2003 USAMO Problems/Problem 4
Let be a triangle. A circle passing through and intersects segments and at and , respectively. Lines and intersect at , while lines and intersect at . Prove that if and only if .
Extend segment through to such that . Then if and only if quadrilateral is a parallelogram, or, . Hence if and only if , that is, .
Because quadrilateral is cyclic, . It follows that if and only if that is, quadrilateral is cyclic, which is equivalent to Because , if and only if triangles and are similar, that is or .
We first assume that . Because and , triangles and are similar. Consequently, . This exact condition can also be visualized through power of M with respect to the circumcircle of triangle and getting the angle condition from the alternate segment theorem. Because quadrilateral is cyclic, . Hence implying that , so . Because quadrilateral is cyclic, . Hence Because and , triangles and are similar. Consequently, , or . Therefore implies .
Now we assume that . Applying Ceva's Theorem to triangle and cevians gives implying that , so .
Consequently, . Because quadrilateral is cyclic, . Hence Because and , triangles and are similar. Consequently, , or .
Combining the above, we conclude that if and only if .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
|2003 USAMO (Problems • Resources)|
|1 • 2 • 3 • 4 • 5 • 6|
|All USAMO Problems and Solutions|