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 $\frac{26}{56+12\pi}=\boxed{\textbf{(E)} \frac{13}{28+6\pi}}$ .

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2021 April MIMC 10 Problems/Problem 24

Solution