2008 AMC 10A Problems/Problem 17

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

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

See also

2008 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 10 Problems and Solutions

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