Difference between revisions of "2011 AMC 10A Problems/Problem 18"

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== Problem 18 ==
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#redirect [[2011 AMC 12A Problems/Problem 11]]
Circles <math>A, B,</math> and <math>C</math> each have radius 1. Circles <math>A</math> and <math>B</math> share one point of tangency. Circle <math>C</math> has a point of tangency with the midpoint of <math>\overline{AB}</math>. What is the area inside Circle <math>C</math> but outside circle <math>A</math> and circle <math>B</math> ?
 
<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>
 
 
 
<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{(E)}\ 1 + \frac{\pi}{2} </math>
 
[[Category: Introductory Geometry Problems]]
 
 
 
== Solution ==
 
 
 
Not specific: Draw a rectangle with vertices at the centers of <math>A</math> and <math>B</math> and the intersection of <math>A, C</math> and <math>B, C</math>. Then, we can compute the shaded area as the area of half of <math>C</math> plus the area of the rectangle minus the area of the two sectors created by <math>A</math> and <math>B</math>. This is <math>\frac{\pi (1)^2}{2}+(2)(1)-2 \cdot \frac{\pi (1)^2}{4}=\boxed{ \mathbf{(C)} 2}</math>.
 
 
 
== Solution 2==
 
 
 
<asy>
 
unitsize(12mm);
 
defaultpen(linewidth(.8pt));
 
 
 
pair A,B,C;
 
A=(0,0);
 
B=(2,0);
 
C=(1,1);
 
</asy>
 
 
 
== See Also ==
 
 
 
 
 
{{AMC10 box|year=2011|ab=A|num-b=17|num-a=19}}
 
{{MAA Notice}}
 

Latest revision as of 19:19, 27 June 2020