imc

The formula for the area of a circle is $\pi r^2$ so the radius of this circle is $\sqrt{156}.$ Because $OA=4\sqrt{3}=\sqrt{48}<\sqrt{156},A$ must be in the interior of circle $O.$ Let $s$ be the side length of the triangle, the unknown value, and let $X$ be the point on $BC$ where $OX\perp BC.$ Since $△ABC$ is equilateral, $BX=\frac{s}{2}$ and $AX=\frac{s\sqrt{3}}{2}.$ We are given $AO=4\sqrt{3}.$ Use the Pythagorean Theorem and solve for $s.$ $$\begin{array}{rl}(\sqrt{156}{)}^2& ={\left(\frac{s}{2}\right)}^2+{\left(\frac{s\sqrt{3}}{2}+4\sqrt{3}\right)}^2\\ 156& =\frac{1}{4}{s}^2+\frac{3}{4}{s}^2+12s+48\\ 0& =s^2+12s-108\\ 0& =(s-6)(s+18)\\ s& =\boxed{6}\end{array}$$