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

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