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Difference between revisions of "2021 Fall AMC 10B Problems/Problem 11"

(→‎Solution 1)
(→‎Solution 1)
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A = origin; B = (-0.5,0.866025); C=(0,1.7320508); D=(1,1.7320508); E=(1.5,0.866025); F=(1,0);
 
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(A--B--C--D--E--F--cycle);
−
path arc((0.5,2.598076), C, D, direction=CCW);
+
draw(arc((0.5,2.598076), C, D));
  
 
</asy>
 
</asy>

Revision as of 14:52, 24 November 2021

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)); [/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 (Problems • Answer Key • Resources)
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

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