Re: [問題] IMO 2013 in Colombia Day 1
: ※ 引述《FAlin (FA（バルシェ應援）)》之銘言：
: : 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.
第一題自己的做法(含思考過程)
首先猜測
2^(k+1) - 1 2^k - 1 1
1 + ----------- = ( 1 + -------)( 1 + ---)
n n m
然後發現m不可能為整數
接著代數字找歸律 (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為偶數時，有 ( 1 + ----------- ) = ( 1 + --------------- )( 1 + ----- )
n 2^(k+1) + n - 2 n/2
2^(k+1) - 1 1 2^(k-1)-1
n為奇數時，有 ( 1 + ----------- ) = ( 1 + --- )( 1 + --------- )
n n (n+1)/2
所以可用二階數學歸納法
歸納後證畢
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◆ From: 124.11.128.7
※ 編輯: FAlin 來自: 124.11.128.7 (07/26 23:33)
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07/27 02:05, , 1F
07/27 02:05, 1F
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