PUMaC 2023 · 代数（B 组） · 第 4 题

PUMaC 2023 — Algebra (Division B) — Problem 4

If θ is the unique solution in (0 , π ) to the equation 2 sin( x )+3 sin +sin(2 x )+3 sin = 0, 2 2 √ a − b then cos( θ ) = for positive integers a, b, c such that a and c are relatively prime. Find c a + b + c .

解析

If θ is the unique solution in (0 , π ) to the equation 2 sin( x )+3 sin +sin(2 x )+3 sin = 0, 2 2 √ a − b then cos( θ ) = for positive integers a, b, c such that a and c are relatively prime. Find c a + b + c . Proposed by Ben Zenker and Nancy Xu Answer: 110 3 x 5 x x Using sum-to-product, we get sin + sin = 2 sin(2 x ) cos . 2 2 2 Factor out a sin( x ) of the whole expression (after using double angle on sin(2 x )), to get: x sin( x ) 2 + 2 cos( x ) + 12 cos( x ) cos = 0 2 1 x 2 sin( x ) > 0 in (0 , π ), so we can safely ignore it. Let u = cos , then cos( x ) = 2 u − 1 using 2 2 2 3 2 double angle. We now solve 2+2(2 u − 1)+12 u (2 u − 1) = 0, which becomes 6 u + u − 3 u = 0. The solution u = 0 corresponds to x = π , so we ignore it as well. √ − 1+ 73 2 We then just need the solution to 6 u + u − 3 = 0, which is u = . 12 √ 1 − 73 2 Compute cos( x ) = 2 u − 1 = , so a + b + c = 1 + 73 + 36 = 110 . 36