Difference between revisions of "1998 AIME Problems/Problem 6"
m (→Solution: downsize slightly) |
|||
(One intermediate revision by one other user not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | Let <math> | + | 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]] | [[Image:AIME_1998-6.png|350px]] | ||
− | There are several [[similar triangles]]. <math>\triangle PAQ | + | 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;"> | ||
Line 11: | Line 12: | ||
</div> | </div> | ||
− | Also, <math>\triangle BRQ | + | Also, <math>\triangle BRQ\sim DRC</math>, so: |
<div style="text-align:center;"> | <div style="text-align:center;"> | ||
− | <math> | + | <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> | ||
Line 24: | Line 25: | ||
<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 /> | ||
− | <math> | + | <math>735RC = (RC + 847)(RC - 112)</math><br /> |
<math>0 = RC^2 - 112\cdot847</math> | <math>0 = RC^2 - 112\cdot847</math> | ||
</div> | </div> | ||
− | Thus, <math> RC = \sqrt{112*847} = 308 \ | + | 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 == | == See also == | ||
Line 34: | Line 39: | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 19:38, 4 July 2013
Problem
Let be a parallelogram. Extend
through
to a point
and let
meet
at
and
at
Given that
and
find
Solution
Solution 1
There are several similar triangles. , so we can write the proportion:
Also, , so:
Substituting,
Thus, .
Solution 2
We have so
. We also have
so
. Equating the two results gives
and so
which solves to
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.