HMMT 十一月 2021 · 冲刺赛 · 第 18 题
HMMT November 2021 — Guts Round — Problem 18
题目详情
- [10] Let x, y, z be real numbers satisfying 1 1 1
- y + z = x + + z = x + y + = 3 . x y z m The sum of all possible values of x + y + z can be written as , where m, n are positive integers and n gcd( m, n ) = 1. Find 100 m + n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2021, November 13, 2021 — GUTS ROUND Organization Team Team ID#
解析
- [10] Let x, y, z be real numbers satisfying 1 1 1
- y + z = x + + z = x + y + = 3 . x y z m The sum of all possible values of x + y + z can be written as , where m, n are positive integers and n gcd( m, n ) = 1. Find 100 m + n . Proposed by: Sean Li Answer: 6106 1 1 1 1 Solution: The equality + y + z = x + + z implies + y = x + , so xy = − 1 or x = y . Similarly, x y x y yz = − 1 or y = z , and zx = − 1 or z = x . 1 If no two elements multiply to − 1, then x = y = z . which implies 2 x + = 3 and so ( x, y, z ) ∈ x 1 1 1 8 { (1 , 1 , 1) , ( , , ) } . Otherwise, we may assume xy = − 1, which implies z = 3 and x + y = , whence 2 2 2 3 1 { x, y, z } = {− , 3 , 3 } . 3 1 1 1 1 61 The final answer is (1 + 1 + 1) + ( + + ) + ( − + 3 + 3) = . 2 2 2 3 6