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

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== Solution ==
 
== Solution ==
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=== Solution 1 ===
 
[[Image:AIME_1998-6.png|350px]]
 
[[Image:AIME_1998-6.png|350px]]
  
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There are several [[similar triangles]]. <math>\triangle PAQ \displaystyle \sim \triangle PDC</math>, so we can write the [[proportion]]:
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There are several [[similar triangles]]. <math>\triangle PAQ\sim \triangle PDC</math>, so we can write the [[proportion]]:
  
 
<div style="text-align:center;">
 
<div style="text-align:center;">
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</div>  
 
</div>  
  
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Also, <math>\triangle BRQ \displaystyle \sim DRC</math>, so:
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Also, <math>\triangle BRQ\sim DRC</math>, so:
  
 
<div style="text-align:center;">
 
<div style="text-align:center;">
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<math>\displaystyle \frac{QR}{RC} = \frac{QB}{CD} = \frac{112}{RC} = \frac{CD - AQ}{CD} = 1 - \frac{AQ}{CD} \displaystyle</math><br />
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<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>
 
<math>\frac{AQ}{CD} = 1 - \frac{112}{RC} = \frac{RC - 112}{RC}</math>
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<math>\frac{AQ}{CD} = \frac{735}{847 + RC} = \frac{RC - 112}{RC}</math><br />
 
<math>\frac{AQ}{CD} = \frac{735}{847 + RC} = \frac{RC - 112}{RC}</math><br />
  
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<math>\displaystyle735RC = (RC + 847)(RC - 112)</math><br />
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<math>735RC = (RC + 847)(RC - 112)</math><br />
 
<math>0 = RC^2 - 112\cdot847</math>
 
<math>0 = RC^2 - 112\cdot847</math>
 
</div>
 
</div>
  
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Thus, <math> RC = \sqrt{112*847} = 308 \displaystyle</math>.
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Thus, <math> RC = \sqrt{112*847} = 308</math>.
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=== 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 ==
 
== See also ==

Revision as of 14:37, 11 August 2012

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

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 $\frac{112}{RC} = \frac{ RC}{847}$ and so $RC^2=112*847$ which solves to $RC=\boxed{308}$

See also

1998 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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All AIME Problems and Solutions
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