$\triangle{ABC}$ 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.
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.