HMMT 二月 2015 · 冲刺赛 · 第 20 题

英文解析

20. [ 11 ] What is the largest real number θ less than π (i.e. θ < π ) such that
    ∏10
    cos(2 θ ) 6 = 0 kk =0
    ( )and
    ∏10
    1 + = 1?1
    cos(2 θ )kk =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 wantk 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 + zk =0 k =0
    Gutswhich partially telescopes to − 110
    ∏
    k kz + z
    2 − 2
    ( z + z ) .
    11 11
    2 − 2
    z + zk =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 andθ 2 ℓπonly if + ℓπ = 1024 θ for some integer ℓ , so θ = . The largest value for ℓ such that θ < π is at
    2 2047
    2046 πℓ = 1023, which gives θ = .3
    2047
    Remark. It’s also possible to do this without complex numbers, but it’s less systematic. The stepsk 2 k − 1
    1+cos 2 θ 2 cos 2 θkare the same, though, first note that 1 + sec 2 θ = = using the identity cos 2 x =
    k kcos 2 θ cos 2 θ
    2 cos x − 1 (what does this correspond to in complex numbers?). hen we telescope using the identity 2
    sin 2 x
    2 cos x = (again, what does this correspond to in complex numbers?).
    sin xn