# 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== | ||

− | + | 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>\fbox{\textbf{(A)} </math>\frac{50\pi+1}{280}<math>}</math>. |

## Revision as of 14:03, 26 April 2021

On a coordinate plane, point denotes the origin which is the center of the diamond shape in the middle of the figure. Point has coordinate , and point , , and are formed through , , and rotation about the origin , respectively. Quarter circle (formed by the arc and line segments and ) has area . Furthermore, another quarter circle formed by arc and line segments , is formed through a reflection of sector across the line . The small diamond centered at is a square, and the area of the little square is . Let denote the area of the shaded region, and denote the sum of the area of the regions (formed by side , arc , and side ), (formed by side , arc , and side ) and sectors and . Find in the simplest radical form.

## Solution

First of all, we know that . Since the area of the quarter circle is , we can get that 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 .

The sum of the area of the regions (formed by side , arc , and side ), (formed by side , arc , and side ) and sectors and can be calculated by turning into a square, and subtract the extra areas. Since has length , we know that the height of the two right triangles are and the based are . . We want to also subtract the shaded quarter circle. The area is . The region enclosed by arc and length is the reflection of the previous area. The area . The region is also the reflection. Therefore, the total area is .

As a result, the ratio is $\fbox{\textbf{(A)}$ (Error compiling LaTeX. ! File ended while scanning use of \fbox .)\frac{50\pi+1}{280}$}$ (Error compiling LaTeX. ! Extra }, or forgotten $.).