HMMT 二月 2004 · 冲刺赛 · 第 37 题

HMMT February 2004 — Guts Round — Problem 37

解析

Simplify sin(2 πk/ 4009). k =1 √ 4009 Solution: 2004 2 ∏ k − k 4008 ζ − ζ 2 πi/ 4009 4009 k Let ζ = e so that sin(2 πk/ 4009) = and x − 1 = ( x − ζ ). Hence k =0 2 i ∏ ∏ 4008 4008 4008 k k 1 + x + · · · + x = ( x − ζ ). Comparing constant coefficients gives ζ = 1, k =1 k =1 ∏ ∏ 4008 4008 k k setting x = 1 gives (1 − ζ ) = 4009, and setting x = − 1 gives (1 + ζ ) = 1. k =1 k =1 Now, note that sin(2 π (4009 − k ) / 4009) = − sin(2 πk/ 4009), so ( ) 2 2004 4008 ∏ ∏ 2004 sin(2 πk/ 4009) = ( − 1) sin(2 πk/ 4009) k =1 k =1 4008 k − k ∏ ζ − ζ = 2 i k =1 4008 2 k ∏ 1 ζ − 1 = 4008 k (2 i ) ζ k =1 4008 ∏ 1 2 k = ( ζ − 1) 4008 2 k =1 4008 ∏ 1 k k = ( ζ − 1)( ζ + 1) 4008 2 k =1 4009 · 1 = . 4008 2 12 However, sin( x ) is nonnegative on the interval [0 , π ], so our product is positive. Hence √ 4009 it is . 2004 2