HMMT 二月 2012 · 冲刺赛 · 第 18 题

HMMT February 2012 — Guts Round — Problem 18

[ 11 ] Let x and y be positive real numbers such that x + y = 1 and (3 x − 4 x )(3 y − 4 y ) = − . 2 Compute x + y . n

解析

[ 11 ] Let x and y be positive real numbers such that x + y = 1 and (3 x − 4 x )(3 y − 4 y ) = − . 2 Compute x + y . √ 6 Answer: Solution 1: Let x = cos( θ ) and y = sin( θ ). Then, by the triple angle formulae, we 2 1 3 3 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 π this 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 2 3 1 1 1 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 Guts n