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Harvard-MIT Mathematics Tournament

Geometry Difficulty 5.5 AIME, harder Prove it United States

Problem:

Let $ABC$ be a triangle with $\angle BAC = 90^\circ$ . A circle is tangent to the sides $AB$ and $AC$ at $X$ and $Y$ respectively, such that the points on the circle diametrically opposite $X$ and $Y$ both lie on the side $BC$ . Given that $AB = 6$ , find the area of the portion of the circle that lies outside the triangle.

Solution

Solution:

Let $O$ be the center of the circle, and $r$ its radius, and let $X'$ and $Y'$ be the points diametrically opposite $X$ and $Y$ , respectively. We have $OX' = OY' = r$ , and $\angle X' O Y' = 90^\circ$ . Since triangles $X' O Y'$ and $BAC$ are similar, we see that $AB = AC$ . Let $X''$ be the projection of $Y'$ onto $AB$ . Since $X'' B Y'$ is similar to $ABC$ , and $X'' Y' = r$ , we have $X'' B = r$ . It follows that $AB = 3r$ , so $r = 2$ .

Then, the desired area is the area of the quarter circle minus that of the triangle $X' O Y'$ . And the answer is $\frac{1}{4} \pi r^2 - \frac{1}{2} r^2 = \pi - 2$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.