Difference between revisions of "2012 AIME II Problems/Problem 15"
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<math>AF^2 = AE^2 + EF^2 - 2 \cdot AE \cdot EF \cdot cos \angle AEF.</math> Adding these two and simplifying we get: | <math>AF^2 = AE^2 + EF^2 - 2 \cdot AE \cdot EF \cdot cos \angle AEF.</math> Adding these two and simplifying we get: | ||
− | <math>EF = AF \cdot cos \angle AFE + AE \cdot cos \angle AEF</math> (2). Ah, but <math>\angle AFE = \angle ACE</math> (since F lies on | + | <math>EF = AF \cdot cos \angle AFE + AE \cdot cos \angle AEF</math> (2). Ah, but <math>\angle AFE = \angle ACE</math> (since F lies on \omega), and we can find <math>cos \angle ACE</math> using the law of cosines: |
<math>AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot cos \angle ACE</math>, and plugging in <math>AE = 8, AC = 3, BE = BC = 7,</math> we get <math>cos \angle ACE = -1/7 = cos \angle AFE</math>. | <math>AE^2 = AC^2 + CE^2 - 2 \cdot AC \cdot CE \cdot cos \angle ACE</math>, and plugging in <math>AE = 8, AC = 3, BE = BC = 7,</math> we get <math>cos \angle ACE = -1/7 = cos \angle AFE</math>. |
Revision as of 00:08, 18 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 the angle bisector theorem to find , , and use the angle bisector formula to find . Use Power of the Point to find , and so . Use law of cosines to find , hence as well, and is equilateral, so .
I'm sure there is a more elegant solution from here, but instead we do'll some hairy law of cosines:
(1)
Adding these two and simplifying we get:
(2). Ah, but (since F lies on \omega), and we can find using the law of cosines:
, and plugging in we get .
Also, , and (since is on the circle with diameter ), so .
Plugging in all our values into equation (2), we get:
, or .
Finally, we plug this into equation (1), yielding:
Thus,
or The answer is .
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 |