What follows is my solution. There are other ways to get the correct answer. All of them, I think, will involve finding areas of little parts of the figure, and adding them up cleverly.

To make things easier to talk about, I'm going to define two areas, A and B:

Area A	Area B

The area of one of the circles is Pi. The total area will be 3 Pi - 3 A + B. When we add the areas of the three circles, we've double-counted each area A, and then when we subtract one of each, we've over-subtracted area B, so we have to add that one back in.

Finding Area B:
The center of Area B is an equilateral triangle of side length 1. Using the formula for the area of a triangle and the Pythagorean theorem, we get that the area of this triangle is sqrt(3)/4.

Area = sqrt(3)/4

Area B also contains a sector of a circle. Because we know that the angle is 60 degrees, we know its area is 1/6 the area of the circle, or Pi/6.

Area = Pi/6

Now we can find the area of the little piece shown below, which will be very useful. Its area will be the difference of the areas we've found so far. We will call this area C

Area of a little piece C = Pi/6-sqrt(3)/4

Now simple addition gives us area B

Area B = Pi/6 + 2C

Finding Area A:
Area A also contains a sector. This one has an angle of 2 * 60 = 120 degrees, or one-third of the circle. So its area is Pi/3.

We can now use simple addition to find area A:

Area B = Pi/3 + 2C

Putting it all together:

Our total area is 3 PI - 3A + B. Using the above formulas gives us our total of 3Pi / 2 + sqrt(3)

Former Iowa representative Leonard L. Boswell has a name that anagrams to "new or old labels"

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