Difference between revisions of "1998 AIME Problems/Problem 6"

Latest revision as of 18:38, 4 July 2013

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 $RC=\boxed{308}$

@@ Line 1: / Line 1: @@
-{{empty}}
 == Problem ==
+Let <math>ABCD</math> be a [[parallelogram]].  Extend <math>\overline{DA}</math> through <math>A</math> to a point <math>P,</math> and let <math>\overline{PC}</math> meet <math>\overline{AB}</math> at <math>Q</math> and <math>\overline{DB}</math> at <math>R.</math>  Given that <math>PQ = 735</math> and <math>QR = 112,</math> find <math>RC.</math>
 == Solution ==
-{{solution}}
+=== Solution 1 ===
+[[Image:AIME_1998-6.png|350px]]
+There are several [[similar triangles]]. <math>\triangle PAQ\sim \triangle PDC</math>, so we can write the [[proportion]]:
+<div style="text-align:center;">
+<math>\frac{AQ}{CD} = \frac{PQ}{PC} = \frac{735}{112 + 735 + RC} = \frac{735}{847 + RC}</math>
+</div>
+Also, <math>\triangle BRQ\sim DRC</math>, so:
+<div style="text-align:center;">
+<math>\frac{QR}{RC} = \frac{QB}{CD} = \frac{112}{RC} = \frac{CD - AQ}{CD} = 1 - \frac{AQ}{CD}</math><br />
+<math>\frac{AQ}{CD} = 1 - \frac{112}{RC} = \frac{RC - 112}{RC}</math>
+</div>
+Substituting,
+<div style="text-align:center;">
+<math>\frac{AQ}{CD} = \frac{735}{847 + RC} = \frac{RC - 112}{RC}</math><br />
+<math>735RC = (RC + 847)(RC - 112)</math><br />
+<math>0 = RC^2 - 112\cdot847</math>
+</div>
+Thus, <math> RC = \sqrt{112*847} = 308</math>.
+=== Solution 2 ===
+We have <math>\triangle BRQ\sim \triangle DRC</math> so <math>\frac{112}{RC} = \frac{BR}{DR}</math>. We also have <math>\triangle BRC \sim \triangle DRP</math> so <math>\frac{ RC}{847} = \frac {BR}{DR}</math>. Equating the two results gives <math>\frac{112}{RC} =  \frac{ RC}{847}</math> and so <math>RC^2=112*847</math> which solves to <math>RC=\boxed{308}</math>
 == See also ==
-* [[1998 AIME Problems/Problem 5 | Previous problem]]
+{{AIME box|year=1998|num-b=5|num-a=7}}
-* [[1998 AIME Problems/Problem 7 | Next problem]]
-* [[1998 AIME Problems]]
+[[Category:Intermediate Geometry Problems]]
+{{MAA Notice}}

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