2001 AIME II Problems/Problem 4

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 $m + n$ .

The coordinates of $P$ can be written as $\left(a, \frac{15a}8\right)$ and the coordinates of point $Q$ can be written as $\left(b,\frac{3b}{10}\right)$ . By the midpoint formula, we have $\frac{a+b}2=8$ and $\frac{15a}{16}+\frac{3b}{20}=6$ . Solving for $b$ gives $b= \frac{80}{7}$ , so the point $Q$ is $\left(\frac{80}7, \frac{24}7\right)$ . The answer is twice the distance from $Q$ to $(8,6)$ , which by the distance formula is $\frac{60}{7}$ . Thus, the answer is $\boxed{067}$ .