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Problem 1346

AIME late

Geometry Difficulty 5.1 Find the answer HMMT February

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 $\triangle A B C$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official 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 $\frac{1}{2} B C \cdot A E \cdot \sin \angle A E B=100 \sqrt{2}$ .

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.