HMMT 二月 2005 · 冲刺赛 · 第 28 题

英文解析

28. 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 andx
    3 2
    setting t = yields 2004( t − 3 t ) − 2005(1 − 3 t ) = 0. A quick check shows that thisyx x x
    1 2 3
    cubic has three real roots. Since the three roots are precisely , , and , we musty 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