Area of shaded region between three circles and a boundary

Extend the sides to make an equilateral triangle, and label it as follows:

Since $AB = AC = 1$ and $\angle BAC = 60°$ , $CQ = DR = \sqrt{3}$ , the area of $\triangle AQC$ is $A_{\triangle AQC} = \frac{\sqrt{3}}{2}$ , the area of sector $ABC$ is $A_{ABC} = \frac{1}{6}\pi$ , and the area of the blue region bounded by $BCQ$ is $A_{BCQ} = A_{\triangle AQC} - A_{ABC} = \frac{\sqrt{3}}{2} - \frac{1}{6}\pi$ .

Since $RQ = RD + DC + CQ = \sqrt{3} + 2 + \sqrt{3} = 2 + 2\sqrt{3}$ , the area of equilateral $\triangle PQR$ is $A_{\triangle PQR} = \frac{\sqrt{3}}{4}(2 + 2\sqrt{3})^2 = 6 + 4\sqrt{3}$ .

The area of the purple region is then $A_{\text{purple}} = A_{\triangle PQR} - 3A_{\text{circle}} - 6A_{BCQ} = (6 + 4\sqrt{3}) - 3\pi - 6 (\frac{\sqrt{3}}{2} - \frac{1}{6}\pi) = 6 + \sqrt{3} - 2\pi$ .

Therefore, $a = 6$ , $b = 3$ , $c = 2$ , and $a + b + c = \boxed{11}$ .

By joining up the centres of the circles, and dropping perpendiculars from each centre to the "outside" lines, we see that the whole region is comprised of

• an equilateral triangle of side length $2$ ,
• three $2 \times 1$ rectangles,
• three sectors of a unit circle, which together form a single circle

Thus the shaded region has area $\sqrt{3} + 3\times2 + \pi - 3\pi = 6 + \sqrt{2} - 2\pi$ , making the answer $6+3+2=\boxed{11}$ .