HMMT February

Geometry Difficulty 5.3 AIME, harder Find the answer

Square $ABCD$ is inscribed in circle $\omega$ with radius 10. Four additional squares are drawn inside $\omega$ but outside $ABCD$ such that the lengths of their diagonals are as large as possible. A sixth square is drawn by connecting the centers of the four aforementioned small squares. Find the area of the sixth square.

Your answer

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

Solution

Let $DEGF$ denote the small square that shares a side with $AB$ , where $D$ and $E$ lie on $AB$ . Let $O$ denote the center of $\omega, K$ denote the midpoint of $FG$ , and $H$ denote the center of $DEGF$ . The area of the sixth square is $2 \cdot \mathrm{OH}^{2}$ . Let $KF=x$ . Since $KF^{2}+OK^{2}=OF^{2}$ , we have $x^{2}+(2x+5\sqrt{2})^{2}=10^{2}$ . Solving for $x$ , we get $x=\sqrt{2}$ . Thus, we have $OH=6\sqrt{2}$ and $2 \cdot OH^{2}=144$ .

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