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

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<asy>
 
<asy>
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unitsize(12mm);
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pathpen = linewidth(.7);  
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defaultpen(linewidth(.6pt));
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pointpen = black;
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pair A=(-1,0), B=-A, C=(0,1);  
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pair A=(0,0);
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fill(arc(C,1,0,180)--arc(A,1,90,0)--arc(B,1,180,90)--cycle, gray(0.5));  
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pair B=(2,0);
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D(CR(D("A",A,SW),1));  
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pair C=(1,1);
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D(CR(D("B",B,SE),1));  
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draw(Circle((0,0),1));
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D(CR(D("C",C,N),1));  
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draw(Circle((2,0),1));
 
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draw(Circle((1,1),1));
 
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dot((0,0));
 
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dot((2,0));
 
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dot((1,1));
 
 
draw((0,0)--(0,1));
 
draw((0,0)--(0,1));
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fill(arc(C,1,0,180)--arc(A,1,90,0)--arc(B,1,180,90)--cycle, gray(0.5));
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draw((0,0)--(1,0));
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label("A",(0,0),SW);
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draw((1,0)--(0,1));
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label("B",(2,0),E);
 
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label("C",(1,1),N);
 
 
</asy>
 
</asy>
  

Revision as of 11:15, 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] 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)); draw((0,0)--(0,1)); draw((0,0)--(1,0)); draw((1,0)--(0,1)); [/asy]

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

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