Difference between revisions of "2008 AMC 10A Problems/Problem 17"
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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? | 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== | ||
Revision as of 11:03, 21 January 2016
Problem
An equilateral triangle has side length 
. What is the area of the region containing all points that are outside the triangle but not more than 
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]](http://latex.artofproblemsolving.com/e/9/a/e9aa15f65931b4eee7d795103170e3680b8a36f3.png)
The region described contains three rectangles of dimensions 
, and three 
degree arcs of circles of radius . Thus the answer is ![\[3(3 \times 6) + 3 \left( \frac{120^{\circ}}{360^{\circ}} \times 3^2 \pi\right) = 54 + 9\pi \Longrightarrow \mathrm{(B)}.\]](http://latex.artofproblemsolving.com/7/3/6/7364eac7efec70a0ea974a671cb20e987912389c.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
