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Difference between revisions of "2020 AIME I Problems/Problem 1"

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=== Solution 2 ===
 
=== Solution 2 ===
  
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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{481}</math>
+
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{13}{RC} =  \frac{ RC}{17797}</math> and so <math>RC^2=13*17797</math> which solves to <math>RC=\boxed{481}</math>

Revision as of 15:58, 27 February 2020

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

AIME 1998-6.png

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 $\frac{13}{RC} = \frac{ RC}{17797}$ and so $RC^2=13*17797$ which solves to $RC=\boxed{481}$

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