Let $M$ be the midpoint of $\overline{AB}.$ Compute the power of $M$ and of $C$ with respect to the incircle: both come out to $\frac{2}{9}CM^2.$

Equal powers give equal tangent lengths from $M$ and $C,$ which forces $AC = AM = 10;$ then combine $MT = \frac{BC - 10}{2}$ with the median length formula

Let $M$ be the midpoint of $\overline{AB}.$ Relabel $A$ and $B$ if necessary so that the incircle’s point of tangency with $\overline{AB}$ lies on $\overline{AM}.$ Let the incircle cut median $\overline{CM}$ at $S$ and $N,$ with $CS = SN = NM = \frac{1}{3}CM,$ and 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 $$\begin{aligned} AC &= AR + RC \\ &= AT + TM \\ &= AM = 10. \end{aligned}$$

Write $a = BC$ and $s = \frac{20 + a + 10}{2} = 15 + \frac{a}{2}.$ The standard tangent length gives $AT = s - a,$ so $$\begin{aligned} MT &= AM - AT \\ &= 10 - \left(15 - \frac{a}{2}\right) \\ &= \frac{a - 10}{2}, \end{aligned}$$ while the median length formula gives $CM^2 = \frac{2 \cdot 10^2 + 2a^2 - 20^2}{4} = \frac{a^2 - 100}{2}.$ The triangle inequality gives $a \gt 20 - 10 = 10,$ so $a - 10$ is nonzero. 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 $m + n = 24 + 14 = 38.$