HMMT 二月 2003 · 冲刺赛 · 第 21 题
HMMT February 2003 — Guts Round — Problem 21
题目详情
- [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
- 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 satisfying both inequalities such that the slope from it to the origin is as low as possible. We find that this point is (2 , 4) (or (3 , 6)), as circled in the figure, so the answer is 2 / 4 = 1 / 2. 5 r 4s = r + 12 3r = 2s - 3 s