: ※ 引述《FAlin (FA（バルシェ应援）)》之铭言： : : 1. Prove that for any two positive integers k , n there exist positive : : integers m_1 , m_2 , ... , m_k such that : : 2^k - 1 1 1 1 : : 1 + ------- = ( 1 + --- )( 1 + --- )...( 1 + --- ) . : : n m_1 m_2 m_k : : 2. Giver 2013 red and 2014 blue points in the plane , no three of them on a : : line. We aim to split plane by lines (not passing through these points) : : into regions such that there are no regions containing points of both the : : colors. What is the least number of lines that always suffice? : : 3. Let ABC be a triangle and that A_1 , B_1 , and C_1 be points of cantact of : : the excircles with the sides BC , AC , and AB , respectively. Prove that if : : the circumcenter of △A_1B_1C_1 lies on the circumcircle of △ABC , then : : △ABC is a right triangle. 第一题自己的做法(含思考过程) 首先猜测 2^(k+1) - 1 2^k - 1 1 1 + ----------- = ( 1 + -------)( 1 + ---) n n m 然後发现m不可能为整数 接着代数字找归律 (n,k) (2,1) = 3/2 (2,2) = 5/2 = 5/4 * 2/1 (2,3) = 9/2 = 9/8 * 2/1 * 2/1 = 3/2 * 3/2 * 2/1 (2,4) = 17/2 = 17/6 * 2/1 * 2/1 * 2/1 (2,5) = 33/2 = 33/32 * 2/1 * 2/1 * 2/1 * 2/1 = 11/10 * 5/4 * 3/2 * 2/1 * 2/1 (3,1) = 4/3 (3,2) = 6/3 = 4/3 * 3/2 = 4/3 * 3/2 (3,3) = 10/3 = 5/4 * 4/3 * 2/1 = 4/3 * 5/2 (3,4) = 18/3 = 4/3 * 3/2 * 3/2 * 2/1 = 4/3 * 9/2 (3,5) = 34/3 = 17/16 * 4/3 * 4/3 * 3/2 * 2/1 = 4/3 * 17/2 (4,1) = 5/4 (4,2) = 7/4 = 7/6 * 3/2 (4,3) = 11/4 = 11/10 * 5/2 (4,4) = 19/4 = 19/18 * 9/2 (4,5) = 35/4 = 35/34 * 17/2 (5,1) = 6/5 (5,2) = 8/5 = 6/5 * 4/3 (5,3) = 12/5 = 6/5 * 6/3 (5,4) = 20/5 = 6/5 * 10/3 得出 2^(k+1) - 1 1 2^k-1 n为偶数时，有 ( 1 + ----------- ) = ( 1 + --------------- )( 1 + ----- ) n 2^(k+1) + n - 2 n/2 2^(k+1) - 1 1 2^(k-1)-1 n为奇数时，有 ( 1 + ----------- ) = ( 1 + --- )( 1 + --------- ) n n (n+1)/2 所以可用二阶数学归纳法 归纳後证毕 -- valuable sheaves 4 FELIDS ╔╦╦═╦╗╔═╦╦╦═╗ storyteller Blessing Card ║║║═║║║╚╣╩║═╣ JESTER ║║║║║╚╬╗║║║═╣ REVOLT ╚═╩╩╩═╩═╩╩╩═╝ PLAY THE JOKER AFFLICT / Fragment --

※ 发信站: 批踢踢实业坊(ptt.cc) ◆ From: 124.11.128.7 ※ 编辑: FAlin 来自: 124.11.128.7 (07/26 23:33)