crux mathematicorum v 34 n 8
64
A(0, 0) B(6, 0) C(6, 4) D(0, 4) P (4, 3) BC M AD N AB Q CD R AP DM BR cos(A + B) + cos(A − B) = 2 cos A cos B A B cos C + cos D = 2 cos C + D 2 cos C − D 2 C D cos 20 ◦ + cos 60 ◦ + cos 100 ◦ + cos 140 ◦ AB 3 C AC =2 AB ACF CBE AKE K FC
Transcript of crux mathematicorum v 34 n 8
449MATHEMATICAL MAYHEMMathemati al Mayhem began in 1988 as a Mathemati al Journal for and byHigh S hool and University Students. It ontinues, with the same emphasis,as an integral part of Crux Mathemati orum with Mathemati al Mayhem.The Mayhem Editor is Ian VanderBurgh (University of Waterloo). Theother sta� members are Monika Khbeis (As ension of Our Lord Se ondaryS hool, Mississauga), Eri Robert (LeoHayesHigh S hool, Frederi ton), LarryRi e (University of Waterloo), and Ron Lan aster (University of Toronto).
Mayhem ProblemsVeuillez nous transmettre vos solutions aux probl �emes du pr �esent num�eroavant le 15 Mars 2009. Les solutions re� ues apr �es ette date ne seront prises en ompte que s'il nous reste du temps avant la publi ation des solutions.Chaque probl �eme sera publi �e dans les deux langues oÆ ielles du Canada(anglais et fran� ais). Dans les num�eros 1, 3, 5 et 7, l'anglais pr �e �edera le fran� ais,et dans les num�eros 2, 4, 6 et 8, le fran� ais pr �e �edera l'anglais.La r �eda tion souhaite remer ier Jean-Mar Terrier, de l'Universit �e deMontr �eal, d'avoir traduit les probl �emes.M369. Propos �e par l' �Equipe de Mayhem.Soit A(0, 0), B(6, 0), C(6, 4) et D(0, 4) les sommets d'un re tangle.Par le point P (4, 3), on tra e d'une part une droite horizontale oupant BCen M et AD en N et d'autre part une droite verti ale oupant AB en Q etCD en R. Montrer que AP , DM et BR passent toutes par le meme point.M370. Propos �e par l' �Equipe de Mayhem.(a) Montrer que cos(A + B) + cos(A − B) = 2 cos A cos B pour tous lesangles A et B.(b) Montrer que cos C + cos D = 2 cos
(
C + D
2
)
cos(
C − D
2
) pour tousles angles C et D.( ) Trouver la valeur exa te de cos 20◦ + cos 60◦ + cos 100◦ + cos 140◦,sans l'aide d'une al ulatri e.M371. Propos �e par Panagiote Ligouras, �E ole Se ondaire L �eonard deVin i, No i, Italie.Un segment AB de longueur 3 ontient un point C tel que AC = 2. On onstruit d'un meme ot �e de AB deux triangles �equilat �eraux ACF et CBE.D �eterminer l'aire du triangle AKE si K est le point milieu de FC.
450M372. Propos �e par l' �Equipe de Mayhem.Soit x un nombre r �eel satisfaisant x3 = x + 1. Trouver des entiers a, bet c de sorte que x7 = ax2 + bx + c.M373. Propos �e par Kunal Singh, �etudiant, Kendriya Vidyalaya S hool,Shillong, Inde.Les ot �es d'un triangle sont mesur �es par trois nombres entiers ons �e u-tifs et le plus grand angle est le double du plus petit. D �eterminer la longueurdes ot �es du triangle.M374. Propos �e par Mih�aly Ben ze, Brasov, Roumanie.Soit p un nombre premier �x �e, ave p ≥ 3. Trouver le nombre de solu-tions de x3 + y3 = x2y + xy2 + p2009, o �u x et y sont des entiers.M375. Propos �e par Ne ulai Stan iu, �E ole Te hnique Sup �erieure de SaintMu eni Sava, Ber a, Roumanie.D �eterminer toutes les solutions r �eelles du syst �eme d' �equations
1
x2+
4
y2+
9
z2= 4 ; x2 + y2 + z2 = 9 ; xyz =
9
2..................................................................M369. Proposed by the Mayhem Sta�.A re tangle has verti es A(0, 0), B(6, 0), C(6, 4), and D(0, 4). A hor-izontal line is drawn through P (4, 3), meeting BC at M and AD at N . Averti al line is drawn through P , meeting AB at Q and CD at R. Prove that
AP , DM , and BR all pass through the same point.M370. Proposed by the Mayhem Sta�.(a) Prove that cos(A + B) + cos(A − B) = 2 cos A cos B for all anglesA and B.(b) Prove that cos C + cos D = 2 cos
(
C + D
2
)
cos(
C − D
2
) for all anglesC and D.( ) Determine the exa t value of cos 20◦ + cos 60◦ + cos 100◦ + cos 140◦,without using a al ulator.M371. Proposed by Panagiote Ligouras, Leonardo da Vin i High S hool,No i, Italy.Suppose that the line segment AB has length 3 and C is on AB with
AC = 2. Equilateral triangles ACF and CBE are onstru ted on the sameside of AB. If K is the midpoint of FC, determine the area of △AKE.
451M372. Proposed by the Mayhem Sta�.A real number x satis�es x3 = x + 1. Determine integers a, b, and cso that x7 = ax2 + bx + c.M373. Proposed by Kunal Singh, student, Kendriya Vidyalaya S hool,Shillong, India.The side lengths of a triangle are three onse utive positive integersand the largest angle in the triangle is twi e the smallest one. Determine theside lengths of the triangle.M374. Proposed by Mih�aly Ben ze, Brasov, Romania.Suppose that p is a �xed prime number with p ≥ 3. Determine thenumber of solutions to x3 + y3 = x2y + xy2 + p2009, where x and y areintegers.M375. Proposed by Ne ulai Stan iu, Saint Mu eni Sava Te hnologi alHigh S hool, Ber a, Romania.Determine all real solutions to the system of equations
1
x2+
4
y2+
9
z2= 4 ; x2 + y2 + z2 = 9 ; xyz =
9
2.
Mayhem SolutionsM332. Proposed by Dionne Bailey, Elsie Campbell, and Charles R.Diminnie, Angelo State University, San Angelo, TX, USA.A losed right ir ular ylinder has an integer radius and an integerheight. The numeri al value of the volume is four times the numeri al valueof its total surfa e area (in luding its top and bottom). Determine the small-est possible volume for the ylinder.Solution by Cao Minh Quang, Nguyen Binh Khiem High S hool, Vinh Long,Vietnam.Let r and h be the radius and the height of the losed right ir ular ylinder. The volume of su h a ylinder is V = πr2h and the surfa e area isA = 2πr2 + 2πrh.From the hypotheses, πr2h = 4(2πr2 + 2πrh), or rh = 8r + 8h, orrh − 8r − 8h + 64 = 64, or (r − 8)(h − 8) = 64. Note that r − 8 > −8 andh − 8 > −8. This gives us the following possibilities:
452r − 8 1 2 4 8 16 32 64h − 8 64 32 16 84 4 2 1
r 9 10 12 16 24 40 72h 72 40 24 16 12 10 9V 5832π 4000π 3456π 4096π 6912π 16000π 46656πThus, the smallest possible volume for the ylinder is 3456π.Also solved by DENISE CORNWELL, student, Angelo State University, San Angelo,TX, USA; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; RICARD PEIR �O, IES \Abastos",Valen ia, Spain; BILLY SUANDITO, Palembang, Indonesia; and TITU ZVONARU, Com�ane�sti,Romania. There was 1 in omplete solution submitted.M333. Proposed by the Mayhem Sta�.Anne and Brenda play a game whi h begins with a pile of n toothpi ks.They alternate turns with Anne going �rst. On ea h player's turn, she mustremove 1, 3, or 6 toothpi ks from the pile. The player who removes the lasttoothpi k wins the game. For whi h of the values of n from 36 to 40 in lusivedoes Brenda have a winning strategy?Solution by Ri hard I. Hess, Ran ho Palos Verdes, CA, USA, modi�ed by theeditor.We an build a table of winning and losing positions for Anne. Herwinning positions start with 1, 3, or 6, sin e she an immediately win byremoving all of the toothpi ks.Starting with 2 toothpi ks, Anne must remove 1 toothpi k, leavingBrenda with 1, and so Brenda wins. Starting with 4 toothpi ks, Anne mustremove 1 or 3 toothpi ks, leaving Brenda with 3 or 1 (respe tively), and soBrenda wins by removing all of the toothpi ks.Starting with 5 toothpi ks, Anne an remove 3 toothpi ks, thus leavingBrenda with 2 toothpi ks. Sin e 2 is a losing position for whoever goes �rst,then Brenda loses, so Anne wins.So far, 1, 3, 5, and 6 are winning positions for Anne, while 2 and 4 arelosing positions for Anne.Starting with a pile of size n, Anne must redu e the pile to one of size
n − 1, n − 3, or n − 6 and pass to Brenda. If the person who goes �rst hasa winning strategy starting with a pile of ea h of these sizes, then Anne willlose. In other words, if Anne has a winning strategy starting with piles ofsize n − 1, n − 3, and n − 6, then Anne will lose starting with a pile of sizen, as Brenda an implement Anne's strategy for the smaller pile and win, nomatter what Anne does. If one or more of these pile sizes are su h that the�rst person does not have a winning strategy, then Anne should redu e tothis size, thus preventing Brenda from being able to win, and so Anne herselfwill win.We an examine the ases from n = 7 to n = 40, obtaining the follow-ing lists:
453Winning positions for Anne: 1, 3, 5, 6, 7, 8, 10, 12, 14, 15, 16, 17, 19, 21,23, 24, 25, 26, 28, 30, 32, 33, 34, 35, 37, 39.Losing positions for Anne: 2, 4, 9, 11, 13, 18, 20, 22, 27, 29, 31, 36, 38, 40.Therefore, Brenda wins for n = 36, 38, 40.Also solved by JACLYN CHANG, student, Western Canada High S hool, Calgary, AB.See the Problem of the Month olumn in [2007 : 15{17℄ for a similar problem with amore detailed explanation.M334. Proposed by the Mayhem Sta�.(a) Determine all integers x for whi h x − 3
3x − 2is an integer.
(b) Determine all integers y for whi h 3y3 + 3
3y2 + y − 2is an integer.
I. Solution by Edin Ajanovi , student, First Bosniak High S hool, Sarajevo,Bosnia and Herzegovina.Let A be an integer su h that A =x − 3
3x − 2. Then 3A is an integer and
3A =3x − 9
3x − 2=
3x − 2 − 7
3x − 2= 1 − 7
3x − 2.Thus, 7
3x − 2is an integer; that is, 3x − 2 is a divisor of 7, so 3x − 2 isone of ±1, ±7. Sin e x is an integer, x = 1 or x = 3. This answers part (a).Now let B be an integer su h that
B =3y3 + 3
3y2 + y − 2= y − y2 − 2y − 3
3y2 + y − 2
= y − (y − 3)(y + 1)
(y + 1)(3y − 2)= y − y − 3
3y − 2.
Sin e y is an integer, y − 3
3y − 2is an integer. From the solution to part (a),
y = 1 or y = 3, whi h answers part (b).II. Solution by Edward T.H. Wang, Wilfrid Laurier University, Waterloo, ON.We show that the only integer solutions to part (a) are x = 1 and x = 3.Let f(x) =x − 3
3x − 2. Then f(0) =
3
2, f(1) = −2, f(2) = −1
4, and
f(3) = 0. Of these, only f(1) and f(3) are integers.If x > 3, then f(x) is not an integer, sin e 3x − 2 > x − 3 > 0 forx > 3 and so 0 <
x − 3
3x − 2< 1.If x ≤ −1, let x = −s where s ≥ 1. Then f(x) = f(−s) =
s + 3
3s + 2.Sin e 3s + 2 > s + 3 > 0 for s ≥ 1, f(−s) is not an integer by a similarargument so, f(x) is not an integer.
454Therefore, f(x) is an integer for integer values of x if and only if x = 1or x = 3.Also solved by RICARD PEIR �O, IES \Abastos", Valen ia, Spain; CAO MINH QUANG,Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; and TITU ZVONARU, Com�ane�sti,Romania. There was one in orre t and one in omplete solution submitted.M335. Proposed by the Mayhem sta�.In a sequen e of four numbers, the se ond number is twi e the �rstnumber. Also, the sum of the �rst and fourth numbers is 9, the sum of these ond and third is 7, and the sum of the squares of the four numbers is 78.Determine all su h sequen es.Solution by Denise Cornwell, student, Angelo State University, San Angelo,TX, USA.Let a, b, c, and d represent the �rst, se ond, third and fourth number,respe tively. We an now write the given information as b = 2a, a + d = 9,
b + c = 7 and a2 + b2 + c2 + d2 = 78.The �rst three equations allow us to rewrite b, c, and d in terms of a,obtaining b = 2a, c = 7 − b = 7 − 2a, and d = 9 − a.Therefore,a2 + (2a)2 + (7 − 2a)2 + (9 − a)2 = 78 ,
a2 + 4a2 + 49 − 28a + 4a2 + 81 − 18a + a2 − 78 = 0 ,5a2 − 23a + 26 = 0 ,
(5a − 13)(a − 2) = 0 ,hen e a =13
5or a = 2.Therefore, the sequen es are a =
13
5, b =
26
5, c =
9
5, d =
32
5and
a = 2, b = 4, c = 3, d = 7. Both sequen es satisfy the given requirements.Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo,Bosnia and Herzegovina; JACLYN CHANG, student, Western Canada High S hool, Calgary,AB; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; CAO MINH QUANG, Nguyen BinhKhiem High S hool, Vinh Long, Vietnam; KUNAL SINGH, student, Kendriya Vidyalaya S hool,Shillong, India; BILLY SUANDITO, Palembang, Indonesia; and TITU ZVONARU, Com�ane�sti,Romania. There was one in orre t and one in omplete solution submitted.M336. Proposed by the Mayhem Sta�.A latti e point is a point (x, y) in the oordinate plane with ea h of xand y an integer. Suppose that n is a positive integer. Determine the numberof latti e points inside the region |x| + |y| ≤ n.Solution by Edin Ajanovi , student, First Bosniak High S hool, Sarajevo,Bosnia and Herzegovina, modi�ed by the editor.We an rewrite the given inequality as the equations |x| + |y| = 0 and
|x| + |y| = k for 1 ≤ k ≤ n, where x, y ∈ Z.
455The equation |x| + |y| = 0 has one integer solution only, namely
(x, y) = (0, 0).Consider next |x| + |y| = k, for an integer k with 1 ≤ k ≤ n. We anremove the absolute values by splitting into four ases:Case 1. The integers x and y satisfy x + y = k, where x ≥ 0 and y ≥ 0.This has solutions (k, 0), (k −1, 1), . . . , (1, k −1), (0, k), for a total of k +1solutions.Case 2. The integers x and y satisfy x − y = k, where x ≥ 0 and y < 0.This has solutions (k − 1, −1), (k − 2, −2), . . . , (1, −(k − 1)), (0, −k), fora total of k solutions.Case 3. The integers x and y satisfy −x + y = k, where x < 0 and y ≥ 0.This ase is the same as Case 3, but with the roles of x and y swit hed, sothere are a total of k solutions here as well.Case 4. The integers x and y satisfy −x − y = k, where x < 0 and y < 0.This has solutions (− 1, −(k − 1)
), (− 2, −(k − 2)), . . . , (− (k − 2), −2
),(
− (k − 1), −1), for a total of k − 1 solutions.Thus, for ea h k with 1 ≤ k ≤ n, the equation |x| + |y| = k has
(k + 1) + k + k + (k − 1) = 4k solutions.Therefore, the number of latti e points inside the region is1 +
n∑
k=1
4k = 1 + 4
n∑
k=1
k = 1 + 4 · (1 + 2 + · · · + n)
= 1 + 4 · n(n + 1)
2= 2n2 + 2n + 1 .Also solved by RICARD PEIR �O, IES \Abastos", Valen ia, Spain. There were one in orre tand two in omplete solutions submitted.M337. Proposed by the Mayhem Sta�.On sides AB and CD of re tangle ABCD with AD < AB, points Fand E are hosen so that AFCE is a rhombus.(a) If AB = 16 and BC = 12, determine EF .(b) If AB = x and BC = y, determine EF in terms of x and y.Solution by Kunal Singh, student, Kendriya Vidyalaya S hool, Shillong, India.We present the solution to (b), whi h is a general version of the spe i� ase in (a).Suppose that AF = FC = CE = EA = m. Let O be the point ofinterse tion of diagonals AC and EF of rhombus AFCE. Note that ACand EF are perpendi ular and bise t ea h other at O.
456By the Pythagorean Theorem,
CF 2 − FB2 = CB2 ,m2 − (x − m)2 = y2 ,
m2 − x2 + 2mx − m2 = y2 ,2mx = x2 + y2 ,
m =x2 + y2
2x.
Now, AF = m =x2 + y2
2xand AC =
√AB2 + BC2 =
√
x2 + y2. Also,OA = 1
2AC. Thus, by the Pythagorean Theorem again,
OF 2 = AF 2 − OA2
=
(
x2 + y2
2x
)2
−(√
x2 + y2
2
)2
=
(
x4 + y4 + 2x2y2
4x2
)
−(
x2 + y2
4
)
=x4 + y4 + 2x2y2 − x4 − x2y2
4x2
=y4 + x2y2
4x2.Therefore,
OF =
√
y4 + x2y2
4x2=
√
y2(y2 + x2)
2x=
y√
x2 + y2
2x,and
EF = 2 OF =2y√
x2 + y2
2x=
y√
x2 + y2
x.In part (a), this yields
EF =12
√162 + 122
16=
12(20)
16= 15 .
Also solved by EDIN AJANOVIC, student, First Bosniak High S hool, Sarajevo, Bosniaand Herzegovina; JACLYN CHANG, student, Western Canada High S hool, Calgary, AB (part(a) only); RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; RICARD PEIR �O, IES \Abastos",Valen ia, Spain; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam;BILLY SUANDITO, Palembang, Indonesia; LUYAN ZHONG-QIAO, Columbia InternationalCollege, Hamilton, ON; and TITU ZVONARU, Com�ane�sti, Romania.
457Problem of the MonthIan VanderBurghHere is a problem that might seem to be not very interesting initially,but turns out to have a whole lot of unexpe ted solutions.Problem (2005 Canadian Open Mathemati sChallenge) In the grid shown, ea h row hasa value assigned to it and ea h olumn has avalue assigned to it. The number in ea h ellis the sum of its row and olumn values. Forexample, the \8" is the sum of the value as-
3 0 5 6 −2−2 −5 0 1 y
5 2 x 8 00 −3 2 3 −5
−4 −7 −2 −1 −9signed to the 3rd row and the value assigned to the 4th olumn. Determinethe values of x and y.It is tempting �rst of all to give labels to the values that are assignedto the rows and olumns in order to be able to dive into some algebra. Let'slabel the values assigned to the �ve olumns A, B, C, D, E and the valuesassigned to the �ve rows a, b, c, d, e.Ea h entry in the table gives us an equation involving two of thesevariables. For example, the −3 in row 4, olumn 2 gives us d + B = −3,and the −9 in row 5, olumn 5, gives us e + E = −9. We ould pro eed andwrite down 25 equations, one for ea h entry in the table. These equationswould in lude 12 variables { the 10 that label the rows and olumns togetherwith x and y. We ould then spend pages and pages wading through algebratrying to ome up with the answers. At this point, we would hope that therehas to be a better way. Maybe we should have looked before we leapt!Here are three neat ways to approa h this. (As a point of interest, Iwas re ently talking about this problem with a friend while driving and soneither of us really wanted to do any algebra, and so were for ed to ome upwith better ways to do it.)Solution 1. If we hoose �ve entries from the table whi h in lude one fromea h row and one from ea h olumn, then the sum of these entries is onstantno matter how we hoose the entries, as it is always equal toA + B + C + D + E + a + b + c + d + e .Can you see why? Here are three ways in whi h this an be done (looking atthe underlined numbers in the two grids below and the grid on the followingpage):
3 0 5 6 −2−2 −5 0 1 y
5 2 x 8 00 −3 2 3 −5
−4 −7 −2 −1 −9
3 0 5 6 −2−2 −5 0 1 y
5 2 x 8 00 −3 2 3 −5
−4 −7 −2 −1 −9
458Therefore,
3 + (−5) + 2 + 8 + (−9)
= (−4) + (−3) + x + 1 + (−2)
= 3 + y + 2 + (−2) + 3 ,or −1 = x − 8 = y + 6. Thus, x = 7 andy = −7.
3 0 5 6 −2−2 −5 0 1 y
5 2 x 8 00 −3 2 3 −5
−4 −7 −2 −1 −9
Solution 2. Consider the �rst two entries in row 1. From the labels above, wehave 3 = A + a and 0 = B + a. Subtra ting these, we obtain the equation3 = 3 − 0 = (A + a) − (B + a) = A − B.Noti e that whenever we take entries in olumns 1 and 2 from the samerow, their di�eren e will always equal A − B, whi h is equal to 3. Similarly,sin e the di�eren e between the 0 and the 5 in the �rst row is 5, then everyentry in olumn 3 will be 5 greater than the entry in olumn 2 from the samerow. In row 3, we see that x = 2 + 5 = 7.Also, sin e the di�eren e between the 6 and the −2 in the �rst row is8, then every entry in olumn 5 is 8 less than the entry in olumn 4 from thesame row. In row 2, we see that y = 1 − 8 = −7. Thus, x = 7 and y = −7.Solution 3. Consider the sub-grid 0 1
x 8.Sin e the 0 is in row 2 and olumn 3, then 0 = b + C. Similarly,
1 = b + D, 8 = c + D, and x = c + C.But then 0 + 8 = (b + C) + (c + D) = (c + C) + (b + D) = x + 1, orx = 7.In a similar way, by looking at the sub-grid 1 y
8 0we an show that
1 + 0 = y + 8, or y = −7. Thus, x = 7 and y = −7.So there are three di�erent but neat solutions to the problem. Onefootnote to the �nal solution is that in fa t, in any sub-grid of the formp qr s
, we must have p + s = q + r. Can you see why?Another interesting point about this problem is that it might be easierfor those who know less! If we repla ed the x and the y with \?" and gaveit to someone who didn't know a lot of algebra, they might �nd the answersfaster than those of us who go immediately to algebra. Sometimes, the extrama hinery that we have an get in the way.As 2008 draws to a lose, the Mayhem Editor has three enormous sets ofthanks to o�er. First, to the Mayhem Sta�, espe ially to Monika Khbeis andEri Robert, for all of their help over the past year. Se ond, to the Editor-in-Chief of CRUX with MAYHEM, V �a lav Linek, for all of his help and en- ouragement over the past year (as well as for his sharp eyes!). Third, tothe Mayhem readership for their support. Please keep those problems andsolutions oming!
459THE OLYMPIAD CORNERNo. 274
R.E. WoodrowWith the Winter break oming up, I have de ided to fo us this issuemainly on providing problems for your puzzling pleasure, and to give sometime for the mails to deliver the solutions to problems from 2008 numbersof the Corner to restore the readers' solutions �le, whi h is parti ularly thinfor the February 2008 number, as you will see later in the olumn.To start you o� we have the problems proposed but not used at the47th IMO in Slovenia 2006. My thanks go to Robert Morewood, CanadianTeam Leader at the IMO for olle ting them for our use.47th INTERNATIONAL MATHEMATICAL OLYMPIADSLOVENIA 2006Problems Proposed But Not UsedContributing Countries. Argentina, Australia, Brazil, Bulgaria, Canada,Colombia, Cze h Republi , Estonia, Finland, Fran e, Georgia, Gree e, HongKong, India, Indonesia, Iran, Ireland, Italy, Japan, Republi of Korea,Luxembourg, Netherlands, Poland, Peru, Romania, Russia, Serbia andMontenegro, Singapore, Slovakia, South Afri a, Sweden, Taiwan, Ukraine,United Kingdom, United States of Ameri a, Venezuela.Problem Sele tion Committee. Andrej Bauer, Robert Gerets hl�ager, G �ezaK �os, Mar in Ku zma, Sventoslav Sav hev.AlgebraA1. Given an arbitrary real number a0, de�ne a sequen e of real numbersa0, a1, a2, . . . by the re ursion
ai+1 = ⌊ai⌋ · {ai} , i ≥ 0 ,where ⌊ai⌋ is the greatest integer not ex eeding ai, and {ai} = ai − ⌊ai⌋.Prove that ai = ai+2 for suÆ iently large i.A2. Let a0 = −1 and de�ne the sequen e of real numbers a0, a1, a2, . . .by the re ursionn∑
k=0
an−k
k + 1= 0for n ≥ 1. Show that an > 0 for n ≥ 1.
460A3. Let c0 = 1, c1 = 0 and de�ne the sequen e c0, c1, c2, . . . by there ursion cn+2 = cn+1 + cn for n ≥ 0. Let S be the set of ordered pairs(x, y) su h that
x =∑
j∈J
cj and y =∑
j∈J
cj−1for some �nite set J of positive integers. Prove that there exist real numbersα, β, m, and M with the property that an ordered pair of non-negativeintegers (x, y) satis�es the inequality
m < αx + βy < Mif and only if (x, y) ∈ S. (By onvention an empty sum is 0.)A4. Let a1, a2, . . . , an be positive real numbers. Prove that∑
i<j
aiaj
ai + aj
≤ n
2(a1 + a2 + · · · + an)
∑
i<j
aiaj .A5. Let a, b, and c be the lengths of the sides of a triangle. Prove that
√b + c − a
√b +
√c − √
a+
√c + a − b
√c +
√a −
√b
+
√a + b − c
√a +
√b − √
c≤ 3 .
Combinatori sC1. There are n ≥ 2 lamps L1, L2, . . . , Ln arranged in a row. Ea h of themis either on or o�. Initially the lamp L1 is on and all of the other lamps areo�. Ea h se ond the state of ea h lamp hanges as follows: if the lamp Liand its neighbours (L1 and Ln ea h have one neighbor, any other lamp hastwo neighbours) are in the same state, then Li is swit hed o�; otherwise, Liis swit hed on. Prove that there are(a) in�nitely many n for whi h all of the lamps will eventually be o�,(b) in�nitely many n for whi h the lamps will never be all o�.C2. Let S be a �nite set of points in the plane su h that no three of themare on a line. For ea h onvex polygon P whose verti es are in S, let a(P )be the number of verti es of P , and let b(P ) be the number of points of Swhi h are outside of P . Prove that for every real number x∑
P
xa(P )(1 − x)b(P ) = 1 ,where the sum is taken over all onvex polygons with verti es in S. (A linesegment, a point, and the empty set are onvex polygons of 2, 1, and 0verti es, respe tively.)
461C3. A ake is shaped as an n × n square with n2 unit squares. Strawberrieslie on some of the unit squares so that ea h row or olumn ontains exa tlyone strawberry; all this arrangement A.Let B be another su h arrangement. Suppose that every grid re tanglewith one vertex at the top left orner of the ake ontains no fewer strawber-ries of arrangement B than of arrangement A. Prove that the arrangementB an be obtained from A by performing a sequen e of swaps, where a swap onsists of sele ting a grid re tangle with only two strawberries, situated atits top right orner and bottom left orner, and then moving these two straw-berries to the other two orners of that re tangle.C4. An (n, k)-tournament is a ompetition with n players held in k roundssu h that(a) Ea h player plays in ea h round, and every two players meet at moston e.(b) If player Ameets player B in round i, player C meets player D in round
i, and player A meets player C in round j, then player B meets playerD in round j.Determine all pairs (n, k) for whi h there exists an (n, k)-tournament.C5. A holey triangle is an upward equilateral triangle of side length n with
n upward unit triangular holes ut out. A diamond is a unit rhombus withangles of 60◦ and 120◦. Prove that a holey triangle T an be tiled withdiamonds if and only if for ea h k = 1, 2, . . . , n every upward equilateraltriangle of side length k in T ontains at most k holes.C6. Let P be a onvex polyhedron with no parallel edges and no edge par-allel to a fa e other than the two fa es it borders. A pair of points on P areantipodal if there exist two parallel planes ea h ontaining one of the pointsand su h that P lies between them. Let A be the number of antipodal pairsof verti es and let B be the number of antipodal pairs of mid-points of edges.Express A − B in terms of the numbers of verti es, edges, and fa es of P.GeometryG1. Let ABCD be a trapezoid withAB||CD and AB > CD. PointsK and L lie on the line segmentsAB and CD, respe tively, su h thatAK : KB = DL : LC. Suppose thatthere are points P and Q on the linesegment KL satisfying ..........................................................................................................................................................................................................................................................................................................................................................................................................................................
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...
..
...
...
...
...
..
...
...
...
...
...
..
...
...
...
...
..
...
...
...
...
..
...
...
...
...
...
..
...
...
...
...
..
...
...
...
...
..
...
...
...
...
...
...............................................................................................................................................................................................................................
p p
pp
p
p
p
p
A B
CD
K
Q
P
L
∠APB = ∠BCD and ∠CQD = ∠ABC .Prove that the points P , Q, B, and C are on y li .
462G2. Let ABCDE be a onvex pen-tagon su h that
∠BAC = ∠CAD = ∠DAE ;∠ABC = ∠ACD = ∠ADE .The diagonalsBD andCE interse t at
P . Prove that the line AP bise ts theside CD. ..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
.......................................................................................................................................................................................................................................................................................
p
p
p
p
p
p
........................
....................
........................
........................
....................
........................
...........
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
A
B
C
D
E
P
G3. A point D is hosen on the sideAC of a triangle ABC with
∠ACB < ∠BAC < 90◦in su h a way that BD = BA. Thein ir le of ABC is tangent to AB andAC at points K and L, respe tively.Let J be the in entre of triangleBCD.Prove that the line KL interse ts theline segment AJ at its mid-point. .......................................................................................................................................................................................................................................................................................................................................................................................................................................
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
............................................................................................................................................................................................................................................................................................................................................................................................. ...........................................................................................................................................................................................................................................................................................
.
..
.
..
.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
....................................................................................
...
..
..
..
..
.
..
..
.
.
..
.
..
.
..
..
..
..
..............................................................
............................................................
.
.
..
.
..
.
..
.
..
.
..
.
..
..
..
..
.
..
..
..
..
.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..................................................................
...........................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................
q q
q
q q q
q
q
q
q
A
B
C D L
J
K
..
..
..
..
..
..
..
..
..
..
..
..
............................................................................................................................................................
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
G4. In triangle ABC, let J be the entre of the ex ir le tangent to sideBC at A1 and to the extensions ofsides AC and AB at B1 and C1, re-spe tively. Suppose that the linesA1B1 and AB are perpendi ular andinterse t at D. Let E be the footof the perpendi ular from C1 to lineDJ. Determine the angles ∠BEA1and ∠AEB1.
.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
..........................................................................................................................................................................................................................................................................................................................................................................................................................................
.....................................................................................
.......................................................................................................................................................................................
.
..
..
.
..
..
.
..
..
.
..
.
..
.
..
..
..
..
..
..
.
..
..
..
.
..
.
..
..
..
.
..
.
..
..
..
.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.............................................................................................................
.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.....................................................................................................................................................................................................................
........................... . . . . . . . . . . . ..............
....................................................
q q q q
q
q q
A D B C1
E
J
A1
C
B1
G5. Cir lesω1 and ω2 with entresO1and O2 are externally tangent at pointD and internaly tangent to a ir le ωat points E and F , respe tively. Linet is the ommon tangent of ω1 and ω2at D. Let AB be the diameter of ωperpendi ular to t, so that A, E, andO1 are on the same side of t. Provethat the lines AO1, BO2, EF , and tare on urrent. q qq
q q
q
q
.......................................................................................................................................................................................................................................................................................................................................................................................................
.......................................................................................................................................................................................................................................................................................................................................................................................................
..
.
.
..
.
..
..
.
..
..
..
..
..
.
..
..
..
..
..
..
..............................
....................................................................................................................................................
..............................................................
..
.
..
.
..
.
..
..
.
..
.
..
.
..
..
.
..
..
.
..
..
.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
....................................................................
..................................................................................................................................................................................................................................................................................................................................
..............................................................................................................................................
........................................................................................................................................................................................................................................................................................................................................................................................................................................
.
..
.
..
..
..
.
..
.
..
.
..
.
..
..
.
..
.
..
.
..
.
..
.
..
.
..
..
.
..
.
..
.
..
.
..
.
..
.
..
..
.
..
.
..
.
..
.
..
.
..
.
..
..
.
..
.
..
.
..
.
..
.
..
.
..
..
.
..
.
..
.
..
.
..
.
..
.
..
..
.
..
.
..
.
..
.
..
.
..
.
..
..
.
..
.
..
.
..
.
..
.
..
.
..
..
.
..
.
..
.
..
.
..
.
..
.
..
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
..
.
..
.
..
.
..
.
..
.
..
.
..
..
.
..
.
..
.
..
.
..
.
..
.
..
..
.
..
.
..
.
..
.
..
.
..
.
..
..
.
..
.
..
.
..
.
..
.
..
.
..
..
.
..
.
..
.
..
.
..
.
..
.
..
..
.
..
.
..
.
..
.
..
.
..
.
..
..
.
..
.
..
.
..
A B
DO2O1
F
E
....................... . . . . . . . . . ............
ω
ω1
ω2t
463G6. In a triangle ABC, let Ma, Mb,Mc be the respe tive mid-points ofthe sides BC, CA, AB and let Ta,Tb, Tc be the mid-points of the ar sBC, CA, AB of the ir um ir le ofABC not ontaining A, B, C, respe -tively. For ea h i ∈ {a,b,c}, let ωibe the ir le with diameter MiTi. Letpi be the ommon external tangent toωj , ωk su h that {i,j,k} = {a,b,c}and su h that ωi lies on one side ofpi while ωj , ωk lie on the other side.Prove that the lines pa, pb, pc form atriangle similar to ABC and �nd theratio of similitude.
..
.
..
..
.
..
.
..
.
..
..
.
..
.
..
.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
..
.
..
.
..
..
.
..
.
..
.
..
.
..
.
..
..
..
..
..
..
..
..
..
.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
............................................................................................................................................
...........................................................................................................................................................................................................................................................
................................................................................................................................................................................................................................................................................................................................................................................................................................................
......................................................................................................................................................................................................................................................................................................................
......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................r r
r
r rr
r
r
r
A
B C
Mb
Ma
Mc
Tc
Tb
Ta
..
.
..
.
.
..
..
..
.
..
..
..
..
..
..
..
..
..
.................................................................................
..
...
..
..
.
..
.
..
.
..
.
..
.
..
..
..
..
..................................................................................................................
..
..
.
..
..
..
.
..
..
..
..
..
.
..
.
..
.
..
.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
............................................................
...........................................................................................................................................................................................................................................................................................................
..................................................................................................................................
.
.
.
.
.
.
.
.
.
.
..
..
.
..
..
..
......................................................
..
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
..
..
..........................................................................
ωa
ωb
ωc
G7. Let ABCD be a onvex quadri-lateral. A ir le passing through A andD and a ir le passing through B andC are externally tangent at the pointP in the interior of the quadrilateral.Prove that if ∠PAB + ∠PDC ≤ 90◦and ∠PBA + ∠PCD ≤ 90◦, thenAB + CD ≥ BC + AD.
.
..
..
.
..
.
..
.
..
.
..
.
..
..
..
.
..
..
..
..
.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
........................................................................
........................................................................................................................................................................................................................................................................................................................................
................................................................................................................................................... ..
.
..
..
..
..
.
..
..
.
..
.
..
.
..
.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
...............................................
...................................................................................................................................................................................................................................................
.....................................................................................................
r
r
r
r
rP
A
B
C
D...........................................................................................................................................................................................................
....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
G8. Points A1, B1, C1 are on the sides BC, CA, AB of a triangle ABC, re-spe tively. The ir um ir les of triangles AB1C1, BC1A1, CA1B1 interse tthe ir um ir le of triangle ABC again at points A2, B2, C2, respe tively(that is, A2 6= A, B2 6= B, C2 6= C). Points A3, B3, C3 are symmetri to A1, B1, C1 with respe t to the mid-points of the sides BC, CA, ABrespe tively. Prove that the triangles A2B2C2 and A3B3C3 are similar.
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
..
.
..
..
..
..
..
..
.
..
..
..
..
..
.
..
..
.
..
.
..
..
..
.
..
.
..
.
..
.
..
.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..........................................................................................................................................................................
.............................................................................................................................................................................................................................................................................................................
...
...
...
....
...
...
...
...
....
...
..
..
..
...
..
..
..
..
...
..
..
..
..
...
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
..
..
...
..
..
..
..
...
..
..
..
..
..
....
...
...
...
...
...
...
....
...
...
...........................................................................................................................................................................................................................................................................................................
........................................................................................................................................................................................................................................................................................................................................................................
.
.
..
.
..
..
.
..
.
..
..
.
..
.
..
..
..
..
..
..
..
..
..
..
..
.
..
..
.
..
.
..
.
..
.
..
..
.
..
.
..
..
..
..
.
..
..
..
..
..
..
..
.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.................................................................................................................................
.....................................................................................................................................................................................................................
...
...
...
...
...
...
...
...
..
..
..
..
..
..
..
...
..
..
.
..
..
.
..
..
.
..
.
..
..
.
..
.
..
.
..
..
.
..
.
..
..
.
..
.
..
..
.
..
..
..
...
..
..
..
..
..
..
..
...
...
...
...
...
...
...
...........................................................................................................................................................................................................................
...................................................................................................................................................................................................................................................................
.
.
..
.
..
.
..
.
..
.
..
.
..
..
..
..
..
.
..
..
..
..
.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
....................................................................
.................................................................................................................................
..
...
..
..
...
..
..
...
.
..
.
..
.
..
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
.
..
..
..
..
...
..
..
..
...
..
..................................................................................................................................................................................................................................................................................
..
.
..
..
.
..
.
..
..
.
..
.
..
..
..
.
..
..
..
..
..
..
..
.
..
.
..
..
.
..
.
..
.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
...........................................................................................................
......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
...........................................................................................................................................................................................................................
...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................r r
r
r
r
r
r
r
r
r
A
B C
B2
C2
B1
A2
C1
A1
464Number TheoryN1. Given x ∈ (0, 1) let y ∈ (0, 1) be the number whose nth digit after thede imal point is the (2n)th digit after the de imal point of x. Prove that if xis a rational number, then y is a rational number.N2. For ea h positive integer n let
f(n) =1
n
(⌊
n
1
⌋
+
⌊
n
2
⌋
+ · · · +
⌊
n
n
⌋) ,where ⌊x⌋ is the greatest integer not ex eeding x.(a) Prove that f(n + 1) > f(n) for in�nitely many n.(b) Prove that f(n + 1) < f(n) for in�nitely many n.N3. Find all solutions in integers of the equationx7 − 1
x − 1= y5 − 1 .
N4. Let a and b be relatively prime integers with 1 < b < a . De�ne theweight of an integer c, denoted by w(c), to be the minimum possible valueof |x| + |y| taken over all pairs of integers x and y su h thatax + by = c .An integer c is alled a lo al hampion if w(c) ≥ max{w(c ± a), w(c ± b)}.Find all lo al hampions and determine their number.N5. Prove that for every positive integer n, there exists an integer m su hthat 2m + m is divisible by n.
A se ond problem set is the 2005/06 Swedish Mathemati al Contest.My thanks go to Robert Morewood, Canadian Team Leader at the IMO, for olle ting them for our use.SWEDISH MATHEMATICAL CONTEST 2005/2006Final RoundNovember 19, 2005 (Time: 5 hours)1. Find all solutions in integers x and y of the equation(x + y2)(x2 + y) = (x + y)3 .
4652. A queue in front of a ounter onsists of 12 persons. The ounter is then losed be ause of a te hni al problem and the 12 people are redire ted toanother one. In how many di�erent ways an the new queue be formed ifea h person maintains the same position as before, or is one step loser tothe front, or is one step farther from the front?3. In the triangle ABC the angle bise tor from A interse ts the side BCin the point D and the angle bise tor from C interse ts the side AB in thepoint E. The angle at B is greater than 60◦. Prove that AE + CD < AC.4. The polynomial f(x) is of degree four. The zeroes of f are real and forman arithmeti progression, that is, the zeroes are a, a+d, a+2d, and a+3dwhere a and d are real numbers. Prove that the three zeroes of f ′(x) alsoform an arithmeti progression.5. Ea h square on a 2005×2005 hessboard is painted either bla k or white.This is done in su h a way that ea h 2 × 2 \sub- hessboard" ontains an oddnumber of bla k squares. Show that the number of bla k squares among thefour orner squares is even. In how many di�erent ways an the hessboardbe painted so that the above ondition is satis�ed?6. All the edges of a regular tetrahedron are of length 1. The tetrahedronis proje ted orthogonally into a plane. Determine the largest possible areaand the least possible area of the image.
Next we look at an alternate solution to problem 2 of the 17th IrishMathemati al Olympiad dis ussed at [2007 : 151; 2008 : 88-89℄.2. Let A and B be distin t points on a ir le T . Let C be a point distin tfrom B su h that |AB| = |AC| and su h that BC is tangent to T at B.Suppose that the bise tor of ∠ABC meets AC at a point D inside T . Showthat ∠ABC > 72◦.Alternate Solution by Luyan Zhong-Qiao, Columbia International College,Hamilton, ON.Let D′ 6= B be the se ond point ofinterse tion of BD with the ir le T .Let ∠ABD = ∠DBC = β. Sin eAB = AC, we have ∠ACB = 2β.Also, BC is tangent to the ir le andAB is a hord, hen e ∠D′AB = β.Let ∠ABO = γ.
.........................................................................................................................................................................................................................................................................................................................................................................
.................................................................................................................................................................................................................................................
...
....
....
....
...
....
...
..
..
..
..
..
..
..
..
..
..
..
..
..
.
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
..
.
..
.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
..
..
..
..
..
..
..
..
..
..
.................................................................................................................................................................................................................................................................................................................................................................................................................
O
BA
C
D′
D
β.......................
..
..
..
..
..
..
..
..
..
..
.
γ..................
ββ
.......................
2β
......
..............
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
..........
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
.....
466Sin e BC is tangent to T , we have 2β + γ = 90◦. Adding the anglesin the isos eles triangle ABC yields 4β + ∠CAB = 180◦. From these twoequations it follows that ∠CAB = 2γ. Sin e D is inside T we have
β = ∠D′AB > ∠CAB = 2γ ,and therefore β
2> γ. This last inequality together with 2β + γ = 90◦ yields
2β +β
2> 90◦, or 5β
2> 90◦. Hen e, β > 36◦ and ∠ABC = 2β > 72◦.
Next we look at a omment on the solution to problem 6 of the 2007Italian Olympiad [2007: 149{150; 208: 84{85℄.6. Let P be a point inside the triangle ABC. Say that the lines AP , BP ,and CP meet the sides of ABC at A′, B′, and C′, respe tively. Letx =
AP
PA′ , y =BP
PB′ , z =CP
PC′ .Prove that xyz = x + y + z + 2.Comment by J. Chris Fisher, University of Regina, Regina, SK.The result is a theorem of Euler featured in the Crux Mathemati orumarti le \Euler's Triangle Theorem" by G.C. Shephard in [1999 : 148{153℄.Euler's proof is very neat, even ni er than any of the solutions found in CruxMathemati orum. It an be lo ated in Edward Sandifer's book, How EulerDid It, and on his webpage: http://www.maa.org/news/howeulerdidit.html li k on \19th Century Triangle Geometry".Now we turn to solutions from our readers to problems given in theFebruary 2008 number of the Corner. First a solution to a problem of the11th Form, Final Round, XXXI Russian Mathemati al Olympiad given in theCorner at [2008: 20{21℄.5. (N. Agakhanov) Does there exist a bounded fun tion f : R → R with
f(1) > 0 su h thatf2(x + y) ≥ f2(x) + 2f(xy) + f2(y)for all x, y ∈ R?Solved by Li Zhou, Polk Community College, Winter Haven, FL, USA.Suppose that f is su h a fun tion. Let a0 = 1 and an = an−1 +
1
an−1for n ≥ 1. Thenf2(a1) ≥ f2(a0) + 2f(1) + f2
(
1
a0
)
≥ 2f(1) .
467As an indu tion step, assume that f2(an) ≥ 2nf(1) for some n ≥ 1. Then
f2(an+1) = f2
(
an +1
an
)
≥ f2(an) + 2f(1) + f2
(
1
an
)
≥ f2(an) + 2f(1) ≥ 2(n + 1)f(1) , ompleting the indu tion. Hen e f2(an) ≥ 2nf(1) for all n ≥ 1, ontra-di ting the fa ts that f(1) > 0 and f is bounded.And to omplete our �les for the Corner, we look at a problem of theTaiwan Mathemati al Olympiad, Sele ted Problems 2005, given in [2008:21{22℄.1. A △ABC is given with side lengths a, b, and c. A point P lies inside
△ABC, and the distan es from P to the three sides are p, q, and r, respe -tively. Prove thatR ≤ a2 + b2 + c2
18 3√
pqr,where R is the ir umradius of △ABC. When does equality hold?Solved by Arkady Alt, San Jose, CA, USA; Mi hel Bataille, Rouen, Fran e;and George Tsapakidis, Agrinio, Gree e. We give Bataille's write-up.Let F denote the area of △ABC. We have the well-known relation
2F =abc
2R, but also from the de�nition of p, q, and r we have the equation
2F = ap + bq + cr. Thus, the proposed inequality is equivalent toabc
2(ap + bq + cr)≤ a2 + b2 + c2
18 3√
pqror(a2 + b2 + c2)(ap + bq + cr) ≥ 9abc 3
√pqr . (1)By the AM{GM Inequality,
a2 + b2 + c2 ≥ 33√
a2b2c2 and ap + bq + cr ≥ 3 3√
abcpqr ,and the inequality (1) now follows from(a2 + b2 + c2)(ap + bq + cr) ≥ 9
3√
a2b2c2 · 3√
abc · 3√
pqr .That ompletes the Corner for this number, and this Volume. As JoanneCanape, who has been translating my s ribbles into beautiful LATEX has de- ided that twenty-plus years is enough, I want to thank her too for all thehelp over the time we've worked together.
468BOOK REVIEWJohn Grant M LoughlinThe Symmetries of ThingsBy John H. Conway, Heidi Burgiel, and Chaim Goodman{Strauss, publishedby AK Peters, Wellesley, MA, 2008ISBN 978-1-56881-220-5, hard over, 423+xxv pages, US$75.00Reviewed by J. Chris Fisher, University of Regina, Regina, SKThe authors set themselves the ambitious goal of produ ing a book thatappeals to everybody. As far as I an tell from a single reading, they havesu eeded admirably. The �rst thing anybody would noti e about the bookis that it is �lled with beautiful and fas inating photographs and omputerdrawings. No spe ial knowledge is required for admiring beauty; this bookwould be as mu h at home on the living room o�ee table as on the oÆ eshelf. Of ourse, it is primarily a mathemati s book.The ontents have been organized into three parts. The �rst of themdes ribes and enumerates the symmetries found in repeating patterns on sur-fa es; it is written at a level suitable for Crux withMayhem readers. This partmight well serve as a textbook for a geometry ourse dire ted at universitystudents spe ializing in mathemati s, edu ation, physi al s ien e, or om-puter s ien e. What makes the authors' approa h both novel and elementaryis the introdu tion of what they all the orbifold signature notation. Groupsare not needed here; the on ept an be easily des ribed and qui kly mas-tered. Here is the idea. A point in a pattern where two mirrors of symmetrymeet at an angle of π
mis alled kaleidos opi and is denoted by ∗m; pointshaving rotational symmetry of order m (but no kaleidos opi symmetry) are alled gyrational and are denoted by m (with no asterisk). If a region hasan oppositely oriented image in the pattern that is not explained by mirrors,then these two regions must be related by a glide re e tion, whi h here is alled a mira le (short for \mirrorless rossing", they say), denoted by ×.To identify the signature of any repeatingplane pattern one writes down the symbolsstarting from the middle and working out-ward. First lo ate mirror lines and ea hkind of kaleidos opi point, if any (where twopoints are of the same kind if they are relatedby a symmetry of the pattern); list them afterthe asterisk in de reasing order. Next lo ateany gyrational points and order them beforeany asterisk. Then look for mira les. Typi alsignatures are ∗632 for a pattern whose sym-metry is explained by three kinds of mirrorsthat meet in pairs at angles of π
6, π
3, and π
2;
632 (having no asterisk) for a pattern with.........................................................................................
................................................................................................................................................................................................................................................................................................................................................................................................
................................................................................................................................................................................................................................................................................................................................................................................................
............................................................................................................................................................
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
....................................................................................................................................................................................................................................................................................................................................................................................
....................................................................................................................................................................................................................................................................................................................................................................................
....................................................................................................................................................................................................................................................................................................................................................................................
∗6
3
2r
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
........................................................
.........................................................
.........................................................
........................................................
............................................................................................................................................
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
........................................................
∗632
4696{fold, 3{fold, and 2{fold gyrational points but no re e tions or mira les;2∗22 for a pattern with two kinds of kaleidos opi points where a pair ofmirrors interse t at right angles, and one point where there is a half-turnsymmetry but no mirror.
632
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
..
.
..
.
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
..
.
..
.
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
..
.
..
.
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
..
.
..
.
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
..
.
..
.
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
..................................................................
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
...................................................................
.
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
.
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
.
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..................................................................
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.
..
.
..
.
.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................
..................................................................................................................................................................
..................................................................................................................................................................
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
q
2
2
2
∗
2∗22Unlike most other notation systems that have been devised for des rib-ing plane symmetry, these orbifold signatures an also be used to des ribefrieze patterns and spheri al patterns. (We learn in Part III that they workequally well for des ribing hyperboli patterns.) But how does one know thatthe resulting lists of 17 signatures for plane patterns, 7 signatures for friezepatterns, and 14 signatures for spheri al patterns are orre t and omplete?There is a \Magi Theorem" that assigns a ost to every symbol in the sig-nature in su h a way that plane patterns and frieze patterns ost exa tly $2while spheri al patterns ost a bit less. That theorem tells us immediatelywhi h signatures are feasible. To establish the Magi Theorem, a patternon the surfa e is asso iated with a folded surfa e they all an orbifold. Theorbifold is obtained by identifying points related by a symmetry of the pat-tern (whereby points of an orbit are folded atop of one another so that asingle representative point of every orbit lives on the orbifold). This soundsa bit s ary, but the authors manage to explain the details in a gentle wayusing suitable pi tures and simple examples. They then state the Classi� a-tion Theorem for Surfa es and provide Conway's elementary and intuitive Zipproof. They also prove that these surfa es an be distinguished using Euler'sformula (involving the numbers of verti es, edges, and fa es of a suitable mapon the surfa e), whi h they also prove. Sin e the orbifolds are easily lassi�edusing Euler's formula, the orresponding patterns are thereby lassi�ed.Remarkably, all the proofs should satisfy the professional mathemati- ian even though they are dire ted at an elementary audien e. The authorsa hieve this feat by repeatedly redu ing te hni al diÆ ulties down to prob-lems that are postponed to the following hapter. This way they presentone on ept at a time, as ompared to the typi al textbook's initial barrageof poorly motivated de�nitions and lemmas. Their proofs are every bit asbrilliant as their notation. The illustrations are not just beautiful, but they
470have been arefully hosen to larify the exposition. I really appre iatedthe authors' de ision to repeat pi tures that they require for illustrating newideas | instead of making the reader turn ba k to a pi ture on an earlierpage, they reprodu ed a smaller version of it whenever needed. The authors learly have fun oining whimsi al new words; their terminology will not ap-peal to everybody, but the informal nature of their dis ussions makes forenjoyable reading. I rather liked the word mira le in pla e of the standard,but awkward and misleading term glide re e tion; however I saw little needfor gyrational in pla e of rotational or wandering in pla e of translation. Wewill have to wait to see whi h words at h on.What I have des ribed so far is the ontent of the 116 pages of the�rst nine hapters. Originally, a ording to the prefa e, this was all thatthe authors had intended to write. But they de ided it was worthwhile toextend the signature to olour symmetry, and the book grew from there.For a areful reading of Part II the reader needs some group theory anda bit of mathemati al maturity. The authors' main goal for this part is topresent their analysis and notation for olour symmetry. They enumerate thep{fold olour types for plane, spheri al, and frieze patterns (for all primesp). The omplete lassi� ations appear in a book for the �rst time. Alongthe way the authors show how their orbifold notation orresponds to previ-ous lassi� ation systems, whi h gives them the opportunity to dis uss theshort omings of those systems. Also in this part, they enumerate the iso-hedral tilings of the sphere and plane, and they extend to n = 2009 theBes he-Ei k-O'Brien table of the number of abstra t groups of ea h order n.The informative lists of Part II an probably be understood by readerswho might not take an interest in the a ompanying te hni al arguments.Similar omments apply to Part III, whi h the authors expe t to be ompletelyunderstood only by a few professional mathemati ians. Still, as they pointout, mu h of Part III an pro�tably be explored by other readers, while manymore will enjoy inspe ting the pretty pi tures. Here, among other things, theauthors dis uss hyperboli groups and Ar himedean polyhedra and tilings;they list the 219 rystallographi spa e groups (and explain why hemistsdistinguish 230 groups), and they provide a omplete list for the �rst time inprint of the 4{dimensional Ar himedean polytopes. Apparently they ouldhave kept writing, but they de ided to leave something for the rest of us todo. Their �nal words are, \A universe awaits | Go forth!"I thank Bru e Shawyer for inviting me to serve as Book Review Editor. I amgrateful to the late Jim Totten for guiding me during his tenure; Bru e Crofootfor insightful ommentary; V �a lav Linek for re ent support; Shawn Godin forleadership with Mayhem. Thanks go to all the reviewers, but espe ially thistrio: Chris Fisher, a dependable sour e of thought provoking reviews usually on erning geometry; Ed Barbeau, an e le ti mathemati ian who is eager tohelp; and my su essor, Amar Sodhi. Amar's passion for mathemati s willshine as he assumes this role. Wel ome Amar! Thanks to the CRUX withMAYHEM ommunity for an enjoyable journey. | John Grant M Loughlin
471Old Idaho Usual HereRobert Israel, Stephen Morris, and Stan WagonN. Kildonan [1℄ raised the following problem. Take an arbitrary wordusing at most 10 distin t letters, su h as DONALD COXETER. Can onesubsititute distin t digits for the 10 letters so as to make the resulting base
10 number divisible by d? The answer depends on d. If d has 100 digits thenthe answer is learly NO. If d = 100 then the answer is again NO, sin e ER annot be 00. If d is 2, the answer is learly YES: just let the units digit beeven; d = 5 or d = 10 are just as easy.Kildonan proved that divisibility by d = 3 an always be a hieved andR. Israel and R. I. Hess extended this to d = 9; the ase of d = 7 wasleft unresolved. In this note we settle all ases. The reader interested in animmediate hallenge should try to prove that divisibility by d = 45 is alwayspossible. This appears to be the hardest ase.To phrase things pre isely, a word is a string made from 10 or fewerdistin t letters; for ea h word and ea h possible substitution of distin t digitsfor the letters, there is an asso iated value: the base 10 number one getsafter making the substitution. If all substitutions yield a value for the wordw that is not divisible by d, then w is alled a blo ker for d. If any wordending (on the right) with w fails to be divisible by d, then w is alled astrong blo ker for d. An integer d is alled attainable if the value of everysuÆ iently long word an be made divisible by d by some substitution ofdistin t digits for letters. Thus d is not attainable if there exist arbitrarilylong blo kers. The use of arbitrarily long strings is important be ause, forexample, AB is a blo ker for 101, but only be ause it is too short. An integerd is strongly attainable if the value of every word an be made divisible by dby an appropriate substitution.In this paper we will �nd all attainable integers; moreover, they areall strongly attainable. Note that any divisor of an attainable number isattainable.Some ases, su h as d = 2, d = 5, or d = 10 are extremely easy toattain, and it is just about as easy to attain d = 4 or d = 8. It takes a littlework to show that d = 3 and d = 9 are attainable (proofs given below). Ourmain theorem resolves the attainability status of all integers.Theorem 1 An integer is attainable if and only if it divides one of the integers18, 24, 45, 50, 60, or 80.To start, we dis uss the ases of d = 3 and d = 9 for ompleteness andto introdu e the ideas needed later. We use the well known fa t that whend is 3 or 9, then d divides a number if and only if d divides the sum of itsdigits.Copyright c© 2008 Canadian Mathemati al So ietyCrux Mathemati orum with Mathemati al Mayhem, Volume 34, Issue 8
472The number d = 3 is attainable (Kildonan [1℄). Given a word, letthe 10 letters be grouped as Ai, Bi, and Ci, where ea h Ai has a multipli ity(perhaps 0) that is divisible by 3, ea hBi has a multipli ity of the form 3k+1,and ea h Ci has a multipli ity of the form 3k+2. Look for one, two, or threepairs among the Bi and repla e them with digits 1 and 2, and 4 and 5 for these ond pair, and 7 and 8 for the third pair. Then look for pairs of the Ciand repla e them with digits in any of the still-available pairs among (1, 2),
(4, 5), and (7, 8). These substitutions take are of Bi ∪ Ci ex ept possiblyfour letters (sin e we used three pairs) and we an substitute 0, 3, 6, and 9for them. The letters Ai an be assigned the remaining digits in any order.Thus, the �nal number has a digit sum divisible by d = 3.The number d = 9 is attainable (Solution II by Israel and Hess [1℄).Suppose a word has length n. Suppose some letter o urs k times, wheren−k is not divisible by 3. Assign 9 to this letter and assign 0 to 8 arbitrarilyto the other letters. Let the value of the resulting number be v (mod 9).Now repla e ea h digit from 0 to 8 by the next higher digit, wrapping ba kto 0 in the ase of 8. This adds n − k to the value modulo 9. But n − k isrelatively prime to 9, so we an do this −v/(n−k) times, where the divisionuses the inverse of n − k modulo 9, in order to obtain the value 0 modulo 9.The other ase is that every letter has a multipli ity k ≡ n (mod 3).If in fa t every multipli ity is ongruent to n (mod 9), then any assignmentwill yield a value ongruent to n(0 + 1 + · · · + 9) = 45n ≡ 0 (mod 9).Otherwise there is a multipli ity k ≡ n (mod 3) but k 6≡ n (mod 9), andthen we pro eed as in the �rst half of the proof: assign 9 to this letter, 0 to8 to the other letters, and then y li ally permute the values 0 to 8. Ea hpermutation adds n − k modulo 9 and this will eventually transform thevalue v, whi h is divisible by 3, to a value divisible by 9, be ause 3 dividesn − k but 9 does not.Now to the proof of Theorem 1, whi h follows from these four lemmas.Lemma 1 Any integer divisible by a prime greater than 5 is not attainable.Lemma 2 The largest attainable powers of 2, 3, and 5 are 16, 9, and 25,respe tively.Lemma 3 The numbers 36, 48, 75, 90, 100, and 120 are not attainable.Lemma 4 The numbers 18, 24, 45, 50, 60, and 80 are attainable.The ordering of these lemmas indi ates how Theorem 1was found. Firstthe ases of d = 7 and d = 11 were settled and that led to the general resultof Lemma 1. It followed that the only andidates for attainability had theform 2a3b5c. On e the powers of 2, 3, and 5 were resolved (Lemma 2), the andidate list was redu ed to the 45 divisors of 3600 = 16 · 9 · 25. Resolvingthe situation for those divisors, with some omputer help, led to Lemmas 3and 4. Finally, the omputer sear hes were eliminated and the whole thingwas redone by hand. Theorem 1 follows from the lemmas be ause Lemmas3 and 4 settle the status of all 45 divisors of 3600.
473A key idea is that the ten digits sumto 45. So we begin with Lemma 3,whi h shows how unattainability isproved. We use the fa t that an in-teger is ongruent modulo 9 (hen emodulo 3) to the sum of its digits.Let A, B, C, D, E, F , G, H, J ,K be the ten letters and let wg bethe on atenation of g opies of wordw. The table at right lists the blo k-ers needed for Lemmas 2 and 3; mostwere found by a omputer sear h.
d blo ker27 AAB32 ABBAB36 (ABCDEFGHJ)5J4K48 (ABCDEFGH)2KKJK75 AABA90 A6(BCDEFGHJ)7JK100 AB120 ABCDEFGHJJJK125 BBAWe show that the words in the table are blo kers. The easiest ase is
d = 100, sin e the value of any word ending in AB is not divisible by 100.Case 1. The number d = 36 an be blo ked. We have(ABCDEFGHJ)5J4K ≡ K + 4J + 5(45 − K) ≡ 4J − 4K (mod 9) .The only way 4(J−K) is divisible by 9 is if JK is either 90 or 09, and neitheris divisible by 4. Extension on the left by A9i preserves the value modulo 36,be ause 111111111 is divisible by 9.Case 2. The number d = 48 an be blo ked. The rightmost 4 digits of theword (ABCDEFGH)2KKJK must be one of 0080, 2272, 4464, 6656, or
8848, as these are the only words of the form KKJK that are divisible by16. However, now the value of the word modulo 3 is one of the entriesbelow, where we work with ve tors and ignore K whi h o urs three times:
2(
(45, 45, 45, 45, 45) − (8, 7, 6, 5, 4) − (0, 2, 4, 6, 8))
+ (8, 7, 6, 5, 4)
= (82, 79, 76, 73, 70) ,and no entry is divisible by 3. Left extension by A3i preserves the valuemodulo 48.Case 3. The number d = 75 an be blo ked. Here we have the ongruen eAABA ≡ 51A + 10B (mod 75). Multiplying by 53 transforms the on-gruen e to 3A + 5B ≡ 0 (mod 75). However, 3 ≤ 3A + 5B ≤ 69, sothe ongruen e is never satis�ed. Left extension by A3i preserves the valuemodulo 75.Case 4. The number d = 90 an be blo ked. We have
A6(BCDEFGHJ)7JK ≡ 6A + 7(45 − A) + J (mod 9) ,be ause K must be 0. The expression simpli�es to J − A modulo 9, whi h annot be divisible by 9 be ause 0 is already assigned to K. Left extensionby A9i preserves the value modulo 9.
474Case 5. The number d = 120 an be blo ked. We have
ABCDEFGHJJJK ≡ 45 + 2J ≡ 2J (mod 3) .However, JJK must be either 440 or 880 to obtain divisibility by 40, there-fore, 2J is either 8 or 16, and so is not divisible by 3. Left extension by A3ipreserves the value modulo 120.Case 6. The number d = 32 an be blo ked. Any word ending in ABBABhas a value satisfying 10010A + 1101B ≡ 26A + 13B (mod 32). If this is ongruent to 0modulo 32, then we may an el 13, leaving 2A+B. However,this sum is between 1 and 18 + 8 = 26, so it is not divisible by 32.Case 7. The number d = 125 an be blo ked. A number is divisible by 125 ifand only if it ends in 125, 250, 375, 500, 625, 750, 875, or 000. Thus, BBAis a strong blo ker for 125.Case 8. The number d = 27 an be blo ked. The value of AAB satis�es the ongruen e 110A + B ≡ 2A + B (mod 27). However, 1 ≤ 2A + B ≤ 26,whi h is not divisible by 27. This shows nonattainability, be ause we an addthe pre�x A27i, whi h leaves the value modulo 27 un hanged.Next we prove Lemma 1. Our �rst proof of this was a little ompli ated(see the Proposition that follows), but when we fo used on words involv-ing two letters only we dis overed Theorem 2, whi h yields Lemma 1 in all ases ex ept d = 7. Re all Euler's theorem, that aφ(d) ≡ 1 (mod d) whengcd(a, d) = 1. It follows that if d is oprime to 10, then there is a smallestpositive integer, denoted by ordd(10), su h that 10ordd(10) ≡ 1 (mod d).Theorem 2 Let d be oprime to 10 and greater than 10 with e = ordd(10).Then w = Ake−1B is a blo ker for d for any positive integer k.Proof: Assume �rst that k = 1 so that w is just Ae−1B. If 3 does not divided then the value of w satis�es the ongruen e
B + A
e−1∑
i=1
10i = B − A + A10e − 1
9≡ B − A (mod d) .
Sin e d > 10, d annot divide B − A. Now suppose that 3 divides d andd > 81. Suppose the value of w, in the formula just given, is a multiple of d.Then multiplying by 9 yields 9(B − A) + A (10e − 1) = 9Kd, and hen e ddivides 9(B − A). However, A 6= B and −81 ≤ 9(B − A) ≤ 81, so d > 81 annot divide 9(B − A), a ontradi tion.There remain the ases where 3 divides d and 11 ≤ d ≤ 81, namelyd ∈ {21, 27, 33, 39, 51, 57, 63, 69, 81}. Suppose that d is one of thesebut d 6= 21, 27, 81; then ordd(10) = ord3d(10). This means that fromB −A+A (10e − 1) /9 = Kd, we have 9(B −A)+A (10e − 1) = 3K(3d),when e 3d divides 9(B − A). Thus, d divides 3(B − A), whi h means thatd ≤ 27, a ontradi tion.
475For d = 21 the value of w modulo 21 is B−A, whi h is not divisible by
21. For d = 27 the value of w modulo 27 is 2A + B and 1 ≤ 2A + B ≤ 26,so the value is not divisible by d. For d = 81 the value of w modulo 81 is8A + B and 1 ≤ 8A + B ≤ 80, so the value is not divisible by d.The extension to the ase of general k is straightforward.The pre eding result blo ks all primes greater than 10. We need to dealalso with d = 7. One an give an alternate onstru tion in the general asethat in ludes d = 7, and we give the following without proof.Proposition Suppose that d is oprime to 10 and d does not divide 9. Let
w = KJKeHKeGKeFKeEKeDKeCKeBKeAKe ,where e = ordd(10) − 1. Then after any substitution the value of w is ongruent to 9(d + 1)
2(mod d), and so is not divisible by d.For d = 7 the word w of the Proposition has length 55. A di�erentapproa h led to the mu h shorter example OLD IDAHO USUAL HERE,its value is always 3 · 45 (mod 7). This 17- hara ter word is thus a blo kerfor d = 7, and it an be made arbitrarily long by prepending E6.On to Lemma 2. The positive results for d = 16 and d = 25 are notdiÆ ult, but they are omitted as they follow from the ases of d = 80 and
d = 50 (proved below); the ase of d = 9 was dis ussed earlier, as were thenegative results for d = 32, 27, and 125. It remains only to prove Lemma 4.Case 1. The number d = 50 is strongly attainable. Just use 00 or 50 for therightmost two digits.Case 2. The number d = 80 is strongly attainable. Assign 0 to the rightmostletter; 8 to the next new letter that o urs reading from the right, 4 to thenext one, and 2 to the next one after that. The value is then divisible by 16and also by 5; divisibility by 80 is only a�e ted by the four rightmost digits.Case 3. The number d = 18 is strongly attainable. Given a word, �nd anassignment that makes it divisible by 9. If the rightmost digit is even, we aredone. Otherwise, repla e this digit y with 9 − y. This preserves divisibilityby 9 and makes the rightmost digit even.Case 4. The number d = 60 is strongly attainable. The word ends in eitherAA or BA. In either ase, assign 0 to A and 6 to B. Let the eight remainingletters be grouped as Ai, Bi, and Ci as in the proof for d = 3. Use the pairs(1, 2), (4, 5), and (7, 8) on whatever pairs of letters an be found within Bior within Ci. This leaves at most two single letters in the B and C groups.Use 3 and 9 for the two singletons, and any remaining digits for the A group.The �nal value is then divisible by 3, 4, and 5.Case 5. The number d = 24 is strongly attainable. The idea is to modify theproof for d = 3 so as to guarantee divisibility by 8. Re alling the proof thatd = 3 is strongly attainable, all two letters mat hed if they are repla ed in
476that proof by 1 and 2, or by 4 and 5, or by 7 and 8. If the word ends in AAAjust make sure A is either 0 or 8. The remaining ases are that the word endsin one of the patterns ABC, ABB, BBA, or ABA.If the ending is ABC with A and C mat hed, then use 152 or 192,depending on whether B is part of a mat hed pair or not. If A and C areunmat hed use 320 or 360 a ording as B is part of a mat hed pair or not.If the ending is ABB with A and B mat hed, then use 488, sin e themat hing an use 4 and 8 as well as 1 and 2. If both A and B are unmat hed,then use 600. If A is mat hed and B is not, use 800. If B is mat hed and Ais not, use 088.If the ending is BBA, then pro eed as if the ending was ABB, but useinstead 448, 336, 008, and 880 for the four sub ases.If the ending is ABA, then pro eed similarly, using 848, 696, 808, and080 for the four sub ases.Case 6. The number d = 45 is strongly attainable. Let m(X) denote themultipli ity of the letter X in the given word redu ed modulo 9; let X denotethe digit assigned to X. Let A be the rightmost letter and assign 0 to it, thusensuring divisibility by 5.Assume �rst that the multipli ities of at least eight of the nine remain-ing digits are all mutually ongruent modulo 3, and assign 9 to the otherletter. Let the m-values of the eight letters be 3ai + c, where ea h ai is anon-negative integer and c ∈ {0, 1, 2}. Let Li be the digits assigned to theseeight letters. Sin e ∑Li = 36, whi h 9 divides, the value modulo 9 of theword is 3
∑
aiLi. So we want ∑ aiLi to be divisible by 3. Assign the pairs(1, 2), (4, 5), and (7, 8) to pairs of letters with equal ai. Assign 3 and 6 tothe remaining two. The total is then divisible by 9 and therefore by 45.In the other ase we an hoose a letter, K say, with m(K) not on-gruent modulo 3 to the length of the word; therefore S, the sum of the mul-tipli ities of the nine letters other than K, is not divisible by 3. Let K = 9.Case 6a. There is a letter, say B, with m(B) = m(A). Then let B = 1 andassign the remaining digits arbitrarily.Case 6b. There is no letter as in Case 6a. Then we an �nd two letters amongB, C, D, E, F , G, H, J , say C and D, with m(C) 6≡ m(D) (mod 3).Consider B; we know m(B) 6= m(A). Set B to be the non-negative residuemodulo 9 of m(D)−m(C) and note that B − A ≡ m(D)−m(C) (mod 9)whi h is not divisible by 3. Assign unused digits to the lettersC andD so thatC − D ≡ m(B) − m(A) (mod 9); there are enough digits left for this to bepossible. Assign the remaining digits arbitrarily.Nowwe an treat both ases to get the result. The assignment produ essome total value, redu ed modulo 9 to v. If v 6= 0 then repla e ea h digitbetween 0 and 8 by the next higher digit, wrapping ba k to 0 in the ase of 8.This adds S to the value modulo 9 and does not alter B−A or C−D modulo9. But S is relatively prime to 9, so we an do this −v/S times, where thedivision uses the inverse modulo 9 of S, in order to a hieve divisibility by 9.If A is 0 or 5, then we are done.
477If A is 3 or 6, then swit h digits of A and B, where we know that B isnot divisible by 3; this is be ause the value modulo 3 of B − A, whi h startsout nonzero, does not hange in the translational step. If we are in Case 6b,then also swit h C and D; the net hange is
(
B − A)(
m(A) − m(B))
+(
C − D)(
m(D) − m(C))
≡(
B − A)(
D − C)
+(
C − D)(
B − A)
≡ 0 (mod 9) ,so divisibility by 9 is preserved.As A is now not divisible by 3, we an multiply ea h digit less than 9by 5/A (mod 9). This preserves divisibility by 9 and makes A = 5. Thetotal is now divisible by 45. The proof that d = 45 is strongly attainableis omplete, as is the proof Lemma 4, and thus the proof of Theorem 1 is omplete.There are several variations to this problem that one might onsider,su h as using bases other than 10. Another variant is to restri t the alphabetto the two letters A and B. We use the terms 2-attainable and 2-blo ker inthis ontext. Using te hniques similar to those presented, we obtained thefollowing result.Theorem 3 A number is 2-attainable if and only if it divides one of 24, 50,60, 70, 80, or 90.The negative part of the proof required �nding a 2-blo ker for ea hd ∈ {28, 36, 48, 120, 175}. The reader might enjoy �nding them; they areall short, of length at most 7.A knowledgmentWe thank Bill Sands for bringing this problem to our attention.Referen es[1℄ N. Kildonan, Problem 1859 (Solution), Crux Mathemati orum, 20:6,June 1994, pp. 168{170.Robert Israel Stephen MorrisUniversity of British Columbia Newbury, Berkshire, EnglandVan ouver, BC, Canada [email protected]
Stan WagonMa alester CollegeSt. Paul, MN, [email protected]
478A Limit of an Improper Integral Dependingon One Parameter
Iesus C. DinizIn this arti le we will al ulate the following limit of an improper inte-gral that depends on one parameter λ ∈ R+
limλ→0+
∫ ∞
0
exp(
−λK(x + l)n)
λKnxn−1 dx ,where K and l are positive real numbers and n is a positive integer.This is an interesting example where one annot inter hange the orderof the limit and the integral. Through a ni e appli ation of elementary toolssu h as hange of variables in integrals and the Binomial Theorem, one isable to obtain this limit.This problem arose in the al ulation of a lower bound for a proba-bility of theoreti al interest in the study of multidimensional Poisson pointpro esses, namelylim
λ→0+
∫ ∞
0
exp(
−λvn(1)[
(r + l)n − rn])
λvn(1)nrn−1 exp(
−λvn(1)rn)
dr ,where vn(1) is the volume of the n{dimensional unit ball, λ is the Poissonintensity, l is the distan e between two distinguished points in Rn, and ris the distan e from the �rst point to the losest o urren e in the Poissonpoint pro ess. We shall show that this limit is equal to 1.Proposition For all positive real numbers K and l, and for ea h positive in-teger n we have
limλ→0+
∫ ∞
0
exp(
−λK[
(x + l)n])
λKnxn−1 dx = 1 .Proof: The ase n = 1 is a straightforward al ulation:lim
λ→0+
∫ ∞
0
exp(
−λK(x + l))
λK dx
= limλ→0+
(
− exp(
−λK(x + l))
∣
∣
∣
∞
0
)
= limλ→0+
exp (−λKl) = 1 .Hen eforth, we take n > 1.Copyright c© 2008 Canadian Mathemati al So ietyCrux Mathemati orum with Mathemati al Mayhem, Volume 34, Issue 8
479Making �rst the hange of variable
v = Kxn , dv = Knxn−1 dx ,in the integral we obtain∫ ∞
0
λ exp[
−λ(
v1n + (lnK)
1n
)n]
dv .Changing variables again, this time a ording tou
1n = v
1n + (lnK)
1n , dv =
(
u
v
)( 1n
−1)du ,we obtain
limλ→0+
∫ ∞
0
λ exp[
−λ(
v1n + (lnK)
1n
)n]
dv
= limλ→0+
∫ ∞
lnKλ exp (−λu)
(
u
v
)( 1n
−1)du
= limλ→0+
∫ ∞
lnKλ exp (−λu)
[
1 −(
lnKu
)
1n
]n−1
du ,whi h by the Binomial Theorem be omeslim
λ→0+
∫ ∞
lnKλ exp (−λu)
n−1∑
k=0
(
n − 1
k
)
(−1)k(
lnKu
)
k
n
du
= 1 + limλ→0+
n−1∑
k=1
(
n − 1
k
)∫ ∞
lnKλ exp (−λu) (−1)k
(
lnKu
)k
n
du .After yet another hange of variable,
w = u1n , dw =
1
nu( 1
n−1) du ,the pre eding expression be omes
1 + limλ→0+
n−1∑
k=1
(
n − 1
k
)
n(
−lK 1n
)k
λ
∫ ∞
lK1n
exp(
−λwn)
wn−1−k dw .It remains to show that
limλ→0+
λ
∫ ∞
lK1n
exp (−λwn) wn−1−k dw (1)is zero for ea h k ∈ {1 , 2, . . . , n − 1}.
480To do this, we make a �nal hange of variables
z = λ1n w , dz = λ
1n dw ,whi h yields
λ
∫ ∞
lK1n
exp (−λwn) wn−1−k dw
= λk
n
∫ ∞
(λlnK)1n
exp(
−zn)
zn−1−k dz
< λk
n
∫ ∞
0
exp(
−zn)
zn−1−k dz . (2)It now suÆ es for us to show that the last integral in (2) is �nite forea h k ∈ {1, 2, . . . , n − 1}.SettingC =
∫ 1
0
exp (−zn) zn−1−k dzand noting that if z ≥ 1, then zn−1−k ≤ zn−1, we �nally obtain∫ ∞
0
exp(
−zn)
zn−1−k dz ≤ C +
∫ ∞
1
exp(
−zn)
zn−1 dz
= C +1
ne,whi h is a �nite number. This ompletes the proof.
AcknowledgmentsThis work was supported by FAPESP grant 0415864-1.The author also thanks J.C.S. de Miranda for many dis ussions aboutthis problem. Iesus C DinizInstitute of Mathemati s and Statisti sUniversity of S~ao PauloS~ao Paulo, [email protected]
481Sliding Down In lines with Fixed Des entTime: a Converse to Galileo's Law of Chords
Je� BabbSuppose that a ve tor is an hored at the origin and lies along the posi-tive x-axis. Consider rotating the ve tor ounter lo kwise about the originthrough an angle A, with 0 < A < π. Consider a parti le of mass m whi his initially at rest and then slides, under gravity and without fri tion, from astarting point on the in lined ve tor down towards the origin. Let DA denotethe distan e of the starting point from the origin. For ea h value of A, sup-pose that DA is hosen to ensure that the parti le requires exa tly T se ondsto rea h the origin. Determine the urve hara terized by the starting pointsof the parti les.For the parti le under onsideration, let dA(t) be the distan e travelledalong the ve tor at time t, vA(t) = d′A(t) be the velo ity along the ve tor attime t, and let aA(t) = v′
A(t) be the a eleration along the ve tor at time t.Note that by de�nition, DA = dA(T ).If aA(t) is some onstant K, thendA(t) =
K
2t2 . (1)This may be on�rmed by integrating aA(t) twi e with respe t to time andapplying the initial onditions dA(0) = 0 and vA(0) = 0 to obtain zero forboth onstants of integration.Sin e the parti le is sliding down a fri tionless in line in the Earth'sgravitational �eld, the omponent of a eleration along the in line is
K = g sin A, where g is the a eleration due to gravity at the Earth's surfa e.If the des ent time is �xed at T se onds, thenDA = dA(T ) =
g
2T 2 sin A . (2)A point (r, θ) in polar oordinates may be expressed in Cartesian oordinatesas (x, y), where x = r cos θ and y = r sin θ. Consider the following equa-tion, whi h is expressed in polar oordinates as
r = 2c sin θ , (3)where r > 0 and 0 < θ < π. Multiplying both sides of equation (3) by ryieldsr2 = 2cr sin θ (4)Copyright c© 2008 Canadian Mathemati al So ietyCrux Mathemati orum with Mathemati al Mayhem, Volume 34, Issue 8
482By setting x = r cos θ and y = r sin θ, equation (4) may be re-expressed inCartesian oordinates as
x2 + y2 = 2cy .Subtra ting 2cy from ea h side and ompleting the square on y2 −2cy yieldsx2 + (y − c)2 = c2 (5)whi h is the equation of a ir le of radius c entred at the point (0, c).Thus, as depi ted in the �gure at right,the lo us of points de�ned by equation (2) isa ir le resting upon the origin (0, 0), entredat (0,
g
4T 2) on the verti al axis, and of radius
g
4T 2.On an histori al note, the motivation forthis problem arose while onsidering the Lawof Chords, whi h was stated and proven byGalileo Galilei in his 1638 masterpie e Dia-logues Con erning Two New S ien es. Galileo onsidered rates of des ent along a verti al ir- le and, with his Proposition VI, establishedthe Law of Chords (see [1℄, p. 212):
..
.
..
..
.
..
.
..
..
.
..
.
..
..
..
..
..
..
..
..
..
.
..
..
..
..
.
..
.
..
.
..
.
..
.
..
.
..
..
..
..
..
..
..
..
..
.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
................................................................................................................
...................................................................................................................................................................................................................
..
...
...
....
...
...
...
..
...
..
..
..
..
..
...
..
..
..
.
..
..
.
..
..
.
..
..
.
..
.
..
.
..
..
.
..
.
..
..
.
..
.
..
..
.
...
..
..
..
..
..
..
..
..
..
...
...
...
...
...
...
...
....................................................................................................................................................................................................................
.....................................................................................................................................................................................................................................................
r
.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
................................................................................................................................................................................................................................
..
.
..
..
..
.
..
..
..
..
.
..
..
.
..
.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
A
(
0,g
4T 2)
If from the highest or lowest point in a verti al ir le there bedrawn any in lined planes meeting the ir umferen e, the timesof des ent along these hords are ea h equal to the other.Galileo's proof of the Law of Chords is presented via a series of geomet-ri propositions, whi h require familiarity with many of Eu lid's theorems.The question the author wished to address was whether the verti al ir le is the only urve with the property that des ent time to the lowestpoint on the urve is onstant for all hords. This paper demonstrates thatthe verti al ir le, or one of its omponent ar s interse ting at the lowestpoint of the ir le, is indeed the only su h urve.Referen es[1℄ Galilei, Galileo. 1638/1952 Dialogues Con erning Two New S ien es,translated by H. Crew and A. de Salvio, En y lopaedia Britanni a,Chi ago, 1952. Je� BabbDepartment of Mathemati s and Statisti sUniversity of WinnipegWinnipeg, MB, R3B [email protected]
483PROBLEMSToutes solutions aux probl �emes dans e num�ero doivent nous parvenir au plustard le 1er juin 2009. Une �etoile (⋆) apr �es le num�ero indique que le probl �eme a �et �esoumis sans solution.Chaque probl �eme sera publi �e dans les deux langues oÆ ielles du Canada(anglais et fran� ais). Dans les num�eros 1, 3, 5 et 7, l'anglais pr �e �edera le fran� ais,et dans les num�eros 2, 4, 6 et 8, le fran� ais pr �e �edera l'anglais. Dans la se tion dessolutions, le probl �eme sera publi �e dans la langue de la prin ipale solution pr �esent �ee.La r �eda tion souhaite remer ier Rolland Gaudet, de Coll �ege universitaire deSaint-Bonifa e, Winnipeg, MB et Jean-Mar Terrier, de l'Universit �e de Montr �eal,d'avoir traduit les probl �emes.
3371. Corre tion. Propos �e par George Tsintsifas, Thessalonique, Gr �e e.Soit ABC un triangle de ot �es respe tifs a, b et c, et soit M un deses points int �erieur. Les droites AM , BM et CM oupent respe tivementles ot �es oppos �es aux points A1, B1 et C1. Les droites passant par M etperpendi ulaires aux ot �es oupent respe tivement BC, CA et AB en A2,B2, et C2. Soit p1, p2 et p3 les distan es respe tives de M aux ot �es BC,CA et AB. Montrer que
[A2B2C2]
[A1B1C1]=
(ap1 + bp2)(bp2 + cp3)(cp3 + ap1)
2a2b2c2
(
a
p1
+b
p2
+c
p3
) ,o �u [KLM ] d �esigne l'aire du triangle KLM .3389. Propos �e par Mih�aly Ben ze, Brasov, Roumanie.Pour a ∈ R, la suite (xn) est d �e�nie par x0 = a et xn+1 = 4xn − x2npour tout n ≥ 0. Montrer qu'il existe un nombre in�ni de valeurs a ∈ Rtelles que la suite (xn) est p �eriodique.3390. Propos �e par Mih�aly Ben ze, Brasov, Roumanie.Montrer que si A, B, C et D sont les solutions de
X2 =
(
3 −55 8
) ,alors A2007 +B2007 +C2007 +D2007 = O, o �u O est la matri e nulle de taille2 × 2.3391. Propos �e par Mi hel Bataille, Rouen, Fran e.Soit ABCD un quadrilat �ere onvexe tel que AC et BD se oupent �aangle droit en P , et soit respe tivement I, J , K et L les milieux de AB, BC,CD et DA. Montrer que les er les (PIJ), (PJK), (PKL) et (PLI) sont ongruents si et seulement si ABCD est y lique.
4843392. Propos �e par Mi hel Bataille, Rouen, Fran e.Soit A, B, C, D et E on y liques ave respe tivement V et W sur lesdroites AB et AD. Montrer que si la droite CE, la parall �ele �a CB par Vet la parall �ele �a CD par W sont on ourantes, alors les triangles EV B etEWD sont semblables. La r �e iproque est-elle vraie ?3393. Propos �e par Dragoljub Milo�sevi� et G. Milanova , Serbie.Soit le triangle ABC, o �u a = BC, b = AC, c = AB et o �u s est ledemi p �erim �etre. Montrer que
y + z
x· A
a(s − a)+
z + x
y· B
b(s − b)+
x + y
z· C
c(c − a)≥ 9
s2,
o �u les angles A, B et C sont mesur �es en radians et x, y et z sont des nombresr �eels positifs quel onques.3394. Propos �e par Dragoljub Milo�sevi� et G. Milanova , Serbie.Soit ABCD un t �etra �edre ; d �esignons par hA et mA les longueurs res-pe tives de la hauteur et de la m�ediane issues du sommet A sur la fa eoppos �ee BCD. Si V est le volume du t �etra �edre, montrer que(hA + hB + hC + hD)(m2
A + m2B + m2
C + m2D) ≥ 128
√3
V .3395. Propos �e par Tai hi Maekawa, Takatsuki, Pr �efe ture d'Osaka, Japon.Soit le triangleABC ; d �enotons parH son ortho entre et parR le rayonde son er le ir ons rit. Montrer que 4R3 − (l2+m2+n2)R − lmn = 0,o �u AH = l, BH = m et CH = n.3396. Propos �e par Neven Juri� , Zagreb, Croatie.Soit n un entier positif et, pour i, j et k dans {1, 2, . . . , n}, posons
aijk = 1 + mod(k − i + j − 1, n) + nmod(i − j + k − 1, n)
+ n2mod(i + j + k − 2, n) ,o �u mod(a, n) est le r �esidu de a modulo n, hoisi parmi 0, 1, . . . , n − 1.Pour quels n le ube portant les valeurs aijk est-il un ube magique ? (I i,le mot \magique" se r �ef �ere au fait que la somme des aijk est onstante sideux des indi es restent �xes et l'autre varie, et de plus que les sommes desdiagonales prin ipales du ube sont aussi �egales �a ette meme onstante.)
4853397. Propos �e par Jos �e Luis D��az-Barrero, Universit �e Polyte hnique deCatalogne, Bar elone, Espagne.�Evaluer
limn→∞
1
n2
∫ n
0
√n2 − x2
2 + x−xdx .3398. Propos �e par Bru e Shawyer, Universit �e Memorial de Terre-Neuve,St. John's, NL.Soit l' �equation
⌊
n
10
⌋
+(
n − 10⌊
n
10
⌋)
· 10⌊log10 n⌋ =2n
3,(a) montrer quen = 5294117647058823 est une solution de ette �equation,(b) ⋆ d �eterminer toute autre solution enti �ere et positive de ette �equation.3399. Proposed by Vin ent�iu R�adules u, Universit �e de Craiova, Craiova,Roumanie.Montrer qu'il n'existe au une fon tion positive et deux fois di� �erentiable
f : [0, ∞) → R telle que f(x)f ′′(x) + 1 ≤ 0 pour tout x ≥ 0.3400. Propos �e par Yakub N. Aliyev, Universit �e d'Etat de Bakou, Bakou,Azerba��djan.Pour des entiers positifs m et k, posons (m)k = m(1 + 10 + 102 +· · · + 10k−1) ; par exemple, (1)2 = 11 et (3)4 = 3333. D �eterminer tous lesnombres r �eels α tels que
⌊
10n√
(1)2n + α⌋
= (3)2n −⌊
5 − 9α
6
⌋
est valide pour tout entier positif n, o �u ⌊x⌋ d �enote le plus grand entier ned �epassant pas x..................................................................3371. Corre tion. Proposed by George Tsintsifas, Thessaloniki, Gree e.Let ABC be a triangle with a, b, and c the lengths of the sides oppositethe verti es A, B, and C, respe tively, and let M be an interior point of△ABC. The lines AM , BM , and CM interse t the opposite sides at thepoints A1, B1, and C1, respe tively. Lines through M perpendi ular to thesides of △ABC interse t BC, CA, and AB at A2, B2, and C2, respe tively.Let p1, p2, and p3 be the distan es from M to the sides BC, CA, and AB,respe tively. Prove that
[A2B2C2]
[A1B1C1]=
(ap1 + bp2)(bp2 + cp3)(cp3 + ap1)
2a2b2c2
(
a
p1
+b
p2
+c
p3
) ,where [KLM ] denotes the area of triangle KLM .
4863389. Proposed by Mih�aly Ben ze, Brasov, Romania.For a ∈ R de�ne a sequen e (xn) by x0 = a and xn+1 = 4xn − x2
nfor all n ≥ 0. Prove that there exist in�nitely many a ∈ R su h that thesequen e (xn) is periodi .3390. Proposed by Mih�aly Ben ze, Brasov, Romania.Prove that if A, B, C, and D are the solutions ofX2 =
(
3 −55 8
) ,then A2007 +B2007 +C2007 +D2007 = O, where O is the 2× 2 zero matrix.3391. Proposed by Mi hel Bataille, Rouen, Fran e.Let ABCD be a onvex quadrilateral su h that AC and BD interse tin right angles at P , and let I, J , K, and L be the mid-points of AB, BC,CD, and DA, respe tively. Show that the ir les (PIJ), (PJK), (PKL),and (PLI) are ongruent if and only if ABCD is y li .3392. Proposed by Mi hel Bataille, Rouen, Fran e.Let A, B, C, D, and E be on y li with V and W on the lines AB andAD, respe tively. Show that if the line CE, the parallel to CB through V ,and the parallel to CD through W are on urrent, then triangles EV B andEWD are similar. Does the onverse hold?3393. Proposed by Dragoljub Milo�sevi� and G. Milanova , Serbia.Let ABC be a triangle with a = BC, b = AC, c = AB, and semiper-imeter s. Prove that
y + z
x· A
a(s − a)+
z + x
y· B
b(s − b)+
x + y
z· C
c(c − a)≥ 9
s2,where the angles A, B, and C are measured in radians and x, y, and z areany positive real numbers.3394. Proposed by Dragoljub Milo�sevi� and G. Milanova , Serbia.Let ABCD be a tetrahedron with hA and mA the lengths of the alti-tude and the median from vertex A to the opposite fa e BCD, respe tively.If V is the volume of the tetrahedron, prove that
(hA + hB + hC + hD)(m2A + m2
B + m2C + m2
D) ≥ 128√3
V .3395. Proposed by Tai hi Maekawa, Takatsuki City, Osaka, Japan.Let triangle ABC have ortho entre H and ir umradius R. Prove that4R3 − (l2 + m2 + n2)R − lmn = 0, where AH = l, BH = m, andCH = n.
4873396. Proposed by Neven Juri� , Zagreb, Croatia.Let n be a positive integer, and for i, j, and k in {1, 2, . . . , n} let
aijk = 1 + mod(k − i + j − 1, n) + nmod(i − j + k − 1, n)
+ n2mod(i + j + k − 2, n) ,where mod(a, n) is the residue of a modulo n in the range 0, 1, . . . , n − 1.For whi h n is the ube with entries aijk a magi ube? (Here "magi " meansthat the sum of aijk is onstant if two indi es are �xed and the third indexvaries, and also the sums along the great diagonals of the ube are equal tothis onstant.)3397. Proposed by Jos �e Luis D��az-Barrero, Universitat Polit �e ni a deCatalunya, Bar elona, Spain.Evaluatelim
n→∞
1
n2
∫ n
0
√n2 − x2
2 + x−xdx .
3398. Proposed by Bru e Shawyer, Memorial University of Newfound-land, St. John's, NL.Given the equation⌊
n
10
⌋
+(
n − 10⌊
n
10
⌋)
· 10⌊log10 n⌋ =2n
3,(a) show that n = 5294117647058823 is a solution,(b) ⋆ �nd all other positive integer solutions of the equation.3399. Proposed by Vin ent�iu R�adules u, University of Craiova, Craiova,Romania.Prove that there does not exist a positive, twi e di�erentiable fun tion
f : [0, ∞) → R su h that f(x)f ′′(x) + 1 ≤ 0 for all x ≥ 0.3400. Proposed by Yakub N. Aliyev, Baku State University, Baku, Azer-baijan.For positive integersm and k let (m)k = m(1+10+102+· · ·+10k−1),for example, (1)2 = 11 and (3)4 = 3333. Find all real numbers α su h that⌊
10n√
(1)2n + α⌋
= (3)2n −⌊
5 − 9α
6
⌋
holds for ea h positive integer n, where ⌊x⌋ is the greatest integer not ex- eeding x.
488SOLUTIONSNo problem is ever permanently losed. The editor is always pleasedto onsider for publi ation new solutions or new insights on past problems.
3229. [2007 : 170, 172; 2007 : 179{181℄ Proposed by Mih�aly Ben ze,Brasov, Romania.(a) Let x and y be positive real numbers, and let n be a positive integer.Prove that(x + y)n
n∑
k=0
1(n
k
)
xn−kyk≥ n + 1 + 2
n∑
i=1
n−i∑
k=0
(nk
)
( nk+i
) ≥ (n + 1)2.(b)⋆ Let x1, x2, . . . , xk be positive real numbers, and let n be a positiveinteger. Determine the minimum value of
(x1 + x2 + · · · + xk)n∑
i1+···+ik=n
i1, . . . , ik ≥ 0
i1!i2! · · · ik!
n!xi11 xi2
2 · · · xik
k
.Solution to (b) by Sergey Sadov, Memorial University of Newfoundland,St. John's, NL.We may assume that x1 +x2 + · · ·+xk = 1 without loss of generality.The minimum o urs when x1 = x2 = · · · = xk =
1
k. We will obtainthis as a onsequen e of a more general proposition. For onvenien e, let
X = (x1, x2, . . . , xk) be a ve tor in Rk+ (that is, xi > 0 for all i) and let
Q = (q1, q2, . . . , qk) be a multi-index with non-negative integer entries qi.Let|Q| =
n∑
i=1
qi ;|X| =
n∑
i=1
xi ;Q! = q1!q2! · · · qk! ;
XQ = xq1
1 xq2
2 · · · xqk
k .Finally, let ∆ = {X ∈ Rk+ : |X| = 1} be the positive unit simplex in R
k ofdimension k − 1.
489Theorem 1 With the above notation, let
P (X) =∑
|Q|=n
cQ
XQbe a homogeneous rational fun tion of degree −n in the variables xi. Sup-pose that cQ ≥ 0 for ea h Q and that P (X) is a symmetri fun tion, thatis, inter hanging any xi and xj does not hange the value of P (X). Thenthe minimum value of P (X) over the simplex ∆ exists and is attained whenx1 = x2 = · · · = xk =
1
k.Proof. If P (X) is identi ally zero, then there is nothing to prove, so we as-sume at least one oeÆ ient cQ is not zero. Note that P (x) has a minimumvalue, sin e P (X) is ontinuous on ∆ and tends to in�nity as any of the
xi approa hes 0. Suppose that the minimum o urs at a point with at leasttwo unequal oordinates. Without loss of generality (due to the symmetry ofP (X)) we may assume that x1 > x2. We will show that by slightly de reas-ing x1 and slightly in reasing x2 (while keeping all other variables and thesum x1 + x2 un hanged) the value of P (X) will be ome smaller, ontrary toour assumption. Fixing x3, x4, . . . , xn auses P (X) to be ome a symmetri fun tion of two variables
F (u, v) = P (u, v, x3, . . . , xn) =∑
j+s≤n
Aj,s
ujvs,where the oeÆ ients Aj,s depend on x3, x4, . . . , xn. Note that ea h Aj,sis non-negative and at least one of these is positive, and that Aj,s = As,j.Thus, F (u, v) is a linear ombination with non-negative oeÆ ients, not allzero, of fun tions of the form
Fj,s(u, v) =1
ujvs+
1
usvj,where j and s are non-negative integers with j+s ≤ n. Now let u(t) = x1−tand v(t) = x2 + t, so that du
dt= −1 and dv
dt= 1. It suÆ es to prove thatthe \time derivative" d
dtFj,s
(
u(t), v(t)) is negative at t = 0. We have
d Fj,s
(
u(t), v(t))
dt=
∂Fj,s
(
u, v)
∂v− ∂Fj,s
(
u, v)
∂u
=
(
j
u− s
v
)
1
ujvs+
(
s
u− j
v
)
1
usvj.Di�erentiating on e again, we obtain
d2Fj,s
(
u(t), v(t))
dt2=
(
j
u2+
s
v2+
(
j
u− s
v
)2)
1
ujvs+
(
j
v2+
s
u2+
(
j
v− s
u
)2)
1
usvj.
490Hen e, d2F
dt2> 0. Sin e dF
dt= 0 when u = v (this o urs when t =
u − v
2),it follows that dF
dt< 0 when t = 0. We have obtained a ontradi tion byassuming x1 > x2 at a point a hieving the minimum. This proves that theminimum o urs when all the xi are equal.It follows that the minimum sought in part (b) is
kn∑
i1+···+ik=n
i1,...,ik≥0
i1!i2! . . . ik!
n!.
No other solutions to part (b) were re eived.Regarding part (a) Sadov remarks thatmin
x+y=1
n∑
k=0
1(n
k
)
xn−kyk= 2n
n∑
j=0
1(n
j
) = n + 1 + 2
n−1∑
j=0
n−j∑
i=0
(nj
)
( ni+j
) ,where the �rst equality follows Theorem 1 and the last expression is the minimum obtainedby Bataille [2007 : 179{181℄. He notes that the �rst equality yields a minimum of at least2n+1, onsiderably improving the lower bound of (n + 1)2 if n > 4. He observes that ifbi =
(ni
)
xn−iyi , then by the AM{HM Inequality(
1
n + 1
∑ 1
bi
)−1
≤ 1
n + 1
∑
bi =(x + y)n
n + 1,hen e (x + y)2
∑
b−1
i ≥ (n + 1)2, whi h yields a qui k proof of part (a).Sadov omments that the fun tion P(X) in Theorem 1 is S hur- onvex, referring to [1℄for the de�nition of this term and appli ations. He indi ates that the AM-GM Inequality and theAM{HM Inequality an be obtained by taking P(X) = (x1x2 · · · xn)−1 and P(X) =∑
xiin Theorem 1, respe tively.He mentions that parti ular ases (and other theorems of a more general nature) ofTheorem 1 an be found in [2℄, Chapter 3, Se tion G, Examples G.1.k and G.1.m; though hebelieves that Theorem 1 is present somewhere in the existing literature.Finally, he refers to [3℄, Se tion 2.18, for a treatment of (the related) Muirhead'sInequality.Referen es[1℄ M.L. Clevenson and W. Watkins, Majorization and the Birthday Inequality, Math.Magazine, vol. 64, No. 3 (1991), pp. 183{188.[2℄ A. Marshall and I. Olkin, Inequalities: Theory of Majorization and Its Appli ations,A ademi Press, 1979.[3℄ G.H. Hardy, J.E. Littlewood, and G. Polya, Inequalities, Cambridge University Press,1952.3289. [2007 : 485, 487℄ Proposed by Virgil Ni ula, Bu harest, Romania.Let ABC be a triangle for whi h there exists a point D in its interiorsu h that ∠DAB = ∠DCA and ∠DBA = ∠DAC. Let E and F be pointson the lines AB and CA, respe tively, su h that AB = BE and CA = AF .Prove that the points A, E, D, and F are on y li .
491A omposite of similar solutions by John G. Heuver, Grande Prairie, AB andGeorge Tsapakidis, Agrinio, Gree e.TrianglesADC andBDA are similar (be ause their angles are assumedto be equal), when e
CD
AD=
CA
AB=
2CA
2AB=
CF
AE.It follows that △DCF ∼ △DAE (∠DCF = ∠DCA = ∠DAB = ∠DAE,while the adja ent sides are proportional from the previous step), so that
∠AFD (= ∠CFD) = ∠AED. Noting that E and F both lie on the sameside of AD, we on lude that A, E, F , and D lie on the same ir le, asdesired.Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; �SEFKETARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; RICARDOBARROSO CAMPOS, University of Seville, Seville, Spain; MICHEL BATAILLE, Rouen,Fran e; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; APOSTOLISK. DEMIS, Varvakeio High S hool, Athens, Gree e; OLIVER GEUPEL, Br �uhl, NRW, Germany;GEOFFREY A. KANDALL, Hamden, CT, USA; ANDREA MUNARO, student, University ofTrento, Trento, Italy; PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU,Com�ane�sti, Romania; and the proposer.Bataille omments that D an be onstru ted as the point inside△ABC where the ir leFAE interse ts the ir le through B, C, and the ir um entre O. It lies on the �rst ir le bythe result of this problem. It lies on the se ond ir le be ause ∠BOC = ∠BDC = 2 ∠BACas follows: on the one hand the angle at A is ins ribed in the ir le BAC that is entred at Oand is therefore 1
2∠BOC; on the other hand the similar triangles ADC and BDA �t togetherat A and at D in su h a way that the two exterior angles at D (that form ∠BDC) sum to twi ethe sum of the two interior angles at A (that form ∠BAC).
3290. [2007 : 485, 487℄ Proposed by Virgil Ni ula, Bu harest, Romania.Let ABCD be a trapezoid with AD ‖ BC. Denote the lengths ofAD and BC by a and b, respe tively. Let M be the mid-point of CD, andlet P and Q be the mid-points of AM and BM , respe tively. If N is theinterse tion of DP and CQ, prove that N belongs to the interior of △ABMif and only if 1
3<
a
b< 3.Solution by Joel S hlosberg, Bayside, NY, USA.There is a misleading subtlety in the statement of the problem: we shallsee that the on lusion fails should AD = BC; in other words, our trapezoid
ABCD must not be a parallelogram.Sin e all of the onditions of the problem are invariant under an aÆnetransformation, we an assume without loss of generality thatAB ⊥ AD and AB = 4 .We therefore introdu e a Cartesian oordinate system with the origin at A,and with B on the positive x{axis and D on the positive y{axis; then A, B,
492C, and D have oordinates
A(0, 0) , B(4, 0) , C(4, b) , D(0, a) ,where a and b are positive. It follows that the oordinates of M , P , and QareM
(
2,a + b
2
) , P
(
1,a + b
4
) , Q
(
3,a + b
4
) ,so that the lines DP and CQ satisfyy =
(
b − 3a
4
)
x + a and y =
(
3b − a
4
)
x + a − 2b ,when e N = DP ∩ CQ has oordinatesN
(
4b
a + b,(a − b)2
a + b
) .A point in the plane is on the same side of AB as M if and only if itsy{ oordinate is positive. The y{ oordinate of N satis�es
(a − b)2
a + b≥ 0 ,with equality if and only if a = b. Sin e P = DP ∩ AM has x{ oordinate
1, a point on line DP is on the same side of AM as B if and only if itsx{ oordinate ex eeds 1. Lastly, sin e Q = CQ ∩ BM has x{ oordinate 3,a point on line CQ is on the same side of BM as A if and only if itsx{ oordinate is less than 3. Therefore, N is within the interior of △ABMif and only if a 6= b and 1 <
4b
a + b< 3. This inequality is equivalent to
1
3<
a
b< 3 ,whi h is the desired inequality; it is equivalent to N belonging to the interiorof △ABM as long as a 6= b.Also solved by ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHELBATAILLE, Rouen, Fran e; CHIP CURTIS,Missouri Southern State University, Joplin, MO, USA;FRANCISCO JAVIER GARC�IA CAPIT �AN, IES �Alvarez Cubero, Priego de C �ordoba, Spain; OLIVERGEUPEL, Br �uhl, NRW, Germany; PETER Y. WOO, Biola University, La Mirada, CA, USA; TITUZVONARU, Com�ane�sti, Romania; and the proposer.
3291. [2007 : 485, 487℄ Proposed by Virgil Ni ula, Bu harest, Romania.Let ABC be an isos eles triangle with AB = AC. Find all points Psu h that the sum of the squares of the distan es of the points A, B, and Cfrom any line through P is onstant.
493Solution by Mi hel Bataille, Rouen, Fran e.We will introdu e a Cartesian oordinate system. Label the points asA(0, a), B(−b, 0), C(b, 0), andP (u, v), where a and b are positive and u andv are variables. The equation of a line ℓ through P is x cos θ + y sin θ = p,where p = u cos θ + v sin θ and θ is an arbitrary real number. Let d(Q, ℓ)denote the distan e between a point Q and the line ℓ, and let
S = d(A, ℓ)2 + d(B, ℓ)2 + d(C, ℓ)2 .We then have d(A, ℓ)2 = (a sin θ − p)2, d(B, ℓ)2 = (p + b cos θ)2, andd(C, ℓ)2 = (p − b cos θ)2. After a simple al ulation, we obtain
S = (cos 2θ)
(
3(u2 − v2)
2+ av − a2
2+ b2
)
+ (sin 2θ)(3uv − au) +3(u2 + v2)
2− av +
a2
2+ b2 .
The number S is independent of θ if and only if3uv − au = 0 and 3(u2 − v2) + 2av − a2 + 2b2 = 0 ,whi h gives
u = 0 and − 3v2 + 2av − a2 + 2b2 = 0 , (1)orv =
a
3and 3
(
u2 − a2
9
)
+2a2
3− a2 + 2b2 = 0 . (2)Now, (1) is satis�ed if and only if 0 < a ≤ b
√3 (that is, ∠B = ∠C ≤ 60◦)and the oordinates of P are
(
0,a +
√
2(3b2 − a2)
3
) or (
0,a −
√
2(3b2 − a2)
3
) .Similarly, (2) is satis�ed if and only if a ≥ b
√3 (that is, ∠B = ∠C ≥ 60◦)and the oordinates of P are
(√
2(a2 − 3b2)
3,
a
3
) or (
−√
2(a2 − 3b2)
3,
a
3
) .In on lusion, if ABC is equilateral, only the entre of ABC is a solutionfor P . Otherwise, there are two solutions for P :P
(
0,a ±
√
2(3b2 − a2)
3
)
(if ∠A > 60◦) ,P
(
±√
2(a2 − 3b2)
3,
a
3
)
(if ∠A < 60◦) .
494In both ases, the two solutions are symmetri about the entroid of thetriangle; in the �rst ase, the two points lie on the median through A andin the se ond ase the two points lie on a line parallel to BC and passingthrough the entroid of △ABC.Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; ROY BARBARA,Lebanese University, Fanar, Lebanon; CHIP CURTIS, Missouri Southern State University,Joplin, MO, USA; FRANCISCO JAVIER GARC�IA CAPIT �AN, IES �Alvarez Cubero, Priego deC �ordoba, Spain; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; V �ACLAVKONE �CN �Y, Big Rapids, MI, USA (2 solutions); SKIDMORE COLLEGE PROBLEM SOLVINGGROUP, Skidmore College, Saratoga Springs, NY, USA; GEORGE TSAPAKIDIS, Agrinio, Gree e;PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU, Com�ane�sti, Romania;and the proposer.
3292. [2007 : 485, 488℄ Proposed by Mih�aly Ben ze, Brasov, Romania.Let a, b, c, and d be arbitrary real numbers. Show that11a2 + 11b2 + 221c2 + 131d2 + 22ab + 202cd + 48c + 6
≥ 98ac + 98bc + 38ad + 38bd + 12a + 12b + 12d .Solution by Oliver Geupel, Br �uhl, NRW, Germany.The proof is by ontradi tion. Assume that11a2 + 11b2 + 221c2 + 131d2 + 22ab + 202cd + 48c + 6
< 98ac + 98bc + 38ad + 38bd + 12a + 12b + 12d .We onsider the quadrati f(x) = 11x2 + px + q withp = 22b − 98c − 38d − 12 ,q = 11b2 + 221c2 + 131d2 + 202cd + 48c + 6 − 98bc − 38bd
− 12b − 12d .Then f(a) < 0, and the leading oeÆ ient of f is positive, hen e f has twodistin t real roots; that is, the dis riminant of f is positive. By omputingthe dis riminant, we �nd p2−44q = −120(c+6d−1)2 ≤ 0, a ontradi tion.Also solved by APOSTOLIS K. DEMIS, Varvakeio High S hool, Athens, Gree e; MICHELBATAILLE, Rouen, Fran e; CHIP CURTIS,Missouri Southern State University, Joplin, MO, USA;RICHARD I. HESS, Ran ho Palos Verdes, CA, �E-U; WALTHER JANOUS, Ursulinengymnasium,Innsbru k, Austria; KEE-WAI LAU, Hong Kong, China; JOEL SCHLOSBERG, Bayside, NY, USA;STAN WAGON, Ma alester College, St. Paul, MN, USA; PETER Y. WOO, Biola University, LaMirada, CA, USA; and the proposer.
4953293. [2007 : 485, 488℄ Proposed by Mih�aly Ben ze, Brasov, Romania.Prove that
n∏
k=1
arcsin
(
9k + 2√27k3 + 54k2 + 36k + 8
)
arctan
(
1√3k + 1
) = 3n .Composite of similar solutions by Mi hel Bataille, Rouen, Fran e andDouglass L. Grant, Cape Breton University, Sydney, NS, modi�ed by theeditor.For ea h k = 1, 2, . . . , n let Pk denote the orresponding fa tor underthe produ t sign. We prove that in fa t, Pk = 3 for ea h k.Note �rst that 27k3 + 54k2 + 36k + 8 = (3k + 2)3. For �xed k, letθ = tan−1
(
1√3k + 1
). Then tan θ =1√
3k + 1implies sin θ =
1√3k + 2
.Sin e tan−1 is an in reasing fun tion, we have0 < θ ≤ tan−1
(
1
2
)
< tan−1
(
1√
3
)
=π
6,
hen e, 0 < 3θ <π
2.From
sin(3θ) = 3 sin θ − 4 sin3 θ
=3
√3k + 2
− 4√
(3k + 2)3=
9k + 2√
(3k + 2)3,we obtain
sin−1
(
9k + 2√
27k3 + 54k2 + 36k + 8
)
= sin−1
(
9k + 2√
(3k + 2)3
)
= sin−1(sin 3θ) = 3θ ,and it follows that Pk =
3θ
θ= 3.Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; �SEFKETARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; DIONNE BAILEY,ELSIE CAMPBELL, CHARLES DIMINNIE, and KARL HAVLAK, Angelo State University, SanAngelo, TX, USA; ROY BARBARA, Lebanese University, Fanar, Lebanon; CHIP CURTIS,Missouri Southern State University, Joplin, MO, USA; OLEH FAYNSHTEYN, Leipzig,Germany; OLIVER GEUPEL, Br �uhl, NRW, Germany; SALEM MALIKI �C, student, SarajevoCollege, Sarajevo, Bosnia and Herzegovina; HENRY RICARDO, Medgar Evers College
(CUNY ), Brooklyn, NY, USA; JOEL SCHLOSBERG, Bayside, NY, USA; GEORGE TSAPAKIDIS,Agrinio, Gree e; PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU,Com�ane�sti, Romania; and the proposer.
4963294. [2007 : 486, 488℄ Proposed by Mih�aly Ben ze, Brasov, Romania.For all positive integers m and n, show thatm(m+1)n2(n+1)2(2n2 +2n−1)−n(n+1)m2(m+1)2(2m2 +2m−1)is divisible by 720.Solution by Chip Curtis, Missouri Southern State University, Joplin, MO,USA. Let f(k) = k(k + 1)
(
2k2 + 2k − 1). Write
A (m, n) = m (m + 1) n2 (n + 1)2(
2n2 + 2n − 1)
− n (n + 1) m2 (m + 1)2(
2m2 + 2m − 1)
= mn (m + 1) (n + 1) [f (n) − f (m)] .Let C(m, n) = mn(m + 1)(n + 1) and D(m, n) = f(n) − f(m), so thatA(m, n) = C(m, n) D(m, n). Sin e 720 = 24325, it suÆ es to show thatA(m, n) is divisible by 16, 9, and 5.(a) Divisibility by 16. The residues of f(n) modulo 4 are given in the fol-lowing table.
n (mod 4) 0 1 2 3f(n) (mod 4) 0 2 2 0(i) If n ≡ 0 or 3 (mod 4), or m ≡ 0 or 3 (mod 4), then C(m, n)is divisible by 8 and D(m, n) is divisible by 2, so that A(m, n) isdivisible by 16.(ii) Otherwise, C(m, n) is divisible by 4 and D(m, n) is divisible by
4, and therefore, A(m, n) is divisible by 16.(b) Divisibility by 9. The residues of f(n) modulo 9 are given in the nexttable.n (mod 9) 0 1 2 3 4 5 6 7 8
f(n) (mod 9) 0 6 3 6 6 6 3 6 0(i) If n ≡ 0 or 8 (mod 9), or m ≡ 0 or 8 (mod 9), then C(m, n) ≡ 0(mod 9), so that A(m, n) is divisible by 9.(ii) If n ≡ 2 or 6 (mod 9), or m ≡ 2 or 6 (mod 9), then C(m, n)and D(m, n) are ea h divisible by 3, and therefore, A(m, n) isdivisible by 9.(iii) Otherwise, f(n) ≡ 6 (mod 9) and f(m) ≡ 6 (mod 9), so thatD(m, n) is divisible by 9. Thus, A(m, n) is divisible by 9.
497( ) Divisibility by 5. The residues of f(n) modulo 5 are given in the tablebelow.
n (mod 9) 0 1 2 3 4f(n) (mod 9) 0 1 1 1 0(i) If n ≡ 0 or 4 (mod 5), or m ≡ 0 or 4 (mod 5), then C(m, n) isdivisible by 5, so that A(m, n) is divisible by 5.(ii) Otherwise, D(m, n) is divisible by 5, and therefore, A(m, n) isdivisible by 5.Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia andHerzegovina; ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE,Rouen, Fran e; BRIAN D. BEASLEY, Presbyterian College, Clinton, SC, USA; OLIVER GEUPEL,Br �uhl, NRW, Germany; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; WALTHER JANOUS,Ursulinengymnasium, Innsbru k, Austria; ANDREA MUNARO, student, University of Trento,Trento, Italy; MICHAEL PARMENTER, Memorial University of Newfoundland, St. John's,NL; SKIDMORE COLLEGE PROBLEM SOLVING GROUP, Skidmore College, Saratoga Springs,NY, USA; PETER Y. WOO, Biola University, La Mirada, CA, USA; KONSTANTINE ZELATOR,University of Toledo, Toledo, OH, USA; TITU ZVONARU, Com�ane�sti, Romania; and theproposer.
3295. [2007 : 486, 488℄ Proposed by Mi hel Bataille, Rouen, Fran e.Let u : R → R be a bounded fun tion. For x > 0, letf(x) = sup {u(t) : t > ln(1/x)}and g(x) = sup {u(t) − xe−t : t ∈ R} .Prove that lim
x→0+f(x) = lim
x→∞g(x).Solution by Peter Y. Woo, Biola University, La Mirada, CA, USA, modi�edby the editor.The proof onsists of showing that both limits are equal to the limit
L = lim supt→∞
u(t), whi h exists sin e the fun tion u(t) is bounded. Let K besu h that |u(t)| < K for all t ∈ R. Given any ǫ > 0 and any t0 > 0, we have(i) there exists some t > t0 su h that L − ǫ < u(t), and(ii) there exists some t1 > t0 su h that for all t > t1, u(t) < L + ǫ.Let ǫ > 0 be given. For a �xed x > 0 we have limt→∞
xe−t = 0, so there existst0 > 0 su h that xe−t < ǫ for any t > t0. By part (i), there exists t > t0su h that L − ǫ < u(t), hen e
L − 2ǫ < u(t) − xe−t ≤ g(x) .Thus, g(x) > L − 2ǫ for ea h x > 0.
498On the other hand, there exists t1 ∈ R su h that u(t) < L + ǫ for all
t > t1. Sin e e−t1 > 0, let M > 0 be su h that K − Me−t1 < L + ǫ. We laim that if x > M , then g(x) ≤ L + ǫ. For this it suÆ es to show thatu(t) − xe−t < L + ǫ whenever x > M and t ∈ R.Indeed, if x > M and t ≤ t1, then u(t)−xe−t < K −Me−t1 < L+ǫ,while if x > M and t > t1, then u(t) − xe−t < u(t) < L + ǫ.Therefore, for x > M we have L − 2ǫ < g(x) < L + 2ǫ, hen elim
x→∞g(x) = L.Finally, writing S(v) = sup {u(t) : t > v} and making two hanges ofvariable in the limit yields
limx→0+
f(x) = limx→0+
S(
ln(1/x))
= limy→∞
S(ln y) = limz→∞
S(z) = L .Also solved by OLIVER GEUPEL, Br �uhl, NRW, Germany; and the proposer.3296. [2007 : 486, 488℄ Proposed by Mi hel Bataille, Rouen, Fran e.Find the greatest onstant K su h that
b2c2
a2(a − b)(a − c)+
c2a2
b2(b − c)(b − a)+
a2b2
c2(c − a)(c − b)> K
for all distin t positive real numbers a, b, and c.Solution by Peter Y. Woo, Biola University, La Mirada, CA, USA, expandedby the editor.We prove that K = 10.Let L denote the left side of the given inequality. Sin e L is ompletelysymmetri in a, b, and c, we may assume without loss of generality thata < b < c.Note �rst that L =
P
Qwhere Q = a2b2c2(b − a)(c − a)(c − b) and
P = b4c4(c − b) − c4a4(c − a) + a4b4(b − a).Observing that P = 0 when c = b or b = a or c = a, we �nd bystraightforward but tedious omputations thatP = b4c4(c − b) − a4
(
c5 − b5)
+ a5(
c4 − b4)
= (c − b)(
b4c4 − a4(
c4 + c3b + c2b2 + cb3 + b4)
+ a5(
c3 + c2b + cb2 + b3)
)
499= (c − b)
(
c4(
b4 − a4)
+ a4(a − b)(
c3 + c2b + cb2 + b3)
)
= (c − b)(b − a)(
c4(
b3 + b2a + ba2 + a3)
− a4(
c3 + c2b + cb2 + b3)
)
= (c − b)(b − a)(
b3(
c4 − a4)
+ cb2a(
c3 − a2)
+ c2ba2(
c2 − a2)
+ c3a3(c − a))
= (c − b)(b − a)(c − a)(
b3(
c3 + c2a + ca2 + a3)
+ cb2a(
c2 + ca + a2)
+ c2ba2(c + a) + c3a3) .
Hen e, P
Q=
W
a2b2c2, where
W = b3(
c3 + c2a + ca2 + a3)
+ cb2a(
c2 + ca + a2)
+ c2ba2(c + a) + c3a3 .By writing W as a sum of 10 terms and using the AM{GM Inequality, wereadily see that W ≥ 10(
a20b20c20)1/10
= 10a2b2c2, from whi h it followsthat L ≥ 10. Sin e a, b, and c are distin t, equality annot hold. Thus,L > 10.Finally, if we set a = 1, b = 1 + ε, and c = 1 + 2ε and let ε → 0+,then the value of L an be made arbitrarily lose to 10 from the right. Hen e,K = 10.Also solved by CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA;OLIVER GEUPEL, Br �uhl, NRW, Germany; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA;and the proposer. There were one in orre t and three in omplete solutions (whi h only showedthat L > 10 and then on luded immediately that K = 10).3297. [2007 : 486, 488℄ Proposed by Stanley Rabinowitz, MathPro Press,Chelmsford, MA, USA.If A, B, and C are the angles of a triangle, prove that
sin A + sin B sin C ≤ 1 +√
5
2.When does equality hold?
500Solution by D.J. Smeenk, Zaltbommel, the Netherlands.The following are equivalent
sin A + sin B sin C ≤ 1 +√
5
2;
2 sin A + cos(B − C) − cos(B + C) ≤ 1 +√
5 ;2 sin A + cos A + cos(B − C) ≤ 1 +
√5 .Let ϕ be the �rst quadrant angle with cos ϕ =
2√5and sin ϕ =
1√5(the angle ϕ is approximately 26.6◦). The last inequality then be omes
√5 sin(A + ϕ) + cos(B − C) ≤ 1 +
√5 .However, sin(A+ϕ) ≤ 1 and cos(B −C) ≤ 1, so the last inequality is true.Equality holds when A = arcsin
2√5
≈ 63.4◦ and B = C ≈ 58.3◦.Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosniaand Herzegovina; ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHELBATAILLE, Rouen, Fran e; BRIAN D. BEASLEY, Presbyterian College, Clinton, SC,USA; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; APOSTOLISK. DEMIS, Varvakeio High S hool, Athens, Gree e; OLEH FAYNSHTEYN, Leipzig,Germany; OLIVER GEUPEL, Br �uhl, NRW, Germany; RICHARD I. HESS, Ran ho Palos Verdes,CA, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; V �ACLAV KONE �CN �Y,Big Rapids, MI, USA; KEE-WAI LAU, Hong Kong, China; THANOS MAGKOS, 3rd High S hoolof Kozani, Kozani, Gree e; SALEM MALIKI �C, student, Sarajevo College, Sarajevo, Bosnia andHerzegovina; JUAN-BOSCO ROMEROM�ARQUEZ, Universidad de Valladolid, Valladolid, Spain;JOEL SCHLOSBERG, Bayside, NY, USA; SKIDMORE COLLEGE PROBLEM SOLVING GROUP,Skidmore College, Saratoga Springs, NY, USA; PANOS E. TSAOUSSOGLOU, Athens, Gree e;GEORGE TSAPAKIDIS, Agrinio, Gree e; PETER Y. WOO, Biola University, La Mirada, CA, USA;KONSTANTINE ZELATOR, University of Toledo, Toledo, OH, USA; TITU ZVONARU, Com�ane�sti,Romania; and the proposer. There was one in orre t solution submitted.Janous proved a more general result, namely, for λ > 0
λ sin A + sin B sin C ≤ 1 +√
4λ2 + 1
2,with equality if and only if A = arccos
(
1√4λ2+1
) and B = C.3298. [2007 : 486, 489℄ Proposed by Stanley Rabinowitz, MathPro Press,Chelmsford, MA, USA.Let ABC be a triangle of area 1
2in whi h a is the length of the sideopposite vertex A. Prove that
a2 + csc A ≥√
5 .[Ed.: The proposer's only proof of this is by omputer. He is hopingthat some CRUX with MAYHEM reader will �nd a simpler solution.℄
501Solution by Kee-Wai Lau, Hong Kong, China.Let b = AC, c = AB and let S denote the area of triangle ABC.Sin e S = 1
2bc sin A = 1
2, we obtain bc = csc A ≥ 1.By the Law of Cosines we have (regardless of the sign of cos A) that
a2 + csc A = a2 + bc = b2 + c2 − 2bc cos A + bc
≥ b2 + c2 − 2bc
√
1 − sin2 A + bc
= b2 + c2 − 2√
b2c2 − (bc sin A)2 + bc
=(
b2 + bc + c2)
− 2√
b2c2 − 1
≥ 3bc − 2√
b2c2 − 1 .For x ≥ 1, let y = 3x − 2√
x2 − 1. Then y is positive and from(y − 3x)2 = 4(x2 − 1) we get 5x2 − 6xy + y2 + 4 = 0. Sin e x is real, thedis riminant of the quadrati polynomial above must be non-negative.Thus, (−6y)2 − 20(y2 + 4) ≥ 0, or 16y2 − 80 ≥ 0, from whi h weobtain y ≥
√5. The result now follows by setting x = bc.Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e (two solutions);�SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia and Herzegovina; MICHELBATAILLE, Rouen, Fran e; ROY BARBARA, Lebanese University, Fanar, Lebanon; CHIPCURTIS, Missouri Southern State University, Joplin, MO, USA; APOSTOLIS K. DEMIS,Varvakeio High S hool, Athens, Gree e (two solutions); OLEH FAYNSHTEYN, Leipzig,Germany; OLIVER GEUPEL, Br �uhl, NRW, Germany; RICHARD I. HESS, Ran ho Palos Verdes,CA, USA; V �ACLAV KONE �CN �Y, Big Rapids, MI, USA; SOTIRIS LOURIDAS, Aegaleo, Gree e;PANOS E. TSAOUSSOGLOU, Athens, Gree e; PETER Y. WOO, Biola University, La Mirada,CA, USA; KONSTANTINE ZELATOR, University of Toledo, Toledo, OH, USA; TITU ZVONARU,Com�ane�sti, Romania; and the proposer. There was one in orre t solution submitted.From the proof given above, it is easy to dedu e that equality holds if and only if b = cand a = 4
√
45, in whi h ase A = cos−1
(
23
)
≈ 48.19◦ . This was pointed out by Barbara,Geupel, Hess, Kone� n �y, Tsapakidis, and Zvonaru.Both Barbara and Demis generalized to an arbitrary triangle of area k > 0, proving thata2 + csc A ≥
√8k + 1 and the lower bound √
8k + 1 is the best possible.3299. [2007 : 487, 489℄ Proposed by Vi tor Oxman, Western Galilee Col-lege, Israel.Given positive real numbers a, b, and wb, show that(a) if a triangle ABC exists with BC = a, CA = b, and the length ofthe interior bise tor of angle B equal to wb, then it is unique up toisomorphism;(b) for the existen e of su h a triangle in (a), it is ne essary and suÆ ientthat
b >2a |a − wb|2a − wb
≥ 0 ;( ) if ha is the length of the altitude to side BC in su h a triangle in (a),we have b > |a − wb| + 12ha.
502Solution by Mi hel Bataille, Rouen, Fran e.(a) For onvenien e we write w = wb. Let the interior bise tor of ∠Bmeet AC at W . We assume that △ABC exists with BC = a, CA = b, andBW = w, and shall produ e a (impli itly de�ned) formula for the third sidec = AB. We re all that w =
2ac cos B2
a + c, so that
2a cosB
2− w =
aw
c.
By the Law of Cosines we have WC2 = a2 + w2 − 2aw cosB
2; using thestandard formula WC =
ab
a + c, we therefore have
a2b2
(a + c)2= a2 − w
(
2a cosB
2− w
)
= a2 − aw2
c.It follows that c must be the unique positive solution of f(x) = a, where fis the de reasing fun tion on (0, ∞) given by
f(x) =ab2
(a + x)2+
w2
x.Thus, if a triangle ABC with the given parameters does exist, then its sidelengths are uniquely determined, and (a) is proved.(b) First, it is easy to see that the two given inequalities are equivalentto the onjun tion of the following three inequalities
w < 2a ; (2a + b)w < 2a(a + b) ; 2a(a − b) < (2a − b)w . (1)Se ond, the existen e of a suitable triangle ABC is equivalent to the fa tthat the solution c of f(x) = a satis�es |a − b| < c < a + b; that is,f(
|a − b|)
> a > f(a + b) . (2)To show that (1) and (2) are equivalent, we �rst suppose that the inequalitiesin (1) hold. The inequality f(a+ b) < a redu es to the equivalent inequality(2a + b)w < 2a(a + b), whi h holds by (1). As for f
(
|a − b|)
> a, it isequivalent toab2 |a − b| + w2
(
a + |a − b|)2
> a |a − b|(
a + |a − b|)2 .This redu es to w2b2 > 0 when a ≤ b, so it holds then; but it also holds if
b < a sin e then it be omes w(2a − b) > 2a(a − b), whi h holds by (1). Wehave proved that (1) implies (2).Conversely, we assume that (2) holds (that is, that a triangle ABCwith the given parameters exists). Then w
2a=
c
a + c· cos
B
2; hen e w < 2a.
503Moreover, from f(a + b) < a we obtain (2a + b)w < 2a(a + b). As for the ondition 2a(a − b) < (2a − b)w, it follows from f
(
|a − b|)
> a if a > b,and from (2a − w)(a − b) < aw if a ≤ b (be ause 2a > w, the left side isnegative). The desired equivalen e follows.( ) Note that b > |a − w| + 12ha is equivalent to
ab > a |a − w| + Area(ABC) ;that is, to ab > a |a − w| +(
a + c
2
)
w sinB
2. Sin e a |a − w| <
(2a − w)b
2(from part (b)), the latter will ertainly hold ifb ≥ (a + c) sin
B
2.This inequality is equivalent to
sin B ≥ (sin A + sin C) sinB
2,or to
2 cosB
2≥ 2 sin
(
A + C
2
)
cos(
A − C
2
) ,or �nally to1 ≥ cos
(
A − C
2
) ,whi h is ertainly true. The result follows.Also solved by CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA;OLIVER GEUPEL, Br �uhl, NRW, Germany; PETER Y. WOO, Biola University, La Mirada, CA,USA (part ( ) only); and the proposer.Parts (a) and (b) of our problem appear on page 11 of D.S. Mitrinovi� et al., Re entAdvan es in Geometri Inequalities, Kluwer A ademi Publishers, 1989 as the �rst of 40 existen eresults from a 1952 paper (in Cze h) by G. Petrov.In addition to his solution, Oxman also addressed the question of onstru tibility. Exer- ise 4 on page 142 of G �unter Ewald's Geometry: An Introdu tion (Wadsworth Publ., 1971) saysthat in general a triangle annot be onstru ted by ruler and ompass given the lengths a, b,and wb, even when that triangle exists. The author suggests that the proof of his laim an besimpli�ed by taking both the given side lengths equal to 1. The formula f(x) = 1 from part(a) of the featured solution (with a = b = 1, and w2 hosen to be rational) is a ubi equationwith rational oeÆ ients. One simply has to hoose a value of w for whi h the resulting u-bi equation has no rational root. The theory of Eu lidean onstru tions then tells us that thepositive root, namely c, annot be onstru ted by using ruler and ompass.3300. [2007 : 487, 489℄ Proposed by Arkady Alt, San Jose, CA, USA.Let a, b, and c be positive real numbers. For any positive integer nde�ne
Fn =
3(an + bn + cn)
a + b + c−∑ y li bn + cn
b + c
.(a) Prove that Fn ≥ 0 for n ≤ 5.(b)⋆ Prove or disprove that Fn ≥ 0 for n ≥ 6.
504Solution by Cao Minh Quang, Nguyen Binh Khiem High S hool, Vinh Long,Vietnam.Sin e F1 = 0, we take n > 1. We note that (xn−1 − yn−1
)
(x−y) ≥ 0for all positive x and y, with equality if and only if x = y. We have(a + b + c)Fn = 3 (an + bn + cn) − (a + b + c)
∑ y li bn + cn
b + c
= (an + bn + cn) −∑ y li a (bn + cn)
b + c
=∑ y li [an − a (bn + cn)
b + c
]
=∑ y li [ab
(
an−1 − bn−1)
(b + c)+
ac(
an−1 − cn−1)
(b + c)
]
=∑ y li ab
(
an−1 − bn−1)
(a − b)
(b + c)(c + a)≥ 0 .
Equality holds if and only if a = b = c.Also solved by �SEFKET ARSLANAGI �C, University of Sarajevo, Sarajevo, Bosnia andHerzegovina; ROY BARBARA, Lebanese University, Fanar, Lebanon; VASILE CIRTOAJE,University of Ploiesti, Romania; CHIP CURTIS, Missouri Southern State University, Joplin, MO,USA; NIKOLAOS DERGIADES, Thessaloniki, Gree e; OLIVER GEUPEL, Br �uhl, NRW, Germany;WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; PANOS E. TSAOUSSOGLOU,Athens, Gree e (part (a) only); STAN WAGON, Ma alester College, St. Paul, MN, USA (part (a)only); TITU ZVONARU, Com�ane�sti, Romania; and the proposer.C�rtoaje mentioned that this problem was posted (together with a solution similar tothe one featured above) by Wolfgang Berndt (Spanferkel) on the Mathlinks Forum websitehttp://www.mathlinks.ro/Forum/viewtopic.php?p=607167 in August 2006. Barbara, C�rtoaje,and Dergiades proved the following generalization: If a1, a2, . . . , am are positive real num-bers, m ≥ 2, and
Fn =m(an
1 + an2 + · · · + an
m)
a1 + a2 + · · · + am−
∑ y li an2 + · · · + an
m
a2 + · · · + am,
then Fn ≥ 0 for all n ≥ 1. Alt ultimately proved that if a, b, c, p, and q are positive realnumbers andF(p, q) =
3(ap + bp + cp)
aq + bq + cq−
∑ y li ap + bp
aq + bq,then (p − q)F(p, q) ≥ 0.
Mathem
atical Association of A
merica P
resents
Problems from
Murray K
lamkin: The C
anadian Collection
Andy Liu &
Bruce S
hawyer, E
ditorsM
urray Klam
kin was a dedicated problem
solver and problem proposer,
who left indelible m
arks on the problemist com
munity. A
fter working in indus-
try and academe in the U
nited States, he spent the last thirty of his eighty
four years in Canada. H
e was fam
ous for his Quickies, problem
s that have quick and neat solutions. In this book you w
ill find all of the problems that he
proposed for CR
UX w
ith MAYH
EM, including all of his Q
uickies. His prob-
lems covered a very w
ide range of topics, and show a great deal of insight
into what is possible in these areas. The problem
s are arranged into sets according to topic,and the lightly edited solutions are as published in C
RU
X w
ith MAYH
EM.
Problem
s from M
urray Klam
kin: The Canadian C
ollection is published by the M
athematical A
ssociation of Am
erica (MA
A) in collaboration w
ith the C
anadian Mathem
atical Society (C
MS
). It is the first volume in the C
anadian C
ollection.
Series: P
roblem B
ooks, C
atalog Code: C
AN
-01280 pp., H
ardbound, 2009, IS
BN
9780-88385-828-8, List: $61.50 M
ember: $49.95
Order your copy today! C
all (301) 617-7800(C
MS m
embers get m
ember price for this book. B
e sure to mention you are a C
MS m
ember w
hen you call.)
506YEAR END FINALEThis marks my �rst year as Editor/Co-Editor of CRUX with MAYHEM, andmy �rst year end �nale. As I write in this spa e, whi h Jim Totten wrote in justone year ago, I am reminded of him and the non-permanen e of things. Just asthe surf swirls sand and pebbles on a bea h, so does time alter words and ourselves.Islands disappear and are built up somewhere else. When I remember Jim's in redibleenthusiasm and passion for mathemati s and his outrea h work, I am moved. Ano ean is there for us to feel its power and its ool refreshing spray, to play in it andbe leansed by its soothing sound.Whi h reminds me that issues of CRUX with MAYHEM are now delayed byseveral weeks! My apologies to all the readers for this onsiderable delay, and in2009 we will work hard to lose the gap.Many people have served this past year (I will borrow Jim's habit of apitalizingtheir names). First, I shall forever be grateful to the late JIM TOTTEN for all thesupport he gave me during the time I Co-Edited with him (until the end of February,2008). After Jim's sudden passing on Mar h 9, 2008, three individuals in parti ular ame to my res ue. I thank CHRIS FISHER for his humour and good spirits in thosetough early days. I thank BRUCE CROFOOT for his solid support asMission Control inthe CRUX oÆ e at Thompson Rivers University and for listening to me when I neededsomeone to talk to. I thank BRUCE SHAWYER for his ontinuing help and advi e onCRUX with MAYHEM, indeed for e�e tively stepping into the role of Editor-at-Large(whi h Jim was going to take up after stepping down as Editor-in-Chief). I also thankBru e S. for his hospitality in Newfoundland and for providing me with a mu h neededsense of ontinuity.I thank the members of the Editorial Board for their hard work and for bring-ing their unique talents to CRUX: JEFF HOOPER, the Asso iate Editor, for his arefulreading and sound judgment; IAN VANDERBURGH, the MAYHEM Editor, whoseProblem of the Month olumn is a joy to read and who is ahead of s hedule (!);ROBERT BILINSKI, for serving as Skoliad Editor for the past four years; ROBERTWOODROW, for handling the Olympiad Corner on top of his immense administrativeduties; JOHN GRANT M LOUGHLIN, for his solid support as Book Reviews Editorand for his fantasti proof reading skills. This is John's last issue as Book Reviews Ed-itor, but he will ontinue as the Guest Editor for the Jim Totten spe ial issue of May,2009. I wel ome AMAR SODHI to the Board who is taking over as Book ReviewsEditor from John. I thank my olleague here in Winnipeg and Arti les Editor JAMESCURRIE, for taking on the position and then turning a ba klog of arti les into a or-nu opia. I thank the Problems Editors for their ontinuing support and hard work:ILIYA BLUSKOV, CHRIS FISHER, MARIA TORRES, and EDWARD WANG. The job ofbeing a Problems Editor is perhaps one of the most diÆ ult on the Board, with tightdeadlines and a deluge of problem proposals and solutions in all areas of mathemat-i s oming from all over the world et hed, s rawled, or digitally re orded on every on eivable medium known to man. A big thank you to the four of you.I thank GRAHAMWRIGHT, theManaging Editor, for his solid support and alsofor providing that vital sense of ontinuity.Others who are not on the Editorial Board but whose work is just as ru ial areMONIKA KHBEIS and ERIC ROBERT who serve on the Mayhem sta�; a big thankyou Monika and Eri . I thank JEAN-MARC TERRIER for translating the problemsthat appear in CRUX with MAYHEM. The Canadian Mathemati al So iety re entlyhonoured Jean-Mar at the 2008 Winter Meeting for his many years of servi e. It is
507well deserved! I look forward to working with Jean-Mar in the New Year. As well,I wel ome ROLLAND GAUDET of Coll �ege Universitaire Saint-Bonifa e, Winnipeg.Rolland is also helping with translations, espe ially in times of risis! I thank ourpast CRUX editor BILL SANDS for his proof reading and sound advi e.My olleagues in the Dept. of Mathemati s and Statisti s have lent their sup-port. Those who have taken pity on this Editor-in-Chief are ANNA STOKKE, ROSSSTOKKE, TERRY VISENTIN, and JEFF BABB. Our se retary, JULIE BEAVER, has alsohelped out in a pin h as little emergen ies have arisen. I thank the previous Deanof S ien e, GABOR KUNSTATTER, for his understanding and foresight in supportingCRUX with MAYHEM at the University of Winnipeg.The LATEX expertise of JOANNE CANAPE at the University of Calgary, and TAOGONG and JUNE ALEONG at Wilfrid Laurier University goes a long way to produ inghigh quality opy. Joanne is �nishing up her work in this area, so after my one yearof servi e here I thank her for giving twenty! I wel ome JILL AINSWORTH on boardwho is taking over from Joanne to prepare the Olympiads.Thanks go to the University of Toronto Press and to Thistle Printing, in parti -ular TAMI EHRLICH. The quality �nished opy and the purple overs are just lovely.I thank youMICHAELDOOB and CRAIG PLATT for te hni al support, and JUDIBORWEIN for putting CRUX on the net.Someone spe ial who has helpedme throughmy �rst year with her areful proofreading and knowledge of geometry is CHARLENE PAWLUCK. Thank you for sharingyour opy of Eu lid with me, and so mu h more.Finally, I thank you, the readers, for all that you have done for me and for thejournal. CRUX with MAYHEM is built from your ontributions and all the time, areand reativity that you put into your submissions is re e ted in these pages. I wish I ould mention all the truly marvellous people that I have gotten to know in the lastyear, but this margin is too small to hold all the names and praises.I lose with a all for a new Skoliad Editor. Please refer anyone to us you maythink is suitable. Skoliad is missing from this issue, but will be ba k next year; fromnow on please send all Skoliad materials dire tly to the Editor-in-Chief (or resendyour past submissions if you did not re eive a reply from us).I wish ea h of you the very best in 2009 in all areas of life, V �a lav (Vazz) Linek
Crux Mathemati orumwith Mathemati al MayhemFormer Editors / An iens R �eda teurs: Bru e L.R. Shawyer, James E. TottenCrux Mathemati orumFounding Editors / R �eda teurs-fondateurs: L �eopold Sauv �e & Frederi k G.B. MaskellFormer Editors / An iens R �eda teurs: G.W. Sands, R.E. Woodrow, Bru e L.R. ShawyerMathemati al MayhemFounding Editors / R �eda teurs-fondateurs: Patri k Surry & Ravi VakilFormer Editors / An iens R �eda teurs: Philip Jong, Je� Higham, J.P. Grossman,Andre Chang, Naoki Sato, Cyrus Hsia, Shawn Godin, Je� Hooper
508INDEX TO VOLUME 34, 2008Contributor Pro�lesFebruary Peter Y. Woo ................................................... 1Skoliad Robert BilinskiFebruary No. 107 ......................................................... 2Mar h No. 108 ....................................................... 65April No. 109 ...................................................... 129May No. 110 ...................................................... 195September No. 111 ...................................................... 257O tober No. 112 ...................................................... 321November No. 113 ...................................................... 385Mathemati al Mayhem Ian VanderBurghFebruary .................................................................. 8Mar h ................................................................. 69April ............................................................... 136May ............................................................... 200September ............................................................... 266O tober ............................................................... 327November ................................................................ 396De ember ............................................................... 449Mayhem ProblemsFebruary M326{M331 .................................................... 8Mar h M332{M337 ................................................... 69April M338{M343 ................................................. 136May M344{M349 ................................................. 200September M350{M356 ................................................. 266O tober M357{M362 ................................................. 327November M363{M368 ................................................. 396De ember M369{M375 ................................................. 449Mayhem SolutionsFebruary M276{M281 ................................................... 10Mar h M282{M287 ................................................... 71April M288{M293 ................................................. 138May M294{M300 ................................................. 202September M301{M312 ................................................. 268O tober M313{M324 ................................................. 329November M325{M331 ................................................. 398De ember M332{M337 ................................................. 451Problem of the Month Ian VanderBurghFebruary ................................................................. 15Mar h ................................................................. 76April ............................................................... 144May ............................................................... 208September ............................................................... 279O tober ............................................................... 338November ................................................................ 404De ember ............................................................... 457
509Mayhem Arti lesAdding UpBru e Shawyer ............................................................ 406The Olympiad Corner R.E. WoodrowFebruary No. 267 ....................................................... 18Mar h No. 268 ....................................................... 79April No. 269 ...................................................... 147May No. 270 ...................................................... 211September No. 271 ...................................................... 282O tober No. 272 ...................................................... 341November No. 273 ...................................................... 408De ember No. 274 ...................................................... 459Book Reviews John Grant M LoughlinHow Euler Did Itby C. Edward SandiferReviewed by J. Chris Fisher ............................................... 34Nonplussed! Mathemati al Proof of Implausible Ideasby Julian HavilReviewed by Robert D. Poodia k .......................................... 36Math Through the Ages: A Gentle History for Tea hers and Othersby William P. Berlingho� and Fernando Q. GouveaReviewed by John Grant M Loughlin....................................... 95Minnesota Math League XXV 1980{2005by A. Wayne RobertsReviewed by Robert L. Crane .............................................. 96The Magi Numbers of the Professorby Owen O'Shea and Underwood DudleyReviewed by Je� Hooper .................................................. 161Geometri Puzzle Designby Stewart CoÆnReviewed by Jim Totten ................................................... 163The Liar Paradox and the Towers of Hanoi:the Ten Greatest Math Puzzles of All Timeby Mar el DanesiReviewed by Amar Sodhi .................................................. 164Problems of the Weekby Jim TottenReviewed by John Grant M Loughlin ..................................... 229Digital Di eby Paul J. NahinReviewed by Amar Sodhi .................................................. 297Cal ulus Gems:Brief Lives and Memorable Mathemati sby George F. SimmonsReviewed by Robert D. Poodia k ......................................... 356From Zero to In�nity:What Makes Numbers Interestingby Constan e ReidReviewed by John Grant M Loughlin ..................................... 357
510Impossible? Surprising Solutions and Counterintuitive Conundrumsby Julian HavilReviewed by Ed Barbeau .................................................. 422The Symmetries of Thingsby John H. Conway, Heidi Burgiel, and Chaim Goodman-StraussReviewed by J. Chris Fisher ............................................... 468Crux Arti les James CurrieA Useful InequalityRoy Barbara ............................................................... 38Sharpening the Hadwiger-Finsler InequalityCezar Lupu and Cosmin Pohoat� �a .......................................... 97Industrial Grade Primes with a Money-Ba k GuaranteeMi hael P. Abramson .................................................... 165The Proof of Three Open InequalitiesVasile C�rtoaje ............................................................ 231The Sum of a Cube and a Fourth PowerThomas Mauts h and Gerhard J. Woeginger ............................. 358Twin Problems on Non-Periodi Fun tionsEugen J. Ionas u .......................................................... 424Old Idaho Usual HereRobert Israel, Stephen Morris, and Stan Wagon ......................... 471A Limit of an Improper Integral Depending on One ParameterIesus C. Diniz ............................................................. 478Sliding Down In lines with Fixed Des ent Time:a Converse to Galileo's Law of ChordsJe� Babb .................................................................. 481ProblemsFebruary 3301{3312 ..................................................... 44Mar h 3313{3325 .................................................... 102April 3326{3337 .................................................... 170May 3282, 3338{3350 ............................................. 239September 3351{3362 .................................................... 298O tober 3363{3375 .................................................... 362November 3376{3388 .................................................... 430De ember 3371, 3389{3400.............................................. 483SolutionsFebruary 3201{3213 ..................................................... 49Mar h 3214{3225, 3227 ............................................. 107April 3226, 3228{3238, 3242 ....................................... 176May Klamkin{05, 3239{3241, 3243{3250 ......................... 244September 3251{3362 .................................................... 303O tober 3263{3275, 3282 ............................................. 367November 3276{3281, 3283{3288 ....................................... 435De ember 3229, 3289{3300 ............................................. 488Mis ellaneousEditorial .................................................................... 193In Memoriam: James Edward Totten ...................................... 194MAA Book Promotion .......................................................505Year End Finale ............................................................. 506
511Proposers and solvers appearing in the SOLUTIONS se tion in 2008:ProposersAnonymous Proposer 3211Gregory Akulov 3285Arkady Alt 3300Mi hel Bataille 3217, 3237, 3238, 3251, 3252, 3266, 3267, 3295, 3296Mih�aly Ben ze 3204, 3205, 3206, 3213, 3214, 3216, 3228, 3229, 3230,3239, 3240, 3253, 3292, 3293, 3294K.S. Bhanu 3284Alper Cay 3258Vasile C�rtoaje 3209, 3210, 3274, 3275, Klamkin{05,M.N. Deshpande 3283, 3284Jos �e Luis D��az-Barrero 3212, 3247, 3263, 3269, 3281, 3282Marian Din a 3205J. Chris Fisher 3224Ovidiu Furdui 3218, 3226, 3227, 3261, 3262, 3277, 3288G.P. Henderson 3201Ignotus 3231Neven Juri� 3259, 3276, 3286Geo�rey A. Kandall 3235V�a lav Kone� n�y 3256Tidor Mitev 3236Virgil Ni ula 3242, 3241, 3260, 3264, 3265, 3270, 3271, 3273, 3278, 3279,3280, 3287, 3289, 3290, 3291
Vi tor Oxman 3299Fran is o Pala ios Qui~nonero 3212Pantelimon George Popes u 3269, 3281, 3282A hilleas Pavlos Poryfyriadis 3223D.E. Prithwijit 3272Stanley Rabinowitz 3297, 3298Juan-Bos o Romero M�arquez 3221Bill Sands 3257, 3268D.J. Smeenk 3202, 3203, 3250Marian Tetiva 3220, 3246George Tsapakidis 3225G. Tsintsifas 3232, 3233, 3234, 3243, 3244, 3254, 3255Pham Van Thuan 3222Dan Vetter 3219Harley Weston 3224Shaun White 3208, 3215John Wiest 3268Peter Y. Woo 3207Paul Yiu 3245Titu Zvonaru 3248, 3249Featured Solvers | IndividualsMohammed Aassila 3248, 3249Arkady Alt 3241, 3263, 3276�Sefket Arslanagi � 3249, 3253, 3257, 3264Dionne Bailey 3228, 3237, 3247, 3269, 3271Roy Barbara 3214, 3232, 3235Ri ardo Barroso Campos 3267Mi hel Bataille 3204, 3205, 3209, 3213, 3217, 3222, 3227, 3229, 3232,3245, 3249, 3256, 3257, 3264, 3269, 3272, 3273, 3276, 3278, 3281, 3283,3285, 3286, 3288, 3291, 3293, 3299Brian D. Beasley 3220Mih�aly Ben ze 3230, 3253Manuel Benito 3253, 3261, 3262Elsie Campbell 3228, 3237, 3247, 3269, 3271Cao Minh Quang 3230, 3253, 3281, 3300�Os ar Ciaurri Ram��rez 3253, 3261, 3262Vasile C�rtoaje Klamkin{05, 3275Chip Curtis 3204, 3221, 3226, 3234, 3261, 3274, 3276, 3294Apostolis K. Demis 3211, 3219, 3224, 3255, 3273Charles R. Diminnie 3221, 3228, 3237, 3247, 3269, 3271Emilio Fern�andez Moral 3253, 3261, 3262J. Chris Fisher 3238Fran is o Javier Gar ��a Capit �an 3221Oliver Geupel 3263, 3273, 3282, 3284, 3292Douglass L. Grant 3293P.C. Hammer 3254Karl Havlak 3271John Hawkins 3240Ri hard I. Hess 3221, 3232, 3283, 3286John G. Heuver 3245, 3289Joe Howard 3287Walther Janous 3245Geo�rey A. Kandall 3221Paula Ko a 3271V�a lav Kone� n �y 3202, 3232, 3244Kee-Wai Lau 3242, 3239, 3246, 3298Tom Leong 3253, 3260Nguyen Thanh Liem 3249
Tai hi Maekawa 3250, 3279Thanos Magkos 3276Salem Maliki � 3214, 3220, 3234, 3245, 3249, 3280Dragoljub Milo�sevi � 3214Andrea Munaro 3253, 3265, 3280Vedula N. Murty 3221Jos �e H. Nieto 3201, 3202, 3257, 3259Mi hael Parmenter 3252, 3284Alex Remorov 3249Henry Ri ardo 3204,Juan-Bos o Romero M�arquez 3221Luz Ron al 3261, 3262Xavier Ros 3212, 3221, 3236, 3250, 3257Sergey Sadov 3229Bill Sands 3268Joel S hlosberg 3215, 3233, 3290D.J. Smeenk 3202, 3221, 3235, 3264, 3279, 3297T. Je�erson Smith 3254David Stone 3240Edmund Swylan 3202, 3206, 3207, 3238Marian Tetiva 3220Phi Thai Thuan 3249Daniel Tsai 3251Panos E. Tsaoussoglou 3249George Tsapakidis 3289G. Tsintsifas 3232Apostolis Vergos 3208Vo Quo Ba Can 3210, 3216, 3223Edward T.H. Wang 3253Shaun White 3208Peter Y. Woo 3203, 3205, 3218, 3221, 3225, 3244, 3245, 3295, 3296Paul Yiu 3245Roger Zarnowski 3237Titu Zvonaru 3243, 3270Featured Solvers | Groups
Missouri State University Problem Solving Group 3259, 3266 Skidmore College Problem Solving Group 3245
512Other Solvers | IndividualsMohammed Aassila 3223, 3237, 3241, 3281, 3283Gregory Akulov 3285Arkady Alt 3221, 3222, 3223, 3226, 3227, 3230, 3236, 3246, 3247, 3248,3249, 3281, 3287, 3300Miguel Amengual Covas 3203, 3223, 3250Christos Anastassiades 3250George Apostolopoulos 3264, 3265, 3270, 3276, 3278, 3284, 3285, 3286,3287, 3289, 3291, 3293, 3298�Sefket Arslanagi � 3203, 3206, 3221, 3222, 3223, 3248, 3249, 3250, 3251,3256, 3260, 3265, 3270, 3271, 3272, 3276, 3281, 3284, 3285, 3287, 3289,3293, 3294, 3297, 3298, 3300Dionne T. Bailey 3222, 3223, 3251, 3253, 3263, 3276, 3285, 3286, 3287,3293Roy Barbara 3201, 3202, 3203, 3213, 3221, 3222, 3223, 3233, 3234, 3236,3237, 3238, 3242, 3241, 3243, 3244, 3247, 3250, 3251, 3252, 3257, 3264,3265, 3266, 3270, 3271, 3272, 3278, 3285, 3286, 3287, 3290, 3291, 3293,3294, 3297, 3298, 3300Ri ardo Barroso Campos 3203, 3250, 3279, 3289,Mi hel Bataille 3201, 3202, 3206, 3207, 3208, 3211, 3221, 3223, 3228,3230, 3233, 3234, 3235, 3236, 3237, 3238, 3242, 3237, 3238, 3239, 3243,3244, 3246, 3247, 3248, 3250, 3251, 3252, 3253, 3255, 3262, 3263, 3265,3266, 3267, 3270, 3271, 3279, 3280, 3282, 3284, 3287, 3289, 3290, 3292,3294, 3295, 3296, 3297, 3298Brian D. Beasley 3294, 3297Fran is o Bellot Rosado 3202, 3203Mih�aly Ben ze 3204, 3205, 3213, 3214, 3216, 3223, 3228, 3229, 3239,3247, 3249, 3269, 3292, 3293, 3294Manuel Benito 3251, 3255, 3257, 3258K.S. Bhanu 3284Paul Bra ken 3222, 3262, 3286Katrina Bri ker 3221Whitney Bullo k 3257Sarah Burnham 3271Elsie M. Campbell 3212, 3222, 3223, 3251, 3253, 3263, 3276, 3285, 3286,3287, 3293Cao Minh Quang 3212, 3214, 3221, 3222, 3223, 3235, 3236, 3242, 3246,3247, 3248, 3249, 3250, 3251, 3260, 3265, 3270, 3276, 3287�Os ar Ciaurri Ram��rez 3251, 3255, 3257, 3258Vasile C�rtoaje 3209, 3210, 3274, 3300Chip Curtis 3202, 3203, 3206, 3216, 3217, 3219, 3222, 3223, 3227, 3230,3235, 3237, 3238, 3239, 3241, 3243, 3244, 3245, 3246, 3247, 3248, 3251,3252, 3256, 3257, 3259, 3262, 3264, 3265, 3266, 3269, 3270, 3271, 3275,3278, 3281, 3283, 3284, 3286, 3287, 3289, 3290, 3291, 3292, 3293, 3296,3297, 3298, 3299, 3300Paul Deiermann 3284, 3285Apostolis K. Demis 3202, 3203, 3205, 3207, 3213, 3214, 3221, 3233, 3234,3235, 3238, 3251, 3256, 3258, 3264, 3265, 3267, 3270, 3289, 3292, 3297,3298Hasan Denker 3222, 3223Nikolaos Dergiades 3300M.N. Deshpande 3284Jos �e Luis D��az-Barrero 3212, 3247, 3263, 3269, 3281, 3286, 3287Charles R. Diminnie 3212, 3222, 3223, 3250, 3251, 3253, 3263, 3276, 3285,3286, 3287, 3293Marian Din a 3205Oleh Faynshteyn 3293, 3297, 3298Emilio Fern�andez Moral 3251, 3255, 3257, 3258Ovidiu Furdui 3218, 3226, 3227, 3261, 3262, 3288Ian June L. Gar es 3251Fran is o Javier Gar ��a Capit �an 3202, 3290, 3291Oliver Geupel 3257, 3264, 3266, 3268, 3276, 3278, 3279, 3281, 3283, 3286,3287, 3289, 3290, 3293, 3294, 3295, 3296, 3297, 3298, 3299, 3300Karl Havlak 3253, 3276, 3285, 3286, 3287, 3293John Hawkins 3212, 3237, 3240, 3241, 3257G.P. Henderson 3201Ri hard I. Hess 3203, 3206, 3208, 3219, 3221, 3222, 3227, 3250, 3251,3253, 3262, 3265, 3270, 3276, 3287, 3292, 3294, 3296, 3297, 3298John G. Heuver 3203, 3221, 3256Joe Howard 3204, 3212, 3221, 3222, 3223, 3227, 3235, 3236, 3237, 3246,3247, 3248, 3251, 3260, 3271, 3276, 3281Peter Hurthig 3250Walther Janous 3239, 3241, 3246, 3247, 3248, 3249, 3250, 3251, 3253,3257, 3260, 3261, 3262, 3263, 3265, 3266, 3267, 3269, 3270, 3271, 3272,3273, 3274, 3291, 3292, 3294, 3297, 3300Neven Juri� 3259, 3276, 3286Natalie Kalmink 3221Geo�rey A. Kandall 3202, 3235, 3250, 3289Paula Ko a 3253, 3276, 3285, 3286V�a lav Kone� n�y 3203, 3221, 3234, 3235, 3243, 3250, 3256, 3258, 3291,3297, 3298
Kee-Wai Lau 3208, 3222, 3223, 3236, 3237, 3241, 3247, 3249, 3248, 3250,3253, 3260, 3271, 3285, 3286, 3292, 3297Tom Leong 3251, 3257Kathleen E. Lewis 3257Carl Libis 3276Sotiris Louridas 3298Tai hi Maekawa 3250Thanos Magkos 3221, 3260, 3286, 3287, 3297SalemMaliki � 3202, 3203, 3207, 3208, 3211, 3216, 3221, 3222, 3223, 3230,3233, 3235, 3236, 3237, 3240, 3241, 3243, 3244, 3246, 3247, 3248, 3250,3251, 3253, 3256, 3257, 3259, 3260, 3266, 3270, 3271, 3272, 3276, 3281,3284, 3287, 3293, 3297Dung Nguyen Manh 3281, 3287Pavlos Maragoudakis 3221, 3251Petrus Martins 3253G. Milanova 3221, 3222, 3223D.M. Milo�sevi � 3222Dragoljub Milo�sevi � 3221, 3222, 3223Todor Mitev 3236Andrea Munaro 3235, 3236, 3251, 3256, 3264, 3276, 3278, 3287, 3289,3294Vedula N. Murty 3214, 3221, 3222, 3223N. Nadeau 3262Virgil Ni ula 3241, 3242, 3260, 3264, 3265, 3270, 3271, 3273, 3278, 3279,3280, 3287, 3289, 3290, 3291Jos �e H. Nieto 3205, 3221, 3237, 3248, 3251, 3252, 3253, 3258Vi tor Oxman 3299Fran is o Pala ios Qui~nonero 3212Mi hael Parmenter 3237, 3238, 3250, 3287, 3294Paolo Perfetti 3281Pantelimon George Popes u 3269, 3281A hilleas Pavlos Porfyriadis 3223D.E. Prithwijit 3272Stanley Rabinowitz 3297, 3298Alex Remorov 3226, 3229, 3230, 3234, 3235, 3236, 3237, 3238, 3241, 3243,3244, 3246, 3247, 3248, 3250Henry Ri ardo 3204, 3212, 3247, 3293Juan-Bos o Romero M�arquez 3221, 3223, 3297Luz Ron al 3251, 3255, 3257, 3258Xavier Ros 3227, 3237, 3247, 3248, 3249, 3251, 3253, 3259, 3260, 3262,3281, 3286, 3287Deborah Salas-Smith 3257Bill Sands 3257Robert P. Sealy 3286Joel S hlosberg 3202, 3203, 3204, 3205, 3207, 3211, 3218, 3221, 3227,3235, 3238, 3244, 3250, 3253, 3255, 3265, 3266, 3270, 3271, 3272, 3276,3278, 3283, 3285, 3287, 3292, 3293, 3297Bob Serkey 3287Andrew Siefker 3253D.J. Smeenk 3203, 3221, 3223, 3233, 3234, 3244, 3250, 3265, 3278, 3287Digby Smith 3286David R. Stone 3212, 3237, 3240, 3241, 3257Edmund Swylan 3203, 3217, 3221, 3234, 3236, 3237, 3243, 3244, 3250,3257Alexandros Sygelakis 3209, 3212Winfer C. Tabares 3251Ashley Tangeman 3253Marian Tetiva 3246Daniel Tsai 3203, 3286Panos E. Tsaoussoglou 3203, 3222, 3223, 3236, 3248, 3250, 3287, 3297,3298, 3300George Tsapakidis 3222, 3223, 3265, 3270, 3276, 3286, 3287, 3291, 3293,3297G. Tsintsifas 3233, 3234, 3243, 3244, 3254, 3255Pham Van Thuan 3222Phi Thai Thuan 3236, 3248Apostolis Vergos 3212, 3248Dan Vetter 3219Vo Quo Ba Can 3214, 3222Stan Wagon 3208, 3222, 3223, 3292, 3300Peter Y. Woo 3201, 3202, 3207, 3211, 3212, 3213, 3214, 3220, 3222, 3223,3225, 3233, 3234, 3235, 3236, 3238, 3241, 3243, 3246, 3247, 3248, 3250,3251, 3252, 3253, 3254, 3255, 3256, 3258, 3259, 3262, 3264, 3265, 3267,3270, 3271, 3273, 3276, 3278, 3279, 3280, 3281, 3284, 3285, 3286, 3287,3289, 3290, 3291, 3292, 3293, 3294, 3297, 3298, 3299Bingjie Wu 3257, 3272, 3281, 3287Konstantine Zelator 3203, 3265, 3266, 3272, 3294, 3297, 3298Titu Zvonaru 3202, 3203, 3221, 3222, 3223, 3233, 3234, 3235, 3236, 3237,3238, 3244, 3246, 3247, 3249, 3250, 3263, 3264, 3265, 3266, 3267, 3271,3272, 3281, 3287, 3289, 3290, 3291, 3293, 3294, 3297, 3298, 3300Other Solvers | GroupsAteneo Problem Solving Group 3221, 3251Missouri State Univeristy Problem Solving Group 3253, 3257, 3272 Skidmore College Problem Solving Group 3221, 3237, 3291, 3294, 3297University of Regina Math Central Consultants 3211
