jmc

Consider point $A$ at the center of the diagram. Drawing in lines as shown below divides the region into 3 parts with equal areas. Because the full circle around point $A$ is divided into 3 angles of equal measure, each of these angles is 120 degrees in measure. Now consider a circle of radius 4 inscribed inside a regular hexagon: Now, the pieces of area inside the hexagon but outside the circle are identical to the pieces of area the original region was divided into. There were 3 pieces in the original diagram, but there are 6 in the hexagon picture. Thus, the area of the original region is the half the area inside the hexagon but outside the circle.

Because $ABO$ is equilateral, $BMO$ is a 30-60-90 right triangle, so $BM=\frac{4}{\sqrt{3}}$. Thus, the side length of the equilateral triangle is $AB=2BM=\frac{8}{\sqrt{3}}$. Now we know the base $AB$ and the height $MO$ so we can find the area of triangle $ABO$ to be $\frac{1}{2}\cdot \frac{8}{\sqrt{3}}\cdot 4=\frac{16}{\sqrt{3}}=\frac{16\sqrt{3}}{3}$. The entirety of hexagon $ABCDEF$ can be divided into 6 such triangles, so the area of $ABCDEF$ is $\frac{16\sqrt{3}}{3}\cdot 6=32\sqrt{3}$. The area of the circle is $\pi 4^2=16\pi$. Thus, the area inside the heagon but outside the circle is $32\sqrt{3}-16\pi$. Thus, the area of the original region is $\frac{32\sqrt{3}-16\pi }{2}=16\sqrt{3}-8\pi$.

Now we have $a=16$, $b=3$ and $c=-8$. Adding, we get $16+3+(-8)=\boxed{11}$.