1998 AIME Problems/Problem 6

Problem

Let $A\displaystyle BCD$ 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 $RC=\boxed{308}$

1998 AIME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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1998 AIME Problems/Problem 6

Contents

Problem

Solution

Solution 1

Solution 2

See also