257James Edward Totten

1947{2008

A photograph of Jim Totten in his oÆ e taken by his olleague Don DesBrisayaround the time of Jim's retirement. Jim loved to demonstrate mathemat-i s using models, su h as the L �en �art SphereTM kit for investigating spheri algeometry that sits in the ba kground. (Photo ourtesy of Don DesBrisay)

258IN MEMORIAMJames Edward Totten, 1947{2008With 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. He obtained a B.S . from the University ofSaskat hewan, and then an M.S . in Computer S ien e and a Ph.D. in Math-emati s from the University of Waterloo. After a two year NRC Postdo toralFellowship in Tubingen, Germany, he joined the fa ulty at Saint Mary's Uni-versity in Halifax. One of us (Robert) �rst got to know Jim when he and Jimshared an oÆ e at the University of Saskat hewan while Jim visited there in1978-1979. That was a long old winter, but Jim's a tive interest and en-thusiasm for mathemati s and the tea hing of mathemati s made the year amemorable one. The next year Jim took a position at Cariboo College, wherehe remained as it evolved into the University College of the Cariboo and theninto Thompson Rivers University, 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 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, sonDean, daughter-in-law Christie and granddaughter Mikayla of Se helt, father Wilf Totten ofEdmonton, sister Judy Totten of Regina, sister Josie Laing (Neil) of Onoway,sister-in-law Marilyn Totten of Regina, mother-in-law Joy e Ladell ofSwansea Point, brother-in-law Brian Ladell (Iris) of Red Deer, sister-in-lawConstan e Ladell (David Dahl) of Kamloops, many ousins, family friends,and mathemati al olleagues; all miss his warmth and love. He was prede- eased by his mother Ali e, brother Gerry, and father-in-law Syd Ladell.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

259Totten Commemorative IssueAfter the sudden and unexpe ted death of Jim Totten onMar h 9, 2008,we announ ed in the May 2008 issue that a spe ial ommemorative issueof CRUX with MAYHEM would appear in Jim's honour. We now have thepleasure of presenting this issue to our readers, for whi h the members ofthe editorial board have made spe ial e�orts.This issue has also drawn upon many in Jim's ommunity, in ludingvarious olleagues at Thompson Rivers University (TRU). The obituary pre-pared by Bru e Shawyer and Robert Woodrow (that appeared in the May2008 CMS Notes) has been adapted with the assistan e of Lynne Totten forpubli ation here. It is pre eded by a pi ture of Jim in his oÆ e.After serving for �ve years as a Problems Editor for CRUX, Jim be ameEditor-in-Chief of CRUX with MAYHEM in 2003 and ran the journal untilhe began handing over the reins just months before his death last year. Healso had an enormous in uen e in his home provin e of British Columbia inspreading his love of mathemati s through his outrea h a tivities. It is nosurprise, then, that this spe ial issue has a distin tive regional emphasis. Apie e about Jim's a tivities and in uen e follows this editorial, whi h readersmay wish to orrelate with the arti le by Clint Lee on the British ColumbiaSe ondary S hool Mathemati s Contest and Jim's inspiring role in its devel-opment. Skoliad 118 in this issue features a ontest sele ted by John Ciriani,a founder of the Cariboo ontest that Jim helped to develop. The Mayhemse tion has a spe ial Problem of the Month taken from one of the ontests.The Mayhem se tion also has a full-length arti le by that master ofmathemati al re reations, Ross Honsberger, who inspired Jim's interest inproblem solving while he and Jim were oÆ e neighbours at the University ofWaterloo. An arti le by Mi hel Bataille on the geometry of bi entri quadri-laterals gra es the arti les se tion, followed by further re reations in AwaniKumar's arti le on knight's tours on the surfa e of a ube.Serendipitously, we found it �tting to reprint a book review writtenby Jim himself in the book reviews se tion. Also in the book reviews is anamusing ane dote of Andy Liu related to Jim.From our readers' submissions we have ompiled 26 spe ial problemsdedi ated to Jim's memory in luding 10 inMayhem and 16 in CRUX, a dozenof whi h make a spe ial se tion and four others that open the usual CRUX omplement. Our solutions se tion is parti ularly ri h this time around withsome hefty problems settled by CRUX readers.For those wishing to learn more about Jim we announ e here that thepro eedings of the May 2009 onferen e Sharing Mathemati s: A Tribute toJim Totten will be published later this year.

260Jim Totten's Rea hJohn Grant M LoughlinShane Rollans opened the onferen e Sharing Mathemati s: A Tributeto Jim Totten with these words: \This onferen e is a tribute to our long-time olleague, Jim Totten ... Jim had at least �ve passions. Foremost washis family. His wife, Lynne, shared his passion for the outdoors. His son,Dean, shared his passion for golf. His other passions were playing ho keyand sharing his love of mathemati s, whi h brings us here."Jim is known for his work with the British Columbia Se ondary S hoolMathemati s Contest (see Clint Lee's arti le on pages 307-309 for details)and over fourteen years of servi e to Crux, in luding nine years as a ProblemsEditor prior to be oming Editor-in-Chief of CRUXwithMayhem in 2003. Thispro�le of Jim o�ers a broader pi ture of Jim's ontributions in mathemati sand spirit, mainly as seen through the eyes of his olleagues.When Jim arrived at Cariboo College in 1979 he promptly organized aPutnam team that he led until 2005. He also started the \Problem of theWeek" tradition, something he had done previously at the University ofSaskat hewan and St. Mary's University. Ea h week he would olle t so-lutions, grade them, and post the results omplete with a solution. For the�rst 27 years no problem was repeated, although Jim did allow himself torepost a few of his favourite problems in his �nal year. Jim ompiled 80 ofthese problems and solutions into a book, Problems of the Week, Volume VIIin the ATOM (A Taste of Mathemati s) Series published by the CanadianMathemati al So iety (CMS) in 2007.Cariboo College be ame the University College of the Cariboo (UCC)and in 1989 o�ered degrees with majors in mathemati s in asso iation withthe University of British Columbia (UBC). Jim then began tea hing upperlevel ourses ranging from Linear Programming to Complex Analysis to GraphTheory and Geometry. He also haired the UCC Dept. of Mathemati s andStatisti s (1994-1998). UCC later be ame Thompson Rivers University (TRU),o�ering an independent degree in 2005.Ri k Brewster shares an unusual perspe tive on Jim's ontributions inKamloops, ranging from his time as a lo al high s hool student, to an under-graduate at UCC, and �nally as a olleague of Jim's. Ri k writes:\It is lear that Jim was well on tra k to a areer as a resear h mathe-mati ian with 14 papers from 1974 to 1980, many solo (e.g., Basi propertiesof restri ted linear spa es, Dis rete Math. 13 (1975), No. 1, 67-74) and oth-ers ollaborative (e.g., On a lass of linear spa es with two onse utive linedegrees, Ars Combin. 10 (1980), 107-114, o-authored with Lynn Batten).He made a de ision to suspend his resear h programme when he moved toCariboo in 1979. He was obviously happy (and at pea e) with his de ision ashe found other outlets for sharing his love of mathemati s."[At http://www.ams.org/mathscinet/search/author.html?mrauthid=298276are links to a list of Jim's resear h publi ations and related information.℄

261\My overwhelming memory of Jim is of his enthusiasm. I ertainlynoti ed it as the high s hool kid visiting Cariboo College, but I thought alltea hers were keen. As his student in 2nd year though, it was lear Jim was ut from a di�erent loth. He had very high expe tations of us. He pushedus hard on assignments, but his enthusiasm balan ed out this work. In otherwords, he a ted as though we would be thrilled to work on tough problems(math and omputing s ien e), be ause problem solving is so mu h fun. Thatsort of keenness never disappeared."\I parti ularly remember se ond year Dis rete Math (Math 222), the�rst time it was taught at Cariboo College. Jim was so ex ited to tea h inhis area. Late in the ourse he gave us a talk about �nite geometry and thehunt for the proje tive plane of order 10. I an say I really didn't understandmu h of what he said that day, but I was stru k by his ex itement in beingable to take us to the `frontier of mathemati s' (his words). This seemed veryimportant to him: namely, that we were part of the mathemati al ommu-nity, and we had the opportunity to ontribute. I was in graduate s hoolwhen the nonexisten e of the proje tive plane of order 10 was established.Upon hearing that result, I remembered Jim's le ture 7 years earlier. Twothoughts: `Ah! That's what Jim was talking about,' and `Wow, he had a lotof faith in a group of se ond year students to share that with us.' "\While I was tea hing Number Theory in my �rst semester at UCC, Jimshowed up in my oÆ e one day with a olle tion of notes (about 30 yearsold) from a ourse he taught in graduate s hool. He was so keen to show mea onstru tion of a ertain ring of fun tions based on number theoreti ideasthat he had presented in the (grad) ourse. It was ni e mathemati s, but abit advan ed for third years without abstra t algebra. Still Jim's enthusiasmwas so typi al. An opportunity to share should not be wasted."The Adrien Pouliot Award is a CMS award honouring \signi� ant andsustained ontributions to mathemati s edu ation in Canada." Members ofthe TRU department nominated Jim in 2007. In a supporting letter, JohnCiriani e hoed Ri k's sentiment: \He was parti ularly pleased to developand tea h a ourse in geometry. Even students who found the ourse diÆ ultre ognized Jim's love of the subje t and were able to relate to his enthusi-asm for it." John added, \I must mention that Jim derives great pleasure inmaking presentations in s ien e fairs and s hools. He is parti ularly pleasedwhen he en ounters talented students who share his enthusiasm for mathe-mati s. I believe he plans to ontinue this a tivity after he retires." In fa t,Park rest Elementary S hool in Kamloops made Jim an honourary tea hingsta� member in 2007 to re ognize his long term volunteer servi e.Don DesBrisay expresses his respe t: \With themath ontests, Problemof the Week, mathemagi shows, Putnam, Crux, his boxes of puzzles (manyhe rafted himself), his journal/book olle tions, and his vast knowledge,intelle t, and wit (ex luding some awful puns), Jim kept us all aware of thejoy and ex itement found inmathemati s and in life. All this as well as family,outdoor lub a tivities, golf and ho key!!" Quoting Kirk Evenrude: \Jimwentat golf with the same determination and dedi ation as he did everything.

262When I �rst met him in 1982 his handi ap was about 13. In 2007 it was 6.One day he shot 69 at the Kamloops lub. ... He had boundless energy whenit ame to golf. ... It was easy to spot Jim on the ourse; all you had to dowas look for a big white Tilley hat. He loved playing in tournaments and wasvery ompetitive, but I think he liked the so ial part as mu h as the golf."Indeed Jim's outrea h had many bran hes. His parti ipation at a 1999CMS Edu ation Session (organized by Bru e Shawyer and Ed Williams) inSt. John's raised the pro�le of the BC ontest. While there Jim visited my lass at Memorial University of Newfoundland sharing his wooden puzzlesand love of re reational mathemati s with an enamored lass of future highs hool math tea hers. An entirely di�erent form of mathemati al outrea hbegan in 1969 when Jim initiated the weekly pi kup ho key games throughthe Fa ulty of Mathemati s at the University of Waterloo.Gra e and gratitude are the words that ome to mind as I re e t uponmy ollaborations with Jim Totten. Jim ombined intelle t and heart in amanner that pla ed the greater good ahead of his own. Jim was one of thoseexemplary people in the a ademi ommunity who ons iously expressed ap-pre iation for the work of others { a gift in itself. This form of outrea h isless visible but equally important. It is not unlike the tea her who, though hallenging, manages to reate a safe spa e for making errors and genuinelyfumbling with mathemati al ideas. Jim was a great tea her. He was hon-oured with separate awards for tea hing and merit, and shortly after his re-tirement in 2007 with a Professor Emeritus designation. Those present atthe Mar h 2008 elebration of Jim's life at the Grand Hall in TRU witnessedthe love and outpouring of appre iation for Jim and his family. Students,ho key players, golfers, olleagues, hikers, and family spoke to the breadthof Jim's passions and a tivities, a taste of whi h has been o�ered here.These losing words are from Fae DeBe k and then, Dennis A reman:\Jim's top priority has always been his students, but his enthusiasmfor mathemati s and his e�orts to promote mathemati al ideas extend farbeyond his own lassroom. He has been an inspiration and a model for allin our department to follow. In the past two years I have a ted as the lo al oordinator for the Math Contest and have begun to appre iate at a deeperlevel Jim's phenomenal ontribution in this parti ular regard. His in uen ethroughout the ollege/university system in the provin e annot be over-stated."\Jim was a wonderful olleague and a good friend. He loved everythingaboutMathemati s and assumed everyone else would too if it was just sharedwith them. His outrea h a tivities showed true dedi ation to that goal fromCrux to years of s hool visits and Problems of the Week. He was a generousperson who loved his family and all aspe ts of his life and we all mourn hisloss but are inspired by his example." John Grant M Loughlin

263SKOLIAD No. 118Lily Yen and Mogens HansenPlease send your solutions to problems in this Skoliad by 1 De ember, 2009.A opy of Crux will be sent to one pre-university reader who sends in solu-tions before the deadline. The de ision of the editors is �nal.

This spe ial issue of CRUX with MAYHEM features a sele tion froma ontest that Jim Totten played a pivotal role in developing. The CaribooCollege High S hool Mathemati s Contest began in 1973 and has sin e de-veloped into the British Columbia Se ondary S hool Math Contest. In 1992a ompilation of the problems was published in a book edited by Jim Totten.In the prefa e he wrote: \One person above all must re eive our thanks forall his time spent proof-reading and o�ering suggestions and en ouragement;John Ciriani ontinues to guide and inspire all of us in this department withhis dedi ation to mathemati s and tea hing."We therefore asked John Ciriani to sele t the problems for this Skoliad.John Ciriani replied: \It was diÆ ult to sele t a ontest whi h wasmemorablein terms of Jim's early ontributions. In the end I hose the Junior Final 1990.This �nal ontains a problem about golfers in a hurry to get to the ourse andre e ts Jim's love of the game of golf."Our thanks go to John GrantM Loughlin, University of New Brunswi k,for suggesting John Ciriani and for onta ting him for us; to John Ciriani,Kamloops, British Columbia, for his hoi e of ontest and the reasons be-hind the hoi e; and to Rolland Gaudet, University College of St. Bonifa e,Winnipeg, MB, for the translation.Cariboo College High S hool Mathemati s Contest1990Junior Final, Part B1. A boy on a bi y le oasts down from the top of a hill. He overs 4 metresin the �rst se ond and in ea h su eeding se ond overs 5 metres more thanin the previous se ond. He rea hes the bottom of the hill in 11 se onds.(a) How long is the hill?(b) What is the boy's average speed in metres per se ond?( ) What distan e did he over in the last se ond?2. Two golfers, on their way to the ourse, rea hed a railway rossing justas a 2.5 km train arrived. Rather than waiting, they de ided to go on to thenext rossing 1km along in the dire tion the train was going. They travelledat 50 km/h while the train travelled at 70 km/h.

264(a) How long did they have to wait for the train to lear the rossing?(b) Rather than travelling at 50 km/h, how fast would they have had totravel to rea h the rossing just as the train was learing the rossing?3. A student asks you to hoose a number from 1 to 9 and multiply it by 109,then asks you to �nd the sum of the digits in the produ t. Knowing the sumof the digits, the student is able to tell you the number with whi h you began.Explain how this an be done.4. Suppose you throw 5 darts at a round board with a radius of 25√

2 m. Ifall 5 darts sti k in the board, show that at least two of them must be within50 m of ea h other.5. The diameter, AB, of a ir leis divided into 4 equal parts by thepoints C, D, and E. Semi ir les aredrawn on AC, AD, AE, and AB asshown. Find the ratio of the area ofthe shaded parts to the area of the un-shaded parts. ...............................................................................................................................................................................................................................................................................................................

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A BC D ECon ours math �ematique du Coll �ege Cariboo 1990Niveau se ondaire, �nale junior, partie B1. Un gar� on roule sans p �edaler, �a partir du haut d'une olline. Il ouvre 4m�etres pendant la premi �ere se onde ; haque se onde subs �equente, il ouvre5 m�etres de plus que pendant la se onde pr �e �edente. Il arrive au bas de la olline apr �es 11 se ondes.(a) Quelle est la distan e totale par ourue ?(b) Quelle est la vitesse moyenne du gar� on, en m�etres �a la se onde ?( ) Quelle est la distan e par ourue pendant la derni �ere se onde ?2. Deux golfeurs arrivent �a un passage �a niveau juste au moment o �u un trainde longueur 2, 5 kilom�etres s'y pr �esente. Au lieu d'attendre, ils d �e ident dese rendre au pro hain passage �a niveau, �a une distan e de 1 kilom�etre du pre-mier passage �a niveau, dans la dire tion o �u le train se dirige. Ils se d �epla ent�a 50 kilom�etres �a l'heure tandis que le train va �a 70 kilom�etres �a l'heure.(a) Combien de temps ont-ils �a attendre au deuxi �eme passage �a niveau,avant que le train ait d �egag �e le passage ?(b) Au lieu de se d �epla er �a 50 kilom�etres �a l'heure, �a quelle vitesse les deuxgolfeurs auraient-ils besoin de se d �epla er a�n d'arriver au deuxi �emepassage �a niveau justement au moment o �u le train d �egage le passage ?

2653. Un �etudiant vous demande de hoisir un entier de 1 �a 9, puis de le mul-tiplier par 109 ; il vous demande alors de lui fournir la somme des hi�resdans e produit. Connaissant ette somme, l' �etudiant peut alors vous direave quel entier vous avez ommen �e. Expliquer omment il le fait.4. Vous lan ez 5 �e hettes vers une ible de rayon 25

√2 entim �etres. Les 5 �e hettes frappent la ible. Montrer qu'au moins deux d'entre elles doiventse retrouver �a une distan e d'au plus 50 entim �etres l'une de l'autre.5. Des demi er les sont tra �esave AC, AD, AE et AB ommediam�etres, o �u le segment AB estdivis �e en 4 parties �egales par lespoints C, D et E. Pour le s h �ema�a droite, d �eterminer le ratio desparties olor �ees aux parties non olor �ees. ...............................................................................................................................................................................................................................................................................................................

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A BC D E

Next follow the solutions to the British Columbia Se ondary S hoolMathemati s Contest, 2008, Junior Final, Part A [2008 : 321{324℄. Notethat problems 7 and 10 below have been adjusted so that the length of asegment is given as |XY | instead of XY .1. Jeeves the valet was promised a salary of $8000 and a ar for a year ofservi e. Jeeves left the job after 7 months of servi e and re eived the ar and$1600 as his orre tly prorated salary. The dollar value of the ar was:(A) 6400 (B) 7200 (C) 7360 (D) 8000 (E) 15360Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto,ON. Let x be the ost of the ar. The total value of everything that Jeevesre eives after seven months equals 712

of what he would have re eived hadhe worked for a whole year. Thus x + 1600 = 712

(x + 8000). It followsthat 12x + 19200 = 7x + 56000, so 5x = 36800. Hen e x = 7360 and theanswer is (C).Also solved by JOCHEM VAN GAALEN, student, Medway High S hool, Arva, ON.2. Re all that n! = n × (n − 1) × (n − 2) × · · · × 2 × 1. The maximumvalue of the integer x su h that 3x divides 30! is:(A) 30 (B) 14 (C) 13 (D) 10 (E) 4Solution by Jo hem van Gaalen, student, Medway High S hool, Arva, ON.The following table lists the multiples of 3 that are less than or equalto 30 along with the number of times ea h is divisible by 3.

266Multiple 3 6 9 12 15 18 21 24 27 30Times 1 1 2 1 1 2 1 1 3 1Thus 30! is divisible by three 1+1+2+1+1+2+1+1+3+1 or fourteentimes, and the answer is (B).Also solved by JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON.3. In the diagram, ABCDEF is a regularhexagon. Line segments AE and FC meet at

Z. The ratio of the area of triangle FZE tothe area of the quadrilateral ABCZ is:(A) 1 : 5 (B) 1 : 4 (C) 4 : 1(D) 5 : 1 (E) 1 : 6

A B

C

DE

F

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Z

Solution by Jo hem van Gaalen, student, Medway High S hool, Arva, ON.Assume without loss of generality that the side length of the hexagonis 2. Angles inside a regular hexagon are 120◦, so∠FEZ = 120◦−90◦ = 30◦and ∠EFZ = 12

(120◦

)= 60◦. Thus △FZE is a 30◦{60◦{90◦ triangle withsides 1, 2, and √

3, namely |FZ| = 1, |EF | = 2, and |EZ| =√

3. Thus,△FZE has area √

3

2.Now draw a line segment from B to D interse ting CF at the point G.Then △CGB ∼= △FZE, so the area of △CGB is also √

3

2. Sin e ABGZ isa re tangle with sides 2 and √

3, it has area 2√

3. Hen e the area of trapezoidABCZ is √

3

2+ 2

√3 =

5√

3

2.Therefore, the desired ratio is √

3

2:

5√

3

2= 1 : 5, and the answer is (A).Also solved by JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON.4. De�ne ⌊x⌋ to be the greatest integer less than or equal to x. For example,

⌊7⌋ = 7, ⌊7.2⌋ = 7, and ⌊−5.5⌋ = −6. If z is a real number that is not aninteger, then the value of ⌊z⌋ + ⌊1 − z⌋ is:(A) −1 (B) 0 (C) 1 (D) 2 (E) zSolution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto,ON. Let x = ⌊z⌋ and α = z−x. Then x is an integer and 0 < α < 1 (sin e zis not an integer, α 6= 0). Evidently, ⌊1−z⌋ = ⌊1−x−α⌋ = ⌊−x+(1−α)⌋.Sin e 0 < 1 − α < 1, you have that ⌊−x + (1 − α)⌋ = −x. Therefore,⌊z⌋ + ⌊1 − z⌋ = x + (−x) = 0 and the answer is (B).As our solver points out, the question implies that the value of ⌊z⌋+⌊1−z⌋ is indepen-dent of z (as long as z is not an integer). Thus you an �nd the answer by simply substitutingsome value, say 1.5, for z. Of ourse this does not prove anything but relies on trust in theformulation of the question.

2675. Examinations in ea h of three subje ts, Anatomy, Biology, and Chemistry,were taken by a group of 41 students. The following table shows how manystudents failed in ea h subje t, as well as in the various ombinations:subje t A B C AB AC BC ABC# failed 12 5 8 2 6 3 1(For instan e, 5 students failed in Biology, among whom there were 3 whofailed both Biology and Chemistry, and just 1 of the 3 who failed all threesubje ts.) The number of students who passed all three subje ts is:(A) 4 (B) 16 (C) 21 (D) 24 (E) 26Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto,ON. The two students who fail Anatomy and Biology in lude the single stu-dent who fails all three subje ts. Therefore just one student fails Anatomyand Biology but not Chemistry. Likewise, �ve students fail Anatomy andChemistry but not Biology, and two students fail Biology and Chemistry butnot Anatomy.The 12 students who fail Anatomy in lude

• the single student who fails Anatomy and Biology but not Chemistry;and• the �ve students who fail Anatomy and Chemistry but not Biology; and• the single student who fails all three subje ts.Therefore the number of students who fail Anatomy but pass the other twosubje ts is 12−1−5−1 = 5. Likewise, one student fails Biology and passesthe other two subje ts, and zero students fail Chemistry but pass the othertwo subje ts.Therefore the total number of students who fail at least one subje t is

5 + 1 + 0 + 1 + 5 + 2 + 1 = 15, when e the number of students who passall three subje ts is 41 − 15 = 26, and the answer is (E).Also solved by JOCHEM VAN GAALEN, student, Medway High S hool, Arva, ON.This type of question is most easily solved by means of a Venn Diagram. The diagramsbelow show how you may sequentially enter information into the Venn Diagram...................................................................................................................................................................

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2686. Six points, A, B, C, D, E, and F are arrangedin the formation shown in the diagram, with A, B,and C on a straight line. Three of these six pointsare sele ted to form a triangle. The number of su htriangles that an be formed is:(A) 12 (B) 14 (C) 16(D) 19 (E) 20

D E

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Solution by Jo hem van Gaalen, student, Medway High S hool, Arva, ON.There are (63

)= 6C3 =

6!

3!(6 − 3)!= 20 ways to hoose three pointsfrom the given six. One of these, {A, B, C}, yields no triangle as A, B, Care ollinear. Thus, you an make 19 triangles, and the answer is (D).Also solved by JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON.7. In the diagram, C is the entre of the ir le and

AD is tangent to the ir le at D. AC is a straightline. If |AD|=10 and |AB|=7, the length of BC is:(A) √151 − 7

2(B) √

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Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto,ON. Sin e AD is tangent to the ir le, ∠CDA = 90◦. Let r be the radius ofthe ir le. Then |CD| = |BC| = r and |AC| = |AB| + |BC| = 7 + r. Bythe Pythagorean Theorem, |AD|2 + |CD|2 = |AC|2, so 102 +r2 = (7+r)2.Thus 100 + r2 = 49 + 14r + r2, when e 51 = 14r, and r =51

14. Then theanswer is (C).Also solved by JOCHEM VAN GAALEN, student, Medway High S hool, Arva, ON.8. When 20082008 is multiplied out, the units digit in the �nal produ t is:(A) 8 (B) 6 (C) 4 (D) 2 (E) 0Solution by Jo hem van Gaalen, student, Medway High S hool, Arva, ON.Sin e the units digit of a produ t depends only on the units digits of thefa tors, we only need to onsider the units digit of powers of 8. The tableshows the �rst of these.Power 81 82 83 84 85 86 87Units digit 8 4 2 6 8 4 2Note that the pattern 8, 4, 2, 6 repeats itself after 84 as it must, sin e onlythe units digits matter.

269Now 2008 mod 4 = 0, that is, the remainder when dividing 2008 by 4is 0. Thus the units digit of 20082008 is 6 and the answer is (B).Also solved by JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON.9. Re all that a prime number is an integer greater than one that is divisibleonly by one and itself. Consider the set of two-digit numbers less than 40 thatare either prime or divisible by only one prime number. From this set sele tthose for whi h the sum of the digits is a prime number, and the positivedi�eren e between the digits is another prime number. The sum of the valuesof the numbers sele ted is:(A) 29 (B) 41 (C) 54 (D) 70 (E) 93Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto,ON. The two-digit primes or powers of primes less than 40 are 11, 13, 16,

17, 19, 23, 25, 27, 29, 31, 32, and 37. Requiring the digit sum to be a primenarrows the list to 11, 16, 23, 25, 29, and 32. Requiring that the di�eren eof the digits be a prime further narrows the list to 16, 25, and 29. The sumof these three numbers is 70, and the answer is (D).10. In the diagram ABCD is a re tangle with|AD| = 1, and both DE and BF perpendi ular to thediagonal AC. Further, |AE| = |EF | = |FC|. Thelength of the side AB is:(A) √

2 (B) √3 (C) 2(D) √

5 (E) 3A B

CD

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Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto,ON. Let x be the ommon length of AE, EF , and FC. Then |EC| = 2x.By the Pythagorean Theorem, |AE|2+ |DE|2 = |AD|2, so x2+ |DE|2 = 12,when e |DE| =√

1 − x2.Note that ∠ECD = 90◦ − ∠EDC = ∠ADE, so △DEA ∼ △CED.Therefore,|EA||DE| =

|ED||CE| ⇐⇒ x√

1 − x2=

√1 − x2

2x.Thus 2x2 = 1 − x2, so x2 = 1

3. Finally, |AB| = |CD| and CD has length√

|DE|2 + |CE|2 =√

1 − x2 + 4x2 =√

1 + 3x2 =√

1 + 3/3 =√

2, andthe answer is (A).That ompletes another Skoliad. This issue's prize for the best solutionsgoes to Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON.We look forward to re eiving more solutions from more readers.

270MATHEMATICAL MAYHEMMathemati al Mayhem began in 1988 as a Mathemati al Journal for and byHigh S hool and University Students. It ontinues, with the same emphasis,as an integral part of Crux Mathemati orum with Mathemati al Mayhem.The Mayhem Editor is Ian VanderBurgh (University of Waterloo). Theother sta� members are Monika Khbeis (As ension of Our Lord Se ondaryS hool, Mississauga) and Eri Robert (Leo Hayes High S hool, Frederi ton).

Mayhem ProblemsPlease send your solutions to the problems in this edition by 15 November2009. Solutions re eived after this date will only be onsidered if there is time beforepubli ation of the solutions.Ea h problem is given in English and Fren h, the oÆ ial languages of Canada.In issues 1, 3, 5, and 7, English will pre ede Fren h, and in issues 2, 4, 6, and 8,Fren h will pre ede English.Ea h of the following Mayhem problems is spe ially dedi ated to the memoryof Jim Totten, hen e the spe ial numbering used below. The numbering of the regularMayhem problems will resume in subsequent issues.The editor thanks Jean-Mar Terrier of the University of Montreal for transla-tions of the problems.Totten{M1. Proposed by Shawn Godin, Cairine Wilson Se ondaryS hool, Orleans, ON.An ient Egyptians wrote all fra tions in terms of distin t unit fra tions(that is, in terms of distin t fra tions with numerators of 1). For example,instead of writing 11

12, they would write 1

2+

1

3+

1

12. The unit fra tion 1

2 anbe written in terms of other unit fra tions as 1

2=

1

3+

1

6. Find an in�nitefamily of unit fra tions ea h of whi h an be written as the sum of two unitfra tions.Totten{M2. Proposed by Bru e Shawyer, Memorial University of New-foundland, St. John's, NL.The boundary of the shadow on the moon isalways a ir ular ar . On a ertain day, the moonis seen with the shadow passing through diamet-ri ally opposite points. If the entre of the ir ularar forming the shadow is on the ir umferen e ofthe moon, determine the exa t proportion of themoon that is not in shadow.

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271Totten{M3. Proposed by John Ciriani, Kamloops, BC.Prove that the quadrati equation ax2 + bx + c = 0 does not have arational root if a, b, and c are odd integers.Totten{M4. Proposed by Bill Sands, University of Calgary, Calgary,AB. In a survey, some students were asked whether they liked the olourorange. Exa tly 2% of the boys in the survey liked orange, while exa tly 59%of the girls in the survey liked orange. Altogether, exa tly 17% of the studentsin the survey liked orange. Find the smallest possible number of students inthe survey.Totten{M5. Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.Let a 6= 1 be a positive real number. Determine all pairs of positiveintegers (x, y) su h that loga x − loga y = loga(x − y).Totten{M6. Proposed by Suzanne Feldberg, Thompson Rivers Univer-sity, Kamloops, BC.It is widely known how to draw a 5-pointed star qui kly. To make itsymmetri , one pla es 5 verti es at 72◦ intervals about a ir le and onne tsthe verti es with line segments of equal length without lifting one's pen. Bystarting from a �xed point and using the same method, one an draw twodi�erent (and symmetri ) 7-pointed stars without lifting one's pen.

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How many di�erent 6-pointed, 8-pointed, or 9-pointed stars an onedraw this way? How many di�erent n-pointed stars an one draw this way?Totten{M7. Proposed by John Grant M Loughlin, University of NewBrunswi k, Frederi ton, NB.An unmarked ruler is known to be exa tly6 m in length. It is possible to exa tly measureall integer lengths from 1 m to 6 m using only 2marks, as shown in the diagram, at 1 m and 4 m,sin e 2 = 6 − 4, 3 = 4 − 1, and 5 = 6 − 1. ..

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272Suppose that an unmarked ruler is known to be exa tly 30 m in length.(a) Find a way of pla ing 9 or fewer marks on the ruler to be able to exa tlymeasure all integer lengths from 1 m to 30 m.(b) Prove that at least 7 marks are needed to be able to exa tly measure allinteger lengths from 1 m to 30 m.( )⋆ Determine the smallest number of marks required on the ruler to beable to exa tly measure all integer lengths from 1 m to 30 m.Totten{M8. Proposed by Edward J. Barbeau, University of Toronto,Toronto, ON.Let T be the set of all ordered triples (a, b, c) of positive integers su hthat a < b < c. We say that two triples (a, b, c) and (u, v, w) are equivalentif a : b : c = u : v : w. We use this relation to partition T into equivalen e lasses. The triple (a, b, c) is geometri if ac = b2 (that is, its terms form ageometri sequen e) and harmoni if 1

a+

1

c=

2

b(that is, the re ipro als ofits terms form an arithmeti sequen e).(a) Verify that if (a, b, c) is geometri , then all triples equivalent to it arealso geometri .(b) Verify that if (a, b, c) is harmoni , then all triples equivalent to it arealso harmoni .( ) LetG be the set of equivalen e lasses of geometri triples and H be theset of equivalen e lasses of harmoni triples. Determine a one-to-one orresponden e between G and H.Totten{M9. Proposed by Kirk Evenrude, Kamloops, BC.A train 900m long, travelling at 90km/h, approa hes a 100m longbridge.(a) How many se onds does it take the train to lear the bridge?(b) Suppose that, just as the train rea hes the bridge, it begins to slowdown at the rate of 0.2m/s2. Now how long does it take to lear thebridge?Totten{M10. Proposed by Ni holas Bu k, College of New Caledonia,Prin e George, BC.Show that if p is a prime number, and A and B are positive integerssu h that p divides A, p2 does not divide A, and p does not divide B, thenthe Diophantine equation Ax2 + By2 = p2008 does not have any solutionsin positive integers x and y..................................................................

273Totten{M1. Propos �e par Shawn Godin, �E ole se ondaire Cairine Wil-son, Orl �eans, ON.Les an iens �egyptiens �e rivaient toutes leurs fra tions en termes defra tions unitaires distin tes, 'est- �a-dire de fra tions distin tes ave 1 ommenum�erateur. Par exemple, au lieu d' �e rire 11

12, ils auraient �e rit 1

2+

1

3+

1

12.La fra tion unitaire 1

2peut etre �e rite en termes d'autres fra tions unitaires omme 1

2=

1

3+

1

6. Trouver une famille in�nie de fra tions unitaires, ha unepouvant etre �e rite omme somme de deux fra tions unitaires.Totten{M2. Propos �e par Bru e Shawyer, Universit �e Memorial deTerre-Neuve, St. John's, NL.La fronti �ere de l'ombre sur la lune est tou-jours un ar de er le. Un ertain jour, onpeut onstater que l'ombre passe par des pointsdiam�etralement oppos �es. Si le entre de l'ar ir- ulaire formant ette ombre se trouve sur la ir- onf �eren e de la lune, d �eterminer la portion ex-a te de la lune qui n'est pas dans l'ombre.

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...........................................................................Totten{M3. Propos �e par John Ciriani, Kamloops, BC.Montrer que l' �equation quadratique ax2 +bx+c = 0 n'a pas de ra inerationnelle si a, b et c sont des entiers impairs.Totten{M4. Propos �e par Bill Sands, Universit �e de Calgary, Calgary,AB. Lors d'un sondage, on a demand �e �a ertains �etudiants s'ils aimaient la ouleur orange. Chez les gar� ons exa tement 2% ont r �epondu par oui, tandisque hez les �lles, la proportion exa te de oui a �et �e de 59%. Comme autotal, exa tement 17% des r �epondants ont dit aimer l'orange, on demande detrouver quel est le plus petit nombre possible de parti ipants au sondage.Totten{M5. Propos �e par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie.Soit a 6= 1 un nombre r �eel positif. Trouver toutes les paires d'entierspositifs (x, y) telles que loga x − loga y = loga(x − y).Totten{M6. Propos �e par Suzanne Feldberg, Universit �e Thompson Rivers,Kamloops, BC.On sait bien omment dessiner rapidement une �etoile �a 5 sommets.Pour en faire une sym�etrique, on pla e les 5 sommets �a intervalles de 72◦ surun er le et on relie les sommets par des segments re tilignes de longueur�egale sans lever son rayon. En partant d'un point �x �e et en utilisant la meme

274m�ethode, on peut dessiner sans lever son rayon deux �etoiles di� �erentes (etsym�etriques) �a 7 sommets.

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Combien d' �etoiles di� �erentes �a 6 sommets, 8 sommets, ou 9 sommetspeut-on ainsi dessiner? Combien d' �etoiles di� �erentes �a n sommets peut-onainsi dessiner?Totten{M7. Propos �e par John GrantM Loughlin, Universit �e duNouveau-Brunswi k, Frederi ton, NB.On utilise une r �egle sans graduation dont onsait qu'elle mesure exa tement 6 m. Il est pos-sible de mesurer exa tement toutes les longueursenti �eres de 1 m �a 6 m ave seulement 2marques, omme indiqu �e dans la �gure i- ontre, �a 1 m et4 m, ar 2 = 6 − 4, 3 = 4 − 1 et 5 = 6 − 1. ..

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Supposons maintenant qu'on utilise une r �egle d'exa tement 30 m delong.(a) Trouver un moyen de pla er 9 marques sur la r �egle a�n d'etre apablede mesurer exa tement toutes les longueurs enti �eres, de 1 m �a 30 m.(b) Montrer qu'au moins 7 marques sont n �e essaires pour etre apable demesurer exa tement toutes les longueurs enti �eres, de 1 m �a 30 m.( )⋆ D�eterminer le plus petit nombre de marques �a pla er sur la r �egle a�nd'etre apable de mesurer exa tement toutes les longueurs enti �eres, de1 m �a 30 m.Totten{M8. Propos �e par Edward J. Barbeau, Universit �e de Toronto,Toronto, ON.Soit T l'ensemble de toutes les triplets ordonn �es d'entiers positifs

(a, b, c) tels que a < b < c. On dit que deux triplets (a, b, c) et (u, v, w)sont �equivalents si a : b : c = u : v : w. On utilise ette �equivalen e pour par-titionner T en lasses d' �equivalen e. Le triplet (a, b, c) est dit g �eom�etriquesi ac = b2 ( .- �a-d. ses �el �ements forment une suite g �eom�etrique) etharmonique si 1

a+

1

c=

2

b( .- �a-d. les inverses de ses �el �ements formentune suite arithm�etique).

275(a) V �eri�er que si (a, b, c) est g �eom�etrique, alors tous les triplets qui luisont �equivalents sont aussi g �eom�etriques.(b) V �eri�er que si (a, b, c) est harmonique, alors tous les triplets qui luisont �equivalents sont aussi harmoniques.( ) Soit G l'ensemble des lasses d' �equivalen e de triplets g �eom�etriques et

H l'ensemble des lasses d' �equivalen e de triplets harmoniques. Trou-ver une orrespondan e biunivoque entre G et H.Totten{M9. Propos �e par Kirk Evenrude, Kamloops, BC.Un train de 900m de long s'appro he d'un pont d'une longueur de100m �a la vitesse de 90 km/h.(a) En ombien de se ondes le train traversera-t-il le pont?(b) Supposons qu'au moment d'atteindre le pont, le train d �e �el �ere de

0.2m/s2. Combien de temps mettra-t-il ette fois pour traverser lepont?Totten{M10. Propos �e par Ni holas Bu k, Coll �ege de New Caledonia,Prin e George, BC.Montrer que si p est un nombre premier, et si A et B sont des entierspositifs tels que p divise A, p2 ne divise pas A, et p ne divise pas B, alorsl' �equation diophantienne Ax2 +By2 = p2008 n'a au une solution en entierspositifs x et y.Mayhem Solutions

M363. Proposed by Bru e Shawyer, Memorial University of Newfound-land, St. John's, NL.Suppose that A is a six-digit positive integer and B is the positive in-teger formed by writing the digits of A in reverse order. Prove that A − Bis a multiple of 9.Solution by Ja lyn Chang, student, Western Canada High S hool, Calgary,AB. Sin e A is a six-digit positive integer, it an be expressed in the form105a+104b+103c+102d+101e+f , where a is a positive integer and b, c,d, e, and f are nonnegative integers. Sin e B is formed by writing the digitsof A in reverse order, B is of the form 105f +104e+103d+102c+101b+a.

276Their di�eren e, A − B, an then be simpli�ed as follows:

A − B =(105a + 104b + 103c + 102d + 101e + f

)

− (105f + 104e + 103d + 102c + 101b + a

)

= 105a − a + 104b − 101b + 103c − 102c

+ 102d − 103d + 101e − 104e + f − 105f

= 99999a + 9990b + 900c − 900d − 9990e − 99999f

= 9(11111a + 1110b + 100c − 100d − 1110e − 11111f) .Sin e the digits of A are integers, sums and di�eren es of multiples of thesedigits are integers too. Thus, the di�eren e of A and B is a multiple of nine.Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; SHAMIL ASGARLI,student, Burnaby South Se ondary S hool, Burnaby, BC; GEORGIOS BASDEKIS, student,1st High S hool of Karditsa, Karditsa, Gree e; CAO MINH QUANG, Nguyen Binh Khiem HighS hool, Vinh Long, Vietnam; PETER CHIEN, student, Central Elgin Collegiate, St. Thomas,ON; COURTIS G. CHRYSSOSTOMOS, Larissa, Gree e; ANTONIO GODOY TOHARIA, Madrid,Spain; JOS �E HERN�ANDEZ SANTIAGO, student, Universidad Te nol �ogi a de la Mixte a,Oaxa a, Mexi o; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; HUGO LUYO S�ANCHEZ,Ponti� ia Universidad Cat �oli a del Peru, Lima, Peru; RICARD PEIR �O, IES \Abastos",Valen ia, Spain; ROBERT SHEETS, Southeast Missouri State University, Cape Girardeau, MO,USA; MRIDUL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; MRINAL SINGH,student, Kendriya Vidyalaya S hool, Shillong, India; ALEX SONG, student, Elizabeth ZieglerPubli S hool, Waterloo, ON; and JIXUAN WANG, student, Don Mills Collegiate Institute,Toronto, ON.M364. Proposed by the Mayhem Sta�.A semi ir le of radius 2 is drawn with diameter AB. The square PQRSis drawn with P and Q on the semi ir le and R and S on AB. Is the area ofthe square less than or greater than one-half of the area of the semi ir le?Solution by Peter Chien, student, Central Elgin Collegiate, St. Thomas, ON,modi�ed by the editor.Pla e AB on the x-axis with themidpoint of AB (that is, the entre ofthe semi ir le) at the origin. The full ir le has radius 2 and entre (0, 0),and so has equation x2 + y2 = 4.Let Q have oordinates (a, b),with a and b positive. Sin e Q is onthe semi ir le, then a2 + b2 = 4.Sin e PQRS is a square, whi h mustsit symmetri ally inside the semi ir le,then we also have b = 2a.....................................................................................................................................................................................................................................................................................

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A B

P Q(a, b)

S R

x

y

Thus, a2 +(2a)2 = 4, hen e a2 =4

5and so a =

2√5

=2√

5

5. Sin e thelength of one side of PQRS is b = 2a =

4√

5

5, then the area of the square

277is (4

√5

5

)2

=80

25=

16

5= 3.2.One-half the area of the semi ir le is 1

2×1

2×π×22 = π < 3.2. The areaof the square is therefore greater than one-half the area of the semi ir le.Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; SHAMIL ASGARLI,student, Burnaby South Se ondary S hool, Burnaby, BC; RICARDO BARROSO CAMPOS, Uni-versity of Seville, Seville, Spain; CAO MINH QUANG, Nguyen Binh Khiem High S hool,Vinh Long, Vietnam; JACLYN CHANG, student, Western Canada High S hool, Calgary, AB;COURTIS G. CHRYSSOSTOMOS, Larissa, Gree e; ANTONIO GODOY TOHARIA, Madrid,Spain; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; RICARD PEIR �O, IES \Abastos",Valen ia, Spain; ROBERT SHEETS, Southeast Missouri State University, Cape Girardeau, MO,USA; MRIDUL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; MRINAL SINGH,student, Kendriya Vidyalaya S hool, Shillong, India; ALEX SONG, student, Elizabeth ZieglerPubli S hool, Waterloo, ON; and JIXUAN WANG, student, Don Mills Collegiate Institute,Toronto, ON.M365. Proposed by Alexander Gurevi h, student, University ofWaterloo,Waterloo, ON.Let D be the family of lines of the form y = nx + n2, with n ≥ 2a positive integer. Let H be the family of lines of the form y = m, where

m ≥ 2 is a positive integer. Prove that a line from H has a prime y-inter eptif and only if this line does not interse t any line from D at a point with anx- oordinate that is a nonnegative integer.Solution by Alex Song, student, Elizabeth Ziegler Publi S hool, Waterloo,ON, modi�ed by the editor.Let m ≥ 2 be a positive integer. Consider a line y = m from H. Itsy-inter ept is m. It suÆ es to prove two things:(a) if m is prime, then for any positive integer n with n ≥ 2, the simulta-neous equations y = m and y = nx + n2 do not have a solution for xthat is a nonnegative integer, and(b) if m is omposite, then there is a positive integer n ≥ 2 su h that thesimultaneous equations y = m and y = nx + n2 do have a solutionfor x that is a nonnegative integer.Putting this another way, wemust prove that ifm is prime, then the equationm = nx + n2 does not have a nonnegative integer solution for x for anyn ≥ 2, and if m is omposite, then the equation m = nx + n2 does have anonnegative integer solution for x for some n ≥ 2.Assume that m is prime and that n is a positive integer with n ≥ 2.Suppose also that m = nx+n2 = n(x+n) has an integer solution for x withx ≥ 0. Note that m has only two positive divisors, namely m and 1. Sin en ≥ 2, then n must equal m, and so x+n = 1, whi h gives x = 1−n ≤ −1.Thus, x is not a positive integer or zero. This is a ontradi tion, so there isno n for whi h m = nx + n2 has nonnegative integer solutions for x.

278Assume next that m is omposite. Then m = pq for some integers pand q with 2 ≤ p ≤ q. Thus, we want to �nd integers n and x with n ≥ 2and x ≥ 0 su h that m = n(x + n) = pq. Setting n = p and x + p = qsatis�es the equation. Here, x = q − p ≥ 0, so the restri tions on x and nare satis�ed. Thus, for some n there is a nonnegative solution for x.Therefore, a line from H has a prime y-inter ept if and only if this linedoes not interse t any line from D at a point with an x- oordinate that is anonnegative integer.Also solved by SHAMIL ASGARLI, student, Burnaby South Se ondary S hool, Burnaby,BC; ANTONIO GODOY TOHARIA, Madrid, Spain; RICHARD I. HESS, Ran ho Palos Verdes, CA,USA; ROBERT SHEETS, Southeast Missouri State University, Cape Girardeau, MO, USA; andJIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON.M366. Proposed by John GrantM Loughlin, University of New Brunswi k,Frederi ton, NB.The roots of the equation x3 + bx2 + cx + d = 0 are p, q, and r. Finda quadrati equation with roots (p2 + q2 + r2) and (p + q + r).Solution by Shamil Asgarli, student, Burnaby South Se ondary S hool, Burn-aby, BC.Sin e p, q, and r are the roots of the ubi equation, we an fa tor theleft side of the equation to get (x − p)(x − q)(x − r) = 0. Expanding yields

x3 − (p+q+r)x2 +(pq+qr+rp)x−pqr = 0. Comparing oeÆ ients withthe original equation, we obtain p + q + r = −b while pq + qr + rp = c.Sin e (p + q + r)2 = p2 + q2 + r2 + 2(pq + qr + rp), then we have(−b)2 = p2 + q2 + r2 + 2c, so p2 + q2 + r2 = b2 − 2c.A possible quadrati equation with the desired roots is therefore

(x − (

b2 − 2c))(

x − (−b))

= 0 ,or x2 +(b − b2 + 2c

)x +

(2bc − b3

)= 0.Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; GEORGIOSBASDEKIS, student, 1st High S hool of Karditsa, Karditsa, Gree e; CAOMINH QUANG, NguyenBinh Khiem High S hool, Vinh Long, Vietnam; JACLYN CHANG, student, Western CanadaHigh S hool, Calgary, AB; COURTIS G. CHRYSSOSTOMOS, Larissa, Gree e; ANTONIO GODOYTOHARIA, Madrid, Spain; JOS �E HERN�ANDEZ SANTIAGO, student, Universidad Te nol �ogi a dela Mixte a, Oaxa a, Mexi o; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; HUGO LUYOS�ANCHEZ, Ponti� ia Universidad Cat �oli a del Peru, Lima, Peru; RICARD PEIR �O, IES \Abastos",Valen ia, Spain; ROBERT SHEETS, Southeast Missouri State University, Cape Girardeau, MO,USA; ALEX SONG, student, Elizabeth Ziegler Publi S hool, Waterloo, ON; and JIXUAN WANG,student, Don Mills Collegiate Institute, Toronto, ON.M367. Proposed by George Tsapakidis, Agrinio, Gree e.For the positive real numbers a, b, and c we have a + b + c = 6.Determine the maximum possible value of a

√bc + b

√ac + c

√ab.

279Solution by Jos �e Hern �andez Santiago, student, Universidad Te nol �ogi a dela Mixte a, Oaxa a, Mexi o.Applying the Arithmeti Mean{Geometri Mean Inequality to positivereal numbers a, b, and cwe obtain abc ≤

(a + b + c

3

)3

= 8 and onsequently√

abc ≤ 2√

2. (Equality holds here if and only if a = b = c and so if andonly if a = b = c = 2.)We an then further on lude thata√

bc + b√

ac + c√

ab =√

a√

abc +√

b√

abc +√

c√

abc

=√

abc(√

a +√

b +√

c)

≤ 2√

2(√

a +√

b +√

c) . (1)Now,

(√a +

√b +

√c)2

= a + b + c + 2(√

ab +√

ac +√

bc)

= 6 + 2(√

ab +√

ac +√

bc) ,sin e a + b + c = 6.Applying the AM{GM Inequality on e more yields 2

√xy ≤ x+y, thuswe obtain 2

(√ab +

√ac +

√bc) ≤ a + b + a + c + b + c = 2(a + b + c).(Again, equality holds if and only if a = b = c = 2.)Hen e, (√a +

√b +

√c)2 ≤ 6 + 2(a + b + c) = 18 and onsequently√

a +√

b +√

c ≤ √18 = 3

√2.From (1), it follows that a√

bc + b√

ac + c√

ab ≤ (2√

2)(

3√

2)

= 12,so the maximum possible value is 12, whi h we have seen is a hieved whena = b = c = 2.Also solved by ARKADY ALT, San Jose, CA, USA; GEORGE APOSTOLOPOULOS,Messolonghi, Gree e; SHAMIL ASGARLI, student, Burnaby South Se ondary S hool,Burnaby, BC; GEORGIOS BASDEKIS, student, 1st High S hool of Karditsa, Karditsa, Gree e;CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; COURTIS G.CHRYSSOSTOMOS, Larissa, Gree e; ANTONIO GODOY TOHARIA, Madrid, Spain; RICHARDI. HESS, Ran ho Palos Verdes, CA, USA; JOE HOWARD, Portales, NM, USA; RICARDPEIR �O, IES \Abastos", Valen ia, Spain; ALEX SONG, student, Elizabeth Ziegler Publi S hool,Waterloo, ON; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; and JIXUANWANG, student, Don Mills Collegiate Institute, Toronto, ON. There was one in orre t solutionsubmitted.M368. Proposed by J. Walter Lyn h, Athens, GA, USA.An in�nite series of positive rational numbers a1 + a2 + a3 + · · · isthe fastest onverging in�nite series with a sum of 1, a1 = 1

2, and ea h aihaving numerator 1. (By \fastest onverging", we mean that ea h term anis su essively hosen to make the sum a1 + a2 + · · · + an as lose to 1 aspossible.) Determine a5 and des ribe a re ursive pro edure for �nding an.

280Solution by Robert Sheets, Southeast Missouri State University, Cape Girar-deau, MO, USA.Let sn be the nth partial sum of the in�nite series. Then a1 =

1

2and

s1 =1

2. Sin e all terms of the in�nite series are positive, none of the terms an be negative or zero, thus no partial sum an be greater than or equal to

1. We therefore need a2 < 1 − s1 =1

2. Seeking the largest su h a2, we �ndthat a2 =

1

3sin e a2 is a fra tion of positive integers with numerator equalto 1. Then s2 =

1

2+

1

3=

5

6whi h gives a3 <

1

6, when e a3 =

1

7. Similarly,we �nd that s3 =

41

42, giving a4 =

1

43, and s4 =

1805

1806, giving a5 =

1

1807.Next, we des ribe a re ursive pro edure to �nd an. We onje ture thatif an =

1

k, then sn =

k(k − 1) − 1

k(k − 1)and an+1 =

1

k(k − 1) + 1. (Note that

2(1)+1 = 3, 3(2)+1 = 7, 7(6)+1 = 43, and 43(42)+1 = 1807, so theseequations hold for n = 1, 2, 3, and 4.)We prove these re ursive relations by indu tion. We have already ver-i�ed the base ases above. Suppose that for some positive integers n and kwe have an =1

k, sn =

k(k − 1) − 1

k(k − 1), and an+1 =

1

k(k − 1) + 1. We provethat the relations will also hold for sn+1 and an+2. Let t = k(k − 1) + 1;this means that an+1 =

1

tand sn =

t − 2

t − 1. Then

sn+1 = sn + an+1

=t − 2

t − 1+

1

t=

t2 − 2t + t − 1

t(t − 1)

=t2 − t − 1

t(t − 1)=

t(t − 1) − 1

t(t − 1).

Finally, sin e an+2 < 1 − sn+1 =1

t(t − 1)and an+2 is the largest fra tionwith numerator 1 satisfying this property, we �nd that an+2 =

1

t(t − 1) + 1,as required.Therefore, the result holds by indu tion and for all positive integers n,if an =

1

kfor some positive integer k, then an+1 =

1

k(k − 1) + 1.Also solved by ARKADY ALT, San Jose, CA, USA; RICHARD I. HESS, Ran ho PalosVerdes, CA, USA; and ALEX SONG, student, Elizabeth Ziegler Publi S hool, Waterloo, ON.There was one in omplete solution submitted.Hess noted that the sequen e of denominators appearing in the series is A000058 in theOn-Line En y lopedia of Integer Sequen es (http://www.research.att.com/~njas/sequences/)and is known as Sylvester's sequen e. It is a urious fa t that two onse utive terms ofSylvester's sequen e di�er by a square, sin e the number k in the sequen e is followed by

k2 − k + 1, yielding a di�eren e of (k2 − k + 1) − k = (k − 1)2.

281Problem of the Month

Ian VanderBurghAmong Jim Totten's many interests was his involvement inmathemati soutrea h. In parti ular, Jim was a driving for e behind the Cariboo CollegeHigh S hool Mathemati s Contest. He edited a volume of problems takenfrom those ontests written between 1973 and 1992. This month, we look atone of the problems from this volume.Problem (1989 Cariboo College HighS hool Mathemati s Contest, SeniorFinal Round, Part B) In the �gure,

AEFB, BGHC, and ACJD aresquares onstru ted on the sides of△ABC. If the ( ombined) area of thetwo shaded squares equals the areaof the rest of the �gure, show thatthe area of △ABC equals the area of△FBG and then �nd the number ofdegrees in ∠ABC.

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A

B C

D

E

F

G H

J

β

This problem a tually appears to be the \poster hild" for this ontest,as it appears on the over of the ompilation book and appears as a kind oflogo elsewhere on the web.Before we look at the solution, there is one really useful formula uponwhi h we should agree. If in △XY Z we have XY = z and Y Z = x, thenthe area of △XY Z is equal to 12xz sin(∠XY Z). This formula is a greatalternative to the standard area formula \1

2bh" if all you have is one angleof a triangle and the lengths of the two sides en losing it. We'll derive thisformula after the solution to the problem.Solution Let BC = a, AC = b, and AB = c. Sin e AEFB is a square, then

BF = AE = c. Sin e BGHC is a square, then BG = CH = a. Sin eACJD is a square, then AD = CJ = b.Sin e ∠ABC = β and ∠ABF = ∠CBG = 90◦, it then follows that∠FBG = 360◦ − β − 90◦ − 90◦ = 180◦ − β.We �rst need to prove that△ABC and △FBG have equal areas. Fromthe formula in the preamble, the area of △ABC is 1

2(BC)(AB) sin(∠ABC)or 1

2ac sin β. Similarly, the area of △FBG is 1

2(BG)(FB) sin(∠FBG) or

12ac sin

(180◦ − β

).But sin β = sin(180◦ − β) for any angle β, so the two areas are equal.While it may not be immediately obvious why this helps, we an stall abit by noting that we an use the same argument to on lude that the area of

282△EAD and the area of △HCJ are ea h equal to the area of △ABC. Canyou see why?At this point, we should probably use the pie e of information that wewere given, namely, that the ombined area of square AEFB and squareBGHC equals the area of the rest of the �gure. We use the short-hand|AEFB| to denote the area of �gure AEFB. Thus, we are told that

|AEFB| + |BGHC| =

|FBG| + |ABC| + |EAD| + |HCJ | + |ACJD| .Using some of what we know so far, this be omesc2 + a2 = 4|ABC| + b2 ,or

c2 + a2 = 4(

1

2ac sin β

)+ b2 .Remember, we're trying to �nd β. We seem to have too many otherpie es of information oating around to have any hope of doing this. But atthis point, the amazing pattern re ognition abilities of the brain might ki kin. This equation looks somewhat similar to a law that we often use. Thismight prompt us to try applying that law. Do you see what I'm getting at?Applying the Law of Cosines in △ABC gives b2 = a2 + c2 −2ac cos β.Substituting this into the last equation, we obtain

c2 + a2 = 2ac sin β + (a2 + c2 − 2ac cos β) ;2ac cos β = 2ac sin β ;

cos β = sin β ,sin e ac > 0. Sin e cos β = sin β, and β is an angle in a triangle, thenβ = 45◦, and we are done.To me, this was quite surprising. Well, the answer itself wasn't so sur-prising sin e these problems almost always have 30◦, 45◦, or 60◦ as an an-swer. But, it was surprising to me that the angle β was ompletely deter-mined from the given information while no other information (side lengthsor angles) an be determined.Before wrapping up the olumn thismonth, we should go ba k and look at theformula from the preamble. Suppose that∠XY Z is a ute. Drop a perpendi ular fromX to F on Y Z.Then the area of △XY Z is equal to12(Y Z)(XF ). But Y Z = x and we also have

XF = XY sin(∠XY Z) = z sin(∠XY Z), so .................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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the area equals 12xz sin(∠XY Z), as required. Can you prove this in the asethat angle Y is obtuse or a right angle?

283The Tanker Problem

Ross HonsbergerWhen I walk my puppy with his retra table leash, he is forever gettingbehind, at hing up and running past, utting a ross in front, and gettingbehind again. When I was having my after-lun h nap the other day, thisreminded me of those old \patrol boat ir ling an o ean liner" problems andI mused over the following situation.A se urity patrol boat repeatedly ir les a supertanker that is a giganti re tangular box 450 metres long and 50 metres a ross. The o ean is alm andthe tanker travels at a onstant speed along a straight path. The patrol boatgoes up the left side, a ross the front, down the right side, and a ross theba k, and keeps doing it over and over.The patrol boat travels in only two di-re tions of the ompass { when going par-allel to the path of the tanker, it travels instraight lanes parallel to the tanker, one onea h side at a distan e of 25 metres from it,and when rossing in front or behind, it goesstraight a ross perpendi ular to the path ofthe tanker.Negle ting the dimensions of the pa-trol boat (that is, onsidering it to be rep-resented geometri ally by a point) and giventhat it goes onstantly at twi e the speed ofthe tanker and that its turns are instanta-neous, what is the shortest distan e that thepatrol boat must travel in ompleting one y le around the tanker?I felt the problem would be an easyre reation with a pleasing analysis and I ex-pe ted a solution along the following lines.Let the tanker and the lanes of the pa-trol boat be labeled as in Figure 1 and let theboat travel towards Q when it is in lane PQ.

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A

B C

DM

P

Q

R

S

s

25-�

50 -�

450

6

?

Figure 1Consider a y le that begins at M on PQ; that is, to begin the boat isabreast of the stern AD of the tanker. While the boat travels 900m up PQ,the tanker will have advan ed 450m and the boat will be abreast of the front

BC of the tanker at the point E on PQ (Figure 2).Copyright c© 2009 Canadian Mathemati al So ietyCrux Mathemati orum with Mathemati al Mayhem, Volume 35, Issue 5

284If the boat were to start rossing be-tween its lanes at this point, it would rashinto the tanker at a point 12.5m down theside. In order to ross in front of the tanker,the boat must get 25 + 50 = 75m a ross thegap to the right hand edge of the tanker beforethe tanker advan es to its level. Therefore, itneeds to have a lead of 37.5m when it startsits turn, for the tanker will advan e 37.5mwhile the boat is going a ross this 75m.In order to get a 37.5m lead, the boatmust ontinue up PQ another 75m beyond E(while the boat is doing this the tanker ad-van es 37.5m to yield a net lead of 37.5m).Summarizing, the boat and the tankerare abreast atE and BC; as the boat advan esthe 75m to F , the tanker advan es 37.5m to

UV ; the boat turns at F and while going the75m along FK, the tanker advan es 37.5m toGK. Thus, the boat and the tanker just miss

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

A

BC

D

Q S

E

FG K

L

N

U V

Y

P RFigure 2ea h other at K. And while the boat ompletes the remaining 25m alongKL to its other lane, the tanker takes a 12.5m lead at Y N .Thus, when the boat turns down SR it is already 12.5m down that sideof the tanker. Now, the boat need go only to a point H that is 12.5m fromthe far end before making its turn a ross the ba k (while it rosses the 25mgap toward the tanker, the tanker moves ahead this 12.5m and the boat justmisses the bottom right orner of the tanker). Hen e the boat needs to overa distan e of only 450 − 12.5 − 12.5 = 425m along this side before turning.Sin e it goes twi e as fast as the tanker, while the boat overs two-thirds ofthis distan e, the tanker moves ahead the other third, and so the boat needtravel down that side only two-thirds of 425m, that is, a distan e of 850

3m.Now, from just missing the tanker at its bottom right orner, the boatstill has to go 75m to get ba k to lane PQ. While the boat is doing this, thetanker gets ahead 37.5m, and so, in order to at h up and omplete its y le,the patrol boat must go another 75m up PQ to get abreast of the tanker.Altogether, then, the patrol boat must travel a grand total of

900 + 75 + 100 +850

3+ 100 + 75 = 1533

1

3m .This solution blithely assumes that the patrol boat and the tanker harm-lessly slide by ea h other when they arrive simultaneously at the tanker'sright hand orners. However, it makes a great deal more sense to interpretsu h an event as a ollision: both vessels an hardly move into the sameposition at the same time without fatal onsequen es for the patrol boat.Fortunately, this nagging un ertainty is easily removed by adding the ondi-tion that the two vessels are never to get loser to ea h other than 25m.

285This is readily a ommodated by having the boat get a lead of 62.5m at

F , before turning along the dangerous 75m stret h FK between the lanes.This gives the boat a net lead of 62.5 − 37.5 = 25m when it rea hes theright hand side of the tanker, thus avoiding any possibility of a ollision andproviding the 25m separation at this riti al jun ture.In order to get 62.5m ahead, the boatmust go another 125m up PQ beyond E toa point F (Figure 3). Thus, when the patrolboat rea hes F , the tanker is at UV , and thepatrol boat is 62.5m ahead.Now the boat turns along FL, and whenit rea hes L, the tanker has advan ed 37.5mto IT , putting the tanker 25m behind. Whilethe boat pro eeds 25m along LG to the otherlane, the tanker uts its lead to 12.5matWN .Thus, when the boat turns down SR, itis already 12.5m ahead. In order to remain25m from the other end of the tanker when itgoes ba k a ross the �rst lane, it will need togo farther down the side to a point H that is12.5m beyond the other end. Then, when itturns and loses the 25m stret h to the nearerside of the tanker, the tanker will have ad-van ed another 12.5m to yield a separationof 25m between them. Hen e the patrol boatmust over 450 + 12.5 + 12.5 = 475m alongthis side before making its turn. As before, itneeds to travel down that side only two-thirdsof this distan e, that is, 950

3m.

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A

BC

D

E

F GL

N

Q

I

S

T

U V

W

P RFigure 3Thus, when passing the right rear orner of the tanker, the boat is al-ready 25m behind, and still has 75m of the gap to negotiate to get ba k tolane PQ. While the boat is doing this, the tanker gets ahead another 37.5m,for a total of 62.5m, and so, in order to at h up and omplete its y le, thepatrol boat must go another 125m up PQ to get abreast of the tanker.Altogether, then, the patrol boat must travel a grand total of

900 + 125 + 100 +950

3+ 100 + 125 = 1666

2

3m .This seemed most satisfa tory until the brilliant solution of my ol-league Larry Ri e (retired from the University of Toronto S hools and theUniversity of Waterloo) revealed another unwarranted assumption in my so-lution. Sin e the patrol boat had a 25m lead when it rea hed the right handedge of the tanker, it went un hallenged that the minimum 25m distan ebetween the boat and the front right orner of the tanker would be main-tained as the boat ompleted the �nal 25m of the rossing to its other lane.

286Unfortunately this is not true; it is not hard to show that the vessels do get loser together than 25m when the boat is ompleting the rossing.So, how mu h of a leadwould the boat have to havewhen it passes the right handedge of the tanker in orderto maintain a 25m separation?Sin e 25m isn't enough, per-haps a lead of 27m would do?Again, it is not diÆ ult to seethat 27m is not enough. So,how about 28m? Bingo: 28mis suÆ ient, with a little tospare. Consequently, any leadgreater than 28m would alsosuÆ e. But we seek the short-est y le of the boat around thetanker. Thus we need to det-

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

d

GL N

J

T

x

2x

(25 + y) − x

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

Figure 4ermine the smallest a eptable lead. We know it's some value, 25 + y, be-tween 27 and 28m. Here y is not a variable, but a well-de�ned �xed value.In Figure 4, let the boat pass the right hand edge of the tanker with alead TL of (25+y) metres. As the tanker advan es x metres to J on TL, theboat advan es 2x metres along LG to the point N . We want to determinethe smallest positive value y su h that the segment d = JN is at least 25mfor all x between 0 and 12.5.By the Pythagorean Theorem, we have

d2 = (2x)2 + (25 + y − x)2

= 4x2 + (25 + y)2 − 2x(25 + y) + x2

= 5x2 + y2 + 50y + 625 − 50x − 2xy ,and for d2 ≥ 625, we need5x2 + y2 + 50y − 50x − 2xy ≥ 0 .Sin e the only variable here is x, we may write

f(x) = 5x2 + y2 + 50y − 50x − 2xy ,where we require f(x) ≥ 0 for 0 ≤ x ≤ 12.5.Di�erentiating, we getf ′(x) = 10x − 50 − 2y = 2(5x − 25 − y) ,whi h vanishes for x = 5+

1

5y. Sin e the graph of f(x) is a parabola openingupwards, its minimum o urs when f ′(x) = 0.

287Hen e

min f(x) = 5(5 +

1

5y)2

+ y2 + 50y − 50(5 +

1

5y)

− 2y(5 +

1

5y)

=4

5y2 + 40y − 125 .Now, we don't want this minimum to be greater than 0, for in that ase dwould always ex eed 25 and a lead of 25 + y would be greater than it needbe. So for minimum y, we require the minimum to equal 0, and

4

5y2 + 40y − 125 = 0 ,or

4y2 + 200y − 625 = 0 .The number y being positive, we obtainy =

−200 +√

2002 − 4(4)(−625)

2(4)=

−200 +√

50000

8

=25

2

√5 − 25 ≈ 2.95085 ,or 2.951 to three de imal pla es of a ura y.Thus a lead of 27.951 would assure a universal 25m separation andprovide a shortest y le to a reasonable degree of a ura y.We have seen that an additional 125m beyond E gives the boat a 25mlead when it rea hes the right hand edge of the tanker. Hen e a lead of

27.951m would require the boat to go a further 2(2.951) = 5.902m for atotal of 130.902m up its left lane before starting its turn a ross the front.Moreover, its lead is now 12.5 + 2.951 = 15.451m when it starts down theright lane.To produ e the same separation at the other end of the tanker, the boatneeds to go this same distan e, 15.451m, farther down this lane than the endof the tanker, for a total passage of 450+2(15.451) = 480.902m. As before,the boat needs to over only two-thirds of this distan e, namely 320.601m.After returning the 100m to its left lane, the boat would be another 50mbehind, for a total of 65.451m, and therefore needs to go another 130.902mup PQ in order to omplete the y le. Altogether, then, the boat travelsapproximately a total of900 + 130.902 + 100 + 320.601 + 100 + 130.902 = 1682.405m .Finally, let's onsider Larry's wonderful solution { a ompletely di�er-ent approa h.The unknown distan es in the y le are those travelled by the patrolboat in the lanes PQ and RS; the distan es travelled in rossing between thelanes is always 100 + 100 = 200m. Larry obtains these unknown distan esfrom the basi formula distan e = speed × time .

288If we denote the speed of the tanker by v, then the speed of the boat is 2v,and it remains only to determine the times t1 and t2 that the boat spends inlanes PQ and RS respe tively, sin e the distan e travelled in lane PQ willbe (2v)t1 and the distan e travelled in lane RS will be (2v)t2.Knowing only the ratio of the speeds of the boat and the tanker, youmight well wonder how he is going to obtain anything of value about thesetimes. Larry very ingeniously enlists the aid of the aptain of the tanker inthis matter.To this end, onsider the observa-tions of the aptain as he wat hes theboat ir le his tanker (Figure 5). First,he sees the boat run up lane PQ fromE to F . At F , the boat turns a rossthe lanes in front of him and, as it pro- eeds, it keeps getting loser to him be- ause the tanker is moving ahead. Uponrea hing the other lane, the boat pro- eeds along it and eventually turns ba kto lane PQ. As it rosses ba k, thetanker, still going forward, moves intothe lead. Finally, the boat gets abreastof the tanker up lane PQ. Thus the ap-tain sees the boat y le the tanker alongthe sides of a trapezoid EFGH.Between two points on a slantedside of the trapezoid, the boat movestwi e as far toward the other lane as itdoes toward, or away from, the tanker,implying that the slopes of these sidesare ±1

2. Thus the trapezoid is isos elesand enjoys the symmetry of su h a �g-ure. Sin e the slope of FG is −1

2and

LC =75m, it follows that LF =37.5m.The symmetri al part EK is therefore

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

E

H

F

G

K

LC

U

V

P

Q

R

S

2525

12.550

37.5

Figure 5also 37.5m, making the total length of side EF = 450 + 2(37.5) = 525m.Similarly, sin e CU = 25m, then UG = HV = 12.5m, making the lengthof side GH = 450 − 2(12.5) = 425m.Now, time equals distan e divided by speed, so the time t1 that it takesthe boat to traverse EF is 525 divided by the boat's speed along EF . Whilethe boat goes at speed of 2v relative to the o ean, the aptain sees it pro eedup EF at the speed at whi h it overtakes the tanker; that is, at the relativespeed of 2v − v = v. Hen e the aptain al ulates t1 =

525

v.

289Along side GH, the aptain observes the boat is going at a relativespeed of 2v + v = 3v, and so he al ulates t2 =

425

3v.It follows that the distan e the boat travels up and down the lanes is

2v

(525

v

)+ 2v

(425

3v

)= 1050 +

850

3= 1333

1

3m .Adding in the 200m travelled in rossing between lanes, we obtain the sameearlier grand total of 15331

3m for the �rst ase, in whi h we generously al-lowed the vessels to slide by ea h other without in ident.Allowing an additional 25m bu�er at ea h rossing between the laneswould in rease ea h of the parallel sides of the trapezoid by 50m and yieldthe orresponding times t1 =

575

vand t2 =

475

3v.The resulting y le, then, would have a total length of

2v

(575

v

)+ 2v

(475

3v

)+ 200 = 1150 + 316

2

3+ 200 = 1666

2

3m ,as found earlier.Finally, (Figure 6), a universalseparation of 25m, would add an extra

2(12.5)(√

5) ≈ 25(2.236) = 55.9mto ea h of the parallel sides of the orig-inal trapezoid EFGH.This is be ause the right-angled

△XY C has legs in the ratio 1 : 2. Canyou see why?Thus, t1 =580.9

vand t2 =

480.9

3vwhi h yields a y le length of approxi-mately2(580.9) + 2

(480.9

3

)+ 200

= 1161.8 + 320.6 + 200

= 1682.4m ,in lose agreement with our previousresult.......................................................................................................................................

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F

GX Y

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12.5

25

Figure 6Ross Honsberger326 Dale Cres entWaterloo, ON, N2J 3Y3

rosshonsberger@rogers.com

290THE OLYMPIAD CORNERNo. 279

R.E. WoodrowAfter a mu h needed break, Joanne Canape has agreed to ta kle trans-forming my s ribbles into LATEX again. Wel ome ba k!We begin this number with problems of the 37th AustrianMathemati alOlympiad of 2006. Thanks go to Robert Morewood, Canadian Team Leaderto the IMO in Slovenia, for olle ting them for our use.37th AUSTRIAN MATHEMATICAL OLYMPIADRegional Competition (Qualifying Round)April 27, 20061. Let 0 < x < y be real numbers andH =

2xy

x + y, G =

√xy , A =

x + y

2, and Q =

√x2 + y2

2be the harmoni , geometri , arithmeti , and quadrati means of x and y,respe tively. It is well known that H < G < A < Q holds. Order theintervals [H, G], [G, A], and [A, Q] by length.2. Let n > 1 be an integer and a a real number. Determine all real solutions(x1, x2, . . . , xn) of the following system of equations:

x1 + ax2 = 0 ,x2 + a2x3 = 0 ,x3 + a3x4 = 0 ,...

xn−1 + an−1xn = 0 ,xn + anx1 = 0 .3. In a nonisos eles triangle ABC, w is the bise tor of the external angle at

C. The extension of AB interse ts w in D. Let kA be the ir um ir le of thetriangle ADC, and kB the ir um ir le of the triangle BDC. Furthermore,let tA be the tangent of kA at A, tB be the tangent of kB at B, and P be thepoint in whi h these two tangents interse t.Assume that the two points A and B are given. Determine the set ofall points P = P (C), su h that ABC is a ute-angled but not isos eles.

2914. Let {hn}∞

n=1 be a harmoni sequen e of positive rational numbers. Inother words, ea h hn is the harmoni mean of its neighbours:hn =

2hn−1hn+1

hn−1 + hn+1

.Prove that if some term hj of the sequen e is the square of a rationalnumber, then the sequen e ontains an in�nite number of terms hk that areea h squares of rational numbers.National Competition (Final Round, Part 1)May 21, 20061. Let k and n be integers with k ≥ 2, n > 10k, and the de imal expansionof n ending with exa tly k zeros. Give the best possible lower bound (in termsof k = k(n) ≥ 2) for the number of ways to represent n as the di�eren e ofsquares of two nonnegative integers.

2. Prove that the sequen e {(n + 1)nn2−n

7n2 + 1

}∞

n=0

is stri tly in reasing.3. The in ir le of triangle ABC tou hes the lines BC and AC at D and E,respe tively. Prove that if AD and BE are of the same length, then thetriangle is isos eles.4. Let ⌊u⌋ denote the greatest integer less than or equal to the real number uand let {u} = u−⌊u⌋. Let f(x) = ⌊x2⌋+{x} for all positive real numbers x.Find an in�nite arithmeti progression of distin t positive rational numberswith denominator 3 (after an ellation) whi h do not lie in the image of f.National Competition (Final Round, Part 2)Day 1 (May 31, 2006)1. Find the number of nonnegative integers n ≤ N with the property thatthe de imal expansion of some multiple of n ontains only the digits 2 and 6(not ne essarily the same number of ea h).2. Prove that

3(a + b + c) ≥ 83√

abc +3

√a3 + b3 + c3

3for all positive real numbers a, b, and c. Determine when equality holds.

2923. Given triangle ABC, let point R be on the extension of AB beyondB with BR = BC, and let point S be on the extension of AC beyond Cwith CS = CB. Let the diagonals of BRSC interse t in the point A′, and onstru t the points B′ and C′ similarly. Prove that the area of the hexagonAC′BA′CB′ is the sum of the areas of triangles ABC and A′B′C′.Day 2 (June 1, 2006)4. Determine all rational numbers x su h that 1 + 105 · 2x is the square ofa rational number.5. Find all monotoni fun tions f : R → R su h that

f(−f(x)

)= f

(f(x)

)= f(x)2 .(A fun tion f is monotoni if either f(a) ≤ f(b) for all a < b or f(a) ≥ f(b)for all a < b.)6. Let A be a nonzero integer. Find all integer solutions of the followingsystem of equations:

x + y2 + z3 = A ,1

x+

1

y2+

1

z3=

1

A,

xy2z3 = A2 .Next we give the Brazilian Mathemati al Olympiad 2005. Thanks againgo to Robert Morewood, Canadian Team Leader to the IMO in Slovenia, for olle ting them for our use.Brazilian Mathemati al Olympiad 2005First Day (O tober 22, 2005)1. A positive integer is a palindrome if reversing its digits leaves it un hanged(for example, 481184, 131, and 2 are palindromes). Find all pairs (m, n) ofpositive integers su h that 111 . . . 1︸ ︷︷ ︸

m ones × 111 . . . 1︸ ︷︷ ︸n ones is a palindrome.

2. Determine the smallest real number C su h thatC(x2005

1 + x20052 + · · · + x2005

5

) ≥ x1x2x3x4x5

(x125

1 + x1252 + · · · + x125

5

)16for all positive real numbers x1, x2, x3, x4, and x5.3. A square is ontained in a ube if ea h of its points is on a fa e or in theinterior of the ube. Determine the largest ℓ su h that there exists a squareof side ℓ ontained in a ube of edge length 1.

293Se ond Day (O tober 23, 2005)4. We have four harged batteries, four un harged batteries, and a radiowhi h needs two harged batteries to work. We do not know whi h batteriesare harged and whi h ones are un harged. What is the least number ofattempts that suÆ es to make sure the radio will work? (An attempt onsistsof putting two batteries in the radio and he king if the radio works or not).5. Let ABC be an a ute triangle and let F be its Fermat point, that is, theinterior point of ABC su h that ∠AFB = ∠BFC = ∠CFA = 120◦. Forea h of the triangles ABF , BCF , and CAF , draw its Euler line, that is, theline onne ting its ir um entre and its entroid.Prove that these three lines are on urrent.6. Let b be an integer and let a and c be positive integers. Prove that thereexists a positive integer x su h that

ax + x ≡ b (mod c) ,that is, prove there exists a positive integer x su h that c divides ax + x − b.Next we look at the problems of the 4th Grade Croatian Mathemati alOlympiad, written April 26{29, 2006. Thanks again go to Robert Morewood,Canadian Team Leader to the IMO in Slovenia, for olle ting them for ouruse. Croatian Mathemati al Olympiad 2006National Competition4th Grade1. Prove that three tangents to a parabola always form the sides of a trianglewhose altitudes interse t on the dire trix of the parabola.2. Let k and n be positive integers. Prove that

(n4 − 1

) (n3 − n2 + n − 1

)k+ (n + 1)n4k−1is divisible by n5 + 1.3. The ir les Γ1 and Γ2 interse t at the points A and B. The tangentline to Γ2 through the point A meets Γ1 again at C and the tangent line to

Γ1 through A meets Γ2 again at D. A half-line through A, interior to theangle ∠CAD, meets Γ1 at M , meets Γ2 at N , and meets the ir um ir le of△ACD at P . Prove that |AM | = |NP |.4. Six islands are onne ted by the Ferryboat and the Hydrofoil Boat om-panies. Any pair of islands is onne ted, in both dire tions, by exa tly one

294of these two ompanies. Prove that it is possible to tour four of the islandsusing exa tly one ompany. That is, prove that there are four islands A, B,C, and D and one ompany whose boats sail on the lines A ↔ B, B ↔ C,C ↔ D, D ↔ A.

Next we give the problems of the Balkan Mathemati al Olympiad 2006written at Ni osia, Cyprus. Again we thank Robert Morewood, CanadianTeam Leader to the IMO in Slovenia, for olle ting them for the Corner.Balkan Mathemati al Olympiad 2006Ni osia, Cyprus1. (Gree e) Let a, b, and c be real numbers. Prove that1

a(1 + b)+

1

b(1 + c)+

1

c(1 + a)≥ 3

1 + abc.

2. (Gree e) Let ABC be a triangle and m a line whi h interse ts the sidesAB and AC at interior points D and F , respe tively, and interse ts the lineBC at a point E su h that C lies between B and E. The lines through pointsA, B, C and parallel to the line m interse t the ir um ir le of triangle ABCagain at the points A1, B1, C1, respe tively. Prove that the lines A1E, B1F ,and C1D are on urrent.3. (Romania) Find all triples of positive rational numbers (m, n, p) su h thatea h of the numbers

m +1

np, n +

1

pm, p +

1

mnis an integer.4. (Bulgaria) Let m be a �xed positive integer. For ea h positive integer alet the sequen e {an}∞n=0 be de�ned by a0 = a and for n ≥ 0 the re ursion

an+1 =

{an

2if an is even ,

an + m otherwise .Find all values of a su h that the sequen e is periodi .As a �nal set of problems for this number we give the problems ofthe Finnish Mathemati al Olympiad 2006, Final Round. Thanks again goto Robert Morewood, Canadian Team Leader at the IMO in Slovenia for ol-le ting them.

295Finnish Mathemati al Olympiad 2006Final Round (February 3, 2006)1. Determine all pairs (x, y) of positive integers su h that

x + y + xy = 2006 .2. For all real numbers a, prove that3(1 + a2 + a4

) ≥ (1 + a + a2

)2

3. The numbers p, 4p2 + 1, and 6p2 + 1 are primes. Determine p.4. Prove that if twomedians of a triangle are perpendi ular, then the trianglewhose sides are ongruent to the medians of the original triangle is a righttriangle.5. The game of Nelipe is played on a 16 × 16grid. At ea h turn a player pi ks a numberfrom {1, 2, . . . , 16} and writes it in one ofthe squares of the grid. At ea h stage of thegame the numbers in ea h row, olumn, andin every one of the 16 four by four subsquaresmust be di�erent. A player loses if he or shehas no legal move. Whi h player wins, if bothplay with an optimal strategy? ...........................................................................................................................................................................................................................................................................................

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Next we turn to solutions from our readers to problems from the Septem-ber 2008 number of the Corner, from the 19th Lithuanian Team Contest inMathemati s written O tober 2, 2004 [2008: 282{284℄.1. Twelve numbers { four 1's, four 5's, and four 6's { are written in someorder around a ir le. Does there always exist a three-digit number om-prised of three neighbouring numbers (its digits an be taken lo kwise or ounter lo kwise) that is divisible by 3?Solved by John Grant M Loughlin, Universityof New Brunswi k, Frederi ton, NB; and TituZvonaru, Com�ane�sti, Romania. We give theresponse of Grant M Loughlin.The arrangement at right shows thatthere is not always a three digit number om-prised of three neighbouring numbers that isdivisible by 3.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................. 5

5

11

6

6

5

5

11

6

6

2. Solve the equation 2 cos(2πx) + cos(3πx) = 0.

296Solved by George Apostolopoulos, Messolonghi, Gree e; and George Tsapakidis,Agrinio, Gree e. We give the write-up of Tsapakidis.Using the double and triple angle formulas, the equation an be writtenequivalently as

4 cos3(πx) + 4 cos2(πx) − 3 cos(πx) − 2 = 0 ,[2 cos(πx) + 1

][2 cos2(πx) + cos(πx) − 2

]= 0 .Therefore cos(πx) = −1

2or 2 cos2(πx) + cos(πx) − 2 = 0, hen e,either cos(πx) = −1

2or cos(πx) =

√17 − 1

4. So, either cos(πx) = cos

2π

3or cos(πx) = cos θ, where θ = arccos

(√17 − 1

4

). Thus,πx = 2kπ ± 2π

3or πx = 2kπ ± arccos

(√17 − 1

4

) , k ∈ Z ,and hen ex = 2k ± 2

3or x = 2k ± 1

πarccos

(√17 − 1

4

) , k ∈ Z .3. Solve the equation 3x⌊x⌋ = 13, where ⌊x⌋ denotes the integer part of thenumber x.Solved by Pavlos Maragoudakis, Pireas, Gree e; Edward T.H. Wang, WilfridLaurier University, Waterloo, ON; and Titu Zvonaru, Com�ane�sti, Romania.We give Wang's write-up.We show that the only solution is x =

√39

3.Sin e 00 is unde�ned, x 6= 0.If x < 0, then ⌊x⌋ = −n for some integer n ≥ 1. Thus, 3x⌊x⌋ =

3

xn. If

n is odd, then 3

xn< 0. If n is even, then n ≥ 2 implies x < −1 or −x > 1.Thus, 3

xn=

3

(−x)n< 3. In either ase, 3x⌊x⌋ 6= 13. Hen e, it remains to onsider x > 0.If x ≤ 2, then x⌊x⌋ ≤ 4 and if x ≥ 3, then x⌊x⌋ ≥ 27. Hen e,

2 < x < 3. Let x = 2 + r where 0 < r < 1. Then 3x⌊x⌋ = 13 be omes3(2+r)2 = 3

(r2+4r+4

)= 13, and we have 3r2+12r−1 = 0. Solving, weobtain r =

−12 ± √156

6= −2±

√39

3. Sin e r > 0, we reje t r = −2−

√39

3.Hen e, r = −2 +

√39

3and it follows that x =

√39

3.4. Let 0 ≤ a ≤ 1 and 0 ≤ b ≤ 1. Prove the inequality

a√2b2 + 5

+b√

2a2 + 5≤ 2√

7.

297Solved by George Apostolopoulos, Messolonghi, Gree e; Mi hel Bataille,Rouen, Fran e; Oliver Geupel, Br �uhl, NRW, Germany; Pavlos Maragoudakis,Pireas, Gree e; and Titu Zvonaru, Com�ane�sti, Romania. We give the solutionof Bataille.Sin e a2 ≤ 1, we have 2b2 + 5 ≥ 2b2 + 2a2 + 3. Similarly, we have2a2 + 5 ≥ 2b2 + 2a2 + 3 and it follows that

a√2b2 + 5

+b√

2a2 + 5≤ a + b√

2b2 + 2a2 + 3.Therefore, it suÆ es to show that √

7(a + b) ≤ 2√

2b2 + 2a2 + 3. Squaringand rearranging terms, this is equvalent to 12ab ≤ (a − b)2 + 12, whi h is ertainly true sin e 12ab ≤ 12 and (a − b)2 + 12 ≥ 12. The result follows.Clearly, equality holds if and only if a = b = 1.5. If a, b, and c are nonzero real numbers, what values an be taken by theexpressiona2 − b2

a2 + b2+

b2 − c2

b2 + c2+

c2 − a2

c2 + a2?Solution by Titu Zvonaru, Com�ane�sti, Romania.We have −1 <

a2 − b2

a2 + b2< 1, sin e −2a2 < 0 < 2b2.Setting α = a2b2 + b2c2 + c2a2, we obtain

(a2 − b2

) (b2 + c2

) (c2 + a2

)+(b2 − c2

) (a2 + b2

) (c2 + a2

)

+(c2 − a2

) (a2 + b2

) (b2 + c2

)

=(a2 − b2

) (c4 + α

)+(b2 − c2

) (a4 + α

)+(c2 − a2

) (b4 + α

)

=(a2 − b2

)c4 +

(b2 − c2

)a4 +

(c2 − a2

)b4

= a4b2 − a2b4 − a4c2 + b4c2 + a2c4 − b2c4

= a2b2(a2 − b2

)− c2(a4 − b4

)+ c4

(a2 − b2

)

=(a2 − b2

) (a2b2 − a2c2 − b2c2 + c4

)

=(a2 − b2

) [a2(b2 − c2

)− c2(b2 − c2

)]

= − (a2 − b2

) (b2 − c2

) (c2 − a2

) .It follows that∣∣∣∣a2 − b2

a2 + b2+

b2 − c2

b2 + c2+

c2 − a2

c2 + a2

∣∣∣∣ =

∣∣∣∣a2 − b2

a2 + b2· b2 − c2

b2 + c2· c2 − a2

c2 + a2

∣∣∣∣ < 1 .Let k ∈ (−1, 1) be a real number. Then, setting a2 = b2(

1 + k

1 − k

), weobtainlimc→0

(a2 − b2

a2 + b2+

b2 − c2

b2 + c2+

c2 − a2

c2 + a2

)= k + 1 − 1 = k .

298It follows that a2 − b2

a2 + b2+

b2 − c2

b2 + c2+

c2 − a2

c2 + a2takes all real values in (−1, 1).[Ed.: It seems the solver assumes a basi fa t about onne tedness,namely, that the image of a onne ted set under a ontinuous, real-valuedfun tion is a onne ted subset of the real line, that is, an interval. Thus, theexpression in a, b, and c a hieves all values between any two of its givenvalues, and sin e the image ontains points arbitrarily lose to 1 and −1 the on lusion follows.℄6. Determine all pairs of real numbers (x, y) su h that

x6 = y4 + 18 ,y6 = x4 + 18 .Solved by George Apostolopoulos, Messolonghi, Gree e; Oliver Geupel, Br �uhl,NRW, Germany; George Tsapakidis, Agrinio, Gree e; Edward T.H. Wang,Wilfrid Laurier University, Waterloo, ON; and Titu Zvonaru, Com�ane�sti, Ro-mania. We give the write-up of Apostolopoulos.Observe that x 6= 0 and y 6= 0. Sin e x6 − y4 = 18 and y6 − x4 = 18,we have

x6 − y6 + x4 − y4 = 0 ,(x2)3 − (

y2)3

+(x2)2 − (

y2)2

= 0 ,(x2 − y2

) (x4 + x2y2 + y4

)+(x2 − y2

) (x2 + y2

)= 0 ,

(x2 − y2

) (x4 + x2y2 + y4 + x2 + y2

)= 0 .However, x4+x2y2+y4+x2+y2 > 0, hen e x2 = y2 and x6−x4−18 = 0.We put x2 = w, then the equation in x be omes w3 − w2 − 18 = 0, or

(w −3)(w2 + 2w + 6

)= 0, hen e w = 3 sin e w2 +2w +6 = x4 +2x2 +6is positive for all real numbers x.Now, x2 = w = 3, hen e x = ±√

3 and there are four solutions(x, y) = (±√

3, ±√3).7. Find all triples (m, n, r) of positive integers su h that

2001m + 4003n = 2002r .Solution by Titu Zvonaru, Com�ane�sti, Romania.We will �nd all triples of nonnegative integers satisfying the equation.Sin e 32k = 9k ≡ 1 (mod 8) and 32k+1 = 9k · 3 ≡ 3 (mod 8), wehave that2001m + 4003n ≡ 1 + 3n ≡ 2, 4 (mod 8) .If r ≥ 3, then 2002n ≡ 0 (mod 8), hen e r ≤ 2 and we have 0 ≤ n < r ≤ 2.

299If r = 2 and n = 0, then the equation be omes 2001m + 1 = 20022,hen e 2001m = 20022 − 1, hen e 2001m = 2001 · 2003, hen e we have

2001m−1 = 2003 and we see that there are no solutions in this ase.If r = 2 and n = 1, then we obtain 2001m + 4003 = 20022, hen e2001m = (2001 + 1)2 − (2 · 2001 + 1), hen e 2001m = 20012 and m = 2.If r = 1 and n = 0, then 2001m + 1 = 2002, hen e m = 1.Therefore, the solutions are (m, n, r) ∈ {(1, 0, 1), (2, 1, 2)}, of whi honly (m, n, r) = (2, 1, 2) onsists entirely of positive integers.8. Assume that m and n are positive integers. Prove that, if mn − 23 isdivisible by 24, then m3 + n3 is divisible by 72.Solved by George Apostolopoulos, Messolonghi, Gree e; Mi hel Bataille,Rouen, Fran e; Edward T.H.Wang, Wilfrid Laurier University, Waterloo, ON;and Titu Zvonaru, Com�ane�sti, Romania. We give Wang's version.By assumption, mn = 24k + 23, k ∈ Z; hen e mn ≡ 7 (mod 8).Clearly, m and n are both odd. Thus, m ≡ 1, 3, 5, or 7 (mod 8)and similarly for n. All of the solutions to mn ≡ 7 (mod 8) are givenby (m, n) ≡ (1, 7), (7, 1), (3, 5), or (5, 3) (mod 8), where the ongruen e(m, n) ≡ (a, b) (mod d) means that m ≡ a (mod d) and n ≡ b (mod d).In the �rst two ases, we have m3 + n3 ≡ 1 + 343 = 344 ≡ 0 (mod 8) andin the other two ases, we have m3 + n3 ≡ 27 + 125 = 152 ≡ 0 (mod 8).Hen e,

8 | (m3 + n3) . (1)It remains to show that

9 | (m3 + n3) . (2)Sin e mn = 24k + 23, we have mn ≡ 2 (mod 3), whi h impliesthat (m, n) ≡ (2, 1) or (1, 2) (mod 3). Thus, m + n ≡ 0 (mod 3). Sin e

m3 + n3 = (m + n)((m + n)2 − 3mn

), the relation (2) follows.The on lusion now follows from (1) and (2).9. Is it possible that, for some a, both expressions 1 − 2a√

35

a2and a +

√35are integers?Solved by Oliver Geupel, Br �uhl, NRW, Germany; Edward T.H. Wang, WilfridLaurier University, Waterloo, ON; and Titu Zvonaru, Com�ane�sti, Romania.We give Geupel's write-up.Yes, this is possible. Let m =

1 − 2a√

35

a2and n = a +

√35; then

m = 1 and n = ±6 if a = ±6 − √35.We prove additionally that there are no other solutions.If m = 0, then a =

√35

70and n =

71√

35

70is not an integer. Hen e,

|m| ≥ 1 be ause m is an integer, and then |1 − 2a√

35| ≥ a2.

300If 1 − 2a

√35 ≥ a2, then n2 =

(a +

√35)2 ≤ 36 and thus |n| ≤ 6.Otherwise, −(1 − 2a

√35) ≥ a2 and a2 − 2

√35a + 1 ≤ 0. Therefore,

2√

35 − √34 ≤ n ≤ 2

√35 +

√34, hen e 7 ≤ n ≤ 17. Altogether, we have

−6 ≤ n ≤ 17. The 24 ases an now be eliminated using a al ulator.[Ed.: Substitute a = n − √35 into ma2 = 1 − 2a

√35 and simplifyto obtain m

(n2 + 35

)− 71 = (m − 1)2n√

35. Sin e ea h side is a rationalnumber, n = 0 or m = 1, whi h qui kly leads to the unique solution.℄11. What is the greatest value that a produ t of positive integers an take iftheir sum is equal to 2004?Solved by George Apostolopoulos, Messolonghi, Gree e; and Oliver Geupel,Br �uhl, NRW, Germany. We �rst give Apostolopoulos' solution.Let α1 + α2 + · · · + αn = 2004, where ea h αj is a positive integer.We want to maximize the produ t α1α2 · · · αn.If αk ≥ 4 for some k, then repla ing αk by 2 and αk − 2 leaves thesum the same and an only in rease the produ t, sin e 2(αk − 2) ≥ ak forak ≥ 4. Therefore, the maximum produ t is a hieved when ea h ak is either2 or 3. Now, 2 + 2 + 2 = 3 + 3 but 23 < 32, so the produ t is maximizedby taking as many 3's in the sum as possible while leaving at most two 2's.Sin e 2004 = 3 · 668 + 0 · 2, the largest possible produ t is 3668.Next we give Geupel's solution to a di�erent reading of the problem.We prove that the greatest value a produ t of two positive integers xand y an take if their sum is equal to a �xed positive integer n, is ⌊n2

4

⌋. Inthe spe ial ase n = 2004 the result is 10022 = 1004004. Indeed, we havexy = x(n − x) = −(x − n

2

)2+

n2

4. For even n, we obtain xy ≤ n2

4, whereequality holds if and only if x = y =

n

2. For odd n we have xy ≤ n2 − 1

4,where equality holds if and only if {x, y} =

{n − 1

2, n + 1

2

}.12. Positive integers a, b, c, u, v, and w satisfy the system of equationsa + u = 21 ,b + v = 31 ,c + w = 667 .Can abc be equal to uvw?Solved by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON; andTitu Zvonaru, Com�ane�sti, Romania. We give Wang's write-up.Yes. If a = 2·7, u = 7, b = 3·5, v = 24, c =

(23)·29, and w = 3·5·29,then the equations are satis�ed and abc = uvw = 24 · 3 · 5 · 7 · 29 = 48720.

30113. Let u be the real root of the equation x3 − 3x2 + 5x − 17 = 0, and letv be the real root of the equation x3 − 3x2 + 5x + 11 = 0. Find u + v.Solved by George Apostolopoulos, Messolonghi, Gree e; Mi hel Bataille,Rouen, Fran e; Jos �e Luis D��az-Barrero, Universitat Polit �e ni a de Catalunya,Bar elona, Spain; George Tsapakidis, Agrinio, Gree e; and Titu Zvonaru,Com�ane�sti, Romania. We give the solution of D��az-Barrero.For real numbers x and t let f(x) = x3 − 3x2 +5x and g(t) = t3 +2t.Then f(x) = (x−1)3+2(x−1)+3 and g is an odd, in reasing, and bije tivefun tion, as an be easily he ked. Furthermore,

g(u − 1) = (u − 1)3 + 2(u − 1) = f(u) − 3 = 17 − 3 = 14 ,g(v − 1) = (v − 1)3 + 2(v − 1) = f(v) − 3 = −11 − 3 = −14 .We have g(u − 1) = −g(v − 1) = g(1 − v) be ause g is odd, andmoreover sin e g is bije tive, we dedu e that u−1 = 1−v, hen e u+v = 2.15. Does there exist a polynomial, P (x), with integer oeÆ ients su h thatfor all x in the interval [ 4

10, 9

10

] the inequality |P (x) − 23| < 10−10 is valid?Solution by Oliver Geupel, Br �uhl, NRW, Germany.We give su h a polynomial expli itly. Let P (x) =2

3

[1 − (

1 − 3x4)n]and note that P (x) has integer oeÆ ients. For 0.4 ≤ x ≤ 0.9 we have

−0.9683 = 1 − 3(0.9)4 ≤ 1 − 3x4 ≤ 1 − 3(0.4)4 = 0.9232 ,hen e ∣∣1 − 3x4∣∣ ≤ 0.9683. Therefore, the ondition ∣∣∣P (x) − 2

3

∣∣∣ < 10−10 issatis�ed for 0.4 < x < 0.9 if 2

3(0.9683)n < 10−10. It now suÆ es to hoose

n =

⌊1 − log

103 − log

102

1 − log10

9.683

⌋+ 1. (Using a al ulator, we �nd that n = 59.)16. Does there exist a positive number a0 su h that all the members ofthe in�nite sequen e a0, a1, a2, . . . , de�ned by the re urren e formula

an =√

an−1 + 1, n ≥ 1, are rational numbers?Solution by Oliver Geupel, Br �uhl, NRW, Germany.We prove by ontradi tion that there is no su h a0. For ea h n ≥ 0 letan =

pn

qn, where pn and qn are oprime positive integers. From

p2n

q2n

= a2n = an−1 + 1 =

pn−1 + qn−1

qn−1

,it follows thatqn−1p2

n = q2n (pn−1 + qn−1) . (1)

302Then qn−1 | q2

n(pn−1 + qn−1) and the numbers qn−1 and pn−1 + qn−1 are oprime, hen e qn−1 | q2n.On the other hand, (1) yields q2

n | qn−1p2n and pn and qn are oprime,hen e q2

n | qn−1.It follows that q2n = qn−1 for all n ≥ 0. We on lude that qn = 1for ea h n. The an are therefore positive integers. It is readily he ked that

an > 1 and an > an+1 for all n > 0. This ontradi tion ompletes theproof.17. Let a, b, and c be the sides of a triangle and let x, y, and z be realnumbers su h that x + y + z = 0. Prove thata2yz + b2zx + c2xy ≤ 0 .Solved by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain; Mi helBataille, Rouen, Fran e; George Tsapakidis, Agrinio, Gree e; and Titu Zvonaru,Com�ane�sti, Romania. We give the write-up of Amengual Covas.More generally, let α, β, γ, x, y, and z be real numbers and supposethat x + y + z = 0.For real numbers u and v we have (u + v)2 ≤ 2

(u2 + v2

), hen e(β + γ)2yz + (γ + α)2zx + (α + β)2xy

≤ 2[(

β2 + γ2)yz +

(γ2 + α2

)zx +

(α2 + β2

)xy]

= 2[α2x(y + z) + β2y(z + x) + γ2z(x + y)

]

= −2(α2x2 + β2y2 + γ2z2

) ≤ 0 .Equality o urs only if α = β = γ and αx = βy = γz = 0.To establish the proposed inequality, we make use of the linear trans-formation a = β + γ, b = γ + α, and c = α + β, where α, β, and γ areuniquely determined positive numbers and apply the above inequality.Equality holds only if the triangle is equilateral and x = y = z = 0.18. Points M and N are on the sides AB and BC of the triangle ABC,respe tively. It is given that AM

MB=

BN

NC= 2 and ∠ACB = 2∠MNB.Prove that ABC is an isos eles triangle.Solved by George Apostolopoulos, Messo-longhi, Gree e; D.J. Smeenk, Zaltbommel,the Netherlands; George Tsapakidis, Agrinio,Gree e; and Titu Zvonaru, Com�ane�sti, Roma-nia. We give the solution of Smeenk.Let γ = ∠ACB, so that ∠MNB = 1

2γ.The line through C parallel to NM meets ABat D. Then 2 : 3 = BN : BC = BM : BD,hen e BD = 3

2BM = 1

2AB. Sin e the bise -tor of ∠ACB is also a median, CA = CB.

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30319. The two diagonals of a trapezoid divide it into four triangles. The areasof three of them are 1, 2, and 4 square units. What values an the area of thefourth triangle have?Solved by George Tsapakidis, Agrinio, Gree e; and Titu Zvonaru, Com�ane�sti,Romania. We give the solution of Tsapakidis.Let [XY Z] be the area of triangle XY Z.Then [ABD] = [ABC], as the two triangleshave equal altitudes from their ommon baseAB. Therefore,

[OAD] + [OAB] = [OBC] + [OAB] ,hen e [OAD] = [OBC]. ....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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

A B

CD

O

Triangles OAB and OCD are similar, so we have OA

OC=

OB

OD, that is

OA

OC· OD

OB= 1, and hen e [OAB]

[OBC]· [ODC]

[OBC]= 1, so [OBC]2 = [OAB][OBC].It follows that [OBC] = 2 = [OAD], [OAB] = 1, and [ODC] = 4, that is,the fourth triangle has area 2 square units.20. The ratio of the lengths of the diagonals of a rhombus is a : b. Find theratio of the area of the rhombus to the area of an ins ribed ir le.Solved by George Apostolopoulos, Messolonghi, Gree e; George Tsapakidis,Agrinio, Gree e; and Titu Zvonaru, Com�ane�sti, Romania. We give the solu-tion of Tsapakidis.Let r be the radius of the ir le ins ribed inthe rhombus. The required ratio is then

4 · 12OA · OB

πr2=

2

π· OA · OB · 1

r2

=2

π· OA · OB

(1

OA2+

1

OB2

)

=2

π

(OB

OA+

OA

OB

)=

2

π

(b

a+

a

b

)

=2(a2 + b2

)

πab,

where 1

r2=

1

OA2+

1

OB2holds sin e r is thealtitude of the right triangle OAB.

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That ompletes the Corner for this spe ial issue in honour of Jim Totten.The Editorial Board de ided to stay with \business as usual" for theOlympiadCorner. Having worked with Jim over many years, I suspe t that is what hewould have opted for.

304BOOK REVIEWSAmar SodhiFor this spe ial issue we feature book reviews by Andy Liu and JohnGrant M Loughlin, former Book Reviews Editors (1991-1998 and 2002-2008,respe tively) who ea h knew Jim Totten for an extended period.The following sentiment is from Andy: \Jim and I went ba k way beforehe be ame formally involved with Crux. His father lived in Edmonton, and he ame to visit twi e a year. Whenever he was in town, we would get togetherfor a Chinese dim-sum. One day, he asked me why I was no longer doing thereview olumn. A tually, I had already stepped down for a ouple of years.However, I replied ippantly I had just been �red by the new editor, littleknowing that I was sitting a ross the table from the new editor himself. Hewas very upset, and despite repeated apology, it remained a sore point withhim. So to set the re ord straight, I now state publi ly that I was never �redby Jim Totten as the Reviews Editor for Crux, and in his memory, it is only�tting that I ontribute a Book Review to this spe ial issue."All-Star Mathlete PuzzlesBy Di k Hess, Sterling Publishing Co. In ., New York, 2009ISBN 978-1-4027-5528-6, soft over, 94+ii pages, US$6.95Reviewed by Andy Liu, University of Alberta, Edmonton, ABSterling has published a wide range of mathemati al puzzle books. Atone end are the de�nitive treatises like Jerry Slo um's The Tangram Book.At the other end are rather prosai o�erings. Nevertheless, they do serve apurpose in gradually attra ting new enthusiasts. They o�er the novi es aneasier path into the hobby than more serious works like Rodolfo Kur han'sMesmerizing Math Puzzles, Serhiy Grabar huk's The New Puzzle Classi s,and the present volume.Di k Hess is well known to the readers of CRUX with MAYHEM as aregular ontributor. His intriguing problems always ome with an aura ofsomething out of the ordinary. Here are a ouple of samples.Problem 89(A) Find an expression equal to 88 whi h uses ea h of the digits

1, 2, and 3 exa tly on e. You may use a ombination of the operations addi-tion, subtra tion, multipli ation, division, exponents, roots, on atenation,de imals, repeating de imals, fa torials, and bra kets.Problem 44 The �gure at right onsists of four unit squaresjoined edge to edge. Find a polygon, not ne essarily on-vex, su h that �ve non-overlapping opies an �t insidethis �gure, and over more than 98% of its area. qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq

qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqIt should be mentioned that Di k Hess is an avid tennis fan, and thereare no books by him without problems related to that game. To �nd outmore about these problems, and the solution to the two problems above,buy the book! It is inexpensive and very highly re ommended.

305John adds: \The se ond edition of Ravi Vakil's book A Mathemati alMosai has been published. The original edition blended biography, prob-lems, and mathemati al onne tions, in an e�e tive manner. This sentimentis aptured in the 1997 CRUX with MAYHEM review by Jim Totten, so wetherefore reprint Jim's review in its entirety pre eded by a brief review ofthe se ond edition. As I was the Book Reviews Editor during Jim's tenureas Editor-in-Chief, the opportunity to here a ompany a review by Jim is anhonour. Jim took a keen interest in the books that were reviewed in CRUXwith MAYHEM. Indeed, Jim wrote enthusiasti ally about Stewart CoÆn'sGeometri Puzzle Design in a review published in April 2008.A Mathemati al Mosai : Patterns & Problem Solving (New ExpandedEdition)By Ravi Vakil, Brendan Kelly Publishing, Burlington, ON, 2008ISBN 978-1-895997-28-6, soft over, 288 pages, US$19.95Reviewed by John Grant M Loughlin, University of New Brunswi k,Frederi ton, NBI highly re ommend the new and expanded edition of A Mathemati alMosai : Patterns & Problem Solving, but only if you do not have a opy of the1996 edition already. Quoting from the foreword: "(T)here are few hangeshere and there from the �rst edition, but this book remains essentially thesame one that appeared in 1996." That omment re e ts this reviewer's takeon the book.The book is about 30 pages longer than the original version. Sele t ad-ditions of se tions su h as "Higher-dimensional versions of Platoni Solids"within the Combinatori s Chapter, insertion of mathemati al portraits, andin lusion of a more thorough index a ount for this di�eren e.Other hanges re e t updating of biographi al information and anno-tated referen es. Also, Ravi Vakil is now a well established algebrai geome-ter. However, with this in mind, the O tober 1997 CRUX with MAYHEMreview by Jim Totten is still relevant today. It is noteworthy that Ravi Vakilis a founding o-editor of Mathemati al Mayhem whi h be ame part of thisjournal in 1997, the same year as the publi ation of Jim's review.

A Mathemati al Mosai By Ravi Vakil, published by Brendan Kelly Publishing In ., 1996,2122 Highview Drive, Burlington, ON L7R3X4ISBN 1-895997-04-6, soft over, 253+ pages, US$16.95 plus handling.Reviewed by Jim Totten, University College of the Cariboo.So, you have a group of students who have de ided they want the extra hallenge of doing some mathemati s ompetitions. You want a sour e ofproblems whi h will pique the students' interest, and whi h also lead to fur-ther exploration. The problem sour e should lend itself well to independentwork. The question is: where do you �nd the appropriate level enri hmentmaterial? Many of us have already tried to answer this question and have a

306 olle tion of su h problem books. Well, here is a book to be added to your olle tion!It is ertainly a problem book, but it is mu h more than that. The au-thor at one moment guides the reader through some very ni e mathemati aldevelopments, and throws out problems as they rop up in the development,and in the next moment uses a problem as a starting point for some inter-esting mathemati al development.With a few ex eptions the problems in this book are not new, nor are thesolutions. They are, however, well organized, both by topi and by level ofmathemati al maturity needed. Answers are NOT always provided; insteadthere is often simply a solution strategy or hint given, and o asionally thereis simply a referen e to some other sour e for a full-blown treatment. Evenwhen answers are provided, they are not tu ked away at the end of the book,but rather they are worked into another topi (usually later in the book, butnot always), where they be ome part of the development of another topi orproblem.The author is a PhD andidate in pure mathemati s at Harvard Univer-sity (at the time the book was written). Being still very young, he knows howto speak to today's teenagers. His sense of humour and general pu kishnessis present throughout: just when you are lulled into some serious ompu-tation in probability, he deviously throws a tri k question at you, that hasa totally non-obvious answer (non-obvious, that is, until you CAREFULLYre-read the question).Many mathemati s books published today in lude short biographies onfamous mathemati ians through history, espe ially those whose names omeup in the theory developed in the book. This book is no ex eption. But whatis unique about this book is the in lusion of Personal Pro�les of young math-emati ians from several ountries that he has met at International Mathe-mati al Olympiads (IMOs) in the past. The pro�les are quite diverse, whi hmeans that most bright students ould �nd one to identify with and to use asa role model. The author and those he pro�les have taken a risk in doing this:they have tried to predi t some of the important mathemati ians in the earlypart of the next entury. It should be interesting to follow their areers andsee if those predi tions an ome true, or if by pla ing them in the spotlight,they �nd too mu h pressure to deal with.The problems range from puzzles that elementary s hool hildren an doto problems that provide training for Putnam andidates (toward the end ofthe book). There are many ross-referen es and onne tions between seem-ingly unrelated problems from di�erent areas of mathemati s, onne tionsthat most students would be unable to make. Many of these onne tionsare new to this reviewer. However, on e made these onne tions are quite lear.As for his redentials, Ravi Vakil pla ed among the top �ve in the Put-nam ompetition in all four of his undergraduate years at the University ofToronto. Before that he won two gold medals and a silver medal in IMOsand oa hed the Canadian team to the IMO from 1989 to 1995.

307The British Columbia Se ondary S hoolMathemati s ContestClint LeeIn 1973 John Ciriani of Cariboo College (whi h be ame Thompson RiversUniversity, TRU) initiated the Cariboo College High S hool Mathemati sContest with the help of olleagues in the Cariboo College Mathemati s andStatisti s Department. The intent of the ontest was to enable high s hoolstudents in the Cariboo College region with an interest in mathemati s to getlo al support and re ognition for their interest and abilities in mathemati s.In 1979 Jim Totten joined the department and qui kly be ame an enthusias-ti ontributor to and supporter of the ontest. In 1992 he ompiled a book ontaining all of the problems from the ontest papers over this period. Healso o-edited (with Leonard Janke) a separate book of solutions.Meanwhile, two other institutions in British Columbia developed highs hool mathemati s ontests. In 1977, at the suggestion of Robin Insley, theMathemati s Department at the College of New Caledonia (CNC), onsistingof Phil Be kman, Robin Insley, Peter Trushell, and myself, developed a highs hool mathemati s ontest for students in the Central Interior region servedby CNC, inspired by the su essful ontest being run at Cariboo College. Itran su essfully and independently from its in eption until 1994. In 1990,shortly after I joined theMathemati s Department at Okanagan College (OC)a third high s hool mathemati s ontest was initiated for students in the OCregion whi h ran independently from 1991 until 1994.The 1993 meeting of the British Columbia Committee in UndergraduatePrograms in Mathemati s (BCCUPM, now BCCUPMs) was held at SelkirkCollege in Castlegar. There Jim Totten approa hed members of the OC andCNC mathemati s departments with a suggestion that the three institutionspool their resour es to o�er a single mathemati s ontest. The ontest wouldserve the three olleges as well as any others in BC interested in su h a on-test, with the name, as suggested by Jim, the British Columbia Colleges HighS hool Mathemati s Contest. This title was intended to make expli it thatthe ontest was intended for olleges, who intera ted dire tly with their re-gions, rather than universities. Both OC and CNC enthusiasti ally agreed.Jim then approa hed other institutions and found further support for theidea. Over the years Colleges evolved into University Colleges and then intoUniversities, so the original name be ame inappropriate and was hanged tothe British Columbia Se ondary S hool Mathemati s Contest (BCSSMC).It took a year to organize the ombined ontest with the �rst beingheld in 1995. Jim's idea was that Cariboo College would prepare the on-test papers, with ontributions from the other institutions, and all parti -ipating institutions would use these for their lo al ontests. The 1995 andCopyright c© 2009 Canadian Mathemati al So ietyCrux Mathemati orum with Mathemati al Mayhem, Volume 35, Issue 5

3081996 ontests were handled this way. The ombined ontest, following theCariboo model, onsists of Preliminary and Final Rounds. Ea h round has aJunior level for grades 8 to 10 and a Senior level for grades 11 and 12. ThePreliminary Round onsisting of twelve multiple hoi e questions is writtenand marked at the individual s hools in early Mar h. Based on these re-sults, ea h s hool typi ally hooses two to four students to parti ipate in theFinal Round at the sponsoring post-se ondary institution in early May. TheFinal Round onsists of ten multiple hoi e questions and �ve questionsrequiring written answers, all of whi h are marked by a panel of fa ultyfrom the sponsoring institution. In addition to the morning ontest writing,Final Round a tivities usually in lude a morning session for tea hers, lun hfor all parti ipants, an afternoon a tivity for the students during marking, andan award eremony. Examples of afternoon student a tivities are le tures bylo al/invited speakers, sports a tivities involving mathemati s, or s avengerhunts. The prizes awarded to winners range from s holarship prizes to ashor book prizes, depending on the sponsoring institution. Note that opies ofthe ontest papers from 1999 to the present, with solutions, are available atthe BCSSMC website at http://people.okanagan.bc.ca/clee/BCSSMC.Following the su ess of the 1995 and 1996 ontests, it was suggestedthat the only improvement would be to have members of more institutionsbe dire tly involved in the ontest preparation pro ess. For the 1997 ontestDon DesBrisay, Kirk Evenrude, and Ja k Bradshaw from Cariboo College;Ni holas Bu k and Edward Dobrowolski from CNC; Dave Murray, JohnGrant M Loughlin, and Clint Lee from OC; Jim Bailey from College of theRo kies (formerly East Kootenay Community College) and Wayne Matthewsfrom Camosun College; met at Cariboo College in Kamloops in August 1996for the inaugural Math Contest Brainstorming Session. Jim Totten had mademost of the lo al arrangements, in luding billets and a party at the end of the�rst day for all of the parti ipants. However, he was absent due to winningan opportunity to play in a pro-am golf event at the Greater Van ouver Open.A year later people ame together in Kamloops again, before shifting to theKalamalka Campus of Okanagan College in Vernon where the session washosted by myself in 1998. Sin e 1999, the Brainstorming Sessions have beenheld in onjun tion with the BCCUPMs arti ulation meetings in mid-May,either before or after the main meeting. The Brainstorming Sessions in ludeabout 8 to 16 representatives from a ross-se tion of institutions a ross theprovin e. Parti ipants usually bring some problems that they have preparedahead of time and materials for generating additional problems as needed.The session is divided into Junior and Senior groups, ea h of whi h preparesrough drafts of Preliminary and Final round papers. Drafts are later typesetinto tentative versions of the ontest papers. These are reviewed and re-vised over several months by as many as twenty reviewers. Finished formsare then produ ed in luding solutions to all of the problems that ultimatelyappear on the ontest papers.From the beginning, until he took the position of Editor-in-Chief ofCRUX with Mayhem, Jim parti ipated a tively in these sessions sharing his

309infe tious enthusiasm for mathemati s and its manifestations in all areas.Jim's humour ontributed to the su ess and the enjoyment experien ed byall parti ipants. He ontributed problems and solutions for the ontest, type-set either the ontest papers or the solutions, and parti ipated in the reviewof the ontest papers as they evolved into their �nished form. His abilities asa proofreader, his attention to detail, and his skills with wording problems learly were invaluable to all of those involved in the pro ess.Also during this period, Jim was a tive in re ruiting institutions to par-ti ipate in the ontest. He made himself available to any institution, es-pe ially smaller institutions just getting started with the ontest, to helpduring the �nal round in any apa ity, espe ially as a speaker. In 2004it was suggested that The Pa i� Institute for the Mathemati al S ien es(PIMS) might be willing to support some of these a tivities. Ri k Brewsterapproa hed PIMS for su h support and as a result PIMS provides supportfor a range of the provin e-wide a tivities asso iated with the ontest. Jim'sre ruiting a tivities brought the number of parti ipating institutions up toa maximum of twelve in one year, with a ore of at least ten institutions.The institutions that have parti ipated in the ontest over the years are:TRU in Kamloops, CNC in Prin e George, OC in Kelowna, Langara Collegein Van ouver, Capilano University in North Van ouver, Camosun College inVi toria, University of the Fraser Valley in Abbotsford, North Island Collegein Campbell River, Malaspina University in Nanaimo, College of the Ro kiesin Cranbrook, Selkirk College in Castlegar, Northwest Community Collegein Prin e Rupert, UBC Okanagan in Kelowna, and Douglas College in NewWestminster. Approximately 2500 grade 8 to 12 students parti ipate in thePreliminary Round of the Contest ea h year and 500 parti ipate again in theFinal Round.While Jim was Editor-in-Chief of CRUX with Mayhem, he s aled ba khis involvement with the ontest. During this period I took over many of theresponsibilities that Jim had undertaken. I typeset the drafts of the ontestpapers, managed the evolution of the ontest papers, saw to the distributionof the papers to the parti ipating institutions, and oversaw and typeset so-lutions. Jim ontinually remained available to provide advi e and ontributehis reviewing skills in doing a �nal run through of the ontest papers. Shortlybefore his death, he onta ted me and indi ated that now that he was wind-ing down his Editor-in-Chief duties and had retired from TRU, he was lookingforward to in reasing his involvement with the ontest.It is lear that without the inspiration and e�orts of Jim Totten, theBritish Columbia Se ondary S hool Mathemati s Contest would not existtoday. His enthusiasm for and ontributions to the ontest were key indeveloping the feelings of involvement and ownership of all of the parti -ipating institutions. Clint LeeOkanagan CollegeVernon, BC

clee@okanagan.bc.ca

310A Duality for Bi entri QuadrilateralsMi hel BatailleIn memory of Jim TottenBi entri quadrilaterals are onvex quadrilaterals that have both an in- ir le and a ir um ir le. As every problem solver probably knows, su h aquadrilateral is easily onstru ted from its future in ir le. Below, we reviewthis property that we formulate as an internal theorem about the determi-nation of a bi entri quadrilateral. The purpose of this note is to state andprove an external theorem whi h is, in a sense, the dual of the internal one.Internal Theorem A bi entri quadrilateral is uniquely determined by threeof its sides tangent to a given ir le.This results immediately from the following well-known hara terization:Let PQRS be a onvex quadrilateral ins ribed in a ir le γ. Thetangents to γ at P , Q, R, and S are the sidelines of a bi entri quadrilateral if and only if QS is perpendi ular to PR.For ompleteness, here is a short proof, in whi h ∠(ℓ, ℓ′) denotes the dire tedangle between lines ℓ and ℓ′, measured modulo π.Proof: Let I be the entre of γ and

A, B, C, and D be the verti es of thequadrilateral formed by the tangents,as in the �gure. Observing that A, P ,I, and S are on y li , we have

∠(AB, AD) = ∠(AP, AS)

= ∠(IP, IS) = 2∠(QP, QS) .Similarly,∠(CB, CD) = 2∠(PQ, PR) ,so that ∠(AB, AD) = ∠(CB, CD) ifand only if ∠(QP, QS) = ∠(PQ, PR)mod π

2, that is, if and only if QS ⊥PR(sin e QS and PR are not parallel).

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γA

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We are guided to the next theorem by onsidering the ir um ir le tobe the dual of the in ir le, and re alling the duality between a tangent to a ir le and its point of tangen y.Copyright c© 2009 Canadian Mathemati al So ietyCrux Mathemati orum with Mathemati al Mayhem, Volume 35, Issue 5

311External Theorem A bi entri quadrilateral is uniquely determined by threeof its verti es lying on a given ir le.As above, the proof rests upon a hara terization of the bi entri quadri-lateral and will provide a onstru tion of this quadrilateral.We begin by showing that the onjun tion of two simple propertiesdistinguishes bi entri quadrilaterals among all quadrilaterals ins ribed in agiven ir le. The following lemma is inspired by (and is an extension of) aproblem posed in this journal (see [2℄, [3℄).Lemma Let ABCD be a onvex quadrilateral ins ribed in a ir le Γ withdiagonals AC and BD meeting at X. Then ABCD is bi entri if and only if(a) B and D are on the same side of the perpendi ular bise tor of AC, and(b) BX =

(cos2

B

2

)BD.Proof: First, we assume that ABCD is bi- entri . Then,

BA − BC = DA − DC (1)so that BA−BC and DA−DC ertainlyhave the same sign and (a) holds.Let us denote area by [·] and for simpli -ity let AB = a, BC = b, CD = c,DA = d, and AC = e. Observing thatD = ∠ADC = π − B, we have

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Γ

A

B

C

D

X

q

q q

q

BX

BD=

[ABX ]

[ABD]=

[CBX ]

[CBD]=

[ABC]

[ABCD]=

ab sin B

ab sin B + cd sin D=

ab

ab + cd.Furthermore, from the Law of Cosines we dedu e that

cos B =a2 + b2 − e2

2ab=

e2 − c2 − d2

2cd=

a2 + b2 − c2 − d2

2(ab + cd).Squaring ea h side of (1), we obtain a2 + b2 − c2 − d2 = 2(ab − cd) and so

cos B =ab − cd

ab + cd= 2 · BX

BD− 1 and (b) follows.Conversely, if (b) holds, then (a − b)2 = (c − d)2 follows readily fromthe above al ulations. Sin e a − b and d − c have the same sign by part (a),we have a − b = d − c, that is, AB + CD = BC + DA, whi h implies that

ABCD has an in ir le (see [1℄).For a proof of the external theorem, onsider a triangle ABC with ir- um ir le Γ. We are looking for a point D on Γ su h that ABCD is bi entri .Let Y be the foot of the perpendi ular from B to AC, and let Y ′ be the pointsu h that −−→BY =

(cos2

B

2

)−−→BY ′; equivalently, −−→

Y Y ′ =(tan2 B

2

)−−→BY .

312The lemma implies that the desired point D must lie both on Γ and onthe line m through Y ′ perpendi ular to

BY and on the same side as B of the per-pendi ular bise tor n ofAC. It remains toshow that m ne essarily ontains a pointof Γ. To this end, let n interse t AC at Mand Γ at U and V , where U and B are onthe same ar AC.Now ∠AUC = B and AV CU is y li , hen e ∠AV C = π−B and we havethe relations UM =1

2

(cot

B

2

)AC and

V M =1

2

(tan

B

2

)AC. Observing that

BY ≤ UM , we obtain....................................................................................................................................................................................................................................................

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Γ

U

V

B

CAM Y

Y ′

n

m

Y Y ′ ≤(tan2 B

2

)· 1

2

(cot

B

2

)AC =

1

2

(tan

B

2

)AC = V M .It follows that m lies between AC and the tangent to Γ at V . In other words,ex ept if B = U , m interse ts Γ intwo points, one of whi h is the desiredpoint D. The lemma then ensures that

ABCD is bi entri .To on lude, the �gure at rightshows how to qui kly onstru t D us-ing K on the bise tor BV of ∠ABCwith KC perpendi ular to BC, sin eBC

BK= cos

B

2=

BK

BC ′implies that

BC

BC ′= cos2

B

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Γ

CA

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C′

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V

D

A knowledgment. The author would like to thank the referee for his arefulreading of the manus ript and for a suggestion whi h greatly improved theoriginal proof of the external theorem.Referen es[1℄ N. Altshiller-Court, College Geometry, Dover reprint (2007), p. 135.[2℄ D.J. Smeenk, Problem 2027, Crux Mathemati orum, 21 (1995) p. 90;Solution in 22 (1996), p. 94.[3℄ Problem 3211, Crux Mathemati orum with Mathemati al Mayhem, 33(2007) p. 43; Solution in 34 (2008), p. 61. Mi hel Bataille12 rue Sainte-Catherine76000 Rouen, Fran emichelbataille@wanadoo.fr

313A Study of Knight's Tours on the Surfa e ofa CubeAwani Kumar

1 IntroductionThe lassi al puzzle of a knight's tour was, is, and will always be fas inat-ing. The reason is its simpli ity as well as its omplexity, whi h have endless harm. For enturies, the traditional study of the knight's tour was mostly on�ned to square boards. Later, S hubert [1℄, Gibbins [2℄, and Stewart[3℄ delved into the knight's tour puzzle in 3-dimensional spa e. More re- ently, Kumar [4℄ looked into the possibilities of knight's tours in ubesand uboids having magi properties. Awani Kumar, Fran is Gaspalou, andGuenter Stertenbrink have dis overed a magi knight's tour inside a 4×4×4 ube. However, a lot remains to be explored and dis overed even on its sur-fa e. A tour of a knight on a square board is alled a magi tour if the sumof the numbers in ea h row and olumn is the same (the magi onstant).Perusal of the literature reveals that knight's tours have been onstru ted onthe surfa es of ubes smaller or larger than 4 × 4 × 4 but, astonishingly, noton a 4 × 4 × 4 ube itself. We know that a knight's tour is not possible on a4 × 4 board, but is it possible on the surfa e of a 4 × 4 × 4 ube? Will su htours be open or losed? Can they have magi properties? What about magi tours on the surfa es of larger ubes? We shall look at these questions.2 Knight’s tours on surfaces of small cubesThe reader an visualize the knight's move on the fa e of a ube in a mu hbetter way when it is unfolded, as shown in Figure 1. On a onventionalboard a knight an over only up to eight squares. However, this need notbe the ase when it is moving on the surfa e of a ube or uboid. In fa t, it an over up to 10 squares, as shown in Figure 2a. Some readers may �ndthis strange be ause the onse utive knight's moves are not equidistant, ason a onventional board. Well, don't worry. It is be ause a ube an beunfolded along its edges in many di�erent ways. Look at Figure 2b and theknight's move to the square labelled \6" is lear. If it is ever un lear, thenappropriately unfold a ube along its edges using a pair of s issors!A knight's tour is even possible on the smallest possible ube measuring1 × 1 × 1, as shown in Figure 3. The dis erning reader will observe that herethe numbers are arranged as on a onventional die. That is, the numbers onopposite fa es add up to 7.Copyright c© 2009 Canadian Mathemati al So ietyCrux Mathemati orum with Mathemati al Mayhem, Volume 35, Issue 5

314

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10 1

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9 Kt. 5

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Figure 5

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

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

63 90 11 30

10 31 62 91

89 64 29 12

32 9 92 61

51 22 33 88 7 28 65 94 13 60 79 42

36 85 50 23 66 93 8 27 78 43 16 57

21 52 87 34 25 6 95 68 59 14 41 80

86 35 24 49 96 67 26 5 44 77 58 15

1 48 69 76

70 73 4 45

47 2 75 72

74 71 46 3

19 54 39 82

38 83 18 55

53 20 81 40

84 37 56 17

Figure 8

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87 66 11 30

10 31 86 67

65 88 29 12

32 9 68 85

21 76 33 64 89 28 7 70 13 84 55 42

62 35 24 73 8 69 90 27 54 43 16 81

75 22 63 34 25 92 71 6 83 14 41 56

36 61 74 23 72 5 26 91 44 53 82 15

93 46 1 52

4 49 94 45

47 96 51 2

50 3 48 95

19 78 57 40

60 37 18 79

77 20 39 58

38 59 80 17

Figure 9

315Watkins [5℄ and Pi kover [6℄ have given a knight's tour on the surfa eof a 2×2×2 ube. Figure 4 and Figure 5 are two examples of losed and opentours, respe tively, on the surfa e of a 2 × 2 × 2 ube. There are millions ofsu h tours and, therefore, these are of little interest. However, tours havingmagi properties are a di�erent story. Figure 6 and Figure 7 are open and losed tours, respe tively, having magi properties. In these tours, the sumof the numbers on ea h fa e is 50, the magi sum. There are thousands ofsu h tours. In fa t, 2 × 2 × 2 is the smallest ube on whi h su h interestingtours with magi fa es are possible. But how an su h tours be onstru ted?Well, the se ret lies in the systemati movement of the knight over the fa esof the ube. Altogether, there are 24 squares. Put them into four groups:

1 to 6, 7 to 12, 13 to 18, and 19 to 24. Start from any fa e and omplete theknight's tour in su h a way that ea h fa e has numbers from all the groups.With a little patien e and perseveran e, tours with magi properties an be onstru ted.Before pro eeding any further, let us prove the following theorem.Theorem For odd n, a knight's tour on the surfa e of an n×n×n ube annothave a magi sum on its fa es.Proof: Assume there is a magi sum. Altogether, there are 6n2 numbers,whi h sum up to 3n2(6n2 + 1). The magi sum (the sum on ea h fa e) isthus n2(6n2 + 1)

2. Sin e n is odd, the numerator is odd. Thus, the sum ofthe numbers on ea h fa e is not an integer, a ontradi tion.

3 Knight’s tours on the surface of a 4×4×4 cubeProfessional and amateur mathemati ians have been studying the knight'stour inside a 4 × 4 × 4 ube for over a entury but, astonishingly, no one haslooked for them on its surfa e! There are billions of tours on the surfa e of a4 × 4 × 4 ube, and onsequently these are of little interest. However, tourshaving magi properties are, again, a di�erent story. We have seen that thefa es of a 2×2×2 ube an have magi properties and this is true for tours ona 4×4×4 ube too. In fa t, this is true for all even-sided ubes. Later, we willsee that a doubly-even ube (say, 8×8×8) an have both rows and olumnsmagi but a singly-even ube (say, 6×6×6) has only the rows or the olumns(but not both) magi . More interesting and hallenging is the onstru tion oftours in whi h all rows and olumns have magi properties. Figure 8 is onesu h tour in whi h �ve fa es are magi in rows and olumns and only 4 lines(2 rows and 2 olumns) are just a hair's breadth away from the magi on-stant. The dis erning reader must have observed that these tours are madeup of regular quads. Here also the numbers are arranged in four groups:1 to 24, 25 to 48, 49 to 72, and 73 to 96. All these groups are represented inea h row and ea h olumn of the fa es. The same holds true for the four 2×2squares on ea h of the fa es. Figure 9 shows a reentrant tour. Curiously, it

316 an be onverted to a magi tour of two knights by inter hanging the squares46 and 48. One knight overs the squares 1 to 48 and the other overs 49 to96. It is a four-fold y li almost magi tour. That is, it retains its almostmagi properties (44 magi lines) when starting from the square 25, 49, or73. Readers are en ouraged to improve on this.4 Knight’s tours on the surface of a 6×6×6 cubeThe knight's tour on the surfa e of 6 × 6 × 6 ube has gotten s antily littleattention. Perusal of the literature reveals that, in spite of billions of possibletours, only Watkins [5℄ has given a solution found by Arden Rzewni ki andJesse Howard. The author has enumerated 88 semimagi tours (only rows or olumns are magi ) on a 6 × 6 board and by arefully sele ting and linkingthese tours, one an obtain hundreds of reentrant tours on the surfa e of a6 × 6 × 6 ube. One su h example is shown in Figure 10. Sin e it is a losedtour, by sele ting a suitable starting point (square 19 as 1, square 20 as 2, andso on), we an get a tour having up to 12magi lines with magi onstant 651.In fa t, by hoosing any of the six starting point squares 19, 55, 91, 127, 163,or 199, we an get tours having 12 magi lines. Jelliss [7℄ has proved thatthere annot be magi tours on singly-even boards (that is, 6×6, 10×10, andso forth). But what about these on the surfa es of singly-even ubes? Theauthor onje tures that they do not exist. The best the author ould a hieveis shown in Figure 11. All six fa es there are semimagi and altogether thereare 40 magi lines. So, the magi sum has been a hieved up to 55%. Readersare hallenged to improve on this.5 Knight’s tours on the surface of an 8×8×8 cubeThis is the oldest problem of the lot, over 200 years old. Dudeney [8℄ writesthat the problem was raised by Vandermonde, an 18th entury musi ian andmathemati ian. Dudeney gave a solution by ompleting a tour on a fa e andthen pro eeding to the next fa e. Subsequent writers, namely Petkovi [9℄,Pi kover [6℄, and Watkins [5℄ have merely reprodu ed Dudeney's solution,giving an erroneous impression it is the only possible tour. In fa t, there aretrillions of su h tours. Using powerful omputers and intelligent programs,the international team of Ma kay, Meyrigna , and Stertenbrink [10℄ enu-merated all of the 280magi tours on the 8×8 board. By judi iously sele tingand linking these tours, one an get hundreds of tours and by rearranging theknight's tour, we an get magi tours on all six fa es (magi onstant= 1540).Figure 12 is one su h tour; all its rows and olumns are magi . The diagonalsums are twi e the magi onstant on two of the fa es but, in spite of intensee�ort, the author ould not obtain any diagonally magi fa e or a reentrantmagi tour. Readers are en ouraged to look for them.

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

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

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

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

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

213 190 187 204 201 196

186 205 212 197 188 203

191 214 189 202 195 200

206 185 198 211 182 209

215 192 183 208 199 194

184 207 216 193 210 181

69 46 59 44 57 52 1 10 35 28 25 12 172 147 180 157 174 145

60 43 68 53 66 37 34 29 2 11 8 27 179 156 173 146 163 158

47 70 45 58 51 56 3 36 9 26 13 24 148 171 162 175 152 167

42 61 54 67 38 65 30 33 20 5 16 7 155 178 153 166 159 164

71 48 63 40 55 50 21 4 31 18 23 14 170 149 176 161 168 151

62 41 72 49 64 39 32 19 22 15 6 17 177 154 169 150 165 160

88 93 80 95 82 105

73 102 89 104 79 96

92 87 94 81 106 83

101 74 103 90 97 78

86 91 76 99 84 107

75 100 85 108 77 98

124 109 128 137 122 111

129 138 123 110 127 136

116 125 130 139 112 121

131 140 117 126 135 144

118 115 142 133 120 113

141 132 119 114 143 134

Figure 10

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

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

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

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

123 116 101 94 125 118

100 95 124 117 102 93

115 122 99 108 119 126

96 107 112 121 92 103

111 114 105 98 109 120

106 97 110 113 104 91

195 30 25 32 197 184 1 212 201 10 13 210 136 81 142 75 144 73

24 189 196 185 26 33 18 11 2 211 202 9 141 90 135 82 127 76

29 194 31 190 183 198 3 200 213 12 209 14 134 137 80 143 74 83

188 23 186 27 34 19 214 17 4 205 8 203 89 140 87 130 77 128

193 28 21 36 191 182 199 206 215 6 15 208 86 133 138 79 84 131

22 187 192 181 20 35 216 5 16 207 204 7 139 88 85 132 129 78

180 165 38 47 178 49

37 52 179 50 39 46

164 173 166 45 48 177

53 44 51 172 169 40

174 163 42 167 176 171

43 54 175 170 41 168

160 157 162 55 60 57

155 68 159 58 149 62

158 161 156 61 56 59

67 154 69 148 63 150

70 147 152 65 72 145

153 66 71 146 151 64

Figure 11

318

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

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

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

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

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

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

167 170 183 220 185 222 195 198

182 219 168 171 194 197 186 223

169 166 217 184 221 188 199 196

218 181 172 165 200 193 224 187

179 216 201 208 173 164 189 210

204 207 180 213 192 209 174 161

215 178 205 202 163 176 211 190

206 203 214 177 212 191 162 175

39 54 331 348 333 350 35 50 15 354 17 384 11 358 23 378 242 159 144 225 240 255 146 129

330 347 38 53 36 51 334 351 18 383 14 355 22 379 10 359 227 142 241 160 145 130 239 256

55 40 345 332 349 34 49 336 353 16 381 20 357 12 377 24 158 243 226 143 254 237 132 147

346 329 56 37 52 335 352 33 382 19 356 13 380 21 360 9 141 228 157 244 131 148 253 238

41 58 327 344 323 62 337 48 3 366 29 372 5 364 25 376 156 245 140 229 152 249 236 133

328 343 42 57 340 45 322 63 32 369 4 365 28 373 8 361 139 230 153 248 135 234 149 252

59 326 341 44 61 324 47 338 367 2 371 30 363 6 375 26 246 155 232 137 250 151 134 235

342 43 60 325 46 339 64 321 370 31 368 1 374 27 362 7 231 138 247 154 233 136 251 150

66 83 302 317 300 315 70 87

303 318 67 84 69 86 299 314

320 65 82 301 316 297 88 71

81 304 319 68 85 72 313 298

94 291 80 305 296 311 74 89

79 306 93 292 73 90 295 312

290 95 308 77 92 293 310 75

307 78 289 96 309 76 91 294

102 283 268 117 288 97 270 115

267 118 101 284 269 116 287 98

120 103 282 265 100 285 114 271

281 266 119 104 113 272 99 286

106 121 264 277 260 127 112 273

263 280 105 124 109 274 257 128

122 107 278 261 276 259 126 111

279 262 123 108 125 110 275 258

Figure 12

3196 Conclusion and an AcknowledgementWe have seen that a knight's tour is possible on the surfa es of ubes ofvarious sizes. Cubes of size 4 × 4 × 4 and larger an have magi rows andmagi olumns on their fa es.The author has felt that in India, getting the referen es is more diÆ ultthan dis overing a magi tour on a ube. The author is grateful to TakayaIwamoto for providing photo opies of [2℄ and [3℄.Referen es[1℄ H. S hubert (1904); Mathematis he Mussestunden Eine Sammlung vonGeduldspielen, Kunstst �u ken und Unterhaltungs-aufgaben mathema-tis her Natur. (Leipzig), p. 235-7, 4×4×4 losed tours, p. 238, 3×4×6 losed tour. [Mathemati al Asso iation Library, Lei ester University℄[2℄ N.M. Gibbins (1944); Chess in 3 and 4 dimensions, Mathemati alGazette, May 1944, pp. 46-50.[3℄ J. Stewart (1971); Solid Knight's Tours, Journal of Re reational Math-emati s Vol. 4, No. 1, January 1971, p. 1.[4℄ A. Kumar (2006); Studies in Tours of the Knight in Three Dimensions,The Games and Puzzles Journal #43 (electronic), January-April 2006.[5℄ J.J. Watkins (2005); A ross the Board, The Mathemati s of Chess-board Problems, Prin eton University Press, 2005, pp. 10-11, pp. 88-89, pp. 93-94.[6℄ C.A. Pi kover (2004); The Zen of Magi Squares, Cir les, and Stars,Prin eton University Press, 2004, pp. 214-216.[7℄ G.P. Jelliss (2003); Existen e Theorems onMagi Tours, The Games andPuzzles Journal #25 (electronic), January-Mar h 2003.[8℄ H.E. Dudeney (1970); Amusements in Mathemati s, Dover Publi a-tions, New York, 1970, p. 103, p. 229.[9℄ M. Petkovi (1997); Mathemati s and Chess, Dover Publi ations, NewYork, 1997, p. 53, p. 64.[10℄ G. Stertenbrink (2003); Computing Magi Knight Tours,

http://magictour.free.fr/ Awani Kumar2/11-C, Vijayant Khand Gomti NagarLu know 226 010Indiaawanieva@gmail.com

320TOTTEN PROBLEMSThe following problems have all been dedi ated by the proposers to the lastingmemory of Jim Totten. Solutions should arrive no later than 1 Mar h 2010. Anasterisk (⋆) after a number indi ates that a problemwas proposed without a solution.The editor thanks Jean-Mar Terrier of the University of Montreal for transla-tions of the problems.

TOTTEN{01. Proposed by Cosmin Pohoat� �a, Tudor Vianu NationalCollege, Bu harest, Romania.Let H be the ortho entre of triangle ABC and let P be the se ondinterse tion of the ir um ir le of triangle AHC with the internal bise tor of∠BAC. If X is the ir um entre of triangle APB and if Y is the ortho entreof triangle APC, prove that the length of XY is equal to the ir umradiusof triangle ABC.TOTTEN{02. Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.Let k ≥ 2 be an integer and let f : [0, ∞) → R be a bounded on-tinuous fun tion. If x is a positive real number, �nd the value of

limn→∞

k√

n

∫ x

0

f(t)(1 + tk

)n dt .TOTTEN{03. Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.Let f : [0, 1] → R be a ontinuous fun tion. Find the limit

limn→∞

∫ 1

0

· · ·∫ 1

0

f

n1

x1

+ · · · +1

xn

dx1 . . . dxn .

TOTTEN{04. Proposed by Juan-Bos o Romero M�arquez, Universidadde Valladolid, Valladolid, Spain.Suppose that 0 < a < b, m0 =√

ab, and m1 =a + b

2. If x ≥ 0, provethat

x

m1(x + m1)≤ 1

b − alog

b(x + a)

a(x + b)≤ x

m0(x + m0).

TOTTEN{05. Proposed by Mi hel Bataille, Rouen, Fran e.Let I be the in entre of triangle ABC. Let the point A′ be su h that−−→AA′ = (cos A)

−→AI, and let points B′ and C′ be de�ned similarly. Find theradius of the ir le passing through A′, B′, and C′ and lo ate its entre.

321TOTTEN{06. Proposed by Bill Sands, University of Calgary, Calgary,AB. Jim and three of his buddies played a round of golf. As usual, Jim wonthe game. In fa t, he beat every two of his three buddies, in the followingsense. Let his three buddies be A, B, and C and let ai be A's s ore on hole i,for all 1 ≤ i ≤ 18, and similarly de�ne bi and ci. Set Sab =

18∑i=1

min(ai, bi),and similarly de�ne Sac and Sbc. Then Jim's total s ore was less than Sab,Sac, andSbc. However, Jim's s ore wasmore thanSabc =

18∑i=1

min(ai, bi, ci).Jim's s ore was 72. What was the minimum possible s ore of any of hisbuddies?TOTTEN{07. Proposed by �Sefket Arslanagi � and Faruk Zejnulahi, Uni-versity of Sarajevo, Sarajevo, Bosnia and Herzegovina.Let a, b, and c be nonnegative real numbers su h that a2 +b2 +c2 = 1.Prove or disprove that(a) 1 ≤ a

1 − ab+

b

1 − bc+

c

1 − ca≤ 3

√3

2,

(b) 1 ≤ a

1 + ab+

b

1 + bc+

c

1 + ca≤ 3

√3

4.

TOTTEN{08. Proposed by Ri hard Hoshino, Government of Canada,Ottawa, ON.In triangle ABC suppose that AB < AC. Let D and M be the pointson side BC for whi h AD is the angle bise tor and AM is the median. LetF be on side AC so that AD is perpendi ular to DF . Finally, let E be theinterse tion of AM and DF . Prove that AB · DE + AB · DF = AC · EF .TOTTEN{09. Proposed by Ri hard Hoshino, Government of Canada,Ottawa, ON.Let n and k be integers with n ≥ 2 and k ≥ 0. Consider n dinner guestssitting around a ir ular table. Let gn(k) be the number of ways that k ofthese n guests an be hosen so that no two hosen guests are sitting next toone another. To illustrate, g6(0) = 1, g6(1) = 6, g6(2) = 9, g6(3) = 2, andg6(4) = 0 for all k ≥ 4. For ea h n ≥ 2, let

fn(x) =∑

k≥0

gn(k)xk .For example, f6(x) = 1+6x+9x2+2x3 = (1+2x)

(1+4x+x2

). Determineall n for whi h (1 + 2x) is a fa tor of fn(x).

322TOTTEN{10. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.Determine all triangles ABC whose side lengths are positive integersand su h that cos C =

4

5.TOTTEN{11. Proposed by Walther Janous, Ursulinengymnasium, Inns-bru k, Austria.(a) Let x, y, and z be positive real numbers su h that x+y+z = 1. Provethat

8√

3

9≤(

1√x

− √x

)(1

√y

− √y

)(1√z

− √z

) .(b) ⋆. Let n ≥ 2 and let x1, x2, . . . , xn be positive real numbers su hthat x1 + x2 + · · · + xn = 1. Prove or disprove that

(n − 1√

n

)n

≤n∏

k=1

(1

√xk

− √xk

) .TOTTEN{12. Proposed by Mih�aly Ben ze, Brasov, Romania.Let w, x, y, and z be positive real numbers with w+x+y+z = wxyz,and let

f(x) =3

√1

2+

√1

4− 1

x3+

3

√1

2−√

1

4− 1

x3.Prove that 3

√wxy + 3

√xyz + 3

√yzw + 3

√zwx ≥ f(w)+f(x)+f(y)+f(z)..................................................................TOTTEN{01. Propos �e par Cosmin Pohoat� �a, Coll �ege National Tudor Vianu,Bu arest, Roumanie.Soit H l'ortho entre du triangle ABC et soit P la se onde interse tiondu er le ir ons rit du triangle AHC ave la bisse tri e int �erieure de l'angle

BAC. Si X est le entre du er le ir ons rit du triangle APB et Y l'ortho- entre du triangle APC, montrer que la longueur de XY est �egale au rayondu er le ir ons rit du triangle ABC.TOTTEN{02. Propos �e par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie.Soit k ≥ 2 un entier et f : [0, ∞) → R une fon tion ontinue et born �ee.Pour un nombre r �eel positif x, trouver la valeur delim

n→∞

k√

n

∫ x

0

f(t)(1 + tk

)n dt .

323TOTTEN{03. Propos �e par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie.Soit f : [0, 1] → R une fon tion ontinue. Trouver la limite

limn→∞

∫ 1

0

· · ·∫ 1

0

f

n1

x1

+ · · · +1

xn

dx1 . . . dxn .

TOTTEN{04. Propos �e par Juan-Bos o Romero M�arquez, Universit �e deValladolid, Valladolid, Espagne.On suppose que 0 < a < b, m0 =√

ab et m1 =a + b

2. Si x ≥ 0,montrer que

x

m1(x + m1)≤ 1

b − alog

b(x + a)

a(x + b)≤ x

m0(x + m0).

TOTTEN{05. Propos �e par Mi hel Bataille, Rouen, Fran e.Soit I le entre du er le ins rit du triangle ABC. Soit A′ le point telque −−→AA′ = (cos A)

−→AI, et soit B′ et C′ les points d �e�nis de mani �ere ana-logue. Trouver le rayon du er le passant parA′,B′ etC′ ainsi que la positionde son entre.TOTTEN{06. Propos �e par Bill Sands, Universit �e de Calgary, Calgary, AB.Jim et trois de ses opains ont jou �e une ronde de golf. Comme d'ha-bitude, Jim a gagn �e la partie. En fait, il a battu haque paire de ses trois opains, dans le sens suivant. Soit A, B et C ses trois opains et soit ai les ore de A au trou i pour tout 1 ≤ i ≤ 18 ; on d �e�nit de meme bi et ci. Soit

Sab =18∑

i=1

min(ai, bi), et de mani �ere analogue, Sac et Sbc. Le s ore totalde Jim a �et �e plus petit que Sab, Sac et Sbc. Par ontre, son r �esultat de 72d �epassait Sabc =18∑

i=1

min(ai, bi, ci). Quel �etait le s ore minimal possible de ha un de ses opains ?TOTTEN{07. Propos �e par �Sefket Arslanagi � et Faruk Zejnulahi, Univer-sit �e de Sarajevo, Sarajevo, Bosnie et Herz �egovine.Soit a, b et c trois nombres r �eels non n �egatifs tels que a2 +b2 +c2 = 1.Etudier la validit �e de(a) 1 ≤ a

1 − ab+

b

1 − bc+

c

1 − ca≤ 3

√3

2,

(b) 1 ≤ a

1 + ab+

b

1 + bc+

c

1 + ca≤ 3

√3

4.

324TOTTEN{08. Propos �e par Ri hard Hoshino, Gouvernement du Canada,Ottawa, ON.Supposons que dans le triangle ABC, on a AB < AC. Soit D etM les points sur le ot �e BC pour lesquels AD est la bisse tri e et AMla m �ediane. Soit F le point sur le ot �e AC tel que AD soit perpendi ulaire �aDF . Soit �nalement E l'interse tion de AM et DF . Montrer que AB · DE+ AB · DF = AC · EF .TOTTEN{09. Propos �e par Ri hard Hoshino, Gouvernement du Canada,Ottawa, ON.Soit n et k deux entiers tels que n ≥ 2 et k ≥ 0. On a n hotes �a d�ner,assis autour d'une table ronde. Soit gn(k) le nombre de possibilit �es que kde es n hotes puissent etre hoisis de telle sorte qu'au une paire d'hotes hoisis soit assis l'un �a ot �e de l'autre. Par exemple, g6(0) = 1, g6(1) = 6,g6(2) = 9, g6(3) = 2, et g6(4) = 0 pour tout k ≥ 4. Pour tout n ≥ 2, soit

fn(x) =∑

k≥0

gn(k)xk .Par exemple, f6(x) = 1+6x+9x2+2x3 = (1+2x)

(1+4x+x2

). D �eterminertous les n pour lesquels (1 + 2x) est un fa teur de fn(x).TOTTEN{10. Propos �e par D.J. Smeenk, Zaltbommel, Pays-Bas.Trouver tous les triangles ABC dont la longueur des ot �es sont desentiers positifs et tels que cos C = 45.TOTTEN{11. Propos �e par Walther Janous, Ursulinengymnasium, Inns-bru k, Autri he.(a) Soit x, y et z trois nombres r �eels positifs tels que x + y + z = 1.Montrer que

8√

3

9≤(

1√x

− √x

)(1√y

− √y

)(1√z

− √z

) .(b) ⋆. Soit n ≥ 2 et soit x1, x2, . . . , xn n nombres r �eels positifs tels quex1 + x2 + · · · + xn = 1. Etudier la validit �e de

(n − 1√

n

)n

≤n∏

k=1

(1√xk

− √xk

) .TOTTEN{12. Propos �e par Mih�aly Ben ze, Brasov, Roumanie.Soit w, x, y et z des nombres r �eels positifs ave w+x+y+z = wxyz,et soit

f(x) =3

√1

2+

√1

4− 1

x3+

3

√1

2−√

1

4− 1

x3.Montrer que 3

√wxy+ 3

√xyz+ 3

√yzw+ 3

√zwx ≥ f(w)+f(x)+f(y)+f(z).

325PROBLEMSSolutions to problems in this issue should arrive no later than 1 Mar h 2010.An asterisk (⋆) after a number indi ates that a problem was proposed without asolution.Ea h problem is given in English and Fren h, the oÆ ial languages of Canada.In issues 1, 3, 5, and 7, English will pre ede Fren h, and in issues 2, 4, 6, and 8,Fren h will pre ede English. In the solutions' se tion, the problem will be stated inthe language of the primary featured solution.Problems 3451, 3452, 3453, and 3454 below are dedi ated by the proposers tothe memory of Jim Totten.The editor thanks Jean-Mar Terrier of the University of Montreal for transla-tions of the problems.

3451. Proposed by Jos �e Luis D��az-Barrero, Universitat Polit �e ni a deCatalunya, Bar elona, Spain.Let (X, < ·, · >) be a real or omplex inner produ t spa e and let x, y,and z be nonzero ve tors in X. Prove that∑ y li ∣∣∣∣< z, x > < x, y >

||x||

∣∣∣∣1/2

≤∑ y li ( ||x||

||y|| ||z||

)1/2

|< y, z >| .3452. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands.Prove the following and generalize these results.(a) tan2 36◦ + tan2 72◦ = 10,(b) tan4 36◦ + tan4 72◦ = 90,( ) tan6 36◦ + tan6 72◦ = 850,(d) tan8 36◦ + tan8 72◦ = 8050.3453. Proposed by S ott Brown, Auburn University, Montgomery, AL,USA. Triangle ABC has side lengths a = BC, b = AC, c = AB; andaltitudes ha, hb, hc from the verti es A, B, C, respe tively. Prove that8

∑ y li h2

a(hb + hc)

+ 16hahbhc ≤ 3

√3

∑ y li a2(b + c)

+ 6

√3abc .

3263454. Proposed by Ri hard Hoshino, Government of Canada, Ottawa,ON. Let a, b, c, and d be positive real numbers su h that a + b+ c + d = 1.Prove that

a3

b + c+

b3

c + d+

c3

d + a+

d3

a + b≥ 1

8.

3455. Proposed by Mi hel Bataille, Rouen, Fran e.Find the minimum value of x2 +y2 +z2 over all triples (x, y, z) of realnumbers su h that13x2 + 40y2 + 45z2 − 36xy + 12yz + 24xz ≥ 2009and hara terize all the triples at whi h the minimum is attained.3456. Proposed by Mi hel Bataille, Rouen, Fran e.Given a triangle ABC with ir um ir le Γ, let ir le Γ′ entred on theline BC interse t Γ at D and D′. Denote by Q and Q′ the proje tions of Dand D′ on the line AB, and by R and R′ their proje tions on AC; assumethat none of these proje tions oin ide with a vertex of the triangle.Show that ifΓ′ is orthogonal to Γ, then BQ

BQ′=

CR

CR′. Does the onversehold?3457. Proposed by Mi hel Bataille, Rouen, Fran e.Let An =

⌊n/2⌋∑j=0

(−3)j

2j + 1aj, where aj =

n∑k=2j

(k2j

)k + 1

2kand n is a posi-tive integer. Prove that A1 + A2 + · · · + An ≥ n with equality for in�nitelymany n.3458. Proposed by Bru e Shawyer, Memorial University of Newfound-land, St. John's, NL.Determine the entral angle of a se tor, su h that the square drawnwithone vertex on ea h radius of the se tor and two verti es on the ir umferen e,has area equal to the square of the radius of the se tor.3459. Proposed by Zafar Ahmed, BARC, Mumbai, India.Let a, b, c and p, q, r be positive real numbers. Prove that if q2 ≤ prand r2 ≤ pq, then

a

pa + qb + rc+

b

pb + qc + ra+

c

pc + qa + rb≤ 3

p + q + r.When does equality hold?

3273460. Proposed by Tran Quang Hung, student, Hanoi National Univer-sity, Vietnam.The triangle ABC has ir um entre O, ortho entre H, and ir umra-dius R. Prove that

3R − 2 OH ≤ HA + HB + HC ≤ 3R + OH3461. Proposed by Tran Quang Hung, student, Hanoi National Univer-sity, Vietnam.Let I be the in entre of triangle ABC and let A′, B′, and C′ be theinterse tions of the rays AI, BI, and CI with the respe tive sides of thetriangle. Prove thatIA + IB + IC ≥ 2(IA′ + IB′ + IC′) .3462. Proposed by Sotiris Louridas, Aegaleo, Gree e.Let x, y, and z be positive real numbers su h that

(x3 + z3 − y3

) (y3 + x3 − z3

) (z3 + y3 − x3

)> 0 .Prove that

(x3+ y3+ z3+ 3xyz

) ∏ y li (x3+ y3− z3+ xyz)

≤ 3∏ y li 3

√x4 (x2 + yz)

4 ..................................................................3451. Propos �e par Jos �e Luis D��az-Barrero, Universit �e Polyte hnique deCatalogne, Bar elone, Espagne.Soit (X, < ·, · >) un espa e ve toriel r �eel ou omplexe muni d'un pro-duit s alaire et soit x, y et z des ve teurs non nuls de X. Montrer que

∑ y lique ∣∣∣∣< z, x > < x, y >

||x||

∣∣∣∣1/2

≤∑ y lique( ||x||

||y|| ||z||

)1/2

|< y, z >| .3452. Propos �e par D.J. Smeenk, Zaltbommel, Pays-Bas.Montrer les �egalit �es suivantes et g �en �eraliser es r �esultats.(a) tan2 36◦ + tan2 72◦ = 10,(b) tan4 36◦ + tan4 72◦ = 90,( ) tan6 36◦ + tan6 72◦ = 850,(d) tan8 36◦ + tan8 72◦ = 8050.

3283453. Propos �e par S ott Brown, Universit �e Auburn, Montgomery, AL,USA. Soit a = BC, b = AC, c = AB les longueurs des ot �es du triangleABC, de hauteurs respe tives ha, hb, hc issues des sommets A, B, C. Mon-trer que8

∑ y lique h2

a(hb + hc)

+ 16hahbhc ≤ 3

√3

∑ y lique a2(b + c)

+ 6

√3abc .

3454. Propos �e par Ri hard Hoshino, Gouvernement du Canada, Ottawa,ON. Soit a, b, c et d des nombres r �eels non n �egatifs ave a + b + c + d = 1.Montrer quea3

b + c+

b3

c + d+

c3

d + a+

d3

a + b≥ 1

8.

3455. Propos �e by Mi hel Bataille, Rouen, Fran e.Trouver la valeur minimale de x2+y2+z2 pour tous les triplets (x, y, z)de nombres r �eels tels que13x2 + 40y2 + 45z2 − 36xy + 12yz + 24xz ≥ 2009et ara t �eriser tous les triplets pour lesquels le minimum est atteint.3456. Propos �e par Mi hel Bataille, Rouen, Fran e.�Etant donn �e un triangle ABC et Γ son er le ir ons rit, soit Γ′ un er le entr �e sur la droiteBC et oupant Γ en D etD′. D �esignons parQ etQ′les proje tions de D et D′ sur la droite AB, et soit R et R′ leurs proje tionssur AC ; supposons qu'au une de es proje tions ne oin ide ave un sommetdu triangle.Montrer que si Γ′ est orthogonal �a Γ, alors BQ

BQ′=

CR

CR′. La r �e iproqueest-elle valide ?3457. Propos �e par Mi hel Bataille, Rouen, Fran e.Soit An =

⌊n/2⌋∑j=0

(−3)j

2j + 1aj, o �u aj =

n∑k=2j

(k2j

)k + 1

2ket n est un entierpositif. Montrer que A1 + A2 + · · · + An ≥ n o �u l'on a �egalit �e pour unein�nit �e de n.

3293458. Propos �e by Bru e Shawyer, Universit �e Memorial de Terre-Neuve,St. John's, NL.D �eterminer l'angle au entre d'un se teur tel que le arr �e, onstruit ave un sommet sur haque rayon du se teur et deux sommets sur la ir onf �eren e,poss �ede une aire �egale au arr �e du rayon du se teur.3459. Propos �e par Zafar Ahmed, BARC, Mumbai, Inde.Soit a, b, c et p, q, r des nombres r �eels positifs. Montrer que si q2 ≤ pret r2 ≤ pq, alors

a

pa + qb + rc+

b

pb + qc + ra+

c

pc + qa + rb≤ 3

p + q + r.Quand y a-t-il �egalit �e ?3460. Propos �e par Tran Quang Hung, �etudiant, Universit �e Nationale deHanoi, Vietnam.Soit O le entre du er le ir ons rit du triangle ABC, de rayon R , et

H son ortho entre. Montrer que3R − 2 OH ≤ HA + HB + HC ≤ 3R + OH

3461. Proposed by Tran Quang Hung, �etudiant, Universit �e Nationale deHanoi, Vietnam.Soit I le entre du er le ir ons rit du triangle ABC et soit A′, B′et C′ les interse tions des rayons AI, BI et CI ave les ot �es respe tifs dutriangle. Montrer queIA + IB + IC ≥ 2(IA′ + IB′ + IC′) .3462. Propos �e par Sotiris Louridas, Aegaleo, Gr �e e.Soit x, y et z trois nombres r �eels positifs tels que

(x3 + z3 − y3

) (y3 + x3 − z3

) (z3 + y3 − x3

)> 0 .Montrer que

(x3+ y3+ z3+ 3xyz

) ∏ y lique (x3+ y3− z3+ xyz)

≤ 3∏ y lique 3

√x4 (x2 + yz)

4 .

330SOLUTIONSNo problem is ever permanently losed. The editor is always pleasedto onsider for publi ation new solutions or new insights on past problems.2557. [2000 : 304℄ Proposed by Gord Sinnamon, University of WesternOntario, London, ON and Hans Heinig, M Master University, Hamilton,ON.(a) Show that for all positive sequen es {xi} and all integers n > 0,

n∑

k=1

k∑

j=1

j∑

i=1

xi ≤ 2

n∑

k=1

k∑

j=1

xj

2

x−1k .

(b)⋆ Does the above inequality remain true without the fa tor of 2?( )⋆ [Proposed by the editors℄ What is the minimum onstant c that anrepla e the fa tor 2 in the above inequality?Solution to part ( ) by Li Chao, student, High S hool AÆliated to RenminUniversity of China, Beijing, China.The given inequality is equivalent ton∑

i=1

(n + 2 − i

2

)xi < c

n∑

i=1

1

xi

i∑

j=1

xj

2 .Making the substitution

sk =

k∑

i=1

xi , k = 0 , 1 , 2 , . . . ,nand then simplifying puts the inequality into another equivalent form:n∑

i=1

(n + 1 − i)si < cn∑

i=1

s2i

si − si−1

.By the Cau hy{S hwartz Inequality we have(

n∑

i=1

(n + 1 − i)si

)2

≤(

n∑

i=1

s2i

si − si−1

)(n∑

i=1

(n + 1 − i)2(si − si−1)

)

=

(n∑

i=1

s2i

si − si−1

)(n∑

i=1

(2n + 1 − 2i)si

)

<

(n∑

i=1

s2i

si − si−1

)(2

n∑

i=1

(n + 1 − i)si

) .

331Hen e,

n∑

i=1

(n + 1 − i)si < 2n∑

i=1

s2i

si − si−1

,so that taking c = 2 ensures that the inequality holds for all positive realnumbers x1, x2, . . . , xn.For ea h n > 4 we will now give a stri tly in reasing sequen e of num-bers s0, s1, . . . , sn su h that s0 = 0 andlim

n→∞

(n∑

i=1

(n + 1 − i)si

)

(n∑

i=1

s2i

si − si−1

) = 2

(this is equivalent to spe ifying x1, x2, . . . , xn). Take n > 4, let s0 = 0 andsi =

(n − 1)(n − 2)

(n − i)(n − i − 1), 1 ≤ i ≤ n − 2 ;

sn = 2sn−1 = 4sn−2 .Then we haven∑

i=1

s2i

si − si−1

= 1 +

n−2∑

i=2

(n − 1)2(n − 2)2

(n − i)2(n − i − 1)2

(n − 1)(n − 2)

[1

(n − i)(n − i − 1)− 1

(n − i + 1)(n − i)

]

+ 6(n − 1)(n − 2)

= 1 +(n − 1)(n − 2)

2

n−2∑

i=2

(2

n − i − 1− 1

n − i

)

+ 6(n − 1)(n − 2)

= 1 +(n − 1)(n − 2)

2

[1 +

1

2+

1

3+ · · · +

1

n − 3+

(1 − 1

n − 2

)]

+ 6(n − 1)(n − 2)

= 1 +(n − 1)(n − 2)

2

[ln n + O(1)

]+ 6(n − 1)(n − 2)

=1

2n2 ln n + O(n2) ,where the notation f(n) = O

(g(n)

) here means that there is a positive onstant C su h that |f(n)| ≤ C|g(n)| holds for suÆ iently large n, and

332we have used the fa t that (1 +

1

2+

1

3+ · · · +

1

n

)− ln n tends to Euler's onstant, γ, as n tends to in�nity.We also have

n∑

i=1

(n + 1 − i)si

=

n−2∑

i=1

(n − i + 1)(n − 1)(n − 2)

(n − i)(n − i − 1)

+ 4(n − 1)(n − 2)

= (n − 1)(n − 2)

n−2∑

i=1

(2

n − i − 1− 1

n − i

)

+ 4(n − 1)(n − 2)

= (n − 1)(n − 2)

[1 +

1

2+ · · · +

1

n − 2+

(1 − 1

n − 1

)]

+ 4(n − 1)(n − 2)

= (n − 1)(n − 2)[ln n + O(1)

]+ 4(n − 1)(n − 2)

= n2 ln n + O(n2) .We now omputelim

n→∞

(n∑

i=1

(n + 1 − i)si

)

(n∑

i=1

s2i

si − si−1

) = limn→∞

n2 ln n + O(n2)1

2n2 ln n + O(n2)

= limn→∞

2 + 2O(n2)/n2 ln n

1 + 2O(n2)/n2 ln n= 2 ,as desired.Therefore, the minimum value of the onstant c is 2.No other solutions to part ( ) were submitted.

3351. [2008 : 298, 300℄ Proposed by Toshio Seimiya, Kawasaki, Japan.Let ABC be a triangle with AB > AC. Let P be a point on the lineAB beyond A su h that AP + PC = AB. Let M be the midpoint of BC,and let Q be the point on the side AB su h that CQ ⊥ AM . Prove thatBQ = 2AP .

333Solution by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain; andTai hi Maekawa, Takatsuki City, Osaka, Japan.LetD be the point on theline PB beyond P su h thatPD = PC. Sin eAD = AP + PD

= AP + PC = AB ,A is the midpoint of the seg-ment BD. It follows thatAM ||DC, be ause M is themidpoint of BC. Hen e∠DCQ = 90◦. Let PN ⊥ DCwith point N on the line CD.Then DN = NC, be ause tri-angle CDP is isos eles. Sin e ..

..

..

..

..

..

.

..

..

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..

..

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..

..

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..

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

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

A

B C

D

M

N

P

Q

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

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

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

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

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

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

N is the midpoint of CD and PN ‖ QC, it follows that P is the midpointof QD. Thus PQ = PD = PC. Hen eAP + AQ = PQ = PC = AB − AP ,and therefore,

2AP = AB − AQ = BQ ,as laimed.Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; �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; CHRIS BROYLES and MATTHEW STEIN,Southeast Missouri State University, Cape Girardeau, MO, USA; CHIP CURTIS, MissouriSouthern State University, Joplin, MO, USA; IAN JUNE L. GARCES, Ateneo de ManilaUniversity, Quezon City, The Philippines; OLIVER GEUPEL, Br �uhl, NRW, Germany (twosolutions); JOHN G. HEUVER, Grande Prairie, AB; PETER HURTHIG, Columbia College,Van ouver, BC; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; GIANNISG. KALOGERAKIS, Canea, Crete, Gree e; V �ACLAV KONE �CN �Y, Big Rapids, MI, USA;KEE-WAI LAU, Hong Kong, China; SALEM MALIKI �C, student, Sarajevo College,Sarajevo, Bosnia and Herzegovina; MISSOURI STATE UNIVERSITY PROBLEM SOLVINGGROUP, Spring�eld, MO, USA; MADHAV R. MODAK, formerly of Sir ParashurambhauCollege, Pune, India; ALEX SONG, student, Elizabeth Ziegler Publi S hool, Waterloo, ON;PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU, Com�ane�sti, Romania;and the proposer.3352. [2008 : 298, 300℄ Proposed by Toshio Seimiya, Kawasaki, Japan.Let ABC be a right-angled triangle with right angle at A. Let I be thein entre of △ABC, and let D and E be the interse tions of BI and CI withAC and AB, respe tively. Prove that

BI · ID

CI · IE=

AB

AC.

334Solution by Salem Maliki� , student, Sarajevo College, Sarajevo, Bosnia andHerzegovina; and Madhav R. Modak, formerly of Sir ParashurambhauCollege, Pune, India.Sin e AI bise ts the right angle at A, we have

∠CID =1

2∠B +

1

2∠C =

1

2

(180◦ − ∠A

)= 45◦ = ∠CAI .Hen e, triangles CAI and CID are similar, so that

ID

CI=

AI

AC.Likewise, triangles BAI and

BIE are similar, and there-fore,BI

IE=

AB

AI.The result now follows by mul-tiplying a ross the last twoequations.

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A B

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45◦45◦

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Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallor a, Spain; GEORGEAPOSTOLOPOULOS, Messolonghi, Gree e; �SEFKET ARSLANAGI �C, University of Sarajevo,Sarajevo, Bosnia and Herzegovina; ROY BARBARA, Lebanese University, Fanar, Lebanon;RICARDO BARROSO CAMPOS, University of Seville, Seville, Spain; MICHEL BATAILLE, Rouen,Fran e; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; CHIPCURTIS, Missouri Southern State University, Joplin, MO, USA; RICHARD I. HESS, Ran hoPalos Verdes, CA, USA; JOHN G. HEUVER, Grande Prairie, AB; JOE HOWARD, Portales, NM,USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; V �ACLAV KONE �CN �Y, BigRapids, MI, USA; TAICHI MAEKAWA, Takatsuki City, Osaka, Japan; MISSOURI STATE UNI-VERSITY PROBLEM SOLVING GROUP, Spring�eld, MO, USA; JOEL SCHLOSBERG, Bayside,NY, USA; D.J. SMEENK, Zaltbommel, the Netherlands; GEORGE TSAPAKIDIS, Agrinio, Gree e;PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU, Com�ane�sti, Romania;and the proposer.3353. [2008 : 298, 301℄ Proposed by Mih�aly Ben ze, Brasov, Romania.Let ABC be a triangle all of whose side lengths are positive integers.(a) Determine all su h triangles where one angle has twi e the measure ofa se ond angle.(b) Determine all su h triangles where two medians are perpendi ular.Composite of solutions by Roy Barbara, Lebanese University, Fanar, Lebanonand Mi hel Bataille, Rouen, Fran e.We denote by α, β, and γ the angles opposite the sides BC = a,CA = b, and AB = c respe tively.

335(a) Without loss of generality we take γ = 2α and show that the solutionsare the triangles with

a = dn2 , b = d(m2 − n2

) , c = dmn , (1)where d, m, andn are positive integers, m is oprime ton, andn < m < 2n.We use an equivalen e proved in the April 2006 issue (see [2006 : 159℄)of this journal: A triangle satis�es γ = 2α if and only if c2 = a(a + b).Now suppose that a, b, and c are the sides of a triangle with γ = 2α.Let d = gcd(a, b). Then a = da′ and b = db′ with a′ oprime to b′. Clearly,d also divides c, so c = dc′.Sin e a′ and b′ are oprime, then a′ and a′ + b′ are also oprime. Sin e(c′)2 = a′(a′ + b′), it follows that there exist oprime positive integers mand n su h that a′ = n2 and a′ + b′ = m2.Thus, (1) holds for positive integers d, m, and n with m oprime to nand n < m.Finally, sin e ABC is a triangle, a + c > b, hen e 2n2 > m(m − n),hen e m < 2n.Conversely, if (1) holds for positive integers d, m, and n with the afore-mentioned restri tions, then the inequality n < m < 2n ensures that a, b,and c are the side lengths of a triangle and also the relation c2 = a(a + b)holds, whi h ensures that γ = 2α.(b) Without loss of generality, we restri t the problem to the ase wherethe medians from B and C are perpendi ular.First we show that the positive real numbers a, b, and c are the sides ofa triangle ABC where the medians from B and C are perpendi ular if andonly if

b2 + c2 = 5a2 and bc > 2a2 . (2)If G is the entroid of ABC, then BG is perpendi ular to CG if andonly if BC2 = BG2 + CG2, or1

9

(2a2 + 2c2 − b2

)+

1

9

(2a2 + 2b2 − c2

)= a2 ,that is, b2 + c2 = 5a2.In addition, 1 > cos α =

b2 + c2 − a2

2bc=

2a2

bc, hen e bc > 2a2.Conversely, if the onditions in (2) hold, then (b+c)2 = 5a2+2bc > a2and (b − c)2 = 5a2 − 2bc < a2. Hen e, |b − c| < a < b + c and there existsa triangle ABC with sides BC = a, AC = b, and AB = c. The relation

b2 + c2 = 5a2 holds for this triangle, so the medians from B and C areperpendi ular.Thus the problem amounts to �nding the positive integers a, b, and cthat satisfy (2). Be ause of the homogeneity of the onditions, we need only�nd the solutions su h that gcd(a, b, c) = 1; the general solution is thenobtained by multiplying these primitive solutions by positive integers.

336Now let a, b, c be a primitive solution. Then there exist oprime posi-tive integers m and n for whi h b − a

2a − c=

2a + c

b + a=

m

n, and thus

a(2m + n) − nb − mc = 0 ,a(m − 2n) + mb − nc = 0 .Solving for a, b, and c yieldsa = λ

(m2 + n2

) ,b = λ

(n2 − m2 + 4mn

) ,c = 2λ

(m2 − n2 + mn

) ,where λ is a positive rational number. The ondition bc > 2a2 redu es to(m2 − n2

) (3mn + 2n2 − 2m2

)> 0 ,whi h holds if and only if n < m < 2n.Writing λ =

u

v, where u and v are positive oprime integers, we obtain

va = u(m2 + n2

) ,vb = u

(n2 − m2 + 4mn

) ,vc = 2u

(m2 − n2 + mn

) .Now, gcd(u, v) = 1 and gcd(a, b, c) = 1, hen e u = 1 and(a, b, c) =

(m2 + n2

v,n2 − m2 + 4mn

v,2(m2 − n2 + mn

)

v

) , (3)where v = gcd(m2 + n2, n2 − m2 + 4mn, 2

(m2 − n2 + mn

)).Conversely, it is easy to show that a triple su h as in (3) with m and n oprime and n < m < 2n is a primitive solution.If p is prime, k is a positive integer, and pk divides v, then pk dividesboth m2 + n2 and 2(n2 − m2 + 4mn

)+ 2

(m2 − n2 + mn

)= 10mn.However, p does not divide n or m, sin e otherwise it would divideboth n and m, and so pk divides 10. Hen e, p = 2 or p = 5 and k = 1,whi h implies that v ∈ {1, 2, 5, 10}.If v is even then, sin e v divides m2 + n2, m, n must be odd.If 5 divides v then 5 divides (m2+n2

)+(n2−m2+4mn

)= 2n(n+2m)and (n2 − m2 +4mn

)− (m2 +n2)

= 2m(2n− m). Sin e 5 does not dividem or n, then 5 divides both n + 2m and 2n − m. Note that n + 2m = 5kand 2n − m = 5l is equivalent to m = 2k − l and n = k + 2l, and that inthis ase n < m < 2n be omes 0 < 3l < k.A omplete des ription of the primitive solutions an now be given:

• (a, b, c) =

(m2 + n2

10,

n2 − m2 + 4mn

10,2(m2 − n2 + mn)

10

), wherem = 2k − l, n = k + 2l; gcd(k, l) = 1; k, l are odd; and 0 < 3l < k.

337• (a, b, c) =

(m2 + n2

5,

n2 − m2 + 4mn

5,2(m2 − n2 + mn)

5

), wherem = 2k − l, n = k + 2l; gcd(k, l) = 1; k, l have opposite parity;and 0 < 3l < k.

• (a, b, c) =

(m2 + n2

2,

n2 − m2 + 4mn

2,2(m2 − n2 + mn)

2

), wheregcd(m, n) = 1; m, n are odd; n + 2m is not divisible by 5; and0 < n < m < 2n.

• (a, b, c) =(m2 + n2, n2 − m2 + 4mn, 2(m2 − n2 + mn

), wheregcd(m, n) = 1; m, n have opposite parity; n + 2m is not divisibleby 5; and 0 < n < m < 2n.Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e (part (a) only); CHIPCURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Br �uhl,NRW, Germany (part (a) only); RICHARD I. HESS, Ran ho Palos Verdes, CA, USA (part (a)only); WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria (part (a) only); RODOLFOLARREA and LUZ RONCAL, Logro ~no, Spain (part (b) only); MADHAV R. MODAK, formerlyof Sir Parashurambhau College, Pune, India (part (a) only); and TITU ZVONARU, Com�ane�sti,Romania (part (a) only). There were three in omplete solutions submitted.Modak remarks that Theorem 44 of [1℄ gives all integer solutions to b2 + c2 = 5a2,while Janous refers to [2℄ where solutions to this equation are obtained by fa toring over theGaussian integers and using the fa t that 5 is a sum of two squares.Geupel notes that Problem 129(a) of [3℄ asks for the smallest integral triangle for whi hone angle is twi e another.Referen es[1℄ L.E. Di kson, Introdu tion to the Theory of Numbers, Dover Publ., 1957.[2℄ L.J. Mordell, Dophantine Equations, A ademi Press, London and New York, 1969.[3℄ D.O. Shklarsky, N.N. Khentzov, and I.M. Yaglom, The USSR Olympiad Problem Book,Dover Publ., New York, 1962.

3354. [2008 : 298, 301℄ Proposed by Jos �e Luis D��az-Barrero, UniversitatPolit �e ni a de Catalunya, Bar elona, Spain.Evaluatelim

n→∞

n∑

k=1

ln

(n2 + k2

n2

)k3/n4.Composite of nearly identi al solutions submitted by all the solvers whosenames appear below.The fun tion f(x) = x3 ln

(1 + x2

) is ontinuous and hen e Riemann

338integrable on [0, 1]. Using integration by parts, we have

limn→∞

n∑

k=1

ln

(n2 + k2

n2

)k3/n4

= limn→∞

1

n

n∑

k=1

(k

n

)3

ln

(1 +

(k

n

)2)

=

∫ 1

0

x3 ln(1 + x2

)dx =

1

4x4 ln

(1 + x2

)∣∣∣∣1

0

− 1

2

∫ 1

0

x5

1 + x2dx

=1

4ln 2 − 1

2

∫ 1

0

(x3 − x +

x

1 + x2

)dx

=1

4ln 2 − 1

2

(1

4x4 − 1

2x2 +

1

2ln(1 + x2

))∣∣∣∣1

0

=1

8.Solvers: GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; DIONNE BAILEY, ELSIECAMPBELL, and CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA;MICHEL BATAILLE, Rouen, Fran e; CHIP CURTIS, Missouri Southern State University, Joplin,MO, USA; OLIVER GEUPEL, Br �uhl, NRW, Germany; JOS �E HERN�ANDEZ SANTIAGO, student,Universidad Te nol �ogi a de la Mixte a, Oaxa a, Mexi o; RICHARD I. HESS, Ran ho PalosVerdes, CA, USA; JOE HOWARD, Portales, NM, USA; PETER HURTHIG, Columbia College,Van ouver, BC; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; KEE-WAI LAU,Hong Kong, China; PHIL M CARTNEY, Northern Kentu ky University, Highland Heights, KY,USA; MISSOURI STATE UNIVERSITY PROBLEM SOLVING GROUP, Spring�eld, MO, USA;MADHAV R. MODAK, formerly of Sir Parashurambhau College, Pune, India; GOTTFRIEDPERZ, Pestalozzigymnasium, Graz, Austria; XAVIER ROS, student, Universitat Polit �e ni a deCatalunya, Bar elona, Spain; OVIDIU FURDUI, Campia Turzii, Cluj, Romania; and PETERY. WOO, Biola University, La Mirada, CA, USA.

3355. [2008 : 299, 301℄ Proposed by Todor Yalamov, So�a University,So�a, Bulgaria.For the triangle ABC let (x, y)ABC denote the line whi h interse tsthe union of the segments AB and BC in X and the segment AC in Y su hthatAX

AB + BC=

AY

AC=

x · AB + y · BC

(x + y)(AB + BC),where AX is either the length of the segment AX if X lies between A and

B, or the sum of the lengths of the segments AB and BX if X lies betweenB and C. Prove that the three lines (x, y)ABC , (x, y)BCA, and (x, y)CBAinterse t in a point dividing the segment NI in the ratio x : y, where N isthe Nagel point and I the in entre of △ABC.Solution outline by Peter Y. Woo, Biola University, La Mirada, CA, USA,expanded by the editor.As usual we let a = BC, b = CA, c = AB, and s =

a + b + c

2. Set

t =x

x + y; thus, 1 − t =

y

x + y, and the ratio of interest be omes a fun tion

339of t, namely

f(t) =tc + (1 − t)a

a + c=

AXt

a + c=

AYt

b,where Xt is the point of AB ∪ BC, and Yt of AC, that orrespond to ourparameter t. In this notation, (x, y)ABC = XtYt. Of ourse, when a = c,

f(t) is onstant and (x, y)ABC is the line NI for all x, y. This is onsistentwith what we are to prove unless a = b = c; when △ABC is equilateral wehave N = I (so there is no line NI), and our three lines interse t at thatpoint. Let us therefore assume that a 6= c, so that f(t) is not onstant andthe lines NI and (x, y)ABC interse t.We shall see that (x, y)ABC is the line in the family of lines parallel tothe bise tor of∠B (where B is the middle vertex in the subs ript) that dividesthe segment NI in the ratio t : (1 − t) = x : y. Note that as a onsequen e,(x, y)ABC and (x, y)CBA represent the same line, so the proposer probablyintended (x, y)CAB for his third line. For this problem we are on ernedwith the domain 0 ≤ t ≤ 1; in parti ular,(a) AY1 = bf(1) =

bc

a + c; CY1 = b − AY1 =

ab

a + c; Y1 is the foot of thebise tor of ∠CBA (be ause it divides the side AC in the ratio c : a);(b) AY0 = bf(0) =

ab

a + c; CY0 =

bc

a + c;( ) X1 = B (be ause f(1) =

c

a + c=

AX1

a + c);(d) AX0 = (a + c)f(0) = a: when c ≥ a, X0 lies on AB (when e

AX0 = a and BX0 = c − a); otherwise, when a ≥ c, X0 lies onBC (when e CX0 = c and BX0 = a − c).By items (a) and ( ), the line X1Y1 bise ts ∠CBA and, therefore, itpasses through the in entre I. We next will see that the line X0Y0 passesthrough the Nagel point N . To determine N we use the points P , Q, Rwhere the ex ir les meet the sides BC, CA, AB of △ABC, when e

BR = CQ = s − a , AR = CP = s − b , and AQ = BP = s − c .The Nagel point is de�ned to be the point ommon to AP , BQ, and CR.Apply Menelaus' theorem to transversal NCR of △BQA to dedu e thatBN

NQ=

CA

QC· RB

AR=

b

s − a· s − a

s − b=

b

s − b. (1)Let N ′ = X0Y0 ∩ BQ; we want to show that N ′ = N . Here we will needthe length

QY0 = |AY0 − AQ| =

∣∣∣∣ab

a + c− (s − c)

∣∣∣∣ =|a − c|(s − b)

a + c.

340...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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s − a

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s−

a

s−

b

When c > a we apply Menelaus' theorem to transversal N ′Y0X0 of△BQA to get

BN ′

N ′Q=

Y0A

QY0

· X0B

AX0

=ab

a + c· a + c

(c − a)(s − b)· c − a

a=

b

s − b.

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

A

B = X1C

I

N

P

QR

X0

Y0

Y1

.......

.......

.......

.......

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a − c c

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.

s − c s − bOtherwise, when a > c we apply Menelaus' theorem to transversalN ′Y0X0 of △BCQ to get

BN ′

N ′Q=

Y0C

QY0

· X0B

CX0

=bc

a + c· a + c

(a − c)(s − b)· a − c

c=

b

s − b.

In both ases BN ′

N ′Qequals the value of BN

NQin equation (1), from whi h

341we on lude that N = N ′, and X0Y0 interse ts IN at N , as laimed.It remains to observe that X0Y0 ‖ X1Y1: when c > a (that is, whenX0 lies on AB),

AX0 : AX1 = a : c =ab

a + c:

bc

a + c= AY0 : AY1 ;when a > c (and X0 lies on BC),

CX0 : CX1 = c : a =bc

a + c:

ab

a + c= CY0 : CY1 .Finally, be ause Xt divides the segment X0X1 in the ratio t : (1 − t) while

Yt divides Y0Y1 in that same ratio, XtYt is parallel to both X0Y0 and X1Y1for all t. It follows that XtYt meets the segment NI in a point that dividesit in that same ratio t : (1 − t) = x : y, as desired.Also solved by MICHEL BATAILLE, Rouen, Fran e; OLIVER GEUPEL, Br �uhl, NRW,Germany; and the proposer.Geupel remarked that the Spieker entre is the midpoint of NI (on the line X1/2Y1/2)while the entroid of the triangle is a trise tor of NI (on the line X2/3Y2/3).3356. [2008 : 299, 301℄ Proposed by Cristinel Morti i, Valahia Universityof Targoviste, Romania.Let f : [0, ∞) → R be integrable on [0, 1] and have period 1 (that is,f(x + 1) = f(x) for all x ∈ [0, ∞)). If {xn}∞

n=0 is any stri tly in reasing,unbounded sequen e with x0 = 0 for whi h (xn+1 − xn) → 0, denoter(n) = max{k ∈ N | xk ≤ n} .(a) Prove that

limn→∞

1

n

r(n)∑

k=1

(xk − xk−1)f(xk) =

∫ 1

0

f(x) dx .(b) Prove that

limn→∞

1

ln n

n∑

k=1

f(ln k)

k=

∫ 1

0

f(x) dx .Solution by Oliver Geupel, Br �uhl, NRW, Germany.Let I =

∫ 1

0f(x) dx and

Sn =

r(n+1)∑

k=r(n)+1

(xk − xk−1)f(xk) .

342We �rst prove that

limn→∞

Sn = I . (1)Consider the following partition of the interval [n, n + 1]

n < xr(n)+1 < xr(n)+2 < · · · < xr(n+1) ≤ n + 1 ,where the lengths of the resulting intervals between the points are denotedby ∆1, ∆2, . . . , ∆m and m = r(n + 1) − r(n) + 1. Consider the Riemannsum Rn =m∑

k=1

f(x∗k)∆k, where x∗

k is the right endpoint of the kth interval.Be ause f is bounded and (xk+1 − xk) → 0, we obtainSn = Rn +

(n − xr(n)

)f(xr(n)+1) − (

n + 1 − xr(n+1)

)f(n + 1) → Ias n → ∞.Next we prove the limit in part (a) in the following equivalent form:

limn→∞

1

n

n−1∑

k=0

Sk = I . (2)Let ǫ > 0 be given. By (1) there is an integer N su h that |Sk − I| <

ǫ

2whenever k ≥ N . Let S =

∣∣∣∣N−1∑k=0

(Sk − I)

∣∣∣∣. Whenever n > max{

N , 2S

ǫ

}we have∣∣∣∣∣1

n

n−1∑

k=0

Sk − I

∣∣∣∣∣ ≤ 1

n

∣∣∣∣∣

N−1∑

k=0

(Sk − I)

∣∣∣∣∣ +1

n

∣∣∣∣∣

n−1∑

k=N

(Sk − I)

∣∣∣∣∣

≤ S

n+

n − N

n· ǫ

2< ǫand (2) is proved.For B > 0, let r(B) = max{k | xk ≤ B}. We next prove that

limB→∞

1

B

r(B)∑

k=1

(xk − xk−1)f(xk) = I . (3)Let n = ⌊B⌋, Un =

n−1∑k=0

Sk, and Vn =r(B)∑

k=r(n)+1

(xk − xk−1)f(xk). Thenr(B)∑k=1

(xk − xk−1)f(xk) = Un + Vn and by (2) we have limn→∞

1

nUn = I;therefore, it suÆ es to prove that 1

B(Un + Vn) − 1

nUn → 0 as B → ∞. Butthis follows from the fa t that Vn is bounded and the al ulation

1

B(Un + Vn) − 1

nUn =

Vn

B−(

B − ⌊B⌋B

)(Un

n

)→ 0 − 0 · I = 0 .

343Finally, we prove the limit in part (b). Let

Tn =1

ln n

n∑

k=1

ln

(1 +

1

k − 1

)f(ln k) ; Wn =

1

ln n

n∑

k=1

f(ln k)

k.

We must show that Wn → I. It follows from (3) with xk = ln k, (k > 0)and n = eB that Tn → I. Therefore, it suÆ es to prove Tn − Wn → 0.Let ǫ > 0 be given and let |f | < C, where C is a onstant. Note thatlim

n→∞

(1 +

1

n − 1

)n

= e, so there is an integer K su h that∣∣∣∣ln(1 +

1

k − 1

)k

− 1

∣∣∣∣ <ǫ

2Cwhenever k > K. Moreover, K an be hosen so that HK > Hn − ln n forea h n, where Hn = 1 +1

2+

1

3+ · · · + 1

nis the nth Harmoni number. Let

T =

∣∣∣∣∣

K∑

k=1

1

k

(ln

(1 +

1

k − 1

)k

− 1

)f(ln k)

∣∣∣∣∣ .Now if n > max{K, e2T/ǫ

}, then|Tn − Wn| ≤ T

ln n+

1

ln n

∣∣∣∣∣∣

n∑

k=K+1

1

k

(ln

(1 +

1

k − 1

)k

− 1

)f(ln k)

∣∣∣∣∣∣

<ǫ

2+

ǫ

2= ǫ .This ompletes the proof.Also solved by MICHEL BATAILLE, Rouen, Fran e (part (a) only); MADHAV R. MODAK,formerly of Sir Parashurambhau College, Pune, India (part (a) only); PETER Y. WOO, BiolaUniversity, La Mirada, CA, USA (part (a) only); and the proposer. There were two in ompletesolutions submitted to part (b).The fa t that Sn → I as n → ∞ implies that 1

n

n−1∑k=0

Sk → I as n → ∞ is part of thetheory of Ces �aro sums, whi h some solvers quoted dire tly.3357. [2008 : 172, 175℄ Proposed by Ovidiu Furdui, Campia Turzii, Cluj,Romania.Let a be a real number su h that −1 < a ≤ 1. Prove that∫ 1

0

x + a

x2 + 2ax + 1ln(1 − x) dx =

1

2ln2

(2 sin

θ

2

)+

θ2

8− θπ

4+

π2

24,where θ is the unique solution in (0, π] of the equation cos θ = −a.

344Solution by Mi hel Bataille, Rouen, Fran e.Let

f(a) =

∫ 1

0

x + a

x2 + 2ax + 1ln(1 − x) dx .For a ∈ (−1, 1] and x ∈ [0, 1), we have

∣∣∣∣x + a

x2 + 2ax + 1ln(1 − x)

∣∣∣∣ ≤ | ln(1 − x)|1 + x

≤ | ln(1 − x)| .Sin e | ln(1−x)| = − ln(1−x) is integrable on [0, 1), we see that f(a) is well-de�ned and ontinuous on (−1, 1]. [Ed.: It seems one must also ompute∣∣∣ x + a

x2 + 2ax + 1− x + b

x2 + 2bx + 1

∣∣∣ = |b − a| `

1 − x2´

(x2 + 2ax + 1) (x2 + 2bx + 1)and note that

x2 + 2ax + 1 on [0, 1) is bounded below by 1 if 0 ≤ a ≤ 1 or by 1 − a2if −1 < a < 0, and then ombine this with the salient observation above.℄For s ∈ (0, 1) we have∫ 1−s

0

2(x + a)

x2 + 2ax + 1ln(1 − x) dx

=

∫ 1−s

0

ln(1 − x) · d[ln(x2 + 2ax + 1

)]

= ln s · ln((1 − s)2 + 2a(1 − s) + 1

)+

∫ 1−s

0

ln(x2 + 2ax + 1

)

1 − xdx

= ln s · ln(s2 − (2 + 2a)s + 2 + 2a

)

+

∫ 1

u

ln(u2 − (2 + 2a)u + 2 + 2a

)

udu

= ln s · ln(2 + 2a) + ln s · ln

(1 − s +

s2

2 + 2a

)

+

∫ 1

s

ln(2 + 2a)

udu +

∫ 1

s

ln

(1 − u +

u2

2 + 2a

)

udu

= ln s · ln(2 + 2a) + ε(s) − ln s · ln(2 + 2a)

+

∫ 1

s

ln

(1 − u +

u2

2 + 2a

)

udu ,

where ε(s) → 0 as s → 0+.It follows that2f(a) =

∫ 1

0

ln

(1 − u +

u2

2 + 2a

)

udu

345or 2f(a) = g(θ) where

g(θ) =

∫ 1

0

ln

(1 − u +

u2

4 sin2(θ/2)

)

udu . (1)First, we onsider the ase a = 1 that is, θ = π. Making the substitution

u = 2v, we havef(1) =

1

2

∫ 1

0

ln

(1 − u +

u2

4

)

udu =

∫ 1

2

0

ln(1 − v)

vdv .Next, integrating by parts and then using the well-known identity

∫ 1

0

ln(1 − t)

tdt = −

∫ 1

0

(∞∑

n=1

tn−1

n

)dt = −

∞∑

n=1

1

n2we obtainf(1) = (ln 2)2 +

∫ 1

2

0

ln(v)

1 − vdv

= (ln 2)2 +

∫ 1

0

ln(v)

1 − vdv −

∫ 1

1

2

ln(v)

1 − vdv

= (ln 2)2 +

∫ 1

0

ln(1 − t)

tdt −

∫ 1

2

0

ln(1 − t)

tdt

= (ln 2)2 − π2

6− f(1) ,

so that f(1) = 12(ln 2)2 − π2

12, in agreement with the given formula.To determine g(θ) for θ ∈ (0, π), we di�erentiate under the integralsign in equation (1):

g′(θ) = −cos(θ/2)

sin(θ/2)

∫ 1

0

u

u2 − 4u sin2(θ/2) + 4 sin2(θ/2)du (2)

(To justify this, note that if β ∈(0,

π

2

) and θ ∈ [β, π − β], then for allu ∈ [0, 1],∣∣∣∣cos(θ/2)

sin(θ/2)· u

u2 − 4u sin2(θ/2) + 4 sin2(θ/2)

∣∣∣∣ ≤ u cot(β/2)

u2 + 4 sin2(β/2)(1 − u),and the dominating fun tion is integrable on [0, 1].) [Ed.: One may rewritethe integrand in (1) as the integral of its partial derivative with respe t to

θ, then hange the order of integration by Fubini's Theorem by the solver'sremark, then di�erentiate ea h side with respe t to θ to obtain (2) .℄

346Writing the numerator u as 1

2

(2u − 4 sin2(θ/2)

)+ 2 sin2(θ/2), the al ulation of the integral in (2) is straightforward and gives

g′(θ) = 2 ·12cos(θ/2)

sin(θ/2)ln

(2 sin

θ

2

)+

θ

2− π

2.It follows that

g(θ) =(ln(2 sin(θ/2)

)2+

θ2

4− πθ

2+ Cfor some onstant C. This onstant is π2

12, easily determined using the on-tinuity of g and the already found value of f(1) = g(π). Finally,

f(a) =1

2g(θ) =

1

2ln2

(2 sin

θ

2

)+

θ2

8− θπ

4+

π2

24.Also solved by the proposer. There was one in orre t submission.

3358. Proposed by Toshio Seimiya, Kawasaki, Japan.The interior bise tor of ∠BAC of triangle ABC meets BC at D. Sup-pose that1

BD2+

1

CD2=

2

AD2.Prove that ∠BAC = 90◦.I. Solution by the proposer.LetM be the se ond interse tion ofAD with the ir um ir le of△ABCand let N be the midpoint of BC. Sin e ∠BAM = ∠MAC, we have that

BM = MC and MN ⊥ BC. Sin eBD · CD = AD · DM , we obtain

2

AD2=

1

BD2+

1

CD2

=BD2 + CD2

BD2 · CD2

=BD2 + CD2

AD2 · DM2,hen e,

2DM2 = BD2 + CD2 . (1)Sin e N is the midpoint of BC,

.

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BD2 + CD2 = 2(DN2 + BN2

) .

347Thus, we have from equation (1) that

DM2 = DN2 + BN2 . (2)Sin e ∠DNM = 90◦, we have DM2 = DN2 + MN2, and omparing thiswith (2) yields BN2 = MN2, that is, BN = MN . We have now dedu edthat ∠MBN = 45◦.Therefore,∠BAC = 2∠MAC = 2∠MBC = 2∠MBN = 90◦ .II. Solution by Oliver Geupel, Br �uhl, NRW, Germany.We shall prove the following generalization:

2

AD2− 1

BD2− 1

CD2

< 0 if A < 90◦ ,= 0 if A = 90◦ ,> 0 if A > 90◦ .Let a = BC, b = CA, c = AB, p = BD, q = CD, and w = AD.It is a well-known fa t that ea h interior angle bise tor divides the oppositeside in the ratio of the other two sides. Hen e, p =

ac

b + cand q =

ab

b + c.Another ommon formula is w2 =

4b2c2 cos2(A/2)

(b + c)2. Using these relations aswell as the Law of Cosines, we derive

11

p2+

1

q2

− 12

w2

=1

(b + c)2

a2c2+

(b + c)2

a2b2

−2b2c2 cos2

A

2

(b + c)2

=b2c2

(b + c)2

(a2

b2 + c2− 2 cos2

A

2

)

=b2c2

(b + c)2· b2 + c2 − 2bc cos A − (

b2 + c2)(1 + cos A)

b2 + c2

= −b2c2 cos A

b2 + c2,whi h ompletes the proof.Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallor a, Spain; GEORGEAPOSTOLOPOULOS, Messolonghi, Gree e; �SEFKET ARSLANAGI �C, University of Sarajevo,Sarajevo, Bosnia and Herzegovina; ROY BARBARA, Lebanese University, Fanar, Lebanon;RICARDO BARROSO CAMPOS, University of Seville, Seville, Spain; MICHEL BATAILLE, Rouen,

348Fran e; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; CHIPCURTIS, Missouri Southern State University, Joplin, MO, USA; RICHARD I. HESS, Ran hoPalos Verdes, CA, USA; JOHN G. HEUVER, Grande Prairie, AB; JOE HOWARD, Portales, NM,USA; PETER HURTHIG, Columbia College, Van ouver, BC; WALTHER JANOUS, Ursulinen-gymnasium, Innsbru k, Austria; GIANNIS G. KALOGERAKIS, Canea, Crete, Gree e; KEE-WAILAU, Hong Kong, China; TAICHI MAEKAWA, Takatsuki City, Osaka, Japan; SALEM MALIKI �C,student, Sarajevo College, Sarajevo, Bosnia andHerzegovina; MADHAV R.MODAK, formerly ofSir Parashurambhau College, Pune, India; NANCY MUELLER and SETH STAHLHEBER, South-east Missouri State University, Cape Girardeau, MO, USA; K.C. SANDEEP, student, SoutheastMissouri State University, Cape Girardeau, MO, USA; JOEL SCHLOSBERG, Bayside, NY,USA; BIKRAM KUMAR SITOULA, student, Southeast Missouri State University, CapeGirardeau, MO, USA; SOUTHEAST MISSOURI STATE UNIVERSITY MATH CLUB;D.J. SMEENK, Zaltbommel, the Netherlands; GEORGE TSAPAKIDIS, Agrinio, Gree e; PETERY. WOO, Biola University, La Mirada, CA, USA; and TITU ZVONARU, Com�ane�sti, Romania.There were two in orre t solutions submitted.3359. [2008 : 300, 302℄ Proposed by Ray Killgrove, Vista, CA, USA andDavid Koster, University of Wis onsin, La Crosse, WI, USA.Consider the sequen e {an}∞

n=1 de�ned by an = n2 + n + 1. Finda subsequen e {bn}∞n=1 su h that b1 = a1, b2 = a2, b3 > a3, every pairof terms from this subsequen e are relatively prime, and there are primeswhi h divide no term of the subsequen e.Solution by Oliver Geupel, Br �uhl, NRW, Germany.De�ne {bn} by b1 = a1 = 3, b2 = a2 = 7, and for n > 1 let

bn+1 =

(n∏

k=1

bk

)2

+

(n∏

k=1

bk

)+ 1 .

Then learly {bn} is a subsequen e of {an} and b3 > a3. Sin e we havebn ≡ 1 (mod bm) for m < n, we see that bm and bn are relatively prime.Finally, sin e all the bn's are odd, it follows that the prime number 2 dividesno term of {bn}.Also solved by ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHELBATAILLE, Rouen, Fran e; BRIAN D. BEASLEY, Presbyterian College, Clinton, SC, USA;WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; SALEM MALIKI �C, student,Sarajevo College, Sarajevo, Bosnia and Herzegovina; MISSOURI STATE UNIVERSITY PROB-LEM SOLVING GROUP, Spring�eld, MO, USA; JOEL SCHLOSBERG, Bayside, NY, USA; and theproposers.Most solutions were similar to the one featured above.Geupel remarks that omputations of the residues of n2+n+1modulo p for 1 ≤ n < pshow that no prime p ∈ {2, 5, 11} divides any term of the sequen e {an}.A prime p 6= 3, 7 an divide at most one term bm , m ≥ 3, of the sequen e {bn} in thefeatured solution, so by deleting at most term from that sequen e the prime p an be avoided.Thus, any �nite set of primes not ontaining 3 and 7 an be avoided by a subsequen e of {an}beginning with b1 = a1 and b2 = a2.

3493360. [2008 : 300, 302℄ Proposed by Mi hel Bataille, Rouen, Fran e.For omplex numbers a, b, and c, not all zero, let N (a, b, c) denote thenumber of solutions (z1, z2, z3) ∈ C3 to the system:

z1z3 = a ,z1z2 + z2z3 = b ,

z21 + z2

2 + z23 = c .For whi h a, b, and c does N (a, b, c) attain its minimal value?Solution by Missouri State University Problem Solving Group, Spring�eld,MO, USA.It is straightforward to see that if (z1, z2, z3) is a solution then

(z1 + z2 + z3)2 = 2a + 2b + c ,

(z1 − z2 + z3)2 = 2a − 2b + c .LetA be a square root of 2a+2b+c andB be a square root of 2a−2b+c.We then obtain the system of linear equations:

z1 + z2 + z3 = ±A ,z1 − z2 + z3 = ±B .Solving yields

(z1 + z3, z2) =(

A + B

2,

A − B

2

) , (A − B

2,

A + B

2

) ,(−A + B

2,

−A − B

2

) , (−A + B

2,

−A − B

2

) . (1)The system of two equations z1 + z3 =±A ± B

2and z1z3 = a alwayshas a solution. When su h a solution is paired with the orresponding z2 asolution to the original system is obtained. Also we note that if (z1, z2, z3)is a solution to the system, then −(z1, z2, z3) is a di�erent solution to thesystem (sin e (a, b, c) 6= (0, 0, 0)), thus the system has at least two solutions.We laim that this is in fa t the minimal value of N (a, b, c).If the terms on the right side of (1) are distin t, then N (a, b, c) ≥ 4.The only way for two of those terms to be equal is if A = 0 or B = 0.If A = 0 and B 6= 0, then B2 = −4b and (z1 +z3, z2) = ±

(B

2, −B

2

).Sin e B 6= 0, in order for the original system to have two solutions, thesystem z1 + z3 = ±B

2; z1z3 = a must have only one solution; that is,

z1 = z3, and hen e −b =(±B

2

)2

= 4a. Therefore, b = −4a, and sin eA = 0 we have c = 6a and (z1, z2, z3) = ±(α, −2α, α), where α is a squareroot of a 6= 0.

350If A = B = 0, then b = 0, c = −2a, and (z1, z2, z3) = ±(α, 0, −α),where α is a square root of −a 6= 0.If A 6= 0 and B = 0, then A2 = 4b and (z1 + z3, z2) = ±

(A

2,

A

2

).Sin e A 6= 0, in order for the original system to have two solutions, thesystem z1 + z3 = ±A

2; z1z3 = a must have only one solution, and hen e

b =(±A

2

)2

= 4a. Therefore, b = 4a, and sin e B = 0 we have c = 6a and(z1, z2, z3) = ±(α, 2α, α), where α is a square root of a 6= 0.In summary, the minimal value of N (a, b, c) is two, whi h is attainedfor triples of the form (a, 4a, 6a), (a, −4a, 6a), or (a, 0, −2a), where a is anonzero omplex number.Also solved by MICHEL BATAILLE, Rouen, Fran e; CHIP CURTIS, Missouri SouthernState University, Joplin, MO, USA; OLIVER GEUPEL, Br �uhl, NRW, Germany; ROY BARBARA,Lebanese University, Fanar, Lebanon; and TITU ZVONARU, Com�ane�sti, Romania. There wasone in omplete solution submitted.3361. [2008 : 300, 302℄ Proposed by Mi hel Bataille, Rouen, Fran e.Let the in ir le of triangle ABC meet the sides CA and AB at E andF , respe tively. For whi h points P of the line segment EF do the areas of△EBC, △PBC, and △FBC form an arithmeti progression?Composite of solutions by Oliver Geupel, Br �uhl, NRW, Germany and by TituZvonaru, Com�ane�sti, Romania.Let e, f , and p be the distan es from the points E, F , and P , respe -tively, to the line BC. The areas of the triangles EBC, PBC, and FBCform an arithmeti progression if and only if their respe tive altitudes e, p,and f satisfy

2p = e + f .If AB = AC, then the lines BC and EF are parallel, when e ea h pointP on EF satis�es the desired ondition. Otherwise, only the midpoint ofEF has the desired property. (This last laim follows immediately from themore familiar theorem that if EFF ′ is a triangle with P on side EF , P ′on side EF ′, and PP ′ ‖ FF ′, then P is the midpoint of EF if and only ifFF ′ = 2PP ′. To prove the laim from the triangle theorem, let the linethrough E that is parallel to BC meet at F ′ and P ′ our altitudes to BCfrom F and P , respe tively. Then P is the midpoint of EF if and only ifFF ′ = 2PP ′, if and only if (FF ′ + e) + e = 2PP ′ + 2e, if and only iff + e = 2p.)Also solved by ROY BARBARA, Lebanese University, Fanar, Lebanon; RICARDOBARROSO CAMPOS, University of Seville, Seville, Spain; CHIP CURTIS, Missouri SouthernState University, Joplin, MO, USA; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; TAICHIMAEKAWA, Takatsuki City, Osaka, Japan; D.J. SMEENK, Zaltbommel, the Netherlands; PETERY. WOO, Biola University, La Mirada, CA, USA; and the proposer.

351Observe that the in ir le plays almost no role in the solution|the on lusion holds forany segment EF that lies entirely on one side of the line BC. However, when E and F arethe points of tangen y of the in ir le as in our problem, then AF = AE so that the lo ation of

P for whi h the areas form an arithmeti progression an alternatively be des ribed as lying onthe bise tor of ∠BAC, or as the foot of the perpendi ular from A to EF .3362. [2008 : 172,175℄ Proposed by Jos �e Luis D��az-Barrero, UniversitatPolit �e ni a de Catalunya, Bar elona, Spain.Prove that

∫ 1

0

3

√ln(1 + x)

xdx

∫ 1

0

3

√ln2(1 + x)

x2dx <

π2

12.

Similar solutions by Kee-Wai Lau, Hong Kong, China; Ovidiu Furdui, CampiaTurzii, Cluj, Romania; andMissouri State University Problem Solving Group,Spring�eld, MO, USA.By H�olders inequality, we have∫ 1

0

3

√ln(1 + x)

xdx <

(∫ 1

0

ln(1 + x)

xdx

)1/3 (∫ 1

0

13/2dx

)2/3

=

(∫ 1

0

ln(1 + x)

xdx

)1/3

and∫ 1

0

3

√ln2(1 + x)

x2dx <

( ∫ 1

0

ln(1 + x)

xdx

)2/3 (∫ 1

0

13dx

)1/3

=

( ∫ 1

0

ln(1 + x)

xdx

)2/3.It now follows that∫ 1

0

3

√ln(1 + x)

xdx

∫ 1

0

3

√ln2(1 + x)

x2dx

<

∫ 1

0

ln(1 + x)

xdx

=

∫ 1

0

∞∑

n=1

(−1)n−1xn−1

ndx

=∞∑

n=1

(−1)n−1

n

∫ 1

0

xn−1 dx

=

∞∑

n=1

(−1)n−1

n2=

π2

12,

352where the integral and the sum an be inter hanged sin e the sum onvergesuniformly on [0, 1], and ∞∑

n=1

(−1)n−1

n2=

π2

12follows by straightforward ma-nipulations from the well-known formula ∞∑

n=1

1

n2=

π2

6.Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; DIONNE BAILEY,ELSIE CAMPBELL, CHARLES DIMINNIE, and KARL HAVLAK, Angelo State University,San Angelo, TX, USA; MICHEL BATAILLE, Rouen, Fran e; OLIVER GEUPEL, Br �uhl, NRW,Germany; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; JOE HOWARD, Portales,NM, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; PHIL M CARTNEY,Northern Kentu ky University, Highland Heights, KY, USA; PAOLO PERFETTI, Dipartimentodi Matemati a, Universit �a degli studi di Tor Vergata Roma, Rome, Italy; XAVIER ROS, student,Universitat Polit �e ni a de Catalunya, Bar elona, Spain; PETER Y. WOO, Biola University, LaMirada, CA, USA; and the proposer.Both Geupel and the proposer identi�ed the last integral omputed above as essentiallya value of the dilogarithm fun tion, one form of whi h is Li2(x) =∫

0

x ln(1 − t)/t dt.Crux Mathemati orumwith Mathemati al MayhemFormer Editors / An iens R �eda teurs: Bru e L.R. Shawyer, James E. TottenCrux Mathemati orumFounding Editors / R �eda teurs-fondateurs: L �eopold Sauv �e & Frederi k G.B. MaskellFormer Editors / An iens R �eda teurs: G.W. Sands, R.E. Woodrow, Bru e L.R. ShawyerMathemati al MayhemFounding Editors / R �eda teurs-fondateurs: Patri k Surry & Ravi VakilFormer Editors / An iens R �eda teurs: Philip Jong, Je� Higham, J.P. Grossman,Andre Chang, Naoki Sato, Cyrus Hsia, Shawn Godin, Je� Hooper