AIME 2001 II · 第 4 题

AIME 2001 II — Problem 4

专题: Contest Math
难度: L4
来源: AIME

题目详情

Problem

Let $R = (8,6)$ . The lines whose equations are $8y = 15x$ and $10y = 3x$ contain points $P$ and $Q$ , respectively, such that $R$ is the midpoint of $\overline{PQ}$ . The length of $PQ$ equals $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .

解析

Solution

$AIME diagram$

The coordinates of $P$ can be written as $\left(8a, 15a\right)$ and the coordinates of point $Q$ can be written as $\left(10b, 3b\right)$ . By the midpoint formula, we have $8a+10b=16$ and $15a+3b=12$ . Substituting $b=4-5a$ we derive $a = \frac{4}{7}$ and $b = \frac{8}{7}$ . Thus $P$ is $\left(\frac{32}7, \frac{60}7\right)$ , Q is $\left(\frac{80}7, \frac{24}7\right)$ , and the coordinates form a 3-4-5 triangle dilated by $\frac{12}7$ . Finally the distance $PQ$ must be $\frac{60}{7}$ so the answer is $\boxed{067}$ .