Difference between revisions of "2008 AMC 10A Problems/Problem 17"

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==Problem==
 
==Problem==
An equilateral triangle has side length 6. What is the area of the region containing all points that are outside the triangle but not more than 3 units from a point of the triangle?
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An [[equilateral triangle]] has side length <math>6</math>. What is the [[area]] of the region containing all points that are outside the triangle but not more than <math>3</math> units from a point of the triangle?
  
 
<math>\mathrm{(A)}\ 36+24\sqrt{3}\qquad\mathrm{(B)}\ 54+9\pi\qquad\mathrm{(C)}\ 54+18\sqrt{3}+6\pi\qquad\mathrm{(D)}\ \left(2\sqrt{3}+3\right)^2\pi\\\mathrm{(E)}\ 9\left(\sqrt{3}{+1\right)^2\pi</math>
 
<math>\mathrm{(A)}\ 36+24\sqrt{3}\qquad\mathrm{(B)}\ 54+9\pi\qquad\mathrm{(C)}\ 54+18\sqrt{3}+6\pi\qquad\mathrm{(D)}\ \left(2\sqrt{3}+3\right)^2\pi\\\mathrm{(E)}\ 9\left(\sqrt{3}{+1\right)^2\pi</math>
  
 
==Solution==
 
==Solution==
[[Image:AMC10A-2008-17.png]]
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<center><asy>
 
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pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("6 6")+linewidth(0.7);
Three rectangles are formed that are <math>3*6</math>.  <math>3*3*6=54</math><math>360-90-90-60=120</math> degrees = <math>1/3</math> the circle's area.  <math>1/3*3= 1</math> circle's area, <math>9 \pi</math>.
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pair A=(0,0),B=(6,0),C=6*expi(-pi/3);
 
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D(arc(A,3,90,210)); D(arc(B,3,-30,90)); D(arc(C,3,210,330));
<math>54+9pi \rightarrow</math> B
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D(arc(A,-3,90,210),d); D(arc(B,-3,-30,90),d); D(arc(C,-3,210,330),d);
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D(D(A)--D(B)--D(C)--cycle,linewidth(1));
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D(A--(0,3)--(6,3)--B); D(A--3*expi(7/6*pi)--C+3*expi(7/6*pi)--C); D(B--B+3*expi(11/6*pi)--C+3*expi(11/6*pi)--C);
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MP("3",(0,1.5),W); MP("6",(3,0),NW);
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</asy></center> <!-- Asymptote replacement for Image:AMC10A-2008-17.png by 1=2 -->
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The region described contains three rectangles of dimensions <math>3 \times 6</math>, and three <math>120^{\circ}</math> degree arcs of circles of [[radius]] <math>3</math>. Thus the answer is <cmath>3(3 \times 6) + 3 \left( \frac{120^{\circ}}{360^{\circ}} \times 3^2 \pi\right) = 54 + 9\pi \Longrightarrow \mathrm{(B)}.</cmath>
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=A|num-b=16|num-a=18}}
 
{{AMC10 box|year=2008|ab=A|num-b=16|num-a=18}}
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[[Category:Introductory Geometry Problems]]

Revision as of 13:24, 16 June 2008

Problem

An equilateral triangle has side length $6$. What is the area of the region containing all points that are outside the triangle but not more than $3$ units from a point of the triangle?

$\mathrm{(A)}\ 36+24\sqrt{3}\qquad\mathrm{(B)}\ 54+9\pi\qquad\mathrm{(C)}\ 54+18\sqrt{3}+6\pi\qquad\mathrm{(D)}\ \left(2\sqrt{3}+3\right)^2\pi\\\mathrm{(E)}\ 9\left(\sqrt{3}{+1\right)^2\pi$ (Error compiling LaTeX. Unknown error_msg)

Solution

[asy] pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("6 6")+linewidth(0.7); pair A=(0,0),B=(6,0),C=6*expi(-pi/3); D(arc(A,3,90,210)); D(arc(B,3,-30,90)); D(arc(C,3,210,330)); D(arc(A,-3,90,210),d); D(arc(B,-3,-30,90),d); D(arc(C,-3,210,330),d); D(D(A)--D(B)--D(C)--cycle,linewidth(1)); D(A--(0,3)--(6,3)--B); D(A--3*expi(7/6*pi)--C+3*expi(7/6*pi)--C); D(B--B+3*expi(11/6*pi)--C+3*expi(11/6*pi)--C); MP("3",(0,1.5),W); MP("6",(3,0),NW); [/asy]

The region described contains three rectangles of dimensions $3 \times 6$, and three $120^{\circ}$ degree arcs of circles of radius $3$. Thus the answer is \[3(3 \times 6) + 3 \left( \frac{120^{\circ}}{360^{\circ}} \times 3^2 \pi\right) = 54 + 9\pi \Longrightarrow \mathrm{(B)}.\]

See also

2008 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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