2010 AIME II Problems/Problem 15
In triangle , , , and . Points and lie on with and . Points and lie on 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 .
Solution 1 (Linearity)
Define the function by for points in the plane. Then is linear, so . But since lies on the radical axis of , thus Let , and . Note that and because they are midpoints, while and by Angle Bisector Theorem. Thus we can rewrite this expression as so .
Official Solution (MAA)
The Angle Bisector Theorem implies that lies on and lies on because and . The Angle Bisector Theorem furthermore implies and Because is cyclic, . Because is cyclic, . Because and , triangles and are similar. Hence Applying the Law of Sines to and gives and thus Thus and .
Let . since . Since quadrilateral is cyclic, and , yielding and . Multiplying these together yields *.
Now we claim that . To prove this, we can use cyclic quadrilaterals.
From , and . So, and .
From , and . Thus, and .
Thus, from AA similarity, .
Therefore, , which can easily be computed by the angle bisector theorem to be . It follows that *, giving us an answer of .
- These two ratios are the same thing and can also be derived from the Ratio Lemma.
Ratio Lemma :, for any cevian AD of a triangle ABC. For the sine ratios use Law of Sines on triangles APM and APN, . The information needed to use the Ratio Lemma can be found from the similar triangle section above.
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.
This problem can be solved with barycentric coordinates. Let triangle be the reference triangle with , , and . Thus, and . Using the Angle Bisector Theorem, we can deduce that and . Plugging the coordinates for triangles and into the circle formula, we deduce that the equation for triangle is and the equation for triangle is . Solving the system of equations, we get that . This equation determines the radical axis of circles and , on which points and lie. Thus, solving for gets the desired ratio of lengths, and and plugging in the lengths and gets . From this we get the desired answer of . -wertguk
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