Answer: 38

Solution:

Let $M$ be the midpoint of $\overline{AB},$ and let the incircle cut median $\overline{CM}$ at $S$ and $N,$ with $CS = SN = NM = \frac{1}{3}CM.$ Let the incircle touch $\overline{AB}$ at $T$ and $\overline{AC}$ at $R.$ By Power of a Point, $$\begin{aligned} MT^2 &= MN \cdot MS \\ &= \frac{CM}{3} \cdot \frac{2\,CM}{3} \\ &= \frac{2}{9}CM^2, \\ CR^2 &= CS \cdot CN \\ &= \frac{2}{9}CM^2, \end{aligned}$$ so $MT = CR.$ Since $AR = AT$ (tangents from $A$), we get $AC = AR + RC$ $= AT + TM$ $= AM = 10.$

Write $a = BC$ and $s = \frac{20 + a + 10}{2} = 15 + \frac{a}{2}.$ The standard tangent length gives $AT = s - a,$ so $MT = AM - AT$ $= 10 - \left(15 - \frac{a}{2}\right)$ $= \frac{a - 10}{2},$ while the median length formula gives $CM^2 = \frac{2 \cdot 10^2 + 2a^2 - 20^2}{4} = \frac{a^2 - 100}{2}.$ Substituting into $MT^2 = \frac{2}{9}CM^2:$ $$\begin{aligned} \frac{(a - 10)^2}{4} &= \frac{a^2 - 100}{9} \\ &\quad\Longrightarrow\quad 9(a - 10) \\ &= 4(a + 10) \\ &\quad\Longrightarrow\quad a = 26. \end{aligned}$$

Then the sides are $20,$ $26,$ $10$ with $s = 28,$ and Heron's formula gives $$\begin{aligned} [ABC] &= \sqrt{28 \cdot 8 \cdot 2 \cdot 18} \\ &= \sqrt{8064} = 24\sqrt{14}, \end{aligned}$$ so $m + n = 24 + 14 = 38.$