HMMT 二月 2010 · TEAM2 赛 · 第 3 题

英文解析

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 . Ifn n − 1 0 i
   | r | > , what is the minimum possible value of n ?15
   Answer: 4 We claim that n = 4 is the answer. First, we show that n > 3. Suppose that n ≤ 3. Let 8
   r be the root of the polynomial with | r | ≥ . Then, by the Triangle Inequality, we have:15
   ∣ ∣ ∣ ∣ ∣ ∣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 | )
   The right-hand side is increasing in n , for | r | > 1, so it is bounded by | r | (2 − | r | ). This expression is 3
   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 4 th 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