Given that $x, y,$ and $z$ are real numbers that satisfy:
\begin{align*} x &= \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}}, \\ y &= \sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}}, \\ z &= \sqrt{x^2 - \frac 1{36}}+\sqrt{y^2-\frac 1{36}}, \end{align*}
and that $x+y+z = \frac{m}{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime, find $m+n.$
Let $\triangle XYZ$ be a triangle with sides of length $x, y$ and $z$, and suppose this triangle is acute (so all [altitudes](https://artofproblemsolving.com/wiki/index.php/Altitude) are in the interior of the triangle).

Let the altitude to the side of length $x$ be of length $h_x$, and similarly for $y$ and $z$. Then we have by two applications of the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem) we that \[x = \sqrt{y^2 - h_x^2} + \sqrt{z^2 - h_x^2}\] 
As a [function](https://artofproblemsolving.com/wiki/index.php/Function) of $h_x$, the [RHS](https://artofproblemsolving.com/wiki/index.php?title=RHS&action=edit&redlink=1) of this equation is strictly decreasing, so it takes each value in its [range](https://artofproblemsolving.com/wiki/index.php/Range) exactly once.  Thus we must have that $h_x^2 = \frac1{16}$ and so $h_x = \frac{1}4$ and similarly $h_y = \frac15$ and $h_z = \frac16$.

The area of the triangle must be the same no matter how we measure it; therefore $x\cdot h_x = y\cdot h_y = z \cdot h_z$ gives us $\frac x4 = \frac y5 = \frac z6 = 2A$ and $x = 8A, y = 10A$ and $z = 12A$.  

The [semiperimeter](https://artofproblemsolving.com/wiki/index.php/Semiperimeter) of the triangle is $s = \frac{8A + 10A + 12A}{2} = 15A$ so by [Heron's formula](https://artofproblemsolving.com/wiki/index.php/Heron%27s_formula) we have \[A = \sqrt{15A \cdot 7A \cdot 5A \cdot 3A} = 15A^2\sqrt{7}\]
Thus, $A = \frac{1}{15\sqrt{7}}$ and $x + y + z = 30A = \frac2{\sqrt{7}}$ and the answer is $2 + 7 = \boxed{009}$.


The justification that there is an acute triangle with sides of length $x, y$ and $z$:
Note that $x, y$ and $z$ are each the sum of two positive [square roots](https://artofproblemsolving.com/wiki/index.php/Square_root) of real numbers, so $x, y, z \geq 0$.  (Recall that, by [AIME](https://artofproblemsolving.com/wiki/index.php/AIME) [ convention](https://artofproblemsolving.com/wiki/index.php/Mathematical_convention), all numbers (including square roots) are taken to be real unless otherwise indicated.)  
Also, $\sqrt{y^2-\frac{1}{16}} < \sqrt{y^2} = y$, so we have $x < y + z$, $y < z + x$ and $z < x + y$.  But these conditions are exactly those of the [triangle inequality](https://artofproblemsolving.com/wiki/index.php/Triangle_inequality), so there does exist such a triangle.