HMMT 二月 2011 · TEAM2 赛 · 第 4 题

英文解析

4. [ 40 ] Let a , b , and c be complex numbers such that | a | = | b | = | c | = 1. If
   2 2 2
   a b c + + = 1
   bc ca abas well, determine the product of all possible values of | a + b + c | .
   Answer: 2 Let s = a + b + c . Then
   3 3 3 3 2 2 2 2 2 2
   s = a + b + c + 3( a b + ab + b c + bc + c a + ca ) + 6 abc
   2 2 2
   a b c a b b c c a = abc ( + + + 3( + + + + + ) + 6)
   bc ca ab b a c b a c
   1 1 1 = abc (1 + (3( a + b + c )( + + ) − 9) + 6)
   a b c
   1 1 1 = abc (3 s ( + + ) − 2)
   a b c
   ¯
   = abc (3 s s ¯ − 2) (because ¯ s = ¯ a + b + ¯ c = 1 /a + 1 /b + 1 /c ) = abc (3 | s | − 2)2
   3 2
   Taking absolute values, we find | s | = | 3 | s | − 2 | . It follows that | s | must be a positive real root of
   3 2 3 2 3 2
   x − 3 x + 2 = 0 or x + 3 x − 2 = 0. However, since the negative real roots of x − 3 x + 2 = 0
   3 2
   are exactly the additive inverses of the positive real roots of x − 3 x + 2 = 0, and all three roots of
   3 2 3 2 2
   x − 3 x + 2 = 0 are real ( x − 3 x + 2 = 0 may be factored as ( x − 1)( x − 2 x − 2) = 0, and the
   2 ndiscriminant of x − 2 x − 2 is positive), the product of all possible values of | s | is ( − 2) · ( − 1) , where
   3 2
   n denotes the number of negative real roots of x − 3 x + 2 = 0. By Descartes’s Rule of Signs, we seethat n is odd, so the answer is 2, as desired.
   Warm Up Your Proof Skills! [40]
   The problems in this section require complete proofs .