返回题库

HMMT 十一月 2008 · 冲刺赛 · 第 33 题

HMMT November 2008 — Guts Round — Problem 33

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

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

英文原题

[ 15 ] The polynomial ax 2 − 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 .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1 st HARVARD-MIT NOVEMBER TOURNAMENT, 8 NOVEMBER 2008 — GUTS ROUND

解析

英文解析

  1. [ 15 ] The polynomial ax − bx + c has two distinct roots p and q , with a , b , and c positive integers andwith 0 < p, q < 1. Find the minimum possible value of a .
    Answer: 5 Let x and y be the roots. Then:
    = x + y < 2 ⇒ b < 2 aba = xy < 1 ⇒ c < a ⇒ a > 1 ca
    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. Soa ≥ 5. But the polynomial 5 x − 5 x + 1 satisfies the condition, so 5 is the answer.2
    st 6
    1 HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 — GUTS ROUND