HMMT 二月 2015 · 团队赛 · 第 1 题

HMMT February 2015 — Team Round — Problem 1

[ 5 ] The complex numbers x, y, z satisfy xyz = − 4 ( x + 1)( y + 1)( z + 1) = 7 ( x + 2)( y + 2)( z + 2) = − 3 . Find, with proof, the value of ( x + 3)( y + 3)( z + 3).

解析

[ 5 ] The complex numbers x, y, z satisfy xyz = − 4 ( x + 1)( y + 1)( z + 1) = 7 ( x + 2)( y + 2)( z + 2) = − 3 . Find, with proof, the value of ( x + 3)( y + 3)( z + 3). Answer: − 28 Solution 1. Consider the cubic polynomial f ( t ) = ( x + t )( y + t )( z + t ). By the theory of finite differences, f (3) − 3 f (2) + 3 f (1) − f (0) = 3! = 6, since f is monic. Thus f (3) = 6 + 3 f (2) − 3 f (1) + f (0) = 6 + 3( − 3) − 3(7) + ( − 4) = − 28. Solution 2. Alternatively, note that the system of equations is a (triangular) linear system in w := xyz , ( ) 27 47 v := xy + yz + zx , and u := x + y + z . The unique solution ( u, v, w ) to this system is − , , − 4 . 2 2 Plugging in yields ( x + 3)( y + 3)( z + 3) = w + 3 v + 9 u + 27 ( ) 47 27 = − 4 + 3 · + 9 · − + 27 2 2 = − 28 . Remark. Since f (0) < 0, f (1) > 0, f (2) < 0, f (+ ∞ ) > 0, the intermediate value theorem tells us the roots − x, − y, − z of f are real numbers in (0 , 1), (1 , 2), and (2 , + ∞ ), in some order. With a little more 3 27 2 47 calculation, one finds that x, y, z are the three distinct zeroes of the polynomial X + X + X + 4. 2 2 The three zeroes are approximately − 11 . 484 , − 1 . 825, and − 0 . 191.