HMMT 二月 2010 · TEAM1 赛 · 第 9 题

英文解析

9. [ 30 ] Let p ( x ) = a x + a x + . . . + a be a polynomial with complex coefficients such that a 6 = 0
   n n − 1 0 ian − 1
   i − 1
   for all i . Prove that | r | ≤ 2 max | | for all roots r of all such polynomials p . Here we let | z |
   i =0
   denote the absolute value of the complex number z .ian n − 1 n
   Solution: If r is a root, then − a r = a r + . . . + a . By the Triangle Inequality, |− a r | ≤
   n n − 1 0 nn − 1 n n − 1
   | a r | + ... + | a | . Rearranging this inequality yields | a r | − | a r | − ... − | a | ≤ 0.
   n − 1 0 n n − 1 0
   i − 1 a
   Now suppose | r | = k max | | . Applying this over values of i ranging from m + 1 to n (assuminga
   | a r |ninm + 1 ≤ n ), we get | a r | ≤ . This, along with the above equation, yields:mm n − m
   Team Round Ak
   ( )
   1 1 1 1
   | a r | · 1 − − − − . . . − = 0 nn
   2 3 nk k k k
   1 1
   This is only true when a = 0, r = 0, or (1 − − − . . . ) = 0. The first option is impossible by then 2
   k kconstraints in the problem. The second option implies k = 0. The third option implies that k < 2;
   n − 1 a
   1 1 1 i − 1
   otherwise (1 − − − . . . − ) would always remain positive. Either way, | r | ≤ 2 max | | .
   2 ni =0
   k k k ai