HMMT 二月 2011 · TEAM2 赛 · 第 4 题

HMMT February 2011 — TEAM2 Round — Problem 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 ab as well, determine the product of all possible values of | a + b + c | . Warm Up Your Proof Skills! [40] The problems in this section require complete proofs .

解析

[ 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 ab as 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 ) 2 = abc (3 | s | − 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 n discriminant 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 see that n is odd, so the answer is 2, as desired. Warm Up Your Proof Skills! [40] The problems in this section require complete proofs .