Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.8 AIME Prove it Ireland

$AB$ is a chord of length $6$ in a circle of radius $5$ and with centre $O$ . A square is symmetrically inscribed in the sector $OAB$ with two vertices on the circumference. Find the area of the square.

Solution

Let $XYZ$ be the square with $Y$ and $Z$ on the circle, $X$ on $OA$ and $Y$ on $OB$ . Let $2a = |YZ|$ . Draw $OD \perp YZ$ , $D$ on $YZ$ . Then $OD \perp AB$ and bisects $AB$ . So $|OA| = 5$ , $|AC| = 3 \Rightarrow |OC| = 4 \Rightarrow \frac{|XE|}{|OE|} = \frac{|AC|}{|OC|} = \frac{3}{4} \Rightarrow \frac{a}{|OE|} = \frac{3}{4} \Rightarrow |OE| = \frac{4a}{3}$ .

Also $|OD|^2 = |OY|^2 - |YD|^2 = 25 - a^2$ . Now $|OD| = |OE| + |ED| = \frac{4a}{3} + 2a = \frac{10a}{3} \Rightarrow |OD|^2 = \frac{100a^2}{9}$ .

So, $\frac{100a^2}{9} = 25 - a^2$ from which we get that the area of the square is equal to $4a^2 = \frac{900}{109}$ .

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