HMMT 二月 2015 · 冲刺赛 · 第 28 题

英文解析

28. [ 17 ] Let w , x , y , and z be positive real numbers such that
    0 6 = cos w cos x cos y cos z
    2 π = w + x + y + z
    3 tan w = k (1 + sec w )
    4 tan x = k (1 + sec x )
    5 tan y = k (1 + sec y )
    6 tan z = k (1 + sec z ) .
    (Here sec t denotes when cos t 6 = 0.) Find k .1
    cos t
    √
    u sin u w x
    Answer: 19 From the identity tan = , the conditions work out to 3 tan = 4 tan =
    2 1+cos u 2 2
    y yz w x z
    5 tan = 6 tan = k . Let a = tan , b = tan , c = tan , and d = tan . Using the identity
    2 2 2 2 2 2
    tan M +tan Ntan( M + N ) = , we obtain
    1 − tan M tan N
    ( ) ( )
    ( )
    y + zw + xtan + tanw + x y + z
    2 2
    ( ) ( )
    tan + =
    y + zw + x
    2 2
    1 − tan tan
    2 2
    a + b c + d
    1 − ab 1 − cd+
    ( ) ( )
    a + b c + d=
    1 −
    1 − ab 1 − cda + b + c + d − abc − abd − bcd − acd = .
    1 + abcd − ab − ac − ad − bc − bd − cd
    ( )
    x + y + z + w
    Because x + y + z + w = π , we get that tan = 0 and thus a + b + c + d = abc + abd + bcd + acd .
    Substituting a, b, c, d corresponding to the variable k , we obtain that k − 19 k = 0. Therefore, k can 32
    √ √ √
    be only 0 , 19 , − 19. However, k = 0 is impossible as w, x, y, z will all be 0. Also, k = − 19 is
    √
    impossible as w, x, y, z will exceed π . Therefore, k = 19.