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Harvard-MIT Mathematics Tournament

Geometry Difficulty 5.5 AIME, harder Prove it United States

Problem:

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.

Solution

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 $2 \cdot OH^{2}$ .

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