Is This Enough Information?

We know that for any right triangle that the median to the hypotenuse is half of the length of the hypotenuse, so we get that $c=2\sqrt[3]{a^{3}b+b^{3}a}=2\sqrt[3]{ab}\sqrt[3]{a^{2}+b^{2}}$ .

By Pythagorean Theorem, $c^{2}=a^{2}+b^{2}$ , and so we know that $\sqrt[3]{a^{2}+b^{2}}=c^{\frac{2}{3}}$

Let the altitude to the hypotenuse be $h$ . By the basic formula for the area of a triangle, we get that $\frac{ab}{2}=A=\frac{ch}{2}$ and so we know that $ab=ch$ , and $\sqrt[3]{ab}=c^{\frac{1}{3}}h^{\frac{1}{3}}$

Substituting this all back in, we get that $c=2c^{\frac{1}{3}}h^{\frac{1}{3}}c^{\frac{2}{3}}$ , or $1=2h^{\frac{1}{3}}$ .

Solving for $h$ gives us $h=\frac{1}{8}$ , so $m+n=\boxed{9}$

Suppose vertex of triangle be ( $0,0$ and the other 2 vertex be $(a,0),(0,b)$ .

Now the mid point of hypotenuse is $(a/2,b/2)$ .

By distance formulae and using the above given only info as well as $c= \sqrt{a^{2} + b^{2}}$ ,

we get $c= 8ab$ .

Now by using perpendicular length formulae from $(0,0)$ to line

$x/a + y/b = 1$ .

We find

$h= 1/8$ .

Thus answer is

$\boxed {9}$ .