JOE HOWARD, Portales,NM, USA; SALEMMALIKI �C, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; CAOMINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam (se ond solution); andSHI CHANGWEI, Xi�an City, Shaan Xi Provin e, China. There were 2 in orre t solutions sub-mitted.M300. Proposed by Geo�rey A. Kandall, Hamden, CT, USA.Let ABC be an arbitrary triangle. Let D and E be points on the sides

AC and AB, respe tively, and let P be the point of interse tion of BDand CE. If AE : EB = r and AD : DC = s, determine the ratio of areas[ABC] : [PBC] in terms of r and s.Solution by Salem Maliki� , student, Sarajevo College, Sarajevo, Bosnia andHerzegovina.Let AP meet BC at point K. Starting with Ceva's Theorem applied to△ABC and substituting the given ratios, we obtain

AE

EB· BK

KC· CD

DA= 1 =⇒ r · BK

KC· 1

s= 1 =⇒ KC

BK=

r

s,and hen e

CB

BK=

KC + BK

BK=

r

s+ 1 =

r + s

s.Applying Menelaus' Theorem to the transversal BPD and △AKC, and sub-stituting the expression just obtained, we have

AD

DC· CB

BK· KP

P A= 1 =⇒ s · r + s

s· KP

P A= 1

=⇒ P A

KP= r + s ,so it follows that

KA

KP=

P A + KP

KP= r + s + 1 .Sin e x

y=

z

timplies x

y=

z

t=

x + z

y + t, we ombine this with the previousresults to obtain

r + s + 1 =KA

KP=

[△BKA]

[△BKP ]=

[△KAC]

[△KP C]

=[△BKA] + [△KAC]

[△BKP ] + [△KP C]=

[△ABC]

[△P BC],where [△BKA] denotes the area of △BKA, and so forth. This shows that

[△ABC] : [△PBC] = r + s + 1.Also solved by HASAN DENKER, Istanbul, Turkey; RICARD PEIR �O, IES \Abastos",Valen ia, Spain; KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; J. SUCK,Essen, Germany; and TITU ZVONARU, Com�ane�sti, Romania. There was 1 in orre t solutionsubmitted.

208Problem of the Month

Ian VanderBurghAs spring arrives inmost parts of the ountry, it's time to dust o� our bi- y les. This problem is a bit reminis ent of the gears and wheels of a bi y le.Problem (1996 Cayley Contest)Two ir ular dis s have radii 8 and 28. The larger dis is �xed while the smaller dis rolls around the outsideof the larger dis . In their original positions, point Aon the smaller dis oin ides with point B on the largerdis . The least number of rotations that the small dis makes about its entre before A and B again oin ide is(A) 5 (B) 2 (C) 9 (D) 4.5 (E) 3.5

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qA

B

A friend ofmine askedme about this problem from the ar hives a oupleof months ago. It stumped me! Here's what I did initially:As the smaller ir le rolls around the larger ir le, we assume thatno slipping o urs. As a result, the length of ir umferen e trav-elled as measured on the smaller ir le equals that as measuredon the larger ir le.Let's suppose that the smaller ir le travels m times around its ir umferen e while rolling n times around the larger ir le. Sin ewe want points A and B to line up at the end of the pro ess, thenea h of m and n should be integers.The length of ir umferen e travelled along the smaller ir le ism2π(8) = 16πm and the length travelled along the larger ir leis n2π(28) = 56πn.For these lengths to be equal, 16πm = 56πn or 2m = 7n. Sin em and n are positive integers and we want m to be as small aspossible, then m = 7 and n = 2.Therefore, the smaller ir le rotates 7 times around its entre be-fore the �rst time that A and B oin ide.Wait... 7 is not one of the possible answers! Is the question wrong?Did we do something wrong mathemati ally? Is our logi wrong?The question is orre t. In fa t, this is indeed the �rst time that Aand B oin ide again, but somehow 7 is the wrong answer. The thing thatwe haven't taken into a ount is that as the smaller ir le rolls, its motionaround the larger ir le produ es extra rotations around its entre.Here are three ways to look at this problem. It's up for you to de idehow valid these solutions are and whi h onvin es you the most.

209Solution 1. 7 + 2 = 9.That's the shortest solution to a POTM in my time here! What is thistrying to say? The smaller ir le rotates 7 times while rolling 2 times aboutthe entre of the larger ir le. So perhaps the smaller ir le rotates an extra2 times around its entre, for a total of 9 times. Do you buy this? (A tually,a good point here is that the total number of rotations of the smaller ir leshould be at least 7, so given the possible answers, (C) 9 must be orre t.)Solution 2. When two ir les are tangent (like the smaller and larger ir les),the line joining their entres passes through the point of tangen y.Thus, the distan e between the entres is the sum of the radii, or 8+28 = 36.Therefore, the entre of the smaller ir le is always a distan e of 36 from the entre of the larger ir le. Hen e, the entre of the smaller ir le travelsaround a ir le of radius 36.The entre of the smaller ir le moves 2 times around this ir le, so travels atotal distan e of 2(2π(36)) = 144π.The ir umferen e of the smaller ir le is 2π(8) = 16π, and so while the entre of this ir le moves a distan e of 144π m, it must rotate 144π

16π= 9times.This feels better than Solution 1, but still leaves a somewhat un ertainfeeling. The ratio of distan e travelled by the entre to the number of ro-tations works on a at surfa e, but should it work rolling around a ir ularsurfa e?The last solution is slightly less intuitive, but a bit more mathemati al.Solution 3. First, we turn the diagram so that the line joining the entres ishorizontal. We don't really have to do this, but you might �nd it easier tovisualize.Next let's suppose that the smaller ir le has rolled so that the point of onta t between the ir les is at an angle of x below the horizontal. (We'lltreat the angles as measured in radians, butfeel free to think in degrees if that's easier foryou.) Call the entre of the larger ir le O,the entre of the smaller ir le C, the pointof tangen y T ; we label the end-point of theradius to the left of C as P . We have de�ned

∠TOB = x. Sin e OB is parallel to PC, thenwe also have ∠TCP = x.We now use the fa t that the length of ar BTequals the length of ar TA (by equal distan-................................................................................................................................................................................

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q

q

O AB

CP

Tx

q q

q

q

es rolled). The length of ar BT as a fra tion of the entire ir umferen eof the ir le with entre O equals the fra tion that ∠TOB is of the angleasso iated with a full revolution. Thus, the length of ar BT is∠TOB

2π

(2π(TO)

) .

210Similarly, the length of ar TA is

∠TCA

2π

(2π(TC)

) .Sin e these ar s are equal in length∠TOB

2π

(2π(TO)

)=

∠TCA

2π

(2π(TC)

) ,while in degrees this equation is∠TOB

360◦(2π(TO)

)=

∠TCA

360◦(2π(TC)

) .The radian version of this equation be omesx

2π

(2π(28)

)=

∠TCA

2π

(2π(8)

) ,while the degree version be omesx

360◦(2π(28)

)=

∠TCA

360◦(2π(8)

) .Both of these give 28x = 8∠TCA, when e ∠TCA = 7

2x. Therefore, wehave

∠PCA = ∠TCP + ∠TCA =9

2x .What does this tell us? When the smaller ir le has rolled twi e around the ir umferen e of the larger ir le, we'll have x = 4π, so this tells us that thetotal angle travelled by B around C is 9

2(4π) = 18π. Thus, the smaller ir lemakes 9 omplete rotations around its entre.I �nd this problem pretty fas inating be ause of its ounterintuitivenature. I hope that these di�erent solutions gave you something to thinkabout!

Notes from the Mayhem EditorCalling all proposers! Do you have a neat problem that you'd like to seein Mayhem? We would appre iate re eiving your problem proposals. They an be sent by email to: mayhem-editors@cms.math.ca or by regular mail to:Ian VanderBurgh; CEMC, University of Waterloo; 200 University Ave. W.;Waterloo, ON, Canada; N2L 3G1. Please keep in mind that we are lookingfor problems at the level of those that have appeared so far in Volume 34.Some of Mayhem's loyal followers have noti ed that a few favouriteproblems from the past have reappeared in re ent issue(s). While this mayhave been a idental, we believe that this is perfe tly a eptable in Mayhem,as its audien e is intended to be students in se ondary and post-se ondarys hools, who tend to move on after a few years.

211THE OLYMPIAD CORNERNo. 270

R.E. WoodrowThe Canadian summer break is oming up. I would like to en ouragereaders to use this time to send me solutions to problems from the Cornerto build up my �le for future numbers of the Corner. For your problempleasure we �rst give the twenty problems of the Mathemati al CompetitionBalti Way 2004. My thanks go to Felix Re io, Canadian Team Leader to theIMO in Mexi o, for olle ting them for our use.MATHEMATICAL COMPETITION BALTIC WAY 2004November 7, 20041. Let a1, a2, a3, . . . be a sequen e of non-negative real numbers su h thatfor ea h n ≥ 1 both an + a2n ≥ 3n and an+1 + n ≤ 2√

(n + 1)an hold.(a) Prove that an ≥ n for ea h n ≥ 1.(b) Give an example of su h a sequen e.2. Let P (x) be a polynomial with non-negative oeÆ ients. Prove that ifP

(1

x

)P (x) ≥ 1for x = 1, then the same inequality holds for ea h positive x.3. Let p, q, and r be positive real numbers and let n be a positive integer.If pqr = 1, prove that

1

pn + qn + 1+

1

qn + rn + 1+

1

rn + pn + 1≤ 1 .

4. Let {x1, x2, . . . , xn} be a set of real numbers with arithmeti mean X.Prove that there is a positive integer K su h that the arithmeti mean of ea hof the sets{x1 ,x2 , . . . ,xK} , {x2 ,x3 , . . . ,xK} , . . . , {xK−1 ,xK} , {xK}is not greater than X.5. For integers k and n let (k)2n+1 be the multiple of 2n + 1 losest to k.Determine the range of the fun tion f(k) = (k)3 + (2k)5 + (3k)7 − 6k.

2126. A positive integer is written on ea h of the six fa es of a ube. For ea hvertex of the ube we ompute the produ t of the numbers on the three ad-ja ent fa es. The sum of these produ ts is 1001. What is the sum of the sixnumbers on the fa es?7. Find all sets X onsisting of at least two positive integers su h that forevery pair m,n ∈ X, where n > m, there exists k ∈ X su h that n = mk2.8. Let f(x) be a non- onstant polynomial with integer oeÆ ients. Provethat there is an integer n su h that f(n) has at least 2004 distin t primefa tors.9. A set S of n − 1 natural numbers is given, where n ≥ 3. There exist atleast two elements in this set whose di�eren e is not divisible by n. Provethat it is possible to hoose a non-empty subset of S so that the sum of itselements is divisible by n.10. Is there an in�nite sequen e of prime numbers p1, p2, p3, . . . su h that|pn+1 − 2pn| = 1 for ea h n ≥ 1?11. An m × n table is given with +1 or −1 written in ea h ell. Initiallythere is exa tly one −1 in the table and all the other ells ontain a +1. Amove onsists of hoosing a ell ontaining −1, repla ing this −1 by a 0, andthen multiplying all the numbers in the neighbouring ells by −1 (two ellsare neighbouring if they share a side). For whi h (m, n) an a sequen e ofsu h moves always redu e the table to all zeroes, regardless of whi h ell ontains the initial −1?12. There are 2n di�erent numbers in a row. By one move we an ex hangeany two numbers, or y li ally permute any three numbers, that is, hoose a,b, and c and repla e b by a, c by b, and a by c. What is the minimum numberof moves that suÆ es to arrange the numbers in in reasing order?13. The 25 member states of the European Union set up a ommittee withthe following rules.(a) The ommittee shall meet every day.(b) At ea h meeting, at least one member state shall be represented.( ) At any two di�erent meetings, a di�erent set of member states shall berepresented.(d) The set of states represented at the nth meeting shall in lude, for every

k < n, at least one state that was represented at the kth meeting.For how many days an the ommittee have its meetings?

21314. A pile of one, two, or three nuts is small, while a pile of four ormore nutsis large. Two persons play a game, starting with a pile of n nuts. A playermoves by taking a large pile of nuts and splitting it into two non-empty piles(either pile an be large or small). If a player annot move, he loses. Forwhi h values of n does the �rst player have a winning strategy?15. A ir le is divided into 13 segments, numbered onse utively from 1 to13. Five eas alled A, B, C, D, and E sit in the segments 1, 2, 3, 4, and5, respe tively. A ea an jump to an empty segment �ve positions away ineither dire tion around the ir le. Only one ea jumps at a time, and two eas annot o upy the same segment. After some jumps, the eas are ba kin the segments 1, 2, 3, 4, and 5, but possibly in some other order than theystarted in. Whi h orders are possible?16. Through a point P exterior to a given ir le pass a se ant and a tangentto the ir le. The se ant interse ts the ir le at A and B, and the tangenttou hes the ir le at C on the same side of the diameter through P as Aand B. The proje tion of C onto the diameter is Q. Prove that QC bise ts∠AQB.17. Consider a re tangle with sides of lengths 3 and 4, and on ea h side pi kan arbitrary point that is not a orner. Let x, y, z, and u be the lengths ofthe sides of the quadrilateral spanned by these points. Prove that

25 ≤ x2 + y2 + z2 + u2 ≤ 50 .18. A ray emanating from the vertex A of the triangle ABC interse ts theside BC at X and the ir um ir le of ABC at Y . Prove that1

AX+

1

XY≥ 4

BC.

19. In triangle ABC let D be the mid-point of BC and let M be a point onthe side BC su h that ∠BAM = ∠DAC. Let L be the se ond interse tionpoint of the ir um ir le of triangle CAM with AB, and let K be the se ondinterse tion point of the ir um ir le of triangle BAM with the side AC.Prove that KL ‖ BC.20. Three ir ular ar s w1, w2, and w3 with ommon end-points A and Bare on the same side of the line AB, and w2 lies between w1 and w3. Tworays emanating from B interse t these ar s at M1, M2, M3 and K1, K2, K3,respe tively. Prove thatM1M2

M2M3

=K1K2

K2K3

.

214We now give the XIX Olimpiada Iberoameri ana de Matem�ati as 2004.Thanks again go to Felix Re io, Canadian Team Leader to the IMO inMexi o,for olle ting them for the Corner.XIX OLIMPIADA IBEROAMERICANA DEMATEM �ATICASCastell �on, Espa~naSeptember 17{26, 20041. Some ells of a 1001×1001 board are oloured a ording to the followingrules.(a) If two ells share a ommon side, then at least one of them is oloured.(b) In every set of six onse utive ells in a row or in a olumn, there are atleast two neighbouring ells that are oloured.Determine the minimum number of ells that are oloured.2. Let A be a �xed exterior point with respe t to a given ir le with entre

O and radius r. Let M be a point on the ir le and let N be diametri allyopposite to M with respe t to O. Find the lo us of the entres of the ir lespassing through A, M , and N , as the point M is varied on the ir le.3. Let n and k be positive integers su h that n is odd or n and k are even.Prove there exist two integers a and b su h that gcd(a, n) = gcd(b, n) = 1and k = a + b.4. Find all pairs of positive integers (a, b) su h that a and b ea h have twodigits, and su h that 100a + b and 201a + b are perfe t squares with fourdigits ea h.5. In a s alene triangle ABC, the interior bise tors of the angles A, B, andC meet the opposite sides at points A′, B′, and C′ respe tively. Let A′′ bethe interse tion of BC with the perpendi ular bise tor of AA′, let B′′ be theinterse tion of AC with the perpendi ular bise tor of BB′, and let C′′ bethe interse tion of AB with the perpendi ular bise tor of CC′. Prove thatA′′, B′′, and C′′ are ollinear.6. Given a subset H of the plane, a point P of the plane is said to be a utpoint of H if there are four points A, B, C, and D in H su h that the linesAB and CD are distin t and meet at P . If H0 is a �nite subset of the plane,then de�ne the sets H1, H2, H3, . . . re ursively by taking Hj+1 to be theunion of Hj and the set of ut points of Hj, j ≥ 0.Prove that, if the union of the Hj is �nite, then Hj = H1 for ea h j.

215As a third group of problems we next give the Quali� ation Round ofthe Swedish Mathemati al Contest 2004/2005. Thanks again to Felix Re iofor obtaining them for our use.SWEDISH MATHEMATICAL CONTEST 2004/2005Quali� ation RoundO tober 5, 2004

1. The ities A, B, C, D, and E are onne ted by straight roads (more thantwo ities may lie on the same road). The distan e from A to B, and fromC to D, is 3 km. The distan e from B to D is 1 km, from A to C it is 5 km,from D to E it is 4 km, and �nally, from A to E it is 8 km. Determine thedistan e from C to E.2. Linda writes the four positive integers a, b, c, and d on a pie e of paper.Sin e she is amused by arithmeti , she adds the numbers in pairs, obtainingthe sums a + b, a + c, . . . , c + d, but she forgets to write down one of thepossible sums. The �ve sums she obtains are 7, 11, 12, 18, and 23. Whi hsum did Linda forget? What are the positive integers a, b, c, d ?3. Determine the greatest and the least value of

mn

(m + n)2,if m and n are positive integers, ea h not greater than 2004.4. Let k and n be integers with 1 < k < n. If a set of n real numbers hasthe property that the mean value of any k of them is an integer, show thatall n numbers are integers.5. At all points with integer oordinates in the plane there are verti al nailsof altitude h and of negligible width. When a spheri al balloon is dropped itwill burst on e it tou hes one of the nails. A balloon is doomed if it bursts nomatter where it is dropped. What is the radius, R, of the smallest doomedballoon?6. Let 2n (where n ≥ 1) points lie in the plane so that no straight line ontains more than two of them. Paint n of the points blue and paint theother n points yellow. Show that there are n segments, ea h with one blueend-point and one yellow end-point, su h that ea h of the 2n points is anend-point of one of the n segments and none of the segments have a pointin ommon.

216The Final Round problems of the SwedishMathemati al Contest 2004/2005 omplete the set. Thanks are due to Felix Re io for obtaining them.SWEDISH MATHEMATICAL CONTEST 2004/2005Final RoundNovember 20, 20041. Two ir les in the plane of the same radius R interse t at a right angle.How large is the area of the region whi h lies inside both ir les?2. The oins in a ertain ountry have the values 1, 2, 3, 4, and 5. Nisse isbuying a pair of shoes. When he is about to pay he tells the salesman that hehas 100 oins, but that he does not know how many there are of ea h kind.\Wonderful, then you have even money," says the salesman. How mu h didthe shoes ost, and how did the salesman know that Nisse had even money?3. The fun tion f satis�es f(x) + xf(1 − x) = x2 for all real numbers x.Determine the fun tion f.4. If tan v = 2v and 0 < v < π

2, then does it follow that sin v < 20

21?5. A square of integer side n, where n ≥ 2, is divided into n2 squares ofside 1. Next, n−1 straight lines are drawn so that the interior of ea h of thesmall squares (the boundary not being in luded in the interior) is interse tedby at least one of the straight lines.(a) Give an example whi h shows that this an be a hieved for some n ≥ 2.(b) Show that among the n − 1 straight lines there are two lines whi hinterse t in the interior of the square of side n.6. Show that any onvex, n-sided polygon of area 1 ontains a quadrilateralof area at least 1

2.

Our last group of problems is the Abel Competition 2004{2005, NorwayFinal. Thanks again to Felix Re io for olle ting them for the Corner.ABEL COMPETITION 2004{2005Norway FinalMar h 10, 20051. (a) A positive integer m is triangular if m = 1 + 2 + · · · + n for someinteger n > 0. Show that m is triangular if 8m + 1 is a perfe t square.(b) The base of a pyramid is a right-angled triangle with sides of integerlengths. The height of the pyramid is also an integer. Show that the volumeof the pyramid is an even integer.

2172. (a) There are nine small �sh in a ubi aquarium with a side of 2 metres,entirely �lled with water. Show that at any given time there an be foundtwo �sh whose distan e apart is not greater than √

3 metres.(b) Let A be the set of all points in spa e with integer oordinates. Showthat for any nine points in A, there are at least two points among them su hthat the mid-point of the segment joining the two is also a point in A.3. (a) Let △ABC be isos eles with AB = AC, and let D be the mid-pointof BC. The points P and Q lie respe tively on the segments AD and ABsu h that PQ = PC and Q 6= B. Show that ∠PQC = 1

2∠BAC.(b) Let ABCD be a rhombus with ∠BAD = 60◦. Let F , G, and H bepoints on the segments AD, CA, and DC respe tively su h that DFGH isa parallelogram. Show that △BHF is equilateral.4. (a) Let a, b, and c be positive real numbers. Prove that

(a + b)(a + c) ≥ 2√

abc(a + b + c) .(b) Let a, b, and c be real numbers su h that ab+bc+ca > a+b+c > 0.Prove that a + b + c ≥ 3.I seem to have mis�led a solution and omment by Miguel AmengualCovas to Problem 1 of the Thai Mathemati al Olympiad (see the last numberof the Corner). He noted that the answer, ∠B = 1

2∠A = 1

270◦ = 35◦,follows from Problem #2559 [2000 : 305℄ and its solution [2001 : 466-467℄.Also, two letters arrived from Pavlos Maragoudakis after we had �-nalized the April Corner. One letter in luded a solution to Problem 8 (notdis ussed in the April Corner) of the Sele ted Problems of the Thai Mathe-mati al Olympiad 2003, given at [2007 : 278℄.8. Find all primes p su h that p2 + 2543 has less than 16 distin t positivedivisors.Solution by Pavlos Maragoudakis, Pireas, Gree e.If p = 2 then p2+2543 = 2547 = 32 ·283 whi h has (2+1)(1+1) = 6positive divisors.If p = 3 then p2 + 2543 = 2552 = 23 · 11 · 29 whi h has pre isely

(3 + 1)(1 + 1)(1 + 1) = 16 positive divisors.If p > 3 then p ≡ ±1 (mod 3) so p2 + 2543 ≡ 2544 ≡ 0 (mod 3).Also p is odd, so p2 ≡ 1 (mod 8) and p2 + 2543 ≡ 0 (mod 8). Thus3 | (p2 + 2543) and 8 | (p2 + 2543), hen e p2 + 2543 = 24x where x > 24.The numbers 1, 2, 3, 4, 6, 8, 12, 24, x, 2x, 3x, 4x, 6x, 8x, 12x, and 24x arethen 16 divisors of p2 + 2543, that are distin t and positive.So p = 2 is the only prime su h that p2 +2543 has less than 16 distin tpositive divisors.

218Next we revisit the 25th Albanian Mathemati al Olympiad, Test 1 andTest 2, given at [2007: 278{279℄. We start with solutions to Test 1.1. There are 20 pupils in a village s hool. Any two of them have the samegrandfather. Show that there exists a grandfather who has at least 14 grand- hildren.Comment byMiguel Amengual Covas, Cala Figuera, Mallor a, Spain. Solvedby Pavlos Maragoudakis, Pireas, Gree e; and Titu Zvonaru, Com�ane�sti,Romania. We give the omment of Amengual Covas.The problem was proposed in the Tournament of Towns, Tournament16, Junior Questions, Autumn 1994, O Level. A solution an be found in:P.J. Taylor and A.M. Storozhev, Tournament of Towns 1993{1997, Book 4,Australian Mathemati al Trust, 1998, p. 57.2. Let M , N , and P be the respe tive mid-points of sides BC, CA, and ABof triangle ABC, and let G be the interse tion point of its medians. Provethat if BN =

√3

2AB and BMGP is a y li polygon, then triangle ABC isequilateral.Solved by Geo�rey A. Kandall, Hamden, CT, USA; Pavlos Maragoudakis,Pireas, Gree e; George Tsapakidis, Agrinio, Gree e; and Titu Zvonaru,Com�ane�sti, Romania. We give the solution of Tsapakidis.Sin e BMGP is a y li polygon we obtain

AG · AM = AP · AB ,2

3AM · AM =

1

2AB · AB ,

AM2 =3

4AB2 = BN2 ,AM = BN ,hen e AC = BC. The segment CP is also analtitude, that is ∠GPB = 90◦, and sin e oppositeangles of a yli quadrilateral add to 180◦, we also

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A

B C

G

M

NP

have ∠GMB = 90◦. Therefore AM is also an altitude, hen e AB = AC,whi h together with AC = BC shows that triangle ABC is equilateral.3. Let xk and yk (for k = 1, 2, . . . , n) be positive real numbers that satisfykxkyk ≥ 1.(a) Prove that n∑

k=1

xk − yk

x2k + y2

k

≤ 1

4n

√n + 1.

(b) When does equality hold in part (a)?

219Solved by Mi hel Bataille, Rouen, Fran e; and Pavlos Maragoudakis, Pireas,Gree e. We give Bataille's solution.(a) First note that for k = 1, 2, . . . , n we have

xk − yk

x2k + y2

k

≤ |xk − yk||xk − yk|2 + 2xkyk

≤ tk

t2k +2

k

,where we have set tk = |xk − yk|.But t2k +

2

k≥ 2tk

√2

k, and hen e tk

t2k +2

k

≤√

k

2√

2. It follows that

n∑

k=1

xk − yk

x2k + y2

k

≤ 1

2√

2(√

1 +√

2 + · · · +√

n) . (1)From an inequality of means,

(√1 +

√2 + · · · +

√n

n

)2

≤ 1 + 2 + · · · + n

n, (2)

hen e√

1 +√

2 + · · · +√

n ≤ n

√n + 1

2.Returning to (1), we obtain

n∑

k=1

xk − yk

x2k + y2

k

≤ 1

2√

2· n

√n + 1

2=

1

4n

√n + 1 .

(b) The inequality is stri t in (2) if n ≥ 2, for then √1, √

2, . . . , √n aredistin t. Thus, equality does not hold in (a) if n ≥ 2. If n = 1 and equalityholds, then

√2

4=

x1 − y1

x21 + y2

1

≤ t1

t21 + 2x1y1

≤ t1

t21 + 2≤

√2

4,so that equality holds throughout. It follows that

x1y1 = 1 and √2

4=

x1 − y1

(x1 − y1)2 + 2,

hen e x1 −y1 =√

2. It is now easy to dedu e that equality holds if and onlyif n = 1 andx1 =

√2 ±

√6

2, y1 =

−√

2 ±√

6

2.

2204. Find prime numbers p and q su h that p2 − p + 1 = q3.Solution by Titu Zvonaru, Com�ane�sti, Romania.We will prove that the only solution to p2 − p + 1 = q3 where p isa prime number and q is a natural number is p = 19 and q = 7 (giving192 − 19 + 1 = 343 = 73).We have

p(p − 1) = (q − 1)(q2 + q + 1) , (1)thus p | (q − 1) or p | (q2 + q + 1). If p | (q − 1), then p < q, whi h leadsto the ontradi tion p2 − p + 1 = q3 > p3. It follows that p | (q2 + q + 1),that is,q2 + q + 1 = kp . (2)By (1) we obtainp − 1 = k(q − 1) . (3)By (2) and (3) we obtain q2 + q + 1 = k(kq − k + 1), or equivalently

q2 + (1 − k2)q + (k2 − k + 1) = 0 .Sin e q is an integer, the dis riminant ∆ = (k2 − 1)2 − 4(k2 − k + 1) ofthe quadrati equation is a perfe t square. Also, sin e ∆ < (k2 − 1)2 and∆ ≡ (k2 − 1)2 (mod 2), it follows that

∆ ≤ (k2 − 3)2 ⇐⇒ (k2 − 1)2 − (k2 − 3)2 ≤ 4(k2 − k + 1)

⇐⇒ 4(k2 − 2) ≤ 4(k2 − k + 1) ,and hen e k ≤ 3. If k = 2, then ∆ = −3, whi h is not a perfe t square.However, if k = 3, then ∆ = 62, and we obtain q = 7 and p = 19.We now turn to our readers' solutions to Test 2 of the AlbanianMathemati alOlympiad.2. Prove the inequality1√

a + 1

b+ 0.64

+1√

b + 1

c+ 0.64

+1√

c + 1

a+ 0.64

≥ 1.2 ,where a > 0, b > 0, c > 0, and abc = 1.Solved by Arkady Alt, San Jose, CA, USA; D. Kipp Johnson, Beaverton, OR,USA; and Pavlos Maragoudakis, Pireas, Gree e. We give Johnson's write-up.We �rst make the standard substitution a =

x

y, b =

y

z, and c =

z

x, forsuitable positive numbers x, y, and z. Then

1√a + 1

b+ 0.64

=1√

xy

+ zy

+ 0.64=

√y

√x + 0.64y + z

,

221and the desired inequality be omes

∑ y li √y

√x + 0.64y + z

≥ 1.2 .This inequality is homogeneous of degree zero, so without loss of generalitywe let x + y + z = 1, where x, y, and z are in the interval (0, 1), and ourgoal is to prove that∑ y li √

y√

x + y + z − 0.36y=

∑ y li √y

√1 − 0.36y

=∑ y li 5

√y

√25 − 9y

≥ 1.2 .Now for 0 < x < 1 we have the su ession of equivalent inequalities

5√

x√

25 − 9x≥ 6x

5,

25 ≥ 6√

x√

25 − 9x ,625 ≥ 36x(25 − 9x) ,

0 ≤ 9x2 − 25x +625

36,

whi h are all true, as the last inequality is just (3x− 25

6)2 ≥ 0. Adding threesu h inequalities we have

5√

x√

25 − 9x+

5√

y√

25 − 9y+

5√

z√

25 − 9z

≥ 6x

5+

6y

5+

6z

5=

6(x + y + z)

5= 1.2 ,and our goal is a hieved. This ompletes the proof.3. Solve the following equation in integers:

y2 = 1 + x + x2 + x3 + x4 .Comments by Mi hel Bataille, Rouen, Fran e; and Titu Zvonaru, Com�ane�sti,Romania. Zvonaru notes that this was problem M242. A solution appearsat [2007 : 141℄. Now we give Bataille's omment.This is an old problem set by T.H. Gronwall and solved by A.A. Bennettin Amer. Math. Monthly, Vol 33, No. 5, May 1926, pp. 291{2. The solvershowed that the solutions are the following six pairs (x, y):(−1, ±1) , (0, ±1) , (3, ±11) .

4. Prove that for any integer n ≥ 2, the number 2n −1 is not divisible by n.

222Solution by Mi hel Bataille, Rouen, Fran e.Suppose on the ontrary that n divides 2n − 1. Sin e 2n − 1 is odd,n and all its prime fa tors are odd as well. Let p be one of these primefa tors. There exist positive integers k, l su h that n = kp and 2n = 1+klp.Re alling that 2p ≡ 2 (mod p) (by Fermat's Little Theorem), we obtain

2k ≡ 1 (mod p) . (1)Now, 2 belongs to the multipli ative group Z/pZ − {0} and the order of 2in this group is an integer m > 1 whi h divides the order p − 1 of the groupand, from (1), also divides k. Thus, hoosing a prime divisor q of m, we get aprime divisor of n satisfying q < p. Repeating with q what we have just donewith p and ontinuing this way, we would onstru t an in�nite sequen e ofprime divisors of n, whi h is learly impossible. This ontradi tion ompletesthe proof.5. In an a ute-angled triangle ABC, Let H be the ortho enter, and letda, db, and dc be the distan es from H to the sides BC, CA, and AB,respe tively. Prove that da + db + dc ≤ 3r, where r is the radius of thein ir le of triangle ABC.Solved by Arkady Alt, San Jose, CA, USA; Miguel Amengual Covas, CalaFiguera, Mallor a, Spain; Pavlos Maragoudakis, Pireas, Gree e; and TituZvonaru, Com�ane�sti, Romania. We give the solution of Amengual Covas.Let a, b, and c be the lengths of the sides opposite A, B, and C, re-spe tively, and let s = 1

2(a + b + c) be the semiperimeter. Let R be the ir umradius of triangle ABC, and let O and A′ be the ir um entre and themid-point of the side BC, respe tively.Letting ha be the length of the altitude from A to BC in △ABC, andusing the well-known relation AH = 2 · OA′, we dedu e that

da = ha − AH = ha − 2 · OA′ = ha − 2R cos A .By symmetry, we have db = hb − 2R cos B and dc = hc − 2R cos C. Itfollows thatda + db + dc = ha + hb + hc − 2R(cos A + cos B + cos C) .We have the relation ha =

2rs

aand its symmetri analogues, and alsowe have the well-known relations

abc = 4Rrs ,ab + bc + ca = s2 + r2 + 4Rr ,

cos A + cos B + cos C =R + r

R.

223Using these we dedu e that

da + db + dc =s2 + r2 − 4Rr

2R,and hen e the given inequality is equivalent to

s2 + r2 − 4R2 ≤ 6Rr . (1)To establish the validity of (1), wemake use of the linear transformationa = y + z, b = z + x, and c = x + y, where x, y, and z are uniquelydetermined positive numbers. Ater applying the transformation, (1) be omes

(x + y + z)2 +xyz

x + y + z− (x + y)2(y + z)2(z + x)2

4xyz(x + y + z)

≤ 6 · (x + y)(y + z)(z + x)

4(x + y + z),or equivalently

4xyz(x + y + z)3 + 4x2y2z2 − (x + y)2(y + z)2(z + x)2

≤ 6xyz(x + y)(y + z)(z + x) .We substitute x3 + y3 + z3 + 3(x + y)(y + z)(z + x) for (x + y + z)3,simplify, and obtain4xyz(x3 + y3 + z3) + 6xyz(x + y)(y + z)(z + x) + 4x2y2z2

≤ [(x + y)(y + z)(z + x)]2 .We substitute 2xyz +∑sym x2y for (x+y)(y + z)(z +x) (the symmetri sum is over the six permutations of x, y, and z) to obtain

4xyz(x3 + y3 + z3) + 6xyz

(2xyz +

∑sym x2y

)+ 4x2y2z2

≤(

2xyz +∑sym)2

x2y =

(∑sym x2y

)2

+ 4xyz

(∑sym x2y

)+ 4x2y2z2 .We have the expansion

(∑sym x2y

)2

=

(∑sym x4y2

)+ 2xyz(x3 + y3 + z3) + 2xyz

(∑sym x2y

)

+ 2(x3y3 + y3z3 + z3x3) + 6x2y2z2 ,whi h we substitute into the above and then simplify to obtain2xyz(x3 +y3 +z3) + 6x2y2z2 ≤

(∑sym x4y2

)+ 2(x3y3 +y3z3 +z3x3) .

224However, this is equivalent to

2[(xy)3 + (yz)3 + (zx)3 − 3x2y2z2

]

+ x4(y − z)2 + y4(z − x)2 + z4(x − y)2 ≥ 0 ,whi h is true, sin e (xy)3 + (yz)3 + (zx)3 ≥ 3x2y2z2 holds by the AM-GMInequality.Equality o urs only if x = y = z, that is, if and only if the triangleABC is equilateral.

Next we turn to solutions from our readers to some problems of the44th Ukrainian Mathemati al Olympiad, 11th Form, Final Round, given at[2007 : 279{280℄.1. (V.M. Leifura) Solve the equationarcsin⌊sin x⌋ = arccos⌊cos x⌋ ,where ⌊a⌋ is the greatest integer not ex eeding a.Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON.For onvenien e, we will write sin−1 x for arcsin x.Let θ = sin−1⌊sin x⌋ = cos−1⌊cos x⌋. From the de�nitions of theinverse trigonometri fun tions, we have −π

2≤ θ ≤ π

2and 0 ≤ θ ≤ π.Hen e, 0 ≤ θ ≤ π

2. We onsider four ases.Case 1. If −1 ≤ sin x < 0, then θ = sin−1(−1) = −π

2, a ontradi tion.Case 2. If 0 < sin x < 1, then θ = sin−1(0) = 0, hen e ⌊cos x⌋ = cos θ = 1and cos x = 1. Thus sin2 x + cos2 x > 1, a ontradi tion.Case 3. If sin x = 0, then cos x = 1 as in Case 2 above, whi h implies that

x = 2kπ, k ∈ Z. Conversely, if x = 2kπ, k ∈ Z, then sin−1⌊sin x⌋ = 0 andalso cos−1⌊cos x⌋ = 0.Case 4. If sin x = 1, then θ = sin−1(1) = π2; hen e ⌊cos x⌋ = cos θ =

cos(

π2

)= 0 and 0 ≤ cos x < 1. Evidently cos x = 0; otherwise we have

sin2 x + cos2 x > 1, a ontradi tion. Solving sin x = 1 and cos x = 0 we�nd that x = π2

+ 2kπ, k ∈ Z. Conversely, if x = π2

+ 2kπ, k ∈ Z,then sin−1⌊sin x⌋ = sin−1(1) = π2and cos−1⌊cos x⌋ = cos−1(0) = π

2.Therefore, the solution set of the given equation is

S =

{2kπ ,π

2+ 2kπ : k ∈ Z

} .

2252. (V.V. Lymanskiy) The a ute-angled triangle ABC is given. Let O bethe entre of its ir um ir le. The perpendi ular bise tor of the side ACinterse ts the side AB and the line BC at the points P and Q, respe tively.Prove that ∠PQB = ∠PBO.Solved by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain; Geo�reyA. Kandall, Hamden, CT, USA; Pavlos Maragoudakis, Pireas, Gree e;D.J. Smeenk, Zaltbommel, the Netherlands; and Titu Zvonaru, Com�ane�sti,Romania. We give Maragoudakis' solution.In triangle BOA we have OA = OB. Thus

∠PBO =180◦ − ∠BOA

2=

180◦ − BKA

2

=180◦ − BK − AK

2

=180◦ − BK − CK

2

=KM − BK − CK

2

=CM − BK

2= ∠PQB .

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A

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Q

6. (O.O. Malakhov) Find the sum of the real roots of the equationx +

x√

x2 − 1= 2004 .Solved by Mi hel Bataille, Rouen, Fran e; Pavlos Maragoudakis, Pireas,Gree e; and Vedula N. Murty, Dover, PA, USA. We give Bataille's solution.The required sum is 2004.Let a = 2004. It is readily seen that the real roots of the given equationare those of P (x) = 0 satisfying 0 < x < a, where

P (x) = (x2 − 1)(a − x)2 − x2 = x4 − 2ax3 + (a2 − 2)x2 + 2ax − a2 .The key observation is that the polynomial P (a−x) is just the same as P (x)(this is easily he ked). Hen e, if x0 is a solution to P (x) = 0, the same istrue of a − x0.Now, P (0) < 0, P (a/2) > 0, P (a) < 0, hen e P (x) has at least tworeal roots in (0, a). If in addition P (x) had two positive or omplex onjugateroots, the produ t of the roots would be positive. However, the produ t ofthe roots is −a2 < 0, therefore P (x) must have a negative root, say x1.Then a − x1 is also a root of P (x) and a − x1 > a. It follows that the fourroots of P (x) are real, with exa tly two of them in (0, a), of the form x2,a − x2 for some x2 ∈ (0, a). These real numbers are the roots of the givenequation and their sum is a = 2004, as announ ed.

2267. (V.M. Rad henko) Does there exist a fun tion f : R → R su h thatf(x2y + f(x + y2)

)= x3 + y3 + f(xy) for all x, y ∈ R ?Solved by Mi hel Bataille, Rouen, Fran e; and Pavlos Maragoudakis, Pireas,Gree e. We give the solution of Bataille.There is no su h fun tion. Assume on the ontrary that f : R → Rsatis�es the given identity. Then, taking x = 0 in the identity, we have

f(f(y2)

)= y3 + f(0) ,and taking y = 0 we have

f(f(x)

)= x3 + f(0) .However, with x = 1 the latter gives f(f(1)

)= 1+f(0), while with y = −1the former yields f

(f(1)

)= −1 + f(0). Thus, we have the ontradi tion

1 + f(0) = −1 + f(0), and the result follows.8. (V.A. Yasinskiy) Let a, b, and c be positive real numbers su h thatabc ≥ 1. Prove that a3 + b3 + c3 ≥ ab + bc + ca.Solved by Arkady Alt, San Jose, CA, USA; Mi hel Bataille, Rouen, Fran e;Pavlos Maragoudakis, Pireas, Gree e; Vedula N. Murty, Dover, PA, USA;George Tsapakidis, Agrinio, Gree e; Panos E. Tsaoussoglou, Athens, Gree e;and Titu Zvonaru, Com�ane�sti, Romania. We give Murty's write-up.We have the two known inequalities

a2 + b2 + c2 ≥ ab + bc + ca ,a3 + b3 + c3 ≥ (a + b + c)(a2 + b2 + c2)

3,from whi h we obtain

a3 + b3 + c3 ≥ (a + b + c)(ab + bc + ca)

3.From the AM-GM Inequality and the given ondition, abc ≥ 1, we have

a + b + c ≥ 3(abc)1/3 ≥ 3 .The desired inequality now follows by ombining the last two inequalities.10. (I.P. Nagel) Let ω be the ins ribed ir le of the triangle ABC. Let L,N , and E be the points of tangen y of ω with the sides AB, BC, and CA,respe tively. Lines LE and BC interse t at the point H, and lines LN andAC interse t at the point J (all the points H, J , N , E lie on the same sideof the line AB). Let O and P be the mid-points of the segments EJ andNH, respe tively. Find S(HJNE) if S(ABOP ) = u2 and S(COP ) = v2.(Here S(F) is the area of �gure F).

227Solution by Titu Zvonaru, Com�ane�sti, Romania.As usual wewrite a = BC, b = CA, c = AB,and s =

1

2(a + b + c).It is well known that AL = AE = s − a,

BL = BN = s − b, and CN = CE = s − c.Assume that c > b. In order that the pointsH, J , N , and E all lie on the same side of the lineAB, we must have c > a.By Menelaus' Theorem applied to △ABCand transversal HEL, we have

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q

q

A

B

C

E

H

J

L

NO

P

HC

HB· LB

LA· EA

EC= 1 ⇒ HC

HC + a=

s − c

s − b⇒ HC =

a(s − c)

c − b,

and by symmetry we have JC =b(s − c)

c − a. Therefore we dedu e

NH = NC + CH = (s − c) +a(s − c)

c − b

=(s − c)(c − b + a)

c − b=

2(s − b)(s − c)

c − b,

and by symmetry we have EJ =2(s − a)(s − c)

c − a. Using the expressions wehave obtained so far, we have

S(HJNE) =NH · JE · sin C

2

=2(s − a)(s − b)(s − c)2

(c − b)(c − a)· sin C , (1)

CP = NP − NC =1

2NH − NC

=(s − b)(s − c)

c − b− (s − c)

=(s − c)(s − b − c + b)

c − b=

(s − c)2

c − b,

and by symmetry, CO =(s − c)2

c − a. It follows that

S(COP ) = v2 =(s − c)4 sin C

2(c − b)(c − a). (2)

228Continuing with our al ulations,

AO = AC + CO = b +(s − c)2

c − a

=−c(2s − c − b) + s2 − ab

c − a=

s2 − ab − ac

c − a

=s2 −

(a(b + c)

)

c − a=

s2 − 2as + a2

c − a=

(s − a)2

c − a,

and by symmetry BP =(s − b)2

c − b, hen e

S(ABOP ) = u2 =(s − a)2(s − b)2

(c − a)(c − b)· sin C

2. (3)Combining (1), (2), and (3) we obtain the desired result

S(HJNE) =2(s − a)(s − b)(s − c)2

(c − b)(c − a)· sin C

= 4

√(s − a)2(s − b)2 · sin C

2(c − b)(c − a)·√

(s − c)4 sin C

2(c − b)(c − a)= 4uv .

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

229BOOK REVIEWJohn Grant M LoughlinProblems of the WeekBy Jim Totten, Canadian Mathemati al So iety, 2007ISBN 0-919558-16-X, oilbound, 60 pages, $12Reviewed by John Grant M Loughlin, University of New Brunswi k,Frederi ton, NBThe author needs no introdu tion to the CRUX readership. Jim Totten ompletes his tenure as Editor with this issue of the journal. As the BookReviews Editor, I am taking the opportunity to a knowledge a re ently pub-lished work of Jim Totten { another way of thanking Jim for his ontributionsto problem solving within CRUX and beyond. The publisher, the CanadianMathemati al So iety, will surely support me taking su h editorial libertywithin the review.Jim Totten taught at Thompson Rivers University (formerly known asthe University College of the Cariboo) from 1979 to 2007. Ea h week ofthe a ademi semesters featured a di�erent posted problem with minimalrepetition of individual problems over the years. This publi ation featuresa sele tion of 80 su h problems used prior to Fall 1986. In fa t, some were�rst used by Jim at Saint Mary's University from 1976-1979.The problems take a range of forms spanning an array of topi s as notedin the Table of Contents. A helpful "Index of Problems by subje t matter" an guide solvers toward topi s of interest. The subje ts (and subheadings)are: Algebra (Equations, Fun tions, Word Problems, Other), Analysis, Com-binatori s, Geometry (Triangles, Cir les, Other), Logi , Numbers, Probabil-ity, and Re reation. Many problems are ross-referen ed to two subje ts.Problems of the Week is Volume VII in the series ATOM (A Taste ofMathemati s). Publi ations in the series are intended for keen high s hoolstudents, mathemati s tea hers, and problem solvers who wish to engagewith a essible material. The ATOM materials lend themselves to indepen-dent study or ollaborative proje ts. This book is no ex eption as it growsout of ollaboration with students though the in lusion of detailed solutionsallows one to learn mu h math independently from problems that mayappear beyond one's grasp.I am reminded of my undergraduate studies at the University ofWaterloo, parti ularly in two separate ourses, ommonly referred to as the"100 Problem Courses." Many of the problems introdu ed to me by DeanHo�man or Ross Honsberger appear in this book. The introdu tion lari�esthat this is not oin idental, as Jim writes in the dedi ation: "For showing methat not only was it a eptable to be ex ited and enthusiasti about math-emati al problem-solving, but that it was to be strongly en ouraged, I amdedi ating this book to Ross Honsberger."A few problems are stated here to pique the interest of problem solvers.

2303. In a ertain lassroom, there are 5 rows with 5 seats per row arrangedin a square. Ea h student is to hange her seat by going either to the seatimmediately in front or behind her, or immediately to the left or right. (Of ourse, not all possibilities are open to all students.) Determine whether this an be done, beginning with a full lass of students.Try to generalize to a re tangular array and �nd the onditions underwhi h su h a hange of seats an be managed.16. The three sides and height of a triangle are four onse utive integers.What is the area of the triangle?48. The Ramada County Department of Highways has just resurfa ed the ounty roads, and now the yellow stripe down the middle of the road mustbe repainted. The tru k used for this purpose is very ineÆ ient as far as gas onsumption is on erned, and thus the Department would like to have thetru k travel the shortestdistan e possible. A roadmap of the ounty isshown (with distan es giv-en in kilometres). The ounty tru k is garaged inMidville, and it must re-turn there when the job isdone. How many kilome-tres must it travel andwhat route should it take? .........................................................................................

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t t t

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W. Midville Midville E. MidvilleS.W. Midville S.E. Midville5

34 30 28 37

821

40

61. Given any two points A and B on the ir umferen e of a ir le, and Ethe mid-point of the ar AB (note that there are really two ar s that ould be alled AB; it does not matter whi h one we hoose as long as the rest of the dis ussion is as-sumed to pertain only to points and ar s lying onthe ar AB that we hose). Let P be any pointon the ar EB and onstru t EN perpendi ularto AP with N on the hord AP . Prove thatAN = NP +PB (we are dealing only with mag-nitudes of line segments here).

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

B

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This book is an ex ellent addition to a library, a departmental o�eeroom, or the math tea hing oÆ e in a s hool. Those who like problem solvingwill enjoy this ompa t resour e, whether a high s hool student aspiring tolearn advan ed math or a seasoned mathemati ian who likes a hallenge.As the Book Reviews Editor through Jim's entire editorial tenure, I wishto express my gratitude for the privilege of working with Jim in this apa ity.Addendum. The review was sent to Jim less than two weeks before hissudden passing. As noted on the inside front over, Jim is a Co-Editor of thisissue. Indeed the loss to the mathemati al ommunity is great. The spirit ofJim Totten is remembered fondly in this review.

231The Proof of Three Open Inequalities

Vasile C�rtoajeThe inequalities below were onje tured in Crux Mathemati orum [2℄and on Mathlinks Forum [5℄ (N.B. inequalities (1) and (2) are in orre tlystated on p. 106 of [2℄):Conje ture. Let x1, x2, . . . , xn be nonnegative real numbers whi h satisfyx1 + x2 + · · · + xn ≥ n. If p > 1, then

∑ y li 1

xp1 + x2 + · · · + xn

≤ 1 , (1)∑ y li x1

xp1 + x2 + · · · + xn

≤ 1 , (2)∑ y li xp

1 − x1

xp1 + x2 + · · · + xn

≥ 0 , (3)where ea h sum is over the n y li permutations of the variables xi.Under the more restri tive ondition x1x2 · · · xn ≥ 1, these inequal-ities had already been stated and proved in [1℄, [2℄, and [4℄, respe tively.On the other hand, the inequality (3) under the ondition x1x2 · · · xn ≥ 1generalizes the following inequality from the IMO (let n = 3 and p =

5

2):Problem 3 (IMO{2005). Let x, y, z be three positive reals su h that xyz ≥ 1.Prove that

x5 − x2

x5 + y2 + z2+

y5 − y2

x2 + y5 + z2+

z5 − z2

x2 + y2 + z5≥ 0 .

In this paper, we prove ea h of these three inequalities by means of theLCF-Theorem, as it is stated in [1℄ and [3℄:Theorem 1 (LCF-Theorem) Let f(u) be a fun tion on an interval I, whi h is on ave for u ≤ s, s ∈ I. If(n − 1)f(x) + f(y) ≤ nf(s)for all x, y ∈ I su h that x ≤ y and (n − 1)x + y = ns, then

f(x1) + f(x2) + · · · + f(xn) ≤ nf

(x1 + x2 + · · · + xn

n

)

for all x1, x2, . . . , xn ∈ I whi h satisfy x1 + x2 + · · · + xn

n≤ s.Copyright c© 2008 Canadian Mathemati al So ietyCrux Mathemati orum with Mathemati al Mayhem, Volume 34, Issue 4

2321 Proof of Inequality (1)Having in view the expression on the left-hand side of (1), it is easy to seethat we need only onsider the ase x1 + x2 + · · · + xn = n. Then, we haveto show that

1

xp1 − x1 + n

+1

xp2 − x2 + n

+ · · · +1

xpn − xn + n

≤ 1 (4)for x1 +x2 + · · ·+xn = n. By Lemma 1 below, it suÆ es to prove (4) for allxi ≥ a = p

11−p whi h satisfy x1 + x2 + · · · + xn = n. To do this, we applythe LCF-Theorem to the fun tion f(u) =

1

up − u + nde�ned on I = [a, ∞),for the ase s = 1.First, we must show that f is on ave on [a, 1]. The se ond derivativeis

f ′′(u) =p(p + 1)u2p−2 − p(p + 3)up−1 − np(p − 1)up−2 + 2

(up − u + n)3.If

g(u) = p(p + 1)u2p−2 − p(p + 3)up−1 + 2 < 0for a ≤ u ≤ 1, then f ′′(u) < 0, and hen e f(u) is on ave on [a, 1]. Indeed,setting t = pup−1 implies 1 ≤ t ≤ p, and hen epg(u) = (p + 1)t2 − p(p + 3)t + 2p

= (p + 1)(t − 1)(t − p) + (1 − p)(t + p) < 0 .By the LCF-Theorem, it suÆ es to show thatn − 1

xp − x + n+

1

yp − y + n≤ 1for any a ≤ x ≤ 1 ≤ y and (n − 1)x + y = n. Sin e this inequality is trivialfor x = y = 1, assume next that x < 1 < y, and write the desired inequalityin su ession as follows

yp − y + n ≥ xp − x + n

xp − x + 1,

yp − y ≥ (n − 1)(x − xp)

xp − x + 1, (5)

yp − y

y − 1≥ x − xp

(1 − x)(xp − x + 1).

Let h(y) =yp − y

y − 1. By the weighted AM-GM Inequality, we have

h′(y) =(p − 1)yp + 1 − pyp−1

(y − 1)2> 0 .

233Therefore, h(y) is stri tly in reasing, and sin e y−1 = (n−1)(1−x) ≥ 1−x,that is, y ≥ 2 − x, we have

h(y) ≥ h(2 − x) =(2 − x)p − 2 + x

1 − x.Then, it suÆ es to prove that

(2 − x)p − 2 + x ≥ x − xp

xp − x + 1.This inequality is equivalent to ea h of the following

(2 − x)p + x − 1 ≥ 1

xp − x + 1,

(1 + t)p − t ≥ 1

(1 − t)p + t,

(1 − t2)p + t(1 + t)p ≥ 1 + t2 + t(1 − t)p ,where t = 1 − x and 0 < t ≤ 1. By Bernoulli's Inequality, we have(1 − t2)p + t(1 + t)p ≥ 1 − pt2 + t(1 + pt) = 1 + t .Thus, it suÆ es to show that t(1 − t) ≥ t(1 − t)p. However, this is learlytrue, and the proof is ompleted. Equality in (1) o urs if and only if we have

x1 = x2 = · · · = xn = 1.Lemma 1. Let p > 1, and let x1, x2, . . . , xn be nonnegative real numbers.If (4) holds for all xi ≥ p1

1−p su h that x1 + x2 + · · · + xn = n, then it holdsfor all xi ≥ 0 su h that x1 + x2 + · · · + xn = n.Proof: Let a = p1

1−p , and let f(x) =1

xp − x + n. From

f ′(x) =1 − pxp−1

(xp − x + n)2,it follows that f ′(a) = 0, f ′(x) > 0 for 0 ≤ x < a, and f ′(x) < 0 for

x > a. Thus, f(x) is stri tly in reasing on [0, a] and stri tly de reasing on[a, ∞). We have to prove that f(x1) + f(x2) + · · · + f(xn) ≤ 1 for allxi ≥ 0 su h that x1 +x2 + · · ·+xn = n. Without loss of generality, assumethat 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn. If x1 ≥ a, then the on lusion followsimmediately by hypothesis. Otherwise, sin e a < 1, there exists an integerk ∈ {1, 2, . . . , n − 1} su h that

0 ≤ x1 ≤ · · · ≤ xk < a ≤ xk+1 ≤ · · · ≤ xn .Let yi = a for i = 1, 2, . . . , k, and let yi = xi for i = k + 1, . . . , n. Wehave yi ≥ a and yi ≥ xi for i = 1, 2, . . . , n; and f(yi) > f(xi) for ea hi = 1, 2, . . . , k, hen e

y1 + y2 + · · · + yn > x1 + x2 + · · · + xn = n

234and

f(y1) + f(y2) + · · · + f(yn) > f(x1) + f(x2) + · · · + f(xn) .Therefore, it suÆ es to show that f(y1) + f(y2) + · · · + f(yn) ≤ 1 for allyi ≥ a with y1 + y2 + · · · + yn > n. By hypothesis, this inequality is truefor all yi ≥ a with y1 + y2 + · · · + yn = n. Sin e f is de reasing on [a, ∞),the inequality is also true for all yi ≥ a su h that y1 + y2 + · · · + yn > n.Remark 1. If 0 ≤ p < 1 and x1 + x2 + · · · + xn ≤ n, then the inequalityin (1) reverses. We an prove this using the AM-HM Inequality together withJensen's Inequality:

∑ y li 1

xp1 + x2 + · · · + xn

≥ n2

∑ y li (xp1 + x2 + · · · + xn)

=n2

n∑i=1

xpi + (n − 1)

n∑i=1

xi

≥ n2

n

(1

n

n∑i=1

xi

)p

+ (n − 1)n∑

i=1

xi

≥ 1 .2 Proof of Inequality (2)Without loss of generality we an assume xi > 0 for ea h i. Furthermore,we need only onsider the ase x1 + x2 + · · · + xn = n. Indeed, if we setr =

x1 + x2 + · · · + xn

nand yi =

xi

rfor i = 1, 2, . . . , n (where r ≥ 1 and

y1 + y2 + · · · + yn = n), then (2) be omes∑ y li y1

rp−1yp1 + y2 + · · · + yn

≤ 1 ,and it suÆ es to prove this for r = 1; that is, to prove that

x1

xp1 − x1 + n

+x2

xp2 − x2 + n

+ · · · +xn

xpn − xn + n

≤ 1 (6)for x1 + x2 + · · · + xn = n. We onsider two ases.Case 1. 1 < p ≤ n + 1. By Bernoulli's Inequality,xp

1 =(1 + (x1 − 1)

)p ≥ 1 + p(x1 − 1) ,and hen e xp1 − x1 + n ≥ n − p + 1 + (p − 1)x1 ≥ 0. Consequently, itsuÆ es to show that

n∑

i=1

xi

n − p + 1 + (p − 1)xi

≤ 1 .

235For p = n + 1, this inequality is an equality. Otherwise, for 1 < p < n + 1,we use

(p − 1)xi

n − p + 1 + (p − 1)xi

= 1 − n − p + 1

n − p + 1 + (p − 1)xito rewrite the inequality asn∑

i=1

1

n − p + 1 + (p − 1)xi

≥ 1 .We then prove it by means of the AM-HM Inequality:

n∑

i=1

1

n − p + 1 + (p − 1)xi

≥ n2

n∑i=1

(n − p + 1 + (p − 1)xi

) = 1 .Case 2. p > n + 1. By Lemma 2 below, it suÆ es to prove (6) for allxi ≥ a = (

n

p − 1)

1p su h that x1 + x2 + · · · + xn = n. We will apply theLCF-Theorem to the fun tion f(u) =

u

up − u + nde�ned on I = [a, ∞), forthe ase s = 1.First, we must show that f is on ave on [a, 1]. The se ond derivativehas the expression

f ′′(u) =−p(p − 1)up−1(up − u + n) + 2(pup−1 − 1)

((p − 1)up − n

)

(up − u + n)3.Sin e (p − 1)up − n ≥ (p − 1)ap − n = 0 for u ≥ a, if

g(u) = −p(p − 1)up−1(up − u + n) + 2pup−1((p − 1)up − n

)< 0for a ≤ u ≤ 1, then f ′′(u) < 0, and hen e f(u) is on ave on [a, 1]. Indeed,we have

g(u)

pup−1= (p − 1)(up + u) − n(p + 1)

≤ 2(p − 1) − 2(p + 1) = −4 < 0 .By the LCF-Theorem, it suÆ es to show that(n − 1)x

xp − x + n+

y

yp − y + n≤ 1for any a ≤ x ≤ 1 ≤ y and (n−1)x+y = n. For x = y = 1, equality o urs.Assume now that x < 1 < y, and write the desired inequality su essivelyas follows

yp − y + n ≥ y(xp − x + n)

xp − nx + n, (7)

236yp − y ≥ (n − 1)x(x − xp)

xp − nx + n. (8)

Inequality (5) states that yp − y ≥ (n − 1)(x − xp)

xp − x + 1. Therefore, it suÆ es toshow that

1

xp − x + 1≥ x

xp − nx + n.This inequality is true, be ause x < 1 and xp−nx+n > xp−x+1. Equalityin (2) o urs if and only if x1 = x2 = · · · = xn = 1.Lemma 2. Let p > n+1, and let x1, x2, . . . , xn be nonnegative real numbers.If (6) holds for all xi ≥ ( n

p−1)

1p su h that x1 + x2 + · · · + xn = n, then itholds for all xi ≥ 0 su h that x1 + x2 + · · · + xn = n.Proof: Let a = (

n

p − 1)

1p , and let f(x) =

x

xp − x + n. From

f ′(x) =n − (p − 1)xp

(xp − x + n)2,

it follows that f ′(a) = 0, f ′(x) > 0 for 0 ≤ x < a, and f ′(x) < 0 for x > a.Consequently, f(x) is stri tly in reasing on [0, a] and stri tly de reasing on[a, ∞). We have to prove that f(x1)+f(x2)+ · · ·+f(xn) ≤ 1 for all xi ≥ 0su h that x1 + x2 + · · · + xn = n. Without loss of generality, assume that0 ≤ x1 ≤ x2 ≤ . . . ≤ xn. If x1 ≥ a, then the on lusion follows by hypoth-esis. Otherwise, sin e a < 1, there exists an integer k ∈ {1, 2, . . . , n − 1}su h that

0 ≤ x1 ≤ · · · ≤ xk < a ≤ xk+1 ≤ · · · ≤ xn .Let yi = a for i = 1, 2, . . . , k, and yi = xi for i = k + 1, k + 2, . . . , n. Wehave yi ≥ a and yi ≥ xi for ea h i; and f(yi) > f(xi) for i = 1, 2, . . . , k,hen ey1 + y2 + · · · + yn > x1 + x2 + · · · + xn = nand

f(y1) + f(y2) + · · · + f(yn) > f(x1) + f(x2) + · · · + f(xn) .Therefore, it suÆ es to show that f(y1) + f(y2) + · · · + f(yn) ≤ 1 for allyi ≥ a su h that y1+y2+· · ·+yn > n. By hypothesis, this inequality is truefor all yi ≥ a su h that y1+y2+· · ·+yn = n. Sin e f is de reasing on [a, ∞),the inequality is also true for all yi ≥ a su h that y1 + y2 + · · · + yn > n.Remark 2. If 0 ≤ p < 1 and x1 + x2 + · · · + xn ≥ n, then the inequality in(2) reverses. By the Cau hy-S hwarz Inequality, we have

237∑ y li x1

xp1 + x2 + · · · + xn

≥

(n∑

i=1

xi

)2

∑ y li x1(xp1 + x2 + · · · + xn)

=

(n∑

i=1

xi

)2

n∑i=1

xp+1

i +

(n∑

i=1

xi

)2

−n∑

i=1

x2i

= 1 +

n∑i=1

x2i −

n∑i=1

xp+1

i

n∑i=1

xp+1

i +

(n∑

i=1

xi

)2

−n∑

i=1

x2i

,and it suÆ es to show that n∑

i=1

x2i −

n∑i=1

xp+1

i ≥ 0. This inequality follows bythe Power Mean Inequality:1

n

n∑

i=1

x2i ≥

(1

n

n∑

i=1

xp+1

i

) 2p+1

≥(

1

n

n∑

i=1

xp+1

i

)(1

n

n∑

i=1

xi

)1−p

≥ 1

n

n∑

i=1

xp+1

i .3 Proof of Inequality (3)Inequality (3) is equivalent to

∑ y li 1

xp1 + x2 + · · · + xn

≤ n

x1 + x2 + · · · + xn

.For x1 + x2 + · · · + xn = n, this inequality immediately follows from (1).Otherwise, if x1 + x2 + · · · + xn > n, then we set r =x1 + x2 + · · · + xn

n,and yi =

xi

rfor i = 1, 2, . . . , n, so that r > 1 and y1 + y2 + · · · + yn = n.Then the inequality be omes

∑ y li 1

rp−1yp1 + y2 + · · · + yn

≤ 1 .This inequality follows from (1), sin e∑ y li 1

rp−1yp1 + y2 + · · · + yn

≤∑ y li 1

yp1 + y2 + · · · + yn

≤ 1 .Equality in (3) o urs if and only if x1 = x2 = · · · = xn = 1.

2384 New ConjecturesReaders an try to extend the inequalities (1) and (2) in di�erent dire tions.One of these dire tions is to �nd a ondition as weak as possible on the pos-itive real number p su h that the inequality below holds for any nonnegativereal numbers x1, x2, . . . , xn with x1 + x2 + · · · + xn ≥ n:

∑ y li x21

xp1 + x2 + · · · + xn

≤ 1 .For n ≥ 2, we onje ture that p ≥ n is su h a suÆ ient ondition.Moreover, we laim that

p ≥ 1 +ln 2

ln n − ln(n − 1)is a ne essary and suÆ ient ondition for the inequality to be true.On the other hand, the study of the reverse inequality for n ≥ 2 andx1 + x2 + · · · + xn ≥ n, is also an interesting problem for readers. We onje ture that the reverse inequality is true for 0 ≤ p ≤ 9

5.Referen es[1℄ V. C�rtoaje, Algebrai Inequalities - Old and New Methods, GIL Pub-lishing House, Romania, 2006.[2℄ V. C�rtoaje, On an Inequality from IMO 2005, Crux Mathemati orumwith Mathemati al Mayhem 32 (2006) 101{106.[3℄ V. C�rtoaje, A generalization of Jensen's Inequality, Gazeta Matemati aA, Issue 2 (2005) 124{138.[4℄ N. Nikolov, A Generalization of an Inequality from IMO 2005

http://arxiv.org/PS\_cache/math/pdf/0512/0512172.pdf[5℄ Mathlinks Website,http://www.mathlinks.ro/Forum/viewtopic.php?p=288276 andhttp://www.mathlinks.ro/Forum/viewtopic.php?t=44480\&start=40Vasile C�rtoajeDepartment of Automation and ComputersUniversity of Ploie�sti39 Bu ure�sti BoulevardPloie�sti, Romania

vcirtoaje@upg-ploiesti.ro

239PROBLEMSToutes solutions aux probl �emes dans e num�ero doivent nous parvenir au plustard le 1er novembre 2008. 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.Oliver Geupel nous a signal �e une erreur dans l' �enon �e du probl �eme #3282[2007 : 429, 431℄, propos �e par Jos �e Luis D��az-Barrero et Pantelimon GeorgePopes u. Nous nous ex usons de ette erreur.

3282. Corre tion. Propos �e par Jos �e Luis D��az-Barrero, Universit �e Poly-te hnique de Catalogne, Bar elone, Espagne et Pantelimon George Popes u,Bu arest, Roumanie.Soit A(z) = zn + an−1zn−1 + · · · + a1z + a0 un polynome uni-taire �a oeÆ ients omplexes. On suppose que a1 = −a0 et que les z �erosz1, z2, . . . , zn de A(z) sont des nombres omplexes non nuls et distin ts.Montrer que

n∑

k=1

ezk

z2k

n∏

j=1

j 6=k

1

zk − zj

= 0 .3338. Propos �e par Toshio Seimiya, Kawasaki, Japon.Un quadrilat �ere ir ulaire onvexe ABCD poss �ede un er le ins rit de entre I. Soit P le point d'interse tion de AC et BD. Montrer que l'on aAP : CP = AI2 : CI2.3339. Propos �e par Toshio Seimiya, Kawasaki, Japon.Soit Γ1 et Γ2 deux er les sans points ommuns et situ �es �a l'ext �erieurl'un de l'autre. Soit ℓ1 et ℓ2 les tangentes externes ommunes �a Γ1 et Γ2.Soit A et B les points d'interse tion de ℓ1 ave Γ1 et Γ2; C et D eux de ℓ2ave Γ1 et Γ2. Notons M et N les points milieu AB et CD, et soit P et Qles points d'interse tion de NA et NB ave Γ1 et Γ2, di� �erents de A et B.Montrer que CP , DQ et MN sont on ourants.3340. Propos �e par Toshio Seimiya, Kawasaki, Japon.Dans un triangle ABC, la bisse tri e de l'angle BAC oupe le er le ir ons rit en un se ond point D. Supposons que AB2 + AC2 = 2AD2.Montrer que AD et BC se oupent �a 45◦.

2403341. Propos �e par Arkady Alt, San Jos �e, CA, �E-U.Soit ABC un triangle quel onque, de ot �es a, b et c; montrer que√

3(Ra + Rb + Rc) ≤ a + b + c, o �u Ra, Rb et Rc sont les distan es res-pe tives du entre du er le ins rit du triangle ABC aux sommets A, B et C.3342. Propos �e par Arkady Alt, San Jos �e, CA, �E-U.Soit respe tivement r et R le rayon des er les ins rit et ir ons rits dutriangle ABC. Montrer que2∑ y lique sin A

2sin

B

2≤ 1 +

r

R.

3343. Propos �e par Stan Wagon, Ma alester College, St. Paul, MN, �E-U.Si l'on supprime toutes les fa torielles dans la s �erie de Ma laurin desin x, on obtient la s �erie de arctan x. Supposons alors qu'on en supprimeseulement une sur deux. La s �erie ainsi obtenue a-t-elle une forme ferm�ee?C'est- �a-dire, peut-on trouver la fon tion dont la s �erie de Ma laurin soit

x − x3

3+

x5

5!− x7

7+

x9

9!− x11

11+ · · · ?

3344. Propos �e par Pham Kim Hung, �etudiant, Universit �e de Stanford,Palo Alto, CA, �E-U.Soit n un entier positif, n ≥ 4, et soit a1, a2, . . . , an des nombres r �eelspositifs tels que a1 + a2 + · · · + an = n. Montrer que1

a1

+1

a2

+ · · · +1

an

− n ≥ 3

n

(a2

1+ a2

2+ · · · + a2

n − n) .

3345. Propos �e par Pham Kim Hung, �etudiant, Universit �e de Stanford,Palo Alto, CA, �E-U.Soit a, b, c et d des nombres r �eels positifs tels que a + b + c + d = 4.Montrer quea

1 + b2c+

b

1 + c2d+

c

1 + d2a+

d

1 + a2b≥ 2 .

3346. Propos �e par Bin Zhao, �etudiant, YunYuan HuaZhong Universit �e deTe hnologie et S ien e, Wuhan, Hubei, Chine.Dans un triangle ABC, montrer queπ∑ y lique 1

A≥

∑ y lique sin A

2

∑ y lique csc A

2

.

2413347. Propos �e par Mih�aly Ben ze, Brasov, Roumanie.Soit A1A2A3A4 un quadrilat �ere onvexe. Soit Bi un point sur AiAi+1pour i ∈ {1, 2, 3, 4}, les indi es �etant pris modulo 4, de sorte que

B1A1

B1A2

=B3A4

B3A3

=A1A4

A2A3

et B2A2

B2A3

=B4A1

B4A4

=A1A2

A3A4

.Montrer que B1B3 ⊥ B2B4 si et seulement si A1A2A3A4 est un quadri-lat �ere ir ulaire.3348. Propos �e par Mih�aly Ben ze, Brasov, Roumanie.Dans un triangle a utangle ABC, soit respe tivement A1, B1 et C1des points sur les ot �es BC, CA et AB, de sorte que ∠AC1B1, ∠BC1A1 et∠ACB soient �egaux. Soit respe tivement M , N et P les entres des er les ir ons rits des triangles AC1B1, BA1C1 et CB1A1. Montrer que AM ,BN et CP sont on ourants si et seulement si AA1, BB1 et CC1 sont leshauteurs du triangle ABC.3349. Propos �e par Mih�aly Ben ze, Brasov, Roumanie.Soit a, b et c trois nombres r �eels positifs. Montrer que6∏ y lique a3 + 1

a2 + 1≥ max

∑ y liquea(1 + bc)(a2 + 1)

a3 + 1, ∑ y liqueab(1 + c)(a2b2 + 1)

a3b3 + 1

.

3350. Propos �e par Panos E. Tsaoussoglou, Ath �enes, Gr �e e.Soit x, y et z trois nombres r �eels positifs tels que x + y + z = 1.Montrer queyz

1 + x+

zx

1 + y+

xy

1 + z≤ 1

4.

.................................................................3282. Corre tion. Proposed by Jos �e Luis D��az-Barrero, Universitat Poli-t �e ni a de Catalunya, Bar elona, Spain and Pantelimon George Popes u,Bu harest, Romania.Let A(z) = zn + an−1zn−1 + · · · + a1z + a0 be a moni polyno-mial with omplex oeÆ ients. Suppose that a1 = −a0, and that the zeroesz1, z2, . . . , zn of A(z) are distin t, non-zero omplex numbers. Prove that

n∑

k=1

ezk

z2k

n∏

j=1

j 6=k

1

zk − zj

= 0 .

2423338. Proposed by Toshio Seimiya, Kawasaki, Japan.A onvex y li quadrilateral ABCD has an in ir le with entre I. LetP be the interse tion of AC and BD. Prove that AP : CP = AI2 : CI2.3339. Proposed by Toshio Seimiya, Kawasaki, Japan.Let Γ1 and Γ2 be two non-interse ting ir les ea h lying in the exteriorof the other. Let ℓ1 and ℓ2 be the ommon external tangents to Γ1 and Γ2.Let ℓ1 meet Γ1 and Γ2 at A and B, respe tively, and let ℓ2 meet Γ1 and Γ2at C and D, respe tively. Let M and N be the mid-points of AB and CD,respe tively, and let P and Q be the interse tions of NA and NB with Γ1and Γ2, respe tively, di�erent from A and B. Prove that CP , DQ, and MNare on urrent.3340. Proposed by Toshio Seimiya, Kawasaki, Japan.The bise tor of ∠BAC interse ts the ir um ir le of△ABC at a se ondpoint D. Suppose that AB2 + AC2 = 2AD2. Prove that the angle ofinterse tion of AD and BC is 45◦.3341. Proposed by Arkady Alt, San Jose, CA, USA.For any triangle ABC with sides of lengths a, b, and c, prove that√

3(Ra + Rb + Rc) ≤ a + b + c, where Ra, Rb, and Rc are the distan esfrom the in entre of △ABC to the verti es A, B, and C, respe tively.3342. Proposed by Arkady Alt, San Jose, CA, USA.Let r and R be the inradius and ir umradius of △ABC, respe tively.Prove that2∑ y li sin A

2sin

B

2≤ 1 +

r

R.

3343. Proposed by Stan Wagon, Ma alester College, St. Paul, MN, USA.If the fa torials are deleted in theMa laurin series for sin x, one obtainsthe series for arctan x. Suppose instead that one alternates fa torials in theseries. Does the resulting series have a losed form? That is, an one �nd anelementary expression for the fun tion whose Ma laurin series isx − x3

3+

x5

5!− x7

7+

x9

9!− x11

11+ · · · ?

3344. Proposed by Hung Pham Kim, student, Stanford University, PaloAlto, CA, USA.Let n be a positive integer, n ≥ 4, and let a1, a2, . . . , an be positivereal numbers su h that a1 + a2 + · · · + an = n. Prove that1

a1

+1

a2

+ · · · +1

an

− n ≥ 3

n

(a2

1+ a2

2+ · · · + a2

n − n) .

2433345. Proposed by Hung Pham Kim, student, Stanford University, PaloAlto, CA, USA.Let a, b, c, and d be positive real numbers su h that a + b+ c + d = 4.Prove that

a

1 + b2c+

b

1 + c2d+

c

1 + d2a+

d

1 + a2b≥ 2 .

3346. Proposed by Bin Zhao, student, YunYuan HuaZhong University ofTe hnology and S ien e, Wuhan, Hubei, China.Given triangle ABC, prove thatπ∑ y li 1

A≥

∑ y li sin A

2

∑ y li csc A

2

.

3347. Proposed by Mih�aly Ben ze, Brasov, Romania.Let A1A2A3A4 be a onvex quadrilateral. Let Bi be a point on AiAi+1for i ∈ {1, 2, 3, 4}, where the subs ripts are taken modulo 4, su h thatB1A1

B1A2

=B3A4

B3A3

=A1A4

A2A3

and B2A2

B2A3

=B4A1

B4A4

=A1A2

A3A4

.Prove that B1B3 ⊥ B2B4 if and only if A1A2A3A4 is a y li quadrilateral.3348. Proposed by Mih�aly Ben ze, Brasov, Romania.Let ABC be an a ute-angled triangle. Let A1, B1, and C1 be pointson the sides BC, CA, and AB, respe tively, su h that the angles ∠AC1B1,∠BC1A1, and ∠ACB are all equal. Let M , N , and P be the ir um entresof △AC1B1, △BA1C1, and △CB1A1, respe tively. Prove that AM , BN ,and CP are on urrent if and only if AA1, BB1, and CC1 are altitudes of△ABC.3349. Proposed by Mih�aly Ben ze, Brasov, Romania.Let a, b, and c be positive real numbers. Show that6∏ y li a3 + 1

a2 + 1≥ max

∑ y li a(1 + bc)(a2 + 1)

a3 + 1, ∑ y li ab(1 + c)(a2b2 + 1)

a3b3 + 1

.

3350. Proposed by Panos E. Tsaoussoglou, Athens, Gree e.Let x, y, and z be positive real numbers su h that x+y+z = 1. Provethatyz

1 + x+

zx

1 + y+

xy

1 + z≤ 1

4.

244SOLUTIONSNo problem is ever permanently losed. The editor is always pleasedto onsider for publi ation new solutions or new insights on past problems.

KLAMKIN{05. [2005 : 328℄ Proposed by Vasile C�rtoaje, Universityof Ploiesti, Romania.Let k and n be positive integers with k < n, and let a1, a2, . . . , an bereal numbers su h that a1 ≤ a2 ≤ · · · ≤ an. Prove that(a1 + a2 + · · · + an)2 ≥ n(a1ak+1 + a2ak+2 + · · · + anan+k)(where the subs ripts are taken modulo n) in the following ases:(a) n = 2k; (b) n = 4k; ( )⋆ 2 <

n

k< 4.Solution to part ( )⋆ by the proposer.Solutions to parts (a) and (b) appeared in [2006 : 315℄. Let n = 4k − j,where j is an integer su h that 1 ≤ j ≤ 2k − 1, and let i =⌊

j2

⌋. Then0 ≤ i ≤ k − 1. Let a be a real number su h that a2k−i ≤ a ≤ a2k−i+1, andlet S1 = a1 + a2 + · · · + an and S2 = a1ak+1 + a2ak+2 + · · · + anan+k.We have to show that

S21

4k − j≥ S2 .We onsider two ases: 2k < n ≤ 3k and 3k ≤ n < 4k.Case 1. Let 2k < n ≤ 3k. Then k ≤ j ≤ 2k−1, and hen e ⌊k

2

⌋≤ i ≤ k−1.Let

S =

3k−i∑

m=k−i+1

am

and P =

2k−i∑

m=k−i+1

amak+m

.

Applying the original inequality in part (b) to the non-de reasing sequen eof 4k real numbers a1, . . . , a2k−i, a, . . . , a︸ ︷︷ ︸j times

, a2k−i+1, . . . , an, we obtain(S1 + ja)2

4k≥ (j − k)a2 + Sa +

(k−i∑

m=1

amak+m

)

+

n∑

m=2k−i+1

amak+m

,

245or equivalently,

(S1 + ja)2

4k≥ (j − k)a2 + Sa + S2 − P .The inequality to be proved follows by adding this inequality to

S21

4k − j+ (j − k)a2 + Sa − P ≥ (S1 + ja)2

4k.The last inequality is true, be ause

S21

4k − j+ ja2 − (S1 + ja)2

4k=

j[S1 − (4k − j)a]2

4k(4k − j)≥ 0and

−ka2 + Sa − P =

2k−i∑

m=k−i+1

(a − am)(ak+m − a) ≥ 0 .Case 2. Let 3k ≤ n < 4k. Then 1 ≤ j ≤ k, and hen e 0 ≤ i ≤

⌊k2

⌋. LetS =

k−i+j∑

m=k−i+1

am

+

3k−i∑

m=3k−i−j+1

am

,

P =

2k−i∑

m=k−i+1

amak+m

, and Q =

2k−i∑

m=k−i+j+1

amak−j+m

.

Applying the original inequality in part (b) to the non-de reasing sequen eof 4k real numbers a1, . . . , a2k−i, a, . . . , a︸ ︷︷ ︸j times

, a2k−i+1, . . . , an, we obtain(S1 + ja)2

4k≥

(k−i∑

m=1

amak+m

)+ a

k−i+j∑

m=k−i+1

am

+

2k−i∑

m=k−i+j+1

amak−j+m

+ a

3k−i∑

m=3k−i−j+1

am

+

n∑

m=2k−i+1

amak+m

,

or equivalently,(S1 + ja)2

4k≥ S2 − P + Q + Sa .

246The required inequality follows by adding this inequality to

S21

4k − j− P + Q + Sa ≥ (S1 + ja)2

4k,whi h is true, be ause

S21

4k − j− (S1 + ja)2

4k+ ja2 =

j[S1 − (4k − j)a]2

4k(4k − j)≥ 0and

−ja2 − P + Q + Sa =

2k−i−j∑

m=k−i+1

(aj+m − am)(ak+m − a)

+

2k−i∑

m=2k−i−j+1

(a − am)(ak+m − a) ≥ 0 .There were no other solutions submitted.

3239. [2007 : 237, 239℄ Proposed by Mih�aly Ben ze, Brasov, Romania.Let n be a positive integer. If α = 1 +1

12(n + 1), prove that

e <

((n + 1)2n+1

(n!)2

) 12n

< eα .Solution by Kee-Wai Lau, Hong Kong, China, modi�ed by the editor.Taking logarithms, the given inequalities are su essively equivalent to

1 <1

2n

[(2n + 1) ln(n + 1) − 2 ln(n!)

]< α ,

2n < (2n + 1) ln(n + 1) − 2 ln(n!) < 2n +n

6(n + 1).Let f(n) = (2n+1) ln(n+1)−2 ln(n!)−2n, and g(n) = f(n)− n

6(n + 1).We are to show that f(n) > 0 and g(n) < 0. For 0 < x < 1, it is wellknown that ln(1 + x) > x − x2

2+

x3

3− x4

4. Hen e,

f(n + 1) − f(n)

= (2n + 3) ln(n + 2) − 2 ln((n + 1)!

)− 2(n + 1)

− (2n + 1) ln(n + 1) + 2 ln(n!) + 2n

= (2n + 3) ln(

n + 2

n + 1

)− 2 = (2n + 3) ln

(1 +

1

n + 1

)− 2

> (2n + 3)

(1

n + 1− 1

2(n + 1)2+

1

3(n + 1)3− 1

4(n + 1)4

)− 2

= An .

247By tedious but straightforward omputations, we �nd that

An =2n2 + 2n − 3

12(n + 1)4> 0 .Hen e, f(n) > f(1) = 3 ln 2 − 2 > 0.Similarly, for 0 < x < 1, it is well known that

ln(1 + x) < x − x2

2+

x3

3− x4

4+

x5

5.Hen e,

g(n + 1) − g(n)

= f(n + 1) − f(n) − n + 1

6(n + 2)+

n

6(n + 1)

= (2n + 3) ln(1 +

1

n + 1

)− 2 − 1

6(n + 1)(n + 2)

< An +2n + 3

5(n + 1)5− 1

6(n + 1)(n + 2)

= Bn .Again, by tedious but straightforward omputations, we �nd thatBn =

−n2 + 19n + 32

60(n + 1)5(n + 2)=

−`

n(n − 19) − 32´

60(n + 1)5(n + 2)< 0for n ≥ 21. On the other hand, using a al ulator, we an readily he k that

g(n) < 0 for n = 1, 2, . . . , 21. It follows that g(n) < 0 for ea h positiveinteger n, and our proof is omplete.Also solved by MICHEL BATAILLE, Rouen, Fran e; WALTHER JANOUS, Ursulinen-gymnasium, Innsbru k, Austria; and the proposer. A partial solution was submitted by SALEMMALIKI �C, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina.3240. [2007 : 237, 239℄ Proposed by Mih�aly Ben ze, Brasov, Romania.Let n be a positive integer. Prove that

⌊√n +

√n + 2 3

√n + 1

⌋=

⌊√4n + 4 3

√n + 2

⌋ ,where ⌊x⌋ denotes the integer part of x.Solution by John Hawkins and David R. Stone, Georgia Southern University,Statesboro, GA, USA.The asserted equality is false in general.For example, when n = 45, we have⌊√

n +

√n + 2 3

√n + 1

⌋= ⌊13.996 · · · ⌋ = 13 ,

248while ⌊√

4n + 4 3√

n + 2

⌋= ⌊14.008 · · · ⌋ = 14 .It also fails for n = 95 and 616.Hawkins and Stone also gave the following omment: It is straightforward to show thatthe fun tion √

n+√

n + 2 3√

n + 1 is less than√4n + 4 3√

n + 2, and the gap between themnarrows as n in reases. But sometimes the values \straddle" an integer, as shown above, sothe desired equality fails. We suspe t that happens in�nitely often (the next instan e is 1249).SALEM MALIKI �C, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina, alsogave the ounterexample n = 45. There were three in orre t \proofs" for the assertion.3241. [2007 : 237, 239℄ Proposed by Virgil Ni ula, Bu harest, Romania.Let a, b, c be any real numbers su h that a2 + b2 + c2 = 9. Prove that

3 · min{a, b, c} ≤ 1 + abc .Solution by Arkady Alt, San Jose, CA, USA, modi�ed by the editor.Without loss of generality, we may assume a ≤ b ≤ c. We want toprove the inequality abc + 1 ≥ 3a.If a ≤ 0, then using the inequality bc ≤ 1

2(b2 + c2), we obtain

abc + 1 − 3a ≥ a|bc| − 3a + 1 ≥ 1

2a(b2 + c2) − 3a + 1

= 1

2a(9 − a2) − 3a + 1 = 1

2(a + 1)2(2 − a) ≥ 0 ,with equality if and only if a = −1 and b = c = 2.Now, let a > 0. Sin e a ≤ b ≤ c, we get 9 = a2 + b2 + c2 ≥ 3a2;when e, a ≤

√3. We have the obvious inequality (c2 − a2)(b2 − a2) ≥ 0,whi h yields

bc ≥ a√

c2 + b2 − a2 = a√

9 − 2a2 .Hen e, it suÆ es to prove the (stronger) inequalitya(a√

9 − 2a2)+ 1 ≥ 3a .If 0 < a ≤ 1, then √

9 − 2a2 ≥√

7 > 9

4; thus, 4a2

√9 − 2a2 > 9a2.Using the inequality (t + 1)2 ≥ 4t, we obtain

(a2√

9 − 2a2 + 1)2 ≥ 4a2

√9 − 2a2 > 9a2 ,whi h yields

a(a√

9 − 2a2)+ 1 ≥ 3a .If 1 < a ≤

√3, then we an prove an equivalent form of our inequality,namely, the one obtained by squaring both sides of

a(a√

9 − 2a2)

≥ 3a − 1 ;

249that is,

2a6 − 9a4 + 9a2 − 6a + 1 ≤ 0 .We have2a6 − 9a4 + 9a2 − 6a + 1 < 2a6 − 9a4 + 9a2 − 5

= (2a2 − 1)(a2 − 1)(a2 − 3) − (a2 + 2)

< 0 ,whi h ompletes the proof.Also solved by MOHAMMED AASSILA, Strasbourg, Fran e; ROY BARBARA, LebaneseUniversity, Fanar, Lebanon; JOHN HAWKINS and DAVID R. STONE, Georgia SouthernUniversity, Statesboro, GA, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k,Austria; KEE-WAI LAU, Hong Kong, China; SALEM MALIKI �C, student, Sarajevo College,Sarajevo, Bosnia and Herzegovina; ALEX REMOROV, student, William Lyon Ma kenzieCollegiate Institute, Toronto, ON; PETER Y. WOO, Biola University, La Mirada, CA, USA; andthe proposer.3243. [2007 : 237, 239℄ Proposed by G. Tsintsifas, Thessaloniki, Gree e.Let ABC be an isos eles triangle with AB = AC, and let P be aninterior point. Let the lines BP and CP interse t the opposite sides at thepoints D and E, respe tively. Find the lo us of P if

PD + DC = PE + EB .Solution by Titu Zvonaru, Com�ane�sti, Romania.The lo us onsists of the points inside the tri-angle on the altitude from A.Note that sin e the triangle is isos eles, thefoot, M , of the altitude is the mid-point of BC, andthe triangle is symmetri about AM . Without lossof generality, label the �gure so that BE ≥ CDand let E′ be the point between A and C for whi hCE′ = BE, and let BE′ and CE interse t at N .Be ause △ABC is isos eles, BCE′E is an isos elestrapezoid that is symmetri about AM ; when e, Nbelongs to AM and NE = NE′. As a onsequen e,the following statements are equivalent:

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

A

B C

D

E E′

M

NP

PD + DC = PE + EB = PN + NE + E′C

= PN + NE′ + E′D + DC ,PD = PN + NE′ + E′D .The last equality says that the quadrilateral PDE′N is degenerate; that is,

D oin ides with E′ and P with N , so that P lies on AM . The lo us of P istherefore the open line segment AM , as laimed.

250Also solved by ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHELBATAILLE, Rouen, Fran e; V �ACLAV KONE �CN �Y, Big Rapids, MI, USA; SALEM MALIKI �C,student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; ALEX REMOROV, student,William Lyon Ma kenzie Collegiate Institute, Toronto, ON; EDMUND SWYLAN, Riga, Latvia;PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer.

3244. [2007 : 238, 239℄ Proposed by G. Tsintsifas, Thessaloniki, Gree e.Let ABC be an isos eles triangle with AB = AC, and let P be aninterior point. Let the lines BP and CP interse t the opposite sides at thepoints D and E, respe tively. Find the lo us of P ifBD + DC = BE + EC .Composite of similar solutions by V�a lav Kone� n�y, Big Rapids, MI, USA; andPeter Y. Woo, Biola University, La Mirada, CA, USA.As with problem 3243, the lo us of P is the portion of the altitude AMinside △ABC.The ondition BD + DC = BE + EC means that points D and Elie on one ellipse of the family of ellipses with fo i B and C, entre M , andaxes along the lines BC and AM . Sin e the ellipse is symmetri about AMand interse ts ea h of the line segments AB and AC in exa tly one point,

BD must interse t CE at P on AM . Conversely, if P lies on AM , then thediagram is symmetri about AM , implying that BD = CE and DC = EB,so that BD + DC = BE + EC.Also solved by ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHELBATAILLE, Rouen, Fran e; SALEM MALIKI �C, student, Sarajevo College, Sarajevo, Bosnia andHerzegovina; ALEX REMOROV, student, William Lyon Ma kenzie Collegiate Institute, Toronto,ON; JOEL SCHLOSBERG, Bayside, NY, USA; D.J. SMEENK, Zaltbommel, the Netherlands;EDMUND SWYLAN, Riga, Latvia; TITU ZVONARU, Com�ane�sti, Romania; and the proposer.Some of the submitted solutions were designed to solve simultaneously this problemand problem 3243 above. For example, a straightforward modi� ation of the featured solutionto 3243 will likewise provide a solution to this problem.3245. [2007 : 238, 240℄ Proposed by Paul Yiu, Florida Atlanti University,Bo a Raton, FL, USA.Suppose that the entre of the nine-point ir le of a triangle lies on thein ir le of the triangle. Show that its antipodal point is the Feuerba h Point;that is, the point where the nine-point ir le and the in ir le are tangent toea h other.I. Composite of similar solutions by Salem Maliki� , student, SarajevoCollege, Sarajevo, Bosnia and Herzegovina; the Skidmore College ProblemSolving Group, Skidmore College, Saratoga Springs, NY, USA; and Peter Y.Woo, Biola University, La Mirada, CA, USA.

251The result holds for tangent ir les in general: the line joining the en-tres of two tangent ir les passes through the point of tangen y. If, moreover,the entre of one lies on the ir umferen e of the other, the entre of the �rst ir le is diametri ally opposed in the se ond ir le to the ommon tangen ypoint.II. Composite of similar solutions by Mi hel Bataille, Rouen, Fran e; JohnG. Heuver, Grande Prairie, AB; Walther Janous, Ursulinengymnasium,Innsbru k, Austria; and the proposer.Let I be the in entre, let N be the nine-point entre, and let r and

R be the inradius and ir umradius. Many standard referen es prove thatthe in ir le is internally tangent to the nine-point ir le, whose radius is 1

2R;hen e, NI = 1

2R − r. Moreover, if N is on the in ir le, we also have

NI = r and, therefore, 1

2R = 2r. The antipodal point of N on the in ir le,say N ′, satis�es NN ′ = 2r, so that NN ′ = 1

2R. Thus, not only is N ′ onthe in ir le, but it is also on the nine-point ir le. As a result, N ′ oin ideswith the unique ommon point to these two ir les, namely, the Feuerba hPoint.There were no other solutions submitted.

3246. [2007 : 238, 240℄ Proposed by Marian Tetiva, B�rlad, Romania.Let a, b, c, d be any positive real numbers with d = min{a, b, c, d}.Prove thata4 + b4 + c4 + d4 − 4abcd

≥ 4d[(a − d)3 + (b − d)3 + (c − d)3 − 3(a − d)(b − d)(c − d)

] .Solution by Kee-Wai Lau, Hong Kong, China.Let x = a − d, y = b − d, and z = c − d. Thus, x ≥ 0, y ≥ 0, andz ≥ 0. It an then be he ked thata4 + b4 + c4 + d4 − 4abcd

− 4d[(a − d)3 + (b − d)3 + (c − d)3 − 3(a − d)(b − d)(c − d)

]

= 2d2[x2 + y2 + z2 + (x − y)2 + (y − z)2 + (z − x)2

]

+ 8xyzd + x4 + y4 + z4

≥ 0 ,and the inequality is proved.Also solved by ARKADY ALT, San Jose, CA, USA; MICHEL BATAILLE, Rouen, Fran e;CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; JOE HOWARD,Portales, NM, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; SALEMMALIKI �C, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; ALEX REMOROV,

252student, William Lyon Ma kenzie Collegiate Institute, Toronto, ON; PETER Y. WOO, BiolaUniversity, La Mirada, CA, USA; TITU ZVONARU, Com�ane�sti, Romania; and the proposer.Janous pointed out that the inequality has the equivalent form

a4 + b4 + c4 + d4 − 4abcd ≥ 4d(a + b + c − 3d)(a2 − ab − ac + b2 − bc + c2) .Sin e the three fa tors on the right side are non-negative, this gives a lever proof for the four-variable AM{GM Inequality.Maliki � ommented that on the right side of the inequality the fa tor 3 an be hangedto 1, resulting in a stronger inequality, namelyx4 +y4 +z4 +2(d2x2 +d2y2 +d2z2)+2

[(dx−dy)2 +(dy−dz)2 +(dz −dx)2

]≥ 0 .Alt also provided the proof for the inequality

a4+b4+c4+d4 −4abcd ≥ 4d[(a−d)3+(b−d)3+(c−d)3 −kd(a−d)(b−d)(c−d)

] ,where k is in the interval [1 − 3/√

2,1 + 3/√

2].

3247. [2007 : 238, 240℄ Proposed by Jos �e Luis D��az-Barrero, UniversitatPolit �e ni a de Catalunya, Bar elona, Spain.Let a1, a2, . . . , an be real numbers, ea h greater than 1. Prove thatn∑

k=1

(1 + logak

(ak+1))2 ≥ 4n ,

where an+1 = a1.Solution by Dionne Bailey, Elsie Campbell, and Charles R. Diminnie, AngeloState University, San Angelo, TX, USA.By the AM{GM Inequality and the hange of base formula, we haven∑

k=1

(1 + logak

(ak+1))2 ≥

n∑

k=1

(2√

logak(ak+1)

)2

= 4n∑

k=1

logak(ak+1) ≥ 4n

(n∏

k=1

logak(ak+1)

)1n

= 4n

(n∏

k=1

ln ak+1

ln ak

)1n

= 4n .For equality to hold, we must have logak

(ak+1) = 1 or ak = ak+1 forall k = 1, 2, . . . , n. Conversely, if a1 = a2 = · · · = an, then learly equalityholds.Also solved by ARKADY ALT, San Jose, CA, USA; ROY BARBARA, Lebanese University,Fanar, Lebanon; MICHEL BATAILLE, Rouen, Fran e; MIH �ALY BENCZE, Brasov, Romania;CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; JOE HOWARD,Portales, NM, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; KEE-WAILAU, Hong Kong, China; SALEM MALIKI �C, student, Sarajevo College, Sarajevo, Bosnia andHerzegovina; ALEX REMOROV, student, William Lyon Ma kenzie Collegiate Institute, Toronto,

253ON; HENRY RICARDO, Medgar Evers College (CUNY ), Brooklyn, NY, USA; XAVIER ROS,student, Universitat Polit �e ni a de Catalunya, Bar elona, Spain; PETER Y. WOO, BiolaUniversity, La Mirada, CA, USA; TITU ZVONARU, Com�ane�sti, Romania; and the proposer.Barbara pointed out that the inequality is a spe ial ase of the more general result thatif x1, x2, . . . , xn are positive reals su h that n∏

k=1

xk = 1, then n∑k=1

(1 + xk)2 ≥ 4n. Ben zeobtained the generalization that if ak > 1 for k = 1, 2, . . . , n, and if P(x) is any polynomialwith positive oeÆ ients, then n∑k=1

P(logak

(ak+1))

≥ nP(1). The urrent problem is thespe ial ase when P(x) = (1 + x)2. The proofs of these two generalizations are virtually thesame as the one featured above.3248. [2007 : 238, 240℄ Proposed by Titu Zvonaru, Com�ane�sti, Romania,and Bogdan Ionit� �a, Bu harest, Romania.If a, b, and c are positive real numbers, prove that

a2(b + c − a)

b + c+

b2(c + a − b)

c + a+

c2(a + b − c)

a + b≤ ab + bc + ca

2.Solution by Mohammed Aassila, Strasbourg, Fran e.The proof is as follows

∑ y li a2(b + c − a)

b + c=

∑ y li a

b + ca(b + c − a)

≤∑ y li a

b + c

(a + (b + c − a)

2

)2

=∑ y li a(b + c)

4=

∑ y li bc

2.

Also solved by ARKADY ALT, San Jose, CA, USA; �SEFKET ARSLANAGI �C, Universityof Sarajevo, Sarajevo, Bosnia and Herzegovina; MICHEL BATAILLE, Rouen, Fran e;JOE HOWARD, Portales, NM, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k,Austria; KEE-WAI LAU, Hong Kong, China; SALEM MALIKI �C, student, Sarajevo College,Sarajevo, Bosnia and Herzegovina; JOS �E H. NIETO, Universidad del Zulia, Mara aibo,Venezuela; PHI THAI THUAN, student, Tran Hung Dao High S hool, Phan Thiet, Vietnam;CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; ALEX REMOROV,student, William Lyon Ma kenzie Collegiate Institute, Toronto, ON; XAVIER ROS, student,Universitat Polit �e ni a de Catalunya, Bar elona, Spain; PANOS E. TSAOUSSOGLOU, Athens,Gree e; APOSTOLIS VERGOS, student, University of Patras, Patras, Gree e; and PETER Y.WOO,Biola University, La Mirada, CA, USA.Janous provided a proof of the more general inequalitya2n(b + c − a)

b + c+

b2n(c + a − b)

c + a+

c2n(a + b − c)

a + b≤ (ab)n + (bc)n + (ca)n

2for any non-negative integern. At the same time he would like to leave the following onje tureto the readers of CRUX with MAYHEM:(an + bn + cn)(abc)n

2≤ a + b − c

a + b(ab)2n +

b + c − a

b + c(bc)2n +

c + a − b

c + a(ca)2nfor any positive integer n.

2543249. [2007 : 238, 240℄ Proposed by Titu Zvonaru, Com�ane�sti, Romania,and Bogdan Ionit� �a, Bu harest, Romania.Let a, b, and c be the lengths of the sides of a triangle. Prove that

(b + c)2

a2 + bc+

(c + a)2

b2 + ca+

(a + b)2

c2 + ab≥ 6 .

I. Composite of solutions by Mohammed Aassila, Strasbourg, Fran e; �SefketArslanagi � , University of Sarajevo, Sarajevo, Bosnia andHerzegovina; Mi helBataille, Rouen, Fran e; Nguyen Thanh Liem, Tran Hung Dao High S hool,Phan Thiet, Vietnam; Salem Maliki� , student, Sarajevo College, Sarajevo,Bosnia and Herzegovina; Alex Remorov, student, William Lyon Ma kenzieCollegiate Institute, Toronto, ON; Phi Thai Thuan, student, Tran HungDao High S hool, Phan Thiet, Vietnam; and Panos E. Tsaoussoglou, Athens,Gree e.We have∑ y li [(b + c)2

a2 + bc− 2

]=

∑ y li b2 + c2 − 2a2

a2 + bc

=∑ y li (a2 − b2)

(1

b2 + ca− 1

a2 + bc

)

=

∑ y li (a − b)2(a + b)(a + b − c)(c2 + ab)

∏ y li (c2 + ab)

≥ 0 .II. Se ond solution by Mohammed Aassila, Strasbourg, Fran e.We prove the result for any positive real numbers a, b, and c. After alengthy omputation, the original inequality∑ y li (b + c)2

a2 + bc− 6 ≥ 0

transforms to the following equivalent form(b − c)2(c − a)2(a − b)2

abc+∑ y li (b + c)2(b − c)2

a≥ 0 ,

whi h is obviously true for any positive real numbers a, b, and c, as laimed.Also solved by ARKADY ALT, San Jose, CA, USA; �SEFKET ARSLANAGI �C, Universityof Sarajevo, Sarajevo, Bosnia and Herzegovina (se ond solution); MIH �ALY BENCZE, Brasov,Romania; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam;WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; KEE-WAI LAU, Hong Kong,China; XAVIER ROS, student, Universitat Polit �e ni a de Catalunya, Bar elona, Spain (threesolutions); and the proposer.

255Janous and Ros have also shown that the inequality is true for any positive real numbers

a, b, and c. Janous has proven that the following generalization holds∑ y li (b + c)2

a2 + µbc≥ 12

µ + 1for any positive real numbers a, b, and c and µ ∈ [0,1] ∪ [2, ∞). Maliki � mentions thatthis is a well-known problem, originally reated by Darij Grinberg and Peter S holze; hegives the web address http://www.mathlinks.ro/viewtopic.php?search_id=7020805&t=60128,where the problem an be found together with several solutions, in luding solutions similar toboth of those featured above.3250. [2007 : 239, 240℄ Proposed by D.J. Smeenk, Zaltbommel, theNetherlands.Let ABC be an isos eles triangle with AB = AC and ∠BAC = 100◦.Let D be the point on the produ tion of AB su h that AD = BC. Find∠ADC.I. Solution by Tai hi Maekawa, Takatsuki City, Osaka, Japan.Constru t the equilateral triangle ADE, asshown in the pi ture. Then △ABC and △CAEare ongruent, sin eAB = CA, BC = AD = AE,and

∠CAE = 100◦ − 60◦ = 40◦ = ∠ABC .Then AC = CE and AD = DE, so that points Dand C both lie on the perpendi ular bise tor of thesegment AE. Let M be the point of interse tionof DC and AE. Then ∠AMD = 90◦, and sin e∠DAE = 60◦, it follows that ∠ADC = 30◦.

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

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

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

A

BC

D

E

M

II. Solution by Xavier Ros, student, Universitat Polit �e ni a de Catalunya,Bar elona, Spain.Let ∠ADC = α. Applying the Law of Sines to triangles ABC andACD, we have

AC

sin 40◦ =BC

sin 100◦ and AC

sin α=

AD

sin(80◦ − α),and, sin e AD = BC, we get

sin(80◦ − α) sin 40◦ = sin α sin 100◦ .Now, sin 100◦ = sin 80◦ = 2 sin 40◦ cos 40◦, so that we obtainsin(80◦ − α) = 2 sin α cos 40◦ ,or

sin 80◦ cos α − sin α cos 80◦ = 2 sin α cos 40◦ .

256The value α = 90◦ is not a solution to this equation, so that we an divideboth sides by cos α to obtain an equation for tan α:

tan α =sin 80◦

2 cos 40◦ + cos 80◦ =cos 10◦

2 cos(30◦ + 10◦) + sin 10◦

=cos 10◦

√3 cos 10◦

=1

√3.Therefore, α = 30◦.Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallor a, Spain; CHRISTOSANASTASSIADES, student, Apostles Peter and Paul Ly eum, Limassol, Cyprus; �SEFKETARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; ROY BARBARA,Lebanese University, Fanar, Lebanon; RICARDO BARROSO CAMPOS, University of Seville,Seville, Spain; MICHEL BATAILLE, Rouen, Fran e; CAO MINH QUANG, Nguyen Binh KhiemHigh S hool, Vinh Long, Vietnam; CHARLES R. DIMINNIE, Angelo State University, SanAngelo, TX, USA; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; PETER HURTHIG,Columbia College, Van ouver, BC; WALTHER JANOUS, Ursulinengymnasium, Innsbru k,Austria; GEOFFREY A. KANDALL, Hamden, CT, USA; V �ACLAV KONE �CN �Y, Big Rapids,MI, USA; KEE-WAI LAU, Hong Kong, China; TAICHI MAEKAWA, Takatsuki City, Osaka,Japan (se ond solution); SALEM MALIKI �C, student, Sarajevo College, Sarajevo, Bosnia andHerzegovina; MICHAEL PARMENTER, Memorial University of Newfoundland, St. John's,NL; ALEX REMOROV, student, William Lyon Ma kenzie Collegiate Institute, Toronto, ON;JOEL SCHLOSBERG, Bayside, NY, USA; EDMUND SWYLAN, Riga, Latvia; PANOS E.TSAOUSSOGLOU, Athens, Gree e; PETER Y. WOO, Biola University, La Mirada, CA, USA; TITUZVONARU, Com�ane�sti, Romania; and the proposer (7 solutions).Two solvers mentioned that the problem is known. Covas supplied the referen es [1℄and [2℄, and Lau gave the earliest referen e [3℄.Referen es[1℄ Ayoub B. Ayoub, \Problem 1104", Pi Mu Epsilon Journal, 12(2005); solution in 13(2005),pp. 186{187.[2℄ Ion Patras u, Probleme de Geometrie Plana, Craiova, 1996, Problem 8.12, p. 51.[3℄ H.J. Webb, \Problem 877", Mathemati sMagazine, 46(1973); solution in 47(1974), p. 172.

Crux Mathemati orumwith Mathemati al MayhemEditor Emeritus / R �eda teur-emeritus: Bru e L.R. ShawyerCrux Mathemati orumFounding Editors / R �eda teurs-fondateurs: L �eopold Sauv �e & Frederi k G.B. MaskellEditors emeriti / R �eda teurs-emeriti: G.W. Sands, R.E. Woodrow, Bru e L.R. ShawyerMathemati al MayhemFounding Editors / R �eda teurs-fondateurs: Patri k Surry & Ravi VakilEditors emeriti / R �eda teurs-emeriti: Philip Jong, Je� Higham, J.P. Grossman,Andre Chang, Naoki Sato, Cyrus Hsia, Shawn Godin, Je� Hooper