Answer: 389

Solution:

Let $x$ be the common radius, and let $O_A,$ $O_B,$ $O_C$ be the centers of $\omega_A,$ $\omega_B,$ $\omega_C.$ Each is at distance $x$ from two sides of the triangle, so each lies on an angle bisector, and the sides of triangle $O_A O_B O_C$ are parallel to those of $ABC$ at distance $x.$ Hence $O_A O_B O_C$ is the image of $ABC$ under the homothety centered at the incenter $I$ with ratio $\frac{r - x}{r},$ where $r$ is the inradius; in particular its circumradius is $R \cdot \frac{r - x}{r},$ where $R$ is the circumradius of $ABC.$

The center of $\omega$ is at distance $2x$ from each of $O_A,$ $O_B,$ $O_C$ (externally tangent equal circles), so it is the circumcenter of $O_A O_B O_C$ and $$2x = R \cdot \frac{r - x}{r}.$$ For the $13$-$14$-$15$ triangle, $s = 21$ and Heron's formula gives area $\sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84,$ so $r = \frac{84}{21} = 4$ and $R = \frac{13 \cdot 14 \cdot 15}{4 \cdot 84} = \frac{65}{8}.$

Then $2x = \frac{65}{8} \cdot \frac{4 - x}{4}$ gives $64x = 260 - 65x,$ so $x = \frac{260}{129}.$ Since $129 = 3 \cdot 43$ shares no factor with $260,$ the answer is $m + n = 260 + 129 = 389.$