2013 USAMO Problems/Problem 1
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[hide]Problem
In triangle , points lie on sides respectively. Let , , denote the circumcircles of triangles , , , respectively. Given the fact that segment intersects , , again at respectively, prove that
Solution 1
In this solution, all lengths and angles are directed.
Firstly, it is easy to see by that concur at a point . Let meet again at and , respectively. Then by Power of a Point, we have Thusly But we claim that . Indeed, and Therefore, . Analogously we find that and we are done.
courtesy v_enhance
Solution 2
Diagram Refer to the Diagram link.
By Miquel's Theorem, there exists a point at which intersect. We denote this point by Now, we angle chase: In addition, we have Now, by the Ratio Lemma, we have (by the Law of Sines in ) (by the Law of Sines in ) by the Ratio Lemma. The proof is complete.
Solution 3
Use directed angles modulo .
Lemma.
Proof.
Now, it follows that (now not using directed angles) using the facts that and , and are similar triangles, and that equals twice the circumradius of the circumcircle of .
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