HMMT-Feb 2011 guts, Problem 1

Let $A B C$ be a triangle with area 1. Let points $D$ and $E$ lie on $A B$ and $A C$, respectively, such that $D E$ is parallel to $B C$ and $D E / B C=1 / 3$. If $F$ is the reflection of $A$ across $D E$, find the area of triangle $F B C$.

Solution: Let $A F$ intersect $B C$ at $H$. Since $D E / B C=1 / 3$ and $F$ and $A$ are equidistant from $D E$, we have $A F=\frac{2}{3} A H$ and $F H=A H-A F=\frac{1}{3} A H$. Furthermore, since $A F$ is perpendicular to $D E$, we have $A H$ and $F H$ are the altitudes of triangles $A B C$ and $F B C$ respectively. Therefore the area of triangle $F B C$ is $\frac{1}{2} \cdot F H \cdot B C=\frac{1}{2} \cdot \frac{1}{3} \cdot A H \cdot B C=\frac{1}{3}$. $\fbox{\frac{1}{3}}$.

Answer: \frac{1}{3}