HMMT 十一月 2022 · 冲刺赛 · 第 21 题

HMMT November 2022 — Guts Round — Problem 21

[11] Let P ( x ) be a quadratic polynomial with real coefficients. Suppose that P (1) = 20, P ( − 1) = 22, and P ( P (0)) = 400. Compute the largest possible value of P (10). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2022, November 12, 2022 — GUTS ROUND Organization Team Team ID# 2

解析

[11] Let P ( x ) be a quadratic polynomial with real coefficients. Suppose that P (1) = 20, P ( − 1) = 22, and P ( P (0)) = 400. Compute the largest possible value of P (10). Proposed by: Luke Robitaille Answer: 2486 2 Solution: Let P ( x ) = ax + bx + c . The given equations give us: a + b + c = 20 a − b + c = 22 2 Hence b = − 1, a + c = 21, and so the final equation gives us ac = 400. Substituting a = 21 − c and solving the cubic in c , we get c = − 4 , 5 , 20. Of these, the smallest value c = − 4 (and hence 2 P ( x ) = 25 x − x − 4) ends up giving the largest value of P (10). 2