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Difference between revisions of "2021 April MIMC 10 Problems/Problem 24"

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

Latest revision as of 14:05, 26 April 2021

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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