HMMT 二月 2007 · 冲刺赛 · 第 31 题

英文解析

31. [ 18 ] A sequence { a } of real numbers satisfies the recursion a = a − 3 a + 3 for all positiven n ≥ 0 n +1
    n nintegers 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 andstrictly 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 αia = ( 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 αiand 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
    so since sin( kx ) = 0 for a positive integer k if and only if x is a multiple of , the solutions α areπk
    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