8. In the tetrahedron $SABC$, the circumradii of $\triangle SAB$, $\triangle SBC$, and $\triangle SCA$ are all 108, and the incenter of the inscribed sphere is $I$, with $SI=125$. Let $R$ be the circumradius of $\triangle ABC$. If $R$ can be expressed in the form $\sqrt{\frac{m}{n}}\left(m, n \in \mathbf{Z}_{+}\right.$, and $(m, n)=1)$, then $m+n=$ $\qquad$
8. 35928845209 .

It is known that the tetrahedron $SABC$ is a regular triangular pyramid. Let the projection of point $S$ on $AB$ be $M$, and $SM=x$. Then 
\[
\left\{
\begin{array}{l}
x(216-x)=\left(\frac{\sqrt{3}}{2} R\right)^{2}, \\ 
\frac{x}{x+\frac{R}{2}}=\frac{125}{\sqrt{x^{2}-\left(\frac{R}{2}\right)^{2}}}.
\end{array}
\right.
\]
Solving this system of equations will yield the conclusion.