1998 AIME Problems/Problem 6: Difference between revisions
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== 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 == | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | |||
Latest revision as of 18:38, 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
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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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
