HMMT 十一月 2013 · 冲刺赛 · 第 27 题

HMMT November 2013 — Guts Round — Problem 27

[ 13 ] Find all triples of real numbers ( a, b, c ) such that a + 2 b − 2 bc = 16 and 2 ab − c = 16. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT NOVEMBER 2013, 9 NOVEMBER 2013 — GUTS ROUND

解析

[ 13 ] Find all triples of real numbers ( a, b, c ) such that a + 2 b − 2 bc = 16 and 2 ab − c = 16. 2 2 2 Answer: (4 , 4 , 4) , ( − 4 , − 4 , − 4) (need both, but order doesn’t matter) a + 2 b − 2 bc and 2 ab − c 2 are both homogeneous degree 2 polynomials in a, b, c , so we focus on the homogeneous equation a + 2 2 2 2 2 2 2 b − 2 bc = 2 ab − c , or ( a − b ) + ( b − c ) = 0. So a = b = c , and a = 2 ab − c = 16 gives the solutions (4 , 4 , 4) and ( − 4 , − 4 , − 4).