2013 USAMO Problems/Problem 1
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
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.
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.
Use directed angles modulo .
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 .
We will use construction arguments to solve the problem.
Let and let We construct lines through the points and that intersect with at the points and respectively, and that intersect each other at We will construct these lines such that
Now we let the intersections of with and be and respectively. This construction is as follows.
We know that Hence, we have,
Since the opposite angles of quadrilateral add up to it must be cyclic. Similarly, we can also show that quadrilaterals and are also cyclic.
Since points and lie on we know that, and that
Hence, the points and coincide with the given points and respectively.
Since quadrilateral is also cyclic, we have,
Similarly, since quadrilaterals and are also cyclic, we have, and,
Since these three angles are of and they are equal to corresponding angles of by AA similarity, we know that
We now consider the point We know that the points and are concyclic. Hence, the points and must also be concyclic.
Hence, quadrilateral is cyclic.
Since the angles and are inscribed in the same arc we have,
Consider by this result, we can deduce that the homothety that maps to will map to Hence, we have that,
Since and hence,
Proof by Negia