HMMT 十一月 2021 · GEN 赛 · 第 10 题
HMMT November 2021 — GEN Round — Problem 10
题目详情
- Real numbers x, y, z satisfy x + xy + xyz = 1 , y + yz + xyz = 2 , z + xz + xyz = 4 . √ a + b c The largest possible value of xyz is , where a, b, c, d are integers, d is positive, c is square-free, d and gcd( a, b, d ) = 1. Find 1000 a + 100 b + 10 c + d .
解析
- Real numbers x, y, z satisfy x + xy + xyz = 1 , y + yz + xyz = 2 , z + xz + xyz = 4 . √ a + b c The largest possible value of xyz is , where a, b, c, d are integers, d is positive, c is square-free, d and gcd( a, b, d ) = 1. Find 1000 a + 100 b + 10 c + d . Proposed by: Sean Li Answer: 5272 Solution 1: Let p = xyz and q = ( x + 1)( y + 1)( z + 1). Then, we get pq = [ x (1 + y )] · [ y (1 + z )] · [ z (1 + x )] = (1 − p )(2 − p )(4 − p ) . Additionally, note that q − p = xy + yz + zx + x + y + z + 1 = ( x + xy ) + ( y + yz ) + ( z + xz ) + 1 = 8 − 3 p. Therefore, we have q = 8 − 2 p . Substituting this into our earlier equation gives us p (8 − 2 p ) = (1 − p )(2 − p )(4 − p ) . √ 2 5 ± 17 We can rearrange this to get (4 − p )(2 − 5 p + p ) = 0. Solving this gives us p = 4 , . Thus, our 2 √ 5+ 17 maximum solution is , which yields an answer of 5272. To show that such a solution exists, see 2 Solution 2. Solution 2: Let r = xyz − 1. Observe that 1 rx = x ( y + yz + xyz ) − ( x + xy + xyz ) = 2 x − 1 ⇐⇒ x = . 2 − r 2 4 Similarly, y = and z = . Therefore 8 = (1 + r )(1 − r )(2 − r )(4 − r ). This factors as 4 − r 1 − r √ 3+ 17 2 r ( r − 3)( r − 3 r − 2) = 0, so the maximum possible value for r is . 2 √ 3+ 17 1 2 Now let’s check that this yields a valid solution for x, y, z . Let r = and let x = , y = , z = 2 2 − r 4 − r 4 8 . Then xyz − 1 = − 1 = 1 + r − 1 = r . Now, we may do our above computations in 1 − r (2 − r )(4 − r )(1 − r ) reverse to get 2 2 x − 1 = 2 x − (2 − r ) x = rx = x yz − x = x ( y + yz + xyz ) − ( x − xy + xyz ) . Repeating the same thing for y and z yields that − 1 x 0 1 − 1 x 0 x + xy + xyz 0 − 1 y 2 0 − 1 y y + yz + xyz = . z 0 − 1 4 z 0 − 1 z + xz + xyz However, since xyz − 1 6 = 0, the determinant of the matrix is nonzero, so we may multiply by its inverse to find that 1 x + xy + xyz 2 = y + yz + xyz . 4 z + xz + xyz Therefore this construction is valid.