More fun with circles

First, we start by drawing an equilateral triangle connecting the centers of each circle. Then we find the area of the 60 degree sector of minor arc VX $\dfrac{6^2\pi(60)}{360}=6\pi$ .

Next we find the area of the sectors outside of the triangle but inside the intersection area. To do this, we subtract the area of the equilateral triangle from the sector that we just found and multiply it by three because there are three sections. $3(6\pi-3*3\sqrt{3}) \Rightarrow 18\pi-27\sqrt{3}$ .

The last step is to add the area of the equilateral triangle $9\sqrt{3}$ and our answer is $18\pi-18\sqrt3$ . Thus a=18, b=18, and c=3 so 18+18+3= $\boxed{39}$ .

The general formula for this problem (circles must intersect at centers or its much tougher)= $\boxed{\dfrac{r^2\pi-r^2\sqrt3}{2}}$ .

I think it would be great if all the problem solution writers on brilliant started to put a general formula at the bottom of their solution whenever possible (I know this isn't really possible for topics such as number theory, but it is quite often for geometry).

Let $A,B$ and $C$ be the centers of the three circles and $S$ be the area of the shaded region.

Connect the centers of the three circles to form an equilateral triangle. Observed that the shaded region is composed of three circular segments and one equilateral triangle.

$A_{segment}=A_{sector}-A_{triangle}=\dfrac{60}{360}\pi (6^2)-\dfrac{\sqrt{3}}{4}(6^2)=6\pi -9\sqrt{3}$

So the area of the shaded region is

$S=3(A_{segment})+A_{triangle}=3(6\pi -9\sqrt{3})+\dfrac{\sqrt{3}}{4}(6^2)=18\pi -27\sqrt{3}+9\sqrt{3}=18\pi -18\sqrt{3}$

The desired answer is $18+18+3=\boxed{39}$