193 EDITORIAL Va lav Linek
Vous avez probablement dej a re u, a travers les orrespondan es, la triste nouvelle que notre oediteur bien aime, Jim Totten, est mort le 9 mars 2008. Une ne rologie suit et edtiorial. En 2009, on onsa ra un numero spe ial a Jim. Les propositions de problemes
onsa res a Jim que nous re evons avant le 1er janvier 2009 seront onsider ees pour un ensemble spe ial de problemes a appara^tre dans le numero spe ial. Les le teurs peuvent egalement envoyer des arti les onsa res a Jim pour le numero spe ial avant le 1er de embre 2008. Les le teurs qui veulent
ontribuer des projets spe iaux devraient nous onsulter aussitot ^ que possible, au as ou une attention parti uliere serait requise pour a
omplir le projet et pour eviter la dupli ation (d'autres peuvent avoir des projets semblables a soumettre). Vue le grand reseau des amis mathemati iens de Jim partout dans le monde, il est fort possible qu'un seul numero spe ial n'est pas suÆsant pour
ontenir tous les propositions et les projets. Dans e as nous distribuerons le materiel supplementaire dans les numeros suivants. Je vous remer ie, le teurs, de votre appui au ours de ette periode diÆ ile. Je suis maintenu a ot par vos mots aimables, et votre en ouragement me donne la motivation pour ontinuer a faire mon travail. Au plaisir de ommuniquer ave vous dans le pro hain futur. ................................................................. Some of you already know the sad news through orresponden e, that our herished Co-Editor, Jim Totten, passed away on Mar h 9, 2008. An obituary follows this editorial. In 2009 a spe ial issue will be dedi ated to Jim. Problem proposals we re eive before January 1, 2009 and dedi ated to Jim will be onsidered for a spe ial set of problems to appear in the spe ial issue. Readers may also send arti les dedi ated to Jim for the spe ial issue before De ember 1, 2008. Readers wishing to ontribute spe ial proje ts should onsult with us as soon as possible, in ase spe ial attention is required to omplete the proje t and to avoid dupli ation (others may have similar proje ts in mind). Given Jim's extensive network of mathemati al friends all over the world, it is quite possible there will be more material than one spe ial issue an hold. In that ase we will distribute the extra material in subsequent issues. I thank you, the readers, for your support during this diÆ ult period. I am buoyed by your kind words, and your en ouragement keeps me going. I very mu h look forward to orresponding with all of you in the future.
194 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. After obtaining his Ba helor's degree at the University of Saskat hewan, Jim then earned a Master's and PhD degree in Mathemati s from the University of Waterloo, after whi h 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 during 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 up a position at Cariboo College, where he remained as it evolved into the University College of the Cariboo and then eventually 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 in 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, his son Dean, daughter-in-law Christie and granddaughter Mikayla of Se helt, his father Wilf Totten of Edmonton, sister Judy Totten of Regina, sister Josie Laing (Neil) of Onoway, sister-in-law Marilyn Totten 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. 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
195 SKOLIAD No. 110 Robert Bilinski
Please send your solutions to the problems in this edition by 1 O tober, . A opy of MATHEMATICAL MAYHEM Vol. 4 will be presented to one pre-university reader who sends in solutions before the deadline. The de ision of the editor is nal. 2008
Nos questions proviennent e mois- i du Con ours de Comte, So iet e Mathematique de Croatie 2007 (niveau se ondaire). Nous remer ions Hanjs Z eljko, Universite de Zagreb, pour la opie du on ours. Con ours de Comte So iet e Mathematique de Croatie er
(niveau se ondaire, 1 grade) 9 mars 2007 . Trouver toutes les solutions entieres de l'equation x2 + 112 = y 2 .
1
. Dans un er le ave entre S et rayon r = 2, deux rayons SA et SB sont tra es. L'angle entre eux est 45◦ . Soit K l'interse tion de la ligne AB et la perpendi ulaire de AS passant par S . Soit L le pied de la hauteur du sommet B dans le triangle ABS . Determiner l'aire du trapeze SKBL. 2
3
. Soient a, b et c trois nombres reels non-nuls donnes. Trouver x, y et z si ay + bx bz + cy cx + az 4a2 + 4b2 + 4c2 = = = xy yz zx x2 + y 2 + z 2
.
. Soient a et b deux nombres reels positifs tels que a > b et ab = 1. Prouvez l'inequation 4
a−b ≤ a2 + b2
√ 2 4
.
Determiner a + b quand l'egalit e tient. 5. Le ratio des longueurs des deux ot ^ es d'un re tangle est 12 : 5. Les diagonales divisent le re tangle en quatre triangles. Les er les ins rits de deux triangles qui ont un ot ^ e ommun sont tra es. Soient r1 et r2 les rayons. Trouver le ratio r1 : r2 .
196 County Competition The Croatian Mathemati al So iety st
(se ondary level, 1 grade) Mar h 9, 2007
. Find all integer solutions to the equation x2 + 112 = y2 .
1
. In a ir le with entre S and radius r = 2, two radii SA and SB are drawn. The angle between them is 45◦ . Let K be the interse tion of the line AB and the perpendi ular to line AS through point S . Let L be the foot of the altitude from vertex B in ∆ABS . Determine the area of trapezoid SKBL. 2
. Let a, b, and c be given nonzero real numbers. Find x, y, and z if
3
ay + bx xy
=
bz + cy yz
=
cx + az
=
zx
4a2 + 4b2 + 4c2 x2 + y 2 + z 2
.
4. Let a and b be positive real numbers su h that a > b and ab = 1. Prove the inequality √
a−b
a2
+
b2
≤
2
4
.
Determine a + b if equality holds. 5. The ratio between the lengths of two sides of a re tangle is 12 : 5. The diagonals divide the re tangle into four triangles. Cir les are ins ribed in two of them having a ommon side. Let r1 and r2 be their radii. Find the ratio r1 : r2 .
Next we give solutions to the 23rd W.J. Blundon Mathemati s Contest given at [2007 : 321-322℄.
23rd
W.J. Blundon Mathemati s Contest
Sponsored by the Canadian Mathemati al So iety in ooperation with The Department of Mathemati s and Statisti s Memorial University of Newfoundland February 22, 2006 . If loga x = logb y, show that ea h is also equal to logab xy.
1
Solution by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto, ON. Let N = loga x = logb y. Then aN = x and bN = y or (ab)N = xy. This means that N is also equal to logab xy.
197 2. In how many ways an 20 dollars be hanged into dimes and quarters, with at least one of ea h oin used?
Solution by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto, ON. An amount of 20 dollars is equal to 2000 ents. The last digit being 0, the number of quarters used is even, sin e the other denomination gives us multiples of 10. The number of nonzero even multiples of 25 whi h are less than 2000 is 39. Therefore there are 39 ways that 20 dollars an be hanged into dimes and quarters, with at least one of ea h used. 3. If one of the women at a party leaves, then 20% of the people remaining at the party are women. If, instead, another woman arrives at the party, then 25% of the people at the party are women. How many men are at the party?
Solution by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto, ON. Fra tions 20% and 25% are equal to 15 and 14 , respe tively. Let x be the number of women at the party and let y be the number of people at the party. A
ording to the problem, we have x−1 = x+1 =
1 (y − 1) , 5 1 (y + 1) . 4
These equations have the solution (x, y) = (7, 31), and therefore the number of men at the party is y − x = 24. . Find two fa tors of 248 − 1 between 60 and 70.
4
Solution by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto, ON. Repeatedly using the formula x2 − y2 = (x − y)(x + y), we have 248 − 1 = (224 + 1)(212 + 1)(26 + 1)(26 − 1) .
Sin e 26 = 64, two fa tors of 248 − 1 between 60 and 70 are 63 and 65. . The yearly hanges in the population ensus of a town for four onse utive years are, respe tively, 25% in rease, 25% in rease, 25% de rease, and 25% de rease. Find the net per ent hange to the nearest per ent over the four years. 5
198 Solution by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto, ON. Let A be the number of people in the town originally. The number of people after four years is A(1.25)(1.25)(0.75)(0.75) = 0.87890625A. Therefore, the net per ent hange to the nearest per ent over four years is a 12% de rease. . If x + y = 5 and xy = 1, nd x3 + y3 .
6
Solution by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto, ON. The steps in the solution are (x + y)3 3
2
2
x + 3x y + 3xy x3 x3 x3
3
+y + y3 + y3 + y3
= 53 ,
125 , 125 − 3xy(x + y) , 125 − 3(1)(5) , 110 .
= = = =
7. The point (4, 1) is on the line that passes through the point (4, 1) and is perpendi ular to the line y = 2x + 1. Find the area of the triangle formed by the line y = 2x + 1, the given perpendi ular line, and the x-axis.
Solution by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto, ON. Let y = mx + b be the equation of the unknown line. Sin e it is perpendi ular to the line y = 2x + 1 and it passes through the point (4, 1), its equation is y = − 21 x + 3. From the equations, we get the oordinates of all the verti es and the dimensions of the triangle. Its base is 6.5 units and its height is 2.6 units. Therefore, its area is 8.45 square units. .. ... .... ... ... ... .. ... . ......... ... 4 13 ... ............ ... ........ 5 5 ....... ................ .... . ......... . ... . . . . . ......... ........ .. ... ......... ... .. ... ......... .... ... .... ......... ... ......... ....... ......... .. .... ......... ............ ......... ... ......... ... ....... . ......... 1 . .. .... . ..... . . 2........................................................................................................................................................................................................................................... . .. ... . . . . . .......... .. . .. . . .. ..... .... .
y = 2x + 1
)
.. ... ....... ......... ..... ....
r (a, 2a + 1) = ( , (4, 1)
r
......... .... .... ...........
r
(− , 0)
(6, 0)
r
. An arbitrary point is sele ted inside an equilateral triangle. From this point perpendi ulars are dropped to ea h side of the triangle. Show that the sum of the lengths of these perpendi ulars is equal to the length of the altitude of the triangle.
8
199 OÆ ial Solution, modi ed by the editor. Let s be the length of the equal sides, let h be the altitude, and let x, y, and z be the three distan es from the interior point to the sides of the triangle. Then,
=
But we also have Can elling
1 s 2
s
1 1 1 sx + sy + sz 2 2 2 1 s(x + y + z) . 2
Area of triangle
=
y
x
............... .....
=
..... ..............
Area of triangle
.. ...... ..... .... ... . ... .. .... ..... . ... .. ... .. . ... ... ... ... .. ... . .. ... . . . .. . . . . ........... . .. . ... . . . . . . . ... . . ...... . . . . . ... . . . . ............. .. ............ . ... . ........ .......... .. ... ........... ... ... . . . . . . . . . ... .. .... ..... .. . . ... . . .... .. ... .. ... . . . . . . .... .. .. . ... . . . . . . . ... ... ... . ... ..... . . . ... .... .. .... ..... . ... . . .... . ... ... . ... . . ... . . .... . ....... . . .... .... ..... ..... . . . .... ..... ... ....... . . . ..... . . ..........................................................................................................................................................................
s
p
z
1 2
s
sh .
from ea h expression for the area, we have x + y + z = h.
Also solved by RUIQI YU, student, Stephen Lea o k Collegiate Institute, Toronto, ON 9
. Find all positive integer triples (x, y, z) satisfying the equations x2 + y − z = 100 and x + y 2 − z = 124 .
Solution by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto, ON. Subtra t the rst equation from the se ond and then fa tor the result to obtain (y − x)(x + y − 1) = 24. Trying out all pairs of fa tors of 24, namely (1, 24), (2, 12), (3, 8), and (4, 6) for ea h of the two expressions leads to two positive integer solutions for x and y, namely (x, y) = (3, 6) or (12, 13). However, the rst solution for x and y leads to a negative z , and must be ex luded. Hen e (x, y, z) = (12, 13, 57). . How many roots are there of the equation sin x =
10
?
1 x 100
Solution by Ruiqi Yu, student, Stephen Lea o k Collegiate Institute, Toronto, ON. If x is measured in degrees, then we have 3 roots, sin e in this ase x the line y = 100 only interse ts one period of y = sin x. If x is measured in radians, then the line has a y-value bigger than 1 for x bigger than 100. x From x = 0 to x = 100, the line y = 100
uts through approximately 15.91 periods of y = sin x. So we see that the line uts the urve 32 times when x is not negative. By symmetry, it uts the urve 32 times when x is not positive. But we ount (0, 0) twi e, hen e the line interse ts the urve 32+32−1 = 63 times. Hen e there are 63 roots when x is measured in radians. That brings us to the end of another issue. This month's winner of a past Volume of Mayhem is Ruiqi Yu. Congratulations Ruiqi! Continue sending in your ontests and solutions.
200 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), Eri Robert (Leo Hayes High S hool, Frederi ton), Larry Ri e (University of Waterloo), and Ron Lan aster (University of Toronto).
Mayhem Problems Veuillez nous transmettre vos solutions aux problemes du present numero avant le 15 juillet 2008. Les solutions re ues apres ette date ne seront prises en
ompte que s'il nous reste du temps avant la publi ation des solutions. Chaque probleme sera publie dans les deux langues oÆ ielles du Canada (anglais et fran ais). Dans les numeros 1, 3, 5 et 7, l'anglais pre edera le fran ais, et dans les numeros 2, 4, 6 et 8, le fran ais pre edera l'anglais. La reda tion souhaite remer ier Jean-Mar Terrier, de l'Universite de Montreal, d'avoir traduit les problemes. . Propose par l'Equipe de Mayhem. On onstruit le tableau arre
M344
1 2 4 5 7 8
3 6 9
forme des nombres de 1 a 9 e rits en ordre roissant, ligne apres ligne. La somme des entiers sur les deux diagonales est 15. Si l'on onstruit un tableau
arre semblable ave les entiers de 1 a 10 000, quelle sera la somme des nombres sur ha une des diagonales ? . Propose par John Grant M Loughlin, Universite du NouveauBrunswi k, Frederi ton, NB. √ L'aire d'un triangle iso ele ABC est q 15. Si AB = 2BC , exprimer le perim etre du triangle ABC en fon tion de q. M345
. Propose par l'Equipe de Mayhem. Determiner le nombre de hi res que omporte l'entier 280 sans l'aide d'une al ulatri e.
M346
201 . Propose par l'Equipe de Mayhem. Quatre nombres entiers positifs a, b, c et d sont tels que
M347
(a + b + c)d (a + c + d)b (a + b + d)c (b + c + d)a
= = = =
420 , 403 , 363 , 228 .
Trouver es quatre nombres. . Propose par l'Equipe de Mayhem. etre in lut les deux Le perim etre d'un se teur de er le est 12 (le perim rayons et l'ar ). Determiner le rayon du er le qui maximise l'aire de e se teur. M348
. Propose par l'Equipe de Mayhem.
M349
(a) Trouver toutes les paires ordonnees d'entiers (x, y) ave
1 1 1 + = . x y 5
(b) Combien y a-t-il de paires ordonnees d'entiers (x, y) ave 1 1 1 + = x y 1200
?
................................................................. . Proposed by the Mayhem sta . Consider the square array
M344
1 2 4 5 7 8
3 6 9
formed by listing the numbers 1 to 9 in order in onse utive rows. The sum of the integers on ea h diagonal is 15. If a similar array is onstru ted using the integers 1 to 10 000, what is the sum of the numbers on ea h diagonal? . Proposed by John Grant M Loughlin, University of New Brunswi k, Frederi ton, NB. √ The area of isos eles △ABC is q 15. Given that AB = 2BC , express the perimeter of △ABC in terms of q.
M345
. Proposed by the Mayhem Sta . Without using a al ulator, nd the number of digits in the integer 280 .
M346
202 . Proposed by the Mayhem Sta . Four positive integers a, b, c, and d are su h that
M347
(a + b + c)d (a + c + d)b (a + b + d)c (b + c + d)a
= 420 , = 403 ,
= 363 , = 228 .
Find the four integers. . Proposed by the Mayhem Sta . The perimeter of a se tor of a ir le is 12 (the perimeter in ludes the two radii and the ar ). Determine the radius of the ir le that maximizes the area of the se tor. M348
. Proposed by the Mayhem Sta .
M349
(a) Find all ordered pairs of integers (x, y) with
1 1 1 + = . x y 5
(b) How many ordered pairs of integers (x, y) are there with 1 1 1 + = x y 1200
?
Mayhem Solutions . Proposed by Bru e Shawyer, Memorial University of Newfoundland, St. John's, NL. Nine ir les of radius 1/2 are externally tangent to a ir le of radius 1 and are tangent to one another, as shown. Determine the distan e between the entres of the rst and last of the ir les of radius 1/2. M294
......................... ... ...................... ...... ...... ... .. .. .... ... . .. ... ............................. .... .. . ... ..... ... . .... . . . .. . . . . . . . . . . . . . . ............................................ .. . ........ ............... . . . . . . ......... ..................... .. ..... ... .. .... .. .... . . ... ..... . ....................................... .. . . .... ... .. ..... ... ... .... ..... .... ............................. .. . . . ........ .............. . . . . ... . ... ...... ... .... . ... ... .. ............ .......... .... .... . ........... . . . . . ... . . . . . . . ...... ...... ..... ... ... ....... ............. ...................................................... ..... . ............................................. .. .... ........ . .. ... ......... .... .. .... ... .... ... ... . . . .......... .......... .... .... . . . . . . . . . . . . . . ........................ .. .... ... ........ .......... ..........
Solution by Cao Minh Quang, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam, adapted by the editor. Label the entre of the rst ir le of radius 12 with A, the entre of the se ond ir le with B , the entre of the last (ninth) ir le with C , and the
entre of the ir le of radius 1 with O. We need to determine the length of AC . We know that AO , BO , and CO are ea h equal to 1 + 12 = 32 , be ause the smaller ir les are tangent to the larger ir le, and AB = 12 + 12 = 1, be ause the smaller ir les with entres A and B are tangent.
203 Let ∠AOB = α. Then ∠AOC = 360◦ − 8α, be ause there are 8 pairs of ir les determining an angle of α at O. Applying the Law of Cosines to △AOC and using cos θ = cos(360◦ − θ), we have AC 2
= = =
AO 2 + CO 2 − 2AO · CO cos ∠AOC 2 2 3 3 3 3 cos(360◦ − 8α) + − 2 2 2 2 2 9 (1 − cos 8α) . 2
To nd cos 8α, we rst nd cos α using the Law of Cosines in △AOB : AB 2 12
= AO 2 + BO 2 − 2AO · BO cos ∠AOB , 2 2 3 3 3 3 = + − 2 cos α , 2 2 2 2
and solving for cos α gives cos α = 79 . Therefore, cos 2α cos 4α cos 8α
2 7
17 , −1 = 9 81 2 5983 17 −1 = − = 2 cos2 2α − 1 = 2 , 81 6561 5983 2 28545857 2 = cos 4α − 1 = 2 −1 = 6561 43046721 2
= 2 cos α − 1 = 2
,
and hen e AC =
r
9 (1 − cos 8α) = 2
s
9 2
28545857 1− 43046721
√ 1904 2 = 2187
.
Also solved by HASAN DENKER, Istanbul, Turkey; ANGELA DREI, Riolo Terme, Italy; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; R. LAUMEN, Deurne, Belgium; MISSOURI STATE UNIVERSITY PROBLEM SOLVING GROUP, Spring eld, MO, USA; RICARD PEIRO, IES \Abastos", Valen ia, Spain; KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; and J. SUCK, Essen, Germany. There was 1 in orre t solution submitted. M295. Proposed by Bru e Shawyer, Memorial University of Newfoundland, St. John's, NL. Square ABCD is ins ribed in one-eighth of a ir le of radius 1 so that there is one vertex on ea h radius and two verti es on the ar . Determine the exa t area of the square in the √ a+b c form , where a, b, c, and d are integers. d . ........ ... ...... .... .... . .... ... ......... . . .......... .......... .... .......... ..... ....................... . .... . . .. .... ............... .... .... ...... .... .... .... . .... . .. .... ... . . . .... .. .. . . .... . . .. .... ... . . . .. .. ... . . . .. ............. .. . . . . . . . . . . . . . . .. .. ...... . . ... . . . . . . .. . . . . .. ...... .. . .. . . . . . . . . . . . . . .. ........................................................................................................................................................
204 Solved independently by Hasan Denker, Istanbul, Turkey; Salem Maliki , student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; R. Laumen, Deurne, Belgium; and Ri ard Peiro, IES \Abastos", Valen ia, Spain. Let N and M represent the midpoints of AD and BC , respe tively. In right triangle B OAN , we have AN = 12 BC and ∠AON = π8 . A Therefore, Sin e we let
1 BC 2
= tan
ON
π 8
. We al ulate
2 tan π8 π 1 = tan = 4 1 − tan2
x = tan
π 8
π 8
tan
O
positive, so by the quadrati formula tan π8 1
√
BC
.
,
and we simplify to obtain
Thus 2ON = 2 − 1 and ON Theorem in triangle BOM we have
π 8
=
....... ... ...... .... .... .... .... .... . . ................ ... ..................... ... ........ . . ..... .. ............. ... . . . .... ... .. ....... .... ............ ... ... .... ...... . ... .... ... ... ... ......... . . . . . . ... ..... ... . ...... . . . . . .... ... .. .. .. ... . . .... . . . M . . .. ... . .... .... ... ... ......... . . ... ... ... . . . . . .. ............. .. ... ... ... . . . . . . . . . . . . . . .. .. ...... . . ...... ... ... ... . . . . . . . N . .. . . . . .. .... .. .. . ............. ........ ... . ...........................................................................................................................................................
D
x + 2x − 1 = 0. Now x √ √ −2 + 8 =x= = 2 − 1. 2
BC , √ 2 2−2
C
is
2
and by the Pythagorean
OB 2 = BM 2 + OM 2 = BM 2 + (ON + N M )2 .
Sin e OB = 1, N M 1 =
= =
1 4
BC
2
+
1 BC 2 + 4 1 4
BC 2 +
hen e the area is
= BC , BM =
1 BC 2
, and ON
BC , 2 2−2
= √
we obtain
!2 √ 2 1 BC 2 2−1 2 = BC + + BC BC 2 √ √ 4 2 2−2 2 2−2 !2 √ √ (2 2 − 1)(2 2 + 2) BC 2 4 √ ! √ ! 11 + 6 2 6+3 2 2 BC = BC 2 , 4 2
2 BC = √ , 6+3 2 2
whi h upon rationalizing is
√ 2− 2 . 3
MEZ MORENO, Universidad de Jaen, Also solved by SAMUEL G O Jaen, Spain; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; and J. SUCK, Essen, Germany.
. Proposed by Daniel Tsai, student, Taipei Ameri an S hool, Taipei, Taiwan. Let n be a positive integer. In the Cartesian plane, onsider the points ak = (k, n) and bk = (k, 0) for k = 1, 2, . . . , n. We onne t ea h pair ak , bk by a straight (verti al) line segment. Then we draw an arbitrary nite number of horizontal line segments, ea h onne ting two adja ent verti al line segments, su h that no one point on any verti al segment is the endpoint of two horizontal segments. M296
205 Let A = {a1 , a2 , . . . , an } and B = {b1 , b2 , . . . , bn }. De ne a map from A to B as follows: starting from ai , travel down the segment until you meet the end-point of a horizontal segment, go to the other end-point of that segment, and ontinue on down the new verti al line, repeating this until there are no more horizontal segments to meet, nally ending at bj for some j . Show that no two points of A map to the same point of B . Solution by Ri hard I. Hess, Ran ho Palos Verdes, CA, USA. The pro ess des ribed is uniquely reversible, so that starting from bj gets you to ai if going from ai got you to bj . Therefore, when going in reverse, there is no possibility of splitting paths from bj to ai , and therefore no two points in A an map to the same point in B . Also solved by JACLYN CHANG, student, Western Canada High S hool, Calgary, AB; and the proposer.
. Proposed by John Grant M Loughlin, University of New Brunswi k, Frederi ton, NB. Numbers su h as 34543 and 713317 whose digits an be reversed without hanging the number are alled palindromes. Show that all four-digit palindromes are multiples of 11. M297
Solved independently by Angela Drei, Riolo Terme, Italy; Samuel Gomez Moreno, Universidad de Jaen, Jaen, Spain; Ri hard I. Hess, Ran ho Palos Verdes, CA, USA; R. Laumen, Deurne, Belgium; Salem Maliki , student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; and Kunal Singh, student, Kendriya Vidyalaya S hool, Shillong, India Every four-digit palindrome an be written in the form abba, where a and b are integers, su h that 1 ≤ a ≤ 9 and 0 ≤ b ≤ 9. We an then on lude that abba = 1000a + 100b + 10b + a = 1001a + 110b = 11(91a + 10b) .
Thus, every four-digit palindrome is a multiple of 11. Also solved by JACLYN CHANG, student, Western Canada High S hool, Calgary, AB; and DANIEL TSAI, student, Taipei Ameri an S hool, Taipei, Taiwan.
. Proposed by John Grant M Loughlin, University of New Brunswi k, Frederi ton, NB. M298
(a) Given that a number is a four-digit palindrome, what is the probability that the number is a multiple of 99? (b) Given that a four-digit number is a multiple of 99, what is the probability that the number is a palindrome?
206 Solution by Samuel Gomez Moreno, Universidad de Jaen, Jaen, Spain. (a) A four-digit palindrome has the form abba, where a and b are integers with 1 ≤ a ≤ 9 and 0 ≤ b ≤ 9. There are therefore 9 · 10 = 90 four-digit palindromes. By Problem M297, we know that all four-digit palindromes are multiples of 11. For a four-digit palindrome, being a multiple of 99 is therefore equivalent to being a multiple of 9. A positive integer is a multiple of 9 if and only if the sum of its digits is also a multiple of 9, so the four-digit palindromes whi h are multiples of 99 are pre isely those for whi h 2(a + b) is a multiple of 9, that is, 2(a + b) equals 18 or 36. A omplete list of these palindromes is 1881, 2772, 3663, 4554, 5445, 6336, 7227, 8118, 9009, and 9999. Thus the probability is 10 = 19 . 90 (b) There are a total of 91 four-digit multiples of 99, namely the multiples from 99(11) = 1089 to 99(101) = 9999. By part (a), we know that among these there are ten four-digit palindromes. Therefore the required probability is 10 . 91 Also solved by ANGELA DREI, Riolo Terme, Italy; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; R. LAUMEN, Deurne, Belgium; and DANIEL TSAI, student, Taipei Ameri an S hool, Taipei, Taiwan. There were 2 in orre t solutions submitted.
. Proposed by Titu Zvonaru, Comane sti, Romania. Let a, b, and c be positive real numbers with ab + bc + ca = 3. Prove
M299
that
ab c2
+1
+
bc a2
+1
ca
+
b2
+1
≥
3
.
2
Solution by Cao Minh Quang, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam. Using the Arithmeti Mean-Geometri Mean Inequality, we have (abc)2/3 =
p ab + bc + ca 3 (ab)(bc)(ac) ≤ = 1, 3
hen e abc ≤ 1. The desired inequality is equivalent to
ab −
ab 2 c +1
+
whi h is the same as
abc2 c2
We have (x − 1)2 is positive. Thus,
+1
≥ 0,
bc − +
bc 2 a +1
bca2 a2
+1
+
+
ca −
cab2 b2
+1
≤
ca 2 b +1 3 2
hen e x2 + 1 ≥ 2x, and hen e
≤
3 2
,
.
1 1 ≤ x2 + 1 2x
when x
abc2 bca2 cab2 abc2 bca2 cab2 3 3 + + ≤ + + = abc ≤ 2 2 2 c +1 a +1 b +1 2c 2a 2b 2 2
.
207 Equality holds if and only if a = b = c = 13 . Also solved by ARKADY ALT, San Jose, CA, USA (two solutions); JOE HOWARD, Portales, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; CAO NM, USA; SALEM MALIKIC, MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam (se ond solution); and SHI CHANGWEI, Xian City, Shaan Xi Provin e, China. There were 2 in orre t solutions submitted.
. Proposed by Geo rey A. Kandall, Hamden, CT, USA. Let ABC be an arbitrary triangle. Let D and E be points on the sides AC and AB , respe tively, and let P be the point of interse tion of BD and CE . If AE : EB = r and AD : DC = s, determine the ratio of areas [ABC] : [P BC] in terms of r and s. M300
Solution by Salem Maliki , student, Sarajevo College, Sarajevo, Bosnia and Herzegovina. Let AP meet BC at point K . Starting with Ceva's Theorem applied to △ABC and substituting the given ratios, we obtain AE BK CD · · = 1 EB KC DA
and hen e
=⇒
r·
BK 1 · = 1 KC s
=⇒
CB KC + BK r r+s = = +1 = BK BK s s
KC r = BK s
,
.
Applying Menelaus' Theorem to the transversal BP D and △AKC , and substituting the expression just obtained, we have AD CB KP · · = 1 DC BK P A
r + s KP · = 1 s PA PA = r + s, KP
=⇒ s · =⇒
so it follows that KA P A + KP = KP KP
Sin e xy = zt implies results to obtain r+s+1
x z x+z = = , y t y+t
= r + s + 1.
we ombine this with the previous
=
KA [△BKA] [△KAC] = = KP [△BKP ] [△KP C]
=
[△BKA] + [△KAC] [△ABC] = [△BKP ] + [△KP C] [△P BC]
where [△BKA] denotes the area of △BKA, [△ABC] : [△P BC] = r + s + 1.
,
and so forth. This shows that
\Abastos", Also solved by HASAN DENKER, Istanbul, Turkey; RICARD PEIRO, IES Valen ia, Spain; KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; J. SUCK, Essen, Germany; and TITU ZVONARU, Comane sti, Romania. There was 1 in orre t solution submitted.
208
Problem of the Month Ian VanderBurgh
As spring arrives in most parts of the ountry, it's time to dust o our bi y les. This problem is a bit reminis ent of the gears and wheels of a bi y le. (1996 Cayley Contest) Two ir ular dis s have radii 8 and 28. The larger dis qA is xed while the smaller dis rolls around the outside B of the larger dis . In their original positions, point A on the smaller dis oin ides with point B on the larger dis . The least number of rotations that the small dis makes about its entre before A and B again oin ide is (A) 5 (B) 2 (C) 9 (D) 4.5 (E) 3.5 A friend of mine asked me about this problem from the ar hives a ouple of months ago. It stumped me! Here's what I did initially:
Problem
........... ......... ............. ..... ... ... .. .. ... ... ..... . . . . . . . . . . . . . . . . . . . . . . . . . . . . .......... .. ... .......... . . . . . . . . . . . . ...... . ... . . . . . . . . . ...... . . ................................ .... .. ... ... .. .. .. .. ... . ... .... .. ... .. ... .. .. ... ... . . . ... ... .... .... .... ..... ..... ......... ..... . . . . . . . ....................................
As the smaller ir le rolls around the larger ir le, we assume that no slipping o
urs. As a result, the length of ir umferen e travelled as measured on the smaller ir le equals that as measured on the larger ir le. Let's suppose that the smaller ir le travels m times around its
ir umferen e while rolling n times around the larger ir le. Sin e we want points A and B to line up at the end of the pro ess, then ea h of m and n should be integers. The length of ir umferen e travelled along the smaller ir le is m2π(8) = 16πm and the length travelled along the larger ir le is n2π(28) = 56πn. For these lengths to be equal, 16πm = 56πn or 2m = 7n. Sin e m and n are positive integers and we want m to be as small as possible, then m = 7 and n = 2. Therefore, the smaller ir le rotates 7 times around its entre before the rst time that A and B oin ide. Wait... 7 is not one of the possible answers! Is the question wrong? Did we do something wrong mathemati ally? Is our logi wrong? The question is orre t. In fa t, this is indeed the rst time that A and B oin ide again, but somehow 7 is the wrong answer. The thing that we haven't taken into a
ount is that as the smaller ir le rolls, its motion around the larger ir le produ es extra rotations around its entre. Here are three ways to look at this problem. It's up for you to de ide how valid these solutions are and whi h onvin es you the most.
209 . 7 + 2 = 9. That's the shortest solution to a POTM in my time here! What is this trying to say? The smaller ir le rotates 7 times while rolling 2 times about the entre of the larger ir le. So perhaps the smaller ir le rotates an extra 2 times around its entre, for a total of 9 times. Do you buy this? (A tually, a good point here is that the total number of rotations of the smaller ir le should be at least 7, so given the possible answers, (C) 9 must be orre t.) Solution 1
. When two ir les are tangent (like the smaller and larger ir les), the line joining their entres passes through the point of tangen y. Thus, the distan e between the entres is the sum of the radii, or 8+28 = 36. Therefore, the entre of the smaller ir le is always a distan e of 36 from the
entre of the larger ir le. Hen e, the entre of the smaller ir le travels around a ir le of radius 36. The entre of the smaller ir le moves 2 times around this ir le, so travels a total distan e of 2(2π(36)) = 144π. The ir umferen e of the smaller ir le is 2π(8) = 16π, and so while the
entre of this ir le moves a distan e of 144π m, it must rotate 144π = 9 16π times. This feels better than Solution 1, but still leaves a somewhat un ertain feeling. The ratio of distan e travelled by the entre to the number of rotations works on a at surfa e, but should it work rolling around a ir ular surfa e? The last solution is slightly less intuitive, but a bit more mathemati al. Solution 3. First, we turn the diagram so that the line joining the entres is horizontal. We don't really have to do this, but you might nd it easier to visualize. Next let's suppose that the smaller ir le has rolled so that the point of
onta t between the ir les is at an angle of x below the horizontal. (We'll treat the angles as measured in radians, but feel free to think in degrees if that's easier for you.) Call the entre of the larger ir le O, the entre of the smaller ir le C , the point q qB of tangen y T ; we label the end-point of the x O qA T radius to the left of C as P . We have de ned ∠T OB = x. Sin e OB is parallel to P C , then q we also have ∠T CP = x. C P We now use the fa t that the length of ar BT equals the length of ar T A (by equal distan es rolled). The length of ar BT as a fra tion of the entire ir umferen e of the ir le with entre O equals the fra tion that ∠T OB is of the angle asso iated with a full revolution. Thus, the length of ar BT is Solution 2
............................. ............... ......... ......... ....... ....... ..... ..... .... . . . .... ... . . ... ... ... . .. .. . .. ... .... .. .... .. ... .................................................................... ... ........ ... . ... ........ .. .. ................. ........ .. . ......... ........ . ... .. ........ .............. .. .. ... ........ .... ... ...... ... ... ... ... ......... ... .... ..... ............... .... . . ...... .. .... .. ... .. ..... .... ................................... .. . . ....... . . . . . .......... ... ... .... . . ................................................ ... . . .... . .. . ..... . . . ............. ................ ....
∠T OB 2π(T O) . 2π
210 Similarly, the length of ar T A is ∠T CA 2π(T C) . 2π
Sin e these ar s are equal in length
∠T CA ∠T OB 2π(T O) = 2π(T C) , 2π 2π
while in degrees this equation is ∠T OB 360◦
∠T CA 2π(T O) = 2π(T C) . ◦ 360
The radian version of this equation be omes
∠T CA 2π(28) = 2π(8) , 2π 2π x
while the degree version be omes x 360◦
Both of these give have
∠T CA 2π(28) = 2π(8) . ◦ 360
28x = 8∠T CA,
when e
∠T CA =
∠P CA = ∠T CP + ∠T CA =
9 2
. Therefore, we
7 x 2
x.
What does this tell us? When the smaller ir le has rolled twi e around the
ir umferen e of the larger ir le, we'll have x = 4π, so this tells us that the total angle travelled by B around C is 92 (4π) = 18π. Thus, the smaller ir le makes 9 omplete rotations around its entre. I nd this problem pretty fas inating be ause of its ounterintuitive nature. I hope that these di erent solutions gave you something to think about! Notes from the Mayhem Editor
Calling all proposers! Do you have a neat problem that you'd like to see in Mayhem? We would appre iate re eiving your problem proposals. They
an be sent by email to:
[email protected] or by regular mail to: Ian VanderBurgh; CEMC, University of Waterloo; 200 University Ave. W.; Waterloo, ON, Canada; N2L 3G1. Please keep in mind that we are looking for problems at the level of those that have appeared so far in Volume 34. Some of Mayhem's loyal followers have noti ed that a few favourite problems from the past have reappeared in re ent issue(s). While this may have been a
idental, we believe that this is perfe tly a
eptable in Mayhem, as its audien e is intended to be students in se ondary and post-se ondary s hools, who tend to move on after a few years.
211 THE OLYMPIAD CORNER No. 270 R.E. Woodrow
The Canadian summer break is oming up. I would like to en ourage readers to use this time to send me solutions to problems from the Corner to build up my le for future numbers of the Corner. For your problem pleasure we rst give the twenty problems of the Mathemati al Competition Balti Way 2004. My thanks go to Felix Re io, Canadian Team Leader to the IMO in Mexi o, for olle ting them for our use. MATHEMATICAL COMPETITION BALTIC WAY 2004
November 7, 2004
. Let a1 , a2 , a3 , . . . be a sequen e of non-negative real p numbers su h that for ea h n ≥ 1 both an + a2n ≥ 3n and an+1 + n ≤ 2 (n + 1)an hold.
1
(a) Prove that an ≥ n for ea h n ≥ 1.
(b) Give an example of su h a sequen e. . Let P (x) be a polynomial with non-negative oeÆ ients. Prove that if
2
P
1 x
P (x) ≥ 1
for x = 1, then the same inequality holds for ea h positive x. . Let p, q, and r be positive real numbers and let n be a positive integer. If pqr = 1, prove that 3
1 pn + q n + 1
+
1 qn + rn + 1
+
1 r n + pn + 1
≤ 1.
. Let {x1 , x2 , . . . , xn } be a set of real numbers with arithmeti mean X . Prove that there is a positive integer K su h that the arithmeti mean of ea h of the sets 4
{x1 ,x2 , . . . ,xK } , {x2 ,x3 , . . . ,xK } , . . .
, {xK−1 ,xK } , {xK }
is not greater than X . 5. For integers k and n let (k)2n+1 be the multiple of 2n + 1 losest to k. Determine the range of the fun tion f (k) = (k)3 + (2k)5 + (3k)7 − 6k.
212 . A positive integer is written on ea h of the six fa es of a ube. For ea h vertex of the ube we ompute the produ t of the numbers on the three adja ent fa es. The sum of these produ ts is 1001. What is the sum of the six numbers on the fa es? 6
7. Find all sets X onsisting of at least two positive integers su h that for every pair m,n ∈ X , where n > m, there exists k ∈ X su h that n = mk2 . 8. Let f (x) be a non- onstant polynomial with integer oeÆ ients. Prove that there is an integer n su h that f (n) has at least 2004 distin t prime fa tors.
. A set S of n − 1 natural numbers is given, where n ≥ 3. There exist at least two elements in this set whose di eren e is not divisible by n. Prove that it is possible to hoose a non-empty subset of S so that the sum of its elements is divisible by n. 9
. Is there an in nite sequen e of prime numbers p1 , p2 , p3 , . . . su h that ea h n ≥ 1?
10
|pn+1 − 2pn | = 1 for
. An m × n table is given with +1 or −1 written in ea h ell. Initially there is exa tly one −1 in the table and all the other ells ontain a +1. A move onsists of hoosing a ell ontaining −1, repla ing this −1 by a 0, and then multiplying all the numbers in the neighbouring ells by −1 (two ells are neighbouring if they share a side). For whi h (m, n) an a sequen e of su h moves always redu e the table to all zeroes, regardless of whi h ell
ontains the initial −1? 11
12. There are 2n di erent numbers in a row. By one move we an ex hange any two numbers, or y li ally permute any three numbers, that is, hoose a, b, and c and repla e b by a, c by b, and a by c. What is the minimum number of moves that suÆ es to arrange the numbers in in reasing order? 13. The 25 member states of the European Union set up a ommittee with the following rules.
(a) The ommittee shall meet every day. (b) At ea h meeting, at least one member state shall be represented. ( ) At any two di erent meetings, a di erent set of member states shall be represented. (d) The set of states represented at the nth meeting shall in lude, for every k < n, at least one state that was represented at the kth meeting. For how many days an the ommittee have its meetings?
213 . A pile of one, two, or three nuts is small, while a pile of four or more nuts is large. Two persons play a game, starting with a pile of n nuts. A player moves by taking a large pile of nuts and splitting it into two non-empty piles (either pile an be large or small). If a player annot move, he loses. For whi h values of n does the rst player have a winning strategy? 14
. A ir le is divided into 13 segments, numbered onse utively from 1 to 13. Five eas alled A, B , C , D , and E sit in the segments 1, 2, 3, 4, and 5, respe tively. A ea an jump to an empty segment ve positions away in
15
either dire tion around the ir le. Only one ea jumps at a time, and two
eas annot o
upy the same segment. After some jumps, the eas are ba k in the segments 1, 2, 3, 4, and 5, but possibly in some other order than they started in. Whi h orders are possible? . Through a point P exterior to a given ir le pass a se ant and a tangent to the ir le. The se ant interse ts the ir le at A and B , and the tangent tou hes the ir le at C on the same side of the diameter through P as A and B . The proje tion of C onto the diameter is Q. Prove that QC bise ts ∠AQB . 16
. Consider a re tangle with sides of lengths 3 and 4, and on ea h side pi k an arbitrary point that is not a orner. Let x, y, z , and u be the lengths of the sides of the quadrilateral spanned by these points. Prove that 17
25 ≤ x2 + y 2 + z 2 + u2 ≤ 50 .
. A ray emanating from the vertex A of the triangle ABC interse ts the side BC at X and the ir um ir le of ABC at Y . Prove that 18
1 1 + AX XY
≥
4 BC
.
19. In triangle ABC let D be the mid-point of BC and let M be a point on the side BC su h that ∠BAM = ∠DAC . Let L be the se ond interse tion point of the ir um ir le of triangle CAM with AB , and let K be the se ond interse tion point of the ir um ir le of triangle BAM with the side AC . Prove that KL k BC . 20. Three ir ular ar s w1 , w2 , and w3 with ommon end-points A and B are on the same side of the line AB , and w2 lies between w1 and w3 . Two rays emanating from B interse t these ar s at M1 , M2 , M3 and K1 , K2 , K3 , respe tively. Prove that
M1 M2 K1 K2 = M2 M3 K2 K3
.
214 We now give the XIX Olimpiada Iberoameri ana de Matemati as 2004. Thanks again go to Felix Re io, Canadian Team Leader to the IMO in Mexi o, for olle ting them for the Corner. XIX OLIMPIADA IBEROAMERICANA DE MATEMATICAS
Castellon, Espa~na
September 17{26, 2004
. Some ells of a 1001 × 1001 board are oloured a
ording to the following rules.
1
(a) If two ells share a ommon side, then at least one of them is oloured. (b) In every set of six onse utive ells in a row or in a olumn, there are at least two neighbouring ells that are oloured. Determine the minimum number of ells that are oloured. . Let A be a xed exterior point with respe t to a given ir le with entre and radius r. Let M be a point on the ir le and let N be diametri ally opposite to M with respe t to O. Find the lo us of the entres of the ir les passing through A, M , and N , as the point M is varied on the ir le.
2
O
. Let n and k be positive integers su h that n is odd or n and k are even. Prove there exist two integers a and b su h that gcd(a, n) = gcd(b, n) = 1 and k = a + b. 3
4. Find all pairs of positive integers (a, b) su h that a and b ea h have two digits, and su h that 100a + b and 201a + b are perfe t squares with four digits ea h.
. In a s alene triangle ABC , the interior bise tors of the angles A, B , and meet the opposite sides at points A′ , B ′ , and C ′ respe tively. Let A′′ be the interse tion of BC with the perpendi ular bise tor of AA′ , let B ′′ be the interse tion of AC with the perpendi ular bise tor of BB ′ , and let C ′′ be the interse tion of AB with the perpendi ular bise tor of CC ′ . Prove that A′′ , B ′′ , and C ′′ are ollinear. 5
C
. Given a subset H of the plane, a point P of the plane is said to be a ut point of H if there are four points A, B , C , and D in H su h that the lines AB and CD are distin t and meet at P . If H0 is a nite subset of the plane, then de ne the sets H1 , H2 , H3 , . . . re ursively by taking Hj+1 to be the union of Hj and the set of ut points of Hj , j ≥ 0. Prove that, if the union of the Hj is nite, then Hj = H1 for ea h j . 6
215 As a third group of problems we next give the Quali ation Round of the Swedish Mathemati al Contest 2004/2005. Thanks again to Felix Re io for obtaining them for our use. SWEDISH MATHEMATICAL CONTEST 2004/2005
Quali ation Round O tober 5, 2004
. The ities A, B , C , D, and E are onne ted by straight roads (more than two ities may lie on the same road). The distan e from A to B , and from C to D , is 3 km. The distan e from B to D is 1 km, from A to C it is 5 km, from D to E it is 4 km, and nally, from A to E it is 8 km. Determine the distan e from C to E . 1
2. Linda writes the four positive integers a, b, c, and d on a pie e of paper. Sin e she is amused by arithmeti , she adds the numbers in pairs, obtaining the sums a + b, a + c, . . . , c + d, but she forgets to write down one of the possible sums. The ve sums she obtains are 7, 11, 12, 18, and 23. Whi h sum did Linda forget? What are the positive integers a, b, c, d ? 3
. Determine the greatest and the least value of mn (m + n)2
,
if m and n are positive integers, ea h not greater than 2004. 4. Let k and n be integers with 1 < k < n. If a set of n real numbers has the property that the mean value of any k of them is an integer, show that all n numbers are integers. 5. At all points with integer oordinates in the plane there are verti al nails of altitude h and of negligible width. When a spheri al balloon is dropped it will burst on e it tou hes one of the nails. A balloon is doomed if it bursts no matter where it is dropped. What is the radius, R, of the smallest doomed balloon? 6. Let 2n (where n ≥ 1) points lie in the plane so that no straight line
ontains more than two of them. Paint n of the points blue and paint the other n points yellow. Show that there are n segments, ea h with one blue end-point and one yellow end-point, su h that ea h of the 2n points is an end-point of one of the n segments and none of the segments have a point in ommon.
216 The Final Round problems of the Swedish Mathemati al Contest 2004/2005
omplete the set. Thanks are due to Felix Re io for obtaining them. SWEDISH MATHEMATICAL CONTEST 2004/2005
Final Round
November 20, 2004
. Two ir les in the plane of the same radius R interse t at a right angle. How large is the area of the region whi h lies inside both ir les?
1
2. The oins in a ertain ountry have the values 1, 2, 3, 4, and 5. Nisse is buying a pair of shoes. When he is about to pay he tells the salesman that he has 100 oins, but that he does not know how many there are of ea h kind. \Wonderful, then you have even money," says the salesman. How mu h did the shoes ost, and how did the salesman know that Nisse had even money?
. The fun tion f satis es f (x) + xf (1 − x) Determine the fun tion f . 3
= x2
for all real numbers x.
. If tan v = 2v and 0 < v < π2 , then does it follow that sin v <
4
20 21
?
. A square of integer side n, where n ≥ 2, is divided into n2 squares of side 1. Next, n − 1 straight lines are drawn so that the interior of ea h of the small squares (the boundary not being in luded in the interior) is interse ted by at least one of the straight lines. (a) Give an example whi h shows that this an be a hieved for some n ≥ 2. (b) Show that among the n − 1 straight lines there are two lines whi h interse t in the interior of the square of side n. 6. Show that any onvex, n-sided polygon of area 1 ontains a quadrilateral of area at least 12 . 5
Our last group of problems is the Abel Competition 2004{2005, Norway Final. Thanks again to Felix Re io for olle ting them for the Corner. ABEL COMPETITION 2004{2005
Norway Final Mar h 10, 2005
. (a) A positive integer m is triangular if m = 1 + 2 + · · · + n for some integer n > 0. Show that m is triangular if 8m + 1 is a perfe t square. (b) The base of a pyramid is a right-angled triangle with sides of integer lengths. The height of the pyramid is also an integer. Show that the volume of the pyramid is an even integer.
1
217 . (a) There are nine small sh in a ubi aquarium with a side of 2 metres, entirely lled with water. Show that at any given√time there an be found two sh whose distan e apart is not greater than 3 metres. (b) Let A be the set of all points in spa e with integer oordinates. Show that for any nine points in A, there are at least two points among them su h that the mid-point of the segment joining the two is also a point in A. 2
3. (a) Let △ABC be isos eles with AB = AC , and let D be the mid-point of BC . The points P and Q lie respe tively on the segments AD and AB su h that P Q = P C and Q 6= B . Show that ∠P QC = 12 ∠BAC . (b) Let ABCD be a rhombus with ∠BAD = 60◦ . Let F , G, and H be points on the segments AD, CA, and DC respe tively su h that DF GH is a parallelogram. Show that △BHF is equilateral.
. (a) Let a, b, and c be positive real numbers. Prove that
4
p (a + b)(a + c) ≥ 2 abc(a + b + c) .
(b) Let a, b, and c be real numbers su h that ab + bc + ca > a + b + c > 0. Prove that a + b + c ≥ 3. I seem to have mis led a solution and omment by Miguel Amengual Covas to Problem 1 of the Thai Mathemati al Olympiad (see the last number of the Corner). He noted that the answer, ∠B = 12 ∠A = 12 70◦ = 35◦ , follows from Problem #2559 [2000 : 305℄ and its solution [2001 : 466-467℄. Also, two letters arrived from Pavlos Maragoudakis after we had nalized the April Corner. One letter in luded a solution to Problem 8 (not dis ussed in the April Corner) of the Sele ted Problems of the Thai Mathemati al Olympiad 2003, given at [2007 : 278℄. 8. Find all primes p su h that p2 + 2543 has less than 16 distin t positive divisors. Solution by Pavlos Maragoudakis, Pireas, Gree e. If p = 2 then p2 + 2543 = 2547 = 32 · 283 whi h has (2 + 1)(1 + 1) = 6 positive divisors. If p = 3 then p2 + 2543 = 2552 = 23 · 11 · 29 whi h has pre isely (3 + 1)(1 + 1)(1 + 1) = 16 positive divisors. If p > 3 then p ≡ ±1 (mod 3) so p2 + 2543 ≡ 2544 ≡ 0 (mod 3). Also p is odd, so p2 ≡ 1 (mod 8) and p2 + 2543 ≡ 0 (mod 8). Thus 3 | (p2 + 2543) and 8 | (p2 + 2543), hen e p2 + 2543 = 24x where x > 24. The numbers 1, 2, 3, 4, 6, 8, 12, 24, x, 2x, 3x, 4x, 6x, 8x, 12x, and 24x are then 16 divisors of p2 + 2543, that are distin t and positive. So p = 2 is the only prime su h that p2 + 2543 has less than 16 distin t positive divisors.
218 Next we revisit the 25th Albanian Mathemati al Olympiad, Test 1 and Test 2, given at [2007: 278{279℄. We start with solutions to Test 1. . There are 20 pupils in a village s hool. Any two of them have the same grandfather. Show that there exists a grandfather who has at least 14 grand hildren.
1
Comment by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain. Solved by Pavlos Maragoudakis, Pireas, Gree e; and Titu Zvonaru, Comane sti, Romania. We give the omment of Amengual Covas. The problem was proposed in the Tournament of Towns, Tournament 16, Junior Questions, Autumn 1994, O Level. A solution an be found in: P.J. Taylor and A.M. Storozhev, Tournament of Towns 1993{1997, Book 4, Australian Mathemati al Trust, 1998, p. 57. 2. Let M , N , and P be the respe tive mid-points of sides BC , CA, and AB of triangle ABC , and let G be the interse tion point of its medians. Prove √ 3 that if BN = 2 AB and BM GP is a y li polygon, then triangle ABC is equilateral.
Solved by Geo rey A. Kandall, Hamden, CT, USA; Pavlos Maragoudakis, Pireas, Gree e; George Tsapakidis, Agrinio, Gree e; and Titu Zvonaru, Comane sti, Romania. We give the solution of Tsapakidis. Sin e BM GP is a y li polygon we obtain A AG · AM
2 AM · AM 3 3 AM 2 = AB 2 4
AM
= AP · AB , =
1 AB · AB , 2
= BN 2 , = BN
.. ....... ...... .. .. .. .. ..... .... . . .. ... ... .. .. .. .. ... .. .. ..... ... .. .. ... . .. .. ... .. .. .. ... .. .. .... .. .. . .. ... .. . .. .. .. .. . . ... ......... ..... . . . ..... .... . .. ...... . . . . . . .. .... ... ..... .. .. . . . .... ... .... .. .. .......... .. .. .. .. .......... .. .... ..... ........ .. . . . .. . ... ... ...... .. .. . . . . ... .. ..... .... .. . .. . . . . ..... .... . ... .. . . . . . . ..... ... ... .. ........ . . ..... ... .. .. ...... . . . . ... ......... ........ . .. . ...............................................................................................................................
P
,
N
G
hen e AC = BC . The segment CP is also an altitude, that is ∠GP B = 90◦ , and sin e opposite C B M angles of a yli quadrilateral add to 180◦ , we also have ∠GM B = 90◦ . Therefore AM is also an altitude, hen e AB = AC , whi h together with AC = BC shows that triangle ABC is equilateral. 3. Let xk and yk kxk yk ≥ 1.
(a) Prove that
(for k = 1, 2, . . . , n) be positive real numbers that satisfy n P xk − yk
k=1
x2k + yk2
≤
1 √ n n 4
+ 1.
(b) When does equality hold in part (a)?
219 Solved by Mi hel Bataille, Rouen, Fran e; and Pavlos Maragoudakis, Pireas, Gree e. We give Bataille's solution. (a) First note that for k = 1, 2, . . . , n we have xk − yk x2k
+
|xk − yk |
≤
yk2
|xk − yk
|2
+ 2xk yk
where we have set tk = |xk − yk |. But
t2k
2 + ≥ 2tk k
r
2 , k
and hen e
t2k
+
√
tk t2k
tk
≤
2 + k
k . 2 2
2 k
,
It follows that
≤ √
n X √ √ xk − yk 1 √ ≤ √ ( 1 + 2 + · · · + n) . 2 2 x + yk 2 2 k=1 k
(1)
From an inequality of means,
√ √ √ !2 1 + 2 + ··· + n n
hen e
√
1+
√
Returning to (1), we obtain n X xk − yk
x2k + yk2
k=1
≤
1 + 2 + ··· + n
,
n
r √ n+1 2+ ··· + n ≤ n 2
(2)
.
r 1 n+1 1 √ ≤ √ ·n = n n + 1. 2 4 2 2 √
√
√
(b) The inequality is stri t in (2) if n ≥ 2, for then 1, 2, . . . , n are distin t. Thus, equality does not hold in (a) if n ≥ 2. If n = 1 and equality holds, then √ 2 4
x1 − y1
=
x21 + y12
≤
t1 t21 + 2x1 y1
≤
t1 t21 + 2
≤
√ 2 4
,
so that equality holds throughout. It follows that x1 y1 = 1
hen e x1 − y1 = if n = 1 and
√
2.
x1
√ 2 x1 − y1 = 4 (x1 − y1 )2 + 2
and
,
It is now easy to dedu e that equality holds if and only
√ √ 2± 6 = 2
,
y1
√ √ − 2± 6 = 2
.
220 . Find prime numbers p and q su h that p2 − p + 1 = q3 .
4
Solution by Titu Zvonaru, Comane sti, Romania. We will prove that the only solution to p2 − p + 1 = q3 where p is a prime number and q is a natural number is p = 19 and q = 7 (giving 192 − 19 + 1 = 343 = 73 ). We have p(p − 1) = (q − 1)(q 2 + q + 1) , (1) 2 thus p | (q − 1) or p | (q + q + 1). If p | (q − 1), then p < q, whi h leads to the ontradi tion p2 − p + 1 = q3 > p3 . It follows that p | (q2 + q + 1), that is, q 2 + q + 1 = kp . (2) By (1) we obtain p − 1 = k(q − 1) . (3) 2 By (2) and (3) we obtain q + q + 1 = k(kq − k + 1), or equivalently q 2 + (1 − k2 )q + (k2 − k + 1) = 0 .
Sin e q is an integer, the dis riminant ∆ = (k2 − 1)2 − 4(k2 − k + 1) of the quadrati equation is a perfe t square. Also, sin e ∆ < (k2 − 1)2 and ∆ ≡ (k2 − 1)2 (mod 2), it follows that ∆ ≤ (k2 − 3)2
⇐⇒ ⇐⇒
(k2 − 1)2 − (k2 − 3)2 ≤ 4(k2 − k + 1) 4(k2 − 2) ≤ 4(k2 − k + 1) ,
and hen e k ≤ 3. If k = 2, then ∆ = −3, whi h is not a perfe t square. However, if k = 3, then ∆ = 62 , and we obtain q = 7 and p = 19. We now turn to our readers' solutions to Test 2 of the Albanian Mathemati al Olympiad. 2. Prove the inequality q a+
1 1 b
+ q + 0.64 b+
1 1 c
+ q + 0.64 c+
1 1 a
+ 0.64
≥ 1.2 ,
where a > 0, b > 0, c > 0, and abc = 1.
Solved by Arkady Alt, San Jose, CA, USA; D. Kipp Johnson, Beaverton, OR, USA; and Pavlos Maragoudakis, Pireas, Gree e. We give Johnson's write-up. We rst make the standard substitution a = xy , b = yz , and c = xz , for suitable positive numbers x, y, and z . Then q a+
1 1 b
+ 0.64
= q
1 x y
+
z y
√ y = √ x + 0.64y + z + 0.64
,
221 and the desired inequality be omes X
y li
√ y ≥ 1.2 . √ x + 0.64y + z
This inequality is homogeneous of degree zero, so without loss of generality we let x + y + z = 1, where x, y, and z are in the interval (0, 1), and our goal is to prove that X
y li
√ √ √ X X y y 5 y = = ≥ 1.2 . √ √ √ x + y + z − 0.36y 1 − 0.36y 25 − 9y
y li
y li
Now for 0 < x < 1 we have the su
ession of equivalent inequalities √ 5 x √ 25 − 9x
25 625
≥ ≥ ≥
0 ≤
6x , 5 √ √ 6 x 25 − 9x , 36x(25 − 9x) , 625 9x2 − 25x + 36
,
whi h are all true, as the last inequality is just (3x − 25 )2 ≥ 0. Adding three 6 su h inequalities we have √ 5 x
√ √ 5 y 5 z + √ + √ √ 25 − 9y 25 − 9x 25 − 9z 6y 6z 6(x + y + z) 6x + + = = 1.2 , ≥ 5 5 5 5
and our goal is a hieved. This ompletes the proof. 3
. Solve the following equation in integers: y 2 = 1 + x + x2 + x3 + x4 .
Comments by Mi hel Bataille, Rouen, Fran e; and Titu Zvonaru, Comane sti, Romania. Zvonaru notes that this was problem M242. A solution appears at [2007 : 141℄. Now we give Bataille's omment. This is an old problem set by T.H. Gronwall and solved by A.A. Bennett in Amer. Math. Monthly, Vol 33, No. 5, May 1926, pp. 291{2. The solver showed that the solutions are the following six pairs (x, y): (−1, ±1) , (0, ±1) , (3, ±11) .
. Prove that for any integer n ≥ 2, the number 2n − 1 is not divisible by n.
4
222 Solution by Mi hel Bataille, Rouen, Fran e. Suppose on the ontrary that n divides 2n − 1. Sin e 2n − 1 is odd, n and all its prime fa tors are odd as well. Let p be one of these prime fa tors. There exist positive integers k, l su h that n = kp and 2n = 1 + klp. Re alling that 2p ≡ 2 (mod p) (by Fermat's Little Theorem), we obtain 2k ≡ 1 (mod p)
.
(1)
Now, 2 belongs to the multipli ative group Z/pZ − {0} and the order of 2 in this group is an integer m > 1 whi h divides the order p − 1 of the group and, from (1), also divides k. Thus, hoosing a prime divisor q of m, we get a prime divisor of n satisfying q < p. Repeating with q what we have just done with p and ontinuing this way, we would onstru t an in nite sequen e of prime divisors of n, whi h is learly impossible. This ontradi tion ompletes the proof. . In an a ute-angled triangle ABC , Let H be the ortho enter, and let and dc be the distan es from H to the sides BC , CA, and AB , respe tively. Prove that da + db + dc ≤ 3r, where r is the radius of the in ir le of triangle ABC . 5
da , db ,
Solved by Arkady Alt, San Jose, CA, USA; Miguel Amengual Covas, Cala Figuera, Mallor a, Spain; Pavlos Maragoudakis, Pireas, Gree e; and Titu Zvonaru, Comane sti, Romania. We give the solution of Amengual Covas. Let a, b, and c be the lengths of the sides opposite A, B , and C , respe tively, and let s = 12 (a + b + c) be the semiperimeter. Let R be the
ir umradius of triangle ABC , and let O and A′ be the ir um entre and the mid-point of the side BC , respe tively. Letting ha be the length of the altitude from A to BC in △ABC , and using the well-known relation AH = 2 · OA′ , we dedu e that da = ha − AH = ha − 2 · OA′ = ha − 2R cos A .
By symmetry, we have follows that
db = hb − 2R cos B
and
dc = hc − 2R cos C .
It
da + db + dc = ha + hb + hc − 2R(cos A + cos B + cos C) .
We have the relation ha = 2rs and its symmetri analogues, and also a we have the well-known relations abc ab + bc + ca cos A + cos B + cos C
= 4Rrs , = s2 + r 2 + 4Rr , R+r = . R
223 Using these we dedu e that da + db + dc =
s2 + r 2 − 4Rr 2R
,
and hen e the given inequality is equivalent to s2 + r 2 − 4R2 ≤ 6Rr .
(1)
To establish the validity of (1), we make use of the linear transformation and c = x + y, where x, y, and z are uniquely determined positive numbers. Ater applying the transformation, (1) be omes
a = y + z , b = z + x, (x + y + z)2
xyz (x + y)2 (y + z)2 (z + x)2 − x+y+z 4xyz(x + y + z)
+
≤ 6·
(x + y)(y + z)(z + x)
,
4(x + y + z)
or equivalently 4xyz(x + y + z)3
+ 4x2 y 2 z 2 − (x + y)2 (y + z)2 (z + x)2 ≤ 6xyz(x + y)(y + z)(z + x) .
We substitute x3 + y3 + z 3 + 3(x + y)(y + z)(z + x) for (x + y + z)3 , simplify, and obtain 4xyz(x3 + y 3 + z 3 ) +
We substitute 2xyz +
P
6xyz(x + y)(y + z)(z + x) + 4x2 y 2 z 2 ≤ [(x + y)(y + z)(z + x)]2 . x2 y
sym
for (x + y)(y + z)(z + x) (the symmetri
sum is over the six permutations of x, y, and z ) to obtain 3
3
3
4xyz(x + y + z ) + 6xyz
≤
2xyz +
X sym
!2
2xyz +
2
x y
sym
!2
X
x2 y =
X
=
2
x y
sym
sym
We have the expansion X
X
sym
4 2
x y
!
x2 y
!2
!
+ 4x2 y 2 z 2 X
+ 4xyz
x2 y
sym
3
3
!
3
+ 4x2 y 2 z 2 .
+ 2xyz(x + y + z ) + 2xyz
X sym
+ 2(x3 y 3 + y 3 z 3 + z 3 x3 ) + 6x2 y 2 z 2 ,
2
x y
!
whi h we substitute into the above and then simplify to obtain 3
3
3
2 2 2
2xyz(x + y + z ) + 6x y z
≤
X sym
4 2
x y
!
+ 2(x3 y 3 + y 3 z 3 + z 3 x3 ) .
224 However, this is equivalent to 2 (xy)3 + (yz)3 + (zx)3 − 3x2 y 2 z 2
+ x4 (y − z)2 + y 4 (z − x)2 + z 4 (x − y)2 ≥ 0 ,
whi h is true, sin e (xy)3 + (yz)3 + (zx)3 ≥ 3x2 y2 z 2 holds by the AM-GM Inequality. Equality o
urs only if x = y = z , that is, if and only if the triangle ABC is equilateral.
Next we turn to solutions from our readers to some problems of the 44th Ukrainian Mathemati al Olympiad, 11th Form, Final Round, given at [2007 : 279{280℄. . (V.M. Leifura) Solve the equation
1
arcsin⌊sin x⌋ = arccos⌊cos x⌋ ,
where ⌊a⌋ is the greatest integer not ex eeding a. Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON. For onvenien e, we will write sin−1 x for arcsin x. Let θ = sin−1 ⌊sin x⌋ = cos−1 ⌊cos x⌋. From the de nitions of the inverse trigonometri fun tions, we have − π2 ≤ θ ≤ π2 and 0 ≤ θ ≤ π. Hen e, 0 ≤ θ ≤ π2 . We onsider four ases. . If −1 ≤ sin x < 0, then θ = sin−1 (−1) = − π2 , a ontradi tion.
Case 1
. If 0 < sin x < 1, then θ = sin−1 (0) = 0, hen e ⌊cos x⌋ = cos θ = 1 and cos x = 1. Thus sin2 x + cos2 x > 1, a ontradi tion. Case 3. If sin x = 0, then cos x = 1 as in Case 2 above, whi h implies that x = 2kπ , k ∈ Z. Conversely, if x = 2kπ , k ∈ Z, then sin−1 ⌊sin x⌋ = 0 and also cos−1 ⌊cos x⌋ = 0. −1 Case 4. If sin x = 1, then θ = sin (1) = π2 ; hen e ⌊cos x⌋ = cos θ = π cos 2 = 0 and 0 ≤ cos x < 1. Evidently cos x = 0; otherwise we have sin2 x + cos2 x > 1, a ontradi tion. Solving sin x = 1 and cos x = 0 we nd that x = π2 + 2kπ, k ∈ Z. Conversely, if x = π2 + 2kπ, k ∈ Z, then sin−1 ⌊sin x⌋ = sin−1 (1) = π2 and cos−1 ⌊cos x⌋ = cos−1 (0) = π2 . Therefore, the solution set of the given equation is
Case 2
S=
2kπ ,
π + 2kπ : k ∈ Z 2
.
225 . (V.V. Lymanskiy) The a ute-angled triangle ABC is given. Let O be the entre of its ir um ir le. The perpendi ular bise tor of the side AC interse ts the side AB and the line BC at the points P and Q, respe tively. Prove that ∠P QB = ∠P BO. 2
Solved by Miguel Amengual Covas, Cala Figuera, Mallor a, Spain; Geo rey A. Kandall, Hamden, CT, USA; Pavlos Maragoudakis, Pireas, Gree e; D.J. Smeenk, Zaltbommel, the Netherlands; and Titu Zvonaru, Comane sti, Romania. We give Maragoudakis' solution. In triangle BOA we have OA = OB . Thus ∠P BO
= = = = =
180◦ − ∠BOA
= 2 d − AK d 180◦ − BK
\ 180◦ − BKA 2
2 d d 180 − BK − CK ◦
2 d − CK d [ − BK KM 2 d [ CM − BK 2
= ∠P QB .
.. ... .... ....... . ... ... . ...... ... .. .. ... .. .. ................................ . . . .............. . ........ ......... ........... ........ ... ........ ........ ... .......... .... ...... . .......... ... ....... . ..... . . . . . .... . ... ...... ... . . . . . . .... ... ..... . .. . . . . . ... . . . . . .. ... .... .... ... .. . . . . . . .. ... .... ... .. .. . . . . . ... ... .. .... .. .. . . . .... .. .. ... . . .. . . . . . . ... .... .. .. .. .. . . . ... ... . .... ...... .. . . .... ... . .... ..... .. ... . . . .. . .... ....... ... .. . .. . . .... ... .. .... .............. ... . . ... ........ . .. .. . . . . . . . . . ........ ... ... . .. .. . . . . . . . . .. . . . ....... ... .. ... .. ........ ... .... ... .. ........ .... .... ............ .... .. . ...... ..................................................................................................................................................... .... . . . . .... ... ..... .... .... ..... ... .... ....... ..... .... ........ ....... . . . . . . .... .......... . ................ ... ........................ ...................... ... ...
Q
K
B
P
O
A
C
M
. (O.O. Malakhov) Find the sum of the real roots of the equation
6
x x+ √ = 2004 . 2 x −1
Solved by Mi hel Bataille, Rouen, Fran e; Pavlos Maragoudakis, Pireas, Gree e; and Vedula N. Murty, Dover, PA, USA. We give Bataille's solution. The required sum is 2004. Let a = 2004. It is readily seen that the real roots of the given equation are those of P (x) = 0 satisfying 0 < x < a, where P (x) = (x2 − 1)(a − x)2 − x2 = x4 − 2ax3 + (a2 − 2)x2 + 2ax − a2 .
The key observation is that the polynomial P (a − x) is just the same as P (x) (this is easily he ked). Hen e, if x0 is a solution to P (x) = 0, the same is true of a − x0 . Now, P (0) < 0, P (a/2) > 0, P (a) < 0, hen e P (x) has at least two real roots in (0, a). If in addition P (x) had two positive or omplex onjugate roots, the produ t of the roots would be positive. However, the produ t of the roots is −a2 < 0, therefore P (x) must have a negative root, say x1 . Then a − x1 is also a root of P (x) and a − x1 > a. It follows that the four roots of P (x) are real, with exa tly two of them in (0, a), of the form x2 , a − x2 for some x2 ∈ (0, a). These real numbers are the roots of the given equation and their sum is a = 2004, as announ ed.
226 . (V.M. Rad henko) Does there exist f x2 y + f (x + y 2 ) = x3 + y 3 + f (xy) 7
a fun tion f : R for all x, y ∈ R ?
→ R
su h that
Solved by Mi hel Bataille, Rouen, Fran e; and Pavlos Maragoudakis, Pireas, Gree e. We give the solution of Bataille. There is no su h fun tion. Assume on the ontrary that f : R → R satis es the given identity. Then, taking x = 0 in the identity, we have f f (y 2 ) = y 3 + f (0) ,
and taking y = 0 we have
f f (x) = x3 + f (0) . However, with x = 1 the latter gives f f (1) = 1+f (0), while with y = −1 the former yields f f (1) = −1 + f (0). Thus, we have the ontradi tion 1 + f (0) = −1 + f (0), and the result follows.
. (V.A. Yasinskiy) Let a, b, and c be positive real numbers su h that Prove that a3 + b3 + c3 ≥ ab + bc + ca.
8
abc ≥ 1.
Solved by Arkady Alt, San Jose, CA, USA; Mi hel Bataille, Rouen, Fran e; Pavlos Maragoudakis, Pireas, Gree e; Vedula N. Murty, Dover, PA, USA; George Tsapakidis, Agrinio, Gree e; Panos E. Tsaoussoglou, Athens, Gree e; and Titu Zvonaru, Comane sti, Romania. We give Murty's write-up. We have the two known inequalities a2 + b2 + c2 a3 + b3 + c3
≥ ab + bc + ca , (a + b + c)(a2 + b2 + c2 ) ≥ 3
,
from whi h we obtain a3 + b3 + c3 ≥
(a + b + c)(ab + bc + ca) 3
.
From the AM-GM Inequality and the given ondition, abc ≥ 1, we have a + b + c ≥ 3(abc)1/3 ≥ 3 .
The desired inequality now follows by ombining the last two inequalities. . (I.P. Nagel) Let ω be the ins ribed ir le of the triangle ABC . Let L, and E be the points of tangen y of ω with the sides AB , BC , and CA, respe tively. Lines LE and BC interse t at the point H , and lines LN and AC interse t at the point J (all the points H , J , N , E lie on the same side of the line AB ). Let O and P be the mid-points of the segments EJ and N H , respe tively. Find S(HJ N E) if S(ABOP ) = u2 and S(COP ) = v 2 . (Here S(F ) is the area of gure F ). 10
N,
227 Solution by Titu Zvonaru, Comane sti, Romania. As usual we write a = BC , b = CA, c = AB , and s = 21 (a + b + c). It is well known that AL = AE = s − a, BL = BN = s − b, and CN = CE = s − c. Assume that c > b. In order that the points H , J , N , and E all lie on the same side of the line AB , we must have c > a. By Menelaus' Theorem applied to △ABC and transversal HEL, we have HC HB
·
LB LA
·
EA EC
HC
= 1 ⇒
HC + a
and by symmetry we have J C NH
=
b(s − c) . c−a
=
s−c
=
(s − c)(c − b + a)
S(HJ N E) = = CP
=
=
and by symmetry, CO =
B
H
J
a(s − c) c−b
,
=
a(s − c)
c−b 2(s − b)(s − c)
2(s − a)(s − c) . c−a
c−b
,
Using the expressions we
N H · J E · sin C
2 2(s − a)(s − b)(s − c)2 (c − b)(c − a)
= NP − NC = =
E C q q P N O
Therefore we dedu e
N C + CH = (s − c) +
and by symmetry we have EJ have obtained so far, we have
L
⇒ HC =
s−b
=
c−b
A .... ....... ...... .. ...... .... ...... . . . . . . . ... ... ...... ... ...... ...... .. . . . . . . . . .. .. ............... ...... .... ............... .. ........ .... ...... .. .. ........ .. ...... .. ...... . . . . . . . . . . . . . .. ..... .. ...... ... .............. .. ...... ........ ... ... ...... ....... .. ...... ................................................................................................................................................... .. .. .. . . .. . . .. .. .. . .. ... .. . .. .. .. .. .... ...
· sin C ,
1 NH − NC 2
(s − b)(s − c) − (s − c) c−b (s − c)(s − b − c + b) (s − c)2 = c−b c−b (s − c)2 . c−a
(1)
,
It follows that
S(COP ) = v 2 =
(s − c)4 sin C
2(c − b)(c − a)
.
(2)
228 Continuing with our al ulations, AO
(s − c)2 c−a 2 −c(2s − c − b) + s − ab s2 − ab − ac = c−a c−a 2 2 2 s − 2as + a (s − a)2 s − a(b + c) = = c−a c−a c−a
= AC + CO = b + = =
and by symmetry BP
=
(s − b)2 , c−b
,
hen e
S(ABOP ) = u2 =
(s − a)2 (s − b)2 sin C · (c − a)(c − b) 2
.
(3)
Combining (1), (2), and (3) we obtain the desired result S(HJ N E) =
2(s − a)(s − b)(s − c)2 s
= 4
(c − b)(c − a)
· sin C
(s − a)2 (s − b)2 · sin C · 2(c − b)(c − a)
s
(s − c)4 sin C = 4uv . 2(c − b)(c − a)
That ompletes the Corner for this issue. Send me your ni e solutions and generalizations.
229 BOOK REVIEW John Grant M Loughlin
Problems of the Week By Jim Totten, Canadian Mathemati al So iety, 2007 ISBN 0-919558-16-X, oilbound, 60 pages, $12 Reviewed by John Grant M Loughlin, University of New Brunswi k, Frederi ton, NB The author needs no introdu tion to the CRUX readership. Jim Totten
ompletes his tenure as Editor with this issue of the journal. As the Book Reviews Editor, I am taking the opportunity to a knowledge a re ently published work of Jim Totten { another way of thanking Jim for his ontributions to problem solving within CRUX and beyond. The publisher, the Canadian Mathemati al So iety, will surely support me taking su h editorial liberty within the review. Jim Totten taught at Thompson Rivers University (formerly known as the University College of the Cariboo) from 1979 to 2007. Ea h week of the a ademi semesters featured a di erent posted problem with minimal repetition of individual problems over the years. This publi ation features a sele tion of 80 su h problems used prior to Fall 1986. In fa t, some were rst used by Jim at Saint Mary's University from 1976-1979. The problems take a range of forms spanning an array of topi s as noted in the Table of Contents. A helpful "Index of Problems by subje t matter"
an guide solvers toward topi s of interest. The subje ts (and subheadings) are: Algebra (Equations, Fun tions, Word Problems, Other), Analysis, Combinatori s, Geometry (Triangles, Cir les, Other), Logi , Numbers, Probability, and Re reation. Many problems are ross-referen ed to two subje ts. Problems of the Week is Volume VII in the series ATOM (A Taste of Mathemati s). Publi ations in the series are intended for keen high s hool students, mathemati s tea hers, and problem solvers who wish to engage with a
essible material. The ATOM materials lend themselves to independent study or ollaborative proje ts. This book is no ex eption as it grows out of ollaboration with students though the in lusion of detailed solutions allows one to learn mu h math independently from problems that may appear beyond one's grasp. I am reminded of my undergraduate studies at the University of Waterloo, parti ularly in two separate ourses, ommonly referred to as the "100 Problem Courses." Many of the problems introdu ed to me by Dean Ho man or Ross Honsberger appear in this book. The introdu tion lari es that this is not oin idental, as Jim writes in the dedi ation: "For showing me that not only was it a
eptable to be ex ited and enthusiasti about mathemati al problem-solving, but that it was to be strongly en ouraged, I am dedi ating this book to Ross Honsberger." A few problems are stated here to pique the interest of problem solvers.
230
3. In a ertain lassroom, there are 5 rows with 5 seats per row arranged
in a square. Ea h student is to hange her seat by going either to the seat immediately in front or behind her, or immediately to the left or right. (Of
ourse, not all possibilities are open to all students.) Determine whether this
an be done, beginning with a full lass of students. Try to generalize to a re tangular array and nd the onditions under whi h su h a hange of seats an be managed. 16. The three sides and height of a triangle are four onse utive integers. What is the area of the triangle? 48. The Ramada County Department of Highways has just resurfa ed the
ounty roads, and now the yellow stripe down the middle of the road must be repainted. The tru k used for this purpose is very ineÆ ient as far as gas
onsumption is on erned, and thus the Department would like to have the tru k travel the shortest distan e possible. A road 40 map of the ounty is shown (with distan es giv- W. Midville t 8 tMidville tE. Midville en in kilometres). The 21
ounty tru k is garaged in 34 30 Midville, and it must re28 37 turn there when the job is done. How many kilomet 5 t tres must it travel and S.W. Midville S.E. Midville what route should it take? 61. Given any two points A and B on the ir umferen e of a ir le, and E the mid-point of the ar AB (note that there are really two ar s that ould be
alled AB ; it does not matter whi h one we E
hoose as long as the rest of the dis ussion is assumed to pertain only to points and ar s lying on P the ar AB that we hose). Let P be any point A N on the ar EB and onstru t EN perpendi ular to AP with N on the hord AP . Prove that AN = N P + P B (we are dealing only with magB nitudes of line segments here). This book is an ex ellent addition to a library, a departmental o ee room, or the math tea hing oÆ e in a s hool. Those who like problem solving will enjoy this ompa t resour e, whether a high s hool student aspiring to learn advan ed math or a seasoned mathemati ian who likes a hallenge. As the Book Reviews Editor through Jim's entire editorial tenure, I wish to express my gratitude for the privilege of working with Jim in this apa ity. .................. ..................... .................. .... ... ..... ...... ......... ... .... ..... ............ ... ... .... . . . .... .. . . .... . ..................................................... ........ . . . . . .............. .. ............ . . . . . . . . . ............ . .... . . .... ..... ............................................................ ..... ... . . .. .. ... . ... ... . . .. .. ... ... .. .... .. .. .. .. ........ . . . . . . .. . .... . . . . . .. . ... .. ... .... ... .. .. ... ... ... ... ..... ... .. .. .. ... . .... . .. ... ... ... .. .. ... ... .. ... .. .. ... .... .. ... .... . . . . ..... . . . . . ..... ......... ........ .......................................................................
.......... ..... .... ..........
.............................................. .......... ........ ........ ......... ..... ... ..... .... ... .... . . . ... ... ... . . ... ... . . .. ... . .. .. ... . . ............................................................................................................................ .... ... .. ..... .. . ... .. ... . . . ... .. .. ... .. .. ... .. . .. ... ... .. ... . ... .. .. .... .. ... .... .. .... .... ....... ...... ...... ........ . . . . . . . ............ .. ......................................
Addendum. The review was sent to Jim less than two weeks before his
sudden passing. As noted on the inside front over, Jim is a Co-Editor of this issue. Indeed the loss to the mathemati al ommunity is great. The spirit of Jim Totten is remembered fondly in this review.
231
The Proof of Three Open Inequalities Vasile C^ rtoaje
The inequalities below were onje tured in Crux Mathemati orum [2℄ and on Mathlinks Forum [5℄ (N.B. inequalities (1) and (2) are in orre tly stated on p. 106 of [2℄): Conje ture. Let x1 , x2 , . . . , xn be nonnegative real numbers whi h satisfy x1 + x2 + · · · + xn ≥ n. If p > 1, then X
+ x2 + · · · + xn
xp1
y li
+ x2 + · · · + xn
1
xp1
y li X X
x1
xp1
y li
xp1 − x1 + x2 + · · · + xn
≤ 1,
(1)
≤ 1,
(2)
≥ 0,
(3)
where ea h sum is over the n y li permutations of the variables xi . Under the more restri tive ondition x1 x2 · · · xn ≥ 1, these inequalities had already been stated and proved in [1℄, [2℄, and [4℄, respe tively. On the other hand, the inequality (3) under the ondition x1 x2 · · · xn ≥ 1 generalizes the following inequality from the IMO (let n = 3 and p = 52 ):
(IMO{2005). Let x, y, z be three positive reals su h that xyz ≥ 1. Prove that Problem 3
x5 − x2
x5 + y 2 + z 2
+
y5 − y2
x2 + y 5 + z 2
+
z5 − z2
x2 + y 2 + z 5
≥ 0.
In this paper, we prove ea h of these three inequalities by means of the LCF-Theorem, as it is stated in [1℄ and [3℄: Theorem 1 (LCF-Theorem) Let f (u) be a fun tion on an interval I , whi h is
on ave for u ≤ s, s ∈ I . If
(n − 1)f (x) + f (y) ≤ nf (s)
for all x, y ∈ I su h that x ≤ y and (n − 1)x + y = ns, then f (x1 ) + f (x2 ) + · · · + f (xn ) ≤ nf
c 2008
x1 + x2 + · · · + xn n
x1 + x2 + · · · + xn ≤ s. n
for all x1 , x2 , . . . , xn ∈ I whi h satisfy Copyright
Canadian Mathemati al So iety
Crux Mathemati orum with Mathemati al Mayhem, Volume 34, Issue 4
232
1
Proof of Inequality (1)
Having in view the expression on the left-hand side of (1), it is easy to see that we need only onsider the ase x1 + x2 + · · · + xn = n. Then, we have to show that 1 xp1
− x1 + n
+
1 xp2
− x2 + n
+ ··· +
1 xpn
− xn + n
(4)
≤ 1
for x1 + x2 + · · · + xn = n. By Lemma 1 below, it suÆ es to prove (4) for all 1 xi ≥ a = p 1−p whi h satisfy x1 + x2 + · · · + xn = n. To do this, we apply the LCF-Theorem to the fun tion f (u) = p 1 de ned on I = [a, ∞), u −u+n
for the ase s = 1. First, we must show that f is on ave on [a, 1]. The se ond derivative is f ′′ (u) =
If
p(p + 1)u2p−2 − p(p + 3)up−1 − np(p − 1)up−2 + 2 (up − u + n)3
.
g(u) = p(p + 1)u2p−2 − p(p + 3)up−1 + 2 < 0
for a ≤ u ≤ 1, then f ′′ (u) < 0, and hen e f (u) is on ave on [a, 1]. Indeed, setting t = pup−1 implies 1 ≤ t ≤ p, and hen e pg(u)
= =
(p + 1)t2 − p(p + 3)t + 2p (p + 1)(t − 1)(t − p) + (1 − p)(t + p) < 0 .
By the LCF-Theorem, it suÆ es to show that xp
n−1
−x+n
+
1 yp
−y+n
≤ 1
for any a ≤ x ≤ 1 ≤ y and (n − 1)x + y = n. Sin e this inequality is trivial for x = y = 1, assume next that x < 1 < y, and write the desired inequality in su
ession as follows yp − y + n
≥
yp − y
≥
yp − y y−1
Let h(y) =
yp − y . y−1
≥
xp − x + n xp − x + 1
,
(n − 1)(x − xp ) xp − x + 1 x − xp
,
(1 − x)(xp − x + 1)
(5) .
By the weighted AM-GM Inequality, we have
h′ (y) =
(p − 1)y p + 1 − py p−1 (y − 1)2
> 0.
233 Therefore, h(y) is stri tly in reasing, and sin e y−1 = (n−1)(1−x) ≥ 1−x, that is, y ≥ 2 − x, we have h(y) ≥ h(2 − x) =
Then, it suÆ es to prove that
(2 − x)p − 2 + x 1−x
(2 − x)p − 2 + x ≥
x − xp
xp − x + 1
.
.
This inequality is equivalent to ea h of the following (2 − x)p + x − 1 ≥ (1 + t)p − t
≥
(1 − t2 )p + t(1 + t)p
≥
1 , −x+1 1 , (1 − t)p + t xp
1 + t2 + t(1 − t)p ,
where t = 1 − x and 0 < t ≤ 1. By Bernoulli's Inequality, we have (1 − t2 )p + t(1 + t)p ≥ 1 − pt2 + t(1 + pt) = 1 + t .
Thus, it suÆ es to show that t(1 − t) ≥ t(1 − t)p . However, this is learly true, and the proof is ompleted. Equality in (1) o
urs if and only if we have x1 = x2 = · · · = xn = 1. Lemma 1. Let p > 1, and let x1 , x2 , . . . , xn be nonnegative real numbers. If (4) holds for all xi ≥ p su h that x1 + x2 + · · · + xn = n, then it holds for all xi ≥ 0 su h that x1 + x2 + · · · + xn = n. Proof: Let a = p , and let f (x) = xp − 1x + n . From 1 1−p
1 1−p
f ′ (x) =
1 − pxp−1 (xp − x + n)2
,
it follows that f ′ (a) = 0, f ′ (x) > 0 for 0 ≤ x < a, and f ′ (x) < 0 for x > a. Thus, f (x) is stri tly in reasing on [0, a] and stri tly de reasing on [a, ∞). We have to prove that f (x1 ) + f (x2 ) + · · · + f (xn ) ≤ 1 for all xi ≥ 0 su h that x1 + x2 + · · · + xn = n. Without loss of generality, assume that 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn . If x1 ≥ a, then the on lusion follows immediately by hypothesis. Otherwise, sin e a < 1, there exists an integer k ∈ {1, 2, . . . , n − 1} su h that 0 ≤ x1 ≤ · · · ≤ xk < a ≤ xk+1 ≤ · · · ≤ xn .
Let yi = a for i = 1, 2, . . . , k, and let yi = xi for i = k + 1, . . . , n. We have yi ≥ a and yi ≥ xi for i = 1, 2, . . . , n; and f (yi ) > f (xi ) for ea h i = 1, 2, . . . , k, hen e y1 + y2 + · · · + yn > x1 + x2 + · · · + xn = n
234 and f (y1 ) + f (y2 ) + · · · + f (yn ) > f (x1 ) + f (x2 ) + · · · + f (xn ) .
Therefore, it suÆ es to show that f (y1 ) + f (y2 ) + · · · + f (yn ) ≤ 1 for all yi ≥ a with y1 + y2 + · · · + yn > n. By hypothesis, this inequality is true for all yi ≥ a with y1 + y2 + · · · + yn = n. Sin e f is de reasing on [a, ∞), the inequality is also true for all yi ≥ a su h that y1 + y2 + · · · + yn > n. . If 0 ≤ p < 1 and x1 + x2 + · · · + xn ≤ n, then the inequality in (1) reverses. We an prove this using the AM-HM Inequality together with Jensen's Inequality: Remark 1
X
1
xp1
y li
+ x2 + · · · + xn
P
≥
y li
=
n P
i=1
≥
2
n2 (xp1
n
+ x2 + · · · + xn ) n2
xpi + (n − 1)
1 n
n P
i=1
xi
p
n P
xi
i=1
n2
+ (n − 1)
n P
xi
≥ 1.
i=1
Proof of Inequality (2)
Without loss of generality we an assume xi > 0 for ea h i. Furthermore, we need only onsider the ase x1 + x2 + · · · + xn = n. Indeed, if we set x + x2 + · · · + xn r = 1 and yi = xri for i = 1, 2, . . . , n (where r ≥ 1 and n y1 + y2 + · · · + yn = n), then (2) be omes X
r p−1 y1p
y li
y1 ≤ 1, + y2 + · · · + yn
and it suÆ es to prove this for r = 1; that is, to prove that x1 xp1
− x1 + n
+
x2 xp2
− x2 + n
+ ··· +
xn xpn
for x1 + x2 + · · · + xn = n. We onsider two ases. Case 1. 1 < p ≤ n + 1. By Bernoulli's Inequality,
− xn + n
p xp1 = 1 + (x1 − 1) ≥ 1 + p(x1 − 1) ,
and hen e xp1 − x1 + n suÆ es to show that
n X
i=1
≥ n − p + 1 + (p − 1)x1 ≥ 0. xi
n − p + 1 + (p − 1)xi
≤ 1.
≤1
(6)
Consequently, it
235 For p = n + 1, this inequality is an equality. Otherwise, for 1 < p < n + 1, we use (p − 1)xi
n − p + 1 + (p − 1)xi
= 1 −
to rewrite the inequality as n X
i=1
n−p+1
n − p + 1 + (p − 1)xi
1 ≥ 1. n − p + 1 + (p − 1)xi
We then prove it by means of the AM-HM Inequality: n X
i=1
1 n − p + 1 + (p − 1)xi
≥ P n
n2
i=1
n − p + 1 + (p − 1)xi
= 1.
. p > n + 1. By Lemma 2 below, it suÆ es to prove (6) for all 1 n ) p su h that x1 + x2 + · · · + xn = n. We will apply the xi ≥ a = (
Case 2
p−1
LCF-Theorem to the fun tion f (u) = up −uu + n de ned on I = [a, ∞), for the ase s = 1. First, we must show that f is on ave on [a, 1]. The se ond derivative has the expression ′′
f (u) =
−p(p − 1)up−1 (up − u + n) + 2(pup−1 − 1) (p − 1)up − n (up − u + n)3
.
Sin e (p − 1)up − n ≥ (p − 1)ap − n = 0 for u ≥ a, if
g(u) = −p(p − 1)up−1 (up − u + n) + 2pup−1 (p − 1)up − n < 0
for a ≤ u ≤ 1, then f ′′ (u) < 0, and hen e f (u) is on ave on [a, 1]. Indeed, we have g(u) pup−1
= ≤
(p − 1)(up + u) − n(p + 1)
2(p − 1) − 2(p + 1) = −4 < 0 .
By the LCF-Theorem, it suÆ es to show that
(n − 1)x y + p ≤ 1 p x −x+n y −y+n
for any a ≤ x ≤ 1 ≤ y and (n−1)x+y = n. For x = y = 1, equality o
urs. Assume now that x < 1 < y, and write the desired inequality su
essively as follows y(xp − x + n) yp − y + n ≥ , (7) p x − nx + n
236 yp − y ≥
(n − 1)x(x − xp ) xp − nx + n
Inequality (5) states that yp − y ≥ show that
(n − 1)(x − xp ) . xp − x + 1
1 x ≥ p xp − x + 1 x − nx + n
.
(8)
Therefore, it suÆ es to .
This inequality is true, be ause x < 1 and xp −nx+n > xp −x+1. Equality in (2) o
urs if and only if x1 = x2 = · · · = xn = 1. Lemma 2. Let p > n+1, and let x1 , x2 , . . . , xn be nonnegative real numbers. n ) su h that x1 + x2 + · · · + xn = n, then it If (6) holds for all xi ≥ ( p−1 holds for all xi ≥ 0 su h that x1 + x2 + · · · + xn = n. n x Proof: Let a = ( p − . From ) , and let f (x) = p 1 x −x+n 1 p
1 p
f ′ (x) =
n − (p − 1)xp (xp − x + n)2
,
it follows that f ′ (a) = 0, f ′ (x) > 0 for 0 ≤ x < a, and f ′ (x) < 0 for x > a. Consequently, f (x) is stri tly in reasing on [0, a] and stri tly de reasing on [a, ∞). We have to prove that f (x1 )+f (x2 )+· · ·+f (xn ) ≤ 1 for all xi ≥ 0 su h that x1 + x2 + · · · + xn = n. Without loss of generality, assume that 0 ≤ x1 ≤ x2 ≤ . . . ≤ xn . If x1 ≥ a, then the on lusion follows by hypothesis. Otherwise, sin e a < 1, there exists an integer k ∈ {1, 2, . . . , n − 1} su h that 0 ≤ x1 ≤ · · · ≤ xk < a ≤ xk+1 ≤ · · · ≤ xn .
Let yi = a for i = 1, 2, . . . , k, and yi = xi for i = k + 1, k + 2, . . . , n. We have yi ≥ a and yi ≥ xi for ea h i; and f (yi ) > f (xi ) for i = 1, 2, . . . , k, hen e y1 + y2 + · · · + yn > x1 + x2 + · · · + xn = n
and f (y1 ) + f (y2 ) + · · · + f (yn ) > f (x1 ) + f (x2 ) + · · · + f (xn ) .
Therefore, it suÆ es to show that f (y1 ) + f (y2 ) + · · · + f (yn ) ≤ 1 for all yi ≥ a su h that y1 +y2 +· · ·+yn > n. By hypothesis, this inequality is true for all yi ≥ a su h that y1 +y2 +· · ·+yn = n. Sin e f is de reasing on [a, ∞), the inequality is also true for all yi ≥ a su h that y1 + y2 + · · · + yn > n. . If 0 ≤ p < 1 and x1 + x2 + · · · + xn ≥ n, then the inequality in (2) reverses. By the Cau hy-S hwarz Inequality, we have
Remark 2
237
X
x1
y li
xp1 + x2 + · · · + xn
y li
=
n P
i=1
=
2
x1 (xp1 + x2 + · · · + xn )
xp+1 i
+
n P
xi
i=1
n P
i=1 n P
i=1
1 +
n P
i=1 n P
xi
i=1
P
≥
n P
n P
xp+1 i
2
xi
2
x2i −
+
− n P
i=1
n P
i=1
xi
n P
i=1
x2i
xp+1 i 2
−
n P
i=1
, x2i
and it suÆ es to show that x2i − xp+1 ≥ 0. This inequality follows by i i=1 i=1 the Power Mean Inequality: n 1 X 2 x n i=1 i
n 1 X p+1 x n i=1 i
≥
n 1 X p+1 x n i=1 i
≥
3
2 ! p+1
!
n 1 X xi n i=1
!1−p
n 1 X p+1 ≥ x . n i=1 i
Proof of Inequality (3)
Inequality (3) is equivalent to X
y li
1 n ≤ xp1 + x2 + · · · + xn x1 + x2 + · · · + xn
.
For x1 + x2 + · · · + xn = n, this inequality immediately follows from (1). Otherwise, if x1 + x2 + · · · + xn > n, then we set r = x1 + x2 +n · · · + xn , and yi = xri for i = 1, 2, . . . , n, so that r > 1 and y1 + y2 + · · · + yn = n. Then the inequality be omes X
r p−1 y1p
y li
1 ≤ 1. + y2 + · · · + yn
This inequality follows from (1), sin e X
r p−1 y1p
y li
1 + y2 + · · · + yn
≤
X
yp
y li 1
1 + y2 + · · · + yn
Equality in (3) o
urs if and only if x1 = x2 = · · · = xn = 1.
≤ 1.
238
4
New Conjectures
Readers an try to extend the inequalities (1) and (2) in di erent dire tions. One of these dire tions is to nd a ondition as weak as possible on the positive real number p su h that the inequality below holds for any nonnegative real numbers x1 , x2 , . . . , xn with x1 + x2 + · · · + xn ≥ n: X
y li
x21 ≤ 1. xp1 + x2 + · · · + xn
For n ≥ 2, we onje ture that Moreover, we laim that p ≥ 1 +
p ≥ n
is su h a suÆ ient ondition.
ln 2 ln n − ln(n − 1)
is a ne essary and suÆ ient ondition for the inequality to be true. On the other hand, the study of the reverse inequality for n ≥ 2 and x1 + x2 + · · · + xn ≥ n, is also an interesting problem for readers. We
onje ture that the reverse inequality is true for 0 ≤ p ≤ 59 . Referen es
[1℄ V. C^rtoaje, Algebrai Inequalities - Old and New Methods, GIL Publishing House, Romania, 2006. [2℄ V. C^rtoaje, On an Inequality from IMO 2005, Crux Mathemati orum with Mathemati al Mayhem 32 (2006) 101{106. [3℄ V. C^rtoaje, A generalization of Jensen's Inequality, Gazeta Matemati a A, Issue 2 (2005) 124{138. [4℄ N. Nikolov, A Generalization of an Inequality from IMO 2005 http://arxiv.org/PS\_cache/math/pdf/0512/0512172.pdf
[5℄ Mathlinks Website,
http://www.mathlinks.ro/Forum/viewtopic.php?p=288276 and http://www.mathlinks.ro/Forum/viewtopic.php?t=44480\&start=40
Vasile C^rtoaje Department of Automation and Computers University of Ploiesti 39 Bu uresti Boulevard Ploiesti, Romania
[email protected]
239 PROBLEMS
Toutes solutions aux problemes dans e numero doivent nous parvenir au plus tard le 1er novembre 2008. Une e toile (⋆) apres le numero indique que le probleme a e te soumis sans solution. Chaque probleme sera publie dans les deux langues oÆ ielles du Canada (anglais et fran ais). Dans les numeros 1, 3, 5 et 7, l'anglais pre edera le fran ais, et dans les numeros 2, 4, 6 et 8, le fran ais pre edera l'anglais. Dans la se tion des solutions, le probleme sera publie dans la langue de la prin ipale solution presentee. La reda tion souhaite remer ier Jean-Mar Terrier, de l'Universite de Montreal, d'avoir traduit les problemes.
Oliver Geupel nous a signale une erreur dans l'enon e du probleme #3282 [2007 : 429, 431℄, propose par Jose Luis Daz-Barrero et Pantelimon George Popes u. Nous nous ex usons de ette erreur.
. Corre tion. Propose par Jose Luis Daz-Barrero, Universite Polyte hnique de Catalogne, Bar elone, Espagne et Pantelimon George Popes u, Bu arest, Roumanie. Soit A(z) = z n + an−1 z n−1 + · · · + a1 z + a0 un polynome ^ unitaire a oeÆ ients omplexes. On suppose que a1 = −a0 et que les zeros z1 , z2 , . . . , zn de A(z) sont des nombres omplexes non nuls et distin ts. Montrer que 3282
n n X ez k Y
k=1
zk2
j=1 j6=k
1 zk − zj
= 0.
. Propose par Toshio Seimiya, Kawasaki, Japon. Un quadrilatere ir ulaire onvexe ABCD possede un er le ins rit de
entre I . Soit P le point d'interse tion de AC et BD. Montrer que l'on a AP : CP = AI 2 : CI 2 . 3338
. Propose par Toshio Seimiya, Kawasaki, Japon. Soit Γ1 et Γ2 deux er les sans points ommuns et situes a l'exterieur l'un de l'autre. Soit ℓ1 et ℓ2 les tangentes externes ommunes a Γ1 et Γ2 . Soit A et B les points d'interse tion de ℓ1 ave Γ1 et Γ2 ; C et D eux de ℓ2 ave Γ1 et Γ2 . Notons M et N les points milieu AB et CD, et soit P et Q les points d'interse tion de N A et N B ave Γ1 et Γ2 , di erents de A et B . Montrer que CP , DQ et M N sont on ourants. 3339
. Propose par Toshio Seimiya, Kawasaki, Japon. Dans un triangle ABC , la bisse tri e de l'angle BAC oupe le er le
ir ons rit en un se ond point D. Supposons que AB 2 + AC 2 = 2AD2 . Montrer que AD et BC se oupent a 45◦ . 3340
240 . Propose par Arkady Alt, San Jose, CA, E-U. Soit ABC un triangle quel onque, de ot ^ es a, b et c; montrer que √ 3(Ra + Rb + Rc ) ≤ a + b + c, ou Ra , Rb et Rc sont les distan es respe tives du entre du er le ins rit du triangle ABC aux sommets A, B et C . 3341
. Propose par Arkady Alt, San Jose, CA, E-U. Soit respe tivement r et R le rayon des er les ins rit et ir ons rits du triangle ABC . Montrer que
3342
2
X
sin
y lique
A 2
sin
B 2
≤ 1+
r R
.
. Propose par Stan Wagon, Ma alester College, St. Paul, MN, E-U. Si l'on supprime toutes les fa torielles dans la serie de Ma laurin de sin x, on obtient la serie de arctan x. Supposons alors qu'on en supprime seulement une sur deux. La serie ainsi obtenue a-t-elle une forme fermee? C'est-a-dire, peut-on trouver la fon tion dont la serie de Ma laurin soit 3343
x−
x3 x5 x7 x9 x11 + − + − + ··· 3 5! 7 9! 11
?
. Propose par Pham Kim Hung, etudiant, Universite de Stanford, Palo Alto, CA, E-U. Soit n un entier positif, n ≥ 4, et soit a1 , a2 , . . . , an des nombres reels positifs tels que a1 + a2 + · · · + an = n. Montrer que 3344
1 1 3 2 1 + + ··· + −n ≥ a1 + a22 + · · · + a2n − n . a1 a2 an n
. Propose par Pham Kim Hung, etudiant, Universite de Stanford, Palo Alto, CA, E-U. Soit a, b, c et d des nombres reels positifs tels que a + b + c + d = 4. Montrer que 3345
a b c d + + + ≥ 2. 2 2 2 1+b c 1+c d 1+d a 1 + a2 b
. Propose par Bin Zhao, etudiant, YunYuan HuaZhong Universite de Te hnologie et S ien e, Wuhan, Hubei, Chine. Dans un triangle ABC , montrer que 3346
π
X
y lique
1 A
≥
X
y lique
sin
A 2
X
y lique
csc
A 2
.
241 . Propose par Mihaly Ben ze, Brasov, Roumanie. Soit A1 A2 A3 A4 un quadrilatere onvexe. Soit Bi un point sur Ai Ai+1 pour i ∈ {1, 2, 3, 4}, les indi es etant pris modulo 4, de sorte que 3347
B1 A1 B1 A2
=
B3 A4 B3 A3
Montrer que B1 B3 latere ir ulaire.
=
A1 A4 A2 A3
⊥ B2 B4
et
B2 A2 B2 A3
=
B4 A1 B4 A4
=
A1 A2 A3 A4
.
si et seulement si A1 A2 A3 A4 est un quadri-
. Propose par Mihaly Ben ze, Brasov, Roumanie. Dans un triangle a utangle ABC , soit respe tivement A1 , B1 et C1 des points sur les ot ^ es BC , CA et AB , de sorte que ∠AC1 B1 , ∠BC1 A1 et ∠ACB soient egaux. Soit respe tivement M , N et P les entres des er les
ir ons rits des triangles AC1 B1 , BA1 C1 et CB1 A1 . Montrer que AM , BN et CP sont on ourants si et seulement si AA1 , BB1 et CC1 sont les hauteurs du triangle ABC .
3348
. Propose par Mihaly Ben ze, Brasov, Roumanie. Soit a, b et c trois nombres reels positifs. Montrer que
3349
6
X a(1 + bc)(a2 + 1) ≥ max 2 a +1 a3 + 1
Y a3 + 1
y lique
y lique
,
X ab(1 + c)(a2 b2 + 1) . a 3 b3 + 1
y lique
. Propose par Panos E. Tsaoussoglou, Athenes, Gre e. Soit x, y et z trois nombres reels positifs tels que x + y Montrer que yz zx xy 1 + + ≤ . 3350
1+x
1+y
1+z
+ z = 1.
4
................................................................. . Corre tion. Proposed by Jose Luis Daz-Barrero, Universitat Polite ni a de Catalunya, Bar elona, Spain and Pantelimon George Popes u, Bu harest, Romania. Let A(z) = z n + an−1 z n−1 + · · · + a1 z + a0 be a moni polynomial with omplex oeÆ ients. Suppose that a1 = −a0 , and that the zeroes z1 , z2 , . . . , zn of A(z) are distin t, non-zero omplex numbers. Prove that
3282
n n X ez k Y 1 = 0. 2 z z − zj k j=1 k k=1 j6=k
242 . Proposed by Toshio Seimiya, Kawasaki, Japan. A onvex y li quadrilateral ABCD has an in ir le with entre I . Let be the interse tion of AC and BD. Prove that AP : CP = AI 2 : CI 2 .
3338
P
. Proposed by Toshio Seimiya, Kawasaki, Japan. Let Γ1 and Γ2 be two non-interse ting ir les ea h lying in the exterior of the other. Let ℓ1 and ℓ2 be the ommon external tangents to Γ1 and Γ2 . Let ℓ1 meet Γ1 and Γ2 at A and B , respe tively, and let ℓ2 meet Γ1 and Γ2 at C and D, respe tively. Let M and N be the mid-points of AB and CD, respe tively, and let P and Q be the interse tions of N A and N B with Γ1 and Γ2 , respe tively, di erent from A and B . Prove that CP , DQ, and M N are on urrent. 3339
. Proposed by Toshio Seimiya, Kawasaki, Japan. The bise tor of ∠BAC interse ts the ir um ir le of △ABC at a se ond point D. Suppose that AB 2 + AC 2 = 2AD2 . Prove that the angle of interse tion of AD and BC is 45◦ . 3340
. Proposed by Arkady Alt, San Jose, CA, USA. For any triangle ABC with sides of lengths a, b, and c, prove that √ 3(Ra + Rb + Rc ) ≤ a + b + c, where Ra , Rb , and Rc are the distan es from the in entre of △ABC to the verti es A, B , and C , respe tively. 3341
. Proposed by Arkady Alt, San Jose, CA, USA. Let r and R be the inradius and ir umradius of △ABC , respe tively. Prove that X A B r 2 sin sin ≤ 1+ .
3342
2
y li
2
R
. Proposed by Stan Wagon, Ma alester College, St. Paul, MN, USA. If the fa torials are deleted in the Ma laurin series for sin x, one obtains the series for arctan x. Suppose instead that one alternates fa torials in the series. Does the resulting series have a losed form? That is, an one nd an elementary expression for the fun tion whose Ma laurin series is 3343
x−
x3 3
+
x5 5!
−
x7 7
+
x9 9!
−
x11 11
+ ···
?
. Proposed by Hung Pham Kim, student, Stanford University, Palo Alto, CA, USA. Let n be a positive integer, n ≥ 4, and let a1 , a2 , . . . , an be positive real numbers su h that a1 + a2 + · · · + an = n. Prove that 3344
1
a1
+
1
a2
+ ··· +
1
an
−n ≥
3
n
a21 + a22 + · · · + a2n − n .
243 . Proposed by Hung Pham Kim, student, Stanford University, Palo Alto, CA, USA. Let a, b, c, and d be positive real numbers su h that a + b + c + d = 4. Prove that 3345
a b c d + + + ≥ 2. 2 2 2 1+b c 1+c d 1+d a 1 + a2 b
. Proposed by Bin Zhao, student, YunYuan HuaZhong University of Te hnology and S ien e, Wuhan, Hubei, China. Given triangle ABC , prove that 3346
π
X 1
y li
A
≥
X
sin
y li
A 2
X
csc
y li
A 2
.
. Proposed by Mihaly Ben ze, Brasov, Romania. Let A1 A2 A3 A4 be a onvex quadrilateral. Let Bi be a point on Ai Ai+1 for i ∈ {1, 2, 3, 4}, where the subs ripts are taken modulo 4, su h that
3347
B1 A1 B1 A2
=
B3 A4 B3 A3
=
A1 A4 A2 A3
and
B2 A2 B2 A3
=
B4 A1 B4 A4
=
A1 A2 A3 A4
.
Prove that B1 B3 ⊥ B2 B4 if and only if A1 A2 A3 A4 is a y li quadrilateral. . Proposed by Mihaly Ben ze, Brasov, Romania. Let ABC be an a ute-angled triangle. Let A1 , B1 , and C1 be points on the sides BC , CA, and AB , respe tively, su h that the angles ∠AC1 B1 , ∠BC1 A1 , and ∠ACB are all equal. Let M , N , and P be the ir um entres of △AC1 B1 , △BA1 C1 , and △CB1 A1 , respe tively. Prove that AM , BN , and CP are on urrent if and only if AA1 , BB1 , and CC1 are altitudes of △ABC .
3348
. Proposed by Mihaly Ben ze, Brasov, Romania. Let a, b, and c be positive real numbers. Show that
3349
6
Y a3 + 1
y li
a2 + 1
y li
a(1 + bc)(a2 + 1) a3 + 1
,
X ab(1 + c)(a2 b2 + 1) a 3 b3 + 1
.
y li
. Proposed by Panos E. Tsaoussoglou, Athens, Gree e. Let x, y, and z be positive real numbers su h that x + y + z = 1. Prove
3350
that
≥ max
X
yz zx xy 1 + + ≤ 1+x 1+y 1+z 4
.
244 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. . [2005 : 328℄ Proposed by Vasile C^rtoaje, University of Ploiesti, Romania. Let k and n be positive integers with k < n, and let a1 , a2 , . . . , an be real numbers su h that a1 ≤ a2 ≤ · · · ≤ an . Prove that KLAMKIN{05
(a1 + a2 + · · · + an )2 ≥ n(a1 ak+1 + a2 ak+2 + · · · + an an+k )
(where the subs ripts are taken modulo n) in the following ases: (a)
n = 2k;
(b)
n = 4k;
( )⋆
2<
n < 4. k
Solution to part ( )⋆ by the proposer. Solutions to parts (a) and (b) appeared in [2006 : 315℄. Let nj =k 4k − j , where j is an integer su h that 1 ≤ j ≤ 2k − 1, and let i = 2j . Then 0 ≤ i ≤ k − 1. Let a be a real number su h that a2k−i ≤ a ≤ a2k−i+1 , and let S1 = a1 + a2 + · · · + an and S2 = a1 ak+1 + a2 ak+2 + · · · + an an+k . We have to show that S12 ≥ S2 . 4k − j
We onsider two ases: 2k < n ≤ 3k and 3k ≤ n < 4k. k Case 1. Let 2k < n ≤ 3k . Then k ≤ j ≤ 2k − 1, and hen e ≤ i ≤ k − 1. 2 Let
S =
3k−i X
m=k−i+1
am
and
P =
2k−i X
m=k−i+1
am ak+m
.
Applying the original inequality in part (b) to the non-de reasing sequen e of 4k real numbers a1 , . . . , a2k−i , a , . . . , a, a , . . . , an , we obtain | {z } 2k−i+1 j times
(S1 + ja)2 4k
2
≥ (j − k)a
+
+ Sa +
k−i X
am ak+m
m=1 n X
m=2k−i+1
am ak+m
,
!
245 or equivalently, (S1 + ja)2 4k
≥ (j − k)a2 + Sa + S2 − P
.
The inequality to be proved follows by adding this inequality to S12 4k − j
(S1 + ja)2
+ (j − k)a2 + Sa − P ≥
4k
.
The last inequality is true, be ause S12 (S1 + ja)2 j[S1 − (4k − j)a]2 + ja2 − = ≥ 0 4k − j 4k 4k(4k − j)
and 2k−i X
−ka2 + Sa − P =
(a − am )(ak+m − a) ≥ 0 .
m=k−i+1
. Let 3k ≤ n < 4k. Then 1 ≤ j ≤ k, and hen e 0 ≤ i ≤
Case 2
S
=
k−i+j X
m=k−i+1
P
=
2k−i X
m=k−i+1
am +
am ak+m
3k−i X
, and
2
. Let
,
am
m=3k−i−j+1
k
Q =
2k−i X
m=k−i+j+1
am ak−j+m
.
Applying the original inequality in part (b) to the non-de reasing sequen e of 4k real numbers a1 , . . . , a2k−i , a , . . . , a, a , . . . , an , we obtain | {z } 2k−i+1 j times
(S1 + ja)2 4k
≥
k−i X
am ak+m
m=1
+
!
2k−i X
m=k−i+j+1
+
n X
m=2k−i+1
or equivalently,
+ a
k−i+j X
m=k−i+1
am
am ak−j+m + a
am ak+m
,
(S1 + ja)2 ≥ S2 − P + Q + Sa . 4k
3k−i X
m=3k−i−j+1
am
246 The required inequality follows by adding this inequality to S12 4k − j
− P + Q + Sa ≥
(S1 + ja)2 4k
,
whi h is true, be ause
S12 (S1 + ja)2 j[S1 − (4k − j)a]2 − + ja2 = ≥ 0 4k − j 4k 4k(4k − j)
and
2
−ja − P + Q + Sa =
2k−i−j X
(aj+m − am )(ak+m − a)
m=k−i+1
+
2k−i X
m=2k−i−j+1
(a − am )(ak+m − a) ≥ 0 .
There were no other solutions submitted.
. [2007 : 237, 239℄ Proposed by Mihaly Ben ze, Brasov, Romania.
3239
Let n be a positive integer. If α = 1 + 12(n1+ 1) , prove that e <
(n + 1)2n+1 (n!)2
1 ! 2n
< eα .
Solution by Kee-Wai Lau, Hong Kong, China, modi ed by the editor. Taking logarithms, the given inequalities are su
essively equivalent to 1 < 2n <
1 (2n + 1) ln(n + 1) − 2 ln(n!) < α , 2n
(2n + 1) ln(n + 1) − 2 ln(n!) < 2n +
n 6(n + 1)
.
Let f (n) = (2n + 1) ln(n + 1) − 2 ln(n!) − 2n, and g(n) = f (n) − 6(nn+ 1) . We are to show that f (n) > 0 and g(n) < 0. For 0 < x < 1, it is well 2 3 4 known that ln(1 + x) > x − x2 + x3 − x4 . Hen e, f (n + 1) − f (n) = (2n + 3) ln(n + 2) − 2 ln (n + 1)! − 2(n + 1) − (2n + 1) ln(n + 1) + 2 ln(n!) + 2n n+2 1 = (2n + 3) ln − 2 = (2n + 3) ln 1 + −2 n+1 n+1 1 1 1 1 > (2n + 3) − + − −2 2 3 4 =
An .
n+1
2(n + 1)
3(n + 1)
4(n + 1)
247 By tedious but straightforward omputations, we nd that An =
2n2 + 2n − 3 > 0. 12(n + 1)4
Hen e, f (n) > f (1) = 3 ln 2 − 2 > 0. Similarly, for 0 < x < 1, it is well known that ln(1 + x) < x −
x2 x3 x4 x5 + − + 2 3 4 5
.
Hen e, g(n + 1) − g(n)
n+1
n
= f (n + 1) − f (n) − + 6(n + 2) 6(n + 1) 1 1 = (2n + 3) ln 1 + −2− <
n+1 6(n + 1)(n + 2) 2n + 3 1 An + − 5(n + 1)5 6(n + 1)(n + 2)
= Bn .
Again, by tedious but straightforward omputations, we nd that Bn =
` ´ − n(n − 19) − 32 −n2 + 19n + 32 = < 0 60(n + 1)5 (n + 2) 60(n + 1)5 (n + 2)
for n ≥ 21. On the other hand, using a al ulator, we an readily he k that g(n) < 0 for n = 1, 2, . . . , 21. It follows that g(n) < 0 for ea h positive integer n, and our proof is omplete. Also solved by MICHEL BATAILLE, Rouen, Fran e; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; and the proposer. A partial solution was submitted by SALEM student, Sarajevo College, Sarajevo, Bosnia and Herzegovina. MALIKIC,
. [2007 : 237, 239℄ Proposed by Mihaly Ben ze, Brasov, Romania. Let n be a positive integer. Prove that
3240
√
n+
q q √ √ n+23n+1 = 4n + 4 3 n + 2
,
where ⌊x⌋ denotes the integer part of x.
Solution by John Hawkins and David R. Stone, Georgia Southern University, Statesboro, GA, USA. The asserted equality is false in general. For example, when n = 45, we have
√ n+
q √ 3 n + 2 n + 1 = ⌊13.996 · · · ⌋ = 13 ,
248 while
q √ 4n + 4 3 n + 2 = ⌊14.008 · · · ⌋ = 14 .
It also fails for n = 95 and 616.
Hawkins andpStone also gave the following It is straightforward to show that p omment: √ √ √ the fun tion n + n + 2 3 n + 1 is less than 4n + 4 3 n + 2, and the gap between them narrows as n in reases. But sometimes the values \straddle" an integer, as shown above, so the desired equality fails. We suspe t that happens in nitely often (the next instan e is 1249). student, Sarajevo College, Sarajevo, Bosnia and Herzegovina, also SALEM MALIKIC, gave the ounterexample n = 45. There were three in orre t \proofs" for the assertion.
. [2007 : 237, 239℄ Proposed by Virgil Ni ula, Bu harest, Romania. Let a, b, c be any real numbers su h that a2 + b2 + c2 = 9. Prove that
3241
3 · min{a, b, c} ≤ 1 + abc .
Solution by Arkady Alt, San Jose, CA, USA, modi ed by the editor. Without loss of generality, we may assume a ≤ b ≤ c. We want to prove the inequality abc + 1 ≥ 3a. If a ≤ 0, then using the inequality bc ≤ 12 (b2 + c2 ), we obtain abc + 1 − 3a ≥ =
a|bc| − 3a + 1 ≥ 1 a(9 2
1 a(b2 2
− a2 ) − 3a + 1 =
+ c2 ) − 3a + 1 1 (a 2
+ 1)2 (2 − a) ≥ 0 ,
with equality if and only if a = −1 and b = c = 2. Now, let√a > 0. Sin e a ≤ b ≤ c, we get 9 = a2 + b2 + c2 ≥ 3a2 ; when e, a ≤ 3. We have the obvious inequality (c2 − a2 )(b2 − a2 ) ≥ 0, whi h yields p p bc ≥ a c2 + b2 − a2 = a 9 − 2a2 . Hen e, it suÆ es to prove the (stronger) inequality p a a 9 − 2a2 + 1 ≥ 3a . √ √ √ If 0 < a ≤ 1, then 9 − 2a2 ≥ 7 > 94 ; thus, 4a2 9 − 2a2 > 9a2 . Using the inequality (t + 1)2 ≥ 4t, we obtain p p 2 a2 9 − 2a2 + 1 ≥ 4a2 9 − 2a2 > 9a2 ,
whi h yields
√
p a a 9 − 2a2 + 1 ≥ 3a .
If 1 < a ≤ 3, then we an prove an equivalent form of our inequality, namely, the one obtained by squaring both sides of p a a 9 − 2a2 ≥ 3a − 1 ;
249 that is, We have
2a6 − 9a4 + 9a2 − 6a + 1 ≤ 0 .
2a6 − 9a4 + 9a2 − 6a + 1 < = <
2a6 − 9a4 + 9a2 − 5 (2a2 − 1)(a2 − 1)(a2 − 3) − (a2 + 2) 0,
whi h ompletes the proof. Also solved by MOHAMMED AASSILA, Strasbourg, Fran e; ROY BARBARA, Lebanese University, Fanar, Lebanon; JOHN HAWKINS and DAVID R. STONE, Georgia Southern University, Statesboro, GA, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, student, Sarajevo College, Austria; KEE-WAI LAU, Hong Kong, China; SALEM MALIKIC, Sarajevo, Bosnia and Herzegovina; ALEX REMOROV, student, William Lyon Ma kenzie Collegiate Institute, Toronto, ON; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer.
. [2007 : 237, 239℄ Proposed by G. Tsintsifas, Thessaloniki, Gree e. Let ABC be an isos eles triangle with AB = AC , and let P be an interior point. Let the lines BP and CP interse t the opposite sides at the points D and E , respe tively. Find the lo us of P if 3243
P D + DC = P E + EB .
Solution by Titu Zvonaru, Comane sti, Romania. The lo us onsists of the points inside the triangle on the altitude from A. Note that sin e the triangle is isos eles, the foot, M , of the altitude is the mid-point of BC , and the triangle is symmetri about AM . Without loss of generality, label the gure so that BE ≥ CD and let E ′ be the point between A and C for whi h CE ′ = BE , and let BE ′ and CE interse t at N . Be ause △ABC is isos eles, BCE ′ E is an isos eles trapezoid that is symmetri about AM ; when e, N belongs to AM and N E = N E ′ . As a onsequen e, the following statements are equivalent: P D + DC PD
= = =
A ..... ......... ... .. .. .. ... .. .. ..... ..... . ... ... .... ... ... .... .. ... ... .. ′ .. . . ..................................................... ... . . ... ...... .... ... .... .. .. ..... ... ...... ... ...... .. .... ... .. ... ..... . . .. .. . . . . .... ...................... ... . . .... .......... .. . . . . . . . . . .... ..... .. .... .... ................... ..... .... .... ... ... .... ... ............ ... .. .............. ......................................................................................................................................
E
.... E ..... ........N .....P......D . ... ..... ... .. C B M
P E + EB = P N + N E + E ′ C P N + N E ′ + E ′ D + DC , P N + N E′ + E′D .
The last equality says that the quadrilateral P DE ′ N is degenerate; that is, D oin ides with E ′ and P with N , so that P lies on AM . The lo us of P is therefore the open line segment AM , as laimed.
250 Also solved by ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHEL Y, Big Rapids, MI, USA; SALEM MALIKIC, BATAILLE, Rouen, Fran e; V ACLAV KONECN student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; ALEX REMOROV, student, William Lyon Ma kenzie Collegiate Institute, Toronto, ON; EDMUND SWYLAN, Riga, Latvia; PETER Y. WOO, Biola University, La Mirada, CA, USA; and the proposer.
. [2007 : 238, 239℄ Proposed by G. Tsintsifas, Thessaloniki, Gree e. Let ABC be an isos eles triangle with AB = AC , and let P be an interior point. Let the lines BP and CP interse t the opposite sides at the points D and E , respe tively. Find the lo us of P if 3244
BD + DC = BE + EC .
Composite of similar solutions by Va lav Kone ny, Big Rapids, MI, USA; and Peter Y. Woo, Biola University, La Mirada, CA, USA. As with problem 3243, the lo us of P is the portion of the altitude AM inside △ABC . The ondition BD + DC = BE + EC means that points D and E lie on one ellipse of the family of ellipses with fo i B and C , entre M , and axes along the lines BC and AM . Sin e the ellipse is symmetri about AM and interse ts ea h of the line segments AB and AC in exa tly one point, BD must interse t CE at P on AM . Conversely, if P lies on AM , then the diagram is symmetri about AM , implying that BD = CE and DC = EB , so that BD + DC = BE + EC . Also solved by ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHEL student, Sarajevo College, Sarajevo, Bosnia and BATAILLE, Rouen, Fran e; SALEM MALIKIC, Herzegovina; ALEX REMOROV, student, William Lyon Ma kenzie Collegiate Institute, Toronto, ON; JOEL SCHLOSBERG, Bayside, NY, USA; D.J. SMEENK, Zaltbommel, the Netherlands; EDMUND SWYLAN, Riga, Latvia; TITU ZVONARU, Comane sti, Romania; and the proposer. Some of the submitted solutions were designed to solve simultaneously this problem and problem 3243 above. For example, a straightforward modi ation of the featured solution to 3243 will likewise provide a solution to this problem.
3245. [2007 : 238, 240℄ Proposed by Paul Yiu, Florida Atlanti University, Bo a Raton, FL, USA. Suppose that the entre of the nine-point ir le of a triangle lies on the in ir le of the triangle. Show that its antipodal point is the Feuerba h Point; that is, the point where the nine-point ir le and the in ir le are tangent to ea h other. I. Composite of similar solutions by Salem Maliki
, student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; the Skidmore College Problem Solving Group, Skidmore College, Saratoga Springs, NY, USA; and Peter Y. Woo, Biola University, La Mirada, CA, USA.
251 The result holds for tangent ir les in general: the line joining the entres of two tangent ir les passes through the point of tangen y. If, moreover, the entre of one lies on the ir umferen e of the other, the entre of the rst
ir le is diametri ally opposed in the se ond ir le to the ommon tangen y point. II. Composite of similar solutions by Mi hel Bataille, Rouen, Fran e; John G. Heuver, Grande Prairie, AB; Walther Janous, Ursulinengymnasium, Innsbru k, Austria; and the proposer. Let I be the in entre, let N be the nine-point entre, and let r and R be the inradius and ir umradius. Many standard referen es prove that the in ir le is internally tangent to the nine-point ir le, whose radius is 12 R; hen e, N I = 12 R − r. Moreover, if N is on the in ir le, we also have N I = r and, therefore, 12 R = 2r . The antipodal point of N on the in ir le, say N ′ , satis es N N ′ = 2r, so that N N ′ = 12 R. Thus, not only is N ′ on the in ir le, but it is also on the nine-point ir le. As a result, N ′ oin ides with the unique ommon point to these two ir les, namely, the Feuerba h Point.
There were no other solutions submitted.
. [2007 : 238, 240℄ Proposed by Marian Tetiva, B^rlad, Romania. Let a, b, c, d be any positive real numbers with d = min{a, b, c, d}. Prove that
3246
a4 + b4 + c4 + d4 − 4abcd ≥ 4d (a − d)3 + (b − d)3 + (c − d)3 − 3(a − d)(b − d)(c − d) .
Solution by Kee-Wai Lau, Hong Kong, China. Let x = a − d, y = b − d, and z = c − d. Thus, x z ≥ 0. It an then be he ked that
≥ 0, y ≥ 0,
and
a4 + b4 + c4 + d4 − 4abcd − 4d (a − d)3 + (b − d)3 + (c − d)3 − 3(a − d)(b − d)(c − d) = 2d2 x2 + y 2 + z 2 + (x − y)2 + (y − z)2 + (z − x)2 + 8xyzd + x4 + y 4 + z 4
≥ 0,
and the inequality is proved. Also solved by ARKADY ALT, San Jose, CA, USA; MICHEL BATAILLE, Rouen, Fran e; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; JOE HOWARD, Portales, NM, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; SALEM student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; ALEX REMOROV, MALIKIC,
252 student, William Lyon Ma kenzie Collegiate Institute, Toronto, ON; PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU, Comane sti, Romania; and the proposer. Janous pointed out that the inequality has the equivalent form a4 + b4 + c4 + d4 − 4abcd ≥ 4d(a + b + c − 3d)(a2 − ab − ac + b2 − bc + c2 ) .
Sin e the three fa tors on the right side are non-negative, this gives a lever proof for the fourvariable AM{GM Inequality. Maliki ommented that on the right side of the inequality the fa tor 3 an be hanged to 1, resulting in a stronger inequality, namely x4 + y4 + z4 + 2(d2 x2 + d2 y2 + d2 z2 ) + 2 (dx − dy)2 + (dy − dz)2 + (dz − dx)2 ≥ 0 .
Alt also provided the proof for the inequality
a + b + c4 + d4 −4abcd ≥ 4d (a−d)3 + (b−d)3 + (c−d)3 −kd(a−d)(b−d)(c−d) , √ √ where k is in the interval 1 − 3/ 2, 1 + 3/ 2 . 4
4
. [2007 : 238, 240℄ Proposed by Jose Luis Daz-Barrero, Universitat Polite ni a de Catalunya, Bar elona, Spain. Let a1 , a2 , . . . , an be real numbers, ea h greater than 1. Prove that 3247
n X
k=1
where an+1 = a1 .
2 ≥ 4n , 1 + logak (ak+1 )
Solution by Dionne Bailey, Elsie Campbell, and Charles R. Diminnie, Angelo State University, San Angelo, TX, USA. By the AM{GM Inequality and the hange of base formula, we have n X
k=1
2 1 + logak (ak+1 )
≥ =
n q 2 X 2 logak (ak+1 )
k=1
4
n X
k=1
=
logak (ak+1 ) ≥ 4n
n Y ln ak+1 4n ln ak k=1
!n1
n Y
k=1
!n1
logak (ak+1 )
= 4n .
For equality to hold, we must have loga (ak+1 ) = 1 or ak = ak+1 for all k = 1, 2, . . . , n. Conversely, if a1 = a2 = · · · = an , then learly equality holds. k
Also solved by ARKADY ALT, San Jose, CA, USA; ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, Fran e; MIHALY BENCZE, Brasov, Romania; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; JOE HOWARD, Portales, NM, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; KEE-WAI student, Sarajevo College, Sarajevo, Bosnia and LAU, Hong Kong, China; SALEM MALIKIC, Herzegovina; ALEX REMOROV, student, William Lyon Ma kenzie Collegiate Institute, Toronto,
253 ON; HENRY RICARDO, Medgar Evers College (CUNY ), Brooklyn, NY, USA; XAVIER ROS, student, Universitat Polite ni a de Catalunya, Bar elona, Spain; PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU, Comane sti, Romania; and the proposer. Barbara pointed out that the inequality is a spe ial ase of the more general result that n n Q P xk = 1, then (1 + xk )2 ≥ 4n. Ben ze if x1 , x2 , . . . , xn are positive reals su h that k=1
k=1
obtained the generalization that if ak > 1 for k = 1, 2, . . . , n, and if P (x) is any polynomial with positive oeÆ ients, then
n P
P loga k (ak+1 ) ≥ nP (1). The urrent problem is the
k=1 + x)2 .
spe ial ase when P (x) = (1 same as the one featured above.
The proofs of these two generalizations are virtually the
. [2007 : 238, 240℄ Proposed by Titu Zvonaru, Comane sti, Romania, and Bogdan Ionita , Bu harest, Romania. If a, b, and c are positive real numbers, prove that 3248
a2 (b + c − a) b+c
+
b2 (c + a − b) c+a
+
c2 (a + b − c)
≤
a+b
ab + bc + ca 2
.
Solution by Mohammed Aassila, Strasbourg, Fran e. The proof is as follows X a2 (b + c − a) b+c
y li
=
≤ =
X
a a(b + c − a) b + c
y li X a a + (b + c − a) 2
y li
b+c
X a(b + c)
y li
4
2
=
X bc
y li
2
.
University Also solved by ARKADY ALT, San Jose, CA, USA; SEFKET ARSLANAGIC, of Sarajevo, Sarajevo, Bosnia and Herzegovina; MICHEL BATAILLE, Rouen, Fran e; JOE HOWARD, Portales, NM, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, student, Sarajevo College, Austria; KEE-WAI LAU, Hong Kong, China; SALEM MALIKIC, Sarajevo, Bosnia and Herzegovina; JOSE H. NIETO, Universidad del Zulia, Mara aibo, Venezuela; PHI THAI THUAN, student, Tran Hung Dao High S hool, Phan Thiet, Vietnam; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; ALEX REMOROV, student, William Lyon Ma kenzie Collegiate Institute, Toronto, ON; XAVIER ROS, student, Universitat Polite ni a de Catalunya, Bar elona, Spain; PANOS E. TSAOUSSOGLOU, Athens, Gree e; APOSTOLIS VERGOS, student, University of Patras, Patras, Gree e; and PETER Y. WOO, Biola University, La Mirada, CA, USA. Janous provided a proof of the more general inequality a2n (b + c − a) b2n (c + a − b) c2n (a + b − c) (ab)n + (bc)n + (ca)n + + ≤ b+c c+a a+b 2 for any non-negative integer n. At the same time he would like to leave the following onje ture
to the readers of CRUX with MAYHEM:
a+b−c b+c−a c+ a−b (an + bn + cn )(abc)n ≤ (ab)2n + (bc)2n + (ca)2n 2 a+b b+c c+a for any positive integer n.
254 . [2007 : 238, 240℄ Proposed by Titu Zvonaru, Comane sti, Romania, and Bogdan Ionita , Bu harest, Romania. Let a, b, and c be the lengths of the sides of a triangle. Prove that 3249
(b + c)2 (c + a)2 (a + b)2 + + ≥ 6. a2 + bc b2 + ca c2 + ab
. Composite of solutions by Mohammed Aassila, Strasbourg, Fran e; Sefket Arslanagi , University of Sarajevo, Sarajevo, Bosnia and Herzegovina; Mi hel Bataille, Rouen, Fran e; Nguyen Thanh Liem, Tran Hung Dao High S hool, Phan Thiet, Vietnam; Salem Maliki , student, Sarajevo College, Sarajevo, Bosnia and Herzegovina; Alex Remorov, student, William Lyon Ma kenzie Collegiate Institute, Toronto, ON; Phi Thai Thuan, student, Tran Hung Dao High S hool, Phan Thiet, Vietnam; and Panos E. Tsaoussoglou, Athens, Gree e. We have I
X (b + c)2
y li
a2 + bc
−2
X b2 + c2 − 2a2
=
a2 + bc X 2 2 (a − b )
y li
=
y li
P
≥
b2 + ca
−
1 a2 + bc
(a − b)2 (a + b)(a + b − c)(c2 + ab)
y li
=
1
Q
(c2 + ab)
y li
0.
. Se ond solution by Mohammed Aassila, Strasbourg, Fran e. We prove the result for any positive real numbers a, b, and c. After a lengthy omputation, the original inequality II
X (b + c)2
y li
a2 + bc
−6 ≥ 0
transforms to the following equivalent form (b − c)2 (c − a)2 (a − b)2 abc
+
X (b + c)2 (b − c)2
y li
a
≥ 0,
whi h is obviously true for any positive real numbers a, b, and c, as laimed. University Also solved by ARKADY ALT, San Jose, CA, USA; SEFKET ARSLANAGIC, of Sarajevo, Sarajevo, Bosnia and Herzegovina (se ond solution); MIHALY BENCZE, Brasov, Romania; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; KEE-WAI LAU, Hong Kong, China; XAVIER ROS, student, Universitat Polite ni a de Catalunya, Bar elona, Spain (three solutions); and the proposer.
255 Janous and Ros have also shown that the inequality is true for any positive real numbers
a, b, and c. Janous has proven that the following generalization holds X (b + c)2 12 ≥ a2 + µbc µ+1
y li
for any positive real numbers a, b, and c and µ ∈ [0, 1] ∪ [2, ∞). Maliki mentions that this is a well-known problem, originally reated by Darij Grinberg and Peter S holze; he gives the web address http://www.mathlinks.ro/viewtopic.php?search_id=7020805&t=60128, where the problem an be found together with several solutions, in luding solutions similar to both of those featured above.
. [2007 : 239, 240℄ Proposed by D.J. Smeenk, Zaltbommel, the Netherlands. Let ABC be an isos eles triangle with AB = AC and ∠BAC = 100◦ . Let D be the point on the produ tion of AB su h that AD = BC . Find ∠ADC . 3250
. Solution by Tai hi Maekawa, Takatsuki City, Osaka, Japan. Constru t the equilateral triangle ADE , as shown in the pi ture. Then △ABC and △CAE are ongruent, sin e AB = CA, BC = AD = AE , and I
A
...... ..... ........ .... .... ........ .... ..... . ... . . . .... ... ... .... .... ... .... ..... ... .... .... . . .... . ... ... .. . . . . .................................................................................................................................... . . . ................. . . . .... . . . . . . . . . . . . . . . . . ... ... ............. . . . . . . . ... . . . . . . . . . . . . . . . . . . . ... ................. ... ... ... ................................ ... ... ........... . ........... ... .. ........... ... .... ........... ... ... ........... . ........... ........... .... ... ................ ...
B
........... .....
∠CAE = 100◦ − 60◦ = 40◦ = ∠ABC .
D
M
C
Then AC = CE and AD = DE , so that points D and C both lie on the perpendi ular bise tor of the segment AE . Let M be the point of interse tion E of DC and AE . Then ∠AM D = 90◦ , and sin e ∠DAE = 60◦ , it follows that ∠ADC = 30◦ . II. Solution by Xavier Ros, student, Universitat Polite ni a de Catalunya, Bar elona, Spain. Let ∠ADC = α. Applying the Law of Sines to triangles ABC and ACD , we have AC sin 40◦
=
BC sin 100◦
and, sin e AD = BC , we get
and
AC sin α
=
AD sin(80◦ − α)
sin(80◦ − α) sin 40◦ = sin α sin 100◦ .
Now, sin 100◦ = sin 80◦ = 2 sin 40◦ cos 40◦ , so that we obtain or
sin(80◦ − α) = 2 sin α cos 40◦ , sin 80◦ cos α − sin α cos 80◦ = 2 sin α cos 40◦ .
,
256 The value α = 90◦ is not a solution to this equation, so that we an divide both sides by cos α to obtain an equation for tan α: tan α
= =
sin 80◦ 2 cos 40◦ + cos 80◦ cos 10◦ 1 = √ √ ◦ 3 cos 10 3
=
cos 10◦ 2 cos(30◦ + 10◦ ) + sin 10◦
.
Therefore, α = 30◦ . Also solved by MIGUEL AMENGUAL COVAS, Cala Figuera, Mallor a, Spain; CHRISTOS ANASTASSIADES, student, Apostles Peter and Paul Ly eum, Limassol, Cyprus; 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; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; CHARLES R. DIMINNIE, Angelo State University, San Angelo, TX, USA; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; PETER HURTHIG, Columbia College, Van ouver, BC; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Y, Big Rapids, Austria; GEOFFREY A. KANDALL, Hamden, CT, USA; V ACLAV KONECN MI, USA; KEE-WAI LAU, Hong Kong, China; TAICHI MAEKAWA, Takatsuki City, Osaka, student, Sarajevo College, Sarajevo, Bosnia and Japan (se ond solution); SALEM MALIKIC, Herzegovina; MICHAEL PARMENTER, Memorial University of Newfoundland, St. John's, NL; ALEX REMOROV, student, William Lyon Ma kenzie Collegiate Institute, Toronto, ON; JOEL SCHLOSBERG, Bayside, NY, USA; EDMUND SWYLAN, Riga, Latvia; PANOS E. TSAOUSSOGLOU, Athens, Gree e; PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU, Comane sti, Romania; and the proposer (7 solutions). Two solvers mentioned that the problem is known. Covas supplied the referen es [1℄ and [2℄, and Lau gave the earliest referen e [3℄. Referen es
[1℄ Ayoub B. Ayoub, \Problem 1104", Pi Mu Epsilon Journal, 12(2005); solution in 13(2005), pp. 186{187. [2℄ Ion Patras u, Probleme de Geometrie Plana, Craiova, 1996, Problem 8.12, p. 51. [3℄ H.J. Webb, \Problem 877", Mathemati s Magazine, 46(1973); solution in 47(1974), p. 172.
Crux Mathemati orum
with Mathemati al Mayhem Editor Emeritus / Reda teur-emeritus: Bru e L.R. Shawyer
Crux Mathemati orum
Founding Editors / Reda teurs-fondateurs: Leopold Sauve & Frederi k G.B. Maskell Editors emeriti / Reda teurs-emeriti: G.W. Sands, R.E. Woodrow, Bru e L.R. Shawyer
Mathemati al Mayhem
Founding Editors / Reda teurs-fondateurs: Patri k Surry & Ravi Vakil Editors emeriti / Reda teurs-emeriti: Philip Jong, Je Higham, J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia, Shawn Godin, Je Hooper
