<s>[INST] A special case of the Fermat-Torricelli point in a triangle $\triangle{ABC}$ is an isosceles right triangle; its legs are of length $s = 30\sqrt{5}$ and hypotenuse is of length $30\sqrt{10}$. The Fermat-Torricelli point $P$ must be along the median from the vertex $C$ of the right angle of the triangle.
$\left\vert\overline{CP}\right\vert = x$, and $\left\vert\overline{AP}\right\vert = y = \left\vert\overline{BP}\right\vert$, and
\begin{equation*}
\mathrm{m}\angle{APC} = 180 - \theta = \mathrm{m}\angle{BPC} .
\end{equation*}
According to the Law of Cosines,
\begin{equation*}
s^{2} = x^2 + y^2 + 2xy\cos(\theta) .
\end{equation*}
According to the Law of Sines,
\begin{equation*}
y = \frac{s/\sqrt{2}}{\sin(\theta)} .
\end{equation*}
According to the Pythagorean Theorem,
\begin{equation*}
\left(s/\sqrt{2} - x\right)^2 + \left(s/\sqrt{2}\right)^2 = y^2 .
\end{equation*}
So,
\begin{equation*}
y = \sqrt{x^{2} - 30\sqrt{10} \, x + 4500} .
\end{equation*}
\begin{equation*}
\left\vert\overline{AP}\right\vert + \left\vert\overline{BP}\right\vert + \left\vert\overline{CP}\right\vert
= x + 2y
= x + 2\sqrt{x^{2} - 30\sqrt{10} \, x + 4500} .
\end{equation*}
The minimum value of this function is
\begin{equation*}
30\sqrt{5\left(2 + \sqrt{3}\right)}
\end{equation*}
at
\begin{equation*}
15\sqrt{10} - 5\sqrt{30} .
\end{equation*}
\begin{equation*}
\left\vert\overline{CP}\right\vert = x
= 15\sqrt{10} - 5\sqrt{30} \approx 20.05
\end{equation*}
and
\begin{equation*}
\left\vert\overline{AP}\right\vert = \left\vert\overline{AP}\right\vert
= y
= \frac{1}{2} \left(30\sqrt{5\left(2 + \sqrt{3}\right)}
- \left(15\sqrt{10} - 5\sqrt{30}\right)\right) \approx 54.77 .
\end{equation*}
My Concern
I am told that the distances between the vertices and the Fermat-Torricelli Point are natural numbers.
 [/INST] I've found no errors in your calculations.

If we can use the fact that the point $P$ satisfies 
$$\angle{APB}=\angle{BPC}=\angle{CPA}=120^\circ$$
then, we can easily see that 
$$|\overline{CP}|=x=15\sqrt{10}-5\sqrt{30},\qquad|\overline{AP}|=|\overline{BP}|=y=10\sqrt{30}$$
which are the same as what you've got.
</s>