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

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<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>
  
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==Solution==
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===Solution 1===
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==Solution 1==
 
<center><asy>
 
<center><asy>
 
pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("6 6")+linewidth(0.7);
 
pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("6 6")+linewidth(0.7);
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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>
 
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>
  
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===Solution 2===
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==Solution 2==
 
After a quick sketch of the problem, one can deduce that there are <math>3</math> rectangles in the figure, each with side length <math>6</math> and width <math>3</math>. Therefore, the combined areas of the rectangles is <math>54</math>. The other three regions are circle-shaped areas, probably expressed in some form of <math>\pi</math>. Answer choices <math>\mathrm{(A)}</math>, <math>\mathrm{(D)}</math>, <math>\mathrm{(E)}</math> are impossible because they either lack an integer or <math>\pi</math> in the answer, and <math>\mathrm{(C)}</math> is impossible since <math>18\sqrt{3}</math> clearly does not belong to the rectangle or the circular areas. We can conclude that the only choice left is <math>\boxed{\mathrm{(B)}}</math>.  
 
After a quick sketch of the problem, one can deduce that there are <math>3</math> rectangles in the figure, each with side length <math>6</math> and width <math>3</math>. Therefore, the combined areas of the rectangles is <math>54</math>. The other three regions are circle-shaped areas, probably expressed in some form of <math>\pi</math>. Answer choices <math>\mathrm{(A)}</math>, <math>\mathrm{(D)}</math>, <math>\mathrm{(E)}</math> are impossible because they either lack an integer or <math>\pi</math> in the answer, and <math>\mathrm{(C)}</math> is impossible since <math>18\sqrt{3}</math> clearly does not belong to the rectangle or the circular areas. We can conclude that the only choice left is <math>\boxed{\mathrm{(B)}}</math>.  
  
 
Solution by Airplane50
 
Solution by Airplane50
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 +
== Video Solution (Based on Solution 1)==
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https://www.youtube.com/watch?v=DfpdV7eCLG4
  
 
==See also==
 
==See also==

Latest revision as of 18:59, 4 June 2021

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?

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


Solution 1

[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 \[3(3 \times 6) + 3 \left( \frac{120^{\circ}}{360^{\circ}} \times 3^2 \pi\right) = 54 + 9\pi \Longrightarrow \mathrm{(B)}.\]

Solution 2

After a quick sketch of the problem, one can deduce that there are $3$ rectangles in the figure, each with side length $6$ and width $3$. Therefore, the combined areas of the rectangles is $54$. The other three regions are circle-shaped areas, probably expressed in some form of $\pi$. Answer choices $\mathrm{(A)}$, $\mathrm{(D)}$, $\mathrm{(E)}$ are impossible because they either lack an integer or $\pi$ in the answer, and $\mathrm{(C)}$ is impossible since $18\sqrt{3}$ clearly does not belong to the rectangle or the circular areas. We can conclude that the only choice left is $\boxed{\mathrm{(B)}}$.

Solution by Airplane50

Video Solution (Based on Solution 1)

https://www.youtube.com/watch?v=DfpdV7eCLG4

See also

2008 AMC 10A (Problems • Answer Key • Resources)
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

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