HMMT 二月 2015 · 代数 · 第 7 题

英文解析

7. Suppose ( a , a , a , a ) is a 4-term sequence of real numbers satisfying the following two conditions:
   1 2 3 4
   • a = a + a and a = a + a ;
   3 2 1 4 3 2
   • there exist real numbers a, b, c such thatan + bn + c = cos( a )2
   for all n ∈ { 1 , 2 , 3 , 4 } .n
   Compute the maximum possible value ofcos( a ) − cos( a )
   1 4
   over all such sequences ( a , a , a , a ).
   1 2 3 4
   √
   Answer: − 9 + 3 13 Let f ( n ) = cos a and m = 1. The second (“quadratic interpolation”)
   condition on f ( m ), f ( m + 1), f ( m + 2), f ( m + 3) is equivalent to having a vanishing third finitendifferencef ( m + 3) − 3 f ( m + 2) + 3 f ( m + 1) − f ( m ) = 0 .
   This is analogous to the “number theoretic” proof of the uniqueness of the base 2 expansion of a nonnegative integer.3
   Algebra
   This is equivalent tof ( m + 3) − f ( m ) = 3 [ f ( m + 2) − f ( m + 1)]
   ⇐⇒ cos( a ) − cos( a ) = 3 (cos( a ) − cos( a ))
   m +3 m m +2 m +1
   ( ) ( )
   a + a a − am +2 m +1 m +2 m +1 = − 6 sin sin
   2 2
   ( ) ( )
   a am +3 m = − 6 sin sin .
   2 2
   ( ) ( )
   m +3 aam
   Set x = sin and y = sin . Then the above rearranges to
   2 2
   2 2 2 2
   (1 − 2 x ) − (1 − 2 y ) = − 6 xy ⇐⇒ x − y = 3 xy.
   √
   − 3 ± 13 2 2
   Solving gives y = x . The expression we are trying to maximize is 2( x − y ) = 6 xy , so we want
   √2 − 3+ 13
   x , y to have the same sign; thus y = x .
   Then | y | ≤ | x | , so since | x | , | y | ≤ 1, to maximize 6 xy we can simply set x = 1, for a maximal value of 2
   √
   √
   − 3+ 13
   6 · = − 9 + 3 13.
   22