HMMT 十一月 2008 · 冲刺赛 · 第 33 题

HMMT November 2008 — Guts Round — Problem 33

[ 15 ] The polynomial ax − bx + c has two distinct roots p and q , with a , b , and c positive integers and with 0 < p, q < 1. Find the minimum possible value of a . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . st 1 HARVARD-MIT NOVEMBER TOURNAMENT, 8 NOVEMBER 2008 — GUTS ROUND

解析

[ 15 ] The polynomial ax − bx + c has two distinct roots p and q , with a , b , and c positive integers and with 0 < p, q < 1. Find the minimum possible value of a . Answer: 5 Let x and y be the roots. Then: b = x + y < 2 ⇒ b < 2 a a c = xy < 1 ⇒ c < a ⇒ a > 1 a 2 2 b > 4 ac > 4 c ⇒ b > 2 c Evaluated at 1, the polynomial must be greater than 0, so a + c > b . Then: 2 c < b < a + c 2 c + 1 ≤ b ≤ a + c − 1 a ≥ c + 2 ≥ 3 If a = 3, then c = 1 and b = 3, by the above bounds, but this polynomial has complex roots. Similarly, if a = 4, then c = 1 and b is forced to be either 3 or 4, again giving either 0 or 1 distinct real roots. So 2 a ≥ 5. But the polynomial 5 x − 5 x + 1 satisfies the condition, so 5 is the answer. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 st 1 HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 — GUTS ROUND