# 1998 AIME Problems/Problem 6

## Problem

Let $ABCD$ be a parallelogram. Extend $\overline{DA}$ through $A$ to a point $P,$ and let $\overline{PC}$ meet $\overline{AB}$ at $Q$ and $\overline{DB}$ at $R.$ Given that $PQ = 735$ and $QR = 112,$ find $RC.$

## Solution

### Solution 1

There are several similar triangles. $\triangle PAQ\sim \triangle PDC$, so we can write the proportion: $\frac{AQ}{CD} = \frac{PQ}{PC} = \frac{735}{112 + 735 + RC} = \frac{735}{847 + RC}$

Also, $\triangle BRQ\sim DRC$, so: $\frac{QR}{RC} = \frac{QB}{CD} = \frac{112}{RC} = \frac{CD - AQ}{CD} = 1 - \frac{AQ}{CD}$ $\frac{AQ}{CD} = 1 - \frac{112}{RC} = \frac{RC - 112}{RC}$

Substituting, $\frac{AQ}{CD} = \frac{735}{847 + RC} = \frac{RC - 112}{RC}$ $735RC = (RC + 847)(RC - 112)$ $0 = RC^2 - 112\cdot847$

Thus, $RC = \sqrt{112*847} = 308$.

### Solution 2

We have $\triangle BRQ\sim \triangle DRC$ so $\frac{112}{RC} = \frac{BR}{DR}$. We also have $\triangle BRC \sim \triangle DRP$ so $\frac{ RC}{847} = \frac {BR}{DR}$. Equating the two results gives $\frac{112}{RC} = \frac{ RC}{847}$ and so $RC^2=112*847$ which solves to $RC=\boxed{308}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 