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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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defaultpen(fontsize(6pt));
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dotfactor=4;
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label("$\circ$",(0,1));
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label("$\circ$",(0.865,0.5));
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label("$\circ$",(-0.865,0.5));
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label("$\circ$",(0.865,-0.5));
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label("$\circ$",(-0.865,-0.5));
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label("$\circ$",(0,-1));
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dot((0,1.5));
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dot((-0.4325,0.75));
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dot((0.4325,0.75));
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dot((-0.4325,-0.75));
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dot((0.4325,-0.75));
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dot((-0.865,0));
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dot((0.865,0));
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dot((-1.2975,-0.75));
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dot((1.2975,-0.75));
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draw(Arc((0,1),0.5,210,-30));
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draw(Arc((0.865,0.5),0.5,150,270));
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draw(Arc((0.865,-0.5),0.5,90,-150));
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draw(Arc((0.865,-0.5),0.5,90,-150));
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draw(Arc((0,-1),0.5,30,150));
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draw(Arc((-0.865,-0.5),0.5,330,90));
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draw(Arc((-0.865,0.5),0.5,-90,30));
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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]]
  
 
== Solution ==
 
== Solution ==
  
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Draw the hexagon between the centers of the circles, and compute its area (6*.5*2*sqrt3 = 6sqrt3). Then add the areas of the three sectors outside the hexagon (2pi) and subtract the areas of the three sectors inside the hexagon (pi) to get the area enclosed in the curved figure (6sqrt3 + pi), which is E.
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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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== 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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