Triangle $ABC$ has $BC=20.$ The [incircle](https://artofproblemsolving.com/wiki/index.php/Incircle) of the triangle evenly [trisects](https://artofproblemsolving.com/wiki/index.php/Trisect) the [ median](https://artofproblemsolving.com/wiki/index.php/Median_of_a_triangle) $AD.$ If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not [ divisible](https://artofproblemsolving.com/wiki/index.php/Divisor) by the [ square](https://artofproblemsolving.com/wiki/index.php/Perfect_square) of a prime, find $m+n.$
Let $E$, $F$ and $G$ be the points of tangency of the incircle with $BC$, $AC$ and $AB$, respectively.  Without loss of generality, let $AC < AB$, so that $E$ is between $D$ and $C$.  Let the length of the median be $3m$.  Then by two applications of the [Power of a Point Theorem](https://artofproblemsolving.com/wiki/index.php/Power_of_a_Point_Theorem), $DE^2 = 2m \cdot m = AF^2$, so $DE = AF$.  Now, $CE$ and $CF$ are two tangents to a circle from the same point, so by the Two Tangent Theorem $CE = CF = c$ and thus $AC = AF + CF = DE + CE = CD = 10$.  Then $DE = AF = AG = 10 - c$ so $BG = BE = BD + DE = 20 - c$ and thus $AB = AG + BG = 30 - 2c$.
Now, by [Stewart's Theorem](https://artofproblemsolving.com/wiki/index.php/Stewart%27s_Theorem) in triangle $\triangle ABC$ with [cevian](https://artofproblemsolving.com/wiki/index.php/Cevian) $\overline{AD}$, we have
\[(3m)^2\cdot 20 + 20\cdot10\cdot10 = 10^2\cdot10 + (30 - 2c)^2\cdot 10.\]
Our earlier result from Power of a Point was that $2m^2 = (10 - c)^2$, so we combine these two results to solve for $c$ and we get
\[9(10 - c)^2 + 200 = 100 + (30 - 2c)^2 \quad \Longrightarrow \quad c^2 - 12c + 20 = 0.\]
Thus $c = 2$ or $= 10$.  We discard the value $c = 10$ as extraneous (it gives us a line) and are left with $c = 2$, so our triangle has area $\sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}$ and so the answer is $24 + 14 = \boxed{038}$.