2004 AMC 10B Problems/Problem 25

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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?

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

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:

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 $4\cdot \left( \frac 16 \cdot \pi\cdot 2^2 \right) - 2 \cdot \left( \frac 12 \cdot 2 \cdot \frac{2\sqrt{3}}2 \right) = \frac 8...$ , and the area of the original shared region is $\left( \frac 83 \pi - 2\sqrt 3 \right) - \pi = \boxed{ \frac 53 \pi - 2\sqrt 3 }$ .

2004 AMC 10B (Problems • Resources)
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2004 AMC 10B Problems/Problem 25

From AoPSWiki

Problem

Solution

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