2004 AMC 10B Problems/Problem 25
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?
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:
We can see that the triangles and are both equilateral with side .
Take a look at the lower circle. The angle is , thus sector is of the circle. The same is true for sector of the lower circle, and sectors and of the upper 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. The area of an equilateral triangle given its side, is
Therefore, the area of the new shaded region is . Lastly, we must subtract the area of the circle that we added earlier, , and we get
Solution 2 (Out Of Time)
Notice that answer choices (C), (D), and (E) are exactly more than answer choices (A), (B). This suggests that answer choices (C), (D), and (E) are meant to trick you into forgetting to subtract the circle with radius (area ) From this, (A) or (B) are probably the answer choices (since you do need to subtract at the end!) Since the angle measures of the sectors are it only makes sense for the answer to have a and not a in them (There's no angle either). So, our answer is .
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