A geometry problem by Benjamin Jones

The total purple area may be found as the "difference" between the areas of two squares of the radii 2+2sgrt2 and 4, because the inside purple area when equally decomposed fills eight white segments of the circles. You will then immediately get the correct answer.

The area can be broken into 3 parts. The area in the corners of ABCD, the area in the pedal curve between the circles, and the area tangent to the square between circles.

Take circle M, and isolate the top left corner. it forms a square with diagonal AM, side length sqrt(2). the area of the square is 2, and the sector separating this is one fourth the area of one circle, or pi/2. Thus, the shaded area in each of the corners is 2- pi/2. There are 4 corners, so those areas occupy 4(2-pi/2) = 8-2 pi.

The pedal curve is more complicated. Half of the area in each pedal is a portion of a sector of the circle. This portion is pi/2 radians, so the sector has area pi/2. Triangle MLQ has area 1, so the area of half of a pedal is pi/2 - 1. The area of each pedal is then double, so pi-2. There are 4 pedals, so the total pedal area is 4 pi -8.

Lastly the area between two circles. Isolating half of this, we look at region formed by the triangle M(3,6), F(4,7), and (3,7), and the rectangle F(4,7), (3,7),(3,7+sqrt(2), and (4,7+sqrt(2)). The sector of circle M in this is pi/4 radians, with area pi/4. the polygonal region has area of the triangle (1/2) plus the area of the rectangle (sqrt(2) - 1). Subtracting the area of the sector from the polygonal region, we have (sqrt(2)-1/2) - (pi/4). There are 8 of these regions, occupying a total of (8sqrt(2) - 4 - 2pi).

The sum of the three basic regions is (8-2pi) + (4pi-8) + (8sqrt(2) - 4 - 2pi) = 8sqrt(2)-4.