HMMT 二月 2006 · 冲刺赛 · 第 20 题

HMMT February 2006 — Guts Round — Problem 20

[8] Compute the number of real solutions ( x, y, z, w ) to the system of equations: x = z + w + zwx z = x + y + xyz y = w + x + wxy w = y + z + yzw 10 9 6 5 4 k

解析

Compute the number of real solutions ( x, y, z, w ) to the system of equations: x = z + w + zwx z = x + y + xyz y = 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 − wz to 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 solutions are ( 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, which is 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