Question

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 $PQ$ equals $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

Solution:

Points on the two lines can be written $P = (8t, 15t)$ and $Q = (10u, 3u).$ Since $R = (8, 6)$ is the midpoint of $\overline{PQ},$ $8t + 10u = 16 \qquad \text{and} \qquad 15t + 3u = 12.$

The second equation gives $u = 4 - 5t;$ substituting into the first, $8t + 40 - 50t = 16,$ so $t = \frac{4}{7}$ and $u = \frac{8}{7}.$ Thus $P = \left(\frac{32}{7}, \frac{60}{7}\right)$ and $Q = \left(\frac{80}{7}, \frac{24}{7}\right).$

Then $PQ = \sqrt{\left(\frac{48}{7}\right)^2 + \left(\frac{36}{7}\right)^2} = \frac{12}{7}\sqrt{4^2 + 3^2} = \frac{60}{7},$ so $m + n = 60 + 7 = 67.$

2001 AIME II Problem 4

Solution:

Problem 4 in Other Years