PUMaC 2011 · 代数（A 组） · 第 4 题

PUMaC 2011 — Algebra (Division A) — Problem 4

[ 4 ] Suppose the polynomial x − x + bx + c has real roots a, b, c . What is the square of the minimum value of abc ?

解析

Vieta’s relations give us a + b + c = 1 ab + bc + ca = b abc = − c From the last equation, ( ab + 1) c = 0, so either c = 0 or ab = − 1. If c = 0, we will have 2 ( abc ) = 0 regardless of the values of a and b . The other case is where c 6 = 0 , ab = − 1 2 (in particular, a, b 6 = 0). From the first two equations, we have ca = b (1 − a − c ) = b . 3 b 3 3 Because ab = − 1, it follows that c = = − b . Hence, 1 = a + b + c = − 1 /b + b + − b , so ab 4 2 b − b + b + 1 = 0. We see that b = − 1 is a root of this quartic, so factoring out ( b + 1), we have that 4 2 3 2 b − b + b + 1 = ( b + 1)( b − b + 1) = 0 . 3 2 2 2 Since b − b + 1 = b ( b − 1) + 1 < − 2 b + 1 < 0 for b < − 1, it follows that b = − 1 is the 3 3 smallest possible real root of the quartic. Then abc = b ≥ ( − 1) = − 1, which is achieved for ( a, b, c ) = (1 , − 1 , 1), and squaring, our answer is 1 .