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HMMT 二月 2003 · 冲刺赛 · 第 21 题

HMMT February 2003 — Guts Round — Problem 21

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

  1. [8] r and s are integers such that
    3 r ≥ 2 s − 3 and 4 s ≥ r + 12 .
    What is the smallest possible value of r/s ?
    HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

英文原题

[8] r and s are integers such that
3 r ≥ 2 s − 3 and 4 s ≥ r + 12 .
What is the smallest possible value of r/s ?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
HARVARD-MIT MATHEMATICS TOURNAMENT, MARCH 15, 2003 — GUTS ROUND

解析

英文解析

  1. r and s are integers such that
    3 r ≥ 2 s − 3 and 4 s ≥ r + 12 .
    What is the smallest possible value of r/s ?
    Solution: 1 / 2
    We simply plot the two inequalities in the sr -plane and find the lattice point satisfyingboth inequalities such that the slope from it to the origin is as low as possible. We findthat this point is (2 , 4) (or (3 , 6)), as circled in the figure, so the answer is 2 / 4 = 1 / 2.
    r 4 s = r + 125
    3 r = 2 s - 3 s