Difference between revisions of "1998 AIME Problems/Problem 6"
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== Solution == | == Solution == | ||
| − | [[Image:AIME_1998-6.png| | + | [[Image:AIME_1998-6.png|350px]] |
There are several [[similar triangles]]. <math>\triangle PAQ \displaystyle \sim \triangle PDC</math>, so we can write the [[proportion]]: | There are several [[similar triangles]]. <math>\triangle PAQ \displaystyle \sim \triangle PDC</math>, so we can write the [[proportion]]: | ||
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</div> | </div> | ||
| − | Thus, <math> RC = \sqrt{112*847} = 308 \displaystyle</math>. | + | Thus, <math> RC = \sqrt{112*847} = 308 \displaystyle</math>. |
== See also == | == See also == | ||
Revision as of 21:03, 7 September 2007
Problem
Let 
be a parallelogram. Extend 
through 
to a point 
and let 
meet 
at 
and 
at 
Given that 
and 
find 
Solution
There are several similar triangles. 
, so we can write the proportion:
Also, 
, so:
Substituting,
Thus, 
.
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 | ||
