2021 April MIMC 10 Problems/Problem 23

On a coordinate plane, point $O$ denotes the origin which is the center of the diamond shape in the middle of the figure. Point $A$ has coordinate $(-12,12)$ , and point $C$ , $E$ , and $G$ are formed through $90^{\circ}$ , $180^{\circ}$ , and $270^{\circ}$ rotation about the origin $O$ , respectively. Quarter circle $BOH$ (formed by the arc $BH$ and line segments $BO$ and $GH$ ) has area $25\pi$ . Furthermore, another quarter circle $DOF$ formed by arc $DF$ and line segments $OF$ , $OD$ is formed through a reflection of sector $BOH$ across the line $y=x$ . The small diamond centered at $O$ is a square, and the area of the little square is $2$ . Let $x$ denote the area of the shaded region, and $y$ denote the sum of the area of the regions $ABH$ (formed by side $AB$ , arc $BH$ , and side $HA$ ), $DFE$ (formed by side $ED$ , arc $DF$ , and side $FE$ ) and sectors $FGH$ and $BCD$ . Find $\frac{x}{y}$ in the simplest radical form.

$\textbf{(A)} ~\frac{50\pi+1}{280} \qquad\textbf{(B)} ~\frac{50\pi\sqrt{2}+\sqrt{2}}{560} \qquad\textbf{(C)} ~\frac{50\pi+1}{140+100\pi} \qquad\textbf{(D)} ~\frac{50\pi+1}{280+100\pi} \qquad\textbf{(E)} ~\frac{50\pi^2+700\pi\sqrt{2}+3001\pi-70\sqrt{2}+60}{2\pi^2+240\pi+6920}\qquad$

Solution

First of all, we know that $BO=OH$ . Since the area of the quarter circle is $25\pi$ , we can get that $BO=OH=\sqrt{\frac{25\pi\cdot4}{\pi}}=10$ Then, we can calculate the area of shaded region. It is made of two quarter circles and two right triangles. The total area would be $2\cdot25\pi+2\cdot\frac{2}{4}=50\pi+1$ .

The sum of the area of the regions $ABH$ (formed by side $AB$ , arc $BH$ , and side $HA$ ), $DFE$ (formed by side $ED$ , arc $DF$ , and side $FE$ ) and sectors $FGH$ and $BCD$ can be calculated by turning $AHBO$ into a square, and subtract the extra areas. Since $BO$ has length $10$ , we know that the height of the two right triangles are $2$ and the based are $12$ . $144-24=120$ . We want to also subtract the shaded quarter circle. The area is $120-25\pi$ . The region enclosed by arc $DF$ and length $DE, FE$ is the reflection of the previous area. The area $HGF=120-100+25\pi=20+25\pi$ . The region $BCD$ is also the reflection. Therefore, the total area is $120-25\pi+120-25\pi+20+25\pi+20+25\pi=280$ .

As a result, the ratio is $\boxed{\textbf{(A)} \frac{50\pi+1}{280}}$ .

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2021 April MIMC 10 Problems/Problem 23

Solution