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Difference between revisions of "2012 AMC 10A Problems/Problem 18"

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== Problem 18 ==
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{{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #14]] and [[2012 AMC 10A Problems|2012 AMC 10A #18]]}}
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== Problem 14 ==
  
 
The closed curve in the figure is made up of 9 congruent circular arcs each of length <math>\frac{2\pi}{3}</math>, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?
 
The closed curve in the figure is made up of 9 congruent circular arcs each of length <math>\frac{2\pi}{3}</math>, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?
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</asy>
  
 
<math>\textbf{(A)}\ 2\pi+6\qquad\textbf{(B)}\ 2\pi+4\sqrt{3}\qquad\textbf{(C)}\ 3\pi+4\qquad\textbf{(D)}\ 2\pi+3\sqrt{3}+2\qquad\textbf{(E)}\ \pi+6\sqrt{3}</math>
 
<math>\textbf{(A)}\ 2\pi+6\qquad\textbf{(B)}\ 2\pi+4\sqrt{3}\qquad\textbf{(C)}\ 3\pi+4\qquad\textbf{(D)}\ 2\pi+3\sqrt{3}+2\qquad\textbf{(E)}\ \pi+6\sqrt{3}</math>
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[[Category: Introductory Geometry Problems]]
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== Solution ==
  
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[[2012 AMC 10A Problems/Problem 18|Solution]]
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Draw the hexagon between the centers of the circles, and compute its area <math>(6)(0.5)(2\sqrt{3})=6\sqrt{3}</math>. Then add the areas of the three sectors outside the hexagon (<math>2\pi</math>) and subtract the areas of the three sectors inside the hexagon but outside the figure(<math>\pi</math>) to get the area enclosed in the curved figure <math>(\pi+6\sqrt{3})</math>, which is <math>\boxed{\textbf{(E)}\ \pi+6\sqrt{3}}</math>.
  
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== Solution ==
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== See Also ==
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{{AMC10 box|year=2012|ab=A|num-b=17|num-a=19}}
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{{AMC12 box|year=2012|ab=A|num-b=13|num-a=15}}
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[[Category:Introductory Geometry Problems]]
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[[Category:Area Problems]]
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{{MAA Notice}}

Revision as of 20:05, 12 July 2020

The following problem is from both the 2012 AMC 12A #14 and 2012 AMC 10A #18, so both problems redirect to this page.

Problem 14

The closed curve in the figure is made up of 9 congruent circular arcs each of length $\frac{2\pi}{3}$, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?

[asy] defaultpen(fontsize(6pt)); dotfactor=4; label("$\circ$",(0,1)); label("$\circ$",(0.865,0.5)); label("$\circ$",(-0.865,0.5)); label("$\circ$",(0.865,-0.5)); label("$\circ$",(-0.865,-0.5)); label("$\circ$",(0,-1)); dot((0,1.5)); dot((-0.4325,0.75)); dot((0.4325,0.75)); dot((-0.4325,-0.75)); dot((0.4325,-0.75)); dot((-0.865,0)); dot((0.865,0)); dot((-1.2975,-0.75)); dot((1.2975,-0.75)); draw(Arc((0,1),0.5,210,-30)); draw(Arc((0.865,0.5),0.5,150,270)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0,-1),0.5,30,150)); draw(Arc((-0.865,-0.5),0.5,330,90)); draw(Arc((-0.865,0.5),0.5,-90,30)); [/asy]

$\textbf{(A)}\ 2\pi+6\qquad\textbf{(B)}\ 2\pi+4\sqrt{3}\qquad\textbf{(C)}\ 3\pi+4\qquad\textbf{(D)}\ 2\pi+3\sqrt{3}+2\qquad\textbf{(E)}\ \pi+6\sqrt{3}$

Solution

Draw the hexagon between the centers of the circles, and compute its area $(6)(0.5)(2\sqrt{3})=6\sqrt{3}$. Then add the areas of the three sectors outside the hexagon ($2\pi$) and subtract the areas of the three sectors inside the hexagon but outside the figure($\pi$) to get the area enclosed in the curved figure $(\pi+6\sqrt{3})$, which is $\boxed{\textbf{(E)}\ \pi+6\sqrt{3}}$.

See Also

2012 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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
2012 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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 12 Problems and Solutions

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

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