Volume 11, Number 3 June 2006 – October 2006 Olympiad Corner Summation 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 integration by parts 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 series. 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 ⎝ i=1 ⎠ ⎝ i=1 ⎠ where Ak = a1+a2+⋯+ak. This formula 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 sequence 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 ⎞ n−1 k n http://www.math.ust.hk/mathematical_excalibur/ 2 = 2 + Ak ⎜ 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 n−1 K. L. Chung’s Inequality. Suppose 1 ⎛ 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 2 Example 2. (1982 USAMO) If x is a (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 |A | ≤ 1/(sin ½) and hence k integer m among 1,2,…, n such that if which yields case n. lim Anbn = 0. π n → ∞ 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.