Given a triangle, show that we can reflect it in some line so that the area of the intersection of the triangle and its reflection has area greater than 2/3 the area of the 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.
© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002