Volume 11, Number 3 June 2006 – October 2006

Olympiad Corner by Parts

The following were the problems of Kin Y. Li the IMO 2006.

Day 1 (July 12, 2006) In calculus, we have a formula called Proof. Applying summation by parts

Problem 1. Let ABC be a triangle with and Cauchy-Schwarz’ inequality, we t incenter I. A point P in the interior of the have f (x)g(x)dx = F(t)g(t) − F(s)g(s) triangle satisfies ∫ n ⎛ n ⎞ n−1 ⎛ k ⎞ s a2 = ⎜ a ⎟a + ⎜ a ⎟(a − a ) t ∑ i ∑ i n ∑∑i k k+1 ∠ PBA + ∠ PCA = ∠ PBC + ∠ PCB . i=1 ⎝ i=1 ⎠ k==11⎝ i ⎠ − F (x)g '(x)dx, ∫ Show that AP ≥ AI, and that equality s n n−1 k holds if and only if P = I. where F(x) is an anti-derivative of f (x). ⎛ ⎞ ⎛ ⎞ ≤ ⎜∑bi ⎟an + ∑∑⎜ bi ⎟(ak − ak +1) Problem 2. Let P be a regular There is a discrete version of this ⎝ i=1 ⎠ k ==11⎝ i ⎠ 2006-gon. A diagonal of P is called formula for . It is called n

good if its endpoints divide the summation by parts, which asserts = ∑ a ibi boundary of P into two parts, each n n−1 i =1 a b =A b − A (b − b ), n 1/ 2 n 1/ 2 composed of an odd number of sides of ∑ k k n n ∑ k k+1 k ⎛ 2 ⎞ ⎛ 2 ⎞ k=1 k=1 ≤ ⎜ a ⎟ ⎜ b ⎟ . P. The sides of P are also called good. ∑ i ∑ i where A = a +a + +a . This formula ⎝ i=1 ⎠ ⎝ i=1 ⎠ k 1 2 ⋯ k Suppose P has been dissected into follows easily by observing that a1 = A1 triangles by 2003 diagonals, no two of Squaring and simplifying, we get and for k > 1, ak = Ak − Ak−1 so that n n which have a common point in the n a 2 ≤ b2 . a b =Ab +(A − A )b + +(A − A )b ∑ i ∑ i interior of P. Find the maximum ∑ k k 1 1 2 1 2 L n n−1 n i=1 i=1 k=1 number of isosceles triangles having two = Anbn − A1(b2 − b1) −L − An−1(bn − bn−1) good sides that could appear in such a Below we will do some more examples configuration. n−1 = Anbn − ∑ Ak (bk +1 − bk ). to illustrate the usefulness of the Problem 3. Determine the least real k =1 summation by parts formula. number M such that the inequality From this identity, we can easily obtain

some famous inequalities. Example 1. (1978 IMO) Let n be a | ab(a2 − b2 ) + bc(b2 − c2 ) + ca(c2 − a2 ) | k 2 2 2 2 positive integer and a , a , , a be a ≤ M (a + b + c ) Abel’s Inequality. Let m ≤ a ≤ M 1 2 ⋯ n ∑ i of distinct positive integers. holds for all real numbers a, b and c. i=1 Prove that for k = 1,2,…,n and b1 ≥ b2 ≥ ⋯ ≥ bn > 0. n n (continued on page 4) a 1 Then k ∑ 2 ≥ ∑ . Editors: ஻ Ի ஶ (CHEUNG Pak-Hong), Munsang College, HK n k =1 k k =1 k

ଽ υ ࣻ (KO Tsz-Mei) b1m ≤ ∑ ak bk ≤ b1M . గ ႀ ᄸ (LEUNG Tat-Wing) k =1 Solution. Since the ai’s are distinct ୊ ፱ (LI Kin-Yin), Dept. of Math., HKUST ؃ Proof. Let Ak= a1+a2+⋯+ak. Applying positive integers, Ak = a1 + a2 + ⋯ + ak ᜢ ݰ (NG Keng-Po Roger), ITC, HKPU ֔ summation by parts, we have is at least 1 + 2 + + k = k(k + 1)/2. ⋯ Artist: ྆ ؾ ़ (YEUNG Sau-Ying Camille), MFA, CU n n−1 a b = A b + A (b − b ). Acknowledgment: Thanks to Elina Chiu, Math. Dept., ∑ k k n n ∑ k k k+1 Applying summation by parts, we have HKUST for general assistance. k =1 k =1 The right side is at least On-line: n a A n−1 ⎛ 1 1 ⎞ http://www.math.ust.hk/mathematical_excalibur/ n−1 k = n + A ⎜ − ⎟ ∑ 2 2 ∑ k ⎜ 2 2 ⎟ mb + m(b − b ) = mb k =1 k n k =1 k (k +1) The editors welcome contributions from all teachers and n ∑ k k+1 1 ⎝ ⎠ k=1 n(n +1) / 2 n−1 k(k +1) (2k +1) students. With your submission, please include your name, ≥ + address, school, email, telephone and fax numbers (if and at most n2 ∑ 2 k 2 (k +1)2 available). Electronic submissions, especially in MS Word, n−1 k =1 are encouraged. The deadline for receiving material for the Mb + M (b − b ) = Mb . 1 ⎛ 1 n−1 2k +1 ⎞ n ∑ k k+1 1 = ⎜1+ + ⎟ next issue is November 25, 2006. k=1 ⎜ ∑ ⎟ 2 ⎝ n k=1 k(k +1) ⎠ For individual subscription for the next five issues for the 05-06 academic year, send us five stamped self-addressed K. L. Chung’s Inequality. Suppose 1 ⎛ 1 n−1 ⎛ 1 1 ⎞⎞ envelopes. Send all correspondence to: k k = ⎜1+ + ∑⎜ + ⎟⎟ a ≥ a ≥ ≥ a > 0 and 2 ⎝ n k=1 ⎝ k k +1⎠⎠ 1 2 ⋯ n ai ≤ bi Dr. Kin-Yin LI ∑ ∑ n n−1 i=1 i=1 1 ⎛ 1 1 ⎞ Department of Mathematics = ⎜ + ⎟ The Hong Kong University of Science and Technology ∑ ∑ for k = 1,2,…,n. Then 2 ⎝ k=1 k k=0 k +1⎠ Clear Water Bay, Kowloon, Hong Kong n n n a2 ≤ b2. 1 Fax: (852) 2358 1643 ∑ i ∑ i = ∑ . Email: [email protected] i=1 i=1 k=1 k Mathematical Excalibur, Vol. 11, No. 3, Jun. 06 – Oct. 06 Page 2

Example 2. (1982 USAMO) If x is a 2 (a1 − a2 )(1− z) + (a2 − a3 )(1− z ) +L always holds. positive real number and n is a positive n integer, then prove that + an (1− z ) = 0. Solution. Let Si = x1 + x2 + ⋯ + xi. Let

Then p = x , [x] [2x] [3x] [nx] ∑ i q = −∑ xi . x > 0 [nx] ≥ + + + L + , (a − a ) + (a − a ) + + a = i xi <0 1 2 3 n 1 2 2 3 L n 2 n where [t] denotes the greatest integer (a1 − a2 )z + (a2 − a3 )z +L+ an z . Then p − q = 0 and p + q = 1. So p = less than or equal to t. However, since |z| = 1 and z ≠ 1, by the 1 1 1 q = . Thus, − ≤ S ≤ for k = Solution. Let a = [kx]/k. Then triangle inequality, 2 2 k 2 k k [ix] 2 n | (a1 − a2 )z + (a2 − a3 )z +L+ an z | 1,2,⋯, n. Ak = ∑ . i=1 i 2 n < (a1 − a2 )z + | (a2 − a3 )z | +L+ | an z | Applying summation by parts, we get In terms of Ak, we are to prove [nx]≥An. = (a1 − a2 ) + (a2 − a3 ) +L+ an , The case n = 1 is easy. Suppose the n n−1 cases 1 to n − 1 are true. Applying which is a contradiction to the last ∑ ai xi = Snan − ∑ Sk (ak+1 − ak ) summation by parts, we have i=1 k=1 displayed equation. So Pn ≠ 0 = P0. n n n−1 n−1

[kx] = ak k = An n − Ak . ∑ ∑ ∑ ≤ Sk (ak − ak+1) k=1 k=1 k=1 Example 4. Show that the series ∑ k=1 Using this and the inductive ∞ sin k converges. n−1 hypothesis, ∑ 1 k =1 k ≤ ∑ (ak − ak+1) n n−1 k=1 2

Ann = ∑[kx] + ∑ Ak Solution. Let ak = sin k and bk = 1/k. k =1 k =1 1 = (a − a ). Using the identity 2 1 n n n−1 1 1 ≤ ∑[kx] + ∑[kx] 1 cos(m − 2) − cos(m + 2) sinmsin = , When x1 = 1/2, xn= −1/2 and all other k=1 k=1 2 2 n−1 = [nx] + ()[kx] + [(n − k)x] xi = 0, we have equality. So the least ∑ we get k=1 such m is 1/2. n−1 cos 1 − cos(k + 1 ) ≤ [nx] + ∑[kx + (n − k)x] A = sin1+ + sink = 2 2 . k=1 k L 1 Example 6. Prove that for all real 2sin 2 = n[nx], numbers a1, a2, …, an, there is an Then |Ak| ≤ 1/(sin ½) and hence integer m among 1,2,…, n such that if which yields case n. lim Anbn = 0. n → ∞ π 0 ≤ θn ≤ θ n−1 ≤ L ≤ θ1 ≤ , Example 3. Consider a polygonal line Applying summation by parts, we get 2

n m P0P1P2…Pn such that ∠ P0P1P2 = ∞ sin k n then ∑ ai sinθi ≤ ∑ ai . = lim ak bk ∑ n→∞ ∑ i=1 i=1 ∠ P1P2P3 = ⋯ = ∠ Pn−2Pn−1Pn, all k=1 k k=1 measure in counterclockwise direction. n−1 Solution. Let Ai = a1+a2+ +ai and bi = lim(Anbn − Ak (bk+1 −bk )) ⋯ n→∞ ∑ k=1 = sin θi, then If P0P1 > P1P2 > > Pn−1Pn, show that 1≥ b1 ≥ b2 ≥L≥ bn ≥ 0. ⋯ ∞ ⎛ 1 1 ⎞ Next let |Am| be the maximum among = A − . P0 and Pn cannot coincide. ∑ k ⎜ ⎟ k=1 ⎝ k k +1⎠ |A1|, |A2|, …, |An|. With an+1 = bn+1 = 0, we apply summation by parts to get Since Solution. Let ak be the length of Pk−1Pk. Consider the complex plane. Each P ∞ ⎛ 1 1 ⎞ 1 ∞ ⎛ 1 1 ⎞ 1 k A ⎜ − ⎟ ≤ ⎜ − ⎟ = , n n+1 ∑ k 1 ∑ 1 k=1 ⎝ k k +1⎠ sin 2 k=1⎝ k k +1⎠ sin 2 corresponds to a complex number. We ∑ai sinθi = ∑aibi i=1 i=1 may set P0 = 0 and P1 = a1. Let θ = ∞ sin k so converges. n ∠ P0P1P2 and z = - cosθ + i sinθ , then ∑ k k =1 = ∑ Ai (bi+1 − bi ) P = n−1 . Applying i=1 n a1 + a2 z +L+ an z summation by parts, we get Example 5. Let a ≥ a ≥ ≥ a with n 1 2 L n ≤ ∑ Am (bi − bi+1 ) n n i=1 Pn = (a1 − a2 ) + (a2 − a3 )(1+ z) +L a1 ≠ an, and Find xi = 0 xi = 1. n−1 ∑ ∑ . i=1 i=1 = A b + an (1+ z + L+ z ) m 1 . the least number m such that ≤ Am If θ = 0, then z = 1 and Pn > 0. If θ ≠ 0, n then assume Pn = 0. We get Pn (1−z) = ∑ ai xi ≤ m(a1 − an ) 0, which implies i=1 Mathematical Excalibur, Vol. 11, No. 3, Jun. 06 – Oct. 06 Page 3

Problem Corner Problem 251. Determine with proof the When k = 0, p divides 2n − 1, which largest number x such that a cubical gift of implies 2n ≡ 1 (mod p). By Fermat’s side x can be wrapped completely by little theorem, 2p ≡ 2 (mod p). Finally, We welcome readers to submit their folding a unit square of wrapping paper when k = 1, we get solutions to the problems posed below (without cutting). for publication consideration. The 1 ≡ 2n+p = 2n 2p ≡ 1·2 = 2 (mod p) solutions should be preceded by the Solution. CHAN Tsz Lung (Math, HKU) solver’s name, home (or email) address and Jeff CHEN (Virginia, USA). implying p divides 2 − 1 = 1, which is a and school affiliation. Please send contradiction. Let A and B be two points inside or on the submissions to Dr. Kin Y. Li, unit square such that the line segment AB Department of Mathematics, The Hong Problem 253. Suppose the bisector of has length d. After folding, the distance Kong University of Science & ∠ BAC intersect the arc opposite the between A and B along the surface of the Technology, Clear Water Bay, Kowloon, angle on the circumcircle of ∆ABC at cube will be at most d because the line Hong Kong. The deadline for A1. Let B1 and C1 be defined similarly. segment AB on the unit square after submitting solutions is November 25, Prove that the area of ∆A1B1C1 is at folding will provide one path between the 2006. least the area of ∆ABC. two points along the surface of the cube,

which may or may not be the shortest Solution. CHAN Tsz Lung (Math, Problem 256. Show that there is a possible. HKU), Jeff CHEN (Virginia, USA) rational number q such that and Anna Ying PUN (STFA Leung Kau Kui College, Form 7). sin1o sin2o sin89o sin90o = q 10. In the case A is the center of the unit L square and B is the point opposite to A on A the surface of the cube with respect to the Problem 257. Let n > 1 be an integer. C center of the cube then the distance along 1 Prove that there is a unique positive , 2 2 the surface of the cube between them is at B integer A < n such that [n /A] + 1 is 1 least 2x. Hence, 2x ≤ d ≤ I divisible by n, where [x] denotes the 2 / 2. Therefore, x ≤ greatest integer less than or equal to x. 2 / 4.

(Source: 1993 Jiangsu Math Contest) The maximum x = 2 / 4 is attainable can B C Problem 258. (Due to Mihaiela be seen by considering the following Vizental and Alfred Eckstein, Arad, picture of the unit square. A1 Romaina) Show that if A, B, C are in the interval (0, π/2), then By a well-known property of the f (A,B,C) + f (B,C,A) + f (C,A,B) ≥ 3, incenter I (see page 1 of Mathematical Excalibur, vol. 11, no. 2), we have AC1 where 4sin x + 3sin y + 2sin z . x f (x, y, z) = = C1I and AB1 = B1I. Hence, ∆AC1B1 ≅ 2sin x + 3sin y + 4sin z ∆IC1B1. Similarly, ∆BA1C1 ≅ ∆IA1C1

and ∆CB1A1 ≅ ∆IB1A1. Letting [⋯] Problem 259. Let AD, BE, CF be the denote area, we have altitudes of acute triangle ABC. Through D, draw a line parallel to line [AB1CA1BC1] = 2[A1B1C1]. EF intersecting line AB at R and line AC at Q. Let P be the intersection of Commended solvers: Alex O Kin-Chit If ∆ABC is not acute, say ∠ BAC is lines EF and CB. Prove that the (STFA Cheng Yu Tung Secondary School) not acute, then and Anna Ying PUN (STFA Leung Kau circumcircle of ⊿PQR passes through 1 Kui College, Form 7). []ABC ≤ []ABA C the midpoint M of side BC. 1 2 (Source: 1994 Hubei Math Contest) Problem 252. Find all polynomials f(x) 1 with integer coefficients such that for ≤ AB CA BC = A B C . Problem 260. In a class of 30 students, [][]1 1 1 1 1 1 every positive integer n, 2n − 1 is divisible 2 number the students 1, 2, …, 30 from by f(n). best to worst ability (no two with the Otherwise, ∆ABC is acute and we can same ability). Every student has the Solution. Jeff CHEN (Virginia, USA) and apply the fact that same number of friends in the class, G.R.A. 20 Math Problem Group (Roma, where friendships are mutual. Call a Italy). 1 student good if his ability is better than []ABC ≤ [][]AB1CA1BC1 = A1B1C1 more than half of his friends. We will prove that the only such 2 Determine the maximum possible polynomials f(x) are the constant (see example 6 on page 2 of number of good students in this class. polynomials 1 and −1. Mathematical Excalibur, vol. 11, no. (Source: 1998 Hubei Math Contest) 2). Assume f(x) is such a polynomial and |f(n)| ≠ 1 for some n > 1. Let p be a prime Commended solvers: Samuel Liló ***************** which divides f(n), then p also divides Abdalla (Brazil) and Koyrtis G. Solutions f(n+ kp) for every integer k. Therefore, p CHRYSSOSTOMOS (Larissa, divides 2n+kp−1 for all integers k ≥ 0. **************** Greece, teacher).

Mathematical Excalibur, Vol. 11, No. 3, Jun. 06 – Oct. 06 Page 4

Problem 254. Prove that if a, b, c > 0, Problem 255. Twelve drama groups are 3 × 4 = 12 slots for performances. then to do a series of performances (with some However, each of groups 4 to 12 has to groups possibly making repeated perform at least twice, yielding at least abc ( a + b + c) + (a + b + c)2 performances) in seven days. Each group 9 × 2 = 18 ( > 12 ) performances, is to see every other group’s performance contradiction. ≥ 4 3abc (a + b + c). at least once in one of its day-offs.

Find with proof the minimum total Case 2: More than 3 groups perform Solution 1. José Luis Díaz-Barrero number of performances by these groups. (Universitat Politècnica de Catalunya, exactly once, say k groups with k > 3. By argument similar to case 1(a), we see Barcelona, Spain) and G.R.A. 20 Math Solution. CHAN Tsz Lung (Math, HKU). Problem Group (Roma, Italy). at most 3 groups can perform on each of Here are three important observations: the remaining 7 − k days (meaning at most 3(7 − k) performance slots). Again, Dividing both sides by (1) Each group perform at least once. the remaining 12 − k groups have to abc(a + b + c) , the inequality is (2) If more than one groups perform on the perform at least twice, yielding 2(12 − k) ≤ 3(7 − k), which implies k ≤ −3, equivalent to same day, then each of these groups will have to perform on another day so the other contradiction. 3 a + b + c ( a + b + c) groups can see its performance in their + ≥ 4 3. Commended solvers: Anna Ying PUN a + b + c abc day-offs. (STFA Leung Kau Kui College, Form 7) By the AM-GM inequality, (3) If a group performs exactly once, on the and Raúl A. SIMON (Santiago, day it performs, it is the only group Chile). 1/ 3 a + b + c ≥ 3( abc) . performing. Therefore, it suffices to show Comments: This was a problem in the We will show the minimum number of 1994 Chinese IMO team training tests. 3( abc)1/ 3 ( a + b + c)3 3 performances is 22. The following In the Chinese literature, there is a + = + t3 ≥ 4 3, a + b + c abc t performance schedule shows the case 22 is solution using the famous Sperner’s possible. theorem which asserts that for a set where again by the AM-GM inequality, with n elements, the number of subsets Day 1: Group 1 a + b + c a + b + c so that no two with one contains the t = = ≥ 3. Day 2: Group 2 1/ 3 1/ 3 n ( abc) (abc) Day 3: Groups 3, 4, 5, 6 other is at most ⎛ ⎞ . We hope to ⎜ ⎟ Day 4: Groups 7, 8, 9, 3 ⎝[]n / 2 ⎠ By the AM-GM inequality a third time, Day 5: Groups 10, 11, 4, 7 present this solution in a future article. 3 3 3 2 3 3 3 t t t 4t Day 6: Groups 12, 5, 8, 10 + t = + + + ≥ ≥ 4 3. t t 3 3 3 3 Day 7: Groups 6, 9, 11, 12.

Solution 2. Alex O Kin-Chit (STFA Assume it is possible to do at most 21 Olympiad Corner Cheng Yu Tung Secondary School). performances. Let k groups perform exactly once, then k + 2(12 − k) ≤ 21 will (continued from page 1) By the AM-GM inequality, we have imply k ≥ 3. Day 2 (July 13, 2006) a + b + c ≥ 3(abc)1/ 3 (1) Case 1: Exactly 3 groups perform exactly and a + b + c ≥ 3(abc)1/ 6 . (2) once, say group 1 on day 1, group 2 on day Problem 4. Determine all pairs (x,y) of integers such that Applying (2), (1), the AM-GM 2 and group 3 on day 3. 1 + 2x + 22x+1 = y2. inequality and (1) in that order below, (a) If at least 4 groups perform on one of we have the remaining 4 days, say groups 4, 5, 6, 7 Problem 5. Let P(x) be a polynomial on day 4, then by (2), each of them has to of degree n > 1 with integer perform on one of the remaining 3 days. By abc ( a + b + c) + (a + b + c)2 coefficients and let k be a positive the pigeonhole principle, two of groups 4, 5, integer. Consider the polynomial Q(x) 3(abc)2 / 3 3(abc)1/ 3 (a b c) 6, 7 will perform on the same day again ≥ + + + = P ( P ( ⋯ P ( P (x) ) ⋯ ) ), where P 1/ 4 later, say groups 4 and 5 perform on day 5. ≥ 4()3(abc)2/3 (abc)(a + b + c)3 occurs k times. Prove that there are at Then they will have to perform separately most n integers t such that Q(t) = t. 1/ 4 ≥ 4()3(abc)2/ 3 (abc)3(abc)1/ 3 (a + b + c)2 on the last 2 days for the other to see. Then groups 1, 2, 3 once each, groups 4, 5 thrice = 4 3abc(a + b + c). Problem 6. Assign to each side b of a each and groups 6, 7, …, 12 twice each at convex polygon P the maximum area least , resulting in at least of a triangle that has b as a side and is Commended solvers: Samuel Liló 3 + 2 × 3 + 7 × 2 = 23 contained in P. Show that the sum of Abdalla (Brazil), CHAN Tsz Lung (Math, HKU), Koyrtis G. the areas assigned to the sides of P is at CHRYSSOSTOMOS (Larissa, performances, contradiction. least twice the area of P. Greece, teacher) and Anna Ying PUN (STFA Leung Kau Kui College, Form (b) If at most 3 groups perform on each of 7). the remaining 4 days, then there are at most