2021 Fall AMC 10B Problems/Problem 11

Problem 11

A regular hexagon of side length $1$ is inscribed in a circle. Each minor arc of the circle determined by a side of the hexagon is reflected over that side. What is the area of the region bounded by these $6$ reflected arcs?

Solution 1

$[asy] import olympiad; unitsize(50); pair A,B,C,D,E,F,O; A = origin; B = (-0.5,0.866025); C=(0,1.7320508); D=(1,1.7320508); E=(1.5,0.866025); F=(1,0); draw(A--B--C--D--E--F--cycle); draw(Circle((0.5,0.866025),1)); [/asy]$

This is the graph of the original Hexagon. After reflecting each minor arc over the sides of the hexagon it will look like this;

$[asy] import olympiad; unitsize(50); pair A,B,C,D,E,F,O; A = origin; B = (-0.5,0.866025); C=(0,1.7320508); D=(1,1.7320508); E=(1.5,0.866025); F=(1,0); draw(A--B--C--D--E--F--cycle); draw(arc((0.5,2.598076), C, D)); draw(arc((2,1.7320508), D, E)); draw(arc((2,0), E, F)); draw(arc((0.5,-0.866025), F, A)); draw(arc((-1,0), A, B)); draw(arc((-1,1.7320508), B, C)); [/asy]$

Solution in Progress

~KingRavi

Solution 2

Let the hexagon described be of area $H$ and let the circle's area be $C$ . Let the area we want to aim for be $A$ . Thus, we have that $C-H=H-A$ , or $A=2H-C$ . By some formulas, $C=\pi{r}^2=\pi$ and $H=6\cdot\frac12\cdot1\cdot(\frac12\cdot\sqrt3)=\frac{3\sqrt3}2$ . Thus, $A=3\sqrt3-\pi$ or $\boxed{(\textbf{B})}$ .

~Hefei417, or 陆畅 Sunny from China

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2021 Fall AMC 10B Problems/Problem 11

Contents

Problem 11

Solution 1

Solution 2

See Also