Difference between revisions of "2012 AIME II Problems/Problem 15"
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== Solution 1== | == Solution 1== | ||
− | Use the angle bisector theorem to find <math>CD=frac{21}{8}</math>, <math>BD=\frac{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 = \frac{\pi} {3}</math>, hence <math>\angle BAD = \frac{\pi}{3}</math> as well, and <math>\triangle BCE</math> is equilateral, so <math>BC=CE=BE=7</math>. | + | Use the angle bisector theorem to find <math>CD=\frac{21}{8}</math>, <math>BD=\frac{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 = \frac{\pi} {3}</math>, hence <math>\angle BAD = \frac{\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'll do some hairy law of cosines: | I'm sure there is a more elegant solution from here, but instead we'll do some hairy law of cosines: |
Revision as of 21:47, 21 November 2016
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 1
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'll do 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 .
Solution 2
Let , , for convenience. We claim that is a symmedian. Indeed, let be the midpoint of segment . Since , it follows that and consequently . Therefore, . Now let . Since is a diameter, lies on the perpendicular bisector of ; hence , , are collinear. From , it immediately follows that quadrilateral is cyclic. Therefore, , implying that is a symmedian, as claimed.
The rest is standard; here's a quick way to finish. From above, quadrilateral is harmonic, so . In conjunction with , it follows that . (Notice that this holds for all triangles .) To finish, substitute , , to obtain as before.
-Solution by thecmd999
Solution 3
First of all, use the Angle Bisector Theorem to find that and , and use Stewart's Theorem to find that . Then use Power of a Point to find that . Then use the circumradius of a triangle formula to find that the length of the circumradius of is .
Since is the diameter of circle , is . Extending to intersect circle at , we find that is the diameter of the circumcircle of (since is ). Therefore, .
Let , , and . Then, by the Pythagorean Theorem,
and
Subtracting the first equation from the second, the term cancels out and we obtain:
By Power of a Point, , so
Since , .
Because and intercept the same arc in circle and the same goes for and , and . Therefore, by AA Similarity. Since side lengths in similar triangles are proportional,
However, the problem asks for , so .
-Solution by TheBoomBox77
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.