HMMT 二月 2012 · 几何 · 第 10 题
HMMT February 2012 — Geometry — Problem 10
题目详情
- Let C denote the set of points ( x, y ) ∈ R such that x + y ≤ 1. A sequence A = ( x , y ) | i ≥ 0 of i i i 2 points in R is ‘centric’ if it satisfies the following properties: • A = ( x , y ) = (0 , 0), A = ( x , y ) = (1 , 0). 0 0 0 1 1 1 • For all n ≥ 0, the circumcenter of triangle A A A lies in C . n n +1 n +2 2 2 Let K be the maximum value of x + y over all centric sequences. Find all points ( x, y ) such 2012 2012 2 2 that x + y = K and there exists a centric sequence such that A = ( x, y ). 2012 th 15 Annual Harvard-MIT Mathematics Tournament Saturday 11 February 2012 Geometry Test Name Team ID# School Team
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解析
- Let C denote the set of points ( x, y ) ∈ R such that x + y ≤ 1. A sequence A = ( x , y ) | i ≥ 0 of i i i 2 points in R is ‘centric’ if it satisfies the following properties: • A = ( x , y ) = (0 , 0), A = ( x , y ) = (1 , 0). 0 0 0 1 1 1 • For all n ≥ 0, the circumcenter of triangle A A A lies in C . n n +1 n +2 2 2 Let K be the maximum value of x + y over all centric sequences. Find all points ( x, y ) such 2012 2012 2 2 that x + y = K and there exists a centric sequence such that A = ( x, y ). 2012 Geometry Test √ √ Answer: ( − 1006 , 1006 3) , ( − 1006 , − 1006 3) Consider any triple of points △ A A A with n n +1 n +2 circumcenter P . By the Triangle Inequality we have A P ≤ A A + A P ≤ A A +1. Since P is the n n n n 0 0 n n 0 n circumcenter, we have P A = P A . Finally we have A a ≤ P A +1 = A P +1 ≤ A A +2. n n n n +2 n +2 0 n n +2 n n n 0 √ √ 2 2 2 2 Therefore x + y ≤ x + y + 2. n +2 n +2 n n It is also clear that equality occurs if and only if A , A , P , A are all collinear and P lies on the n 0 n n +2 n unit circle. √ 2 2 2 From the above inequality it is clear that x + y ≤ 2012. So the maximum value of K is 2012 . 2012 2012 Now we must find all points A that conforms to the conditions of the equality case. P must lie on 2 0 the unit circle, so it lies on the intersection of the unit circle with the perpendicular bisector of A A , 0 1 ( ) √ 1 1 3 which is the line x = . Thus P must be one of , ± . From now on we assume that we take the 0 2 2 2 positive root, as the negative root just reflects all successive points about the x -axis. ( ) √ √ 1 3 If P = , then A , P , A must be colinear, so A = (1 , 3). 0 0 0 2 2 2 2 Then since we must have A , P , A , A colinear and P on the unit circle, it follows that 0 2 n 2 n 2 n +2 2 n ( ) √ √ n 1 3 n +1 P = ( − 1) , . Then by induction we have A = ( − 1) ( n, n 3). To fill out the rest of the 2 n 2 n 2 2 ( ) √ √ n n +1 1 3 sequence, we may take A = ( − 1) ( n + 1 , − n 3) and P = ( − 1) , − so that we get 2 n +1 2 n +1 2 2 the needed properties. Therefore the answer is √ √ A ∈ { ( − 1006 , 1006 3) , ( − 1006 , − 1006 3) } 2012 . Geometry Test