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

m (Problem 11)
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bounded by these <math>6</math> reflected arcs?
 
bounded by these <math>6</math> reflected arcs?
  
==Solution==
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==Solution 1==
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<asy>
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import olympiad;
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unitsize(50);
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Pair A,B,C,D,E,F,O;
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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);
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draw(A--B--C--D--E--F--cycle);
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==Solution 2==
 
Let the hexagon described be of area <math>H</math> and let the circle's area be <math>C</math>.  
 
Let the hexagon described be of area <math>H</math> and let the circle's area be <math>C</math>.  
 
Let the area we want to aim for be <math>A</math>.  
 
Let the area we want to aim for be <math>A</math>.  

Revision as of 14:19, 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);


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

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