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2004 AMC 10B Problems/Problem 25

Problem

A circle of radius $1$ is internally tangent to two circles of radius $2$ at points $A$ and $B$, where $AB$ is a diameter of the smaller circle. What is the area of the region, shaded in the picture, that is outside the smaller circle and inside each of the two larger circles?

$\mathrm{(A) \ } \frac{5}{3} \pi - 3\sqrt 2 \qquad \mathrm{(B) \ } \frac{5}{3} \pi - 2\sqrt 3 \qquad \mathrm{(C) \ } \frac{8}{3} \pi - 3\sqrt 3 \qquad \mathrm{(D) \ } \frac{8}{3} \pi - 3\sqrt 2 \qquad \mathrm{(E) \ } \frac{8}{3} \pi - 2\sqrt 3$


[asy] unitsize(1cm); defaultpen(0.8); pair O=(0,0), A=(0,1), B=(0,-1); path bigc1 = Circle(A,2), bigc2 = Circle(B,2), smallc = Circle(O,1); pair[] P = intersectionpoints(bigc1, bigc2); filldraw( arc(A,P[0],P[1])--arc(B,P[1],P[0])--cycle, lightgray, black ); draw(bigc1); draw(bigc2); unfill(smallc); draw(smallc); dot(O); dot(A); dot(B); label("$A$",A,N); label("$B$",B,S); draw( O--dir(30) ); draw( A--(A+2*dir(30)) ); draw( B--(B+2*dir(210)) ); label("$1$", O--dir(30), N ); label("$2$", A--(A+2*dir(30)), N ); label("$2$", B--(B+2*dir(210)), S ); [/asy]

Solution

The area of the small circle is $\pi$. We can add it to the shaded region, compute the area of the new region, and then subtract the area of the small circle from the result.

Let $C$ and $D$ be the intersections of the two large circles. Connect them to $A$ and $B$ to get the picture below:

[asy] unitsize(1.5cm); defaultpen(0.8); pair O=(0,0), A=(0,1), B=(0,-1); path bigc1 = Circle(A,2), bigc2 = Circle(B,2), smallc = Circle(O,1); pair[] P = intersectionpoints(bigc1, bigc2); filldraw( arc(A,P[0],P[1])--arc(B,P[1],P[0])--cycle, lightgray, black ); draw(bigc1); draw(bigc2); dot(A); dot(B); label("$A$",A,N); label("$B$",B,S); /* dot(O); unfill(smallc); draw(smallc); draw( O--dir(30) ); draw( A--(A+2*dir(30)) ); draw( B--(B+2*dir(210)) ); label("$1$", O--dir(30), N ); label("$2$", A--(A+2*dir(30)), N ); label("$2$", B--(B+2*dir(210)), S ); */ label("$C$",P[0],W); label("$D$",P[1],E); draw( P[0]--A--P[1] ); draw( P[0]--B--P[1] ); draw( A--B ); [/asy]

Now obviously the triangles $\triangle ABC$ and $\triangle ABD$ are equilateral with side $2$.

Take a look at the bottom circle. The angle $ABC$ is $60^\circ$, hence the sector $ABC$ is $1/6$ of the circle. The same is true for the sector $ABD$ of the bottom circle, and sectors $CAB$ and $BAD$ of the top circle.

If we now sum the areas of these four sectors, we will almost get the area of the new shaded region - except that each of the two equilateral triangles will be counted twice.

Hence the area of the new shaded region is $4\cdot \left( \frac 16 \cdot \pi\cdot 2^2 \right) - 2 \cdot \left( \frac 12 \cdot 2 \cdot \frac{2\sqrt{3}}2 \right) = \frac 83 \pi - 2\sqrt 3$, and the area of the original shared region is $\left( \frac 83 \pi - 2\sqrt 3 \right) - \pi = \boxed{ \frac 53 \pi - 2\sqrt 3 }$.

See also

2004 AMC 10B (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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