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, $$MT^2 = MN \cdot MS = \frac{CM}{3} \cdot \frac{2\,CM}{3} = \frac{2}{9}CM^2, \qquad CR^2 = CS \cdot CN = \frac{2}{9}CM^2,$$ 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 $MT^2 = \frac{2}{9}CM^2:$ $$\frac{(a - 10)^2}{4} = \frac{a^2 - 100}{9} \quad\Longrightarrow\quad 9(a - 10) = 4(a + 10) \quad\Longrightarrow\quad a = 26.$$

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