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Difference between revisions of "2021 April MIMC 10 Problems/Problem 23"

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<math>\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</math>
 
<math>\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</math>
 
==Solution==
 
==Solution==
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To be Released on April 26th.
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First of all, we know that <math>BO=OH</math>. Since the area of the quarter circle is <math>25\pi</math>, we can get that <math>BO=OH=\sqrt{\frac{25\pi\cdot4}{\pi}}=10</math> 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 <math>2\cdot25\pi+2\cdot\frac{2}{4}=50\pi+1</math>.
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The sum of the area of the regions <math>ABH</math> (formed by side <math>AB</math>, arc <math>BH</math>, and side <math>HA</math>), <math>DFE</math> (formed by side <math>ED</math>, arc <math>DF</math>, and side <math>FE</math>) and sectors <math>FGH</math> and <math>BCD</math> can be calculated by turning <math>AHBO</math> into a square, and subtract the extra areas. Since <math>BO</math> has length <math>10</math>, we know that the height of the two right triangles are <math>2</math> and the based are <math>12</math>. <math>144-24=120</math>. We want to also subtract the shaded quarter circle. The area is <math>120-25\pi</math>. The region enclosed by arc <math>DF</math> and length <math>DE, FE</math> is the reflection of the previous area. The area <math>HGF=120-100+25\pi=20+25\pi</math>. The region <math>BCD</math> is also the reflection. Therefore, the total area is <math>120-25\pi+120-25\pi+20+25\pi+20+25\pi=280</math>.
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As a result, the ratio is <math>\boxed{\textbf{(A)} \frac{50\pi+1}{280}}</math>.

Latest revision as of 13:04, 26 April 2021

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.

19 (1).png

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

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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