257
James Edward Totten 1947{2008
A photograph of Jim Totten in his oÆ e taken by his olleague Don DesBrisay around the time of Jim's retirement. Jim loved to demonstrate mathemati s using models, su h as the Len art SphereTM kit for investigating spheri al geometry that sits in the ba kground. (Photo ourtesy of Don DesBrisay)
258 IN MEMORIAM James Edward Totten, 1947{2008
With the sudden death of Jim (James Edward Totten) on Mar h 9, 2008, the Mathemati s Community lost someone who was dedi ated to mathemati s edu ation and to mathemati al outrea h. Jim was born August 9, 1947 in Saskatoon and raised in Regina. He obtained a B.S . from the University of Saskat hewan, and then an M.S . in Computer S ien e and a Ph.D. in Mathemati s from the University of Waterloo. After a two year NRC Postdo toral Fellowship in Tubingen, Germany, he joined the fa ulty at Saint Mary's University in Halifax. One of us (Robert) rst got to know Jim when he and Jim shared an oÆ e at the University of Saskat hewan while Jim visited there in 1978-1979. That was a long old winter, but Jim's a tive interest and enthusiasm for mathemati s and the tea hing of mathemati s made the year a memorable one. The next year Jim took a position at Cariboo College, where he remained as it evolved into the University College of the Cariboo and then into Thompson Rivers University, retiring as Professor Emeritus in 2007. During his years in Kamloops, Jim was a mainstay of the Cariboo Contest, an annual event whi h brought students to the ollege and whi h featured a keynote speaker, often drawn from Jim's list of mathemati al friends. This on e in luded an invitation to Robert, whi h featured a talk on publi key en ryption mostly memorable for the failure of te hnology at a key moment, mu h to Jim's amusement. Jim be ame a member of the CMS in 1981, and joined the editorial board of Crux in 1994. When Bru e was looking for someone to su
eed him as Editor-in-Chief, there was no doubt in his mind whom to approa h. Bru e spent a week in Kamloops staying at Jim's home and working with Jim and Bru e Crofoot to smooth the transition. Jim's attention to detail and are was 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 Jim through 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 playing ho key or golf, he was a tive with the Kamloops Outdoors Club. Jim was never just a parti ipant, always an a tive volunteer. Jim is survived by his loving wife of 40 years, Lynne, son Dean, daughterin-law Christie and granddaughter Mikayla of Se helt, father Wilf Totten of Edmonton, sister Judy Totten of Regina, sister Josie Laing (Neil) of Onoway, sister-in-law Marilyn Totten of Regina, mother-in-law Joy e Ladell of Swansea Point, brother-in-law Brian Ladell (Iris) of Red Deer, sister-in-law Constan 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 University Foundation 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
259
Totten Commemorative Issue After the sudden and unexpe ted death of Jim Totten on Mar h 9, 2008, we announ ed in the May 2008 issue that a spe ial ommemorative issue of CRUX with MAYHEM would appear in Jim's honour. We now have the pleasure of presenting this issue to our readers, for whi h the members of the editorial board have made spe ial e orts. This issue has also drawn upon many in Jim's ommunity, in luding various olleagues at Thompson Rivers University (TRU). The obituary prepared by Bru e Shawyer and Robert Woodrow (that appeared in the May 2008 CMS Notes) has been adapted with the assistan e of Lynne Totten for publi 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 ame Editor-in-Chief of CRUX with MAYHEM in 2003 and ran the journal until he began handing over the reins just months before his death last year. He also had an enormous in uen e in his home provin e of British Columbia in spreading his love of mathemati s through his outrea h a tivities. It is no surprise, then, that this spe ial issue has a distin tive regional emphasis. A pie e about Jim's a tivities and in uen e follows this editorial, whi h readers may wish to orrelate with the arti le by Clint Lee on the British Columbia Se ondary S hool Mathemati s Contest and Jim's inspiring role in its development. 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 Mayhem se 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 of mathemati al re reations, Ross Honsberger, who inspired Jim's interest in problem solving while he and Jim were oÆ e neighbours at the University of Waterloo. An arti le by Mi hel Bataille on the geometry of bi entri quadrilaterals gra es the arti les se tion, followed by further re reations in Awani Kumar's arti le on knight's tours on the surfa e of a ube. Serendipitously, we found it tting to reprint a book review written by Jim himself in the book reviews se tion. Also in the book reviews is an amusing ane dote of Andy Liu related to Jim. From our readers' submissions we have ompiled 26 spe ial problems dedi ated to Jim's memory in luding 10 in Mayhem and 16 in CRUX, a dozen of 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 with some hefty problems settled by CRUX readers. For those wishing to learn more about Jim we announ e here that the pro eedings of the May 2009 onferen e Sharing Mathemati s: A Tribute to Jim Totten will be published later this year.
260 Jim Totten's Rea h
John Grant M Loughlin
Shane Rollans opened the onferen e Sharing Mathemati s: A Tribute to Jim Totten with these words: \This onferen e is a tribute to our longtime olleague, Jim Totten ... Jim had at least ve passions. Foremost was his family. His wife, Lynne, shared his passion for the outdoors. His son, Dean, shared his passion for golf. His other passions were playing ho key and sharing his love of mathemati s, whi h brings us here." Jim is known for his work with the British Columbia Se ondary S hool Mathemati 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 Problems Editor prior to be oming Editor-in-Chief of CRUX with Mayhem in 2003. This pro le of Jim o ers a broader pi ture of Jim's ontributions in mathemati s and spirit, mainly as seen through the eyes of his olleagues. When Jim arrived at Cariboo College in 1979 he promptly organized a Putnam team that he led until 2005. He also started the \Problem of the Week" tradition, something he had done previously at the University of Saskat hewan and St. Mary's University. Ea h week he would olle t solutions, grade them, and post the results omplete with a solution. For the rst 27 years no problem was repeated, although Jim did allow himself to repost a few of his favourite problems in his nal year. Jim ompiled 80 of these problems and solutions into a book, Problems of the Week, Volume VII in the ATOM (A Taste of Mathemati s) Series published by the Canadian Mathemati 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 with the University of British Columbia (UBC). Jim then began tea hing upper level ourses ranging from Linear Programming to Complex Analysis to Graph Theory and Geometry. He also haired the UCC Dept. of Mathemati s and Statisti 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 in Kamloops, ranging from his time as a lo al high s hool student, to an undergraduate 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 mathemati ian with 14 papers from 1974 to 1980, many solo (e.g., Basi properties of restri ted linear spa es, Dis rete Math. 13 (1975), No. 1, 67-74) and others ollaborative (e.g., On a lass of linear spa es with two onse utive line degrees, 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 to Cariboo in 1979. He was obviously happy (and at pea e) with his de ision as he found other outlets for sharing his love of mathemati s." [At http://www.ams.org/mathscinet/search/author.html?mrauthid=298276 are 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 ertainly noti ed it as the high s hool kid visiting Cariboo College, but I thought all tea 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 pushed us hard on assignments, but his enthusiasm balan ed out this work. In other words, 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. That sort 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 in his area. Late in the ourse he gave us a talk about nite geometry and the hunt for the proje tive plane of order 10. I an say I really didn't understand mu h of what he said that day, but I was stru k by his ex itement in being able to take us to the `frontier of mathemati s' (his words). This seemed very important to him: namely, that we were part of the mathemati al ommunity, and we had the opportunity to ontribute. I was in graduate s hool when 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. Two thoughts: `Ah! That's what Jim was talking about,' and `Wow, he had a lot of 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, Jim showed up in my oÆ e one day with a olle tion of notes (about 30 years old) from a ourse he taught in graduate s hool. He was so keen to show me a onstru tion of a ertain ring of fun tions based on number theoreti ideas that he had presented in the (grad) ourse. It was ni e mathemati s, but a bit advan ed for third years without abstra t algebra. Still Jim's enthusiasm was so typi al. An opportunity to share should not be wasted." The Adrien Pouliot Award is a CMS award honouring \signi ant and sustained ontributions to mathemati s edu ation in Canada." Members of the TRU department nominated Jim in 2007. In a supporting letter, John Ciriani e hoed Ri k's sentiment: \He was parti ularly pleased to develop and tea h a ourse in geometry. Even students who found the ourse diÆ ult re ognized Jim's love of the subje t and were able to relate to his enthusiasm for it." John added, \I must mention that Jim derives great pleasure in making presentations in s ien e fairs and s hools. He is parti ularly pleased when he en ounters talented students who share his enthusiasm for mathemati 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 hing sta member in 2007 to re ognize his long term volunteer servi e. Don DesBrisay expresses his respe t: \With the math ontests, Problem of the Week, mathemagi shows, Putnam, Crux, his boxes of puzzles (many he 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 the joy and ex itement found in mathemati s and in life. All this as well as family, outdoor lub a tivities, golf and ho key!!" Quoting Kirk Evenrude: \Jim went at golf with the same determination and dedi ation as he did everything.
262 When 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 when it ame to golf. ... It was easy to spot Jim on the ourse; all you had to do was look for a big white Tilley hat. He loved playing in tournaments and was very 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 1999 CMS Edu ation Session (organized by Bru e Shawyer and Ed Williams) in St. John's raised the pro le of the BC ontest. While there Jim visited my
lass at Memorial University of Newfoundland sharing his wooden puzzles and love of re reational mathemati s with an enamored lass of future high s hool math tea hers. An entirely di erent form of mathemati al outrea h began in 1969 when Jim initiated the weekly pi kup ho key games through the 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 upon my ollaborations with Jim Totten. Jim ombined intelle t and heart in a manner that pla ed the greater good ahead of his own. Jim was one of those exemplary people in the a ademi ommunity who ons iously expressed appre iation for the work of others { a gift in itself. This form of outrea h is less 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 genuinely fumbling with mathemati al ideas. Jim was a great tea her. He was honoured with separate awards for tea hing and merit, and shortly after his retirement in 2007 with a Professor Emeritus designation. Those present at the Mar h 2008 elebration of Jim's life at the Grand Hall in TRU witnessed the love and outpouring of appre iation for Jim and his family. Students, ho key players, golfers, olleagues, hikers, and family spoke to the breadth of 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 enthusiasm for mathemati s and his e orts to promote mathemati al ideas extend far beyond his own lassroom. He has been an inspiration and a model for all in 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 deeper level Jim's phenomenal ontribution in this parti ular regard. His in uen e throughout the ollege/university system in the provin e annot be overstated." \Jim was a wonderful olleague and a good friend. He loved everything about Mathemati s and assumed everyone else would too if it was just shared with them. His outrea h a tivities showed true dedi ation to that goal from Crux to years of s hool visits and Problems of the Week. He was a generous person who loved his family and all aspe ts of his life and we all mourn his loss but are inspired by his example." John Grant M Loughlin
263
SKOLIAD No. 118 Lily Yen and Mogens Hansen
Please 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 solutions before the deadline. The de ision of the editors is nal. This spe ial issue of CRUX with MAYHEM features a sele tion from a ontest that Jim Totten played a pivotal role in developing. The Cariboo College High S hool Mathemati s Contest began in 1973 and has sin e developed into the British Columbia Se ondary S hool Math Contest. In 1992 a 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 for all 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 with his 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 was memorable in 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 and re e ts Jim's love of the game of golf." Our thanks go to John Grant M 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 behind 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 Contest 1990
Junior Final, Part B . A boy on a bi y le oasts down from the top of a hill. He overs 4 metres in the rst se ond and in ea h su
eeding se ond overs 5 metres more than in 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?
1
. Two golfers, on their way to the ourse, rea hed a railway rossing just as a 2.5 km train arrived. Rather than waiting, they de ided to go on to the next rossing 1 km along in the dire tion the train was going. They travelled at 50 km/h while the train travelled at 70 km/h. 2
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 to travel 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 sum of the digits, the student is able to tell you the number with whi h you began. Explain how this an be done.
√
. Suppose you throw 5 darts at a round board with a radius of 25 2 m. If all 5 darts sti k in the board, show that at least two of them must be within 50 m of ea h other.
4
.
The diameter, AB , of a ir le is divided into 4 equal parts by the points C , D, and E . Semi ir les are drawn on AC , AD, AE , and AB as shown. Find the ratio of the area of the shaded parts to the area of the unshaded parts. 5
.......................................... .......... ...... . . . . . . ..... ....................................... . . ..... . ....... .......... . ... . ..... ................................ ... .... ...... .......... ... ... .... ................... ... ... ... ........ ..... . ................................................................................................................ A
C
D
E
B
Con ours mathematique du College Cariboo 1990
Niveau se ondaire, nale junior, partie B
1. Un gar on roule sans pedaler, a partir du haut d'une olline. Il ouvre 4 metres pendant la premiere se onde ; haque se onde subsequente, il ouvre 5 metres de plus que pendant la se onde pre edente. Il arrive au bas de la
olline apres 11 se ondes. (a) Quelle est la distan e totale par ourue ? (b) Quelle est la vitesse moyenne du gar on, en metres a la se onde ? ( ) Quelle est la distan e par ourue pendant la derniere se onde ?
. Deux golfeurs arrivent a un passage a niveau juste au moment ou un train de longueur 2, 5 kilometres s'y presente. Au lieu d'attendre, ils de ident de du prese rendre au pro hain passage a niveau, a une distan e de 1 kilometre mier passage a niveau, dans la dire tion ou le train se dirige. Ils se depla ent a 50 kilometres a l'heure tandis que le train va a 70 kilometres a l'heure. (a) Combien de temps ont-ils a attendre au deuxieme passage a niveau, avant que le train ait degag e le passage ? (b) Au lieu de se depla er a 50 kilometres a l'heure, a quelle vitesse les deux golfeurs auraient-ils besoin de se depla er a n d'arriver au deuxieme passage a niveau justement au moment ou le train degage le passage ?
2
265 . Un etudiant vous demande de hoisir un entier de 1 a 9, puis de le multiplier par 109 ; il vous demande alors de lui fournir la somme des hi res dans e produit. Connaissant ette somme, l'etudiant peut alors vous dire ave quel entier vous avez ommen e. Expliquer omment il le fait. 3
√
. Vous lan ez 5 e hettes vers une ible de rayon 25 2 entimetres. Les 5
e hettes frappent la ible. Montrer qu'au moins deux d'entre elles doivent se retrouver a une distan e d'au plus 50 entimetres l'une de l'autre. 4
. Des demi er les sont tra es ave AC , AD, AE et AB omme diametres, ou le segment AB est divise en 4 parties egales par les points C , D et E . Pour le s hema a droite, determiner le ratio des parties olorees aux parties non
olorees. 5
....................................... ........... ....... . . . . . . ..... ...................................... . . . ..... ....... ................ .... ..... ...................................... .... ... ..... ... .. .. ... ... . ........................... ... ... ... ... ....... . ............................................................................................................. A
C
D
E
B
Next follow the solutions to the British Columbia Se ondary S hool Mathemati s Contest, 2008, Junior Final, Part A [2008 : 321{324℄. Note that problems 7 and 10 below have been adjusted so that the length of a segment is given as |XY | instead of XY .
. Jeeves the valet was promised a salary of $8000 and a ar for a year of servi 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: (B) 7200 (C) 7360 (D) 8000 (E) 15360 (A) 6400 1
Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON. Let x be the ost of the ar. The total value of everything that Jeeves 7 re eives after seven months equals 12 of what he would have re eived had 7 he worked for a whole year. Thus x + 1600 = 12 (x + 8000). It follows that 12x + 19200 = 7x + 56000, so 5x = 36800. Hen e x = 7360 and the answer is (C). Also solved by JOCHEM VAN GAALEN, student, Medway High S hool, Arva, ON.
. Re all that n! = n × (n − 1) × (n − 2) × · · · × 2 × 1. The maximum value of the integer x su h that 3x divides 30! is: (A) 30 (B) 14 (C) 13 (D) 10 (E) 4
2
Solution 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 equal to 30 along with the number of times ea h is divisible by 3.
266 Multiple 3 6 9 12 15 18 21 24 27 30 Times 1 1 2 1 1 2 1 1 3 1 Thus 30! is divisible by three 1 + 1 + 2 + 1 + 1 + 2 + 1 + 1 + 3 + 1 or fourteen times, and the answer is (B). Also solved by JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON.
. In the diagram, ABCDEF is a regular hexagon. Line segments AE and F C meet at Z . The ratio of the area of triangle F ZE to the area of the quadrilateral ABCZ is: (A) 1 : 5 (B) 1 : 4 (C) 4 : 1 (D) 5 : 1 (E) 1 : 6
A
3
F
B
...................................................... ...... ... ... ... .... ... ... .. ... .. ..... . . ... ... ..... . ... .. .. ... . . . .... Z ... . . ................................................................................................................ ... ... ... . ... . . . . .. ... ... ... .... .. ... ... . . . ... ... ... ... .. ... ....... ...........................................................
E
C
D
Solution by Jo hem van Gaalen, student, Medway High S hool, Arva, ON. Assume without loss of generality that the side length of the hexagon is 2. Angles inside a regular hexagon are 120◦ , so ∠F EZ = 120◦ −90◦ = 30◦ 1 ◦ and ∠EF Z = 2√120 = 60◦ . Thus △F ZE is a 30◦ { 60◦ { 90◦ triangle with √ sides 1, 2, and √ 3, namely |F Z| = 1, |EF | = 2, and |EZ| = 3. Thus, 3 △F ZE has area . 2 Now draw a line segment from B to D interse ting√CF at the point G. 3 Then △CGB ∼ . Sin e ABGZ is = △F ZE , so the area of △CGB is also 2 √ √ a re tangle√with sides 2 and√ 3, it has area 2 3. Hen e the area of trapezoid √ 3 5 3 ABCZ is +2 3= . 2 2 Therefore, the desired ratio is
√
√ 3 5 3 : = 1 : 5, and the answer is (A). 2 2
Also solved by JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON.
. 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 an integer, then the value of ⌊z⌋ + ⌊1 − z⌋ is: 4
(A) −1
(B) 0
(C) 1
(D) 2
(E) z
Solution 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 z is 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 independent of z (as long as z is not an integer). Thus you an nd the answer by simply substituting some value, say 1.5, for z. Of ourse this does not prove anything but relies on trust in the formulation of the question.
267 . 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 many students failed in ea h subje t, as well as in the various ombinations: 5
subje t A # failed 12
B
5
C
8
AB
2
AC
6
BC
3
ABC
1
(For instan e, 5 students failed in Biology, among whom there were 3 who failed both Biology and Chemistry, and just 1 of the 3 who failed all three subje ts.) The number of students who passed all three subje ts is: (B) 16 (C) 21 (D) 24 (E) 26 (A) 4 Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON. The two students who fail Anatomy and Biology in lude the single student who fails all three subje ts. Therefore just one student fails Anatomy and Biology but not Chemistry. Likewise, ve students fail Anatomy and Chemistry but not Biology, and two students fail Biology and Chemistry but not 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 two subje ts is 12 − 1 − 5 − 1 = 5. Likewise, one student fails Biology and passes the other two subje ts, and zero students fail Chemistry but pass the other two 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 pass all 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 diagrams below show how you may sequentially enter information into the Venn Diagram. A............................................................................................... B ... ... . . .. ... ... ... .. .. .. ... .. .. ................................. ..... . ....... .. . ... .. . . . . . . ...... .. ... .... .... . . . . . . ... ... . ... ... ... . . . . .... . .... ... .. .... ........ ....... ...... ................................................................ ... .. .. ... . . .... .. ..... .... . . . . .......... . ........................
1
C
A................................................................................................ B ... ... . . .. ... ... ... .. ... ... .. 1 .. .. .................................. ..... ........ ..... . ... .. . . . . . ..... .. ... .... ... . . . . . . ... 1 ... ... ... ... ... . . . ..... .. 5 .... .... 2 ... .... ....... ........ ...... ................................................................. ... ... ... .. . .... ... ..... ... . . . ......... . . ..........................
C
A............................................................................................... B . . ... ... . ... ... . .. 5 ... 1 .... 1 .... .. . . .. .... ... ........................... ........ ... .. ......... . .... .. ... ..... .. ... ..... ..... ... ..... ..... . . . . .... .. .... ... .. ... ....... ...... ........ ................................................................ .. .. ... .. . . ... .. .... ... . ...... . . . . . ...............................
5 1 2 0
C
268 6. Six points, A, B , C , D , E , and F are arranged in the formation shown in the diagram, with A, B , and C on a straight line. Three of these six points are sele ted to form a triangle. The number of su h triangles that an be formed is: (A) 12 (B) 14 (C) 16 (D) 19 (E) 20
Dr r A
Er r B
r C
................................................................................................................
r F
Solution by Jo hem van Gaalen, student, Medway High S hool, Arva, ON.
There are 63 = 6 C3 = 3!(66!− 3)! = 20 ways to hoose three points from the given six. One of these, {A, B , C}, yields no triangle as A, B , C are ollinear. Thus, you an make 19 triangles, and the answer is (D). Also solved by JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON.
. In the diagram, C is the entre of the ir le and is tangent to the ir le at D. AC is a straight line. If |AD| = 10 and |AB| = 7, the length of BC is:
7
AD
(A) (D)
√
151 − 7 2 √ 51 2
(B) (E)
√ 14 7 2
(C)
........................... ............. ....... ....... ...... ..... .... .... ... ... ... . .. .... .. .. .... .. .. . . ... . . . . .. .. ..... . .. . . . . .. .. ... . . . . . .. . . . . ... .......... . . ............ . .. ... ........... .... ....... ............ .... ... ....... ........ ..........................................................................................
C p
51 14
B p
A
p
p D
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 of the ir le. Then |CD| = |BC| = r and |AC| = |AB| + |BC| = 7 + r. By the 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 . Then the 14 answer 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) 0
Solution 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 the fa tors, we only need to onsider the units digit of powers of 8. The table shows the rst of these. Power 81 82 83 84 85 86 87 Units digit 8 4 2 6 8 4 2 Note that the pattern 8, 4, 2, 6 repeats itself after 84 as it must, sin e only the units digits matter.
269 Now 2008 mod 4 = 0, that is, the remainder when dividing 2008 by 4 is 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 divisible only by one and itself. Consider the set of two-digit numbers less than 40 that are either prime or divisible by only one prime number. From this set sele t those for whi h the sum of the digits is a prime number, and the positive di eren e between the digits is another prime number. The sum of the values of the numbers sele ted is: (A) 29 (B) 41 (C) 54 (D) 70 (E) 93
Solution 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 prime narrows the list to 11, 16, 23, 25, 29, and 32. Requiring that the di eren e of the digits be a prime further narrows the list to 16, 25, and 29. The sum of these three numbers is 70, and the answer is (D). D.......................................................................................................C ... ... .... .. ... .... ....... .. ... ... ... .... F............................ .......... ... ... .... ......... ....... ......... .
. ... ... . ...... ...... ... ... ..... .... ............. ... ... ... ... ... ........ ... .... . . . . ... ... ..... ..... . . . . ... . . . ... ... ... ............ . ... ... ... ........ . . ......................................................................................................
.... ......... ...... .......... .
. In the diagram ABCD is a re tangle with |AD| = 1, and both DE and BF perpendi ular to the diagonal AC . Further, |AE| = |EF | = |F C|. The length of the side AB is: √ √ (A) 2 (B) 3 (C) 2 √ (D) 5 (E) 3 10
E
A
B
Solution by Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON. Let x be the ommon length of AE , EF , and F C . Then |EC| = 2x. 2 2 2 2 2 2 By the Pythagorean √ Theorem, |AE| +|DE| = |AD| , so x +|DE| = 1 , when e |DE| = 1 − x2 . Note that ∠ECD = 90◦ − ∠EDC = ∠ADE , so △DEA ∼ △CED. Therefore, |ED| |EA| = |DE| |CE|
⇐⇒
√
x = 1 − x2
√
1 − x2 2x
.
Thus 2x2 = 1 − x2 , so x2 = 13 . Finally, |AB| = |CD| and CD has length p p √ √ √ |DE|2 + |CE|2 = 1 − x2 + 4x2 = 1 + 3x2 = 1 + 3/3 = 2, and the answer is (A).
That ompletes another Skoliad. This issue's prize for the best solutions goes to Jixuan Wang, student, Don Mills Collegiate Institute, Toronto, ON. We look forward to re eiving more solutions from more readers.
270
MATHEMATICAL MAYHEM Mathemati al Mayhem began in 1988 as a Mathemati al Journal for and by High S hool and University Students. It ontinues, with the same emphasis, as an integral part of Crux Mathemati orum with Mathemati al Mayhem. The Mayhem Editor is Ian VanderBurgh (University of Waterloo). The other sta members are Monika Khbeis (As ension of Our Lord Se ondary S hool, Mississauga) and Eri Robert (Leo Hayes High S hool, Frederi ton).
Mayhem Problems
Please send your solutions to the problems in this edition by 15 November . Solutions re eived after this date will only be onsidered if there is time before publi 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 memory of Jim Totten, hen e the spe ial numbering used below. The numbering of the regular Mayhem problems will resume in subsequent issues. The editor thanks Jean-Mar Terrier of the University of Montreal for translations of the problems. 2009
. Proposed by Shawn Godin, Cairine Wilson Se ondary S 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, 1 instead of writing 11 , they would write 12 + 13 + 12 . The unit fra tion 12 an 12 be written in terms of other unit fra tions as 12 = 13 + 16 . Find an in nite family of unit fra tions ea h of whi h an be written as the sum of two unit fra tions. Totten{M1
. Proposed by Bru e Shawyer, Memorial University of Newfoundland, St. John's, NL. Totten{M2
The boundary of the shadow on the moon is always a ir ular ar . On a ertain day, the moon is seen with the shadow passing through diametri ally opposite points. If the entre of the ir ular ar forming the shadow is on the ir umferen e of the moon, determine the exa t proportion of the moon that is not in shadow.
.................................................. ........ .... ........... ....... ........ ... ..... ...... ... . . . . .... ... .. . . .... . . ... ... .... . . .. ... ... . . .. ... . ... ... .. .. ... ... .. .. ... ... .. .. ... ... .. .... . ... .. ..... .. .. . . . . .. .. .. .. . . . .. .. .. ... .. ... .. ... .. ... ... .. .... ... . . . . . .... .. ..... .... ... ..... ....... ... ........ ....... .... ............ ........ ..............................................
271 . Proposed by John Ciriani, Kamloops, BC.
Totten{M3
Prove that the quadrati equation ax2 + bx + c rational root if a, b, and c are odd integers. .
Totten{M4
AB.
= 0
does not have a
Proposed by Bill Sands, University of Calgary, Calgary,
In a survey, some students were asked whether they liked the olour orange. 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 students in the survey liked orange. Find the smallest possible number of students in the survey. . Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania.
Totten{M5
Let a 6= 1 be a positive real number. Determine all pairs of positive integers (x, y) su h that loga x − loga y = loga (x − y). . Proposed by Suzanne Feldberg, Thompson Rivers University, Kamloops, BC. Totten{M6
It is widely known how to draw a 5-pointed star qui kly. To make it symmetri , one pla es 5 verti es at 72◦ intervals about a ir le and onne ts the verti es with line segments of equal length without lifting one's pen. By starting from a xed point and using the same method, one an draw two di erent (and symmetri ) 7-pointed stars without lifting one's pen. ..r .... ........ . . ....r.....................................................r .... .. ... .... .... . ...... ..... ... . . . .... . . . .. ..........r ..r.......... .... . ......... .......... ................. .................... ... ....................... .. ...r .r.
............................. ...... ..... ..... .... .... . .... . . ... ... ... ... ... . .. .. . .. .. . .. .. .... .. .. .. .... .. .. .. ... ... . ... . . .. . .. .. ... .. ... .. ... ... . .... . ... .... .... ..... ..... ...... .............................
.r ...... . ..r......... .. .... ..........r .... .......... .. .. ......... ... .... .................. ... .... ........ ............ ......... ................. . . . . . . . ..r......................................................................................r .. ......... ... .......... ............ ..r ..r
............................. ...... ..... ..... .... .... . .... . . ... ... ... ... ... . . .. . .. .. . .. .. .... .. .. .. .... . ... .. ... ... . ... . . .. . .. .. ... .. ... .. ... ... . .... . ... .... .... ..... ..... ...... .............................
How many di erent 6-pointed, 8-pointed, or 9-pointed stars an one draw this way? How many di erent n-pointed stars an one draw this way? . Proposed by John Grant M Loughlin, University of New Brunswi k, Frederi ton, NB. An unmarked ruler is known to be exa tly 1 4 ............................................................................ 6 m in length. It is possible to exa tly measure .. .. all integer lengths from 1 m to 6 m using only 2 .................................................................... marks, as shown in the diagram, at 1 m and 4 m, 6 m sin e 2 = 6 − 4, 3 = 4 − 1, and 5 = 6 − 1. Totten{M7
... .................................................. .... .
... .................................................... ...
272 Suppose 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 tly measure 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 all integer lengths from 1 m to 30 m.
⋆ Determine the smallest number of marks required on the ruler to be
( )
able to exa tly measure all integer lengths from 1 m to 30 m.
. 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 h that a < b < c. We say that two triples (a, b, c) and (u, v, w) are equivalent if 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 a geometri sequen e) and harmoni if a1 + 1c = 2b (that is, the re ipro als of its terms form an arithmeti sequen e). Totten{M8
(a) Verify that if (a, b, c) is geometri , then all triples equivalent to it are also geometri . (b) Verify that if (a, b, c) is harmoni , then all triples equivalent to it are also harmoni . ( ) Let G be the set of equivalen e lasses of geometri triples and H be the set of equivalen e lasses of harmoni triples. Determine a one-to-one
orresponden e between G and H . . Proposed by Kirk Evenrude, Kamloops, BC. A train 900 m long, travelling at 90 km/h, approa hes a bridge. Totten{M9
100 m
long
(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 slow down at the rate of 0.2 m/s2 . Now how long does it take to lear the bridge? 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 integers su h that p divides A, p2 does not divide A, and p does not divide B , then the Diophantine equation Ax2 + By2 = p2008 does not have any solutions in positive integers x and y. .................................................................
273 . Propose par Shawn Godin, E ole se ondaire Cairine Wilson, Orleans, ON. Les an iens egyptiens e rivaient toutes leurs fra tions en termes de fra tions unitaires distin tes, 'est-a-dire de fra tions distin tes ave 1 omme 11 1 1 1 numerateur. Par exemple, au lieu d'e rire , ils auraient e rit . + + 12 2 3 12 1 La fra tion unitaire 2 peut e^ tre e rite en termes d'autres fra tions unitaires 1 1 1
omme 2 = 3 + 6 . Trouver une famille in nie de fra tions unitaires, ha une pouvant e^ tre e rite
omme somme de deux fra tions unitaires. Totten{M1
. Propose par Bru e Shawyer, Universite Memorial de Terre-Neuve, St. John's, NL.
Totten{M2
La frontiere de l'ombre sur la lune est toujours un ar de er le. Un ertain jour, on peut onstater que l'ombre passe par des points diametralement opposes. Si le entre de l'ar ir ulaire formant ette ombre se trouve sur la ir onferen e de la lune, determiner la portion exa te de la lune qui n'est pas dans l'ombre.
............................................... ........... ......... .... ........ ....... ... ....... ..... ... .... ... .... . . . .... ... ... . . ... .. .. . ... . . .. ... .. . . ... .. .. . ... .. .. .... .. .. ... ... .. ... .. ... ... .. . .. ... .. .. . ... .. . . . . . .. .. .. .. . . . ... .. .. .. ... .. .. .... .. ... ... .... ... .. .... . . . ..... ... .. ... ....... ..... ... ......... ....... .............. ....... ..................... .......................
. Propose par John Ciriani, Kamloops, BC. Montrer que l'equation quadratique ax2 + bx + c = 0 n'a pas de ra ine rationnelle si a, b et c sont des entiers impairs. Totten{M3
.
Totten{M4
AB.
Propose par Bill Sands, Universite de Calgary, Calgary,
Lors d'un sondage, on a demande a ertains etudiants s'ils aimaient la par oui, tandis
ouleur orange. Chez les gar ons exa tement 2% ont repondu que hez les lles, la proportion exa te de oui a et e de 59%. Comme au total, exa tement 17% des repondants ont dit aimer l'orange, on demande de trouver quel est le plus petit nombre possible de parti ipants au sondage. . Propose par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie. Soit a 6= 1 un nombre reel positif. Trouver toutes les paires d'entiers positifs (x, y) telles que loga x − loga y = loga (x − y). Totten{M5
Totten{M6. Propose par Suzanne Feldberg, Universite Thompson Rivers, Kamloops, BC. On sait bien omment dessiner rapidement une etoile a 5 sommets. Pour en faire une symetrique, on pla e les 5 sommets a intervalles de 72◦ sur un er le et on relie les sommets par des segments re tilignes de longueur egale sans lever son rayon. En partant d'un point xe et en utilisant la m^eme
274 methode, on peut dessiner sans lever son rayon deux etoiles di erentes (et symetriques) a 7 sommets. ...r. ... ........ . . . ....r.......................................................r .... . ... .... ...... ....... ... . . . .......... .. ... . . . ..r........ ... ................r .......... . . . . . .............. .......... .. ... ...................... .. .......r ..r....
............................. ...... ..... ..... .... .... ... .... ... . ... .. . . ... ... .. . . .. . .. .. ... ... ... .. .. ... . ... ... .. .. ... .. ... . .. .. .. ... ... ... ... ... . . ... .... .... .... .... ...... .... ......... ...... ...................
..r ...... . ..r.......... .. ... ...........r .... ............ ............ ... .... .................. ... ... ..... .... .. .................. ............................ . . . . . . . . .r.......................................................................r . .. ......... .. ......... ............ ....r ..r..
............................. ...... ..... ..... .... .... ... .... ... . ... .. . . ... .. .. . . .. . .. .. ... ... ... .. .. .. ... ... .. .. ... ... ... .. .. .. ... ... ... . . ... ... ... ... .... ... .... .... ...... ..... . . . ......... . ...................
Combien d'etoiles di erentes a 6 sommets, 8 sommets, ou 9 sommets peut-on ainsi dessiner? Combien d'etoiles di erentes a n sommets peut-on ainsi dessiner? Totten{M7. Propose par John Grant M Loughlin, Universite du NouveauBrunswi k, Frederi ton, NB. On utilise une regle sans graduation dont on 1 4 sait qu'elle mesure exa tement 6 m. Il est pos............................................................................ .. sible de mesurer exa tement toutes les longueurs .. .................................................................... entieres de 1 m a 6 m ave seulement 2 marques, 6 m
omme indique dans la gure i- ontre, a 1 m et 4 m, ar 2 = 6 − 4, 3 = 4 − 1 et 5 = 6 − 1. Supposons maintenant qu'on utilise une regle d'exa tement 30 m de long. ... .................................................. ... ..
... .................................................... ...
a n d'^etre apable (a) Trouver un moyen de pla er 9 marques sur la regle de mesurer exa tement toutes les longueurs entieres, de 1 m a 30 m. (b) Montrer qu'au moins 7 marques sont ne essaires pour e^ tre apable de mesurer exa tement toutes les longueurs entieres, de 1 m a 30 m.
⋆ Determiner le plus petit nombre de marques a pla er sur la regle a n
( )
d'^etre apable de mesurer exa tement toutes les longueurs entieres, de 1 m a 30 m.
. Propose par Edward J. Barbeau, Universite de Toronto, Toronto, ON. Soit T l'ensemble de toutes les triplets ordonnes 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 partitionner T en lasses d'equivalen e. Le triplet (a, b, c) est dit geom etrique si ac = b2 ( .-a-d. ses el ements forment une suite geom etrique) et harmonique si a1 + 1c = 2b ( .-a-d. les inverses de ses el ements forment une suite arithmetique). Totten{M8
275 (a) Veri er que si (a, b, c) est geom etrique, alors tous les triplets qui lui sont equivalents sont aussi geom etriques. (b) Veri er que si (a, b, c) est harmonique, alors tous les triplets qui lui sont equivalents sont aussi harmoniques. ( ) Soit G l'ensemble des lasses d'equivalen e de triplets geom etriques et H l'ensemble des lasses d'equivalen e de triplets harmoniques. Trouver une orrespondan e biunivoque entre G et H . . Propose par Kirk Evenrude, Kamloops, BC. Un train de 900 m de long s'appro he d'un pont d'une longueur de 100 m a la vitesse de 90 km/h. Totten{M9
(a) En ombien de se ondes le train traversera-t-il le pont? (b) Supposons qu'au moment d'atteindre le pont, le train de el ere de 0.2 m/s2 . Combien de temps mettra-t-il ette fois pour traverser le pont? Totten{M10. Propose par Ni holas Bu k, College de New Caledonia, Prin e George, BC. Montrer que si p est un nombre premier, et si A et B sont des entiers positifs tels que p divise A, p2 ne divise pas A, et p ne divise pas B , alors l'equation diophantienne Ax2 + By2 = p2008 n'a au une solution en entiers positifs x et y.
Mayhem Solutions
. Proposed by Bru e Shawyer, Memorial University of Newfoundland, St. John's, NL. Suppose that A is a six-digit positive integer and B is the positive integer formed by writing the digits of A in reverse order. Prove that A − B is a multiple of 9. M363
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 form 105 a + 104 b + 103 c + 102 d + 101 e + 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 digits of A in reverse order, B is of the form 105 f + 104 e + 103 d + 102 c + 101 b + a.
276 Their di eren e, A − B , an then be simpli ed as follows: A−B
= = = =
105 a + 104 b + 103 c + 102 d + 101 e + f
− 105 f + 104 e + 103 d + 102 c + 101 b + a
105 a − a + 104 b − 101 b + 103 c − 102 c + 102 d − 103 d + 101 e − 104 e + f − 105 f 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 these digits 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 High S hool, Vinh Long, Vietnam; PETER CHIEN, student, Central Elgin Collegiate, St. Thomas, ON; COURTIS G. CHRYSSOSTOMOS, Larissa, Gree e; ANTONIO GODOY TOHARIA, Madrid, Spain; JOSE HERNANDEZ SANTIAGO, student, Universidad Te nologi a de la Mixte a, Oaxa a, Mexi o; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; HUGO LUYO SANCHEZ, IES \Abastos", Ponti ia Universidad Catoli a del Peru, Lima, Peru; RICARD PEIRO, 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 Ziegler Publi S hool, Waterloo, ON; and JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON.
. Proposed by the Mayhem Sta . A semi ir le of radius 2 is drawn with diameter AB . The square P QRS is drawn with P and Q on the semi ir le and R and S on AB . Is the area of the square less than or greater than one-half of the area of the semi ir le? M364
Solution by Peter Chien, student, Central Elgin Collegiate, St. Thomas, ON, modi ed by the editor. Pla e AB on the x-axis with the y midpoint of AB (that is, the entre of 6 the semi ir le) at the origin. The full Q(a, b) P
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 on x the semi ir le, then a2 + b2 = 4. Sin e P QRS is a square, whi h must A S R B sit symmetri ally inside the semi ir le, then we also have b = 2a. √ 4 2 2 5 2 2 2 Thus, a + (2a) = 4, hen e a = 5 and so a = √ = 5 . Sin e the .... .... .. .. ................................. . . . . . . . . . . . . . . . . . . . . . . . ........... ........ . . . ... . .. . . . . ......... ................................................................................................. . . . . . . .. . .... ......... ...... ..... .... ..... . .... . .... . . . .... . . . . .... ... ... .... .... .... .... .... . . . . . . . ... ... ... ... ... .... . . . . ... ... . .. ... . . . . . ... ... . .. .. . . . . . . .. . .. ... ... ..... . ... . .. ... ... ..... .. .... .. .... .... .... ... .. ..... .. .... .... ... .. ... .. ... ... . . . ... . . . . ................................................................................................................................................................................................................................................ .... ..... ...
length of one side of P QRS is
√ 4 5 b = 2a = , 5
5
then the area of the square
277 is
√ 2 4 5 5
=
80 16 = = 3.2. 25 5
One-half the area of the semi ir le is 12 × 21 ×π×22 = π < 3.2. The area of 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, University 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, IES \Abastos", Spain; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; RICARD PEIRO, 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 Ziegler Publi S hool, Waterloo, ON; and JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON. M365. Proposed by Alexander Gurevi h, student, University of Waterloo, Waterloo, ON. Let D be the family of lines of the form y = nx + n2 , with n ≥ 2 a 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 ept if and only if this line does not interse t any line from D at a point with an x- 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 . Its y -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 simultaneous equations y = m and y = nx + n2 do not have a solution for x that is a nonnegative integer, and (b) if m is omposite, then there is a positive integer n ≥ 2 su h that the simultaneous equations y = m and y = nx + n2 do have a solution for x that is a nonnegative integer. Putting this another way, we must prove that if m is prime, then the equation m = nx + n2 does not have a nonnegative integer solution for x for any n ≥ 2, and if m is omposite, then the equation m = nx + n2 does have a nonnegative 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 with x ≥ 0. Note that m has only two positive divisors, namely m and 1. Sin e n ≥ 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 is no n for whi h m = nx + n2 has nonnegative integer solutions for x.
278 Assume next that m is omposite. Then m = pq for some integers p and q with 2 ≤ p ≤ q. Thus, we want to nd integers n and x with n ≥ 2 and x ≥ 0 su h that m = n(x + n) = pq. Setting n = p and x + p = q satis es the equation. Here, x = q − p ≥ 0, so the restri tions on x and n are 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 line does not interse t any line from D at a point with an x- oordinate that is a nonnegative 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; and JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON.
. Proposed by John Grant M Loughlin, University of New Brunswi k, Frederi ton, NB. The roots of the equation x3 + bx2 + cx + d = 0 are p, q, and r. Find a quadrati equation with roots (p2 + q2 + r2 ) and (p + q + r).
M366
Solution by Shamil Asgarli, student, Burnaby South Se ondary S hool, Burnaby, BC. Sin e p, q, and r are the roots of the ubi equation, we an fa tor the left 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 with the 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 2 (−b) = p2 + q 2 + r 2 + 2c, so p2 + q 2 + r 2 = b2 − 2c. A possible quadrati equation with the desired roots is therefore
or x2 +
x − b2 − 2c x − −b = 0,
b − b2 + 2c x + 2bc − b3 = 0.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; GEORGIOS BASDEKIS, student, 1st High S hool of Karditsa, Karditsa, Gree e; 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; JOSE HERNANDEZ SANTIAGO, student, Universidad Te nologi a de la Mixte a, Oaxa a, Mexi o; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; HUGO LUYO IES \Abastos", SANCHEZ, Ponti ia Universidad Catoli a del Peru, Lima, Peru; RICARD PEIRO, 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.
. Proposed by George Tsapakidis, Agrinio, Gree e. For the positive real numbers a, b, and have a + √ c we √ √b + c Determine the maximum possible value of a bc + b ac + c ab.
M367
= 6.
279 Solution by Jose Hernandez Santiago, student, Universidad Te nologi a de la Mixte a, Oaxa a, Mexi o. Applying the Arithmeti Mean{Geometri Mean Inequality to positive a+b+c 3 real numbers a, b, and c we obtain abc ≤ = 8 and onsequently 3 √ √ abc ≤ 2 2. (Equality holds here if and only if a = b = c and so if and only if a = b = c = 2.) We an then further on lude that √ √ √ a bc + b ac + c ab
Now, √ √ √ 2 a+ b+ c
= =
√ √ √ √ √ √ a abc + b abc + c abc √ √ √ √ abc a+ b+ c = √ √ √ √ ≤ 2 2 a+ b+ c . =
(1)
√ √ √ 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, thus √ √ √ we 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.) √ √ √ 2 Hen e, a + b + c ≤ 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 when a = 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; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; JOE HOWARD, Portales, NM, USA; RICARD IES \Abastos", Valen ia, Spain; ALEX SONG, student, Elizabeth Ziegler Publi S hool, PEIRO, Waterloo, ON; EDWARD T.H. WANG, Wilfrid Laurier University, Waterloo, ON; and JIXUAN WANG, student, Don Mills Collegiate Institute, Toronto, ON. There was one in orre t solution submitted.
. Proposed by J. Walter Lyn h, Athens, GA, USA. An in nite series of positive rational numbers a1 + a2 + a3 + · · · is the fastest onverging in nite series with a sum of 1, a1 = 12 , and ea h ai having numerator 1. (By \fastest onverging", we mean that ea h term an is su
essively hosen to make the sum a1 + a2 + · · · + an as lose to 1 as possible.) Determine a5 and des ribe a re ursive pro edure for nding an . M368
280 Solution by Robert Sheets, Southeast Missouri State University, Cape Girardeau, MO, USA. Let sn be the nth partial sum of the in nite series. Then a1 = 12 and 1 s1 = . Sin e all terms of the in nite series are positive, none of the terms 2
an be negative or zero, thus no partial sum an be greater than or equal to 1 1. We therefore need a2 < 1 − s1 = . Seeking the largest su h a2 , we nd 2 that a2 = 13 sin e a2 is a fra tion of positive integers with numerator equal to 1. Then s2 = 12 + 13 = 56 whi h gives a3 < 16 , when e a3 = 71 . Similarly, 41 1 1 we nd that s3 = 42 , giving a4 = 43 , and s4 = 1805 , giving a5 = 1807 . 1806 Next, we des ribe a re ursive pro edure to nd an . We onje ture that − 1) − 1 if an = k1 , then sn = k(kk(k and an+1 = k(k −11) + 1 . (Note that − 1) 2(1) + 1 = 3, 3(2) + 1 = 7, 7(6) + 1 = 43, and 43(42) + 1 = 1807, so these equations hold for n = 1, 2, 3, and 4.) We prove these re ursive relations by indu tion. We have already veri ed the base ases above. Suppose that for some positive integers n and k − 1) − 1 we have an = k1 , sn = k(kk(k , and an+1 = k(k −11) + 1 . We prove − 1) that the relations will also hold for sn+1 and an+2 . Let t = k(k − 1) + 1; 2 this means that an+1 = 1t and sn = tt − . Then −1 sn+1
Finally, sin e an+2
= sn + an+1 t2 − 2t + t − 1
=
t−2
=
t2 − t − 1 t(t − 1) − 1 = t(t − 1) t(t − 1)
t−1
+
< 1 − sn+1 =
1 t
=
1 t(t − 1)
t(t − 1)
.
and an+2 is the largest fra tion
with numerator 1 satisfying this property, we nd that an+2 = t(t − 11) + 1 , as required. Therefore, the result holds by indu tion and for all positive integers n, if an = k1 for some positive integer k, then an+1 = k(k −11) + 1 .
Also solved by ARKADY ALT, San Jose, CA, USA; RICHARD I. HESS, Ran ho Palos Verdes, 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 the On-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 of Sylvester'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 .
281 Problem of the Month
Ian VanderBurgh
Among Jim Totten's many interests was his involvement in mathemati s outrea h. In parti ular, Jim was a driving for e behind the Cariboo College High S hool Mathemati s Contest. He edited a volume of problems taken from those ontests written between 1973 and 1992. This month, we look at one of the problems from this volume. (1989 Cariboo College High S hool Mathemati s Contest, Senior Final Round, Part B) In the gure, AEF B , BGHC , and ACJ D are squares onstru ted on the sides of △ABC . If the ( ombined) area of the two shaded squares equals the area of the rest of the gure, show that the area of △ABC equals the area of △F BG and then nd the number of degrees in ∠ABC .
Problem
D
..... ................. .......... . . . . . . . . . . . . ..... ............ ...... ..... E ................ ... J . . ......A . .. . ......... ...... . .. ..... . .. .. . . .. .......... ...... ... F ............... . ... .. . ... ............ β ... ............................................. ... .. C .. ... B .. ... . ... .. ... ... ... .. ... . ... .. .... ..... .......................................... G
H
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 of logo elsewhere on the web. Before we look at the solution, there is one really useful formula upon whi h we should agree. If in △XY Z we have XY = z and Y Z = x, then the area of △XY Z is equal to 12 xz sin(∠XY Z). This formula is a great alternative to the standard area formula \ 12 bh" if all you have is one angle of a triangle and the lengths of the two sides en losing it. We'll derive this formula after the solution to the problem. Solution Let BC = a, AC = b, and AB = c. Sin e AEF B is a square, then BF = AE = c. Sin e BGHC is a square, then BG = CH = a. Sin e ACJ D is a square, then AD = CJ = b. Sin e ∠ABC = β and ∠ABF = ∠CBG = 90◦ , it then follows that ∠F BG = 360◦ − β − 90◦ − 90◦ = 180◦ − β . We rst need to prove that △ABC and △F BG have equal areas. From the formula in the preamble, the area of △ABC is 12 (BC)(AB) sin(∠ABC) or 12 ac sin β . Similarly, the area of △F BG is 12 (BG)(F B) sin(∠F BG) or 1 ac sin 180◦ − β . 2 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 a bit 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 . Can you see why? At this point, we should probably use the pie e of information that we were given, namely, that the ombined area of square AEF B and square BGHC equals the area of the rest of the gure. We use the short-hand |AEF B| to denote the area of gure AEF B . Thus, we are told that |AEF B| + |BGHC| = |F BG| + |ABC| + |EAD| + |HCJ | + |ACJ D| .
Using some of what we know so far, this be omes c2 + a2 = 4|ABC| + b2 ,
or
c2 + a2 = 4
1 ac sin β + b2 . 2
Remember, we're trying to nd β . We seem to have too many other pie es of information oating around to have any hope of doing this. But at this point, the amazing pattern re ognition abilities of the brain might ki k in. This equation looks somewhat similar to a law that we often use. This might 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 cos β cos β
sin e
= 2ac sin β + (a2 + c2 − 2ac cos β) ; = 2ac sin β ; = sin β ,
ac > 0. Sin e cos β = sin β , β = 45◦ , and we are done.
and
β
is an angle in a triangle, then
To me, this was quite surprising. Well, the answer itself wasn't so surprising sin e these problems almost always have 30◦ , 45◦ , or 60◦ as an answer. But, it was surprising to me that the angle β was ompletely determined from the given information while no other information (side lengths or angles) an be determined. Before wrapping up the olumn this X month, we should go ba k and look at the ............... . formula from the preamble. Suppose that ... .. ......... z ..... .. ∠XY Z is a ute. Drop a perpendi ular from ..... . . ..... X to F on Y Z . ...... ... ..... . . . . Then the area of △XY Z is equal to .......................................................................... . 1 (Y Z)(XF ). But Y Z = x and we also have Y Z 2 F XF = XY sin(∠XY Z) = z sin(∠XY Z), so the area equals 12 xz sin(∠XY Z), as required. Can you prove this in the ase that angle Y is obtuse or a right angle? .................. ................
283 The Tanker Problem
Ross Honsberger
When I walk my puppy with his retra table leash, he is forever getting behind, at hing up and running past, utting a ross in front, and getting behind again. When I was having my after-lun h nap the other day, this reminded me of those old \patrol boat ir ling an o ean liner" problems and I 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 and the tanker travels at a onstant speed along a straight path. The patrol boat goes up the left side, a ross the front, down the right side, and a ross the ba k, and keeps doing it over and over. The patrol boat travels in only two dire tions of the ompass { when going parQ allel to the path of the tanker, it travels in S .. .. straight lanes parallel to the tanker, one on .. .. .. . ea h side at a distan e of 25 metres from it, 50 B .. C ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . and when rossing in front or behind, it goes ... .. .. .. ... .. .. .. straight a ross perpendi ular to the path of 6 .. . . .. . . . . the tanker. ... .. .. .. ... .. .. .. Negle ting the dimensions of the pa.. . . .. . 450 . . . trol boat (that is, onsidering it to be rep... .. .. .. ... .. ..25 .. resented geometri ally by a point) and given .. -. . . .. . . . that it goes onstantly at twi e the speed of ... .. .. .. ... .. .. .. the tanker and that its turns are instanta.. . . .. . . . . neous, what is the shortest distan e that the .. .. .. .. ... .. .. .. patrol boat must travel in ompleting one ? .. .................................. M ...s
y le around the tanker? A D .. . .. .. I felt the problem would be an easy .. .. re reation with a pleasing analysis and I ex. . pe ted a solution along the following lines. P R Let the tanker and the lanes of the paFigure 1 trol boat be labeled as in Figure 1 and let the boat travel towards Q when it is in lane P Q. Consider a y le that begins at M on P Q; that is, to begin the boat is abreast of the stern AD of the tanker. While the boat travels 900 m up P Q, the tanker will have advan ed 450 m and the boat will be abreast of the front BC of the tanker at the point E on P Q (Figure 2). Copyright
c 2009
Canadian Mathemati al So iety
Crux Mathemati orum with Mathemati al Mayhem, Volume 35, Issue 5
284 If the boat were to start rossing beQ S tween its lanes at this point, it would rash .. Y . N ...................................................................... into the tanker at a point 12.5 m down the ... . . . side. In order to ross in front of the tanker, F ......................................................................... L .. G ... .. K .. the boat must get 25 + 50 = 75 m a ross the . . . .. . . . gap to the right hand edge of the tanker before .................................... .. . . .. V ... .. U .... the tanker advan es to its level. Therefore, it .. .. ... B .. needs to have a lead of 37.5 m when it starts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... E . C ... . its turn, for the tanker will advan e 37.5 m . .. .. ... .. ... .. .. while the boat is going a ross this 75 m. . . . .. . . . . . In order to get a 37.5 m lead, the boat ... .. .. . . ... . .. .. must ontinue up P Q another 75 m beyond E . ... . . .. . . ... (while the boat is doing this the tanker ad.. .. .. . ... . .. .. van es 37.5 m to yield a net lead of 37.5 m). . ... . . .. . . Summarizing, the boat and the tanker ... .. .. . . ................................... .. .. are abreast at E and BC ; as the boat advan es . A D . the 75 m to F , the tanker advan es 37.5 m to P R U V ; the boat turns at F and while going the 75 m along F K , the tanker advan es 37.5 m to Figure 2 GK . Thus, the boat and the tanker just miss ea h other at K . And while the boat ompletes the remaining 25 m along KL to its other lane, the tanker takes a 12.5 m lead at Y N . Thus, when the boat turns down SR it is already 12.5 m down that side of the tanker. Now, the boat need go only to a point H that is 12.5 m from the far end before making its turn a ross the ba k (while it rosses the 25 m gap toward the tanker, the tanker moves ahead this 12.5 m and the boat just misses the bottom right orner of the tanker). Hen e the boat needs to over a distan e of only 450 − 12.5 − 12.5 = 425 m along this side before turning. Sin e it goes twi e as fast as the tanker, while the boat overs two-thirds of this distan e, the tanker moves ahead the other third, and so the boat need travel down that side only two-thirds of 425 m, that is, a distan e of 850 m. 3 Now, from just missing the tanker at its bottom right orner, the boat still has to go 75 m to get ba k to lane P Q. While the boat is doing this, the tanker gets ahead 37.5 m, and so, in order to at h up and omplete its y le, the patrol boat must go another 75 m up P Q to get abreast of the tanker. Altogether, then, the patrol boat must travel a grand total of .... .... .... ....
900 + 75 + 100 +
1 850 + 100 + 75 = 1533 3 3
m.
This solution blithely assumes that the patrol boat and the tanker harmlessly slide by ea h other when they arrive simultaneously at the tanker's right hand orners. However, it makes a great deal more sense to interpret su h an event as a ollision: both vessels an hardly move into the same position at the same time without fatal onsequen es for the patrol boat. Fortunately, this nagging un ertainty is easily removed by adding the ondition that the two vessels are never to get loser to ea h other than 25 m.
285 This is readily a
ommodated by having the boat get a lead of 62.5 m at before turning along the dangerous 75 m stret h F K between the lanes. This gives the boat a net lead of 62.5 − 37.5 = 25 m when it rea hes the right hand side of the tanker, thus avoiding any possibility of a ollision and providing the 25 m separation at this riti al jun ture. In order to get 62.5 m ahead, the boat must go another 125 m up P Q beyond E to Q a point F (Figure 3). Thus, when the patrol S ... ... boat rea hes F , the tanker is at U V , and the L . . . patrol boat is 62.5 m ahead. F .................................................................... G .. W ................................. N .. Now the boat turns along F L, and when . . .. . it rea hes L, the tanker has advan ed 37.5 m .. I ...................................... T ... ... .. .. .. to IT , putting the tanker 25 m behind. While . . . . . . . the boat pro eeds 25 m along LG to the other .. U .................................... V ... ... .. .. .. lane, the tanker uts its lead to 12.5 m at W N . .. . . . .. . . .. Thus, when the boat turns down SR, it .....B .. . . .. ............................... C ... is already 12.5 m ahead. In order to remain E .. ... .. .. .. 25 m from the other end of the tanker when it ... .. .. .. ... .. .. .. goes ba k a ross the rst lane, it will need to . . .. . . . . . go farther down the side to a point H that is .. .. .. .. ... .. .. .. 12.5 m beyond the other end. Then, when it .. . . .. . . . .. turns and loses the 25 m stret h to the nearer .. .. .. . ... .. .. .. side of the tanker, the tanker will have ad. . .. . . . . . van ed another 12.5 m to yield a separation .. .. .................................. . A of 25 m between them. Hen e the patrol boat D . must over 450 + 12.5 + 12.5 = 475 m along P R this side before making its turn. As before, it Figure 3 needs to travel down that side only two-thirds of this distan e, that is, 950 m. 3 Thus, when passing the right rear orner of the tanker, the boat is already 25 m behind, and still has 75 m of the gap to negotiate to get ba k to lane P Q. While the boat is doing this, the tanker gets ahead another 37.5 m, for a total of 62.5 m, and so, in order to at h up and omplete its y le, the patrol boat must go another 125 m up P Q to get abreast of the tanker. Altogether, then, the patrol boat must travel a grand total of F,
... ... ..
.... .... .... ....
900 + 125 + 100 +
950 2 + 100 + 125 = 1666 3 3
m.
This seemed most satisfa tory until the brilliant solution of my olleague Larry Ri e (retired from the University of Toronto S hools and the University of Waterloo) revealed another unwarranted assumption in my solution. Sin e the patrol boat had a 25 m lead when it rea hed the right hand edge of the tanker, it went un hallenged that the minimum 25 m distan e between the boat and the front right orner of the tanker would be maintained as the boat ompleted the nal 25 m of the rossing to its other lane.
286
................... . .... .... ... .....................
Unfortunately this is not true; it is not hard to show that the vessels do get
loser together than 25 m when the boat is ompleting the rossing. So, how mu h of a lead would the boat have to have L 2x N .................................................................. G when it passes the right hand .. ... edge of the tanker in order .. .. .. . to maintain a 25 m separation? (25 + y) − x ... ... . . . . Sin e 25 m isn't enough, per.. ... d .. .... haps a lead of 27 m would do? .. ... Again, it is not diÆ ult to see ..... that 27 m is not enough. So, J ... .. how about 28 m? Bingo: 28 m .. is suÆ ient, with a little to x .. .. spare. Consequently, any lead Figure 4 .. . greater than 28 m would also suÆ e. But we seek the shortT est y le of the boat around the tanker. Thus we need to determine the smallest a
eptable lead. We know it's some value, 25 + y, between 27 and 28 m. 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 a lead T L of (25+ y) metres. As the tanker advan es x metres to J on T L, the boat advan es 2x metres along LG to the point N . We want to determine the smallest positive value y su h that the segment d = J N is at least 25 m for 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 + y 2 + 50y + 625 − 50x − 2xy ,
and for d2 ≥ 625, we need 5x2 + y 2 + 50y − 50x − 2xy ≥ 0 .
Sin e the only variable here is x, we may write f (x) = 5x2 + y 2 + 50y − 50x − 2xy ,
where we require f (x) ≥ 0 for 0 ≤ x ≤ 12.5. Di erentiating, we get f ′ (x) = 10x − 50 − 2y = 2(5x − 25 − y) ,
whi h vanishes for x = 5 + 15 y. Sin e the graph of f (x) is a parabola opening upwards, its minimum o
urs when f ′ (x) = 0.
287 Hen e min f (x)
1 2 1 1 5 5 + y + y 2 + 50y − 50 5 + y − 2y 5 + y
=
5 4 2 y + 40y − 125 . 5
=
5
5
Now, we don't want this minimum to be greater than 0, for in that ase d would always ex eed 25 and a lead of 25 + y would be greater than it need be. So for minimum y, we require the minimum to equal 0, and 4 2 y + 40y − 125 = 0 , 5
or
4y 2 + 200y − 625 = 0 .
The number y being positive, we obtain y
=
−200 +
=
25 √ 2
p 2002 − 4(4)(−625) 2(4)
=
−200 +
√ 50000
8
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 25 m separation and provide a shortest y le to a reasonable degree of a
ura y. We have seen that an additional 125 m beyond E gives the boat a 25 m lead when it rea hes the right hand edge of the tanker. Hen e a lead of 27.951 m would require the boat to go a further 2(2.951) = 5.902 m for a total of 130.902 m up its left lane before starting its turn a ross the front. Moreover, its lead is now 12.5 + 2.951 = 15.451 m when it starts down the right lane. To produ e the same separation at the other end of the tanker, the boat needs to go this same distan e, 15.451 m, farther down this lane than the end of the tanker, for a total passage of 450+ 2(15.451) = 480.902 m. As before, the boat needs to over only two-thirds of this distan e, namely 320.601 m. After returning the 100 m to its left lane, the boat would be another 50 m behind, for a total of 65.451 m, and therefore needs to go another 130.902 m up P Q in order to omplete the y le. Altogether, then, the boat travels approximately a total of 900 + 130.902 + 100 + 320.601 + 100 + 130.902 = 1682.405 m . Finally, let's onsider Larry's wonderful solution { a ompletely di erent approa h. The unknown distan es in the y le are those travelled by the patrol boat in the lanes P Q and RS ; the distan es travelled in rossing between the lanes is always 100 + 100 = 200 m. Larry obtains these unknown distan es from the basi formula distan e = speed × time .
288 If 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 in lanes P Q and RS respe tively, sin e the distan e travelled in lane P Q will be (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, you might well wonder how he is going to obtain anything of value about these times. Larry very ingeniously enlists the aid of the aptain of the tanker in this matter. To this end, onsider the observations of the aptain as he wat hes the boat ir le his tanker (Figure 5). First, Q S he sees the boat run up lane P Q from . .. E to F . At F , the boat turns a ross F ... .. .. the lanes in front of him and, as it pro- 37.5 ... . .
eeds, it keeps getting loser to him be........................ 25 ... U L ... 25 ....................50 .. 12.5 C ....
ause the tanker is moving ahead. Upon .. .. .. G . . . ... rea hing the other lane, the boat pro.. .. .. . ... .. .. ..
eeds along it and eventually turns ba k ... .. .. .. ... to lane P Q. As it rosses ba k, the .. .. .. . ... .. .. .. tanker, still going forward, moves into ... .. .. .. ... the lead. Finally, the boat gets abreast .. .. .. . ... .. .. .. of the tanker up lane P Q. Thus the ap... .. .. .. ... tain sees the boat y le the tanker along .. .. .. ... .. .. .. the sides of a trapezoid EF GH . ... .. .. .. . Between two points on a slanted ... .. .. .. ... .. .. .. H side of the trapezoid, the boat moves .. ........................................... .. twi e as far toward the other lane as it K. .. V .. .. does toward, or away from, the tanker, .. .. . implying that the slopes of these sides .. .. E 1 . . are ± 2 . Thus the trapezoid is isos eles P R and enjoys the symmetry of su h a gure. Figure 5 Sin e the slope of F G is − 12 and LC = 75 m, it follows that LF = 37.5 m. The symmetri al part EK is therefore also 37.5 m, making the total length of side EF = 450 + 2(37.5) = 525 m. Similarly, sin e CU = 25 m, then U G = HV = 12.5 m, making the length of side GH = 450 − 2(12.5) = 425 m. Now, time equals distan e divided by speed, so the time t1 that it takes the boat to traverse EF is 525 divided by the boat's speed along EF . While the boat goes at speed of 2v relative to the o ean, the aptain sees it pro eed up EF at the speed at whi h it overtakes the tanker; that is, at the relative speed of 2v − v = v. Hen e the aptain al ulates t1 = 525 . v ............ ............ ............ ............ ............ ............ ............ ............ ............ ............ ................ ..... ..... ..... ..... ..... ..... ..... .......... ............ ............ .....
..... ............ ........... .......... ............. ..... ..... ..... . . . . . . . . . . . . ............ ............ ............ ............ ............ . . . . . . . . . . . ........... ............ ............ ..... ..... ..... .....
289 Along side GH , the aptain observes the boat is going at a relative . speed 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
475 3v
425 3v
= 1050 +
850 1 = 1333 3 3
m.
Adding in the 200 m travelled in rossing between lanes, we obtain the same earlier grand total of 1533 13 m for the rst ase, in whi h we generously allowed the vessels to slide by ea h other without in ident. Allowing an additional 25 m bu er at ea h rossing between the lanes would in rease ea h of the parallel sides of the trapezoid by 50 m and yield the orresponding times t1 = 575 and t2 = 475 . v 3v The resulting y le, then, would have a total length of 575 v
+ 2v
+ 200 = 1150 + 316
as found earlier. Finally, (Figure 6), a universal separation √ of 25 m, would add an extra 2(12.5) 5 ≈ 25(2.236) = 55.9 m to ea h of the parallel sides of the original trapezoid EF GH . This is be ause the right-angled △XY C has legs in the ratio 1 : 2. Can you see why? Thus, t1 = 580.9 and t2 = 480.9 v 3v whi h yields a y le length of approximately 2(580.9) + 2 = =
480.9 3
+ 200
1161.8 + 320.6 + 200 1682.4 m ,
in lose agreement with our previous result.
2 2 + 200 = 1666 3 3
m,
.. .. .. .. .. F .............. .. .. .............. .. .......... .. .. .......... .. .......... .. .. .......... .. .. ..........12 ...........5. Y ... .. X... ............ .. .. ... . 25 ....... G .. .... .. .. ................................................. .. .. . C. .. .. .. . ... .. .. .. .. .. . .. . ... .. .. .. .. .. . .. . ... .. .. .. . .. .. . .. . ... .. . .. . . ............................................. . ........ ............... .
2v
Figure 6
Ross Honsberger 326 Dale Cres ent Waterloo, ON, N2J 3Y3
[email protected]
290
THE OLYMPIAD CORNER No. 279 R.E. Woodrow
After a mu h needed break, Joanne Canape has agreed to ta kle transforming my s ribbles into LATEX again. Wel ome ba k! We begin this number with problems of the 37th Austrian Mathemati al Olympiad of 2006. Thanks go to Robert Morewood, Canadian Team Leader to the IMO in Slovenia, for olle ting them for our use. th
37
AUSTRIAN MATHEMATICAL OLYMPIAD
Regional Competition (Qualifying Round) April 27, 2006
. Let 0 < x < y be real numbers and
1
H =
2xy x+y
,
G =
√
xy ,
A =
x+y 2
,
and
Q =
s
x2 + y 2 2
be the harmoni , geometri , arithmeti , and quadrati means of x and y, respe tively. It is well known that H < G < A < Q holds. Order the intervals [H, G], [G, A], and [A, Q] by length. . Let n > 1 be an integer and a a real number. Determine all real solutions (x1 , x2 , . . . , xn ) of the following system of equations:
2
x1 + ax2 x2 + a2 x3 x3 + a3 x4
= 0, = 0, = 0,
.. .
xn−1 + an−1 xn xn + an x1
= 0, = 0.
. In a nonisos eles triangle ABC , w is the bise tor of the external angle at The extension of AB interse ts w in D. Let kA be the ir um ir le of the triangle 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 the point in whi h these two tangents interse t. Assume that the two points A and B are given. Determine the set of all points P = P (C), su h that ABC is a ute-angled but not isos eles. 3
C.
291 . Let {hn }∞ n=1 be a harmoni sequen e of positive rational numbers. In other words, ea h hn is the harmoni mean of its neighbours:
4
hn =
2hn−1 hn+1 hn−1 + hn+1
.
Prove that if some term hj of the sequen e is the square of a rational number, then the sequen e ontains an in nite number of terms hk that are ea h squares of rational numbers.
National Competition (Final Round, Part 1) May 21, 2006
. Let k and n be integers with k ≥ 2, n > 10k , and the de imal expansion of n ending with exa tly k zeros. Give the best possible lower bound (in terms of k = k(n) ≥ 2) for the number of ways to represent n as the di eren e of squares of two nonnegative integers. 1
2
. Prove that the sequen e
(
(n + 1)n n2−n 7n2 + 1
)∞
is stri tly in reasing.
n=0
. 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 the triangle is isos eles. 3
. Let ⌊u⌋ denote the greatest integer less than or equal to the real number u and 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 numbers with denominator 3 (after an ellation) whi h do not lie in the image of f . 4
National Competition (Final Round, Part 2) Day 1 (May 31, 2006)
. Find the number of nonnegative integers n ≤ N with the property that the de imal expansion of some multiple of n ontains only the digits 2 and 6 (not ne essarily the same number of ea h). 1
. Prove that
2
√ 3 3(a + b + c) ≥ 8 abc +
s 3
a3 + b3 + c3 3
for all positive real numbers a, b, and c. Determine when equality holds.
292 . Given triangle ABC , let point R be on the extension of AB beyond with BR = BC , and let point S be on the extension of AC beyond C with 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 hexagon AC ′ BA′ CB ′ is the sum of the areas of triangles ABC and A′ B ′ C ′ . 3
B
Day 2 (June 1, 2006)
. Determine all rational numbers x su h that 1 + 105 · 2x is the square of a rational number. 4
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.) . Let A be a nonzero integer. Find all integer solutions of the following system of equations:
6
x + y2 + z3
=
A,
1 1 1 + 2 + 3 x y z
=
1 A
xy 2 z 3
=
A2 .
,
Next we give the Brazilian Mathemati al Olympiad 2005. Thanks again go to Robert Morewood, Canadian Team Leader to the IMO in Slovenia, for
olle ting them for our use. Brazilian Mathemati al Olympiad 2005
First Day (O tober 22, 2005)
. 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) of positive integers su h that 111 . . . 1} × 111 . . . 1} is a palindrome. | {z | {z m ones n ones 2. Determine the smallest real number C su h that 1
125 125 16 C x2005 + x2005 + · · · + x2005 ≥ x1 x2 x3 x4 x5 x125 1 2 5 1 + x2 + · · · + x5
for 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 the interior of the ube. Determine the largest ℓ su h that there exists a square of side ℓ ontained in a ube of edge length 1.
293
Se ond Day (O tober 23, 2005) . We have four harged batteries, four un harged batteries, and a radio whi h needs two harged batteries to work. We do not know whi h batteries are harged and whi h ones are un harged. What is the least number of attempts that suÆ es to make sure the radio will work? (An attempt onsists of putting two batteries in the radio and he king if the radio works or not). 4
. Let ABC be an a ute triangle and let F be its Fermat point, that is, the interior point of ABC su h that ∠AF B = ∠BF C = ∠CF A = 120◦ . For ea h of the triangles ABF , BCF , and CAF , draw its Euler line, that is, the line onne ting its ir um entre and its entroid. Prove that these three lines are on urrent.
5
. Let b be an integer and let a and c be positive integers. Prove that there exists a positive integer x su h that
6
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 al Olympiad, written April 26{29, 2006. Thanks again go to Robert Morewood, Canadian Team Leader to the IMO in Slovenia, for olle ting them for our use. Croatian Mathemati al Olympiad 2006
National Competition th Grade
4
. Prove that three tangents to a parabola always form the sides of a triangle whose altitudes interse t on the dire trix of the parabola.
1
. Let k and n be positive integers. Prove that
2
k n4 − 1 n3 − n2 + n − 1 + (n + 1)n4k−1
is divisible by n5 + 1.
3. The ir les Γ1 and Γ2 interse t at the points A and B . The tangent line 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 the angle ∠CAD, meets Γ1 at M , meets Γ2 at N , and meets the ir um ir le of △ACD at P . Prove that |AM | = |N P |. 4. Six islands are onne ted by the Ferryboat and the Hydrofoil Boat ompanies. Any pair of islands is onne ted, in both dire tions, by exa tly one
294 of these two ompanies. Prove that it is possible to tour four of the islands using 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 2006 written at Ni osia, Cyprus. Again we thank Robert Morewood, Canadian Team Leader to the IMO in Slovenia, for olle ting them for the Corner. Balkan Mathemati al Olympiad 2006
Ni osia, Cyprus
. (Gree e) Let a, b, and c be real numbers. Prove that
1
1 1 1 3 + + ≥ a(1 + b) b(1 + c) c(1 + a) 1 + abc
.
. (Gree e) Let ABC be a triangle and m a line whi h interse ts the sides AB and AC at interior points D and F , respe tively, and interse ts the line BC at a point E su h that C lies between B and E . The lines through points A, B , C and parallel to the line m interse t the ir um ir le of triangle ABC again at the points A1 , B1 , C1 , respe tively. Prove that the lines A1 E , B1 F , and C1 D are on urrent. 2
. (Romania) Find all triples of positive rational numbers (m, n, p) su h that ea h of the numbers
3
m+
1 np
,
n+
1 pm
,
p+
1 mn
is an integer. . (Bulgaria) Let m be a xed positive integer. For ea h positive integer a let the sequen e {an }∞ n=0 be de ned by a0 = a and for n ≥ 0 the re ursion 4
an+1 =
( a n 2
an + m
if an is even , 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 of the Finnish Mathemati al Olympiad 2006, Final Round. Thanks again go to Robert Morewood, Canadian Team Leader at the IMO in Slovenia for olle ting them.
295 Finnish Mathemati al Olympiad 2006
Final Round (February 3, 2006)
1
. Determine all pairs (x, y) of positive integers su h that x + y + xy = 2006 .
. For all real numbers a, prove that
2
3 1 + a2 + a4
≥
1 + a + a2
2
. The numbers p, 4p2 + 1, and 6p2 + 1 are primes. Determine p.
3
. Prove that if two medians of a triangle are perpendi ular, then the triangle whose sides are ongruent to the medians of the original triangle is a right triangle. 4
. The game of Nelipe is played on a 16 × 16 grid. At ea h turn a player pi ks a number from {1, 2, . . . , 16} and writes it in one of the squares of the grid. At ea h stage of the game the numbers in ea h row, olumn, and in every one of the 16 four by four subsquares must be di erent. A player loses if he or she has no legal move. Whi h player wins, if both play with an optimal strategy? 5
................................................................................................................................................................................................................................................................................................................................................................................................................... ........................................................................................................................................................................................................................................ ................................................................................................................................................................................................................................ ........................................................................................................................................................................................................................................................................................................ ............................................................................................................................................................................................................................................................................................................................................................................ .............................................................................................................................................................................................................................................. ........................................................................................................................................................................................................................................................................................ .................................................................................................................................................................................................................................................................................................................................................................................... ........................................................................................................................................................................................ ............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... ....................................................................................................................................................................................................................... .................................................................................................................................................................................................... .........................................................................................................................................................................................................................................................................................
Next we turn to solutions from our readers to problems from the September 2008 number of the Corner, from the 19th Lithuanian Team Contest in Mathemati s written O tober 2, 2004 [2008: 282{284℄. . Twelve numbers { four 1's, four 5's, and four 6's { are written in some order around a ir le. Does there always exist a three-digit number omprised of three neighbouring numbers (its digits an be taken lo kwise or
ounter lo kwise) that is divisible by 3? 1
Solved by John Grant M Loughlin, University of New Brunswi k, Frederi ton, NB; and Titu Zvonaru, Comane sti, Romania. We give the response of Grant M Loughlin. The arrangement at right shows that there is not always a three digit number omprised of three neighbouring numbers that is divisible by 3.
....... .... ... 5 .. ... .. .. 5 .... ... 5 ... ... .. . ... 6 5 ........ . . . . ............ ............ ......... 1 6
. Solve the equation 2 cos(2πx) + cos(3πx) = 0.
2
1 6 .......................... 1
.. ......
6 ......
1
296 Solved 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 written equivalently 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 2
or
2 cos2 (πx) + cos(πx) − 2 = 0,
πx = 2kπ ±
2π 3
or
and hen e x = 2k ±
2 3
or
hen e,
17 − 1 . So, either cos(πx) = cos 2π 4 3 √ 17 − 1 θ = arccos . Thus, 4 √ 17 − 1 πx = 2kπ ± arccos , k ∈ Z, 4
either cos(πx) = − 12 or cos(πx) = or cos(πx) = cos θ, where
√
x = 2k ±
1 arccos π
√
17 − 1 4
,
k ∈ Z.
. Solve the equation 3x⌊x⌋ = 13, where ⌊x⌋ denotes the integer part of the number x. 3
Solved by Pavlos Maragoudakis, Pireas, Gree e; Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON; and Titu Zvonaru, Comane sti, Romania. We give Wang's write-up. √
We show that the only solution is x = 339 . Sin e 00 is unde ned, x 6= 0. If x < 0, then ⌊x⌋ = −n for some integer n ≥ 1. Thus, 3x⌊x⌋ = x3n . If 3 n is odd, then n < 0. If n is even, then n ≥ 2 implies x < −1 or −x > 1. x 3 Thus, x3n = (−x) < 3. In either ase, 3x⌊x⌋ 6= 13. Hen e, it remains to n
onsider x > 0. If x ≤ 2, then x⌊x⌋ ≤ 4 and if x ≥ 3, then x⌊x⌋ ≥ 27. Hen e, ⌊x⌋ 2 < x < 3. Let x = 2 + = 13 be omes r where 0 < r < 1. 2Then 3x 2 2 3(2+r) = 3 r +4r +4 = 13, and we have 3r +12r −1 = 0. Solving, we √ √ √ obtain r = −12 ±6 156 = −2± 339 . Sin e r > 0, we reje t r = −2− 339 . Hen e, r = −2 +
√
39 3
and it follows that x =
√
39 . 3
. Let 0 ≤ a ≤ 1 and 0 ≤ b ≤ 1. Prove the inequality
4
√
a
2b2
b 2 + √ ≤ √ 2 +5 2a + 5 7
.
297 Solved by George Apostolopoulos, Messolonghi, Gree e; Mi hel Bataille, Rouen, Fran e; Oliver Geupel, Bruhl, NRW, Germany; Pavlos Maragoudakis, Pireas, Gree e; and Titu Zvonaru, Comane sti, Romania. We give the solution of Bataille. Sin e a2 ≤ 1, we have 2b2 + 5 ≥ 2b2 + 2a2 + 3. Similarly, we have 2a2 + 5 ≥ 2b2 + 2a2 + 3 and it follows that
a b a+b . + √ ≤ √ √ 2 2 2 2b + 5 2a + 5 2b + 2a2 + 3 √ √ Therefore, it suÆ es to show that 7(a + b) ≤ 2 2b2 + 2a2 + 3. Squaring and 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.
. If a, b, and c are nonzero real numbers, what values an be taken by the expression a2 − b2 b2 − c2 c2 − a2 + + ? 2 2 2 2 2 2
5
a +b
b +c
c +a
Solution by Titu Zvonaru, Comane sti, Romania. 2
2
−b We have −1 < aa2 + < 1, sin e −2a2 < 0 < 2b2 . b2 Setting α = a2 b2 + b2 c2 + c2 a2 , we obtain
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
a2 − b2
= a4 b2 − a2 b4 − a4 c2 + b4 c2 + a2 c4 − b2 c4 = a2 b2 a2 − b2 − c2 a4 − b4 + c4 a2 − b2 = a2 − b2 a2 b2 − a2 c2 − b2 c2 + c4 = a2 − b2 a2 b2 − c2 − c2 b2 − c2 = − a2 − b2 b2 − c2 c2 − a2 .
It follows that
2 2 a − b2 a − b2 b2 − c2 c2 − a2 b2 − c2 c2 − a2 a2 + b2 + b2 + c2 + c2 + a2 = a2 + b2 · b2 + c2 · c2 + a2 < 1 . Let k ∈ (−1, 1) be a real number. Then, setting a2 = b2 1 + k , we 1−k
obtain
lim
c→0
a2 − b2 b2 − c2 c2 − a2 + + a2 + b2 b2 + c2 c2 + a2
= k +1−1 = k.
298 It follows that
a 2 − b2 b2 − c2 c2 − a 2 + 2 + 2 2 2 2 a +b b +c c + a2
takes 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-valued fun tion is a onne ted subset of the real line, that is, an interval. Thus, the expression in a, b, and c a hieves all values between any two of its given values, 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 y6
= =
y 4 + 18 , x4 + 18 .
Solved by George Apostolopoulos, Messolonghi, Gree e; Oliver Geupel, Bruhl, NRW, Germany; George Tsapakidis, Agrinio, Gree e; Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON; and Titu Zvonaru, Comane sti, Romania. 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 − y 6 + x4 − y 4 2 2 3 3 x2 − y 2 + x2 − y 2 x2 − y 2 x4 + x2 y 2 + y 4 + x2 − y 2 x2 + y 2 x2 − y 2 x4 + x2 y 2 + y 4 + x2 + y 2
=
0,
=
0,
= =
0,
0.
However, x4 +x2 y2 +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) w 2 + 2w + 6 = 0, hen e w = 3 sin e w 2 + 2w + 6 = x4 + 2x2 + 6 is positive for all real numbers x. √ x2 = √ w = 3, hen e x = ± 3 and there are four solutions Now, √ (x, y) = (± 3, ± 3). . Find all triples (m, n, r) of positive integers su h that
7
2001m + 4003n = 2002r .
Solution by Titu Zvonaru, Comane 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), we have that 2001m + 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.
299 If 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 e m 2001 = (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 h only (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 is divisible 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, Comane 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 given by (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) and in 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 implies that (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). . Is it possible that, for some a, both expressions are integers? 9
1 − 2a a2
√
35
and a +
√ 35
Solved by Oliver Geupel, Bruhl, NRW, Germany; Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON; and Titu Zvonaru, Comane sti, Romania. We give Geupel's write-up. √
√
35 Yes, this is possible. Let m = 1 − 2a and n = a + 35; then 2 a √ m = 1 and n = ±6 if a = ±6 − 35. We prove additionally√that there are no√other solutions. If m = 0, then a = 7035 and n = 717035 is not an integer. Hen e, √ |m| ≥ 1 be ause m is an integer, and then |1 − 2a 35| ≥ a2 .
300 √
√
If 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 simplify to obtain m n2 + 35 − 71 = (m − 1)2n 35. Sin e ea h side is a rational number, n = 0 or m = 1, whi h qui kly leads to the unique solution.℄ . What is the greatest value that a produ t of positive integers an take if their sum is equal to 2004? 11
Solved by George Apostolopoulos, Messolonghi, Gree e; and Oliver Geupel, Bruhl, 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 α1 α2 · · · αn . If αk ≥ 4 for some k, then repla ing αk by 2 and αk − 2 leaves the sum the same and an only in rease the produ t, sin e 2(αk − 2) ≥ ak for ak ≥ 4. Therefore, the maximum produ t is a hieved when ea h ak is either 2 or 3. Now, 2 + 2 + 2 = 3 + 3 but 23 < 32 , so the produ t is maximized by 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 x 2 and y an take if their sum is equal to a xed positive integer n, is n4 . In the spe ial ase n = 2004 the result is 10022 = 1004004. Indeed, we have n 2 n2 n2 xy = x(n − x) = − x − + . For even n, we obtain xy ≤ , where 2 4 4
equality holds if and only if x = y = n2 . For odd n we have xy 1 n + 1 , 2 . where equality holds if and only if {x, y} = n − 2
≤
n2 − 1 , 4
. Positive integers a, b, c, u, v, and w satisfy the system of equations
12
a+u b+v c+w
= 21 , = 31 ,
= 667 .
Can abc be equal to uvw? Solved by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON; and Titu Zvonaru, Comane 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.
301 . Let u be the real root of the equation x3 − 3x2 + 5x − 17 = 0, and let v be the real root of the equation x3 − 3x2 + 5x + 11 = 0. Find u + v . 13
Solved by George Apostolopoulos, Messolonghi, Gree e; Mi hel Bataille, Rouen, Fran e; Jose Luis Daz-Barrero, Universitat Polite ni a de Catalunya, Bar elona, Spain; George Tsapakidis, Agrinio, Gree e; and Titu Zvonaru, Comane sti, Romania. We give the solution of Daz-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 tive fun tion, as an be easily he ked. Furthermore, g(u − 1) g(v − 1)
= (u − 1)3 + 2(u − 1) = f (u) − 3 = 17 − 3 = 14 , = (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, and moreover sin e g is bije tive, we dedu e that u − 1 = 1 − v, hen e u + v = 2. . Does there exist a polynomial, P (x), with integer oeÆ ients su h that 4 9 , 10 for all x in the interval 10 the inequality |P (x) − 23 | < 10−10 is valid?
15
Solution by Oliver Geupel, Bruhl, NRW, Germany.
We give su h a polynomial expli itly. Let P (x) = 23 1 − 1 − 3x4 and note that P (x) has integer oeÆ ients. For 0.4 ≤ x ≤ 0.9 we have
n
−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 < 10−10
satis ed for 0.4 < x < 0.9 if n=
1 − log10 3 − log10 2 1 − log10 9.683
2 (0.9683)n < 10−10 . 3
+ 1.
3
is
It now suÆ es to hoose
(Using a al ulator, we nd that n = 59.)
16. Does there exist a positive number a0 su h that all the members of the in nite sequen e a0 , a1 , a2 , . . . , de ned by the re urren e formula p an = an−1 + 1, n ≥ 1, are rational numbers?
Solution by Oliver Geupel, Bruhl, NRW, Germany. We prove by ontradi tion that there is no su h a0 . For ea h n ≥ 0 let p an = n , where pn and qn are oprime positive integers. From q n
p2n 2 qn
it follows that
= a2n = an−1 + 1 =
pn−1 + qn−1 qn−1
2 qn−1 p2n = qn (pn−1 + qn−1 )
.
, (1)
302 Then qn−1 | qn2 (pn−1 + qn−1 ) and the numbers qn−1 and pn−1 + qn−1 are
oprime, hen e qn−1 | qn2 . On the other hand, (1) yields qn2 | qn−1 p2n and pn and qn are oprime, hen e qn2 | qn−1 . It follows that qn2 = qn−1 for all n ≥ 0. We on lude that qn = 1 for 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 the proof. 17. Let a, b, and c be the sides of a triangle and let x, y , and numbers su h that x + y + z = 0. Prove that a2 yz + b2 zx + c2 xy ≤ 0 .
z
be real
Solved by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain; Mi hel Bataille, Rouen, Fran e; George Tsapakidis, Agrinio, Gree e; and Titu Zvonaru, Comane sti, Romania. We give the write-up of Amengual Covas. More generally, let α, β , γ , x, y, and z be real numbers and suppose that x + y + z = 0. For real numbers u and v we have (u + v)2 ≤ 2 u2 + v2 , hen e (β + γ)2 yz + (γ + α)2 zx + (α + β)2 xy ≤ 2 β 2 + γ 2 yz + γ 2 + α2 zx + α2 + β 2 xy = 2 α2 x(y + z) + β 2 y(z + x) + γ 2 z(x + y) = −2 α2 x2 + β 2 y 2 + γ 2 z 2 ≤ 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 transformation a = β + γ , b = γ + α, and c = α + β , where α, β , and γ are uniquely determined positive numbers and apply the above inequality. Equality holds only if the triangle is equilateral and x = y = z = 0. . Points M and N are on the sides AB and BC of the triangle ABC , AM BN respe tively. It is given that M = = 2 and ∠ACB = 2∠M N B . B NC Prove that ABC is an isos eles triangle. 18
C .... ....... ... .. .. ... ... ... ... ... .... . . .. ... ... . ... ... .. ... .. ... ... .. ... .. . . ..... .. .. . .. ... . . . ... .... ... .. . . . . . . . . .. ... ... γ .... ... ... ... ... ... .. ... . ... ... 2 .... ... .. . . . . . ... .. ... .. . . . . .. . .. . . . . . . .. . ... .. ... ... .. .. ... ... . . . . . .. .. .. .. . ... . . . . . .. ... .. . . . . . .. . . .. . . . . . ... .. .. .. . . . . . . .................................................................................................................................................. .. .. . ... ...
γ
N
...... ....
Solved by George Apostolopoulos, Messolonghi, Gree e; D.J. Smeenk, Zaltbommel, the Netherlands; George Tsapakidis, Agrinio, Gree e; and Titu Zvonaru, Comane sti, Romania. We give the solution of Smeenk. Let γ = ∠ACB , so that ∠M N B = 12 γ . The line through C parallel to N M meets AB at D. Then 2 : 3 = BN : BC = BM : BD, hen e BD = 23 BM = 12 AB . Sin e the bise tor of ∠ACB is also a median, CA = CB .
A
DM
B
303 19. The two diagonals of a trapezoid divide it into four triangles. The areas of three of them are 1, 2, and 4 square units. What values an the area of the fourth triangle have?
Solved by George Tsapakidis, Agrinio, Gree e; and Titu Zvonaru, Comane sti, Romania. We give the solution of Tsapakidis. Let [XY Z] be the area of triangle XY Z . Then [ABD] = [ABC], as the two triangles B A have equal altitudes from their ommon base AB . Therefore, .................................................................................................. .... . ... ................ ......... ............... .... ... ............. ... .... . . .... ....................... .. . . . . . . . . . ......... .... ..... . .. . . . . . . . .... . . . . . ......... ..... . .. . . .... . . . . . . . . . . ......... ..... .. . . . . . . . . . . . . . ......... ...... .. ............. . . ......... .... . .......... ............... . . ...................................................................................................................................................................
O
[OAD] + [OAB] = [OBC] + [OAB] ,
D
hen e [OAD] = [OBC]. Triangles OAB and OCD are similar, so we have
C OA OB , = OC OD
that is
OA OD [OAB] [ODC] · = 1, and hen e · = 1, so [OBC]2 = [OAB][OBC]. OC OB [OBC] [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 the ratio 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, Comane sti, Romania. We give the solution of Tsapakidis. Let r be the radius of the ir le ins ribed in A the rhombus. The required ratio is then 1 · OA · OB · 2 r 1 1 · OA · OB + OA2 OB 2 OB OA 2 b a + = + OA OB π a b a2 + b2 πr 2
= = =
2 π 2 π 2
πab
=
2
π
,
1 1 where r12 = OA + holds sin e r is the 2 OB 2 altitude of the right triangle OAB .
. ........... .... ... .... .... ... .... ... ..... ...... . . . .... ... . .... ... .... .... .. .... .. .... ....................... ...................... .... .......................... . . . . ......... ... ........ . ....... . . . . . .... .... .... . . . . .... ... ........ .... ... . ....... .......... . . . . . . . . ... .... .... . ...... . . . . . . . . . ... .... ... ..... . ...... . . . . . . . .. .... .. . . ... ... .... .......... .... .. ....................................................................................................................................................................... .... .. . . . . ... .... .... ... .... . . . . .... .. . .. .. .... .... ....... ....... ....... . .... ... .... ... ..... .... .... .... .... .... ........ .... ......... . .......... . . . . . .. ................ .......... .... ...................................................... ... .... ... .... .... .... . .... .. . . . .... .. ..... .... .... .... .... .. .... .......... .....
.. ............... .....
4 · 12 OA · OB
D
r
O
B
C
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 the Olympiad Corner. Having worked with Jim over many years, I suspe t that is what he would have opted for.
304
BOOK REVIEWS Amar Sodhi
For this spe ial issue we feature book reviews by Andy Liu and John Grant 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 before he 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 together for a Chinese dim-sum. One day, he asked me why I was no longer doing the review 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, little knowing that I was sitting a ross the table from the new editor himself. He was very upset, and despite repeated apology, it remained a sore point with him. So to set the re ord straight, I now state publi ly that I was never red by 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 Puzzles By Di k Hess, Sterling Publishing Co. In ., New York, 2009 ISBN 978-1-4027-5528-6, soft over, 94+ii pages, US$6.95 Reviewed by Andy Liu, University of Alberta, Edmonton, AB Sterling has published a wide range of mathemati al puzzle books. At one 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 a purpose in gradually attra ting new enthusiasts. They o er the novi es an easier path into the hobby than more serious works like Rodolfo Kur han's Mesmerizing 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 a regular ontributor. His intriguing problems always ome with an aura of something 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 addition, subtra tion, multipli ation, division, exponents, roots, on atenation, de imals, repeating de imals, fa torials, and bra kets. qqqqqqqqqqqqqqq qq qq Problem 44 The gure at right onsists of four unit squares qq qq qq joined edge to edge. Find a polygon, not ne essarily on- qqqqqqqqqqqqqqqqqqqqqqqqqqqq qq qq q qq vex, su h that ve non-overlapping opies an t inside qq q q qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq this gure, and over more than 98% of its area. It should be mentioned that Di k Hess is an avid tennis fan, and there are no books by him without problems related to that game. To nd out more about these problems, and the solution to the two problems above, buy the book! It is inexpensive and very highly re ommended.
305 John adds: \The se ond edition of Ravi Vakil's book A Mathemati al Mosai has been published. The original edition blended biography, problems, and mathemati al onne tions, in an e e tive manner. This sentiment is aptured in the 1997 CRUX with MAYHEM review by Jim Totten, so we therefore reprint Jim's review in its entirety pre eded by a brief review of the se ond edition. As I was the Book Reviews Editor during Jim's tenure as Editor-in-Chief, the opportunity to here a
ompany a review by Jim is an honour. Jim took a keen interest in the books that were reviewed in CRUX with MAYHEM. Indeed, Jim wrote enthusiasti ally about Stewart CoÆn's Geometri Puzzle Design in a review published in April 2008. A Mathemati al Mosai : Patterns & Problem Solving (New Expanded Edition) By Ravi Vakil, Brendan Kelly Publishing, Burlington, ON, 2008 ISBN 978-1-895997-28-6, soft over, 288 pages, US$19.95 Reviewed by John Grant M Loughlin, University of New Brunswi k, Frederi ton, NB I highly re ommend the new and expanded edition of A Mathemati al Mosai : Patterns & Problem Solving, but only if you do not have a opy of the 1996 edition already. Quoting from the foreword: "(T)here are few hanges here and there from the rst edition, but this book remains essentially the same one that appeared in 1996." That omment re e ts this reviewer's take on the book. The book is about 30 pages longer than the original version. Sele t additions of se tions su h as "Higher-dimensional versions of Platoni Solids" within the Combinatori s Chapter, insertion of mathemati al portraits, and in lusion of a more thorough index a
ount for this di eren e. Other hanges re e t updating of biographi al information and annotated referen es. Also, Ravi Vakil is now a well established algebrai geometer. However, with this in mind, the O tober 1997 CRUX with MAYHEM review by Jim Totten is still relevant today. It is noteworthy that Ravi Vakil is a founding o-editor of Mathemati al Mayhem whi h be ame part of this journal 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 L7R 3X4 ISBN 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 of problems whi h will pique the students' interest, and whi h also lead to further exploration. The problem sour e should lend itself well to independent work. The question is: where do you nd the appropriate level enri hment material? 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 author at one moment guides the reader through some very ni e mathemati al developments, 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 interesting mathemati al development. With a few ex eptions the problems in this book are not new, nor are the solutions. They are, however, well organized, both by topi and by level of mathemati al maturity needed. Answers are NOT always provided; instead there is often simply a solution strategy or hint given, and o
asionally there is simply a referen e to some other sour e for a full-blown treatment. Even when 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, but not always), where they be ome part of the development of another topi or problem. The author is a PhD andidate in pure mathemati s at Harvard University (at the time the book was written). Being still very young, he knows how to speak to today's teenagers. His sense of humour and general pu kishness is present throughout: just when you are lulled into some serious omputation in probability, he deviously throws a tri k question at you, that has a totally non-obvious answer (non-obvious, that is, until you CAREFULLY re-read the question). Many mathemati s books published today in lude short biographies on famous mathemati ians through history, espe ially those whose names ome up in the theory developed in the book. This book is no ex eption. But what is unique about this book is the in lusion of Personal Pro les of young mathemati ians from several ountries that he has met at International Mathemati al Olympiads (IMOs) in the past. The pro les are quite diverse, whi h means that most bright students ould nd one to identify with and to use as a 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 early part of the next entury. It should be interesting to follow their areers and see 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 do to problems that provide training for Putnam andidates (toward the end of the book). There are many ross-referen es and onne tions between seemingly unrelated problems from di erent areas of mathemati s, onne tions that most students would be unable to make. Many of these onne tions are 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 Putnam ompetition in all four of his undergraduate years at the University of Toronto. Before that he won two gold medals and a silver medal in IMOs and oa hed the Canadian team to the IMO from 1989 to 1995.
307 The British Columbia Se ondary S hool Mathemati s Contest
Clint Lee
In 1973 John Ciriani of Cariboo College (whi h be ame Thompson Rivers University, TRU) initiated the Cariboo College High S hool Mathemati s Contest with the help of olleagues in the Cariboo College Mathemati s and Statisti s Department. The intent of the ontest was to enable high s hool students in the Cariboo College region with an interest in mathemati s to get lo 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 enthusiasti 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. He also o-edited (with Leonard Janke) a separate book of solutions. Meanwhile, two other institutions in British Columbia developed high s hool mathemati s ontests. In 1977, at the suggestion of Robin Insley, the Mathemati s Department at the College of New Caledonia (CNC), onsisting of Phil Be kman, Robin Insley, Peter Trushell, and myself, developed a high s hool mathemati s ontest for students in the Central Interior region served by CNC, inspired by the su
essful ontest being run at Cariboo College. It ran su
essfully and independently from its in eption until 1994. In 1990, shortly after I joined the Mathemati s Department at Okanagan College (OC) a third high s hool mathemati s ontest was initiated for students in the OC region whi h ran independently from 1991 until 1994. The 1993 meeting of the British Columbia Committee in Undergraduate Programs in Mathemati s (BCCUPM, now BCCUPMs) was held at Selkirk College in Castlegar. There Jim Totten approa hed members of the OC and CNC mathemati s departments with a suggestion that the three institutions pool their resour es to o er a single mathemati s ontest. The ontest would serve the three olleges as well as any others in BC interested in su h a ontest, with the name, as suggested by Jim, the British Columbia Colleges High S hool Mathemati s Contest. This title was intended to make expli it that the ontest was intended for olleges, who intera ted dire tly with their regions, rather than universities. Both OC and CNC enthusiasti ally agreed. Jim then approa hed other institutions and found further support for the idea. Over the years Colleges evolved into University Colleges and then into Universities, so the original name be ame inappropriate and was hanged to the British Columbia Se ondary S hool Mathemati s Contest (BCSSMC). It took a year to organize the ombined ontest with the rst being held in 1995. Jim's idea was that Cariboo College would prepare the ontest papers, with ontributions from the other institutions, and all parti ipating institutions would use these for their lo al ontests. The 1995 and Copyright
c 2009
Canadian Mathemati al So iety
Crux Mathemati orum with Mathemati al Mayhem, Volume 35, Issue 5
308 1996 ontests were handled this way. The ombined ontest, following the Cariboo model, onsists of Preliminary and Final Rounds. Ea h round has a Junior level for grades 8 to 10 and a Senior level for grades 11 and 12. The Preliminary Round onsisting of twelve multiple hoi e questions is written and marked at the individual s hools in early Mar h. Based on these results, ea h s hool typi ally hooses two to four students to parti ipate in the Final Round at the sponsoring post-se ondary institution in early May. The Final Round onsists of ten multiple hoi e questions and ve questions requiring written answers, all of whi h are marked by a panel of fa ulty from the sponsoring institution. In addition to the morning ontest writing, Final Round a tivities usually in lude a morning session for tea hers, lun h for all parti ipants, an afternoon a tivity for the students during marking, and an award eremony. Examples of afternoon student a tivities are le tures by lo al/invited speakers, sports a tivities involving mathemati s, or s avenger hunts. The prizes awarded to winners range from s holarship prizes to ash or book prizes, depending on the sponsoring institution. Note that opies of the ontest papers from 1999 to the present, with solutions, are available at the BCSSMC website at http://people.okanagan.bc.ca/clee/BCSSMC. Following the su
ess of the 1995 and 1996 ontests, it was suggested that the only improvement would be to have members of more institutions be dire tly involved in the ontest preparation pro ess. For the 1997 ontest Don DesBrisay, Kirk Evenrude, and Ja k Bradshaw from Cariboo College; Ni holas Bu k and Edward Dobrowolski from CNC; Dave Murray, John Grant M Loughlin, and Clint Lee from OC; Jim Bailey from College of the Ro kies (formerly East Kootenay Community College) and Wayne Matthews from Camosun College; met at Cariboo College in Kamloops in August 1996 for the inaugural Math Contest Brainstorming Session. Jim Totten had made most 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 winning an 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 the Kalamalka Campus of Okanagan College in Vernon where the session was hosted by myself in 1998. Sin e 1999, the Brainstorming Sessions have been held in onjun tion with the BCCUPMs arti ulation meetings in mid-May, either before or after the main meeting. The Brainstorming Sessions in lude about 8 to 16 representatives from a ross-se tion of institutions a ross the provin e. Parti ipants usually bring some problems that they have prepared ahead of time and materials for generating additional problems as needed. The session is divided into Junior and Senior groups, ea h of whi h prepares rough drafts of Preliminary and Final round papers. Drafts are later typeset into tentative versions of the ontest papers. These are reviewed and revised over several months by as many as twenty reviewers. Finished forms are then produ ed in luding solutions to all of the problems that ultimately appear on the ontest papers. From the beginning, until he took the position of Editor-in-Chief of CRUX with Mayhem, Jim parti ipated a tively in these sessions sharing his
309 infe tious enthusiasm for mathemati s and its manifestations in all areas. Jim's humour ontributed to the su
ess and the enjoyment experien ed by all parti ipants. He ontributed problems and solutions for the ontest, typeset either the ontest papers or the solutions, and parti ipated in the review of the ontest papers as they evolved into their nished form. His abilities as a 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 parti ipate in the ontest. He made himself available to any institution, espe ially smaller institutions just getting started with the ontest, to help during the nal round in any apa ity, espe ially as a speaker. In 2004 it 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 Brewster approa hed PIMS for su h support and as a result PIMS provides support for a range of the provin e-wide a tivities asso iated with the ontest. Jim's re ruiting a tivities brought the number of parti ipating institutions up to a 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 College in Van ouver, Capilano University in North Van ouver, Camosun College in Vi toria, University of the Fraser Valley in Abbotsford, North Island College in Campbell River, Malaspina University in Nanaimo, College of the Ro kies in Cranbrook, Selkirk College in Castlegar, Northwest Community College in Prin e Rupert, UBC Okanagan in Kelowna, and Douglas College in New Westminster. Approximately 2500 grade 8 to 12 students parti ipate in the Preliminary Round of the Contest ea h year and 500 parti ipate again in the Final Round. While Jim was Editor-in-Chief of CRUX with Mayhem, he s aled ba k his involvement with the ontest. During this period I took over many of the responsibilities that Jim had undertaken. I typeset the drafts of the ontest papers, managed the evolution of the ontest papers, saw to the distribution of the papers to the parti ipating institutions, and oversaw and typeset solutions. Jim ontinually remained available to provide advi e and ontribute his reviewing skills in doing a nal run through of the ontest papers. Shortly before his death, he onta ted me and indi ated that now that he was winding down his Editor-in-Chief duties and had retired from TRU, he was looking forward to in reasing his involvement with the ontest. It is lear that without the inspiration and e orts of Jim Totten, the British Columbia Se ondary S hool Mathemati s Contest would not exist today. His enthusiasm for and ontributions to the ontest were key in developing the feelings of involvement and ownership of all of the parti ipating institutions. Clint Lee Okanagan College Vernon, BC
[email protected]
310 A Duality for Bi entri Quadrilaterals
Mi hel Bataille
In memory of Jim Totten Bi 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 a quadrilateral is easily onstru ted from its future in ir le. Below, we review this property that we formulate as an internal theorem about the determination of a bi entri quadrilateral. The purpose of this note is to state and prove an external theorem whi h is, in a sense, the dual of the internal one. A bi entri quadrilateral is uniquely determined by three of its sides tangent to a given ir le. Internal Theorem
This results immediately from the following well-known hara terization: Let P QRS be a onvex quadrilateral ins ribed in a ir le γ . The tangents 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 P R. For ompleteness, here is a short proof, in whi h ∠(ℓ, ℓ′ ) denotes the dire ted angle between lines ℓ and ℓ′ , measured modulo π. : Let I be the entre of γ and and D be the verti es of the quadrilateral formed by the tangents, as in the gure. Observing that A, P , I , and S are on y li , we have
Proof
A, B , C ,
Pq
..... ........ ...... .........
A q
γ
qQ
qC
qI
q
Similarly, ∠(CB, CD) = 2∠(P Q, P R) ,
so that ∠(AB, AD) = ∠(CB, CD) if and only if ∠(QP, QS) = ∠(P Q, P R) mod π2 , that is, if and only if QS ⊥ P R (sin e QS and P R are not parallel).
R
. ........ ........ ........ ....
=
∠(AB, AD) = ∠(AP, AS) ∠(IP, IS) = 2∠(QP, QS) .
B q
........ ....... ...... ...... .... ....... . . .... . . . . .. ....... ................................................... ...... . . . . . . . .......... ... ........... . . . ........... . . . ......... ....................................... . . . . . ...... ................ .............................................. .......... ....................... ............ .... ....... . . . . . . . . .... ........ ..... ......... . . . . . . .... ...... . . .... ........ ... .... .......... ...... .. ....... .... ....... .... ... ...... .... . . . . . . . . . . . . . . . . .... ....... ... ... .... ... .. ....... ....... .... ....... ... ... .... .. ......... ...... .... ... .. .. ... ....... .. .......... . . . . . . . . . .. ... ... . . . . . ....... ... ... ... .......... . . . . . . . ....... ..... ....... ....... ... .. ..... . . . . ....... . . . .. ....... ........... . .... . ....... . . . . . . . .. .............. ....... .. .... .... ....... ... .... .... ....... .... ..... ........... ....... .. . .... . . . ....... .... ...... ..... . . . . .......... . . ...... ......... ... . ....... ........... ... .. ........ .... ... ....... ...... ... ............ ....... . . . . .. . . . . ........................ . . .. ....... .............................. ... ....... .. ....... .. ....... . . ....... ....... ... .. ....... .. ....... ....... .... ....... .. ......
q
S
q D
We are guided to the next theorem by onsidering the ir um ir le to be 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 iety
Crux Mathemati orum with Mathemati al Mayhem, Volume 35, Issue 5
311 A bi entri quadrilateral is uniquely determined by three of its verti es lying on a given ir le. External Theorem
As above, the proof rests upon a hara terization of the bi entri quadrilateral and will provide a onstru tion of this quadrilateral. We begin by showing that the onjun tion of two simple properties distinguishes bi entri quadrilaterals among all quadrilaterals ins ribed in a given ir le. The following lemma is inspired by (and is an extension of) a problem posed in this journal (see [2℄, [3℄). Let ABCD be a onvex quadrilateral ins ribed in a ir le Γ with diagonals AC and BD meeting at X . Then ABCD is bi entri if and only if
Lemma
(a)
B
(b)
B BX = cos2 BD .
and D are on the same side of the perpendi ular bise tor of AC , and 2
: First, we assume that ABCD is bi entri . Then, Proof
BA − BC = DA − DC
(1)
so that BA − BC and DA − DC ertainly have 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 that D = ∠ADC = π − B , we have
......................................... ........... ................ .......... ......... ......... ....... ....... . . . . . ............. .... . ............ .................. . . ............ .. . . ... ........ . . . . . . . . . . . . ... ........ ....... . . . ... . . . . . . . . . . .. ... ........ ............ .... ... ....... ............ . . . . . .. . ...... . . ... . . . . ...... ..... ... .... ...................... ..... ... ............. .... . ... . ............................................................................................................................................................................................... ......... . .... ... ... ......... .. .. ... ... ......... .. ... ......... . . .... . ... . ......... ... ... ... .. ......... ... ... .. .. ......... ......... .. ... .... .... ... ......... ... ... ... ......... ... ... .. .. ......... .. ... .. ... ......... .. ......... ... ... .. ... ......... .. .. .. ... ......... ......... ............. ... ......... ....... ... .......... .... .... .. .... .... ..... .... ...... ..... . . . . ....... . . ......... ...... ........... ........ .................. ........... .....................................
qB
X
A q
qC
q
Γ
D
BX [ABX ] [CBX ] [ABC] ab sin B ab = = = = = BD [ABD] [CBD] [ABCD] ab sin B + cd sin D ab + cd
.
Furthermore, from the Law of Cosines we dedu e that cos B =
a 2 + b2 − e2 e2 − c2 − d2 a 2 + b2 − c2 − d2 = = 2ab 2cd 2(ab + cd)
.
Squaring ea h side of (1), we obtain a2 + b2 − c2 − d2 = 2(ab − cd) and so ab − cd BX cos B = =2· − 1 and (b) follows. ab + cd BD Conversely, if (b) holds, then (a − b)2 = (c − d)2 follows readily from the 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 footof the perpendi ular from B to AC , and the point −−→ let Y be − − → − − → − − → B B 2 2 ′ ′ su h that BY = cos 2 BY ; equivalently, Y Y = tan 2 BY .
312 The lemma implies that the desired point D must lie both on Γ and on the line m through Y ′ perpendi ular to U BY and on the same side as B of the perq pendi ular bise tor n of AC . It remains to qB show that m ne essarily ontains a point of Γ. To this end, let n interse t AC at M M Y qC A q and Γ at U and V , where U and B are on the same ar AC . n Now ∠AU C = B and AV CU is Γ
y li , hen e ∠AV C = π−B and we have Y′ m 1 B the relations U M = 2 cot 2 AC and ......................... ............... ......... ................ .......... ........ ..... ........ ................. ..... . ... ........ . ..... ........................... ....... ..... ... . . . . . . . ......... .. ..... ... ... .... ... ....................... ....... ..... ............. ..... .... . . . . . . . . . . . . . . . . ........ ........ .. .. .............. ....... ........ ... ....... ............ .... ... ...... ........ ... ... ...... .................... .... ..... .................. .................................. ............... ........ ... . . . . .. ............................................................................................................................................................................ ... ... .... ... .. .... ... ... . .. .. ... ... ... ... .. .... . . ... .. ... ..... ... ... ... .. ... .. ... .. ... .. .. .. .... . . . . . . ... ... .. ..... ... .. .. .. .. ... .. .. .... .. .. .. ... .. .. .. .... .. .... ... . . . . . . . . . ... .. ... .. .. . ... ...................................................................................................................................................................................................................... .. .... ... . .... .. ..... .. .. .... . . ...... . . . . . . . .. ... . ....... ... .. ... .. ......... ....... ........ ........... .. .. .. .......... ...........................................................................................................................................................................................................................
1 B q VM = tan AC . Observing that 2 2 V BY ≤ U M , we obtain B 1 B 1 B Y Y ′ ≤ tan2 · cot AC = tan AC = V M 2
2
2
2
2
.
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 in two points, one of whi h is the desired qB point D. The lemma then ensures that ABCD is bi entri . qC To on lude, the gure at right A q shows how to qui kly onstru t D using K on the bise tor BV of ∠ABC Kq with KC perpendi ular to BC , sin e ′ BC B BK q qC = cos = implies that Γ D BK 2 BC ′ ...................................... .......... ................ .......... ........ ........ ...... ....... . . . ................. . .... ............ ... .............. . . . . . . . . . ....... . . . ... . .. ........... . . . . . . . . . ... ... . .. ......... ... ... .. .... ............ ... .. .. ............ ... ..... .. ............ . . . . .. . . . . ..... . . . . . ..... . . ...................................................................................................................................................................................... ... ... .. ....... ........... . . . . . . . ... . ... ..... ....... ... . ....... ... ..... . . .. . ... . . . . ... ...... .. ... ...... ... .. ............. . ... ... . . .. ... .. ............. ... . . ... ... .. .................... .. .. . . ... .. ............ . . .. ... .. . . . ............ . ... ... ............. .. . ... ... .. .. ................... . . ............ ...... ... .. .. . .... . . . ............ .... .. . .. .... . . . ................................................................................................ . .... . . . ..... .. .. ...... ..... .. ....... ...... .. ......... ....... .. ............ ......... ........................................................ .. ................ ........
. .......... .............
B BC = cos2 . BC ′ 2
q V
. The author would like to thank the referee for his areful reading of the manus ript and for a suggestion whi h greatly improved the original proof of the external theorem. A knowledgment
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 Bataille 12 rue Sainte-Catherine 76000 Rouen, Fran e
[email protected]
313 A Study of Knight's Tours on the Surfa e of a Cube
Awani Kumar
1
Introduction
The lassi al puzzle of a knight's tour was, is, and will always be fas inating. 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 ubes and uboids having magi properties. Awani Kumar, Fran is Gaspalou, and Guenter 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 surfa e. A tour of a knight on a square board is alled a magi tour if the sum of 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 on the surfa es of ubes smaller or larger than 4 × 4 × 4 but, astonishingly, not on a 4 × 4 × 4 ube itself. We know that a knight's tour is not possible on a 4 × 4 board, but is it possible on the surfa e of a 4 × 4 × 4 ube? Will su h tours 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 cubes
The reader an visualize the knight's move on the fa e of a ube in a mu h better way when it is unfolded, as shown in Figure 1. On a onventional board a knight an over only up to eight squares. However, this need not be 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 nd this strange be ause the onse utive knight's moves are not equidistant, as on a onventional board. Well, don't worry. It is be ause a ube an be unfolded along its edges in many di erent ways. Look at Figure 2b and the knight's move to the square labelled \6" is lear. If it is ever un lear, then appropriately unfold a ube along its edges using a pair of s issors! A knight's tour is even possible on the smallest possible ube measuring 1 × 1 × 1, as shown in Figure 3. The dis erning reader will observe that here the numbers are arranged as on a onventional die. That is, the numbers on opposite fa es add up to 7. Copyright
c 2009
Canadian Mathemati al So iety
Crux Mathemati orum with Mathemati al Mayhem, Volume 35, Issue 5
314 ............................................................ .... .... .... ... .. .. .......................................................... .... .... .... .... .... .... ... ... ... .. . . . . .............................................................................................................................................................. . . . . . . . .... .... .... .... .... .... .... .... .... .... .... ... ... ... ............................................................................................................................................................... .... .... .... .... .... .... .... .... .... .... .... .... .... .... ... .... .... .... .... .... .... .. . . . . . . . . . . ...................................................................................................................................................... . . . ..... ..... ..... .... ... ... ........................................................ .... .... .... .... .... .... ... .. .. .......................................................... .... .... .... ... ... ... .... ... ... .......................................................... .... .... .... ... ... ... .... .. .. .......................................................
Figure 2a ............................ . ... .... ..... .... .. .. ....................................................................................... .. .. .... .... ..... .... .... ... ................................................................................. ... .. . ..... ..... .... ... .............................. ... .... ... .... .... ... ... ..........................
Figure 1
9 4
........................................................... ... ... ... ... .... ... ........................................................ ... .... .... ... .... .... ..... ... ... . . . . . .............................................................................................................................................................. . . . . . . . .... .... .... .... .... .... .... .... .... .... ... .... .... ... ............................................................................................................................................................. .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... ... ... ... ... ... ... .................................................................................................................................................................. .... .... .... .... .... .... .... ... ... ........................................................ .... .... .... .... .... .... ... .. .. .......................................................... .... .... .... ... ... ... .... ... ... .......................................................... .... .... .... ... ... ... .... .. .. .......................................................
10 1
Figure 2b
6
9
5
Kt.
8
4
2
7
6
1
............................ . ... .... ..... .... .. .. ....................................................................................... .. .. .... .... ..... .... .... ... ................................................................................. ... .. . ..... ..... .... ... .............................. ... .... ... .... .... ... ... ..........................
1
5
Kt.
8
2
2
7
3
10
3
5
6
3
4
Figure 3
......................................................... ......................................................... .... .... .... .... .... .... .... .... ... ... ... ... ......................................................... ......................................................... .... .... ... ... .... .... ... ... ... ... ... ... .... .... .... .... .... .... .. .. .. .. .. . . . . . . ....................................................................................................................................................... ......................................................................................................................................................... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... .... .... .... .... .... .... .... .... .... ... .... ... .... ... . . . . . . . ...................................................................................................................................................................... ...................................................................................................................................................................... .................................................. . . . ...................................................... . . . . . . ... ..... .... ..... .... ..... .... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... .... .... .... .... ... .... ... ... .... .... .... .... .... .... .... .... .... .... .... .... .... .. . . . . . . . . . . . . . . . . . ......................................................... . . . . . . . . . . . . . . . .................................................... ........................................................................................................................................................ ........................................................................................................................................................ . . . . . .... . . . . . . . . . . . ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... .... ... ... ... ... .. ... ... ... .. ... ... ... .. .. .. .. .. .. .. .. . ........................................................... . . . . . . . . ................................................... ....................................................................................................................................................... ......................................................................................................................................................... . .... . ... .... . . . . . . . . . . . . . . . . . .... ..... .... ..... .... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ..... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. ... .. ... .. . . . . . . . ......................................................... ......................................................... ...................................................................................................................................................................... ...................................................................................................................................................................... . . . . .... . ... . .... . . . . . . . .... . . . . . . . . . . . . . . . . ... .... ... .... ... .... .... .... .... .... .... .... .... .... .... .... .... .... .... ... .... .... ... ... ... ... .... .... ... ... ... ... ... ... ... ... ... ... ... ... ......................................................... ......................................................... ......................................................................................................................................................................... ......................................................................................................................................................................... .... .... ... ... ... ... ... ... .... .... ... ... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... ... ... ... ... ... ... ... ... ........................................................... ........................................................... ........................................................... ........................................................... ... ... ... ... ... ... .... .... .... .... .... .... .... .... ... ... ... ... ........................................................ ........................................................ .... .... .... .... .... .... ... ... ... ... ... ... .... .... ... ... ... ... ........................................................... ........................................................... .... .... .... .... .... .... .... .... .... .... .... .... ... ... .. .. .. .. ........................................................... ...........................................................
1 22
2
14 10
8
17 23
15 11 24 2 13 9
Figure 5
18 3 12 8 23 5
1 22
11 20 18 24 6 13
2 24
16 4
14 10
5 12
7 17
7 20
15 11 23 2 13 9
14 19
13 23 1
21 17
18 3 12 8 24 5
21 3
6
19 6
16 4
16 10
15 19
Figure 4
7 20
Figure 6
12 4
Figure 8
7
9 22
Figure 7
9 21 3
8 18 22 16 10
21 17
14 20
19 6
11 5
........................................................................................... .. .. .. .. ..... .... 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 ..... ...................................................................................
............. ............. ... ............ ... .. ... ... ................ ..... ......... ..........
4 15 1
........................................................................................... .. .. .. .. ..... .... 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
................................ ... ... ... ... ... ... .. .............................
315 Watkins [5℄ and Pi kover [6℄ have given a knight's tour on the surfa e of a 2×2×2 ube. Figure 4 and Figure 5 are two examples of losed and open tours, respe tively, on the surfa e of a 2 × 2 × 2 ube. There are millions of su h tours and, therefore, these are of little interest. However, tours having magi 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 sum of the numbers on ea h fa e is 50, the magi sum. There are thousands of su h tours. In fa t, 2 × 2 × 2 is the smallest ube on whi h su h interesting tours 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 es of 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 the knight'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 annot have 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) is 2 2 thus n (6n2 + 1) . Sin e n is odd, the numerator is odd. Thus, the sum of the 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 cube
Professional and amateur mathemati ians have been studying the knight's tour inside a 4 × 4 × 4 ube for over a entury but, astonishingly, no one has looked for them on its surfa e! There are billions of tours on the surfa e of a 4 × 4 × 4 ube, and onsequently these are of little interest. However, tours having magi properties are, again, a di erent story. We have seen that the fa es of a 2×2×2 ube an have magi properties and this is true for tours on a 4×4×4 ube too. In fa t, this is true for all even-sided ubes. Later, we will see that a doubly-even ube (say, 8 × 8 × 8) an have both rows and olumns magi 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 of tours in whi h all rows and olumns have magi properties. Figure 8 is one su 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 onstant. The dis erning reader must have observed that these tours are made up 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 in ea h row and ea h olumn of the fa es. The same holds true for the four 2×2 squares 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 squares 46 and 48. One knight overs the squares 1 to 48 and the other overs 49 to 96. It is a four-fold y li almost magi tour. That is, it retains its almost magi properties (44 magi lines) when starting from the square 25, 49, or 73. Readers are en ouraged to improve on this.
4
Knight’s tours on the surface of a 6 × 6 × 6 cube
The knight's tour on the surfa e of 6 × 6 × 6 ube has gotten s antily little attention. Perusal of the literature reveals that, in spite of billions of possible tours, only Watkins [5℄ has given a solution found by Arden Rzewni ki and Jesse 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 linking these tours, one an obtain hundreds of reentrant tours on the surfa e of a 6 × 6 × 6 ube. One su h example is shown in Figure 10. Sin e it is a losed tour, by sele ting a suitable starting point (square 19 as 1, square 20 as 2, and so on), we an get a tour having up to 12 magi 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 that there annot be magi tours on singly-even boards (that is, 6×6, 10×10, and so forth). But what about these on the surfa es of singly-even ubes? The author onje tures that they do not exist. The best the author ould a hieve is shown in Figure 11. All six fa es there are semimagi and altogether there are 40 magi lines. So, the magi sum has been a hieved up to 55%. Readers are hallenged to improve on this.
5
Knight’s tours on the surface of an 8×8×8 cube
This is the oldest problem of the lot, over 200 years old. Dudeney [8℄ writes that the problem was raised by Vandermonde, an 18th entury musi ian and mathemati ian. Dudeney gave a solution by ompleting a tour on a fa e and then 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 are trillions of su h tours. Using powerful omputers and intelligent programs, the international team of Ma kay, Meyrigna , and Stertenbrink [10℄ enumerated all of the 280 magi tours on the 8×8 board. By judi iously sele ting and linking these tours, one an get hundreds of tours and by rearranging the knight'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 diagonal sums are twi e the magi onstant on two of the fa es but, in spite of intense e ort, the author ould not obtain any diagonally magi fa e or a reentrant magi tour. Readers are en ouraged to look for them.
317 ............................................................................................................................... ... 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 .... ..................................................................................................................... .............................................................................................................................. .... 123 .... 116 .... 101 .... 94 .... 125 .... 118 .... ... .... .... .... .... .... .... ... ... ... ... ... ... .... ... 129 ... 138 ... 123 ... 110 ... 127 ... 136 .... .............................................................................................................................. .............................................................................................................................. ... ... ... ... ... ... .... ... ... ... ... ... ... ... ... 100 ..... 95 ..... 124 .... 117 ..... 102 .... 93 ..... .... 116 .... 125 .... 130 .... 139 .... 112 .... 121 .... ........................................................................................................................... .............................................................................................................................. .... 115 .... 122 .... 99 .... 108 .... 119 .... 126 .... ... ... ... ... ... ... ... .... 131 .... 140 .... 117 .... 126 .... 135 .... 144 .... .. .. .. .. .. .. ... ................................................................................................................................. .............................................................................................................................. .... 96 .... 107 .... 112 .... 121 .... 92 .... 103 .... .. .. .. .. .. .. ... .. ... ... ... ... ... ... .... 118 ..... 115 ..... 142 ..... 133 ..... 120 ..... 113 ..... ............................................................................................................................ .............................................................................................................................. ... 111 ... 114 ... 105 ... 98 ... 109 ... 120 ... .. .. .. .. .. .. ... ... .... .... .... .... .... .... .... 141 ..... 132 ..... 119 ..... 114 ..... 143 ..... 134 ..... ............................................................................................................................ ................................................................................................................ ... 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 10
Figure 11
318
......................................................................................................................................................................................................... .... 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
319
6
Conclusion and an Acknowledgement
We have seen that a knight's tour is possible on the surfa es of ubes of various sizes. Cubes of size 4 × 4 × 4 and larger an have magi rows and magi olumns on their fa es. The author has felt that in India, getting the referen es is more diÆ ult than dis overing a magi tour on a ube. The author is grateful to Takaya Iwamoto for providing photo opies of [2℄ and [3℄. Referen es
[1℄ H. S hubert (1904); Mathematis he Mussestunden Eine Sammlung von Geduldspielen, Kunststu ken und Unterhaltungs-aufgaben mathematis 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 Gazette, May 1944, pp. 46-50.
3
and
4
dimensions, Mathemati al
[3℄ J. Stewart (1971); Solid Knight's Tours, Journal of Re reational Mathemati 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 Chessboard Problems, Prin eton University Press, 2005, pp. 10-11, pp. 8889, 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 on Magi Tours, The Games and Puzzles Journal #25 (electronic), January-Mar h 2003. [8℄ H.E. Dudeney (1970); Amusements in Mathemati s, Dover Publi ations, New York, 1970, p. 103, p. 229. [9℄ M. Petkovi (1997); Mathemati s and Chess, Dover Publi ations, New York, 1997, p. 53, p. 64. [10℄ G. Stertenbrink (2003); Computing Magi Knight Tours, http://magictour.free.fr/
Awani Kumar 2/11-C, Vijayant Khand Gomti Nagar Lu know 226 010 India
[email protected]
320
TOTTEN PROBLEMS The following problems have all been dedi ated by the proposers to the lasting memory of Jim Totten. Solutions should arrive no later than 1 Mar h 2010. An asterisk (⋆) after a number indi ates that a problem was proposed without a solution. The editor thanks Jean-Mar Terrier of the University of Montreal for translations of the problems. . Proposed by Cosmin Pohoata , Tudor Vianu National College, Bu harest, Romania. Let H be the ortho entre of triangle ABC and let P be the se ond interse 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 AP B and if Y is the ortho entre of triangle AP C , prove that the length of XY is equal to the ir umradius of triangle ABC . TOTTEN{01
. Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania. Let k ≥ 2 be an integer and let f : [0, ∞) → R be a bounded ontinuous fun tion. If x is a positive real number, nd the value of
TOTTEN{02
lim
n→∞
√ k
n
Z
x
f (t) 1 + tk
0
n dt .
. Proposed by Ovidiu Furdui, Campia Turzii, Cluj, Romania. : [0, 1] → R be a ontinuous fun tion. Find the limit
TOTTEN{03
Let f
lim
n→∞
Z
1 0
···
Z
1 0
n
f 1
dx1 . . . dxn . 1 + ··· + x1 xn
. Proposed by Juan-Bos o Romero Marquez, Universidad de Valladolid, Valladolid, Spain. TOTTEN{04
that
Suppose that 0 < a < b, m0 = x m1 (x + m1 )
≤
1 b−a
√ a+b ab, and m1 = .
log
2
b(x + a) a(x + b)
≤
If x ≥ 0, prove
x m0 (x + m0 )
.
. 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 the radius of the ir le passing through A′ , B ′ , and C ′ and lo ate its entre. TOTTEN{05
321 .
Proposed by Bill Sands, University of Calgary, Calgary,
TOTTEN{06
AB.
Jim and three of his buddies played a round of golf. As usual, Jim won the game. In fa t, he beat every two of his three buddies, in the following sense. Let his three buddies be A, B , and C and let ai be A's s ore on hole i, 18 P for all 1 ≤ i ≤ 18, and similarly de ne bi and ci . Set Sab = min(ai , bi ), i=1 and similarly de ne Sac and Sbc . Then Jim's total s ore was less than Sab , 18 P Sac , and Sbc . However, Jim's s ore was more than Sabc = min(ai , bi , ci ). i=1 Jim's s ore was 72. What was the minimum possible s ore of any of his buddies? . Proposed by Sefket Arslanagi and Faruk Zejnulahi, University 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 TOTTEN{07
(a) (b)
1≤
a 1 − ab
+
b 1 − bc
+
c 1 − ca
≤
√ 3 3
,
2 √ a b c 3 3 1≤ + + ≤ . 1 + ab 1 + bc 1 + ca 4
. Proposed by Ri hard Hoshino, Government of Canada, Ottawa, ON. In triangle ABC suppose that AB < AC . Let D and M be the points on side BC for whi h AD is the angle bise tor and AM is the median. Let F be on side AC so that AD is perpendi ular to DF . Finally, let E be the interse tion of AM and DF . Prove that AB · DE + AB · DF = AC · EF . TOTTEN{08
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 guests sitting around a ir ular table. Let gn (k) be the number of ways that k of these n guests an be hosen so that no two hosen guests are sitting next to one another. To illustrate, g6 (0) = 1, g6 (1) = 6, g6 (2) = 9, g6 (3) = 2, and g6 (4) = 0 for all k ≥ 4. For ea h n ≥ 2, let
fn (x) =
X
gn (k)xk .
k≥0
For example, f6 (x) = 1+6x+9x2 +2x3 = (1+2x) all n for whi h (1 + 2x) is a fa tor of fn (x).
1+4x+x2
. Determine
322 . Proposed by D.J. Smeenk, Zaltbommel, the Netherlands. Determine all triangles ABC whose side lengths are positive integers and su h that cos C = 45 . TOTTEN{10
. Proposed by Walther Janous, Ursulinengymnasium, Inns-
TOTTEN{11
bru k, Austria.
(a) Let x, y, and z be positive real numbers su h that x + y + z = 1. Prove that √ √ √ 1 1 8 3 1 √ ≤ √ − x √ − z . √ − y 9
(b)
x
y
z
⋆
. Let n ≥ 2 and let x1 , x2 , . . . , xn be positive real numbers su h that x1 + x2 + · · · + xn = 1. Prove or disprove that
n−1 √ n
n
n Y 1 √ x − √ k xk k=1
≤
.
. Proposed by Mihaly Ben ze, Brasov, Romania. Let w, x, y, and z be positive real numbers with w +x+y +z = wxyz , and let s s TOTTEN{12
r r 1 1 1 1 1 3 1 f (x) = + − 3 + − − 3. 2 4 x 2 4 x √ √ √ √ 3 wxy + 3 xyz + 3 yzw + 3 zwx ≥ f (w) + f (x) + f (y) + f (z). 3
Prove that
.................................................................
TOTTEN{01. Propose par Cosmin Pohoata , College National Tudor Vianu, Bu arest, Roumanie. Soit H l'ortho entre du triangle ABC et soit P la se onde interse tion du er le ir ons rit du triangle AHC ave la bisse tri e interieure de l'angle BAC . Si X est le entre du er le ir ons rit du triangle AP B et Y l'ortho entre du triangle AP C , montrer que la longueur de XY est egale au rayon du er le ir ons rit du triangle ABC .
. Propose par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie. Soit k ≥ 2 un entier et f : [0, ∞) → R une fon tion ontinue et bornee. Pour un nombre reel positif x, trouver la valeur de
TOTTEN{02
lim
n→∞
√ k
n
Z
x
0
f (t) 1 + tk
n dt .
323 . Propose par Ovidiu Furdui, Campia Turzii, Cluj, Roumanie. : [0, 1] → R une fon tion ontinue. Trouver la limite
TOTTEN{03
Soit f
lim
n→∞
Z
1 0
···
Z
1 0
f 1
x1
n + ··· +
dx1 . . . dxn . 1 xn
. Propose par Juan-Bos o Romero Marquez, Universite de Valladolid, Valladolid, Espagne. TOTTEN{04
On suppose que montrer que
0 < a < b, m0 =
x
≤
m1 (x + m1 )
1 b−a
log
√ ab
b(x + a) a(x + b)
a+b . 2
et
m1 =
≤
m0 (x + m0 )
x
Si
x ≥ 0,
.
. Propose par Mi hel Bataille, Rouen, Fran e. Soit I le entre du er le ins rit du triangle ABC . Soit A′ le point tel −−→′ − → que AA = (cos A)AI , et soit B ′ et C ′ les points de nis de maniere analogue. Trouver le rayon du er le passant par A′ , B ′ et C ′ ainsi que la position de son entre. TOTTEN{05
. Propose par Bill Sands, Universite de Calgary, Calgary, AB. Jim et trois de ses opains ont joue une ronde de golf. Comme d'habitude, Jim a gagne 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 le s ore de A au trou i pour tout 1 ≤ i ≤ 18 ; on de nit de m^eme bi et ci . Soit 18 P Sab = min(ai , bi ), et de maniere analogue, Sac et Sbc . Le s ore total i=1 de Jim a et e plus petit que Sab , Sac et Sbc . Par ontre, son resultat de 72 18 P depassait Sabc = min(ai , bi , ci ). Quel etait le s ore minimal possible de i=1
ha un de ses opains ? TOTTEN{06
. Propose par Sefket Arslanagi et Faruk Zejnulahi, Universite de Sarajevo, Sarajevo, Bosnie et Herzegovine. non negatifs tels que a2 + b2 + c2 = 1. Soit a, b et c trois nombres reels Etudier la validite de TOTTEN{07
(a)
1≤
(b)
1≤
a 1 − ab a
1 + ab
+ +
b 1 − bc b
1 + bc
+ +
c 1 − ca c
1 + ca
≤ ≤
√ 3 3 2 √ 3 3 4
, .
324 . Propose par Ri hard Hoshino, Gouvernement du Canada, Ottawa, ON. Supposons que dans le triangle ABC , on a AB < AC . Soit D et M les points sur le ot ^ e BC pour lesquels AD est la bisse tri e et AM la mediane. Soit F le point sur le ot ^ e AC tel que AD soit perpendi ulaire a DF . Soit nalement E l'interse tion de AM et DF . Montrer que AB · DE + AB · DF = AC · EF . TOTTEN{08
. Propose 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 possibilites que k de es n hotes ^ puissent e^ tre 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 TOTTEN{09
fn (x) =
X
gn (k)xk .
k≥0
Par exemple, f6 (x) = 1+6x+9x +2x3 = (1+2x) 1+4x+x2 . Determiner tous les n pour lesquels (1 + 2x) est un fa teur de fn (x). 2
. Propose par D.J. Smeenk, Zaltbommel, Pays-Bas. Trouver tous les triangles ABC dont la longueur des ot ^ es sont des entiers positifs et tels que cos C = 54 .
TOTTEN{10
. Propose par Walther Janous, Ursulinengymnasium, Innsbru k, Autri he. positifs tels que x + y + z = 1. (a) Soit x, y et z trois nombres reels Montrer que
TOTTEN{11
√ √ √ √ 8 3 1 1 1 ≤ √ − x √ − z √ − y 9 y x z
(b)
⋆. Soit n ≥ 2 et soit x , x , . . . , x
1 2 n n nombres reels x1 + x2 + · · · + xn = 1. Etudier la validite de n n Y √ n−1 1 ≤ − xk . √ √ n
k=1
.
positifs tels que
xk
. Propose par Mihaly Ben ze, Brasov, Roumanie. Soit w, x, y et z des nombres reels positifs ave w + x + y + z = wxyz , et soit s s r r TOTTEN{12
3
f (x) =
1 + 2
1 1 − 3 + 4 x
Montrer que √wxy+ √xyz+ √yzw+ 3
3
3
3
1 − 2
1 1 − 3 4 x
.
√ 3 zwx ≥ f (w)+f (x)+f (y)+f (z).
325
PROBLEMS Solutions 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 a solution. 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 in the language of the primary featured solution. Problems 3451, 3452, 3453, and 3454 below are dedi ated by the proposers to the memory of Jim Totten. The editor thanks Jean-Mar Terrier of the University of Montreal for translations of the problems. . Proposed by Jose Luis Daz-Barrero, Universitat Polite ni a de Catalunya, 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 3451
X ||x|| 1/2 X < z, x > < x, y > 1/2 ≤ |< y, z >| ||x|| ||y|| ||z||
y li
y li
.
. Proposed by D.J. Smeenk, Zaltbommel, the Netherlands. Prove the following and generalize these results.
3452
(a)
tan2 36◦ + tan2 72◦ = 10,
(b)
tan4 36◦ + tan4 72◦ = 90,
( )
tan6 36◦ + tan6 72◦ = 850,
(d)
tan8 36◦ + tan8 72◦ = 8050.
.
3453
USA.
Proposed by S ott Brown, Auburn University, Montgomery, AL,
Triangle ABC has side lengths a = BC , b = AC , c = AB ; and altitudes ha , hb , hc from the verti es A, B , C , respe tively. Prove that
X
8 h2a (hb + hc ) + 16ha hb hc
y li
X √ √ ≤ 3 3 a2 (b + c) + 6 3abc .
y li
326 .
3454
ON.
Proposed by Ri hard Hoshino, Government of Canada, Ottawa,
Let a, b, c, and d be positive real numbers su h that a + b + c + d = 1. Prove that b3 c3 d3 1 a3 + + + ≥ . b+c
c+d
d+a
a+b
8
. Proposed by Mi hel Bataille, Rouen, Fran e. Find the minimum value of x2 + y2 + z 2 over all triples (x, y, z) of real numbers su h that 3455
13x2 + 40y 2 + 45z 2 − 36xy + 12yz + 24xz ≥ 2009
and hara terize all the triples at whi h the minimum is attained. . Proposed by Mi hel Bataille, Rouen, Fran e. Given a triangle ABC with ir um ir le Γ, let ir le Γ′ entred on the line BC interse t Γ at D and D′ . Denote by Q and Q′ the proje tions of D and D′ on the line AB , and by R and R′ their proje tions on AC ; assume that none of these proje tions oin ide with a vertex of the triangle. BQ CR . Does the onverse Show that if Γ′ is orthogonal to Γ, then BQ = ′ CR ′ hold? 3456
3457
. Proposed by Mi hel Bataille, Rouen, Fran e.
Let An =
⌊n/2⌋ P j=0
(−3)j aj , 2j + 1
where aj
=
n P
k=2j
k k+1 2j 2k
and n is a posi-
tive integer. Prove that A1 + A2 + · · · + An ≥ n with equality for in nitely many n. . Proposed by Bru e Shawyer, Memorial University of Newfoundland, St. John's, NL. Determine the entral angle of a se tor, su h that the square drawn with one 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.
3458
. Proposed by Zafar Ahmed, BARC, Mumbai, India. Let a, b, c and p, q, r be positive real numbers. Prove that if q2 2 and r ≤ pq, then 3459
a b c 3 + + ≤ pa + qb + rc pb + qc + ra pc + qa + rb p+q+r
When does equality hold?
≤ pr
.
327 3460. Proposed by Tran Quang Hung, student, Hanoi National University, Vietnam. The triangle ABC has ir um entre O, ortho entre H , and ir umradius R. Prove that
3R − 2 OH ≤ HA + HB + HC ≤ 3R + OH 3461. Proposed by Tran Quang Hung, student, Hanoi National University, Vietnam. Let I be the in entre of triangle ABC and let A′ , B ′ , and C ′ be the interse tions of the rays AI , BI , and CI with the respe tive sides of the triangle. Prove that
IA + IB + IC ≥ 2(IA′ + IB ′ + IC ′ ) .
. Proposed by Sotiris Louridas, Aegaleo, Gree e. Let x, y, and z be positive real numbers su h that
3462
Prove that
x3 + z 3 − y 3
y 3 + x3 − z 3
x3 + y 3 + z 3 + 3xyz
Y
y li
≤ 3
z 3 + y 3 − x3
x3 + y 3 − z 3 + xyz
> 0.
Y q 3 4 x4 (x2 + yz) .
y li
................................................................. . Propose par Jose Luis Daz-Barrero, Universite Polyte hnique de Catalogne, Bar elone, Espagne. Soit (X, < ·, · >) un espa e ve toriel reel ou omplexe muni d'un produit s alaire et soit x, y et z des ve teurs non nuls de X . Montrer que 3451
X ||x|| 1/2 X < z, x > < x, y > 1/2 ≤ |< y, z >| ||x||
y lique
y lique ||y|| ||z||
. Propose par D.J. Smeenk, Zaltbommel, Pays-Bas. Montrer les egalit es suivantes et gen eraliser
es resultats. 2 2 ◦ ◦ tan 36 + tan 72 = 10, tan4 36◦ + tan4 72◦ = 90, tan6 36◦ + tan6 72◦ = 850, tan8 36◦ + tan8 72◦ = 8050.
3452
(a) (b) ( ) (d)
.
328 . Propose par S ott Brown, Universite Auburn, Montgomery, AL,
3453
USA.
Soit a = BC , b = AC , c = AB les longueurs des ot ^ es du triangle ABC , de hauteurs respe tives ha , hb , hc issues des sommets A, B , C . Mon-
trer que
X
X √ √ ≤ 3 3 a2 (b + c) + 6 3abc .
y lique
8 h2a (hb + hc ) + 16ha hb hc
y lique
. Propose par Ri hard Hoshino, Gouvernement du Canada, Ottawa,
3454
ON.
Soit a, b, c et d des nombres reels non negatifs ave a + b + c + d = 1. Montrer que a3 b+c
+
b3 c+d
+
c3 d+a
+
d3 a+b
≥
1 8
.
. Propose by Mi hel Bataille, Rouen, Fran e.
3455
Trouver la valeur minimale de x2 +y2 +z 2 pour tous les triplets (x, y, z) de nombres reels tels que 13x2 + 40y 2 + 45z 2 − 36xy + 12yz + 24xz ≥ 2009
et ara teriser tous les triplets pour lesquels le minimum est atteint. . Propose par Mi hel Bataille, Rouen, Fran e.
3456
Etant donne un triangle ABC et Γ son er le ir ons rit, soit Γ′ un
er le entre sur la droite BC et oupant Γ en D et D′ . Designons par Q et Q′ ′ ′ les proje tions de D et D sur la droite AB , et soit R et R leurs proje tions sur AC ; supposons qu'au une de es proje tions ne oin ide ave un sommet du triangle. BQ CR Montrer que si Γ′ est orthogonal a Γ, alors BQ = . La re iproque ′ CR ′ est-elle valide ? 3457
. Propose par Mi hel Bataille, Rouen, Fran e.
Soit An
=
k k+1 et n est 2j 2k k=2j A1 + A2 + · · · + An ≥ n ou l'on a egalit e
⌊n/2⌋ P j=0
positif. Montrer que in nite de n.
(−3)j aj , 2j + 1
ou aj
=
n P
un entier pour une
329 . Propose by Bru e Shawyer, Universite Memorial de Terre-Neuve, St. John's, NL. Determiner l'angle au entre d'un se teur tel que le arre, onstruit ave un sommet sur haque rayon du se teur et deux sommets sur la ir onferen e, possede une aire egale au arre du rayon du se teur. 3458
. Propose par Zafar Ahmed, BARC, Mumbai, Inde. Soit a, b, c et p, q, r des nombres reels positifs. Montrer que si q2 ≤ pr et r2 ≤ pq, alors 3459
a pa + qb + rc
+
b
+
pb + qc + ra
c pc + qa + rb
≤
3 p+q+r
.
Quand y a-t-il egalit e ? . Propose par Tran Quang Hung, etudiant, Universite Nationale de Hanoi, Vietnam. Soit O le entre du er le ir ons rit du triangle ABC , de rayon R , et H son ortho entre. Montrer que 3460
3R − 2 OH ≤ HA + HB + HC ≤ 3R + OH
. Proposed by Tran Quang Hung, etudiant, Universite Nationale de Hanoi, 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 du triangle. Montrer que 3461
IA + IB + IC ≥ 2(IA′ + IB ′ + IC ′ ) .
. Propose par Sotiris Louridas, Aegaleo, Gre e. Soit x, y et z trois nombres reels positifs tels que
3462
x3 + z 3 − y 3
Montrer que
y 3 + x3 − z 3
x3 + y 3 + z 3 + 3xyz
Y
y lique
≤ 3
z 3 + y 3 − x3
x3 + y 3 − z 3 + xyz
> 0.
Y q 3 4 x4 (x2 + yz) .
y lique
330
SOLUTIONS No problem is ever permanently losed. The editor is always pleased to onsider for publi ation new solutions or new insights on past problems. 2557. [2000 : 304℄ Proposed by Gord Sinnamon, University of Western Ontario, London, ON and Hans Heinig, M Master University, Hamilton, ON. (a) Show that for all positive sequen es {xi } and all integers n > 0, j n X k X X
k=1 j=1 i=1
xi ≤ 2
n X
k=1
k X
j=1
2
xj x−1 k .
⋆ Does the above inequality remain true without the fa tor of 2? ( )⋆ [Proposed by the editors℄ What is the minimum onstant c that an
(b)
repla e the fa tor 2 in the above inequality?
Solution to part ( ) by Li Chao, student, High S hool AÆliated to Renmin University of China, Beijing, China. The given inequality is equivalent to 2 i n n X X X n+2−i 1 xj xi < c x 2 j=1 i=1 i=1 i
.
Making the substitution
sk =
k X
xi , k = 0 , 1 , 2 , . . . , n
i=1
and then simplifying puts the inequality into another equivalent form: n X
(n + 1 − i)si < c
i=1
n X
i=1
s2i si − si−1
.
By the Cau hy{S hwartz Inequality we have n X
(n + 1 − i)si
i=1
!2
≤ = <
n X
i=1 n X i=1 n X i=1
s2i si − si−1 s2i si − si−1 s2i si − si−1
! ! !
n X
2
(n + 1 − i) (si − si−1 )
i=1 n X
(2n + 1 − 2i)si
i=1 n X
2
!
(n + 1 − i)si
i=1
!
!
.
331 Hen e,
n X
(n + 1 − i)si < 2
i=1
n X
i=1
s2i si − si−1
,
so that taking c = 2 ensures that the inequality holds for all positive real numbers x1 , x2 , . . . , xn . For ea h n > 4 we will now give a stri tly in reasing sequen e of numbers s0 , s1 , . . . , sn su h that s0 = 0 and
lim
n→∞
n X
(n + 1 − i)si
i=1 n X
i=1
s2i si − si−1
!
!
= 2
(this is equivalent to spe ifying x1 , x2 , . . . , xn ). Take n > 4, let s0 = 0 and si
=
sn
=
(n − 1)(n − 2) (n − i)(n − i − 1)
,
1 ≤ i ≤ n − 2;
2sn−1 = 4sn−2 .
Then we have n X
i=1
s2i si − si−1
= 1 +
n−2 X i=2
(n − 1)(n − 2)
+ 6(n − 1)(n − 2) = 1 +
(n − 1)(n − 2)
2 + 6(n − 1)(n − 2)
= 1 + =
1 2
1 1 − (n − i)(n − i − 1) (n − i + 1)(n − i)
n−2 (n − 1)(n − 2) X 2 1 − 2 n−i−1 n−i i=2
+ 6(n − 1)(n − 2) = 1 +
(n − 1)2 (n − 2)2 (n − i)2 (n − i − 1)2
1+
1 2
+
1 3
+ ··· +
1 n−3
+ 1−
1 n−2
(n − 1)(n − 2) ln n + O(1) + 6(n − 1)(n − 2) 2
n2 ln n + O(n2 ) ,
where the notation f (n) =
onstant C su h that |f (n)|
O g(n) here means that there ≤ C|g(n)| holds for suÆ iently
is a positive large n, and
332
we have used the fa t that 1 + 21 + 13 + · · · + n1
onstant, γ , as n tends to in nity. We also have n X
− ln n
tends to Euler's
(n + 1 − i)si
i=1
=
n−2 X i=1
=
(n − i + 1)
(n − 1)(n − 2)
(n − i)(n − i − 1)
+ 4(n − 1)(n − 2) n−2 X (n − 1)(n − 2) i=1
=
= =
2 1 − n−i−1 n−i
+ 4(n − 1)(n − 2) 1 1 1 (n − 1)(n − 2) 1 + + · · · + + 1− 2 n−2 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 ompute
lim
n→∞
n X
(n + 1 − i)si
i=1 n X
i=1
s2i si − si−1
!
!
=
n2 ln n + O(n2 ) lim 1 n→∞ n2 ln n + O(n2 ) 2
=
lim
n→∞
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.
. [2008 : 298, 300℄ Proposed by Toshio Seimiya, Kawasaki, Japan. Let ABC be a triangle with AB > AC . Let P be a point on the line AB beyond A su h that AP + P C = AB . Let M be the midpoint of BC , and let Q be the point on the side AB su h that CQ ⊥ AM . Prove that BQ = 2AP . 3351
333 Solution by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain; and Tai hi Maekawa, Takatsuki City, Osaka, Japan.
AD
= AP + P D = AP + P C = AB ,
...... ...... .... .. ..... .... . . . ... . ..... .... . ..... ..... ... ..... .. . . . .. .. . . . . . . ..... ... .... ... ..... .. ..... .. .... . . . . . ... ... ..... ..... ... .. ..... .. ......................... . . . . . ......... ... .. ......... .... ..... ......... .... ..... ... ........... ....... ... .. ............. . . . . . . ... .... . .. .. ... ..... .. .. .. ... ..... .... .... .. . .. ... .... ... .. .. ..... . . . . . . . . . . . .. .. ... .. .. .. ... ..... .. ... ... ... ..... .. .. ... .. .. ..... ... .. .. .. ......... . . . . . . . . . . . ... ... . ... ................. .. ... ... .......... .... ..... .. ........... .. ... ... ..... .. .. .. . .... .... .... .. .. ................... .... . . . . . . . . .......... .......... ............ .. ..... .. ... .. .... ....................................................................................................................................................
D
P
A
.......... .............
Let D be the point on the P B beyond P su h that P D = P C . Sin e line
N is the midpoint of the segment BD. It follows that AM ||DC , be ause M is the Q midpoint of BC . Hen e ∠DCQ = 90◦ . Let P N ⊥ DC with point N on the line CD. B C M Then DN = N C , be ause triangle CDP is isos eles. Sin e N is the midpoint of CD and P N k QC , it follows that P is the midpoint of QD. Thus P Q = P D = P C . Hen e AP + AQ = P Q = P C = AB − AP , and therefore, 2AP = AB − AQ = BQ , as laimed. ......... ............. .
.......... ..............
A
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; SEFKET University of Sarajevo, Sarajevo, Bosnia and Herzegovina; ROY BARBARA, ARSLANAGIC, 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, Missouri Southern State University, Joplin, MO, USA; IAN JUNE L. GARCES, Ateneo de Manila University, Quezon City, The Philippines; OLIVER GEUPEL, Bruhl, NRW, Germany (two solutions); JOHN G. HEUVER, Grande Prairie, AB; PETER HURTHIG, Columbia College, Van ouver, BC; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; GIANNIS Y, Big Rapids, MI, USA; G. KALOGERAKIS, Canea, Crete, Gree e; V ACLAV KONECN student, Sarajevo College, KEE-WAI LAU, Hong Kong, China; SALEM MALIKIC, Sarajevo, Bosnia and Herzegovina; MISSOURI STATE UNIVERSITY PROBLEM SOLVING GROUP, Spring eld, MO, USA; MADHAV R. MODAK, formerly of Sir Parashurambhau College, Pune, India; ALEX SONG, student, Elizabeth Ziegler Publi S hool, Waterloo, ON; PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU, Comane sti, Romania; and the proposer.
. [2008 : 298, 300℄ Proposed by Toshio Seimiya, Kawasaki, Japan. Let ABC be a right-angled triangle with right angle at A. Let I be the in entre of △ABC , and let D and E be the interse tions of BI and CI with AC and AB , respe tively. Prove that 3352
BI · ID AB = CI · IE AC
.
334 Solution by Salem Maliki , student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; and Madhav R. Modak, formerly of Sir Parashurambhau College, Pune, India. Sin e AI bise ts the right angle at A, we have ∠CID =
1 2
∠B +
1 2
∠C =
180◦ − ∠A = 45◦ = ∠CAI .
1 2
Hen e, triangles CAI and CID are similar, so that .
Likewise, and BIE are similar, and therefore, IE
=
AB AI
D
.
The result now follows by multiplying a ross the last two equations.
............ ............... .... .... ............... ........ ... ... ........ . .. ........ ..... ..... ........ ... ... ........ ... ........ .... ........ ... ........ ... .... ........ ... ... ........ ... ........ .... ... ........ ... ........ ◦ .... .... ........ ... ........ ....................... . ........ ............... .... ... ........ . . . . ................ ........ ... . . . . . . . . . . . . ........ .. . ........ . . ..... ............................. . ..... ........ ............... . ... . .. ◦ . ........ ............... ... ........ ............... ... .... ◦ . . ............... ... ........ ... . . . ........ . . ... ............... .. .... .... ............... ... ............... ............... ◦ ... ..... ..... ............... ....... ... ......... ..... .... . ... ...........................................................................................................................................................................................................................................................
45
I
....... ..
BI
C ........... ..... ....
AC triangles BAI
....... ..
AI
45 45
45
... ...
CI
=
. ......... .............. .... .........
ID
A
E
B
Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallor a, Spain; GEORGE University of Sarajevo, APOSTOLOPOULOS, Messolonghi, Gree e; SEFKET ARSLANAGIC, Sarajevo, Bosnia and Herzegovina; ROY BARBARA, Lebanese University, Fanar, Lebanon; 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; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; JOHN G. HEUVER, Grande Prairie, AB; JOE HOWARD, Portales, NM, Y, Big USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; V ACLAV KONECN Rapids, MI, USA; TAICHI MAEKAWA, Takatsuki City, Osaka, Japan; MISSOURI STATE UNIVERSITY 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, Comane sti, Romania; and the proposer.
. [2008 : 298, 301℄ Proposed by Mihaly Ben ze, Brasov, Romania. Let ABC be a triangle all of whose side lengths are positive integers.
3353
(a) Determine all su h triangles where one angle has twi e the measure of a se ond angle. (b) Determine all su h triangles where two medians are perpendi ular. Composite of solutions by Roy Barbara, Lebanese University, Fanar, Lebanon and 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 are the triangles with a = dn2 ,
= 2α and show that the solutions
b = d m2 − n2
,
c = dmn ,
(1)
where d, m, and n are positive integers, m is oprime to n, and n < 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 ′ 2 (c ) = a′ (a′ + b′ ), it follows that there exist oprime positive integers m and n su h that a′ = n2 and a′ + b′ = m2 . Thus, (1) holds for positive integers d, m, and n with m oprime to n and 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 aforementioned 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 where the medians from B and C are perpendi ular. First we show that the positive real numbers a, b, and c are the sides of a triangle ABC where the medians from B and C are perpendi ular if and only if b2 + c2 = 5a2 and bc > 2a2 . (2) If G is the entroid of ABC , then BG is perpendi ular to CG if and only if BC 2 = BG2 + CG2 , or 1 1 2a2 + 2c2 − b2 + 2a2 + 2b2 − c2 = a2 , 9 9
that is, b2 + c2 = 5a2 . 2 c2 − a 2 2a2 In addition, 1 > cos α = b +2bc = , hen e bc > 2a2 . bc Conversely, if the onditions in (2) hold, then (b+c)2 = 5a2 +2bc > a2 and (b − c)2 = 5a2 − 2bc < a2 . Hen e, |b − c| < a < b + c and there exists a 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 are perpendi ular. Thus the problem amounts to nding the positive integers a, b, and c that 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 then obtained by multiplying these primitive solutions by positive integers.
336 Now let a, b, c be a primitive solution. Then there exist oprime posib−a 2a + c m tive integers m and n for whi h 2a = = , and thus −c b+a n a(2m + n) − nb − mc a(m − 2n) + mb − nc
Solving for a, b, and c yields
a = b = c =
λ m2 + n2
= 0, = 0.
,
λ n − m + 4mn 2λ m2 − n2 + mn 2
2
, ,
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 uonly if n < m < 2n. Writing λ = v , where u and v are positive oprime integers, we obtain va vb vc
= u m2 + n2
,
= u n − m + 4mn = 2u m2 − n2 + mn 2
2
, .
Now, gcd(u, v) = 1 and gcd(a, b, c) = 1, hen e u = 1 and
! m2 + n2 n2 − m2 + 4mn 2 m2 − n2 + mn (a, b, c) = , , , (3) v v v 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. k If p is prime, k is a positive integer, v , then pk divides and p2 divides 2 2 2 2 2 both m + n and 2 n − m + 4mn + 2 m − n + mn = 10mn. However, p does not divide n or m, sin e otherwise it would divide both 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. 2 2 2 2 If 5 divides v then 5 divides m +n + n −m +4mn = 2n(n+2m) and n2 − m2 + 4mn − m2 + n2 = 2m(2n − m). Sin e 5 does not divide m or n, then 5 divides both n + 2m and 2n − m. Note that n + 2m = 5k and 2n − m = 5l is equivalent to m = 2k − l and n = k + 2l, and that in this ase n < m < 2n be omes 0 < 3l < k.
A omplete des ription of the primitive solutions an now be given:
m 2 + n2 n2 − m 2 + 4mn 2(m 2 − n2 + mn) , , 10 10 10
, where m = 2k − l, n = k + 2l; gcd(k, l) = 1; k, l are odd; and 0 < 3l < k.
• (a, b, c)
=
337 • (a, b, c)
=
m 2 + n2 n2 − m 2 + 4mn 2(m 2 − n2 + mn) , , 5 5 5
m 2 + n2 n2 − m 2 + 4mn 2(m 2 − n2 + mn) , , 2 2 2
m = 2k − l, n = k + 2l; gcd(k, l) = and 0 < 3l < k. • (a, b, c)
=
gcd(m, n) = 1; m, n 0 < n < m < 2n.
are odd;
, where 1; k, l have opposite parity;
n + 2m
, where is not divisible by 5; and
• (a, b, c) = m2 + n2 , n2 − m2 + 4mn, 2(m2 − n2 + mn , where gcd(m, n) = 1; m, n have opposite parity; n + 2m is not divisible by 5; and 0 < n < m < 2n.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e (part (a) only); CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Bruhl, 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); RODOLFO LARREA and LUZ RONCAL, Logrono, ~ Spain (part (b) only); MADHAV R. MODAK, formerly of Sir Parashurambhau College, Pune, India (part (a) only); and TITU ZVONARU, Comane 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 the Gaussian 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 h one 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.
. [2008 : 298, 301℄ Proposed by Jose Luis Daz-Barrero, Universitat Polite ni a de Catalunya, Bar elona, Spain. Evaluate 3354
lim
n→∞
n X
k=1
ln
n2 + k2 n2
k3 /n4
.
Composite of nearly identi al solutions submitted by all the solvers whose names appear below. The fun tion f (x) = x3 ln 1 + x2 is ontinuous and hen e Riemann
338 integrable on [0, 1]. Using integration by parts, we have 2 ! n k 1 X k 3 lim ln = lim ln 1 + n→∞ n→∞ n n n k=1 k=1 1 Z 1 Z 1 1 4 1 x5 = x3 ln 1 + x2 dx = x ln 1 + x2 − dx 4 2 0 1 + x2 0 0 Z 1 1 1 x = ln 2 − x3 − x + dx 4 2 0 1 + x2 1 1 1 4 1 1 1 1 = ln 2 − x − x2 + ln 1 + x2 = . 4 2 4 2 2 8 0 n X
n2 + k2 n2
k3 /n4
Solvers: GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; DIONNE BAILEY, ELSIE CAMPBELL, 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, Bruhl, NRW, Germany; JOSE HERNANDEZ SANTIAGO, student, Universidad Te nologi a de la Mixte a, Oaxa a, Mexi o; RICHARD I. HESS, Ran ho Palos Verdes, 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; GOTTFRIED PERZ, Pestalozzigymnasium, Graz, Austria; XAVIER ROS, student, Universitat Polite ni a de Catalunya, Bar elona, Spain; OVIDIU FURDUI, Campia Turzii, Cluj, Romania; and PETER Y. 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 ts the union of the segments AB and BC in X and the segment AC in Y su h that
g AX AY x · AB + y · BC = = AB + BC AC (x + y)(AB + BC)
,
g is either the length of the segment AX if X lies between A and where AX B , or the sum of the lengths of the segments AB and BX if X lies between B and C . Prove that the three lines (x, y)ABC , (x, y)BCA , and (x, y)CBA interse t in a point dividing the segment N I in the ratio x : y, where N is the 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. t=
As usual we let a = BC , b = CA, c = AB , and s = a + 2b + c . Set x y ; thus, 1 − t = x + , and the ratio of interest be omes a fun tion x+y y
339 of t, namely f (t) =
tc + (1 − t)a a+c
=
gt AX
a+c
=
AYt b
,
where Xt is the point of AB ∪ BC , and Yt of AC , that orrespond to our parameter t. In this notation, (x, y)ABC = Xt Yt . Of ourse, when a = c, f (t) is onstant and (x, y)ABC is the line N I for all x, y . This is onsistent with what we are to prove unless a = b = c; when △ABC is equilateral we have N = I (so there is no line N I ), and our three lines interse t at that point. Let us therefore assume that a 6= c, so that f (t) is not onstant and the lines N I and (x, y)ABC interse t. We shall see that (x, y)ABC is the line in the family of lines parallel to the bise tor of ∠B (where B is the middle vertex in the subs ript) that divides the segment N I 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 probably intended (x, y)CAB for his third line. For this problem we are on erned with the domain 0 ≤ t ≤ 1; in parti ular, (a)
AY1 = bf (1) =
bc ; CY1 = b − AY1 = a ab ; Y1 a+c +c
(b)
AY0 = bf (0) =
ab ; CY0 = a bc ; a+c +c
( )
X1 = B
(d)
g 0 = (a + c)f (0) = a: when c ≥ a, X0 lies on AB (when e AX AX0 = a and BX0 = c − a); otherwise, when a ≥ c, X0 lies on BC (when e CX0 = c and BX0 = a − c).
is the foot of the bise tor of ∠CBA (be ause it divides the side AC in the ratio c : a);
(be ause f (1) =
c AX1 = ); a+c a+c
By items (a) and ( ), the line X1 Y1 bise ts ∠CBA and, therefore, it passes through the in entre I . We next will see that the line X0 Y0 passes through the Nagel point N . To determine N we use the points P , Q, R where 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 N CR of △BQA to dedu e that BN NQ
=
CA QC
·
RB AR
=
b s−a
·
s−a s−b
Let N ′ = X0 Y0 ∩ BQ; we want to show that N ′ the length QY0
=
b s−b
= N.
.
(1)
Here we will need
ab |a − c|(s − b) = |AY0 − AQ| = − (s − c) = a+c a+c
.
340 ....... ........ ... . .. ....... ... . . .. .. ... . ... ... . ... ... ... ... . . . . . ... ... .... . . . . .. .. ........ ... ... ... ... ....... .... ... ... . . . . . . .. ... . ... ... ... ..... ..... ... ... ... ... ...... ... ... . . . . . . . ... . .. .. .. .... .. .. ... ... ... ... .. .. .. ... ... ... . . . . . . . . . .. .. . .. .... ... ... ... .. ............ ... .. .. .. ... . ...... . . . . . . . ... ... . ... .... ... .. ..................... ... ... . . . . . . . . .. . . . .... . .. ... . . . . . . . . . . . . ... . .. ... .. .... ... ..................... ... ... . . . . . ... . . . . . .... ........... .. ... . . . . .. . . ... ... .. .. .. ............................ .. ... ... . . . . . . . . . . ... ........ . ........ . ... . . . . . . . . . . . . . . . .. .. ........ .... . .. . . . . . . . . . . . . . . . ........ ..... .... ... .. . . .. ... . . . . . . . . . . . . . . . . . ....... .. . .. .. . ... ....... .. ... ......... .. ... ... . ....... . . ... . . . 0 . . . . . . ... . . . . . . ... .. .. ............. . .. .. .. .. ... .... ... ... ................. ... .. ... .. .. . .... ... ... . . . . . . . . . .. . . . ...... .. . . ... . . . . . . . . . . . . . .. . . . . .. .. .... . .. ... . . . . . . . . . . . . ... . . . ..... . .. . . . . . . . . . . . . . . . . . .... .. . .. ... . . . . . . . . . . . . . ... . . . ... .. ............. .. ... . .. . . . . . . .. ... ... ... ...... ................... . .. .. ... ... . . . . . . . . . ........... .... ... .. ........ ...... ... ..... . ............ ... ...................... ............ ..... ............ ... ..................... . . . . . . . . ....... . . . . . . . .. ............. . . . . . . . . ...... .... ....... . ............... ................ ...................... .....................
a
s−
b
A
Q Y0
c− s− a a
R
.
B = X1
N
X
Y1
I
................ .. ... .... .... ... .... ... ................. .... .... ... .... ... .... .... ... .... .... ... .... .... ... .... ... .... ... .... .... .... ... ... .... .... .... .... ... ... ................
s−c
s−a
................. .. ... .... .... ... .... .... ... .... .... ... .... ... . ................... . .... .. ..... ... .... .... ... .... ... .... ... .... .... .... ... ... .... .... .... .... ... ... .................
ab a+c
bc a+c
................. .. ... .... .... ... .... .... ... .... .... ... .... ... .... .... ... .... .... ... .... .... ... ... .................. ... .. ..... .... .... ... ... .... .... .... .... ... ... .................
bc a+c
ab a+c
C
P
When c > get
a
△BQA to
we apply Menelaus' theorem to transversal
N ′ Y0 X0
BN ′ Y0 A X0 B ab a+c c−a b = · = · · = ′ N Q QY0 AX0 a + c (c − a)(s − b) a s−b
of
.
A
.................. .... ...................... . . .. ..... . ........ .. ... ... ........ ... ... ........ Y1 . ... . ... . . ... ............. . ... ............... .......... . . . ........ .. . . . . . . . . . . . . ........ ...... ... . . . . . . . I . . ........ ... ......... ... .......... ... ........ . . . . . . . . . . . . . . ... ..... ......... ........ Y0 . . . . . . . . . . .......... ... .......... ...... . . . . . . . . . .......... R ...... .... .... .... .... ...................... ... ........ .......... . ... . . ........ . .... .... .... .. . . ........ Q . . . . . . . . . . . . . . . . . . . . . . . . .... .... .... .......... ...N ............. . ..... . . . . . . . . . . . . . . . . . . . .... .... .... .... ......... ........ .. .............. .... .................................. .... .... .... .... .... .... .... .. ........ ... ......................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ........ . . . . . . . . . . . . . . . . ... .... .... .... .. . .... .... .... .... . ..... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ...........................................................................................................................................................................................................................................................................................................................................
B = X1
X0
P
C
.. . . ............................................................................................................................................................................................................................................................................................................................................. .... .... ....
a−c
c
.. .. ... ............................................................................................................................................................................................................................................................................................................................................ .... .... ....
s−c
Otherwise, when a > of △BCQ to get
N ′ Y0 X0
c
s−b
we apply Menelaus' theorem to transversal
BN ′ Y0 C X0 B bc a+c a−c b = · = · · = N ′Q QY0 CX0 a + c (a − c)(s − b) c s−b ′
In both ases BN equals the value of N ′Q
BN NQ
.
in equation (1), from whi h
341 we on lude that N = N ′ , and X0 Y0 interse ts IN at N , as laimed. It remains to observe that X0 Y0 k X1 Y1 : when c > a (that is, when X0 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 X0 X1 in the ratio t : (1 − t) while divides Y0 Y1 in that same ratio, Xt Yt is parallel to both X0 Y0 and X1 Y1 for all t. It follows that Xt Yt meets the segment N I in a point that divides it in that same ratio t : (1 − t) = x : y, as desired. Yt
Also solved by MICHEL BATAILLE, Rouen, Fran e; OLIVER GEUPEL, Bruhl, NRW, Germany; and the proposer. Geupel remarked that the Spieker entre is the midpoint of NI (on the line X1/2 Y1/2 ) while the entroid of the triangle is a trise tor of NI (on the line X2/3 Y2/3 ).
3356. [2008 : 299, 301℄ Proposed by Cristinel Morti i, Valahia University of 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, denote
r(n) = max{k ∈ N | xk ≤ n} .
(a) Prove that lim
n→∞
Z 1 r(n) 1 X (xk − xk−1 )f (xk ) = f (x) dx . n k=1 0
(b) Prove that lim
n→∞
Z 1 n 1 X f (ln k) = f (x) dx . ln n k=1 k 0
Solution by Oliver Geupel, Bruhl, NRW, Germany. R1 Let I = 0 f (x) dx and r(n+1)
Sn =
X
(xk − xk−1 )f (xk ) .
k=r(n)+1
342 We rst prove that
lim Sn = I .
(1)
n→∞
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 denoted by ∆1 , ∆2 , . . . , ∆m and m = r(n + 1) − r(n) + 1. Consider the Riemann m P f (x∗k )∆k , where x∗k is the right endpoint of the kth interval. sum Rn = k=1 Be ause f is bounded and (xk+1 − xk ) → 0, we obtain Sn = Rn + n − xr(n) f (xr(n)+1 ) − n + 1 − xr(n+1) f (n + 1) → I
as n → ∞. Next we prove the limit in part (a) in the following equivalent form: lim
n→∞
Let
ǫ > 0
n−1 1 X Sk = I . n k=0
be given. By (1) there is an integer
whenever k we have
≥ N.
N −1 P = (Sk − I) .
Let S
n−1 1 X Sk − I n k=0
k=0
≤ ≤
B→∞
S
n
Let n r(B) P k=1
= ⌊B⌋, Un =
r(B) 1 X
B
n−1 P k=0
N
+
n−N n
k=1
Sk ,
·
Whenever
ǫ
2
< ǫ
≤ B}.
ǫ
su h that
N −1 1 X 1 (Sk − I) + n k=0 n
and (2) is proved. For B > 0, let r(B) = max{k | xk lim
(2) |Sk − I| < 2 n o 2S n > max N , ǫ
n−1 X (Sk − I) k=N
We next prove that
(xk − xk−1 )f (xk ) = I .
and Vn
(xk − xk−1 )f (xk ) = Un + Vn
=
r(B) P
(3)
(xk − xk−1 )f (xk ).
k=r(n)+1
and by (2) we have
lim
n→∞
Then
1 Un = I ; n
therefore, it suÆ es to prove that B1 (Un + Vn ) − n1 Un → 0 as B → ∞. But this follows from the fa t that Vn is bounded and the al ulation 1 1 Vn (Un + Vn ) − Un = − B n B
B − ⌊B⌋ B
Un n
→ 0 − 0 ·I = 0.
343 Finally, we prove the limit in part (b). Let Tn =
1 ln n
n X
ln 1 +
k=1
1 k−1
f (ln k) ;
Wn =
1 ln n
n X f (ln k)
k
k=1
.
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 n and let |f | < C , where C is a onstant. Note that 1 = e, so there is an integer K su h that lim 1 + n→∞ n−1 ln 1 +
1 k−1
k
− 1 <
ǫ 2C
whenever k > K . Moreover, K an be hosen so that HK > Hn − ln n for ea h n, where Hn = 1 + 12 + 13 + · · · + n1 is the nth Harmoni number. Let K X 1 T = k k=1
Now if n > max |Tn − Wn | ≤ <
K , e2T /ǫ
ln 1 +
1 k−1
k
, then
n T 1 X 1 + ln n ln n k=K+1 k ǫ ǫ + = ǫ. 2 2
− 1 f (ln k)
!
ln 1 +
1 k−1
k
.
− 1 f (ln k) !
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, Biola University, La Mirada, CA, USA (part (a) only); and the proposer. There were two in omplete solutions submitted to part (b). The fa t that Sn
→I
as n → ∞ implies that
1 n
n−1 P k=0
Sk → I
as n → ∞ is part of the
theory of Cesaro sums, whi h some solvers quoted dire tly.
. [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 3357
Z
1
0
x+a 1 2 θ θ2 θπ π2 ln(1 − x) dx = ln 2 sin + − + 2 2 2 8 4 24 x + 2ax + 1
where θ is the unique solution in (0, π] of the equation cos θ = −a.
,
344 Solution by Mi hel Bataille, Rouen, Fran e. Let Z 1
f (a) =
x2
0
x+a ln(1 − x) dx . + 2ax + 1
For a ∈ (−1, 1] and x ∈ [0, 1), we have
x+a | ln(1 − x)| ≤ | ln(1 − x)| . x2 + 2ax + 1 ln(1 − x) ≤ 1+x
Sin e | ln(1−x)| = − ln(1−x) is integrable on [0, 1), we see that f (a) is wellde ned and ontinuous on (−1, 1]. [Ed.: It seems one´ must also ompute ` |b − a| 1 − x2 x + a x + b and note that − 2 = 2 2 x + 2ax + 1 x + 2bx + 1 (x + 2ax + 1) (x2 + 2bx + 1) 2 x + 2ax + 1 on [0, 1) is bounded below by 1 if 0 ≤ a ≤ 1 or by 1 − a2 if −1 < a < 0, and then ombine this with the salient observation above.℄ For s ∈ (0, 1) we have Z
1−s
2(x + a) ln(1 − x) dx + 2ax + 1 0 Z 1−s = ln(1 − x) · d ln x2 + 2ax + 1 x2 0
= =
=
=
Z 1−s ln x2 + 2ax + 1 ln s · ln (1 − s) + 2a(1 − s) + 1 + dx 1−x 0 ln s · ln s2 − (2 + 2a)s + 2 + 2a Z 1 ln u2 − (2 + 2a)u + 2 + 2a + du u u s2 ln s · ln(2 + 2a) + ln s · ln 1 − s + 2 + 2a u2 Z 1 Z 1 ln 1 − u + ln(2 + 2a) 2 + 2a + du + du u u s s ln s · ln(2 + 2a) + ε(s) − ln s · ln(2 + 2a) u2 Z 1 ln 1 − u + 2 + 2a + du , u s 2
where ε(s) → 0 as s → 0+ . It follows that 2f (a) =
Z
1
0
ln
u2 1−u+ 2 + 2a
u
du
345 or 2f (a) = g(θ) where g(θ) =
Z
ln 1 − u +
1
u
0
First, we onsider the ase a we have f (1) =
1 2
1
that is, θ
= 1
u = 2v ,
Z
u2 4 sin2 (θ/2)
u2 ln 1 − u + 4
u
0
= π.
du .
(1)
Making the substitution
Z
du =
1 2
ln(1 − v)
0
v
dv .
Next, integrating by parts and then using the well-known identity Z
1
ln(1 − t)
0
t
dt = −
Z
1
0
we obtain f (1)
= =
Z
2
(ln 2) +
1 2
0 Z 1
(ln 2)2 +
0
Z
1
∞ X tn−1
n=1
ln(v) 1−v ln(v)
dt = −
∞ X 1
n=1
n2
dv dv −
Z
1
ln(v)
dv 1−v Z 12 ln(1 − t) ln(1 − t) dt − dt t t 0 1−v
=
(ln 2)2 +
=
π2 (ln 2)2 − − f (1) , 6
0
n
!
1 2
2
so that f (1) = 21 (ln 2)2 − π12 , in agreement with the given formula. To determine g(θ) for θ ∈ (0, π), we di erentiate under the integral sign in equation (1): ′
g (θ) = −
cos(θ/2) sin(θ/2)
(To justify this, note that if
Z
1
u
du − 4u sin (θ/2) + 4 sin2 (θ/2) π β ∈ 0, and θ ∈ [β, π − β], then 0
2
u2
2
(2) for all
u ∈ [0, 1], cos(θ/2) u u cot(β/2) sin(θ/2) · u2 − 4u sin2 (θ/2) + 4 sin2 (θ/2) ≤ u2 + 4 sin2 (β/2)(1 − u)
and the dominating fun tion is integrable on [0, 1].) [Ed.: One may rewrite the 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's remark, then di erentiate ea h side with respe t to θ to obtain (2) .℄
,
346
Writing the numerator u as 12 2u − 4 sin2 (θ/2) + 2 sin2 (θ/2), the
al ulation of the integral in (2) is straightforward and gives θ θ π g (θ) = 2 · ln 2 sin + − sin(θ/2) 2 2 2 1 2
′
cos(θ/2)
.
It follows that
g(θ) =
2 θ2 πθ ln(2 sin(θ/2) + − +C 4 2 2
for some onstant C . This onstant is π12 , easily determined using the ontinuity of g and the already found value of f (1) = g(π). Finally, θ θ2 θπ π2 f (a) = g(θ) = + ln 2 sin − + 2 2 2 8 4 24 1
1
2
.
Also solved by the proposer. There was one in orre t submission.
. Proposed by Toshio Seimiya, Kawasaki, Japan. The interior bise tor of ∠BAC of triangle ABC meets BC at D. Suppose that 1 1 2 + = . 2 BD 2 AD 2
3358
Prove that ∠BAC
CD
= 90◦ .
. Solution by the proposer. Let M be the se ond interse tion of AD with the ir um ir le of △ABC and let N be the midpoint of BC . Sin e ∠BAM = ∠M AC , we have that BM = M C and M N ⊥ BC . Sin e BD · CD = AD · DM , we obtain I
= = =
hen e,
1 1 + 2 BD CD 2 BD 2 + CD 2 BD 2 · CD 2 BD 2 + CD 2 AD 2 · DM 2
A.........................................................................................................
B
,
2DM 2 = BD 2 + CD 2 .
(1)
....... ......... ...... ........ ............ ..... ..... .. ... ............... ..... ........ .... .... .... .... . . . ........ ... .... .. ... . . . . . . . . ........ ... ... ... ... . .... . . . . . . ........ ... .. ... ... . . . . . . ........ ... ... .. ... . . . . . . ........ ... . . .. . . . . ... ... . .. ........ ... . .. .. . . . . . ... ........ .. ........ ... .... .... ... ........ ... . ... ... . ........ ... ........ ... .. ... ........ .. ... ...... . ............................................................................................................................................................................................................. .. ... .... .. ... ... ... .. .. .. .. .. .. .. ... .... ... ... ... . ... .. .. . . .. . . . . ... . .. .. .. ... .. ... .. .. ... .. .. ... .. ... ... . . ... . . .. ... ... ... ... .. ... ... ... ... ... .. .. .... ... ... ... .. .. .... ... . . . . . . . . . .... ... .. .. ... .... ... ... . . .... ..... ... ... ... ... ..... ...... .. ...... ....... ... .. .. ... ....... ........ . . . . . . . . . . . . . . . .......... . . ............. .......... .... ..................... ..............................................
N
............ ............
2 AD 2
D
Sin e N is the midpoint of BC , BD 2 + CD 2 = 2 DN 2 + BN 2
M
.
C
347 Thus, we have from equation (1) that DM 2 = DN 2 + BN 2 .
(2)
Sin e ∠DN M = 90◦ , we have DM 2 = DN 2 + M N 2 , and omparing this with (2) yields BN 2 = M N 2 , that is, BN = M N . We have now dedu ed that ∠M BN = 45◦ . Therefore, ∠BAC = 2∠M AC = 2∠M BC = 2∠M BN = 90◦ .
. Solution by Oliver Geupel, Bruhl, NRW, Germany. We shall prove the following generalization:
II
2 AD 2
−
1 BD 2
−
1
<0 =0 >0
CD 2
if if if
A < 90◦ , A = 90◦ , 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 opposite side in the ratio of the other two sides. Hen e, p = b ac and q = b ab . +c +c 2 2
2
(A/2) Another ommon formula is w2 = 4b c(bcos . Using these relations as + c)2 well as the Law of Cosines, we derive
1
=
=
=
=
1
−
1 1 + 2 p2 q
2 w2
1 (b + c)2 (b + c)2 + a 2 c2 a 2 b2
b2 c2 (b + c)2 b2 c2 (b + c)2 −
·
a2 b2 + c2
−
2b2 c2 cos2
A 2
(b + c)2
− 2 cos2
A 2
b2 + c2 − 2bc cos A − b2 + c2 (1 + cos A)
b2 c2 cos A b2 + c2
b2 + c2
,
whi h ompletes the proof. Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallor a, Spain; GEORGE University of Sarajevo, APOSTOLOPOULOS, Messolonghi, Gree e; SEFKET ARSLANAGIC, Sarajevo, Bosnia and Herzegovina; ROY BARBARA, Lebanese University, Fanar, Lebanon; RICARDO BARROSO CAMPOS, University of Seville, Seville, Spain; MICHEL BATAILLE, Rouen,
348 Fran e; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; JOHN G. HEUVER, Grande Prairie, AB; JOE HOWARD, Portales, NM, USA; PETER HURTHIG, Columbia College, Van ouver, BC; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; GIANNIS G. KALOGERAKIS, Canea, Crete, Gree e; KEE-WAI LAU, Hong Kong, China; TAICHI MAEKAWA, Takatsuki City, Osaka, Japan; SALEM MALIKIC, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; MADHAV R. MODAK, formerly of Sir Parashurambhau College, Pune, India; NANCY MUELLER and SETH STAHLHEBER, Southeast Missouri State University, Cape Girardeau, MO, USA; K.C. SANDEEP, student, Southeast Missouri State University, Cape Girardeau, MO, USA; JOEL SCHLOSBERG, Bayside, NY, USA; BIKRAM KUMAR SITOULA, student, Southeast Missouri State University, Cape Girardeau, MO, USA; SOUTHEAST MISSOURI STATE UNIVERSITY MATH CLUB; D.J. SMEENK, Zaltbommel, the Netherlands; GEORGE TSAPAKIDIS, Agrinio, Gree e; PETER Y. WOO, Biola University, La Mirada, CA, USA; and TITU ZVONARU, Comane sti, Romania. There were two in orre t solutions submitted.
. [2008 : 300, 302℄ Proposed by Ray Killgrove, Vista, CA, USA and David Koster, University of Wis onsin, La Crosse, WI, USA. 2 Consider the sequen e {an }∞ n=1 de ned by an = n + n + 1. Find ∞ a subsequen e {bn }n=1 su h that b1 = a1 , b2 = a2 , b3 > a3 , every pair of terms from this subsequen e are relatively prime, and there are primes whi h divide no term of the subsequen e. 3359
Solution by Oliver Geupel, Bruhl, NRW, Germany. De ne {bn } by b1 = a1 = 3, b2 = a2 = 7, and for n > 1 let bn+1 =
n Y
k=1
Then learly
bk
!2
+
n Y
k=1
bk
!
+ 1.
{bn } is a subsequen e of {an } and b3 > a3 . Sin e we have bn ≡ 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 divides no term of {bn }. Also solved by ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, Fran e; BRIAN D. BEASLEY, Presbyterian College, Clinton, SC, USA; student, WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; SALEM MALIKIC, Sarajevo College, Sarajevo, Bosnia and Herzegovina; MISSOURI STATE UNIVERSITY PROBLEM SOLVING GROUP, Spring eld, MO, USA; JOEL SCHLOSBERG, Bayside, NY, USA; and the proposers. Most solutions were similar to the one featured above. Geupel remarks that omputations of the residues of n2 +n+1 modulo p for 1 ≤ n < p show 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 the featured 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 .
349 . [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 the number of solutions (z1 , z2 , z3 ) ∈ C3 to the system: 3360
z1 z3 z1 z2 + z2 z3 z12 + z22 + z32
= a, = b, = 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 (z1 − z2 + z3 )2
= 2a + 2b + c , = 2a − 2b + c .
Let A be a square root of 2a+2b+c and B be a square root of 2a−2b+c. We then obtain the system of linear equations: z1 + z2 + z3 z1 − z2 + z3
Solving yields (z1 + z3 , z2 ) =
= ±A , = ±B .
A+B A−B B A+B , , A− , , 2 2 2 2 −A + B −A − B −A + B −A − B , , , 2 2 2 2
.
(1)
The system of two equations z1 + z3 = ±A2± B and z1 z3 = a always has a solution. When su h a solution is paired with the orresponding z2 a solution 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 the system (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 B 2 = −4b and (z1 + z3 , z2 ) = ± B2 , − B2 . Sin e B 6= 0, in order for the original system to have two solutions, the system z1 + z3 = ± B2 ; z1 z3 = a must have only one solution; that is, B 2 z1 = z3 , and hen e −b = ± = 4a. Therefore, b = −4a, and sin e 2 A = 0 we have c = 6a and (z1 , z2 , z3 ) = ±(α, −2α, α), where α is a square root of a 6= 0.
350 If 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. A If A 6= 0 and B = 0, then A2 = 4b and (z1 + z3 , z2 ) = ± A , . 2 2 Sin e A 6= 0, in order for the original system to have two solutions, the system z1 + z3 = ± A ; z1 z3 = a must have only one solution, and hen e 2 A 2 b= ± = 4a. Therefore, b = 4a, and sin e B = 0 we have c = 6a and 2 (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 attained for triples of the form (a, 4a, 6a), (a, −4a, 6a), or (a, 0, −2a), where a is a
nonzero omplex number.
Also solved by MICHEL BATAILLE, Rouen, Fran e; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Bruhl, NRW, Germany; ROY BARBARA, Lebanese University, Fanar, Lebanon; and TITU ZVONARU, Comane sti, Romania. There was one in omplete solution submitted.
. [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 and F , respe tively. For whi h points P of the line segment EF do the areas of △EBC , △P BC , and △F BC form an arithmeti progression? 3361
Composite of solutions by Oliver Geupel, Bruhl, NRW, Germany and by Titu Zvonaru, Comane 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 , P BC , and F BC form 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 point P on EF satis es the desired ondition. Otherwise, only the midpoint of EF has the desired property. (This last laim follows immediately from the more familiar theorem that if EF F ′ is a triangle with P on side EF , P ′ on side EF ′ , and P P ′ k F F ′ , then P is the midpoint of EF if and only if F F ′ = 2P P ′ . To prove the laim from the triangle theorem, let the line through E that is parallel to BC meet at F ′ and P ′ our altitudes to BC from F and P , respe tively. Then P is the midpoint of EF if and only if F F ′ = 2P P ′ , if and only if (F F ′ + e) + e = 2P P ′ + 2e, if and only if f + e = 2p.) Also solved by ROY BARBARA, Lebanese University, Fanar, Lebanon; RICARDO BARROSO CAMPOS, University of Seville, Seville, Spain; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; TAICHI MAEKAWA, Takatsuki City, Osaka, Japan; D.J. SMEENK, Zaltbommel, the Netherlands; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer.
351 Observe that the in ir le plays almost no role in the solution|the on lusion holds for any segment EF that lies entirely on one side of the line BC. However, when E and F are the 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 on the bise tor of ∠BAC, or as the foot of the perpendi ular from A to EF . 3362. [2008 : 172,175℄ Proposed by Jose Luis Daz-Barrero, Universitat Polite ni a de Catalunya, Bar elona, Spain. Prove that
Z
1
r 3
0
Z
ln(1 + x) dx x
1
r 3
0
ln2 (1 + x) π2 dx < 2 x 12
.
Similar solutions by Kee-Wai Lau, Hong Kong, China; Ovidiu Furdui, Campia Turzii, Cluj, Romania; and Missouri State University Problem Solving Group, Spring eld, MO, USA. By Holders inequality, we have Z
1
s 3
0
ln(1 + x) x
Z
dx <
1
ln(1 + x) x
0
Z
=
1
ln(1 + x) x
0
and Z
1
0
s 3
ln2 (1 + x) x2
dx
<
Z
1
0
=
Z
1
0
It now follows that Z
1
0
s 3
1/3 Z dx
3/2
1 0
1/3 dx
ln(1 + x) x ln(1 + x) x
1
2/3 Z dx 2/3 dx .
=
∞ X (−1)n−1 π2 = n2 12 n=1
,
1
0
s Z 1 2 ln(1 + x) 3 ln (1 + x) dx dx x x2 0 Z 1 ln(1 + x) < dx x 0 Z 1X ∞ (−1)n−1 xn−1 = dx n 0 n=1 Z 1 ∞ X (−1)n−1 = xn−1 dx n 0 n=1
2/3 dx
3
1 dx
1/3
352 where the integral and the sum an be inter hanged sin e the sum onverges ∞ P (−1)n−1 π2 uniformly on [0, 1], and follows by straightforward ma= 2 n 12 n=1
nipulations from the well-known formula
∞ P 1
n=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, Bruhl, 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, Dipartimento di Matemati a, Universita degli studi di Tor Vergata Roma, Rome, Italy; XAVIER ROS, student, Universitat Polite ni a de Catalunya, Bar elona, Spain; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer. Both Geupel and the proposer identi ed the last integral omputed above as essentially R a value of the dilogarithm fun tion, one form of whi h is Li2 (x) = x0 ln(1 − t)/t dt.
Crux Mathemati orum with Mathemati al Mayhem
Former Editors / An iens Reda teurs: Bru e L.R. Shawyer, James E. Totten Crux Mathemati orum
Founding Editors / Reda teurs-fondateurs: Leopold Sauve & Frederi k G.B. Maskell Former Editors / An iens Reda teurs: G.W. Sands, R.E. Woodrow, Bru e L.R. Shawyer Mathemati al Mayhem
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