HMMT 二月 2010 · TEAM2 赛 · 第 3 题

HMMT February 2010 — TEAM2 Round — Problem 3

[ 15 ] Let p ( x ) = a x + a x + . . . + a , where each a is either 1 or − 1. Let r be a root of p . If n n − 1 0 i 15 | r | > , what is the minimum possible value of n ? 8

解析

[ 15 ] Let p ( x ) = a x + a x + . . . + a , where each a is either 1 or − 1. Let r be a root of p . If n n − 1 0 i 15 | r | > , what is the minimum possible value of n ? 8 Answer: 4 We claim that n = 4 is the answer. First, we show that n > 3. Suppose that n ≤ 3. Let 15 r be the root of the polynomial with | r | ≥ . Then, by the Triangle Inequality, we have: 8 ∣ ∣ ∣ ∣ ∣ ∣ n n − 1 n − 2 n − 1 n − 2 ∣ ∣ ∣ ∣ ∣ ∣ | a r | = a r + a r + . . . + a ≤ a r + a r + . . . + | a | n n − 1 n − 2 0 n − 1 n − 2 0 n ∣ ∣ ∣ ∣ | r | − 1 n n − 1 n − 2 ∣ ∣ ∣ ∣ | r | ≤ r + r + . . . + | 1 | = | r | − 1 n +1 n n | r | − 2 | r | + 1 ≤ 0 ⇒ 1 ≤ | r | (2 − | r | ) 3 The right-hand side is increasing in n , for | r | > 1, so it is bounded by | r | (2 − | r | ). This expression is 3 15 decreasing in r for r ≥ . When | r | = , then the right-hand side is less than 1, which violates the 2 8 inequalities. Therefore n > 3. Now, we claim that there is a 4th degree polynomial with a root r with 15 15 4 3 2 | r | ≥ . Let p ( x ) = x − x − x − x − 1. Then p ( ) < 0 and p (2) > 2. By the Intermediate Value 8 8 Theorem, p ( x ) has such a root r . Team Round B