2008 AMC 10A Problems/Problem 17
Contents
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?
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
Solution 2
After a quick sketch of the problem, one can deduce that there are rectangles in the figure, each with side length
and width
. Therefore, the combined areas of the rectangles is
. The other three regions are circle-shaped areas, probably expressed in some form of
. Answer choices
,
,
are impossible because they either lack an integer or
in the answer, and
is impossible since
clearly does not belong to the rectangle or the circular areas. We can conclude that the only choice left is
.
Solution by Airplane50
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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