Answer: 41

Solution:

Let $O$ be the common center and label the triangle $ABC$ with $OA = 3,$ $OB = 4,$ $OC = 5.$ Rotate the plane by $60^\circ$ about $A$ so that $B$ maps to $C,$ and let $P$ be the image of $O.$ Then triangle $AOP$ is equilateral, so $OP = OA = 3,$ and $PC,$ the image of $OB,$ has length $4.$

Triangle $OPC$ has sides $3,$ $4,$ $5,$ so $\angle OPC = 90^\circ.$ In the configuration giving the largest triangle, $O$ lies inside $ABC$ and $\angle APC$ $= \angle APO + \angle OPC$ $= 60^\circ + 90^\circ$ $= 150^\circ,$ so by the Law of Cosines $$\begin{aligned} s^2 &= AC^2 = 3^2 + 4^2 \\ &\quad {}- 2 \cdot 3 \cdot 4\cos 150^\circ \\ &= 25 + 12\sqrt{3}. \end{aligned}$$ (If $O$ lies outside the triangle, the triangle fits in a half-disk of radius $5,$ so its altitude is at most $5$ and $s^2 \le \frac{100}{3},$ which is smaller.)

The area is $\frac{\sqrt{3}}{4}\,s^2 = \frac{\sqrt{3}}{4}\left(25 + 12\sqrt{3}\right)$ $= 9 + \frac{25}{4}\sqrt{3},$ so $a + b + c + d$ $= 9 + 25 + 4 + 3$ $= 41.$