2013 USAMO Problems/Problem 6
Let be a triangle. Find all points on segment satisfying the following property: If and are the intersections of line with the common external tangent lines of the circumcircles of triangles and , then
Let circle (i.e. the circumcircle of ), be with radii , and centers , respectively, and be the distance between their centers.
Proof. Let the external tangent containing meet at and at , and let the external tangent containing meet at and at . Then clearly and are parallel (for they are both perpendicular ), and so is a trapezoid.
Now, by Power of a Point, and so is the midpoint of . Similarly, is the midpoint of . Hence, Let , meet s at , respectively. Then by similar triangles and the Pythagorean Theorem we deduce that and . But it is clear that , is the midpoint of , , respectively, so as desired.
Lemma 2. Triangles and are similar.
Proof. and similarly , so the triangles are similar by AA Similarity.
Also, let intersect at . Then obviously is the midpoint of and is an altitude of triangle .Thus, we can simplify our expression of : where is the length of the altitude from in triangle . Hence, substituting into our condition and using gives Using by Heron's Formula (where is the area of triangle , our condition becomes which by becomes Let ; then . The quadratic in is which factors as Hence, or , and so the corresponding to these lengths are our answer. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.