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

Revision as of 11:00, 29 December 2019

Problem 18

Circles $A, B,$ and $C$ each have radius 1. Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}$ . What is the area inside Circle $C$ but outside circle $A$ and circle $B$ ? $[asy] pathpen = linewidth(.7); pointpen = black; pair A=(-1,0), B=-A, C=(0,1); fill(arc(C,1,0,180)--arc(A,1,90,0)--arc(B,1,180,90)--cycle, gray(0.5)); D(CR(D("A",A,SW),1)); D(CR(D("B",B,SE),1)); D(CR(D("C",C,N),1)); [/asy]$

$\textbf{(A)}\ 3 - \frac{\pi}{2} \qquad\textbf{(B)}\ \frac{\pi}{2} \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ \frac{3\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\pi}{2}$

Solution

Not specific: Draw a rectangle with vertices at the centers of $A$ and $B$ and the intersection of $A, C$ and $B, C$ . Then, we can compute the shaded area as the area of half of $C$ plus the area of the rectangle minus the area of the two sectors created by $A$ and $B$ . This is $\frac{\pi (1)^2}{2}+(2)(1)-2 \cdot \frac{\pi (1)^2}{4}=\boxed{ \mathbf{(C)} 2}$ .

Solution 2

[asy] pair A,B,C; A=(0,0); B=(2,0); C=(1,1); [/asy]

2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AMC 10 Problems and Solutions

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

Revision as of 11:00, 29 December 2019

Contents

Problem 18

Solution

Solution 2

See Also

@@ Line 13: / Line 13: @@
 [asy]
-pair A,B,C,D;
+pair A,B,C;
 A=(0,0);
 B=(2,0);