2003 USAMO Problems/Problem 4
Problem
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 .
Solutions
Solution 1
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 .
Solution 2
We first assume that . Because and , triangles and are similar. Consequently, .
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 .
Solution 3
We will construct a series of equivalent statements. First: where the second equivalence follows from SAS similarity (because ).
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See also
2003 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.