Difference between revisions of "2012 AIME II Problems/Problem 15"
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== Solution == | == Solution == | ||
| − | + | Use angle bisector theorem to find <math>CD=21/8</math>, <math>BD=35/8</math>, and <math>AD=15/8</math>. Use Power of the Point to find <math>ED=49/8</math>, and so <math>AE=8</math>. Use law of cosines to find \angle CAD = \pi/3 } | |
== See Also == | == See Also == | ||
{{AIME box|year=2012|n=II|num-b=14|after=Last Problem}} | {{AIME box|year=2012|n=II|num-b=14|after=Last Problem}} | ||
Revision as of 23:15, 17 April 2012
Problem 15
Triangle 
is inscribed in circle 
with 
, 
, and 
. The bisector of angle 
meets side 
at 
and circle at a second point 
. Let 
be the circle with diameter 
. Circles and meet at and a second point 
. Then 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
Use angle bisector theorem to find 
, 
, and 
. Use Power of the Point to find 
, and so 
. Use law of cosines to find \angle CAD = \pi/3 }
See Also
| 2012 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Last Problem | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
