# 2021 April MIMC 10 Problems/Problem 24

One semicircle is constructed with diameter $AH=4$ and let the midpoint of $AH$ be $M$. Construct a point $O$ on the side of segment $AH$ (closer to segment $AH$ than arc $AH$) such that the distance from $A$ to $O$ is $2\sqrt{5}$, and that $OM$ is perpendicular to the diameter $AH$. Three more such congruent semicircles are formed through multiple $90^{\circ}$rotations around the point $O$. Name the $6$ endpoints of the diameters $B$, $C$, $D$, $E$, $F$, $G$ in a circular direction from $A$ to $H$. Another four congruent semicircles are constructed with diameters $AB, CD, EF, GH$, and that the distance from the diameters to the point $O$ are less than the distance from the arcs to the point $O$. Connect $AC$, $CD$, $DO$, $OG$, and $GA$. Find the ratio of the area of the pentagon $ACDOG$ to the total area of the shape formed by arcs $AB$, $BC$, $CD$, $DE$, $EF$, $FG$, $GH$, $HA$.

$\textbf{(A)} ~\frac{14+10\pi}{17} \qquad\textbf{(B)} ~\frac{13+\sqrt{2}}{28} \qquad\textbf{(C)} ~\frac{4+\sqrt{2}}{7+3\pi} \qquad\textbf{(D)} ~\frac{13}{28+6\pi} \qquad\textbf{(E)} ~\frac{13}{30\pi}\qquad$

## Solution

We can first draw the diagram with the instructions. Since $AH=4$ and $AO=2\sqrt{5}$, we know that $OM=4$. We know that $GO=OC=2\sqrt{5}$, we can calculate the area of triangle $AGC$. $[AGC]=\frac{1}{2}\cdot 4\sqrt{5}\cdot 2\sqrt{5}=20$. The final hard step for this problem is to solve for $CD$, or the diameter of the small circle. We can use coordinate geometry to finish this problem. Set $E$ as the origin point and $ED$ as the positive $x$ direction. We then can get that $D=(4,0)$ since $DE=AH=4$. We can then solve for $O$, and it is easy as well because we can split it into two parts. $O=(2,4)$. Since $OC$ is perpendicular to $OE$, their slopes must be opposite reciprocal. Therefore, $OC$ is $4$ to the right and $2$ down. Thus, the coordinate of $C$ would be $(6,2)$. We can then use the distance formula on $CD$. $CD=\sqrt{(2-0)^2+(6-4)^2}=2\sqrt{2}$. Then, we can draw a height from $O$ to $CD$ and call that $N$. $ON=\sqrt{(2\sqrt{5})^2-(\sqrt{2})^2}=3\sqrt{2}$. Thus, the area of the remaining shape of the shaded part would be $\frac{1}{2}\cdot2\sqrt{2}\cdot3\sqrt{2}=6$. The whole shaded area would have a area of $20+6=26$. We can connect the diameters to get a octagon. It is formed by four big triangles like $DOE$ and four small triangles like $COD$. Therefore, $A_{octagon}=4\cdot\frac{1}{2}\cdot4\cdot4+4\cdot6=56$ The area of the outer semicircles would form two big circles and two small circles. $A_{semicircles}=2\cdot(\sqrt{2})^2\cdot\pi+2\cdot(2)^2\cdot\pi=12\pi$. Thus, the ratio would be $\frac{26}{56+12\pi}=\boxed{\textbf{(E)} \frac{13}{28+6\pi}}$.