HMMT-Feb 2020 guts, Problem 17

Let $A B C$ be a triangle with incircle tangent to the perpendicular bisector of $B C$. If $B C=A E=$ 20 , where $E$ is the point where the $A$-excircle touches $B C$, then compute the area of $\triangle A B C$.

Solution: Solution: Let the incircle and $B C$ touch at $D$, the incircle and perpendicular bisector touch at $X, Y$ be the point opposite $D$ on the incircle, and $M$ be the midpoint of $B C$. Recall that $A, Y$, and $E$ are collinear by homothety at $A$. Additionally, we have $M D=M X=M E$ so $\angle D X Y=\angle D X E=90^{\circ}$. Therefore $E, X$, and $Y$ are collinear. Since $M X \perp B C$, we have $\angle A E B=45^{\circ}$. The area of $A B C$ is \[ \frac{1}{2} B C \cdot A E \cdot \sin \angle A E B=100 \sqrt{2} \] $\fbox{100 \sqrt{2}}$.

Answer: 100 \sqrt{2}