HMMT 二月 2026 · 冲刺赛 · 第 25 题

HMMT February 2026 — Guts Round — Problem 25

[14] Let p ( x ) be the unique polynomial of degree at most 8 and with rational coefficients such that √ √ √ 3 3 3 p ( 2 + 3) = 6 . Compute p (1) .

解析

[14] Let p ( x ) be the unique polynomial of degree at most 8 and with rational coefficients such that √ √ √ 3 3 3 p ( 2 + 3) = 6 . Compute p (1) . Proposed by: Pitchayut Saengrungkongka 42 Answer: = 1 . 68 25 ©2026 HMMT √ √ 3 3 Solution: Let a = 2 + 3 . We have 2 √ √ a 5 3 3 3 a = 5 + 3 6 a = ⇒ 6 = − . 3 3 a Now we just need to invert a . To do this, we find the minimal polynomial of a : 3 3 3 9 6 3 ( a − 5) = 27 · 6 · a = ⇒ a − 15 a − 87 a − 125 = 0 , 8 5 2 1 a − 15 a − 87 a which means that = , so a 125 ( ) 8 5 2 8 5 2 √ 1 a − 15 a − 87 a − a + 15 a + 112 a 3 2 6 = a − = . 3 25 75 Thus, we conclude that 8 5 2 − x + 15 x + 112 x p ( x ) = , 75 which implies that 126 42 p (1) = = . 75 25