HMMT 二月 2007 · 冲刺赛 · 第 25 题

HMMT February 2007 — Guts Round — Problem 25

[ 12 ] Two real numbers x and y are such that 8 y + 4 x y + 4 xy + 2 x + 2 y + 2 x = x + 1 . Find all 2 possible values of x + 2 y .

解析

[ 12 ] Two real numbers x and y are such that 8 y + 4 x y + 4 xy + 2 x + 2 y + 2 x = x + 1 . Find all 2 possible values of x + 2 y . 1 2 2 2 2 Answer: . Writing a = x + 2 y , the given quickly becomes 4 y a + 2 x a + a + x = x + 1. We can 2 2 2 2 rewrite 4 y a for further reduction to a (2 a − 2 x ) + 2 x a + a + x = x + 1, or 2 2 2 2 a + (2 x − 2 x + 1) a + ( − x + x − 1) = 0 . ( ∗ ) The quadratic formula produces the discriminant 2 2 2 2 2 (2 x − 2 x + 1) + 8( x − x + 1) = (2 x − 2 x + 3) , 2 2 − 2 x +2 x − 1 ± (2 x − 2 x +3) an identity that can be treated with the difference of squares, so that a = = 4 1 2 , − x + x − 1. Now a was constructed from x and y , so is not free. Indeed, the second expression flies 2 2 2 2 in the face of the trivial inequality: a = − x + x − 1 < − x + x ≤ x + 2 y = a. On the other hand, a = 1 / 2 is a bona fide solution to (*), which is identical to the original equation. 7