Difference between revisions of "2011 AMC 10A Problems/Problem 18"
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<math> \textbf{(A)}\ 3 - \frac{\pi}{2} \qquad\textbf{(B)}\ \frac{\pi}{2} \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ \frac{3\pi}{4} \qquad\textbf{(B)}\ 1 + \frac{\pi}{2} </math> | <math> \textbf{(A)}\ 3 - \frac{\pi}{2} \qquad\textbf{(B)}\ \frac{\pi}{2} \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ \frac{3\pi}{4} \qquad\textbf{(B)}\ 1 + \frac{\pi}{2} </math> | ||
| + | [[Category: Introductory Geometry Problems]] | ||
== Solution == | == Solution == | ||
Revision as of 10:42, 13 August 2014
Problem 18
Circles 
and 
each have radius 1. Circles 
and 
share one point of tangency. Circle has a point of tangency with the midpoint of 
. What is the area inside Circle but outside circle and circle ?
![[asy] pathpen = linewidth(.7); pointpen = black; pair A=(-1,0), B=-A, C=(0,1); fill(arc(C,1,0,180)--arc(A,1,90,0)--arc(B,1,180,90)--cycle, gray(0.5)); D(CR(D("A",A,SW),1)); D(CR(D("B",B,SE),1)); D(CR(D("C",C,N),1)); [/asy]](http://latex.artofproblemsolving.com/5/2/2/522afa350478b6fce4bc85fcf3c6173d8a3dcedd.png)
Solution
Draw a rectangle with vertices at the centers of and and the intersection of 
and 
. Then, we can compute the shaded area as the area of half of plus the area of the rectangle minus the area of the two sectors created by and . This is 
.
See Also
| 2011 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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. 
