HMMT 十一月 2025 · 冲刺赛 · 第 16 题

HMMT November 2025 — Guts Round — Problem 16

[10] Let a , a , a , a , and a be the five distinct complex solutions of x − 20 x + 25 = 0. Compute 1 2 3 4 5 4 4 4 4 4 a + a + a + a + a . 1 2 3 4 5 ◦

解析

[10] Let a , a , a , a , and a be the five distinct complex solutions of x − 20 x + 25 = 0. Compute 1 2 3 4 5 4 4 4 4 4 a + a + a + a + a . 1 2 3 4 5 Proposed by: Pitchayut Saengrungkongka Answer: 80 © 2025 HMMT 25 5 4 5 Solution: Note that since a − 20 a +25 = 0, we have a = 20 − as 0 is not a root of x − 20 x +25 = 0. i i i a i Therefore, 25 25 25 25 25 4 4 4 4 4 a + a + a + a + a = 20 − + 20 − + 20 − + 20 − + 20 − 1 2 3 4 5 a a a a a 1 2 3 4 5 1 1 1 1 1 = 100 − 25 + + + + a a a a a 1 2 3 4 5 a a a a + a a a a + a a a a + a a a a + a a a a 2 3 4 5 1 3 4 5 1 2 4 5 1 2 3 5 1 2 3 4 = 100 − 25 · a a a a a 1 2 3 4 5 By Vieta’s formulas, a a a a + a a a a + a a a a + a a a a + a a a a = − 20 2 3 4 5 1 3 4 5 1 2 4 5 1 2 3 5 1 2 3 4 a a a a a = − 25 1 2 3 4 5 Substituting these values in, we get − 20 4 4 4 4 4 a + a + a + a + a = 100 − 25 = 80 . 1 2 3 4 5 − 25 ◦