HMMT 二月 2007 · TEAM2 赛 · 第 7 题
HMMT February 2007 — TEAM2 Round — Problem 7
题目详情
- [ 20 ] Three positive reals x, y, and z are such that 2 x + 2( y − 1)( z − 1) = 85 2 y + 2( z − 1)( x − 1) = 84 2 z + 2( x − 1)( y − 1) = 89 . Compute x + y + z. ( ) ( ) ( ) 1 2 2 2
解析
- [ 20 ] Three positive reals x, y, and z are such that 2 x + 2( y − 1)( z − 1) = 85 2 y + 2( z − 1)( x − 1) = 84 2 z + 2( x − 1)( y − 1) = 89 . Compute x + y + z. Answer: 18 . Add the three equations to obtain 2 2 2 x + y + z + 2 xy + 2 yz + 2 zx − 4 x − 4 y − 4 z + 6 = 258 , 2 which rewrites as ( x + y + z − 2) = 256 . Evidently, x + y + z = 2 ± 16 . Since x, y, and z are positive, x + y + z > 0 so x + y + z = 2 + 16 = 18 . ( ) ( ) ( ) 1 2 2 2