jmc

Consider a cross-section of the cone that passes through the apex of the cone and the center of the circular base. It looks as follows: Let $O$ be the center of the sphere (or the center of the circle in the cross-section), let the triangle be $△ABC$, so that $D$ is the midpoint of $BC$ and $A$ is the apex (as $△ABC$ is isosceles, then $\overline{AD}$ is an altitude). Let $P$ be the point of tangency of the circle with $\overline{AC}$, so that $OP\perp AC$. It follows that $△AOP\sim △ACD$. Let $r$ be the radius of the circle. It follows that $$\frac{OP}{AO}=\frac{CD}{AC}⟹OP\cdot AC=AO\cdot CD.$$We know that $CD=12$, $AC=\sqrt{12^2+24^2}=12\sqrt{5}$, $OP=r$, and $AO=AD-OP=24-r$. Thus, $$12r\sqrt{5}=12(24-r)=12^2\cdot 2-12r⟹12r(1+\sqrt{5})=12^2\cdot 2.$$Thus, $r=\frac{24}{1+\sqrt{5}}$. Multiplying the numerator and denominator by the conjugate, we find that $$r=\frac{24}{1+\sqrt{5}}\cdot \frac{\sqrt{5}-1}{\sqrt{5}-1}=\frac{24(\sqrt{5}-1)}{5-1}=6\sqrt{5}-6.$$It follows that $a+c=\boxed{11}$.