Place $B = (0,0),$ $C = (56,0),$ $A = (28,96);$ both centers lie on angle bisectors, and $\tan\frac{B}{2} = \tan\frac{C}{2} = \frac{3}{4}$

The centers are $\left(56 - \frac{64}{3},\, 16\right)$ and $\left(\frac{4q}{3},\, q\right);$ set the distance between them equal to $q + 16$

Place $B = (0, 0)$ and $C = (56, 0);$ the altitude from $A$ has length $\sqrt{100^2 - 28^2} = 96,$ so $A = (28, 96).$ Then $\sin B = \frac{24}{25},$ $\cos B = \frac{7}{25},$ and $$\begin{aligned} \tan\frac{B}{2} &= \frac{\sin B}{1 + \cos B} \\ &= \frac{3}{4} = \tan\frac{C}{2}. \end{aligned}$$ A circle of radius $r$ tangent to $\overline{BC}$ and to a slanted side has its center on the bisector from that base vertex, at height $r$ and horizontal distance $\frac{r}{\tan(\frac{C}{2})} = \frac{4r}{3}$ from the vertex. Thus $P = \left(56 - \frac{64}{3},\, 16\right)$ and $Q = \left(\frac{4q}{3},\, q\right),$ where $q$ is the radius of circle $Q.$

External tangency means $PQ = q + 16:$ $$\begin{aligned} &\left(\frac{104 - 4q}{3}\right)^2 + (16 - q)^2 \\ &= (16 + q)^2. \end{aligned}$$ Since $(16 + q)^2 - (16 - q)^2 = 64q,$ this becomes $(104 - 4q)^2 = 576q,$ i.e. $(26 - q)^2 = 36q,$ which simplifies to $q^2 - 88q + 676 = 0,$ so $q = 44 \pm 6\sqrt{35}.$

The root $44 + 6\sqrt{35} \approx 79.5$ would make circle $Q$ extend outside the triangle, so $q = 44 - 6\sqrt{35}.$ Here $m = 44,$ $n = 6,$ and $k = 35 = 5 \cdot 7,$ giving $m + nk = 44 + 210 = 254.$