PUMaC 2022 · 团队赛 · 第 11 题

PUMaC 2022 — Team Round — Problem 11

For the function π π g ( a ) = max cos x + cos x + + cos x + + cos( x + a ) , x ∈ R 6 4 √ √ √ m + n + p − q 2 let b ∈ R be the input that maximizes g . If cos b = for positive integers 24 m, n, p, q , find m + n + p + q . 2

解析

For the function π π g ( a ) = max cos x + cos x + + cos x + + cos( x + a ) , x ∈ R 6 4 √ √ √ m + n + p − q 2 let b ∈ R be the input that maximizes g . If cos b = for positive integers 24 m, n, p, q , find m + n + p + q . Proposed by Ben Zenker Answer: 54 By the addition formula for cosine, we may rewrite f ( x ) as π π π π f ( x ) = (1 + cos + cos + cos a ) cos x − (sin + sin + sin a ) sin x = A sin x − B sin x 6 4 6 4 √ √ A 2 2 2 2 √ Factoring out A + B , we find f ( x ) = A + B cos( x − θ ), where cos θ = . It 2 2 A + B √ 2 2 2 2 follows that g ( a ) = A + B and it suffices to maximize A + B . Expanding this expression, we find π π π π 2 2 g ( a ) = (1 + cos + cos + cos a ) + (sin + sin + sin a ) 6 4 6 4 2 2 2 2 = ( α + cos a ) + ( β + sin a ) = ( α + β + 1) + (2 α cos a + 2 β sin a ) p 2 2 2 2 = ( α + β + 1) + 2 α + β cos( a − φ ) √ √ √ 2+ 3+ 2 1+ 2 α √ where α = , β = , and cos φ = . It follows that a maximizes g iff 2 2 2 2 α + β α √ a = φ + 2 πk , k ∈ Z , where φ is any angle satisfying cos φ = . Hence the desired 2 2 α + β 2 2 quantity cos a equals cos φ , which equals √ √ √ √ √ √ √ 1 2 2 (2 + 2 + 3) (2 + 2 + 3) 18 + 2 3 + 6 − 3 2 2 4 cos φ = √ √ √ = √ √ = 1 24 ( 6 + 2 3 + 3 2 + 6) 2(2 + 2)(3 + 3) 2 Thus m = 18 , n = 12 , p = 6 , q = 18 and our answer is 18 + 12 + 6 + 18 = 54. 2