HMMT 二月 2008 · 冲刺赛 · 第 31 题
HMMT February 2008 — Guts Round — Problem 31
题目详情
- [ 18 ] Let C be the hyperbola y − x = 1. Given a point P on the x -axis, we construct a sequence
0
of points ( P ) on the x -axis in the following manner: let
be the line with slope 1 passing passing n n through P , then P is the orthogonal projection of the point of intersection ofand C onto the n n +1 n x -axis. (If P = 0, then the sequence simply terminates.) n Let N be the number of starting positions P on the x -axis such that P = P . Determine the 0 0 2008 remainder of N when divided by 2008.
解析
- [ 18 ] Let C be the hyperbola y − x = 1. Given a point P on the x -axis, we construct a sequence
0
of points ( P ) on the x -axis in the following manner: let
be the line with slope 1 passing passing n n 8 through P , then P is the orthogonal projection of the point of intersection ofand C onto the n n +1 n x -axis. (If P = 0, then the sequence simply terminates.) n Let N be the number of starting positions P on the x -axis such that P = P . Determine the 0 0 2008 remainder of N when divided by 2008. Answer: 254 Let P = ( x , 0). Then the ` meet C at ( x , x − x ). Since this point lies on n n n n +1 n +1 n 2 2 the hyperbola, we have ( x − x ) − x = 1. Rearranging this equation gives n +1 n n +1 2 x − 1 n x = . n +1 2 x n n Choose a θ ∈ (0 , π ) with cot θ = x , and define θ = 2 θ . Using the double-angle formula, we have 0 0 0 n 0 2 cot θ − 1 n cot θ = cot(2 θ ) = . n +1 n 2 cot θ n ( ) 2008 It follows by induction that x = cot θ . Then, P = P corresponds to cot θ = cot 2 θ n n 0 2008 0 0 n (assuming that P is never at the origin, or equivalently, 2 θ is never an integer multiple of π ). So, we 0 2008 need to find the number of θ ∈ (0 , π ) with the property that 2 θ − θ = kπ for some integer k . We 0 0 0 kπ 2008 have θ = , so k can be any integer between 1 and 2 − 2 inclusive (and note that since the 0 2008 2 − 1 denominator is odd, the sequence never terminates). It follows that the number of starting positions 2008 is N = 2 − 2. 3 Finally, we need to compute the remainder when N is divided by 2008. We have 2008 = 2 × 251. ( ) 4 2008 250 4 Using Fermat’s Little Theorem with 251, we get 2 ≡ 2 · 256 ≡ 1 · 5 = 5 (mod 251). So we have N ≡ 3 (mod 251) and N ≡ − 2 (mod 8). Using Chinese Remainder Theorem, we get N ≡ 254 (mod 2008).