2010 AIME II Problems/Problem 15
In triangle , , , and . Points and lie on with and . Points and lie on B with and . Let be the point, other than , of intersection of the circumcircles of and . Ray meets at . The ratio can be written in the form , where and are relatively prime positive integers. Find .
Let . since . Since quadrilateral is cyclic, and , yielding and . Multiplying these together yields .
. Also, is the center of spiral similarity of segments and , so . Therefore, , which can easily be computed by the angle bisector theorem to be . It follows that , giving us an answer of .
Note: Spiral similarities may sound complex, but they're really not. The fact that is really just a result of simple angle chasing.
Source:  by Zhero
The work done in this problem leads to a nice extension of this problem:
Given a and points , , , , , , such that , , , , and , , then let be the circumcircle of and be the circumcircle of . Let be the intersection point of and distinct from . Define and similarly. Then , , and concur.
This can be proven using Ceva's theorem and the work done in this problem, which effectively allows us to compute the ratio that line divides the opposite side into and similarly for the other two sides.
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