1996 USAMO Problems/Problem 3
Problem
Let 
be a triangle. Prove that there is a line 
(in the plane of triangle ) such that the intersection of the interior of triangle and the interior of its reflection 
in has area more than 
the area of triangle .
Solution
Let the triangle be ABC. Assume A is the largest angle. Let AD be the altitude. Assume AB ≤ AC, so that BD ≤ BC/2. If BD > BC/3, then reflect in AD. If B' is the reflection of B', then B'D = BD and the intersection of the two triangles is just ABB'. But BB' = 2BD > 2/3 BC, so ABB' has more than 2/3 the area of ABC.
If BD < BC/3, then reflect in the angle bisector of C. The reflection of A' is a point on the segment BD and not D. (It lies on the line BC because we are reflecting in the angle bisector. A'C > DC because ∠CAD < ∠CDA = 90o. Finally, A'C ≤ BC because we assumed ∠B does not exceed ∠A). The intersection is just AA'C. But area AA'C/area ABC = CA'/CB > CD/CB ≥ 2/3.
See Also
| 1996 USAMO (Problems • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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. 
