2021 Fall AMC 10B Problems/Problem 11

Revision as of 15:55, 24 November 2021 by Kingravi (talk | contribs) (Solution 1)

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));   [/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

See Also

2021 Fall AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png