HMMT 二月 2007 · 冲刺赛 · 第 31 题

HMMT February 2007 — Guts Round — Problem 31

[ 18 ] A sequence { a } of real numbers satisfies the recursion a = a − 3 a + 3 for all positive n n ≥ 0 n +1 n n integers n . For how many values of a does a = a ? 0 2007 0

解析

[ 18 ] A sequence { a } of real numbers satisfies the recursion a = a − 3 a + 3 for all positive n n ≥ 0 n +1 n n integers n . For how many values of a does a = a ? 0 2007 0 2007 3 2 Answer: 3 . If x appears in the sequence, the next term x − 3 x + 3 is the same if and only if 3 2 0 = x − 3 x − x + 3 = ( x − 3)( x − 1)( x + 1). Moreover, that next term is strictly larger if x > 3 and strictly smaller if x < − 1. It follows that no values of a with | a − 1 | > 2 yield a = a . 0 0 0 2007 αi − αi Now suppose a = a and write a = 1 + e + e ; the values a we seek will be in bijective 0 2007 0 0 correspondence with solutions α where 0 ≤ α ≤ π . Then 3 3 αi αi − αi − 3 αi αi − αi 3 αi − 3 αi a = ( a − 1) − 3 a + 4 = e + 3 e + 3 e + e − 3 e − 3 e − 3 + 4 = e + e + 1 , 1 0 0 2007 2007 3 αi − 3 αi and an easy inductive argument gives a = e + e + 1 . It follows that a = a is 2007 0 2007 ( ) 2007 equivalent to cos( α ) = cos 3 α . Now, (( ) ) (( ) ) 2007 2007 ( ) 3 + 1 3 − 1 2007 cos 3 α − cos( α ) = 2 sin α sin α , 2 2 π so since sin( kx ) = 0 for a positive integer k if and only if x is a multiple of , the solutions α are k 2 π 4 π 2 π { 0 , , , . . . , π }∪{ 0 , , . . . , π } . Because our values k are consecutive, these sets overlap 2007 2007 2007 3 − 1 3 − 1 3 +1 2007 only at 0 and π , so there are 3 distinct α . 10