Find the ratio between these areas.

Let the center of the circle be $O$ and its radius $r$ . Then the diagonal of the square is $2r$ and its side length is $\sqrt 2 r$ . The area of the square $A_\square = (\sqrt 2 r)^2 = 2 r^2$ .

Note that the center of the circle is also the centroid of the equilateral triangle. Then we have $OD:OB = 1:2$ , implying that the height of the triangle $h=BD = 3r$ and the side length of the triangle $a = 2h \tan 30^\circ = \dfrac {6r}{\sqrt 3} = 2\sqrt 3 r$ and the area of the triangle $A_\triangle = \dfrac {ah}2 = 3 \sqrt 3 r^2$ .

Therefore, the ratio $\dfrac {A_\square}{A_\triangle} = \dfrac {2r^2}{3\sqrt 3 r^2} = \dfrac 2{3\sqrt 3} = \dfrac {2\sqrt 3}9$ . Then $a+b+c = 2+3+9 = \boxed{14}$ .