Dart-Throwing David

Construct circular sectors of radius 2 from each of the vertices. If David throws a dart in the region of any of these these sectors (shaded red in the diagram), then David will not win. Thus, David wins if he throws a dart in any other part of the triangle (shaded green). The probability that David wins is the ratio of the green area to the area of the entire triangle.

By Heron's formula, the area of the entire triangle is: $\sqrt{(9)(4)(3)(2)} = \sqrt{216} = 6\sqrt{6}$ .

The area of the green part is the area of the entire triangle minus the area of the three red areas. Since the angles of a triangle sum to $180^{\circ}$ , the three sectors will form a half-circle with area $\frac{2^2 \pi}{2} = 2\pi$ . Thus, the area of the green part is $6\sqrt{6} - 2\pi$ .

Thus, the probability that David wins is: $\frac{6\sqrt{6} - 2\pi}{6\sqrt{6}} = \frac{18 - \sqrt{6} \pi}{18}$ . Thus, $a=18, b=6, c=18$ , and $a+b+c = \fbox{42.}$