2012 USAJMO Problems/Problem 1
Contents
Problem
Given a triangle , let and be points on segments and , respectively, such that . Let and be distinct points on segment such that lies between and , , and . Prove that , , , are concyclic (in other words, these four points lie on a circle).
Solution
Since , the circumcircle of triangle is tangent to at . Similarly, since , the circumcircle of triangle is tangent to at .
For the sake of contradiction, suppose that the circumcircles of triangles and are not the same circle. Since , lies on the radical axis of both circles. However, both circles pass through and , so the radical axis of both circles is . Hence, lies on , which is a contradiction.
Therefore, the two circumcircles are the same circle. In other words, , , , and all lie on the same circle.
Solution 2
Note that (as in the first solution) the circumcircle of triangle is tangent to at . Similarly, since , the circumcircle of triangle is tangent to at .
Now, suppose these circumcircles are not the same circle. They already intersect at and , so they cannot intersect anymore. Thus, AS must touch the two circumcircles at points and , with on the circumcircle of triangle . By Power of a Point, and . Hence, because , , a contradiction because then, as they lie on the same line segment, M and N must be the same point! (Note line segment, not line.) Hence, the two circumcircles are the same circle.
See also
2012 USAJMO (Problems • Resources) | ||
First Problem | Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.