# 1996 USAMO Problems/Problem 3

## Problem

Let $ABC$ be a triangle. Prove that there is a line $l$ (in the plane of triangle $ABC$) such that the intersection of the interior of triangle $ABC$ and the interior of its reflection $A'B'C'$ in $l$ has area more than $\frac{2}{3}$ the area of triangle $ABC$.

## Solution

Let the triangle be $ABC$. Assume $A$ is the largest angle. Let $AD$ be the altitude. Assume $AB \le AC$, so that $BD \le BC/2$. If $BD > \frac{BC}{3}$, 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 $BB' = 2BD > \frac{2}{3} BC$, so $ABB'$ has more than $\frac{2}{3}$ 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 $\angle{CAD} < \angle{CDA} = 90^{\circ}$. Finally, $A'C \le BC$ because we assumed $\angle B$ does not exceed $\angle A$). The intersection is just $AA'C$. But $\frac{[AA'C]}{[ABC]} = \frac{CA'}{CB} > CD/CB \ge 2/3$.