8. Given that the three sides of $\triangle A B C$ are $13, 14$, and $15$. Four circles $\odot \mathrm{O}, \odot \mathrm{O}_{1}, \odot \mathrm{O}_{2}$, and $\odot O_{3}$ with the same radius $r$ are placed inside $\triangle A B C$, and $\odot O_{1}$ is tangent to sides $A B$ and $A C$, $\odot O_{2}$ is tangent to sides $B A$ and $B C$, $\odot O_{3}$ is tangent to sides $C B$ and $C A$, and $\odot O$ is tangent to $\odot O_{1}, \odot O_{2}, \odot O_{3}$. Then $r=$ $\qquad$
8. $\frac{260}{129}$.

Let's assume $a=13, b=14, c=15$.
It is easy to know that $\triangle A B C \sim \triangle \mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3}$, and $O$ is the circumcenter of $\triangle O_{1} O_{2} O_{3}$, with the circumradius being $2 r$.
Then $\cos \angle O_{1} O_{3} O_{2}=\cos C=\frac{5}{13}$,
$$
\sin \angle O_{1} O_{3} O_{2}=\frac{12}{13} \text {. }
$$

By the Law of Sines, we get
$$
O_{1} O_{2}=4 r \sin \angle O_{1} O_{3} O_{2}=\frac{48 r}{13} \text {. }
$$

Similarly, $\cos A=\frac{3}{5}, \sin A=\frac{4}{5}$,
$$
\begin{array}{l}
\cos B=\frac{33}{65}, \sin B=\frac{56}{65} . \\
\text { Also, } O_{1} O_{2}=15-r \cot \frac{A}{2}-r \cot \frac{B}{2} \\
=15-r \cdot \frac{1+\cos A}{\sin A}-r \cdot \frac{1+\cos B}{\sin B} \\
=15-\frac{15}{4} r, \\
\text { then } \frac{48 r}{13}=15-\frac{15 r}{4} \Rightarrow r=\frac{260}{129} .
\end{array}
$$