HMMT 二月 2015 · 冲刺赛 · 第 20 题
HMMT February 2015 — Guts Round — Problem 20
题目详情
- [ 11 ] What is the largest real number θ less than π (i.e. θ < π ) such that 10 ∏ k cos(2 θ ) 6 = 0 k =0 and ( ) 10 ∏ 1 1 + = 1? k cos(2 θ ) k =0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT FEBRUARY 2015, 21 FEBRUARY 2015 — GUTS ROUND Organization Team Team ID# n
解析
- [ 11 ] What is the largest real number θ less than π (i.e. θ < π ) such that 10 ∏ k cos(2 θ ) 6 = 0 k =0 and ( ) 10 ∏ 1 1 + = 1? k cos(2 θ ) k =0 2046 π Answer: For equality to hold, note that θ cannot be an integer multiple of π (or else sin = 0 2047 and cos = ± 1). iθ/ 2 Let z = e 6 = ± 1. Then in terms of complex numbers, we want k k 10 10 2 − 2 2 ∏ ∏ 2 ( z + z ) (1 + ) = , k +1 k +1 k +1 k +1 2 − 2 2 − 2 z + z z + z k =0 k =0 Guts which partially telescopes to 10 − 1 ∏ k k z + z 2 − 2 ( z + z ) . 11 11 2 − 2 z + z k =0 Using a classical telescoping argument (or looking at binary representation; if you wish we may note − 1 2 that z − z 6 = 0, so the ultimate telescoping identity holds ), this simplifies to 11 11 − 1 2 − 2 10 z + z z − z tan(2 θ ) = . 11 11 − 1 2 − 2 z + z z − z tan( θ/ 2) Since tan x is injective modulo π (i.e. π -periodic and injective on any given period), θ works if and θ 2 ℓπ only if + ℓπ = 1024 θ for some integer ℓ , so θ = . The largest value for ℓ such that θ < π is at 2 2047 2046 π 3 ℓ = 1023, which gives θ = . 2047 Remark. It’s also possible to do this without complex numbers, but it’s less systematic. The steps k 2 k − 1 1+cos 2 θ 2 cos 2 θ k are the same, though, first note that 1 + sec 2 θ = = using the identity cos 2 x = k k cos 2 θ cos 2 θ 2 2 cos x − 1 (what does this correspond to in complex numbers?). hen we telescope using the identity sin 2 x 2 cos x = (again, what does this correspond to in complex numbers?). sin x n