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 } | + | 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 <math>\angle CAD = \pi /3</math>, hence <math>\angle BAD = \pi /3</math> as well, and <math>\triangle BCE</math> is equilateral. |
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+ | I'm sure there is a more elegant solution from here, but instead we do'll some hairy law of cosines: | ||
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+ | <math>AE^2 = AF^2 + EF^2 - 2 \cdot AF \cdot EF \cdot cos \angle AFE.</math> (1) | ||
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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: | ||
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+ | <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 the same circle), and we can find <math>cos \angle ACE</math> using the law of cosines: | ||
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+ | <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</math>. | ||
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+ | Also, <math>\angle AEF = \angle DEF</math>, and <math>\angle DFE = \pi/2</math> (since <math>F</math> is on the circle with diameter <math>DE</math>), so <math>cos \angle AEF = EF/DE = 8 \cdot EF/49</math>. | ||
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+ | Plugging in all our values into equation (2), we get: | ||
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+ | <math>EF = -\frac{AF}{7} + 8 \cdot \frac{8EF}{49}</math>, or <math>EF = \frac{7}{15} \cdot AF</math>. | ||
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+ | Finally, we plug this into equation (1), yielding: | ||
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+ | <math>8^2 = AF^2 + \frac{49}{225} \cdot AF^2 - 2 \cdot AF \cdot \frac{7AF}{15} \cdot EF \cdot \frac{-1}{7}</math> Thus, | ||
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+ | <math>64 = \frac{AF^2}{225} \cdot (225+49+30),</math> or <math>AF^2 = \frac{900}{19}.</math> The answer is <math>919</math>. | ||
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== 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 23:58, 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 , hence as well, and is equilateral.
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 the same circle), 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 |