HMMT 二月 2006 · 冲刺赛 · 第 20 题

英文解析

20. Compute the number of real solutions ( x, y, z, w ) to the system of equations:
    x = z + w + zwx z = x + y + xyzy = w + x + wxy w = y + z + yzw
    Answer: 5
    w + z
    Solution: The first equation rewrites as x = , which is a fairly strong reason
    1 − wzto consider trigonometric substitution. Let x = tan( a ) , y = tan( b ) , z = tan( c ), and
    ° ° °
    w = tan( d ), where − 90 < a, b, c, d < 90 . Under modulo 180 , we find a ≡ c + d ; b ≡
    d + a ; c ≡ a + b ; d ≡ b + c . Adding all of these together yields a + b + c + d ≡ 0.
    Then a ≡ c + d ≡ − a − b so b ≡ − 2 a . Similarly, c ≡ − 2 b ; d ≡ − 2 c ; d ≡ − 2 a . Hence,
    c ≡ − 2 b ≡ 4 a , d ≡ − 2 c ≡ − 8 a , and a ≡ − 2 d ≡ 16 a , so the only possible solutionsare ( a, b, c, d ) ≡ ( t, − 2 t, 4 t, − 8 t ) where 15 t ≡ 0. Checking, these, we see that actually
    5 t ≡ 0, which yields 5 solutions. Our division by 1 − yz is valid since 1 − yz = 0
    iff yz = 1, but x = y + z + xyz so y = − z , which implies that yz ≤ 0 < 1, whichis impossible. (The solutions we have computed are in fact (0,0,0,0) and the cyclic
    ° ° ° °
    permutations of (tan(36 ) , tan( − 72 ) , tan( − 36 ) , tan(72 )).)
    10 9 6 5 4