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HMMT 十一月 2019 · 冲刺赛 · 第 18 题

HMMT November 2019 — Guts Round — Problem 18

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. [10] The polynomial x − 3 x + 1 has three real roots r , r , and r . Compute 1 2 3 √ √ √ 3 3 3 3 r − 2 + 3 r − 2 + 3 r − 2 . 1 2 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2019, November 9, 2019 — GUTS ROUND Organization Team Team ID# 1
解析
  1. [10] The polynomial x − 3 x + 1 has three real roots r , r , and r . Compute 1 2 3 √ √ √ 3 3 3 3 r − 2 + 3 r − 2 + 3 r − 2 . 1 2 3 Proposed by: Milan Haiman Answer: 0 Let r be a root of the given polynomial. Then √ 3 3 2 3 2 r − 3 r + 1 = 0 = ⇒ r − 3 r + 3 r − 1 = 3 r − 2 = ⇒ r − 1 = 3 r − 2 . Now by Vieta the desired value is r + r + r − 3 = 3 − 3 = 0. 1 2 3 1