HMMT 二月 2005 · 冲刺赛 · 第 28 题

HMMT February 2005 — Guts Round — Problem 28

[10] There are three pairs of real numbers ( x , y ), ( x , y ), and ( x , y ) that satisfy 1 1 2 2 3 3 ( ) ( ) ( ) x x x 3 2 3 2 1 2 3 both x − 3 xy = 2005 and y − 3 x y = 2004. Compute 1 − 1 − 1 − . y y y 1 2 3

解析

There are three pairs of real numbers ( x , y ), ( x , y ), and ( x , y ) that satisfy both 1 1 2 2 3 3 ( ) ( ) ( ) x x x 3 2 3 2 1 2 3 x − 3 xy = 2005 and y − 3 x y = 2004. Compute 1 − 1 − 1 − . y y y 1 2 3 Solution: 1/1002 3 2 3 2 3 By the given, 2004( x − 3 xy ) − 2005( y − 3 x y ) = 0. Dividing both sides by y and x 3 2 setting t = yields 2004( t − 3 t ) − 2005(1 − 3 t ) = 0. A quick check shows that this y x x x 1 2 3 cubic has three real roots. Since the three roots are precisely , , and , we must y y y 1 2 3 ( ) ( ) ( ) x x x 3 2 1 2 3 have 2004( t − 3 t ) − 2005(1 − 3 t ) = 2004 t − t − t − . Therefore, y y y 1 2 3 ( ) ( ) ( ) 3 2 x x x 2004(1 − 3(1)) − 2005(1 − 3(1) ) 1 1 2 3 1 − 1 − 1 − = = . y y y 2004 1002 1 2 3