Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
(prob / sol)
Line 1: Line 1:
−
{{empty}}
 
 
== Problem ==
 
== Problem ==
 +
Let <math>A\displaystyle BCD</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}}
+
[[Image:AIME_1998-6.png|500px]]
 +
 
 +
There are several [[similar triangles]]. <math>\triangle PAQ \displaystyle \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 \displaystyle \sim DRC</math>, so:
 +
 
 +
<div style="text-align:center;">
 +
<math>\displaystyle \frac{QR}{RC} = \frac{QB}{CD} = \frac{112}{RC} = \frac{CD - AQ}{CD} = 1 - \frac{AQ}{CD} \displaystyle</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>\displaystyle735RC = (RC + 847)(RC - 112)</math><br />
 +
<math>0 = RC^2 - 112\cdot847</math>
 +
</div>
 +
 
 +
Thus, <math> RC = \sqrt{112*847} = 308 \displaystyle</math>.
 +
 
 
== See also ==
 
== 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]]

Revision as of 21:01, 7 September 2007

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

AIME 1998-6.png

There are several similar triangles. $\triangle PAQ \displaystyle \sim \triangle PDC$, so we can write the proportion:

$\frac{AQ}{CD} = \frac{PQ}{PC} = \frac{735}{112 + 735 + RC} = \frac{735}{847 + RC}$

Also, $\triangle BRQ \displaystyle \sim DRC$, so:

$\displaystyle \frac{QR}{RC} = \frac{QB}{CD} = \frac{112}{RC} = \frac{CD - AQ}{CD} = 1 - \frac{AQ}{CD} \displaystyle$

$\frac{AQ}{CD} = 1 - \frac{112}{RC} = \frac{RC - 112}{RC}$

Substituting,

$\frac{AQ}{CD} = \frac{735}{847 + RC} = \frac{RC - 112}{RC}$

$\displaystyle735RC = (RC + 847)(RC - 112)$
$0 = RC^2 - 112\cdot847$

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

See also

1998 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1998_AIME_Problems/Problem_6&oldid=15882"