2012 AIME II Problems/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 .
Quick Solution using Olympiad Terms
Take a force-overlaid inversion about and note and map to each other. As was originally the diameter of , is still the diameter of . Thus is preserved. Note that the midpoint of lies on , and and are swapped. Thus points and map to each other, and are isogonal. It follows that is a symmedian of , or that is harmonic. Then , and thus we can let for some . By the LoC, it is easy to see so . Solving gives , from which by Ptolemy's we see . We conclude the answer is .
Side Note: You might be wondering what the motivation for this solution is. Most of the people who've done EGMO Chapter 8 should recognize this as problem 8.32 (2009 Russian Olympiad) with the computational finish afterwards. Now if you haven't done this, but still know what inversion is, here's the motivation. We'd see that it's kinda hard to angle chase, and if we could, it would still be a bit hard to apply (you could use trig, but it won't be so clean most likely). If you give up after realizing that angle chasing won't work, you'd likely go in a similar approach to Solution 1 (below) or maybe be a bit more insightful and go with the elementary solution above.
Finally, we notice there's circles! Classic setup for inversion! Since we're involving an angle-bisector, the first thing that comes to mind is a force overlaid inversion described in Lemma 8.16 of EGMO (where we invert with radius and center , then reflect over the -angle bisector, which fixes ). We try applying this to the problem, and it's fruitful - we end up with this solution. -MSC
Use the angle bisector theorem to find , , and use Stewart's Theorem to find . Use Power of Point to find , and so . Use law of cosines to find , hence as well, and is equilateral, so . In triangle , let be the foot of the altitude from ; then , where we use signed lengths. Writing and , we get Note , and the Law of Cosines in gives . Also, , and ( is a diameter), so .
Plugging in all our values into equation , we get: The Law of Cosines in , with and gives Thus . The answer is .
Let , , for convenience. Let be the midpoint of segment . We claim that .
. Since is the angle bisector, it follows that and consequently . Therefore, . Now let . Since , is a diameter, so lies on the perpendicular bisector of ; hence , , are collinear. From , quadrilateral is cyclic. Therefore, . But and are both subtended by arc in , so they are equal. Thus , as claimed. As a result, . Combined with , we get and therefore By Stewart's Theorem on (with cevian ), we get so , so the answer is .
-Solution by thecmd999
Use the angle bisector theorem to find , , and use Stewart's Theorem to find . Use Power of Point to find , and so . Then use the Extended Law of Sine 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 (since is ). Therefore, .
Let , , and . Then , so we get which simplifies to By Power of Point , . Combining with above, we get Note that and the ratio of similarity is . Then and The answer is .
-Solution by TheBoomBox77
Use Law of Cosines in to get . Because bisects , is the midpoint of major arc so and Thus is equilateral. Notice now that But so bisects Thus, Let Use Law of Cosines on to get Use Ptolemy's Theorem on , to get so and the answer is
Denote Let M be midpoint BC. Let be the circle centered at with radius
We calculate the length of some segments. The median The bisector One can use Stewart's Theorem in both cases.
is bisector of We use Power of Point and get We consider the inversion with respect
swap is symmetric to with respect to
swap lies on line is symmetric to with respect to
Points and lies on swap
is diameter Therefore is crosspoint of and
Let be circumcircle is image of line Point maps into
Points and are symmetric to and respectively.
Point lies on which is symmetric with respect to and on which is symmetric to with respect to
is symmetric with respect to
We use Power of Point and get
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