HMMT 二月 2016 · 冲刺赛 · 第 15 题

英文解析

15. [ 9 ] Compute tan tan tan .
    7 7 7
    Proposed by: Alexander Katz
    √
    Answer: 7
    7 ix
    Consider the polynomial P ( z ) = z − 1. Let z = e = cos x + i sin x . Then
    ( ( ) ( ) ( ) )
    7 7 7
    2 4 6
    7 7 5 3
    z − 1 = cos x − cos x sin x + cos x sin x − cos x sin x − 1
    2 4 6
    ( ( ) ( ) ( ) )
    7 7 7
    7 5 2 3 4 + i − sin x + sin x cos x − sin x cos x + sin x cos 6 x
    2 4 6
    Consider the real part of this equation. We may simplify it to 64 cos x − . . . − 1, where the middle 7
    ( )
    ∏
    2 π 4 π 2 πk 17
    terms are irrelevant. The roots of P are x = , , . . . , so cos = . Butk =1
    7 7 7 64
    ( )
    ( ) ( )2
    7 3
    ∏ ∏
    2 πk kπcos = cos
    7 7
    k =1 k =1
    ( )
    ∏
    kπ 13
    so cos = .
    k =1
    7 8
    Now consider the imaginary part of this equation. We may simplify it to − 64 sin x + . . . + 7 sin x ,11
    where again the middle terms are irrelevant. We can factor out sin x to get − 64 sin x + . . . + 7, and 10
    ( )
    ∏
    2 π 12 π 2 πk 76
    this polynomial has roots x = , . . . , (but not 0). Hence sin = − . But, like before,
    k =1
    7 7 7 64
    we have
    ( )
    ( ) ( )2
    6 3
    ∏ ∏
    2 πk 2 πksin = − sin
    7 7
    k =1 k =1
    √
    √
    √
    ( )7
    ∏
    kπ 73
    hence sin = . As a result, our final answer is = 7 .8
    k =11
    7 8
    n 8