PUMaC 2017 · 代数（B 组） · 第 5 题
PUMaC 2017 — Algebra (Division B) — Problem 5
题目详情
- Let Γ be the maximum possible value of a + 3 b + 9 c among all triples ( a, b, c ) of positive real numbers such that log ( a + b + c ) = log (3 a ) = log (3 b ) = log (3 c ) . 30 8 27 125 p If Γ = where p and q are relatively prime positive integers, then find p + q . q
解析
- Suppose log ( a + b + c ) = log (3 a ) = log (3 b ) = log (3 c ) = x 30 8 27 125 x x x 3 x 3 x 3 x for some x , so that 2 3 5 = a + b + c , 2 = 3 a , 3 = 3 b , and 5 = 3 c . Then, 3 3 x 3 x 3 x ( a + b + c ) = 2 3 5 = 27 abc, 1 so equality holds in AM-GM, so a = b = c = . Therefore, the maximum possible value of 3 1 13 a + 3 b + 9 c is + 1 + 3 = , which makes our final answer 16 . 3 3 Problem written by Matt Tyler