HMMT 二月 2012 · 冲刺赛 · 第 18 题

英文解析

18. [ 11 ] Let x and y be positive real numbers such that x + y = 1 and (3 x − 4 x )(3 y − 4 y ) = − .
    Compute x + y .2
    √
    Answer: Solution 1: Let x = cos( θ ) and y = sin( θ ). Then, by the triple angle formulae, we 6
    3 312
    have that 3 x − 4 x = − cos(3 θ ) and 3 y − 4 y = sin(3 θ ), so − sin(3 θ ) cos(3 θ ) = − . We can write
    ( ) ( )2
    1 − 1 π π πthis as 2 sin(3 θ ) cos(3 θ ) = sin(6 θ ) = 1, so θ = sin (1) = . Thus, x + y = cos + sin =
    6 12 12 12
    √ √ √ √ √
    6+ 2 6 − 2 6 + = .
    4 4 2
    3 3 3 3 3 2 2 1
    Solution 2: Expanding gives 9 xy + 16 x y − 12 xy − 12 x y = 9( xy ) + 16( xy ) − 12( xy )( x + y ) = − ,
    2 2 3 1 1 12
    and since x + y = 1, this is − 3( xy ) + 16( xy ) = − , giving xy = − , . However, since x and y are
    2 2 4
    √ √
    √
    √
    1 1 3 6
    2 2
    positive reals, we must have xy = . Then, x + y = x + y + 2 xy = 1 + 2 · = = .
    4 4 2 2
    Gutsn