449
MATHEMATICAL MAYHEM Mathemati al Mayhem began in 1988 as a Mathemati al Journal for and by High S hool and University Students. It ontinues, with the same emphasis,
as an integral part of Crux Mathemati orum with Mathemati al Mayhem. The Mayhem Editor is Ian VanderBurgh (University of Waterloo). The other sta members are Monika Khbeis (As ension of Our Lord Se ondary S hool, Mississauga), Eri Robert (Leo Hayes High S hool, Frederi ton), Larry Ri e (University of Waterloo), and Ron Lan aster (University of Toronto).
Mayhem Problems
Veuillez nous transmettre vos solutions aux problemes du present numero avant le 15 Mars 2009. Les solutions re ues apres ette date ne seront prises en
ompte que s'il nous reste du temps avant la publi ation des solutions. Chaque probleme sera publie dans les deux langues oÆ ielles du Canada (anglais et fran ais). Dans les numeros 1, 3, 5 et 7, l'anglais pre edera le fran ais, et dans les numeros 2, 4, 6 et 8, le fran ais pre edera l'anglais. La reda tion souhaite remer ier Jean-Mar Terrier, de l'Universite de Montreal, d'avoir traduit les problemes.
M369. Propose 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 BC en M et AD en N et d'autre part une droite verti ale oupant AB en Q et CD en R. Montrer que AP , DM et BR passent toutes par le m^eme point. M370. Propose par l'Equipe de Mayhem. (a) Montrer que cos(A + B) + cos(A − B) = 2 cos A cos B pour tous les angles A et B . D C−D (b) Montrer que cos C + cos D = 2 cos C + cos pour tous 2 2 les 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. Propose par Panagiote Ligouras, E ole Se ondaire Leonard de
Vin i, No i, Italie. Un segment AB de longueur 3 ontient un point C tel que AC = 2. On
onstruit d'un m^eme ot ^ e de AB deux triangles equilat eraux ACF et CBE . Determiner l'aire du triangle AKE si K est le point milieu de F C .
450 M372. Propose par l'Equipe de Mayhem.
Soit x un nombre reel satisfaisant x3 = x + 1. Trouver des entiers a, b 7 et c de sorte que x = ax2 + bx + c.
M373. Propose par Kunal Singh, etudiant, Kendriya Vidyalaya S hool, Shillong, Inde. Les ot ^ es d'un triangle sont mesures par trois nombres entiers onse u tifs et le plus grand angle est le double du plus petit. Determiner la longueur des ot ^ es du triangle.
M374. Propose par Mihaly Ben ze, Brasov, Roumanie.
Soit p un nombre premier xe, ave p ≥ 3. Trouver le nombre de solutions de x3 + y3 = x2 y + xy2 + p2009 , ou x et y sont des entiers. M375. Propose par Ne ulai Stan iu, E ole Te hnique Superieure de Saint Mu eni Sava, Ber a, Roumanie. Determiner toutes les solutions reelles du systeme d'equations 1 x2
+
4 y2
+
9 z2
= 4;
x2 + y 2 + z 2 = 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 horizontal line is drawn through P (4, 3), meeting BC at M and AD at N . A verti 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) A and B . (b) Prove that cos C + cos D = 2 cos C and D .
= 2 cos A cos B
C +D 2
cos
C−D 2
for all angles for all angles
( ) 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 same side of AB . If K is the midpoint of F C , determine the area of △AKE .
451
M372. Proposed by the Mayhem Sta . A real number x satis es x3 so that x7 = ax2 + bx + c.
= x + 1.
Determine integers a, b, and c
M373. Proposed by Kunal Singh, student, Kendriya Vidyalaya S hool, Shillong, India. The side lengths of a triangle are three onse utive positive integers and the largest angle in the triangle is twi e the smallest one. Determine the side lengths of the triangle.
M374. Proposed by Mihaly Ben ze, Brasov, Romania. Suppose that p is a xed prime number with p ≥ 3. Determine the number of solutions to x3 + y3 = x2 y + xy2 + p2009 , where x and y are integers.
M375. Proposed by Ne ulai Stan iu, Saint Mu eni Sava Te hnologi al High S hool, Ber a, Romania. Determine all real solutions to the system of equations 1 x2
+
4 y2
+
9 z2
= 4;
x2 + y 2 + z 2 = 9 ;
xyz =
9 2
.
Mayhem Solutions
M332.
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 integer height. The numeri al value of the volume is four times the numeri al value of its total surfa e area (in luding its top and bottom). Determine the smallest 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 = πr2 h and the surfa e area is A = 2πr 2 + 2πrh. From the hypotheses, πr2 h = 4(2πr2 + 2πrh), or rh = 8r + 8h, or rh − 8r − 8h + 64 = 64, or (r − 8)(h − 8) = 64. Note that r − 8 > −8 and h − 8 > −8. This gives us the following possibilities:
452 r−8 h−8 r h V
1 64 9 72 5832π
2 32 10 40 4000π
4 16 12 24 3456π
8 84 16 16 4096π
16 4 24 12 6912π
32 2 40 10 16000π
64 1 72 9 46656π
Thus, the smallest possible volume for the ylinder is 3456π. Also solved by DENISE CORNWELL, student, Angelo State University, San Angelo, \Abastos", TX, USA; RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; RICARD PEIRO, IES Valen ia, Spain; BILLY SUANDITO, Palembang, Indonesia; and TITU ZVONARU, Comane 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 must remove 1, 3, or 6 toothpi ks from the pile. The player who removes the last toothpi k wins the game. For whi h of the values of n from 36 to 40 in lusive does Brenda have a winning strategy? Solution by Ri hard I. Hess, Ran ho Palos Verdes, CA, USA, modi ed by the editor. We an build a table of winning and losing positions for Anne. Her winning positions start with 1, 3, or 6, sin e she an immediately win by removing all of the toothpi ks. Starting with 2 toothpi ks, Anne must remove 1 toothpi k, leaving Brenda with 1, and so Brenda wins. Starting with 4 toothpi ks, Anne must remove 1 or 3 toothpi ks, leaving Brenda with 3 or 1 (respe tively), and so Brenda wins by removing all of the toothpi ks. Starting with 5 toothpi ks, Anne an remove 3 toothpi ks, thus leaving Brenda 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 are losing 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 has a winning strategy starting with a pile of ea h of these sizes, then Anne will lose. In other words, if Anne has a winning strategy starting with piles of size n − 1, n − 3, and n − 6, then Anne will lose starting with a pile of size n, as Brenda an implement Anne's strategy for the smaller pile and win, no matter 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 to this size, thus preventing Brenda from being able to win, and so Anne herself will win. We an examine the ases from n = 7 to n = 40, obtaining the following lists:
453 Winning 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 a more detailed explanation.
M334. Proposed by the Mayhem Sta . (a) Determine all integers x for whi h
x−3 3x − 2
(b) Determine all integers y for whi h
3y 3 + 3 3y 2 + y − 2
is an integer. is 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
3x − 9 3x − 2 − 7 7 3A = = =1− . 3x − 2 3x − 2 3x − 2 7 Thus, 3x − 2 is an integer; that is, 3x − 2
is a divisor of 7, so 3x − 2 is one 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
3y 3 + 3 y 2 − 2y − 3 = y − 3y 2 + y − 2 3y 2 + y − 2 (y − 3)(y + 1) y−3 = y − = y − (y + 1)(3y − 2) 3y − 2 =
.
y−3 is an integer, 3y is an integer. From the solution to part (a), −2 y = 1 or y = 3, whi h answers part (b).
Sin e
y
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. x−3 Let f (x) = 3x . Then f (0) = 32 , f (1) = −2, f (2) = − 14 , and −2 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 for x−3 x > 3 and so 0 < < 1. 3x − 2
s+3 where s ≥ 1. Then f (x) = f (−s) = 3s . +2 Sin e 3s + 2 > s + 3 > 0 for s ≥ 1, f (−s) is not an integer by a similar argument so, f (x) is not an integer.
If x
≤ −1,
let x
= −s
454 Therefore, f (x) is an integer for integer values of x if and only if x = 1 or x = 3.
\Abastos", Valen ia, Spain; CAO MINH QUANG, Also solved by RICARD PEIRO, IES Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; and TITU ZVONARU, Comane 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 rst number. Also, the sum of the rst and fourth numbers is 9, the sum of the se 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 a2 + 4a2 + 49 − 28a + 4a2 + 81 − 18a + a2 − 78 5a2 − 23a + 26 (5a − 13)(a − 2)
= = = =
78 , 0, 0, 0,
hen e a = 13 or a = 2. 5 Therefore, the sequen es are a = 13 , b = 26 , c = 95 , d = 32 and 5 5 5 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 Binh Khiem High S hool, Vinh Long, Vietnam; KUNAL SINGH, student, Kendriya Vidyalaya S hool, Shillong, India; BILLY SUANDITO, Palembang, Indonesia; and TITU ZVONARU, Comane 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 x and y an integer. Suppose that n is a positive integer. Determine the number of 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.
455 The equation
(x, y) = (0, 0).
|x| + |y| = 0
has one integer solution only, namely
Consider next |x| + |y| = k, for an integer k with 1 ≤ k ≤ n. We an remove 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 + 1 solutions. 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), for a 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, so there are a total of k solutions here as well. Case 4. The integers x and y satisfy x < 0 and y < 0. −x − y = k, where 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 (k + 1) + k + k + (k − 1) = 4k solutions.
equation
|x| + |y| = k
has
Therefore, the number of latti e points inside the region is 1+
n X
4k
=
1 + 4
k=1
n X
k=1
=
1 + 4·
k = 1 + 4 · (1 + 2 + · · · + n)
n(n + 1) = 2n2 + 2n + 1 . 2
\Abastos", Valen ia, Spain. There were one in orre t Also solved by RICARD PEIRO, IES and two in omplete solutions submitted.
M337. Proposed by the Mayhem Sta . On sides AB and CD of re tangle ABCD with AD and E are hosen so that AF CE is a rhombus. (a) If AB = 16 and BC (b) If AB = x and BC
= 12,
= y,
< AB ,
points F
determine EF .
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 = F C = CE = EA = m. Let O be the point of interse tion of diagonals AC and EF of rhombus AF CE . Note that AC and EF are perpendi ular and bise t ea h other at O.
456 By the Pythagorean Theorem, CB 2 , y2 , y2 , x2 + y 2 , x2 + y 2 = . 2x
CF 2 − F B 2 m2 − (x − m)2 2 m − x2 + 2mx − m2 2mx
= = = =
m
Now, AF OA =
√
x2 + y 2 2x
and AC = AB 2 + BC 2 = 1 AC . Thus, by the Pythagorean Theorem again, 2 = m =
OF 2
= AF 2 − OA2 2 2 x + y2 = − 2x
Also,
!2 p x2 + y 2 2
x4 + y 4 + 2x2 y 2
x2 + y 2
=
=
x4 + y 4 + 2x2 y 2 − x4 − x2 y 2
4x2
−
4
4x2
4
=
p x2 + y 2 .
2 2
y +x y 4x2
.
Therefore, OF =
s
y 4 + x2 y 2 4x2
and EF = 2 OF =
In part (a), this yields
=
2y
p y 2 (y 2 + x2 ) 2x
p
x2 + y 2 2x
=
=
y
y
p x2 + y 2 2x
p x2 + y 2 x
,
.
√ 12 162 + 122 12(20) EF = = = 15 . 16 16 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 (part \Abastos", (a) only); RICHARD I. HESS, Ran ho Palos Verdes, CA, USA; RICARD PEIRO, IES Valen ia, Spain; CAO MINH QUANG, Nguyen Binh Khiem High S hool, Vinh Long, Vietnam; BILLY SUANDITO, Palembang, Indonesia; LUYAN ZHONG-QIAO, Columbia International College, Hamilton, ON; and TITU ZVONARU, Comane sti, Romania.
457 Problem of the Month
Ian VanderBurgh Here 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 s 3 0 5 6 −2 Challenge) In the grid shown, ea h row has −2 −5 0 1 y a value assigned to it and ea h olumn has a 5 2 x 8 0 value assigned to it. The number in ea h ell 0 −3 2 3 −5 is the sum of its row and olumn values. For −4 −7 −2 −1 −9 example, the \8" is the sum of the value assigned to the 3 row and the value assigned to the 4 olumn. Determine the values of x and y. It is tempting rst of all to give labels to the values that are assigned to the rows and olumns in order to be able to dive into some algebra. Let's label the values assigned to the ve olumns A, B , C , D, E and the values assigned to the ve rows a, b, c, d, e. Ea h entry in the table gives us an equation involving two of these variables. 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 and write down 25 equations, one for ea h entry in the table. These equations would in lude 12 variables { the 10 that label the rows and olumns together with x and y. We ould then spend pages and pages wading through algebra trying to ome up with the answers. At this point, we would hope that there has 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, I was re ently talking about this problem with a friend while driving and so neither of us really wanted to do any algebra, and so were for ed to ome up with better ways to do it.) Solution 1. If we hoose ve entries from the table whi h in lude one from ea h row and one from ea h olumn, then the sum of these entries is onstant no matter how we hoose the entries, as it is always equal to rd
th
A + 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 at the underlined numbers in the two grids below and the grid on the following page): 3 −2 5 0 −4
0 −5 2 −3 −7
5 0 x 2 −2
6 1 8 3 −1
−2 y 0 −5 −9
3 −2 5 0 −4
0 −5 2 −3 −7
5 0 x 2 −2
6 1 8 3 −1
−2 y 0 −5 −9
458 Therefore, = =
3 −2 5 0 −4
3 + (−5) + 2 + 8 + (−9) (−4) + (−3) + x + 1 + (−2) 3 + y + 2 + (−2) + 3 ,
0 −5 2 −3 −7
5 0 x 2 −2
6 1 8 3 −1
−2 y 0 −5 −9
or −1 = x − 8 = y + 6. Thus, x = 7 and y = −7. Solution 2. Consider the rst two entries in row 1. From the labels above, we have 3 = A + a and 0 = B + a. Subtra ting these, we obtain the equation 3 = 3 − 0 = (A + a) − (B + a) = A − B . Noti e that whenever we take entries in olumns 1 and 2 from the same row, 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 every entry in olumn 3 will be 5 greater than the entry in olumn 2 from the same row. 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 is 8, then every entry in olumn 5 is 8 less than the entry in olumn 4 from the same row. In row 2, we see that y = 1 − 8 = −7. Thus, x = 7 and y = −7. Solution 3. Consider the sub-grid
0 x
1 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, or x = 7. In a similar way, by looking at the sub-grid 18 y0 we 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. One footnote to the nal solution is that in fa t, in any sub-grid of the form p q , we must have p + s = q + r. Can you see why? r s Another interesting point about this problem is that it might be easier for those who know less! If we repla ed the x and the y with \?" and gave it to someone who didn't know a lot of algebra, they might nd the answers faster than those of us who go immediately to algebra. Sometimes, the extra ma hinery that we have an get in the way. As 2008 draws to a lose, the Mayhem Editor has three enormous sets of thanks to o er. First, to the Mayhem Sta , espe ially to Monika Khbeis and Eri Robert, for all of their help over the past year. Se ond, to the Editorin-Chief of CRUX with MAYHEM, Va lav Linek, for all of his help and en ouragement over the past year (as well as for his sharp eyes!). Third, to the Mayhem readership for their support. Please keep those problems and solutions oming!
459
THE OLYMPIAD CORNER
No. 274 R.E. Woodrow With the Winter break oming up, I have de ided to fo us this issue mainly on providing problems for your puzzling pleasure, and to give some time for the mails to deliver the solutions to problems from 2008 numbers of the Corner to restore the readers' solutions le, whi h is parti ularly thin for 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 the 47 IMO in Slovenia 2006. My thanks go to Robert Morewood, Canadian Team Leader at the IMO for olle ting them for our use. th
47th INTERNATIONAL MATHEMATICAL OLYMPIAD SLOVENIA 2006
Problems Proposed But Not Used
Contributing Countries. Argentina, Australia, Brazil, Bulgaria, Canada, Colombia, Cze h Republi , Estonia, Finland, Fran e, Georgia, Gree e, Hong Kong, India, Indonesia, Iran, Ireland, Italy, Japan, Republi of Korea, Luxembourg, Netherlands, Poland, Peru, Romania, Russia, Serbia and Montenegro, Singapore, Slovakia, South Afri a, Sweden, Taiwan, Ukraine, United Kingdom, United States of Ameri a, Venezuela. Problem Sele tion Committee. Andrej Bauer, Robert Gerets hlager, Geza Kos, Mar in Ku zma, Sventoslav Sav hev.
Algebra A1. Given an arbitrary real number a0 , de ne a sequen e of real numbers a0 , a1 , a2 , . . .
by the re ursion
ai+1 = ⌊ai ⌋ · {ai } ,
i ≥ 0,
where ⌊ai ⌋ is the greatest integer not ex eeding ai , and {ai } Prove that ai = ai+2 for suÆ iently large i.
A2. Let a0
= −1
by the re ursion
= ai − ⌊ai ⌋.
and de ne the sequen e of real numbers a0 , a1 , a2 , . . . n X an−k
k=0
k+1
for n ≥ 1. Show that an > 0 for n ≥ 1.
= 0
460
A3. Let c0 = 1, c1 = 0 and de ne the sequen e c0 , c1 , c2 , . . . by the re ursion cn+2 = cn+1 + cn for n ≥ 0. Let S be the set of ordered pairs (x, y) su h that X X x = cj and y = cj−1 j∈J
j∈J
for some nite set J of positive integers. Prove that there exist real numbers α, β , m, and M with the property that an ordered pair of non-negative integers (x, y) satis es the inequality m < αx + βy < M
if and only if (x, y) ∈ S . (By onvention an empty sum is 0.)
A4. Let a1 , a2 , . . . , an be positive real numbers. Prove that X i<j
X ai aj n ≤ ai aj . ai + aj 2(a1 + a2 + · · · + an ) i<j
A5. Let a, b, and c be the lengths of the sides of a triangle. Prove that √ √ √ b+c−a c+a−b a+b−c √ √ + √ √ √ √ + √ √ √ ≤ 3. b+ c− a c+ a− b a+ b− c
Combinatori s C1. There are n ≥ 2 lamps L1 , L2 , . . . , Ln arranged in a row. Ea h of them
is either on or o . Initially the lamp L1 is on and all of the other lamps are o . Ea h se ond the state of ea h lamp hanges as follows: if the lamp Li and its neighbours (L1 and Ln ea h have one neighbor, any other lamp has two neighbours) are in the same state, then Li is swit hed o ; otherwise, Li is 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 them
are 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 S whi h are outside of P . Prove that for every real number x X P
xa(P ) (1 − x)b(P ) = 1 ,
where the sum is taken over all onvex polygons with verti es in S . (A line segment, a point, and the empty set are onvex polygons of 2, 1, and 0 verti es, respe tively.)
461
C3. A ake is shaped as an n × n square with n2 unit squares. Strawberries
lie on some of the unit squares so that ea h row or olumn ontains exa tly one strawberry; all this arrangement A. Let B be another su h arrangement. Suppose that every grid re tangle with one vertex at the top left orner of the ake ontains no fewer strawberries of arrangement B than of arrangement A. Prove that the arrangement B 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 at its top right orner and bottom left orner, and then moving these two strawberries to the other two orners of that re tangle.
C4. An (n, k)-tournament is a ompetition with n players held in k rounds
su h that (a) Ea h player plays in ea h round, and every two players meet at most on e. (b) If player A meets 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 player D 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
upward unit triangular holes ut out. A diamond is a unit rhombus with angles of 60◦ and 120◦ . Prove that a holey triangle T an be tiled with diamonds if and only if for ea h k = 1, 2, . . . , n every upward equilateral triangle of side length k in T ontains at most k holes. n
C6. Let P be a onvex polyhedron with no parallel edges and no edge parallel to a fa e other than the two fa es it borders. A pair of points on P are antipodal if there exist two parallel planes ea h ontaining one of the points and su h that P lies between them. Let A be the number of antipodal pairs of 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 .
Geometry
G1. Let ABCD be a trapezoid with
AB||CD and AB > CD . Points K and L lie on the line segments AB and CD , respe tively, su h that AK : KB = DL : LC . Suppose that there are points P and Q on the line segment KL satisfying
D ...p..........................L C ....p........................................................p .. ...p ....... ....... ... .... ...... ..................... .............. ....... . . . . . . . . . . ... .. ....... . ......... P .......................................... ... .... .... .... .........
A
... ... ...... ... ................. ...... ..... ....... ... ....... .... .... .. ....... ..... ... .. .... ..................... ....... . . . . . . . . . . . . . ..... .. ........ . .......... ... . . . . . . . p . . .... .. ..... ...................... .... . . . . ..... ... ............. . . . . . Q ..... ... ........... . . . . . . .. .. .. ..p...............................................................p.....................................................................................p
and ∠CQD = Prove that the points P , Q, B , and C are on y li . ∠AP B = ∠BCD
K ∠ABC .
B
462
G2. Let ABCDE be a onvex pentagon su h that = =
∠CAD = ∠DAE ; ∠ACD = ∠ADE .
The diagonals BD and CE interse t at P . Prove that the line AP bise ts the side CD.
B
G3. A point D is hosen on the side
B
of a triangle ABC with
∠ACB < ∠BAC < 90◦
in su h a way that BD = BA. The in ir le of ABC is tangent to AB and AC at points K and L, respe tively. Let J be the in entre of triangle BCD. Prove that the line KL interse ts the line segment AJ at its mid-point.
C
....q .......... ... ............. ..... . . . ............................................................ ....... ..... ....... ..... . ..... ... . . ... . .... . ....q K ......... ........ . ... . . . . . ... ..... . ... . .. . . . . . . . .. ....... .. . ... . . . . . . .. .. . . .. . . . . . . q . . . . .. . . . . ... . ........................... ..... ..... ... .... .. .. .. ... ........ .. . ... ... .. .... .. . . .. . . . . . . .. ..... .. ... . ... .. .. ..... .......J ..q........................................................ ...... .......... ... . . . . . . . . . ....q........ ...... ... ...... . ... ...... .. . . . . . . . . . . . . . . . . . . . ..q....................................................q.....................q..............................................................q............................................................................................q
D
G4. In triangle ABC , let J be the
entre of the ex ir le tangent to side BC at A1 and to the extensions of sides AC and AB at B1 and C1 , respe tively. Suppose that the lines A1 B1 and AB are perpendi ular and interse t at D. Let E be the foot of the perpendi ular from C1 to line DJ . Determine the angles ∠BEA1 and ∠AEB1 .
A
... .... .......................................................... ........ ............... ....... ............ ..... .......... . . .... . 1...... .... .... ..... . . . ... ... .... . . . ... . . ...... .... .. . . .. ..... ... . . .. . .. ... ... . .. . . ... .. ... ... . . . . ... . . .. . . . . . .. . . .. . .... . . . . . . .. . . . . ... . .. .... . . . . ... . . . . . .. .. .. ..... .... .. . . . . . . . . . . .... .. . . . . ... . . . . . . . . . . .... .. .. . . .. ....... ... .. .... .... ...... ... .... .... ... ....... ... .. .... .... ...... ... ... ... ....... . . . . . . . . . . . ..... .... .. ... ... .... ............ ... ... ..... .... 1 ....... ........................... ................. ..... .... ....... ....... . .. ..... ................. . .... . . . . . . . . . . . .... . . . .. ..... ...... ............................................................................................................................................................................................................................................
B q
qJ
Cq
A
q
q
E
q
q q D B
A
G5. Cir les ω1 and ω2 with entres O1
and O2 are externally tangent at point D and internaly tangent to a ir le ω at points E and F , respe tively. Line t is the ommon tangent of ω1 and ω2 at D. Let AB be the diameter of ω perpendi ular to t, so that A, E , and O1 are on the same side of t. Prove that the lines AO1 , BO2 , EF , and t are on urrent.
L
.......... .........
AC
E
C
q C1
.... ...
..... E ... . ............q..................................................................... ..... . . . . . . . . . . . . .... ... .... ...
..... ... ......... ..... ........ ....
...... .... .... .... ................ .... .. ....... .. ... ... ... ... ............................ ...... ..... ........ . . .... . . . . . . ....... . .. ..... . . . . . . . . ..... . ........ ... . . . . . .. ... . ... . . . ... ... . ... ..... .. .. ... . . . . ... ........ .. .... . . .. 1 . . . . . . .. . .. . ... .. .... ....................2 .. ........ . . ... . .. ... ... ... .. .. .... ... .... ... 2 ... ..... .... .... . ... ...... . . . .... . .... ......... . . . ... . . . . . . . . ... 1 . ................................ ... .... .. . ............................................................................................................................................................................................................................. .. ... . .. ... .. .. . . .. ..... ... .. ... .... .... .... .
ω.................. .........
qF
q O
D
t
ω
A q
q
..............
∠BAC ∠ABC
.....p .................................... ......................... .. ... . A ...p....................................................... . . ... .. ..... ... .. ......................... ... . ... ... .... ...... ....................... . ........ ... . .... ... ........ ... .. ........ ... ... ... ... ........ ... . .. ........ . . .... . ... ........ .. .. ........ ... .. .. . ... ... ......... ... ... ... ............... .. . . .. . . . . P .... .. .. .. ..................p .. ... ... .. .. .. .. ...... . .. .. .. .. ....... p.... .. ...... D .. .. .. ... .. .. .. .. .. . . ...p..................... .. .. .. .. .. .. .. ....... ......... .. . . ............. ........ . ............. ....... ..................... ............. ... ........... ......p...
q O ω
qB
463
G6. In a triangle ABC , let Ma , Mb ,
be the respe tive mid-points of the sides BC , CA, AB and let Ta , Tb , Tc be the mid-points of the ar s BC , CA, AB of the ir um ir le of ABC not ontaining A, B , C , respe tively. For ea h i ∈ {a,b,c}, let ωi be the ir le with diameter Mi Ti . Let pi be the ommon external tangent to ωj , ωk su h that {i,j ,k} = {a,b,c} and su h that ωi lies on one side of pi while ωj , ωk lie on the other side. Prove that the lines pa , pb , pc form a triangle similar to ABC and nd the ratio of similitude. Mc
.................................................................. ......... ........ ........................................... ........ b .................... b ..... . . . . . .... .. .. ... ..... . . . . . . ... .... .. ... .. . . ....... . . . . . .... . . ... . .. . . . . . . . . . .... ...... . . .. .... .. .. .. ... .. .... .. .. ... ... . . .... ... .... ..... .. .. ... ... .. .... .. .. .. .. . . . . .. .... .... ... . . . .....................c . ...... ... ..... .. . . ...... ..... ...... . .. . ..... . ..... .. ... b ..... ... ... ... .. ...... ... .. ..... .... . . . . . . . ......... . . . ... . . . ................. .... . . . ... ....................... . . . . . . . . . . . . . . . . . . . . . . . .. . . . c .... ... . .. . . . . .... ... .. . .. . .. . . .... . . .. . . . . . . .... ... .... ... ... .. .... .. .. ... .... .. .. ... .. .... ... .. .... ... .... . . .... ... ... .... ..... a ... .. ... .. ..... ........................................................................................................................................................................................ . .... . .... . . . . . ... .... ... ... ... ..... .... .. ... ..... a .............. ..... ....... .. ........ .. .... . . .. . . . ......... . ... . . . ...... ........... ........... .... ........................ ...................................................
A.................................. .... ......r ........ ....
Tc r
ω
rT
ω
M r
r M
B
M r
r
r
ω
C
r Ta
G7. Let ABCD be a onvex quadrilateral. A ir le passing through A and D and a ir le passing through B and C are externally tangent at the point P in the interior of the quadrilateral. Prove that if ∠P AB + ∠P DC ≤ 90◦ and ∠P BA + ∠P CD ≤ 90◦ , then AB + CD ≥ BC + AD .
D r
.................................... ........... ........ ........ ..... ... ................................................................................... ..... . .. ................... . .... ..... . ... . . ... ........ ......... . .. .. ... .. ... .... ... .. .. . ... ...... . ... .... ... . ..... . ... .... .... . ... ... ..... . ... .... ... .... . . ...... . ... .. ... . . ... . .... . . . . . . . . . ..... ... ..... ..... . ... ........................ ........... .... .. .... . . . . . . . . . . . . . .... . . . . . . . . .................. ..... ... ......... .. ................ ........ .............. ............................ ............
rC
P r
r B
..r A
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 AB1 C1 , BC1 A1 , CA1 B1 interse t the 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, AB respe tively. Prove that the triangles A2 B2 C2 and A3 B3 C3 are similar. .............. .................... ................................................... A ............................ ........... ....... . ......... .................. ........................... . . ................................ . . ................. . . ..... ......................................... . . ............ . . . . . ... ..... ....... . A2............. ... ... ..... ....... ... ..... ... ... ...... . ....... . . . . . . ... .... ..... ..... . . . . . . . . .... ... ... .... ... .... ... . .. .. . ... ... ... . . . . .......................... ....... . . . . . . . . . . . . .. ... . . . . . . . . . . . . ............ B1 ..... . ....... . . . . . . . . . . . . . .... ... . . . ... .............. . ...... . . . . . . . . . . . . .. . . . . . .. ... .. .. ... ... .. .... ..... ... .. .. .. .. .... .... ... ..... ... .. .... ...... .... . .... . . . . ... ... . . . . ... .... .... . ... .. ..... . . .... . ... .... .... .. .. . . .. C1 ....................................... ... ... ... . .......... .. . .. .......... ...... .. ... ....... .. .... ................. .... ..... .... ... ....... .. ... ........ ... .... .. ... ... ........... .. . A . . . 1 . . . . .................................................................................................................................................................................. . . . .... . ...... B....... .. . . . . . ...... . . . ..... ...... ... . ....... . . ..... ..... . ..... . .. .. ........ .... . . ....... .... ......... ..... ........ . . . . ..... .... . . . . . . . . . . . . . . . . . . ................ .. . ... .. .... ................ ...... ... ... .......................... . .. .. ..... ... .. ... .... ... ..... ...... .. ...... ... . . . ... . . . .... .... ...... .. .... ..... ... ..... ..... .... ...... ..... ... ..... ......... C2 ....... .......... ........ ..... ......... ...... ................. .................. ...................... ............. . . . . . . . . . . . . ................................ .............. .. ..... ....... B2 ....................... ....... ........ .......... ......... . . . ............ . . . . . . . ..... ........................ ........................................
r
r
r
r
r
rC
r
r
r
r
464
Number Theory N1. Given x ∈ (0, 1) let y ∈ (0, 1) be the number whose n digit after the th
de imal point is the (2 ) digit after the de imal point of x. Prove that if x is a rational number, then y is a rational number. n
th
N2. For ea h positive integer n let 1 f (n) = n
n n n + + ··· + 1 2 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 equation x7 − 1 = y5 − 1 . x−1
N4. Let a and b be relatively prime integers with 1 < b < a . De ne the weight of an integer c, denoted by w(c), to be the minimum possible value of |x| + |y| taken over all pairs of integers x and y su h that ax + 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 h that 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 2005/2006
Final Round
November 19, 2005 (Time: 5 hours)
1. Find all solutions in integers x and y of the equation (x + y 2 )(x2 + y) = (x + y)3 .
465
2. 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 to another one. In how many di erent ways an the new queue be formed if ea h person maintains the same position as before, or is one step loser to the front, or is one step farther from the front? 3. In the triangle ABC the angle bise tor from A interse ts the side BC in the point D and the angle bise tor from C interse ts the side AB in the point 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 form an arithmeti progression, that is, the zeroes are a, a + d, a + 2d, and a + 3d where a and d are real numbers. Prove that the three zeroes of f ′ (x) also form 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 odd number of bla k squares. Show that the number of bla k squares among the four orner squares is even. In how many di erent ways an the hessboard be painted so that the above ondition is satis ed?
6. All the edges of a regular tetrahedron are of length 1. The tetrahedron
is proje ted orthogonally into a plane. Determine the largest possible area and the least possible area of the image.
Next we look at an alternate solution to problem 2 of the 17 Irish Mathemati al Olympiad dis ussed at [2007 : 151; 2008 : 88-89℄. th
2. Let A and B be distin t points on a ir le T . Let C be a point distin t from 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 . Show that ∠ABC > 72◦ . C ........ ........ .... ........ ... ........ . . . . .. . . .. ........ .. ..................................................................... . . . . . . . . . . . .. . . . . ............. ............. ....... .. . . .. . . . . . . . . . . . . . . . ........ .. ..... ..... ...... .. . . . . . . . . . . . . . .. . . . . . . . . . ....... ..... .. ..... ...... ... . . . . . . . . . . . . . . . . . ..... . ..... ... ... ............. . . . . . . . . .... .... ..... ... ............ ... . . . . . . . .... ... ..... . ......... . . ... . . . . . . .... .... ... .. ............ . .. . . . . . . . . .... . ... . .. . ..... ......... ... ..... ......... .... ..... .. ................ ..... ..... ... .................... . .... ... .. ........... ......... .......... .... ...................................................................................................................................................................................................................................................... .............. .. .. .......... . . . . ............. . . . . . . . ..... ... . . .............. ...... ... .............. .............. .. .. .............. ............. ..... ............................. ... .. ... .. ... .. . . .. ..
D
′
2β
D
β
γ
O
β
......
A
β
.. ...
Alternate Solution by Luyan ZhongQiao, Columbia International College, Hamilton, ON. Let D′ 6= B be the se ond point of interse tion of BD with the ir le T . Let ∠ABD = ∠DBC = β . Sin e AB = AC , we have ∠ACB = 2β . Also, BC is tangent to the ir le and AB is a hord, hen e ∠D ′ AB = β . Let ∠ABO = γ .
B
466 Sin e BC is tangent to T , we have 2β + γ = 90◦ . Adding the angles in the isos eles triangle ABC yields 4β + ∠CAB = 180◦ . From these two equations it follows that ∠CAB = 2γ . Sin e D is inside T we have β = ∠D ′ AB > ∠CAB = 2γ , β > γ . This last inequality together with 2β + γ = 90◦ yields 2 β 5β 2β + > 90◦ , or > 90◦ . Hen e, β > 36◦ and ∠ABC = 2β > 72◦ . 2 2
and therefore
Next we look at a omment on the solution to problem 6 of the 2007 Italian 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. Let x =
AP P A′
,
y =
Prove that xyz = x + y + z + 2.
BP P B′
,
z =
CP P C′
.
Comment by J. Chris Fisher, University of Regina, Regina, SK. The result is a theorem of Euler featured in the Crux Mathemati orum arti 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 Crux Mathemati orum. It an be lo ated in Edward Sandifer's book, How Euler Did It, and on his webpage: http://www.maa.org/news/howeulerdidit.html
li k on \19 Century Triangle Geometry". th
Now we turn to solutions from our readers to problems given in the February 2008 number of the Corner. First a solution to a problem of the 11 Form, Final Round, XXXI Russian Mathemati al Olympiad given in the Corner at [2008: 20{21℄. th
5. (N. Agakhanov) Does there exist a bounded fun tion f f (1) > 0 su h
that
for all x, y ∈ R?
: R → R
with
f 2 (x + y) ≥ f 2 (x) + 2f (xy) + f 2 (y)
Solved by Li Zhou, Polk Community College, Winter Haven, FL, USA. Suppose that f is su h a fun tion. Let a0 for n ≥ 1. Then 2
2
f (a1 ) ≥ f (a0 ) + 2f (1) + f
2
=1 1 a0
and an
= an−1 +
≥ 2f (1) .
1 an−1
467 As an indu tion step, assume that f 2 (an ) ≥ 2nf (1) for some n ≥ 1. Then 1 f 2 (an+1 ) = f 2 an + an
1 an 2 ≥ f (an ) + 2f (1) ≥ 2(n + 1)f (1) ,
≥ f 2 (an ) + 2f (1) + f 2
ompleting the indu tion. Hen e f 2 (an ) ≥ 2nf (1) for all n di ting the fa ts that f (1) > 0 and f is bounded.
≥ 1,
ontra-
And to omplete our les for the Corner, we look at a problem of the Taiwan 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 that a2 + b2 + c2 R ≤ , √ 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 abc 2F = , but also from the de nition of p, q, and r we have the equation 2R 2F = ap + bq + cr . Thus, the proposed inequality is equivalent to abc 2(ap + bq + cr)
or
≤
a2 + b2 + c2 √ 18 3 pqr
(a2 + b2 + c2 )(ap + bq + cr) ≥ 9abc
By the AM{GM Inequality, √ 3
and and the inequality (1) now follows from a2 + b2 + c2 ≥ 3
a2 b2 c2
√ 3
pqr .
ap + bq + cr ≥ 3
(a2 + b2 + c2 )(ap + bq + cr) ≥ 9
(1)
p 3 abcpqr ,
√ √ √ 3 3 a2 b2 c2 · abc · 3 pqr .
That ompletes the Corner for this number, and this Volume. As Joanne Canape, who has been translating my s ribbles into beautiful LTEX has de ided that twenty-plus years is enough, I want to thank her too for all the help over the time we've worked together. A
468
BOOK REVIEW
John Grant M Loughlin The Symmetries of Things By John H. Conway, Heidi Burgiel, and Chaim Goodman{Strauss, published by AK Peters, Wellesley, MA, 2008 ISBN 978-1-56881-220-5, hard over, 423+xxv pages, US$75.00 Reviewed by J. Chris Fisher, University of Regina, Regina, SK The authors set themselves the ambitious goal of produ ing a book that appeals to everybody. As far as I an tell from a single reading, they have su
eeded admirably. The rst thing anybody would noti e about the book is that it is lled with beautiful and fas inating photographs and omputer drawings. No spe ial knowledge is required for admiring beauty; this book would be as mu h at home on the living room o ee table as on the oÆ e shelf. Of ourse, it is primarily a mathemati s book. The ontents have been organized into three parts. The rst of them des ribes and enumerates the symmetries found in repeating patterns on surfa es; it is written at a level suitable for Crux with Mayhem readers. This part might well serve as a textbook for a geometry ourse dire ted at university students spe ializing in mathemati s, edu ation, physi al s ien e, or omputer s ien e. What makes the authors' approa h both novel and elementary is the introdu tion of what they all the orbifold signature notation. Groups are not needed here; the on ept an be easily des ribed and qui kly mastered. Here is the idea. A point in a pattern where two mirrors of symmetry π meet at an angle of m is alled kaleidos opi and is denoted by ∗m; points having rotational symmetry of order m (but no kaleidos opi symmetry) are
alled gyrational and are denoted by m (with no asterisk). If a region has an 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 repeating plane pattern one writes down the symbols ... starting from the middle and working out... ... ward. First lo ate mirror lines and ea h .. ... kind of kaleidos opi point, if any (where two 3 .................... ........... points are of the same kind if they are related ∗6 ..................................... ..... .... . . . . . . ..r . ................................ by a symmetry of the pattern); list them after .. .. .. ... ... .. 2 the asterisk in de reasing order. Next lo ate any gyrational points and order them before any asterisk. Then look for mira les. Typi al signatures are ∗632 for a pattern whose symmetry is explained by three kinds of mirrors that meet in pairs at angles of π6 , π3 , and π2 ; ∗632 632 (having no asterisk) for a pattern with ... ... ... . . . ... ... ... ... ... ... ... .... ... .... ... .... ... ... ... .. ...... . . . . .................................................................................................................................................................................................... . . .. . .. ... ... ... ..... ... ... .. ... ... .... ... ..... ... .. .. ... ... ... . . . . ... . ... ... ... ... ... ... ... ... .. ... ... ... .. .. ... . . ... . . . . ... ... . .. ... . . . . . ... . ... . . . . . . . ... . ... .. ... ... ... ... . . . . . . . ........................................................................................................................................................................................... . . . . . . . .. . .. ... .... ... .... ... ... .. ... .. ... .. ... ... ... ... ... ... . .. ... . . ... . ... .. . . .. . . . . . . ... ... . ... ... ... ... ... ... .. .. ... .. ... ... ..... ... .... ... ..... ... .. ... ... ... .. ...... . . . . .......................................................................................................................................................................................................... . ... ... ... ... ... ... ... ..... ... ..... ... ..... .. .. .. ... ... ...
469 6{fold, 3{fold, and 2{fold gyrational points but no re e tions or 2∗22 for a pattern with two kinds of kaleidos opi points where
mira les; a pair of mirrors interse t at right angles, and one point where there is a half-turn symmetry but no mirror. ....................................................................................................................................................................................................................................... ... ... ... ... .... .... .... .... ... ... ... ... .... .... .... .... .. .... .... .... .......................................................................................................................................................................................................................................... ..... ..... ..... ..... ..... .... .... .... .... .... ... ... ... ... ... .. . .... .... .... . . . . . . . . . . . ............................................................................................................................................................................................................................................. .. .. . . . . ... ... .... .... . . . .... ..... .................................................................... ... ... .. ... ... .. .. ...................................................................................................................................................................................................................................................... .. .. . .... .... . .... .... .... ......................................................... ..... .... .... .... .... .... .... . .... .... . .... . . .. . . . . . . . . ....................................................................................................................................................................................................................................... ... ... . . . . .... .... .... .... . . . ... ... .... .... .... .... .... .... .... .. .. .. .. .. ............................................................................................................................................................................................................................................. ... ..... ..... ..... ..... . .. . .... .... . .... ... ... ... .... ... .... .... .... .... .... . . . . . . . . ..............................................................................................................................................................................................................................
∗
2
q
2 2
632
2∗22
Unlike most other notation systems that have been devised for des ribing plane symmetry, these orbifold signatures an also be used to des ribe frieze patterns and spheri al patterns. (We learn in Part III that they work equally well for des ribing hyperboli patterns.) But how does one know that the resulting lists of 17 signatures for plane patterns, 7 signatures for frieze patterns, 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 signature in su h a way that plane patterns and frieze patterns ost exa tly $2 while spheri al patterns ost a bit less. That theorem tells us immediately whi h signatures are feasible. To establish the Magi Theorem, a pattern on the surfa e is asso iated with a folded surfa e they all an orbifold. The orbifold is obtained by identifying points related by a symmetry of the pattern (whereby points of an orbit are folded atop of one another so that a single representative point of every orbit lives on the orbifold). This sounds a bit s ary, but the authors manage to explain the details in a gentle way using suitable pi tures and simple examples. They then state the Classi ation Theorem for Surfa es and provide Conway's elementary and intuitive Zip proof. They also prove that these surfa es an be distinguished using Euler's formula (involving the numbers of verti es, edges, and fa es of a suitable map on the surfa e), whi h they also prove. Sin e the orbifolds are easily lassi ed using 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 authors a hieve this feat by repeatedly redu ing te hni al diÆ ulties down to problems that are postponed to the following hapter. This way they present one on ept at a time, as ompared to the typi al textbook's initial barrage of poorly motivated de nitions and lemmas. Their proofs are every bit as brilliant as their notation. The illustrations are not just beautiful, but they
470 have been arefully hosen to larify the exposition. I really appre iated the authors' de ision to repeat pi tures that they require for illustrating new ideas | instead of making the reader turn ba k to a pi ture on an earlier page, they reprodu ed a smaller version of it whenever needed. The authors
learly have fun oining whimsi al new words; their terminology will not appeal to everybody, but the informal nature of their dis ussions makes for enjoyable 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 need for gyrational in pla e of rotational or wandering in pla e of translation. We will 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 that the authors had intended to write. But they de ided it was worthwhile to extend the signature to olour symmetry, and the book grew from there. For a areful reading of Part II the reader needs some group theory and a bit of mathemati al maturity. The authors' main goal for this part is to present their analysis and notation for olour symmetry. They enumerate the p{fold olour types for plane, spheri al, and frieze patterns (for all primes p). The omplete lassi ations appear in a book for the rst time. Along the way the authors show how their orbifold notation orresponds to previous lassi ation systems, whi h gives them the opportunity to dis uss the short omings of those systems. Also in this part, they enumerate the isohedral tilings of the sphere and plane, and they extend to n = 2009 the Bes 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 readers who 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 ompletely understood only by a few professional mathemati ians. Still, as they point out, mu h of Part III an pro tably be explored by other readers, while many more will enjoy inspe ting the pretty pi tures. Here, among other things, the authors dis uss hyperboli groups and Ar himedean polyhedra and tilings; they list the 219 rystallographi spa e groups (and explain why hemists distinguish 230 groups), and they provide a omplete list for the rst time in print of the 4{dimensional Ar himedean polytopes. Apparently they ould have kept writing, but they de ided to leave something for the rest of us to do. Their nal words are, \A universe awaits | Go forth!" I thank Bru e Shawyer for inviting me to serve as Book Review Editor. I am grateful to the late Jim Totten for guiding me during his tenure; Bru e Crofoot for insightful ommentary; Va lav Linek for re ent support; Shawn Godin for leadership with Mayhem. Thanks go to all the reviewers, but espe ially this trio: 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 to help; and my su
essor, Amar Sodhi. Amar's passion for mathemati s will shine as he assumes this role. Wel ome Amar! Thanks to the CRUX with MAYHEM ommunity for an enjoyable journey. | John Grant M Loughlin
471 Old Idaho Usual Here
Robert Israel, Stephen Morris, and Stan Wagon
N. Kildonan [1℄ raised the following problem. Take an arbitrary word using at most 10 distin t letters, su h as DON ALD COXET ER. Can one subsititute 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 then the 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 be even; d = 5 or d = 10 are just as easy. Kildonan proved that divisibility by d = 3 an always be a hieved and R. Israel and R. I. Hess extended this to d = 9; the ase of d = 7 was left unresolved. In this note we settle all ases. The reader interested in an immediate hallenge should try to prove that divisibility by d = 45 is always possible. This appears to be the hardest ase. To phrase things pre isely, a word is a string made from 10 or fewer distin t letters; for ea h word and ea h possible substitution of distin t digits for the letters, there is an asso iated value: the base 10 number one gets after making the substitution. If all substitutions yield a value for the word w that is not divisible by d, then w is alled a blo ker for d. If any word ending (on the right) with w fails to be divisible by d, then w is alled a strong blo ker for d. An integer d is alled attainable if the value of every suÆ iently long word an be made divisible by d by some substitution of distin t digits for letters. Thus d is not attainable if there exist arbitrarily long blo kers. The use of arbitrarily long strings is important be ause, for example, AB is a blo ker for 101, but only be ause it is too short. An integer d is strongly attainable if the value of every word an be made divisible by d by an appropriate substitution. In this paper we will nd all attainable integers; moreover, they are all strongly attainable. Note that any divisor of an attainable number is attainable. Some ases, su h as d = 2, d = 5, or d = 10 are extremely easy to attain, and it is just about as easy to attain d = 4 or d = 8. It takes a little work to show that d = 3 and d = 9 are attainable (proofs given below). Our main theorem resolves the attainability status of all integers. Theorem 1 An integer is attainable if and only if it divides one of the integers
18, 24, 45, 50, 60,
or 80.
To start, we dis uss the ases of d = 3 and d = 9 for ompleteness and to introdu e the ideas needed later. We use the well known fa t that when d is 3 or 9, then d divides a number if and only if d divides the sum of its digits. Copyright
c 2008
Canadian Mathemati al So iety
Crux Mathemati orum with Mathemati al Mayhem, Volume 34, Issue 8
472 The number d = 3 is attainable (Kildonan [1℄). Given a word, let the 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 h Bi 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 three pairs among the Bi and repla e them with digits 1 and 2, and 4 and 5 for the se ond pair, and 7 and 8 for the third pair. Then look for pairs of the Ci and 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 possibly four letters (sin e we used three pairs) and we an substitute 0, 3, 6, and 9 for 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, where n − k is not divisible by 3. Assign 9 to this letter and assign 0 to 8 arbitrarily to 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 k to 0 in the ase of 8. This adds n − k to the value modulo 9. But n − k is relatively prime to 9, so we an do this −v/(n − k) times, where the division uses 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 assignment will 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), and then we pro eed as in the rst half of the proof: assign 9 to this letter, 0 to 8 to the other letters, and then y li ally permute the values 0 to 8. Ea h permutation adds n − k modulo 9 and this will eventually transform the value v, whi h is divisible by 3, to a value divisible by 9, be ause 3 divides n − 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 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.
25,
The ordering of these lemmas indi ates how Theorem 1 was found. First the ases of d = 7 and d = 11 were settled and that led to the general result of Lemma 1. It followed that the only andidates for attainability had the form 2a 3b 5c . 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. Resolving the situation for those divisors, with some omputer help, led to Lemmas 3 and 4. Finally, the omputer sear hes were eliminated and the whole thing was redone by hand. Theorem 1 follows from the lemmas be ause Lemmas 3 and 4 settle the status of all 45 divisors of 3600.
473 A key idea is that the ten digits sum blo ker to 45. So we begin with Lemma 3, d AAB whi h shows how unattainability is 27 ABBAB proved. We use the fa t that an in- 32 (ABCDEF GHJ )5 J 4 K teger is ongruent modulo 9 (hen e 36 modulo 3) to the sum of its digits. 48 (ABCDEF GH)2 KKJ K AABA Let A, B , C , D, E , F , G, H , J , 75 90 A6 (BCDEF GHJ )7 J K K be the ten letters and let w g be AB the on atenation of g opies of word 100 120 ABCDEF GHJ J J K w . The table at right lists the blo kBBA ers needed for Lemmas 2 and 3; most 125 were found by a omputer sear h. We 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 (ABCDEF GHJ )5 J 4 K ≡ K + 4J + 5(45 − K) ≡ 4J − 4K (mod 9)
.
The only way 4(J −K) is divisible by 9 is if J K is either 90 or 09, and neither is 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 the word (ABCDEF GH)2 KKJ K must be one of 0080, 2272, 4464, 6656, or 8848, as these are the only words of the form KKJ K that are divisible by 16. However, now the value of the word modulo 3 is one of the entries below, 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 value modulo 48. Case 3. The number d = 75 an be blo ked. Here we have the ongruen e AABA ≡ 51A + 10B (mod 75). Multiplying by 53 transforms the ongruen e to 3A + 5B ≡ 0 (mod 75). However, 3 ≤ 3A + 5B ≤ 69, so the ongruen e is never satis ed. Left extension by A3i preserves the value modulo 75. Case 4. The number d = 90 an be blo ked. We have A6 (BCDEF GHJ )7 J K ≡ 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 extension by A9i preserves the value modulo 9.
474 Case 5. The number d = 120 an be blo ked. We have ABCDEF GHJ J J K ≡ 45 + 2J ≡ 2J (mod 3)
.
However, J J K must be either 440 or 880 to obtain divisibility by 40, therefore, 2J is either 8 or 16, and so is not divisible by 3. Left extension by A3i preserves the value modulo 120. Case 6. The number d = 32 an be blo ked. Any word ending in ABBAB has a value satisfying 10010A + 1101B ≡ 26A + 13B (mod 32). If this is
ongruent to 0 modulo 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 if and only if it ends in 125, 250, 375, 500, 625, 750, 875, or 000. Thus, BBA is 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 add the 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 involving 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) when gcd(a, d) = 1. It follows that if d is oprime to 10, then there is a smallest positive integer, denoted by ordd (10), su h that 10ord (10) ≡ 1 (mod d). d
Theorem 2 Let d be oprime to 10 and greater than 10 with e = ordd (10).
Then w = Ake−1 B is a blo ker for d for any positive integer k.
Proof: Assume rst that k = 1 so that w is just Ae−1 B . If 3 does not divide d then the value of w satis es the ongruen e B+A
e−1 X i=1
Sin e
10i = B − A + A
10e − 1 ≡ B − A (mod d) 9
.
d > 10, d annot divide B − A. Now suppose that 3 divides d and d > 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 d divides 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, namely d ∈ {21, 27, 33, 39, 51, 57, 63, 69, 81}. Suppose that d is one of these but d 6= 21, 27, 81; then ordd (10) = ord3d (10). This means that from B − 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 that d ≤ 27, a ontradi tion.
475 For d = 21 the value of w modulo 21 is B − A, whi h is not divisible by 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 is 8A + B and 1 ≤ 8A + B ≤ 80, so the value is not divisible by d. The extension to the ase of general k is straightforward. 21.
The pre eding result blo ks all primes greater than 10. We need to deal also with d = 7. One an give an alternate onstru tion in the general ase that 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 = KJ K e HK e GK e F K e EK e DK e CK e BK e AK e ,
where e = ordd (10)
ongruent to 9(d 2+ 1)
− 1.
Then after any substitution the value of (mod d), and so is not divisible by d.
w
is
For d = 7 the word w of the Proposition has length 55. A di erent approa 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 ker for d = 7, and it an be made arbitrarily long by prepending E 6 . On to Lemma 2. The positive results for d = 16 and d = 25 are not diÆ 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 the negative 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 the rightmost two digits. Case 2. The number d = 80 is strongly attainable. Assign 0 to the rightmost letter; 8 to the next new letter that o
urs reading from the right, 4 to the next one, and 2 to the next one after that. The value is then divisible by 16 and 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 an assignment that makes it divisible by 9. If the rightmost digit is even, we are done. Otherwise, repla e this digit y with 9 − y. This preserves divisibility by 9 and makes the rightmost digit even. Case 4. The number d = 60 is strongly attainable. The word ends in either AA or BA. In either ase, assign 0 to A and 6 to B . Let the eight remaining letters 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 Bi or 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 the proof for d = 3 so as to guarantee divisibility by 8. Re alling the proof that d = 3 is strongly attainable, all two letters mat hed if they are repla ed in
476 that proof by 1 and 2, or by 4 and 5, or by 7 and 8. If the word ends in AAA just make sure A is either 0 or 8. The remaining ases are that the word ends in 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 are unmat 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 the mat 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 A is not, use 088. If the ending is BBA, then pro eed as if the ending was ABB , but use instead 448, 336, 008, and 880 for the four sub ases. If the ending is ABA, then pro eed similarly, using 848, 696, 808, and 080 for the four sub ases. Case 6. The number d = 45 is strongly attainable. Let m(X) denote the multipli ity of the letter X in the given word redu ed modulo 9; let Xˆ denote the digit assigned to X . Let A be the rightmost letter and assign 0 to it, thus ensuring divisibility by 5. Assume rst that the multipli ities of at least eight of the nine remaining digits are all mutually ongruent modulo 3, and assign 9 to the other letter. Let the m-values of the eight letters be 3ai + c, where ea h ai is a non-negative integerPand c ∈ {0, 1, 2}. Let Li be the digits assigned to these eight letters. Li = 36P , whi h 9 divides, the value modulo 9 of the P Sin e word is 3 ai Li . So we want ai Li 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 to the 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 ongruent modulo 3 to the length of the word; therefore S , the sum of the mulˆ = 9. tipli ities of the nine letters other than K , is not divisible by 3. Let K ˆ = 1 and Case 6a. There is a letter, say B , with m(B) = m(A). Then let B assign the remaining digits arbitrarily. Case 6b. There is no letter as in Case 6a. Then we an nd two letters among B , 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 Bˆ to be the non-negative residue modulo 9 of m(D) − m(C) and note that Bˆ − Aˆ ≡ m(D) − m(C) (mod 9) whi h is not divisible by 3. Assign unused digits to the letters C and D so that ˆ −D ˆ ≡ m(B) − m(A) (mod 9); there are enough digits left for this to be C possible. Assign the remaining digits arbitrarily. Now we an treat both ases to get the result. The assignment produ es some total value, redu ed modulo 9 to v. If v 6= 0 then repla e ea h digit between 0 and 8 by the next higher digit, wrapping ba k to 0 in the ase of 8. ˆ modulo This adds S to the value modulo 9 and does not alter Bˆ − Aˆ or Cˆ − D 9. But S is relatively prime to 9, so we an do this −v/S times, where the division uses the inverse modulo 9 of S , in order to a hieve divisibility by 9. If Aˆ is 0 or 5, then we are done.
477 If Aˆ is 3 or 6, then swit h digits of A and B , where we know that Bˆ is not divisible by 3; this is be ause the value modulo 3 of Bˆ − Aˆ, whi h starts out nonzero, does not hange in the translational step. If we are in Case 6b, ˆ ; the net hange is then also swit h Cˆ and D ≡
ˆ −A ˆ m(A) − m(B) + C ˆ−D ˆ m(D) − m(C) B ˆ −A ˆ D ˆ −C ˆ + C ˆ−D ˆ B ˆ −A ˆ ≡ 0 (mod 9) , B
so divisibility by 9 is preserved. As Aˆ is now not divisible by 3, we an multiply ea h digit less than 9 by 5/Aˆ (mod 9). This preserves divisibility by 9 and makes Aˆ = 5. The total is now divisible by 45. The proof that d = 45 is strongly attainable is 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 alphabet to the two letters A and B . We use the terms 2-attainable and 2-blo ker in this ontext. Using te hniques similar to those presented, we obtained the following 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 h reader might enjoy nding them; they are
d ∈ {28, 36, 48, 120, 175}. The all short, of length at most 7.
A knowledgment
We 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 University of British Columbia Van ouver, BC, Canada
[email protected]
Stan Wagon Ma alester College St. Paul, MN, USA
[email protected]
Stephen Morris Newbury, Berkshire, England
[email protected]
478 A Limit of an Improper Integral Depending on One Parameter
Iesus C. Diniz In this arti le we will al ulate the following limit of an improper integral that depends on one parameter λ ∈ R+ lim
λ→0+
Z
∞
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 order of the limit and the integral. Through a ni e appli ation of elementary tools su h as hange of variables in integrals and the Binomial Theorem, one is able to obtain this limit. This problem arose in the al ulation of a lower bound for a probability of theoreti al interest in the study of multidimensional Poisson point pro esses, namely lim
λ→0+
Z
∞
0
exp −λvn (1) (r + l)n − r n λvn (1)nr n−1 exp −λvn (1)r n dr ,
where vn (1) is the volume of the n{dimensional unit ball, λ is the Poisson intensity, l is the distan e between two distinguished points in Rn , and r is the distan e from the rst point to the losest o
urren e in the Poisson point 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
lim
λ→0+
Z
∞
0
exp −λK (x + l)n λKnxn−1 dx = 1 .
Proof: The ase n = 1 is a straightforward al ulation: lim
λ→0+
= =
Z
∞ 0
exp −λK(x + l) λK dx
∞ lim − exp −λK(x + l)
λ→0+
0
lim exp (−λKl) = 1 .
λ→0+
Hen eforth, we take n > 1. Copyright
c 2008
Canadian Mathemati al So iety
Crux Mathemati orum with Mathemati al Mayhem, Volume 34, Issue 8
479 Making rst the hange of variable v = Kxn ,
dv = Knxn−1 dx ,
in the integral we obtain Z
∞ 0
h 1 n i 1 λ exp −λ v n + (ln K) n dv .
Changing variables again, this time a
ording to 1
1
1
u n = v n + (ln K) n
,
dv =
we obtain lim
λ→0+
=
lim
λ→0+
=
lim
λ→0+
Z
∞
0 ∞
Z
v
du ,
h 1 n i 1 λ exp −λ v n + (ln K) n dv λ exp (−λu)
ln K
Z
( n1 −1) u
( n1 −1) u v
du
n n1 n−1 l K λ exp (−λu) 1 − du ,
∞
u
ln K
whi h by the Binomial Theorem be omes
=
∞
n−1 X
n k n−1 k l K n (−1) du u λ→0+ ln K k k=0 n nk n−1 X n − 1 Z ∞ l K 1 + lim λ exp (−λu) (−1)k du . λ→0+ u k ln K k=1 lim
Z
λ exp (−λu)
After yet another hange of variable, 1
w = un
,
dw =
1 ( n1 −1) u du , n
the pre eding expression be omes 1 + lim
λ→0+
n−1 X k=1
k Z ∞ n−1 1 n n −lK λ exp −λw n w n−1−k dw . 1 k lK n
It remains to show that
lim λ
λ→0+
Z
∞ 1
exp (−λw n ) w n−1−k dw
lK n
is zero for ea h k ∈ {1 , 2, . . . ,
n − 1}.
(1)
480 To do this, we make a nal hange of variables z = λn w ,
dz = λ n dw ,
1
1
whi h yields λ = λ
Z k n
∞
exp (−λw n ) w n−1−k dw
1 lK n
Z
∞
Z
∞
1 (λln K) n
k
< λn
0
exp −z n z n−1−k dz
exp −z n z n−1−k dz .
(2)
It now suÆ es for us to show that the last integral in (2) is nite for ea h k ∈ {1, 2, . . . , n − 1}. Setting Z 1
C =
exp (−z n ) z n−1−k dz
0
and noting that if z ≥ 1, then z n−1−k Z
∞
0
exp −z n z n−1−k dz
≤ z n−1 , we nally obtain Z ∞ ≤ C + exp −z n z n−1 dz 1
= C +
1 ne
,
whi h is a nite number. This ompletes the proof.
Acknowledgments This work was supported by FAPESP grant 0415864-1. The author also thanks J.C.S. de Miranda for many dis ussions about this problem. Iesus C Diniz Institute of Mathemati s and Statisti s University of S~ao Paulo S~ao Paulo, Brazil
[email protected]
481 Sliding Down In lines with Fixed Des ent Time: a Converse to Galileo's Law of Chords
Je Babb Suppose that a ve tor is an hored at the origin and lies along the positive x-axis. Consider rotating the ve tor ounter lo kwise about the origin through an angle A, with 0 < A < π. Consider a parti le of mass m whi h is initially at rest and then slides, under gravity and without fri tion, from a starting point on the in lined ve tor down towards the origin. Let DA denote the distan e of the starting point from the origin. For ea h value of A, suppose that DA is hosen to ensure that the parti le requires exa tly T se onds to rea h the origin. Determine the urve hara terized by the starting points of the parti les. For the parti le under onsideration, let dA (t) be the distan e travelled along the ve tor at time t, vA (t) = d′A (t) be the velo ity along the ve tor at ′ time t, and let aA (t) = vA (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 , then dA (t) =
K 2 t . 2
(1)
This may be on rmed by integrating aA (t) twi e with respe t to time and applying the initial onditions dA (0) = 0 and vA (0) = 0 to obtain zero for both onstants of integration. Sin e the parti le is sliding down a fri tionless in line in the Earth's gravitational 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, then DA = dA (T ) =
g 2
T 2 sin A .
(2)
A point (r, θ) in polar oordinates may be expressed in Cartesian oordinates as (x, y), where x = r cos θ and y = r sin θ. Consider the following equation, whi h is expressed in polar oordinates as r = 2c sin θ ,
where r yields Copyright
> 0
and 0
< θ < π.
Multiplying both sides of equation (3) by r
r 2 = 2cr sin θ c 2008
(3)
Canadian Mathemati al So iety
Crux Mathemati orum with Mathemati al Mayhem, Volume 34, Issue 8
(4)
482 By setting x = r cos θ and y Cartesian oordinates as
= r sin θ,
equation (4) may be re-expressed in
x2 + y 2 = 2cy .
Subtra ting 2cy from ea h side and ompleting the square on y2 − 2cy yields x2 + (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) is a ir le resting upon the origin (0, 0), entred g 2 at 0, 4 T on the verti al axis, and of radius g 2 T . 4 On an histori al note, the motivation for r 0, g T 2 this problem arose while onsidering the Law 4 of Chords, whi h was stated and proven by Galileo Galilei in his 1638 masterpie e Dialogues Con erning Two New S ien es. Galileo
onsidered rates of des ent along a verti al irA
le and, with his Proposition VI, established the Law of Chords (see [1℄, p. 212): If from the highest or lowest point in a verti al ir le there be drawn any in lined planes meeting the ir umferen e, the times of 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 geometri 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 lowest point on the urve is onstant for all hords. This paper demonstrates that the verti al ir le, or one of its omponent ar s interse ting at the lowest point 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 Babb Department of Mathemati s and Statisti s University of Winnipeg Winnipeg, MB, R3B 2E9 Canada
[email protected]
483
PROBLEMS Toutes solutions aux problemes dans e numero doivent nous parvenir au plus tard le 1er juin 2009. Une etoile (⋆) apres le numero indique que le probleme a et e soumis sans solution. Chaque probleme sera publie dans les deux langues oÆ ielles du Canada (anglais et fran ais). Dans les numeros 1, 3, 5 et 7, l'anglais pre edera le fran ais, et dans les numeros 2, 4, 6 et 8, le fran ais pre edera l'anglais. Dans la se tion des solutions, le probleme sera publie dans la langue de la prin ipale solution present ee. La reda tion souhaite remer ier Rolland Gaudet, de College universitaire de Saint-Bonifa e, Winnipeg, MB et Jean-Mar Terrier, de l'Universite de Montreal, d'avoir traduit les problemes.
3371. Corre tion. Propose par George Tsintsifas, Thessalonique, Gre e.
^ es respe tifs a, b et c, et soit M un de Soit ABC un triangle de ot ses points interieur. Les droites AM , BM et CM oupent respe tivement les ot ^ es opposes aux points A1 , B1 et C1 . Les droites passant par M et perpendi 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 [A2 B2 C2 ] [A1 B1 C1 ]
=
(ap1 + bp2 )(bp2 + cp3 )(cp3 + ap1 ) 2a2 b2 c2
ou [KLM ] designe l'aire du triangle KLM .
a p1
+
b p2
+
c p3
,
3389. Propose par Mihaly Ben ze, Brasov, Roumanie.
Pour a ∈ R, la suite (xn ) est de nie par x0 = a et xn+1 = 4xn − x2n pour tout n ≥ 0. Montrer qu'il existe un nombre in ni de valeurs a ∈ R telles que la suite (xn ) est periodique.
3390. Propose par Mihaly Ben ze, Brasov, Roumanie. Montrer que si A, B , C et D sont les solutions de X
2
=
3 −5 5 8
,
alors A2007 + B 2007 + C 2007 + D2007 = O, ou O est la matri e nulle de taille 2 × 2.
3391. Propose par Mi hel Bataille, Rouen, Fran e.
onvexe tel que AC et BD se oupent a Soit ABCD un quadrilatere angle 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 (P IJ ), (P J K), (P KL) et (P LI) sont
ongruents si et seulement si ABCD est y lique.
484
3392. Propose par Mi hel Bataille, Rouen, Fran e. Soit A, B , C , D et E on y liques ave respe tivement V et W sur les a CB par V droites AB et AD. Montrer que si la droite CE , la parallele et la parallele a CD par W sont on ourantes, alors les triangles EV B et EW D sont semblables. La re iproque est-elle vraie ?
3393. Propose par Dragoljub Milosevi et G. Milanova , Serbie. Soit le triangle ABC , ou demi perim etre. Montrer que
a = BC , b = AC , c = AB
et ou s est le
y+z A z+x B x+y C 9 · + · + · ≥ 2 x a(s − a) y b(s − b) z c(c − a) s
,
ou les angles A, B et C sont mesures en radians et x, y et z sont des nombres reels positifs quel onques.
3394. Propose par Dragoljub Milosevi et G. Milanova , Serbie. edre ; designons par hA et mA les longueurs resSoit ABCD un tetra pe tives de la hauteur et de la mediane issues du sommet A sur la fa e opposee BCD. Si V est le volume du tetra edre, montrer que 128 (hA + hB + hC + hD )(m2A + m2B + m2C + m2D ) ≥ √ V 3
.
3395. Propose par Tai hi Maekawa, Takatsuki, Prefe ture d'O¯saka, Japon. Soit le triangle ABC ; denotons par H son ortho entre et par R le rayon de son er le ir ons rit. Montrer que 4R3 − (l2 +m2 +n2 )R − lmn = 0, ou AH = l, BH = m et CH = n.
3396. Propose 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) + n2 mod(i + j + k − 2, n) ,
ou mod(a, n) est le residu 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 ref ere au fait que la somme des aijk est onstante si deux des indi es restent xes et l'autre varie, et de plus que les sommes des diagonales prin ipales du ube sont aussi egales a ette m^eme onstante.)
485
3397. Propose par Jose Luis Daz-Barrero, Universite Polyte hnique de Catalogne, Bar elone, Espagne. Evaluer
1 lim n→∞ n2
Z
n
0
√ n2 − x2 dx . 2 + x−x
3398. Propose par Bru e Shawyer, Universite Memorial de Terre-Neuve, St. John's, NL. Soit l'equation j
n 10
k
+
j k n 2n n − 10 · 10⌊log10 n⌋ = 10
3
,
(a) montrer que n = 5294117647058823 est une solution de ette equation, (b) ⋆ determiner toute autre solution entiere et positive de ette equation.
3399. Proposed by Vin entiu Radules u, Universite 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. Propose par Yakub N. Aliyev, Universite d'Etat de Bakou, Bakou, Azerbadjan. 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. Determiner tous les nombres reels α tels que j
10n
k j k p 5 − 9α (1)2n + α = (3)2n − 6
est valide pour tout entier positif n, ou ⌊x⌋ denote le plus grand entier ne depassant 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 opposite the 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 the points A1 , B1 , and C1 , respe tively. Lines through M perpendi ular to the sides 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 [A2 B2 C2 ] [A1 B1 C1 ]
=
(ap1 + bp2 )(bp2 + cp3 )(cp3 + ap1 ) 2a2 b2 c2
where [KLM ] denotes the area of triangle KLM .
a p1
+
b p2
+
c p3
,
486
3389. Proposed by Mihaly Ben ze, Brasov, Romania.
For a ∈ R de ne a sequen e (xn ) by x0 = a and xn+1 = 4xn − x2n for all n ≥ 0. Prove that there exist in nitely many a ∈ R su h that the sequen e (xn ) is periodi .
3390. Proposed by Mihaly Ben ze, Brasov, Romania. Prove that if A, B , C , and D are the solutions of X
2
=
3 −5 5 8
,
then A2007 + B 2007 + C 2007 + 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 t in 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 (P IJ ), (P J K), (P KL), and (P LI) 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 and 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 and EW D are similar. Does the onverse hold? AD ,
3393. Proposed by Dragoljub Milosevi and G. Milanova , Serbia.
Let ABC be a triangle with a = BC , b = AC , c = AB , and semiperimeter s. Prove that y+z A z+x B x+y C 9 · + · + · ≥ 2, x a(s − a) y b(s − b) z c(c − a) s
where the angles A, B , and C are measured in radians and x, y, and z are any positive real numbers.
3394. Proposed by Dragoljub Milosevi and G. Milanova , Serbia.
Let ABCD be a tetrahedron with hA and mA the lengths of the altitude and the median from vertex A to the opposite fa e BCD, respe tively. If V is the volume of the tetrahedron, prove that 128 3
(hA + hB + hC + hD )(m2A + m2B + m2C + m2D ) ≥ √ V
.
3395. Proposed by Tai hi Maekawa, Takatsuki City, Osaka, Japan.
Let triangle ABC have ortho entre H and ir umradius R. Prove that 4R3 − (l2 + m2 + n2 )R − lmn = 0, where AH = l, BH = m, and CH = n.
487
3396. 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) + n2 mod(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 " means that the sum of aijk is onstant if two indi es are xed and the third index varies, and also the sums along the great diagonals of the ube are equal to this onstant.)
3397.
Proposed by Jose Luis Daz-Barrero, Universitat Polite ni a de Catalunya, Bar elona, Spain. Evaluate Z n√ 2 1 n − x2 lim 2 dx . −x n→∞ n
0
2+x
3398. Proposed by Bru e Shawyer, Memorial University of Newfoundland, St. John's, NL. Given the equation j
n 10
k
+
j k n 2n n − 10 · 10⌊log10 n⌋ = 10
3
,
(a) show that n = 5294117647058823 is a solution,
(b) ⋆ nd all other positive integer solutions of the equation.
3399. Proposed by Vin entiu Radules 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 integers m 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 j
10n
k j k p 5 − 9α (1)2n + α = (3)2n − 6
holds for ea h positive integer n, where ⌊x⌋ is the greatest integer not ex eeding x.
488
SOLUTIONS No problem is ever permanently losed. The editor is always pleased to onsider for publi ation new solutions or new insights on past problems.
3229. [2007 : 170, 172; 2007 : 179{181℄ Proposed by Mihaly Ben ze, Brasov, Romania. (a) Let x and y be positive real numbers, and let n be a positive integer. Prove that n
(x + y)
n X
k=0
1 n n−k k x y k
≥ n+1+2
n n−i X X
i=1 k=0
n k n k+i
≥ (n + 1)2 .
(b)⋆ Let x1 , x2 , . . . , xk be positive real numbers, and let n be a positive integer. Determine the minimum value of (x1 + x2 + · · · + xk )n
X
i1 +···+ik =n
i1 , . . . , ik ≥ 0
i1 !i2 ! · · · ik !
n!xi11 xi22 · · · xikk
.
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 = k1 . We will obtain this 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|
=
|X|
=
Q! XQ
Finally, let ∆ = {X dimension k − 1.
n X
i=1 n X
qi ; xi ;
i=1
= q1 !q2 ! · · · qk ! ;
= xq11 xq22 · · · xqkk .
∈ Rk+ : |X| = 1}
be the positive unit simplex in Rk of
489 Theorem 1 With the above notation, let X
P (X) =
|Q|=n
cQ XQ
be a homogeneous rational fun tion of degree −n in the variables xi . Suppose that cQ ≥ 0 for ea h Q and that P (X) is a symmetri fun tion, that is, inter hanging any xi and xj does not hange the value of P (X). Then the minimum value of P (X) over the simplex ∆ exists and is attained when 1 x1 = x2 = · · · = xk = . k Proof. If P (X) is identi ally zero, then there is nothing to prove, so we assume at least one oeÆ ient cQ is not zero. Note that P (x) has a minimum value, 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 least two unequal oordinates. Without loss of generality (due to the symmetry of P (X)) we may assume that x1 > x2 . We will show that by slightly de reasing x1 and slightly in reasing x2 (while keeping all other variables and the sum x1 + x2 un hanged) the value of P (X) will be ome smaller, ontrary to our assumption. Fixing x3 , x4 , . . . , xn auses P (X) to be ome a symmetri fun tion of two variables X
F (u, v) = P (u, v, x3 , . . . , xn ) =
j+s≤n
Aj,s uj v s
,
where the oeÆ ients Aj,s depend on x3 , x4 , . . . , xn . Note that ea h Aj,s is 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 all zero, of fun tions of the form Fj,s (u, v) =
1
+
1
,
us v j where j and s are non-negative integers with j+s ≤ n.
and v(t)
= x2 + t,
so that
the \time derivative" d Fj,s
uj v s
du = −1 dt
and
d Fj,s u(t), v(t) dt
u(t), v(t) dt
= =
dv = 1. dt
Now let u(t) = x1 −t It suÆ es to prove that
is negative at t = 0. We have
∂Fj,s u, v ∂Fj,s u, v − ∂u ∂v j s 1 s j 1 − + − u v uj v s u v us v j
.
Di erentiating on e again, we obtain d2 Fj,s u(t), v(t)
= dt2 ! j s j s 2 1 + 2 + − + 2 j u v u v u vs
j s + 2 + 2 v u
j s − v u
2 !
1 us v j
.
490 d2 F dF u−v > 0. Sin e = 0 when u = v (this o
urs when t = ), dt2 dt 2 dF follows that dt < 0 when t = 0. We have obtained a ontradi tion by
Hen e,
it assuming x1 > x2 at a point a hieving the minimum. This proves that the minimum o
urs when all the xi are equal. It follows that the minimum sought in part (b) is kn
i1 !i2 ! . . . ik !
X
n!
i1 +···+ik =n
.
i1 ,...,ik ≥0
No other solutions to part (b) were re eived. Regarding part (a) Sadov remarks that min
x+y=1
n X
1
k=0
n k
xn−k yk
n
= 2
n X
j =0
1
n = n + 1 + 2 j
n−1 X n−j X j =0 i=0
n j n i+j
,
where the rst equality follows Theorem 1 and the last expression is the minimum obtained by Bataille [2007 : 179{181℄. He notes that the rst equality yields a minimum of at least 2n+1 , onsiderably improving the lower bound of (n + 1)2 if n > 4. He observes that if bi = ni xn−i yi , then by the AM{HM Inequality
(x + y)n 1 X 1 −1 1 X , ≤ bi = n+1 bi n+1 n+1
hen e (x + y)2 b−1 ≥ (n + 1)2 , whi h yields a qui k proof of part (a). i 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 Pthe AM{HM Inequality an be obtained by taking P (X) = (x1 x2 · · · xn )−1 and P (X) = xi in Theorem 1, respe tively. He mentions that parti ular ases (and other theorems of a more general nature) of Theorem 1 an be found in [2℄, Chapter 3, Se tion G, Examples G.1.k and G.1.m; though he believes 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's Inequality. P
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 interior su h that ∠DAB = ∠DCA and ∠DBA = ∠DAC . Let E and F be points on 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 .
491 A omposite of similar solutions by John G. Heuver, Grande Prairie, AB and George Tsapakidis, Agrinio, Gree e. Triangles ADC and BDA are similar (be ause their angles are assumed to be equal), when e CD CA 2CA CF = = = AD AB 2AB AE
.
It follows that △DCF ∼ △DAE (∠DCF = ∠DCA = ∠DAB = ∠DAE , while the adja ent sides are proportional from the previous step), so that ∠AF D (= ∠CF D) = ∠AED . Noting that E and F both lie on the same side of AD, we on lude that A, E , F , and D lie on the same ir le, as desired.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; SEFKET University of Sarajevo, Sarajevo, Bosnia and Herzegovina; RICARDO ARSLANAGIC, BARROSO CAMPOS, University of Seville, Seville, Spain; MICHEL BATAILLE, Rouen, Fran e; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; APOSTOLIS K. DEMIS, Varvakeio High S hool, Athens, Gree e; OLIVER GEUPEL, Bruhl, NRW, Germany; GEOFFREY A. KANDALL, Hamden, CT, USA; ANDREA MUNARO, student, University of Trento, Trento, Italy; PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU, Comane sti, Romania; and the proposer. Bataille omments that D an be onstru ted as the point inside △ABC where the ir le F AE interse ts the ir le through B, C, and the ir um entre O. It lies on the rst ir le by the result of this problem. It lies on the se ond ir le be ause ∠BOC = ∠BDC = 2 ∠BAC as follows: on the one hand the angle at A is ins ribed in the ir le BAC that is entred at O and is therefore 12 ∠BOC; on the other hand the similar triangles ADC and BDA t together at A and at D in su h a way that the two exterior angles at D (that form ∠BDC) sum to twi e the 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 k BC . Denote the lengths of AD and BC by a and b, respe tively. Let M be the mid-point of CD , and let P and Q be the mid-points of AM and BM , respe tively. If N is the interse tion of DP and CQ, prove that N belongs to the interior of △ABM if and only if 1 < a < 3. 3
b
Solution by Joel S hlosberg, Bayside, NY, USA. There is a misleading subtlety in the statement of the problem: we shall see 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 aÆne transformation, we an assume without loss of generality that AB ⊥ 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 ,
492 C , 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 Q are a+b a+b a+b M 2, , P 1, , Q 3, , 2
4
4
so that the lines DP and CQ satisfy y =
when e N
b − 3a 4
x + a
and
y =
= DP ∩ CQ has oordinates 4b (a − b)2 , N a+b a+b
3b − a 4
x + a − 2b ,
.
A point in the plane is on the same side of AB as M if and only if its is positive. The y{ oordinate of N satis es
y { oordinate
(a − b)2 ≥ 0, a+b
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 its x{ 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 its x{ oordinate is less than 3. Therefore, N is within the interior of △ABM if and only if a 6= b and 1 <
4b < 3. a+b
This inequality is equivalent to
1 a < < 3, 3 b
whi h is the desired inequality; it is equivalent to N belonging to the interior of △ABM as long as a 6= b. Also solved by ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, Fran e; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; IES Alvarez FRANCISCO JAVIER GARC I A CAPITAN, Cubero, Priego de Cordoba, Spain; OLIVER GEUPEL, Bruhl, NRW, Germany; PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU, Comane 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 P su h that the sum of the squares of the distan es of the points A, B , and C from any line through P is onstant.
493 Solution by Mi hel Bataille, Rouen, Fran e. We will introdu e a Cartesian oordinate system. Label the points as A(0, a), B(−b, 0), C(b, 0), and P (u, v), where a and b are positive and u and v 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 , d(C, ℓ) = (p − b cos θ)2 . After a simple al ulation, we obtain 3(u2 − v 2 ) a2 2 S = (cos 2θ) + av − +b 2 2 3(u2 + v 2 ) a2 + (sin 2θ)(3uv − au) + − av + + b2 . 2 2 2
and
The number S is independent of θ if and only if and
3uv − au = 0
whi h gives
u = 0
or
and
3(u2 − v 2 ) + 2av − a2 + 2b2 = 0 , − 3v 2 + 2av − a2 + 2b2 = 0 ,
(1)
a2 2a2 2 v = and 3 u − (2) + − a2 + 2b2 = 0 . 3 9 3 √ Now, (1) is satis ed if and only if 0 < a ≤ b 3 (that is, ∠B = ∠C ≤ 60◦ ) and the oordinates of P are ! ! p p a + 2(3b2 − a2 ) a − 2(3b2 − a2 ) or . 0, 0, 3 3 √ Similarly, (2) is satis ed if and only if a ≥ b 3 (that is, ∠B = ∠C ≥ 60◦ ) and the oordinates of P are ! ! p p 2(a2 − 3b2 ) a 2(a2 − 3b2 ) a , or − , . 3 3 3 3 a
In on lusion, if ABC is equilateral, only the entre of ABC is a solution for P . Otherwise, there are two solutions for P : P
! p 2(3b2 − a2 ) 3
(if ∠A > 60◦ ) ,
! p 2(a2 − 3b2 ) a ± , 3 3
(if ∠A < 60◦ ) .
0,
P
a±
494 In both ases, the two solutions are symmetri about the entroid of the triangle; in the rst ase, the two points lie on the median through A and in the se ond ase the two points lie on a line parallel to BC and passing through 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 I A CAPITAN, IES Alvarez Cubero, Priego de Cordoba, Spain; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; V ACLAV Y, Big Rapids, MI, USA (2 solutions); SKIDMORE COLLEGE PROBLEM SOLVING KONECN GROUP, Skidmore College, Saratoga Springs, NY, USA; GEORGE TSAPAKIDIS, Agrinio, Gree e; PETER Y. WOO, Biola University, La Mirada, CA, USA; TITU ZVONARU, Comane sti, Romania; and the proposer.
3292. [2007 : 485, 488℄ Proposed by Mihaly Ben ze, Brasov, Romania. Let a, b, c, and d be arbitrary real numbers. Show that 11a2 + 11b2 + 221c2 + 131d2 + 22ab + 202cd + 48c + 6 ≥ 98ac + 98bc + 38ad + 38bd + 12a + 12b + 12d .
Solution by Oliver Geupel, Bruhl, NRW, Germany. The proof is by ontradi tion. Assume that 11a2 + 11b2 + 221c2 + 131d2 + 22ab + 202cd + 48c + 6 < 98ac + 98bc + 38ad + 38bd + 12a + 12b + 12d .
We onsider the quadrati f (x) = 11x2 + px + q with p q
= =
22b − 98c − 38d − 12 , 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 two distin t real roots; that is, the dis riminant of f is positive. By omputing the 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; MICHEL BATAILLE, 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, La Mirada, CA, USA; and the proposer.
495
3293. [2007 : 485, 488℄ Proposed by Mihaly Ben ze, Brasov, Romania. Prove that
n arcsin Y
k=1
9k + 2 √ 3 27k + 54k 2 + 36k + 8
arctan
1 √ 3k + 1
= 3n .
Composite of similar solutions by Mi hel Bataille, Rouen, Fran e and Douglass L. Grant, Cape Breton University, Sydney, NS, modi ed by the editor. For ea h k = 1, 2, . . . , n let Pk denote the orresponding fa tor under the produ t sign. We prove that in fa t, Pk = 3 for ea h k. 3 Note rst that 27k + 54k2 + 36k + 8 = (3k + 2)3 . For xed k, let 1 θ = tan−1 √ . Then tan θ = √ 1 implies sin θ = √ 1 . 3k + 1
3k + 1
3k + 2
Sin e tan−1 is an in reasing fun tion, we have 0 < θ ≤ tan
−1
1 2
−1
< tan
1 √ 3
=
π 6
,
hen e, 0 < 3θ < π2 . From sin(3θ)
we obtain
= 3 sin θ − 4 sin3 θ 4 9k + 2 3 − p = p = √ 3 3k + 2 (3k + 2) (3k + 2)3 sin−1
=
sin−1
and it follows that Pk =
,
9k + 2 √ 27k3 + 54k2 + 36k + 8 ! 9k + 2 p = sin−1 (sin 3θ) = 3θ , (3k + 2)3 3θ = 3. θ
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e; SEFKET University of Sarajevo, Sarajevo, Bosnia and Herzegovina; DIONNE BAILEY, ARSLANAGIC, ELSIE CAMPBELL, CHARLES DIMINNIE, and KARL HAVLAK, Angelo State University, San Angelo, TX, USA; ROY BARBARA, Lebanese University, Fanar, Lebanon; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLEH FAYNSHTEYN, Leipzig, student, Sarajevo Germany; OLIVER GEUPEL, Bruhl, NRW, Germany; SALEM MALIKIC, College, 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, Comane sti, Romania; and the proposer.
496
3294. [2007 : 486, 488℄ Proposed by Mihaly Ben ze, Brasov, Romania. For all positive integers m and n, show that m(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 A(m, n) = C(m, n) D(m, n). Sin e 720 = 24 32 5, it suÆ es to show A(m, n) is divisible by 16, 9, and 5.
that that
(a) Divisibility by 16. The residues of f (n) modulo 4 are given in the following table. n (mod 4) 0 f (n) (mod 4) 0
1 2 2 2
3 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) is divisible 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 next table. n (mod 9) f (n) (mod 9)
0 1 0 6
2 3 3 6
4 5 6 6
6 7 3 6
8 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) is divisible by 9. (iii) Otherwise, f (n) ≡ 6 (mod 9) and f (m) ≡ 6 (mod 9), so that D(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 table below. n (mod 9) f (n) (mod 9)
0 1 0 1
2 3 1 1
4 0
(i) If n ≡ 0 or 4 (mod 5), or m ≡ 0 or 4 (mod 5), then C(m, n) is divisible 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) is divisible by 5. University of Sarajevo, Sarajevo, Bosnia and Also solved by SEFKET ARSLANAGIC, Herzegovina; ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, Fran e; BRIAN D. BEASLEY, Presbyterian College, Clinton, SC, USA; OLIVER GEUPEL, Bruhl, 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, Comane sti, Romania; and the proposer.
3295. [2007 : 486, 488℄ Proposed by Mi hel Bataille, Rouen, Fran e. Let u : R → R be a bounded fun tion. For x > 0, let and Prove that
f (x) g(x)
= =
sup {u(t) : t > ln(1/x)} sup {u(t) − xe−t : t ∈ R} .
lim f (x) = lim g(x).
x→0+
x→∞
Solution by Peter Y. Woo, Biola University, La Mirada, CA, USA, modi ed by the editor. The proof onsists of showing that both limits are equal to the limit L = lim sup u(t), whi h exists sin e the fun tion u(t) is bounded. Let K be t→∞ su 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 t→∞ lim xe−t = 0, so there exists t0 > 0 su h that xe−t < ǫ for any t > t0 . By part (i), there exists t > t0 su 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.
498 On the other hand, there exists t1 ∈ R su h that u(t) < L + ǫ for all Sin e e−t > 0, let M > 0 be su h that K − M e−t < L + ǫ. We
laim that if x > M , then g(x) ≤ L + ǫ. For this it suÆ es to show that u(t) − xe−t < L + ǫ whenever x > M and t ∈ R. Indeed, if x > M and t ≤ t1 , then u(t) − xe−t < K − M e−t < 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 e lim g(x) = L. x→∞ Finally, writing S(v) = sup {u(t) : t > v} and making two hanges of variable in the limit yields t > t1 .
1
1
1
lim f (x) =
x→0+
lim S ln(1/x) = lim S(ln y) = lim S(z) = L .
x→0+
y→∞
z→∞
Also solved by OLIVER GEUPEL, Bruhl, 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 b2 c2 c2 a2 a2 b2 + + > K a2 (a − b)(a − c) b2 (b − c)(b − a) c2 (c − a)(c − b)
for all distin t positive real numbers a, b, and c. Solution by Peter Y. Woo, Biola University, La Mirada, CA, USA, expanded by the editor. We prove that K = 10. Let L denote the left side of the given inequality. Sin e L is ompletely symmetri in a, b, and c, we may assume without loss of generality that a < b < c. Note rst that L = P where Q = a2 b2 c2 (b − a)(c − a)(c − b) and Q P = b4 c4 (c − b) − c4 a4 (c − a) + a4 b4 (b − a). Observing that P = 0 when c = b or b = a or c = a, we nd by straightforward but tedious omputations that P
= b4 c4 (c − b) − a4 c5 − b5 + a5 c4 − b4 = (c − b) b4 c4 − a4 c4 + c3 b + c2 b2 + cb3 + b4 + a5 c3 + c2 b + cb2 + b3
499
= =
=
=
(c − b) c4 b4 − a4 + a4 (a − b) c3 + c2 b + cb2 + b3 (c − b)(b − a) c4 b3 + b2 a + ba2 + a3 − a4 c3 + c2 b + cb2 + b3 (c − b)(b − a) b3 c4 − a4 + cb2 a c3 − a2 + c2 ba2 c2 − a2 + c3 a3 (c − a) (c − b)(b − a)(c − a) b3 c3 + c2 a + ca2 + a3 + cb2 a c2 + ca + a2 + c2 ba2 (c + a) + c3 a3 .
Hen e, W
P W = 2 2 2, Q a b c
=
where
b3 c3 + c2 a + ca2 + a3
+ cb2 a c2 + ca + a2
+ c2 ba2 (c + a) + c3 a3 .
By writing W as a sum of 10 terms and using the AM{GM Inequality, we readily see that W ≥ 10 a20 b20 c20 1/10 = 10a2 b2 c2 , from whi h it follows that 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, Bruhl, 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 showed that 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 √ 1+ 5 sin A + sin B sin C ≤ 2
When does equality hold?
.
500 Solution by D.J. Smeenk, Zaltbommel, the Netherlands. The following are equivalent √ 1+ 5 sin A + sin B sin C ≤ ; 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 and sin ϕ = 5 (the angle ϕ is approximately 26.6◦ ). The last inequality then be omes
1 √ 5
√ √ 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 ≈ 63.4◦ and B = C ≈ 58.3◦ . 5
University of Sarajevo, Sarajevo, Bosnia Also solved by SEFKET ARSLANAGIC, and Herzegovina; ROY BARBARA, Lebanese University, Fanar, Lebanon; MICHEL BATAILLE, Rouen, Fran e; BRIAN D. BEASLEY, Presbyterian College, Clinton, SC, USA; CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; APOSTOLIS K. DEMIS, Varvakeio High S hool, Athens, Gree e; OLEH FAYNSHTEYN, Leipzig, Germany; OLIVER GEUPEL, Bruhl, NRW, Germany; RICHARD I. HESS, Ran ho Palos Verdes, Y, CA, USA; WALTHER JANOUS, Ursulinengymnasium, Innsbru k, Austria; V ACLAV KONECN Big Rapids, MI, USA; KEE-WAI LAU, Hong Kong, China; THANOS MAGKOS, 3rd High S hool student, Sarajevo College, Sarajevo, Bosnia and of Kozani, Kozani, Gree e; SALEM MALIKIC, Herzegovina; JUAN-BOSCO ROMERO MARQUEZ, 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, Comane 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 ≤
with equality if and only if A = arccos √
1 4λ 2 +1
1+
p
4λ2 + 1 2
,
and B = C.
3298. [2007 : 486, 489℄ Proposed by Stanley Rabinowitz, MathPro Press, Chelmsford, MA, USA. Let ABC be a triangle of area opposite vertex A. Prove that
1 2
in whi h
a2 + csc A ≥
a
is the length of the side
√ 5.
[Ed.: The proposer's only proof of this is by omputer. He is hoping that some CRUX with MAYHEM reader will nd a simpler solution.℄
501 Solution by Kee-Wai Lau, Hong Kong, China. Let b = AC , c = AB and let S denote the area of triangle ABC . Sin e S = 12 bc sin A = 12 , 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 q 2 2 ≥ b + c − 2bc 1 − sin2 A + bc p = b2 + c2 − 2 b2 c2 − (bc sin A)2 + bc p = b2 + bc + c2 − 2 b2 c2 − 1 p ≥ 3bc − 2 b2 c2 − 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 + y 2 + 4 = 0. Sin e x is real, the
dis riminant of the quadrati polynomial above must be non-negative. (−6y)2 − 20(y 2 + 4) ≥ 0, or 16y 2 − 80 ≥ 0, from whi h we Thus, √ obtain y ≥ 5. The result now follows by setting x = bc.
Also solved by GEORGE APOSTOLOPOULOS, Messolonghi, Gree e (two solutions); University of Sarajevo, Sarajevo, Bosnia and Herzegovina; MICHEL SEFKET ARSLANAGIC, BATAILLE, Rouen, Fran e; ROY BARBARA, Lebanese University, Fanar, Lebanon; CHIP CURTIS, 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, Bruhl, NRW, Germany; RICHARD I. HESS, Ran ho Palos Verdes, Y, Big Rapids, MI, USA; SOTIRIS LOURIDAS, Aegaleo, Gree e; CA, USA; V ACLAV KONECN 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, Comane sti, Romania; and the proposer. There was one in orre t solution submitted. From dedu e that equality holds if and only if b = c q the proof given above, it is easy to and a = 4 45 , in whi h ase A = cos−1 23 ≈ 48.19◦ . This was pointed out by Barbara, Geupel, Hess, Kone ny, Tsapakidis, and Zvonaru. Both Barbara and Demis generalized to √ an arbitrary triangle of area k > 0, proving that √ a2 + 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 College, 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 of the interior bise tor of angle B equal to wb , then it is unique up to isomorphism; (b) for the existen e of su h a triangle in (a), it is ne essary and suÆ ient that 2a |a − wb | b > ≥ 0; 2a − wb
( ) if ha is the length of the altitude to side BC in su h a triangle in (a), we have b > |a − wb | + 12 ha .
502 Solution by Mi hel Bataille, Rouen, Fran e. (a) For onvenien e we write w = wb . Let the interior bise tor of ∠B meet AC at W . We assume that △ABC exists with BC = a, CA = b, and BW = w , and shall produ e a (impli itly de ned) formula for the third side 2ac cos B 2 c = AB . We re all that w = , so that a+c B aw −w = 2 c
2a cos
By the Law of Cosines we have standard formula
ab WC = , a+c
a2 b2 (a + c)2
.
W C 2 = a2 + w 2 − 2aw cos
we therefore have
B ; 2
B aw 2 = a2 − w 2a cos − w = a2 − 2 c
using the
.
It follows that c must be the unique positive solution of f (x) = a, where f is the de reasing fun tion on (0, ∞) given by f (x) =
ab2 w2 + (a + x)2 x
.
Thus, if a triangle ABC with the given parameters does exist, then its side lengths are uniquely determined, and (a) is proved. (b) First, it is easy to see that the two given inequalities are equivalent to 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 t that 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 inequalities in (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 is equivalent to ab2 |a − b| + w 2 a + |a − b|
2
> a |a − b| a + |a − b|
2
.
This redu es to w2 b2 > 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). We have proved that (1) implies (2). Conversely, we assume that (2) holds (that is, that a triangle ABC w c B with the given parameters exists). Then 2a = · cos ; hen e w < 2a. a+c 2
503 Moreover, 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 is negative). The desired equivalen e follows. ( ) Note that b > |a − w| + 21 ha is equivalent to ab > a |a − w| + Area(ABC) ; c B that is, to ab > a |a − w| + a + w sin . Sin e a |a − w| < 2 2 (from part (b)), the latter will ertainly hold if
B
b ≥ (a + c) sin
.
2
This inequality is equivalent to
sin B ≥ (sin A + sin C) sin
or to
2 cos
or nally to
B ≥ 2 sin 2
A+C 2
1 ≥ cos
cos
A−C 2
whi h is ertainly true. The result follows.
(2a − w)b 2
B 2
,
A−C 2
,
,
Also solved by CHIP CURTIS, Missouri Southern State University, Joplin, MO, USA; OLIVER GEUPEL, Bruhl, 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 ent Advan es in Geometri Inequalities, Kluwer A ademi Publishers, 1989 as the rst of 40 existen e results 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 Gunter Ewald's Geometry: An Introdu tion (Wadsworth Publ., 1971) says that 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 be simpli 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 equation with rational oeÆ ients. One simply has to hoose a value of w for whi h the resulting ubi equation has no rational root. The theory of Eu lidean onstru tions then tells us that the positive 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 de ne
c
be positive real numbers. For any positive integer
Fn =
X bn + cn 3(an + bn + cn ) − a+b+c b+c
y li
(a) Prove that Fn ≥ 0 for n ≤ 5. (b)⋆ Prove or disprove that Fn ≥ 0 for n ≥ 6.
.
n
504 Solution 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) ≥ 0 for 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)
=
(an + bn + cn ) −
=
= =
X bn + cn b+c
y li
X a (bn + cn ) b+c
y li X a (bn + cn ) an − b+c
y li " # X ab an−1 − bn−1 ac an−1 − cn−1 + (b + c) (b + c)
y li X ab an−1 − bn−1 (a − b) ≥ 0. (b + c)(c + a)
y li
Equality holds if and only if a = b = c. University of Sarajevo, Sarajevo, Bosnia and Also solved by SEFKET ARSLANAGIC, ^ Herzegovina; 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, Bruhl, 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, Comane sti, Romania; and the proposer. C^rtoaje mentioned that this problem was posted (together with a solution similar to the one featured above) by Wolfgang Berndt (Spanferkel) on the Mathlinks Forum website http://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 numbers, m ≥ 2, and Fn =
n n X an + · · · + an m(an m 1 + a2 + · · · + am ) 2 − a1 + a2 + · · · + am a2 + · · · + am
,
y li
then Fn ≥ 0 for all n ≥ 1. Alt ultimately proved that if a, b, c, p, and q are positive real numbers and F (p, q) =
X ap + bp 3(ap + bp + cp ) − aq + bq + cq aq + bq
y li
then (p − q)F (p, q) ≥ 0.
,
Mathematical Association of America Presents
Problems from Murray Klamkin: The Canadian Collection Andy Liu & Bruce Shawyer, Editors
Murray Klamkin was a dedicated problem solver and problem proposer, who left indelible marks on the problemist community. After working in industry and academe in the United States, he spent the last thirty of his eighty four years in Canada. He was famous for his Quickies, problems that have quick and neat solutions. In this book you will find all of the problems that he proposed for CRUX with MAYHEM, including all of his Quickies. His problems covered a very wide range of topics, and show a great deal of insight into what is possible in these areas. The problems are arranged into sets according to topic,and the lightly edited solutions are as published in CRUX with MAYHEM. Problems from Murray Klamkin: The Canadian Collection is published by the Mathematical Association of America (MAA) in collaboration with the Canadian Mathematical Society (CMS). It is the first volume in the Canadian Collection.
Order your copy today! Call (301) 617-7800
Series: Problem Books,
Catalog Code: CAN-01
280 pp., Hardbound, 2009,
ISBN 9780-88385-828-8,
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(CMS members get member price for this book. Be sure to mention you are a CMS member when you call.)
506
YEAR END FINALE This marks my rst year as Editor/Co-Editor of CRUX with MAYHEM, and my rst year end nale. As I write in this spa e, whi h Jim Totten wrote in just one year ago, I am reminded of him and the non-permanen e of things. Just as the 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 redible enthusiasm and passion for mathemati s and his outrea h work, I am moved. An o ean is there for us to feel its power and its ool refreshing spray, to play in it and be leansed by its soothing sound. Whi h reminds me that issues of CRUX with MAYHEM are now delayed by several weeks! My apologies to all the readers for this onsiderable delay, and in 2009 we will work hard to lose the gap. Many people have served this past year (I will borrow Jim's habit of apitalizing their names). First, I shall forever be grateful to the late JIM TOTTEN for all the support 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 those tough early days. I thank BRUCE CROFOOT for his solid support as Mission Control in the CRUX oÆ e at Thompson Rivers University and for listening to me when I needed someone to talk to. I thank BRUCE SHAWYER for his ontinuing help and advi e on CRUX 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 thank Bru e S. for his hospitality in Newfoundland and for providing me with a mu h needed sense of ontinuity. I thank the members of the Editorial Board for their hard work and for bringing their unique talents to CRUX: JEFF HOOPER, the Asso iate Editor, for his areful reading and sound judgment; IAN VANDERBURGH, the MAYHEM Editor, whose Problem 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; ROBERT WOODROW, for handling the Olympiad Corner on top of his immense administrative duties; JOHN GRANT M LOUGHLIN, for his solid support as Book Reviews Editor and for his fantasti proof reading skills. This is John's last issue as Book Reviews Editor, 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 Reviews Editor from John. I thank my olleague here in Winnipeg and Arti les Editor JAMES CURRIE, for taking on the position and then turning a ba klog of arti les into a ornu opia. I thank the Problems Editors for their ontinuing support and hard work: ILIYA BLUSKOV, CHRIS FISHER, MARIA TORRES, and EDWARD WANG. The job of being a Problems Editor is perhaps one of the most diÆ ult on the Board, with tight deadlines and a deluge of problem proposals and solutions in all areas of mathemati 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 GRAHAM WRIGHT, the Managing Editor, for his solid support and also for providing that vital sense of ontinuity. Others who are not on the Editorial Board but whose work is just as ru ial are MONIKA KHBEIS and ERIC ROBERT who serve on the Mayhem sta ; a big thank you Monika and Eri . I thank JEAN-MARC TERRIER for translating the problems that appear in CRUX with MAYHEM. The Canadian Mathemati al So iety re ently honoured Jean-Mar at the 2008 Winter Meeting for his many years of servi e. It is
507 well deserved! I look forward to working with Jean-Mar in the New Year. As well, I wel ome ROLLAND GAUDET of College Universitaire Saint-Bonifa e, Winnipeg. Rolland is also helping with translations, espe ially in times of risis! I thank our past 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 support. Those who have taken pity on this Editor-in-Chief are ANNA STOKKE, ROSS STOKKE, TERRY VISENTIN, and JEFF BABB. Our se retary, JULIE BEAVER, has also helped out in a pin h as little emergen ies have arisen. I thank the previous Dean of S ien e, GABOR KUNSTATTER, for his understanding and foresight in supporting CRUX with MAYHEM at the University of Winnipeg. The LATEX expertise of JOANNE CANAPE at the University of Calgary, and TAO GONG and JUNE ALEONG at Wilfrid Laurier University goes a long way to produ ing high quality opy. Joanne is nishing up her work in this area, so after my one year of servi e here I thank her for giving twenty! I wel ome JILL AINSWORTH on board who 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 you MICHAEL DOOB and CRAIG PLATT for te hni al support, and JUDI BORWEIN for putting CRUX on the net. Someone spe ial who has helped me through my rst year with her areful proof reading and knowledge of geometry is CHARLENE PAWLUCK. Thank you for sharing your 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 the journal. CRUX with MAYHEM is built from your ontributions and all the time, are and 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 last year, 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 may think is suitable. Skoliad is missing from this issue, but will be ba k next year; from now on please send all Skoliad materials dire tly to the Editor-in-Chief (or resend your 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, Va lav (Vazz) Linek
Crux Mathemati orum with Mathemati al Mayhem Former Editors / An iens Reda teurs: Bru e L.R. Shawyer, James E. Totten Crux Mathemati orum Founding Editors / Reda teurs-fondateurs: Leopold Sauve & Frederi k G.B. Maskell Former Editors / An iens Reda teurs: G.W. Sands, R.E. Woodrow, Bru e L.R. Shawyer Mathemati al Mayhem Founding Editors / Reda teurs-fondateurs: Patri k Surry & Ravi Vakil Former Editors / An iens Reda teurs: Philip Jong, Je Higham, J.P. Grossman, Andre Chang, Naoki Sato, Cyrus Hsia, Shawn Godin, Je Hooper
508
INDEX TO VOLUME 34, 2008 Contributor Pro les February Peter Y. Woo ................................................... 1 Skoliad Robert Bilinski February No. 107 ......................................................... 2 Mar h No. 108 ....................................................... 65 April No. 109 ...................................................... 129 May No. 110 ...................................................... 195 September No. 111 ...................................................... 257 O tober No. 112 ...................................................... 321 November No. 113 ...................................................... 385 Mathemati al Mayhem Ian VanderBurgh February .................................................................. 8 Mar h ................................................................. 69 April ............................................................... 136 May ............................................................... 200 September ............................................................... 266 O tober ............................................................... 327 November ................................................................ 396 De ember ............................................................... 449 Mayhem Problems February M326{M331 .................................................... 8 Mar h M332{M337 ................................................... 69 April M338{M343 ................................................. 136 May M344{M349 ................................................. 200 September M350{M356 ................................................. 266 O tober M357{M362 ................................................. 327 November M363{M368 ................................................. 396 De ember M369{M375 ................................................. 449 Mayhem Solutions February M276{M281 ................................................... 10 Mar h M282{M287 ................................................... 71 April M288{M293 ................................................. 138 May M294{M300 ................................................. 202 September M301{M312 ................................................. 268 O tober M313{M324 ................................................. 329 November M325{M331 ................................................. 398 De ember M332{M337 ................................................. 451 Problem of the Month Ian VanderBurgh February ................................................................. 15 Mar h ................................................................. 76 April ............................................................... 144 May ............................................................... 208 September ............................................................... 279 O tober ............................................................... 338 November ................................................................ 404 De ember ............................................................... 457
509 Mayhem Arti les Adding Up Bru e Shawyer ............................................................ 406 The Olympiad Corner R.E. Woodrow February No. 267 ....................................................... 18 Mar h No. 268 ....................................................... 79 April No. 269 ...................................................... 147 May No. 270 ...................................................... 211 September No. 271 ...................................................... 282 O tober No. 272 ...................................................... 341 November No. 273 ...................................................... 408 De ember No. 274 ...................................................... 459 Book Reviews John Grant M Loughlin How Euler Did It by C. Edward Sandifer Reviewed by J. Chris Fisher ............................................... 34 Nonplussed! Mathemati al Proof of Implausible Ideas by Julian Havil Reviewed by Robert D. Poodia k .......................................... 36 Math Through the Ages: A Gentle History for Tea hers and Others by William P. Berlingho and Fernando Q. Gouv^ea Reviewed by John Grant M Loughlin....................................... 95 Minnesota Math League XXV 1980{2005 by A. Wayne Roberts Reviewed by Robert L. Crane .............................................. 96 The Magi Numbers of the Professor by Owen O'Shea and Underwood Dudley Reviewed by Je Hooper .................................................. 161 Geometri Puzzle Design by Stewart CoÆn Reviewed by Jim Totten ................................................... 163 The Liar Paradox and the Towers of Hanoi: the Ten Greatest Math Puzzles of All Time by Mar el Danesi Reviewed by Amar Sodhi .................................................. 164 Problems of the Week by Jim Totten Reviewed by John Grant M Loughlin ..................................... 229 Digital Di e by Paul J. Nahin Reviewed by Amar Sodhi .................................................. 297 Cal ulus Gems: Brief Lives and Memorable Mathemati s by George F. Simmons Reviewed by Robert D. Poodia k ......................................... 356 From Zero to In nity: What Makes Numbers Interesting by Constan e Reid Reviewed by John Grant M Loughlin ..................................... 357
510 Impossible? Surprising Solutions and Counterintuitive Conundrums by Julian Havil Reviewed by Ed Barbeau .................................................. 422 The Symmetries of Things by John H. Conway, Heidi Burgiel, and Chaim Goodman-Strauss Reviewed by J. Chris Fisher ............................................... 468 Crux Arti les James Currie A Useful Inequality Roy Barbara ............................................................... 38 Sharpening the Hadwiger-Finsler Inequality Cezar Lupu and Cosmin Pohoata .......................................... 97 Industrial Grade Primes with a Money-Ba k Guarantee Mi hael P. Abramson .................................................... 165 The Proof of Three Open Inequalities Vasile C^rtoaje ............................................................ 231 The Sum of a Cube and a Fourth Power Thomas Mauts h and Gerhard J. Woeginger ............................. 358 Twin Problems on Non-Periodi Fun tions Eugen J. Ionas u .......................................................... 424 Old Idaho Usual Here Robert Israel, Stephen Morris, and Stan Wagon ......................... 471 A Limit of an Improper Integral Depending on One Parameter Iesus C. Diniz ............................................................. 478 Sliding Down In lines with Fixed Des ent Time: a Converse to Galileo's Law of Chords Je Babb .................................................................. 481 Problems February 3301{3312 ..................................................... 44 Mar h 3313{3325 .................................................... 102 April 3326{3337 .................................................... 170 May 3282, 3338{3350 ............................................. 239 September 3351{3362 .................................................... 298 O tober 3363{3375 .................................................... 362 November 3376{3388 .................................................... 430 De ember 3371, 3389{3400 .............................................. 483 Solutions February 3201{3213 ..................................................... 49 Mar h 3214{3225, 3227 ............................................. 107 April 3226, 3228{3238, 3242 ....................................... 176 May Klamkin{05, 3239{3241, 3243{3250 ......................... 244 September 3251{3362 .................................................... 303 O tober 3263{3275, 3282 ............................................. 367 November 3276{3281, 3283{3288 ....................................... 435 De ember 3229, 3289{3300 ............................................. 488 Mis ellaneous Editorial .................................................................... 193 In Memoriam: James Edward Totten ...................................... 194 MAA Book Promotion .......................................................505 Year End Finale ............................................................. 506
511 Proposers and solvers appearing in the SOLUTIONS se tion in 2008: Proposers Anonymous Proposer 3211 Gregory Akulov 3285 Arkady Alt 3300 Mi hel Bataille 3217, 3237, 3238, 3251, 3252, 3266, 3267, 3295, 3296 Mihaly Ben ze 3204, 3205, 3206, 3213, 3214, 3216, 3228, 3229, 3230, 3239, 3240, 3253, 3292, 3293, 3294
K.S. Bhanu 3284 Alper Cay 3258 Vasile C^rtoaje 3209, 3210, 3274, 3275, Klamkin{05, M.N. Deshpande 3283, 3284 Jose Luis Daz-Barrero 3212, 3247, 3263, 3269, 3281, 3282 Marian Din a 3205 J. Chris Fisher 3224 Ovidiu Furdui 3218, 3226, 3227, 3261, 3262, 3277, 3288 G.P. Henderson 3201 Ignotus 3231 Neven Juri 3259, 3276, 3286 Geo rey A. Kandall 3235 Va lav Kone ny 3256 Tidor Mitev 3236 Virgil Ni ula 3242, 3241, 3260, 3264, 3265, 3270, 3271, 3273, 3278, 3279,
Vi tor Oxman 3299 Fran is o Pala ios Qui~nonero 3212 Pantelimon George Popes u 3269, 3281, 3282 A hilleas Pavlos Poryfyriadis 3223 D.E. Prithwijit 3272 Stanley Rabinowitz 3297, 3298 Juan-Bos o Romero Marquez 3221 Bill Sands 3257, 3268 D.J. Smeenk 3202, 3203, 3250 Marian Tetiva 3220, 3246 George Tsapakidis 3225 G. Tsintsifas 3232, 3233, 3234, 3243, 3244, 3254, 3255 Pham Van Thuan 3222 Dan Vetter 3219 Harley Weston 3224 Shaun White 3208, 3215 John Wiest 3268 Peter Y. Woo 3207 Paul Yiu 3245 Titu Zvonaru 3248, 3249
3280, 3287, 3289, 3290, 3291
Featured Solvers | Individuals Mohammed Aassila 3248, 3249 Arkady Alt 3241, 3263, 3276 Sefket Arslanagi 3249, 3253, 3257, 3264 Dionne Bailey 3228, 3237, 3247, 3269, 3271 Roy Barbara 3214, 3232, 3235 Ri ardo Barroso Campos 3267 Mi 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, 3299
Brian D. Beasley 3220 Mihaly Ben ze 3230, 3253 Manuel Benito 3253, 3261, 3262 Elsie Campbell 3228, 3237, 3247, 3269, 3271 Cao Minh Quang 3230, 3253, 3281, 3300 s ar Ciaurri Ramrez 3253, 3261, 3262 O Vasile C^rtoaje Klamkin{05, 3275 Chip Curtis 3204, 3221, 3226, 3234, 3261, 3274, 3276, 3294 Apostolis K. Demis 3211, 3219, 3224, 3255, 3273 Charles R. Diminnie 3221, 3228, 3237, 3247, 3269, 3271 Emilio Fernandez Moral 3253, 3261, 3262 J. Chris Fisher 3238 Fran is o Javier Gar a Capita n 3221 Oliver Geupel 3263, 3273, 3282, 3284, 3292 Douglass L. Grant 3293 P.C. Hammer 3254 Karl Havlak 3271 John Hawkins 3240 Ri hard I. Hess 3221, 3232, 3283, 3286 John G. Heuver 3245, 3289 Joe Howard 3287 Walther Janous 3245 Geo rey A. Kandall 3221 Paula Ko a 3271 Va lav Kone ny 3202, 3232, 3244 Kee-Wai Lau 3242, 3239, 3246, 3298 Tom Leong 3253, 3260 Nguyen Thanh Liem 3249
Tai hi Maekawa 3250, 3279 Thanos Magkos 3276 Salem Maliki 3214, 3220, 3234, 3245, 3249, 3280 Dragoljub Milosevi 3214 Andrea Munaro 3253, 3265, 3280 Vedula N. Murty 3221 Jose H. Nieto 3201, 3202, 3257, 3259 Mi hael Parmenter 3252, 3284 Alex Remorov 3249 Henry Ri ardo 3204, Juan-Bos o Romero Marquez 3221 Luz Ron al 3261, 3262 Xavier Ros 3212, 3221, 3236, 3250, 3257 Sergey Sadov 3229 Bill Sands 3268 Joel S hlosberg 3215, 3233, 3290 D.J. Smeenk 3202, 3221, 3235, 3264, 3279, 3297 T. Je erson Smith 3254 David Stone 3240 Edmund Swylan 3202, 3206, 3207, 3238 Marian Tetiva 3220 Phi Thai Thuan 3249 Daniel Tsai 3251 Panos E. Tsaoussoglou 3249 George Tsapakidis 3289 G. Tsintsifas 3232 Apostolis Vergos 3208 Vo Quo Ba Can 3210, 3216, 3223 Edward T.H. Wang 3253 Shaun White 3208 Peter Y. Woo 3203, 3205, 3218, 3221, 3225, 3244, 3245, 3295, 3296 Paul Yiu 3245 Roger Zarnowski 3237 Titu Zvonaru 3243, 3270
Featured Solvers | Groups Missouri State University Problem Solving Group 3259, 3266
Skidmore College Problem Solving Group 3245
512 Other Solvers | Individuals Mohammed Aassila 3223, 3237, 3241, 3281, 3283 Gregory Akulov 3285 Arkady Alt 3221, 3222, 3223, 3226, 3227, 3230, 3236, 3246, 3247, 3248, 3249, 3281, 3287, 3300
Miguel Amengual Covas 3203, 3223, 3250 Christos Anastassiades 3250 George 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, 3300
Dionne T. Bailey
Kee-Wai Lau 3208, 3222, 3223, 3236, 3237, 3241, 3247, 3249, 3248, 3250, 3253, 3260, 3271, 3285, 3286, 3292, 3297
Tom Leong 3251, 3257 Kathleen E. Lewis 3257 Carl Libis 3276 Sotiris Louridas 3298 Tai hi Maekawa 3250 Thanos Magkos 3221, 3260, 3286, 3287, 3297 Salem Maliki 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,
3222, 3223, 3251, 3253, 3263, 3276, 3285, 3286, 3287,
3284, 3287, 3293, 3297
3230, 3233, 3234, 3235, 3236, 3237, 3238, 3242, 3237, 3238, 3239, 3243,
Dung Nguyen Manh 3281, 3287 Pavlos Maragoudakis 3221, 3251 Petrus Martins 3253 G. Milanova 3221, 3222, 3223 D.M. Milosevi 3222 Dragoljub Milosevi 3221, 3222, 3223 Todor Mitev 3236 Andrea Munaro 3235, 3236, 3251, 3256,
3244, 3246, 3247, 3248, 3250, 3251, 3252, 3253, 3255, 3262, 3263, 3265,
3294
3293
Roy 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, 3300
Ri ardo Barroso Campos 3203, 3250, 3279, 3289, Mi hel Bataille 3201, 3202, 3206, 3207, 3208, 3211,
3221, 3223, 3228,
3266, 3267, 3270, 3271, 3279, 3280, 3282, 3284, 3287, 3289, 3290, 3292, 3294, 3295, 3296, 3297, 3298
Brian D. Beasley 3294, 3297 Fran is o Bellot Rosado 3202, 3203 Mihaly Ben ze 3204, 3205, 3213, 3214,
3280, 3287, 3289, 3290, 3291 3216, 3223, 3228, 3229, 3239,
3247, 3249, 3269, 3292, 3293, 3294
Manuel Benito 3251, 3255, 3257, 3258 K.S. Bhanu 3284 Paul Bra ken 3222, 3262, 3286 Katrina Bri ker 3221 Whitney Bullo k 3257 Sarah Burnham 3271 Elsie M. Campbell 3212, 3222, 3223, 3251, 3253, 3263, 3276, 3285, 3286, 3287, 3293
Cao Minh Quang 3212, 3214,
3264, 3276, 3278, 3287, 3289,
Vedula N. Murty 3214, 3221, 3222, 3223 N. Nadeau 3262 Virgil Ni ula 3241, 3242, 3260, 3264, 3265, 3270, 3271, 3273, 3278, 3279,
3221, 3222, 3223, 3235, 3236, 3242, 3246,
Jose H. Nieto 3205, 3221, 3237, 3248, 3251, 3252, 3253, 3258 Vi tor Oxman 3299 Fran is o Pala ios Qui~nonero 3212 Mi hael Parmenter 3237, 3238, 3250, 3287, 3294 Paolo Perfetti 3281 Pantelimon George Popes u 3269, 3281 A hilleas Pavlos Porfyriadis 3223 D.E. Prithwijit 3272 Stanley Rabinowitz 3297, 3298 Alex Remorov 3226, 3229, 3230, 3234, 3235, 3236, 3237, 3238, 3241, 3243, 3244, 3246, 3247, 3248, 3250
s ar Ciaurri Ramrez 3251, 3255, 3257, 3258 O Vasile C^rtoaje 3209, 3210, 3274, 3300 Chip Curtis 3202, 3203, 3206, 3216, 3217, 3219, 3222, 3223, 3227, 3230,
Henry Ri ardo 3204, 3212, 3247, 3293 Juan-Bos o Romero Marquez 3221, 3223, 3297 Luz Ron al 3251, 3255, 3257, 3258 Xavier Ros 3227, 3237, 3247, 3248, 3249, 3251, 3253, 3259, 3260, 3262,
3235, 3237, 3238, 3239, 3241, 3243, 3244, 3245, 3246, 3247, 3248, 3251,
3281, 3286, 3287
3247, 3248, 3249, 3250, 3251, 3260, 3265, 3270, 3276, 3287
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, 3300
Paul Deiermann 3284, 3285 Apostolis K. Demis 3202, 3203, 3205, 3207, 3213, 3214, 3221, 3233, 3234, 3235, 3238, 3251, 3256, 3258, 3264, 3265, 3267, 3270, 3289, 3292, 3297,
Deborah Salas-Smith 3257 Bill Sands 3257 Robert P. Sealy 3286 Joel 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, 3297
Hasan Denker 3222, 3223 Nikolaos Dergiades 3300 M.N. Deshpande 3284 Jose Luis Daz-Barrero 3212, 3247, 3263, 3269, 3281, 3286, 3287 Charles R. Diminnie 3212, 3222, 3223, 3250, 3251, 3253, 3263, 3276, 3285,
Bob Serkey 3287 Andrew Siefker 3253 D.J. Smeenk 3203, 3221, 3223, 3233, 3234, 3244, 3250, 3265, 3278, 3287 Digby Smith 3286 David R. Stone 3212, 3237, 3240, 3241, 3257 Edmund Swylan 3203, 3217, 3221, 3234, 3236, 3237, 3243, 3244, 3250,
3286, 3287, 3293
3257
3298
Marian Din a 3205 Oleh Faynshteyn 3293, 3297, 3298 Emilio Fernandez Moral 3251, 3255, 3257, 3258 Ovidiu Furdui 3218, 3226, 3227, 3261, 3262, 3288 Ian June L. Gar es 3251 Fran is o Javier Gar a Capita n 3202, 3290, 3291 Oliver Geupel 3257, 3264, 3266, 3268, 3276, 3278, 3279, 3281, 3283, 3286, 3287, 3289, 3290, 3293, 3294, 3295, 3296, 3297, 3298, 3299, 3300
Karl Havlak 3253, 3276, 3285, 3286, 3287, 3293 John Hawkins 3212, 3237, 3240, 3241, 3257 G.P. Henderson 3201 Ri hard I. Hess 3203, 3206, 3208, 3219, 3221,
3223, 3236, 3248, 3250, 3287, 3297,
3298, 3300
George Tsapakidis 3222, 3223, 3265, 3270, 3276, 3286, 3287, 3291, 3293, 3297
3222, 3227, 3250, 3251,
3253, 3262, 3265, 3270, 3276, 3287, 3292, 3294, 3296, 3297, 3298
John G. Heuver 3203, 3221, 3256 Joe Howard 3204, 3212, 3221, 3222, 3223, 3227, 3235, 3236, 3237, 3246, 3247, 3248, 3251, 3260, 3271, 3276, 3281
Peter Hurthig 3250 Walther Janous 3239,
Alexandros Sygelakis 3209, 3212 Winfer C. Tabares 3251 Ashley Tangeman 3253 Marian Tetiva 3246 Daniel Tsai 3203, 3286 Panos E. Tsaoussoglou 3203, 3222,
3241, 3246, 3247, 3248, 3249, 3250, 3251, 3253,
G. Tsintsifas 3233, 3234, 3243, 3244, 3254, 3255 Pham Van Thuan 3222 Phi Thai Thuan 3236, 3248 Apostolis Vergos 3212, 3248 Dan Vetter 3219 Vo Quo Ba Can 3214, 3222 Stan Wagon 3208, 3222, 3223, 3292, 3300 Peter Y. Woo 3201, 3202, 3207, 3211, 3212, 3213, 3214, 3220, 3222, 3223, 3225, 3233, 3234, 3235, 3236, 3238, 3241, 3243, 3246, 3247, 3248, 3250,
3257, 3260, 3261, 3262, 3263, 3265, 3266, 3267, 3269, 3270, 3271, 3272,
3251, 3252, 3253, 3254, 3255, 3256, 3258, 3259, 3262, 3264, 3265, 3267,
3273, 3274, 3291, 3292, 3294, 3297, 3300
3270, 3271, 3273, 3276, 3278, 3279, 3280, 3281, 3284, 3285, 3286, 3287,
Neven Juri 3259, 3276, 3286 Natalie Kalmink 3221 Geo rey A. Kandall 3202, 3235, 3250, 3289 Paula Ko a 3253, 3276, 3285, 3286 Va lav Kone ny 3203, 3221, 3234, 3235, 3243,
3289, 3290, 3291, 3292, 3293, 3294, 3297, 3298, 3299
Bingjie Wu 3257, 3272, 3281, 3287 Konstantine Zelator 3203, 3265, 3266, 3272, 3294, 3297, 3298 Titu Zvonaru 3202, 3203, 3221, 3222, 3223, 3233, 3234, 3235, 3236, 3237, 3250, 3256, 3258, 3291,
3297, 3298
3238, 3244, 3246, 3247, 3249, 3250, 3263, 3264, 3265, 3266, 3267, 3271, 3272, 3281, 3287, 3289, 3290, 3291, 3293, 3294, 3297, 3298, 3300
Other Solvers | Groups Ateneo Problem Solving Group 3221, 3251 Missouri State Univeristy Problem Solving Group 3253, 3257, 3272
Skidmore College Problem Solving Group 3221, 3237, 3291, 3294, 3297 University of Regina Math Central Consultants 3211
