Difference between revisions of "2012 AIME II Problems/Problem 15"
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− | It can be verified with law of cosines that <math>\angle BAC=120^\circ.</math> Also, <math>E</math> is the midpoint of major arc <math>BC</math> so <math>BE=CE,</math> and <math>\angle BEC=60.</math> Thus <math>CBE</math> is equilateral. Notice now that <math>\angle | + | It can be verified with law of cosines that <math>\angle BAC=120^\circ.</math> Also, <math>E</math> is the midpoint of major arc <math>BC</math> so <math>BE=CE,</math> and <math>\angle BEC=60.</math> Thus <math>CBE</math> is equilateral. Notice now that <math>\angle CEB=\angle BFE= 60.</math> But <math>\angle DFE=90</math> so <math>FD</math> bisects <math>\angle BFC.</math> Thus, <math>\frac{BF}{CF}=\frac{BD}{CD}=\frac{BA}{CA}=\frac{5}{3}.</math> |
Let <math>BF=5a, CF=3a.</math> By law of cosines on <math>BFC</math> we find <math>a\sqrt{5^2+3^2-5*3}=a\sqrt{19}=7.</math> But by ptolemy on <math>BFCA</math>, <math>15a+15a=7*AF,</math> so <math>AF= \frac{30}{\sqrt{19}},</math> so <math>AF^2=\frac{900}{19}</math> and the answer is <math>900+19=\boxed{919}</math> | Let <math>BF=5a, CF=3a.</math> By law of cosines on <math>BFC</math> we find <math>a\sqrt{5^2+3^2-5*3}=a\sqrt{19}=7.</math> But by ptolemy on <math>BFCA</math>, <math>15a+15a=7*AF,</math> so <math>AF= \frac{30}{\sqrt{19}},</math> so <math>AF^2=\frac{900}{19}</math> and the answer is <math>900+19=\boxed{919}</math> |
Revision as of 23:59, 29 April 2018
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
Solution 4
It can be verified with law of cosines that Also, is the midpoint of major arc so and Thus is equilateral. Notice now that But so bisects Thus,
Let By law of cosines on we find But by ptolemy on , so so and the answer is
~abacadaea
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.