2004 AMC 10B Problems/Problem 25
Problem
A circle of radius is internally tangent to two circles of radius at points and , where 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?
Solution
The area of the small circle is . 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 and be the intersections of the two large circles. Connect them to and to get the picture below:
Now the triangles and are equilateral with side .
Take a look at the bottom circle. The angle is , hence the sector is of the circle. The same is true for the sector of the bottom circle, and sectors and 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. These triangles have a base of and a height of .
Hence the area of the new shaded region is , and the area of the original shared region is .
See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
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