Difference between revisions of "2012 AIME II Problems/Problem 15"
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== Solution == | == Solution == | ||
− | Use the angle bisector theorem to find <math>CD=21/8</math>, <math>BD=35/8</math>, and use the | + | Use the angle bisector theorem to find <math>CD=21/8</math>, <math>BD=35/8</math>, and use the Stewart's Theorem to find <math>AD=15/8</math>. Use Power of the Point to find <math>DE=49/8</math>, and so <math>AE=8</math>. Use law of cosines to find <math>\angle CAD = \pi /3</math>, hence <math>\angle BAD = \pi /3</math> as well, and <math>\triangle BCE</math> is equilateral, so <math>BC=CE=BE=7</math>. |
I'm sure there is a more elegant solution from here, but instead we do'll some hairy law of cosines: | I'm sure there is a more elegant solution from here, but instead we do'll some hairy law of cosines: |
Revision as of 11:54, 1 February 2014
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 Stewart's Theorem 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 lies on ), 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 |
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