Smallest square to cover a right triangle I believe smallest possible triangle to cover a square is well studied... But how about square covering a triangle? I read the post here, but I think that I don't want such a general question (and complicated solution), just right triangle.
I first encounter the problem, which is, find the square with minimum length that covers a triangle with sides $5,12,13$ completely. The answer is a square with length$\frac{144}{\sqrt{193}}$ and I believe so, using the method here.
But the problem is, why is it the smallest possible square covering the triangle?
My attempts:


If a square covers the triangle and at the same time they have some vertex(s) coinciding:

If it is $\angle BAC$ touching vertex of square, see here, the square covering it should have side length of at least 12.

If it is $\angle BCA$ touching the vertex of square, see here, the square covering it should also have side length of at least 12.

This leaves us to the problem of $\angle ABC$, see here or below:
It is the smallest because whatever it is rotated clockwise/anti-clockwise through the origin, the side length of square will be larger.
For every square covering a triangle, there is a smaller square covering the same triangle while touching at least a vertex using its vertex(s). But how does it implies that the side length of square is at least $\frac{144}{\sqrt{193}}$? I mean, a square which touches the triangle with its vertex may have a shorter side length, but why is there no square with side length even shorter (doesn't touch the triangle using its vertex)? Why do we need only to consider these three cases?
Any help will be appreciate! Thank you.

This answer shows that $\frac{144}{\sqrt{193}}$ is the smallest possible side length.
(This is an answer to the question "why is it the smallest possible square covering the triangle?", but I have to say that this is not an answer to the question "Why do we need only to consider these three cases?". Anyway, I hope this helps.)

Let $A(0,0), B(5,12), C(5,0)$.
First, let us consider the condition that three points $A,B,C$ exist between two parallel lines $L_1$ and $L_2$ whose slope is $\tan\theta\ (0\lt \theta\lt\frac{\pi}{2})$ where the distance between $L_1$ and $L_2$ is $d$.
We may suppose that $L_1$ passes through $C$, so we may write
$$L_1 : y=\tan\theta\ (x-5),\qquad L_2 : y=\tan\theta\ x+\frac{d}{\cos\theta}-5\tan\theta$$
The condition that three points $A,B,C$ exist between $L_1$ and $L_2$ is 
$$0\le \tan\theta\times 0+\frac{d}{\cos\theta}-5\tan\theta\quad \text{and}\quad 12\le\tan\theta\times 5+\frac{d}{\cos\theta}-5\tan\theta,$$
i.e.
$$\sin\theta\le\frac d5\quad\text{and}\quad \cos\theta\le\frac{d}{12}\tag1$$

Next, let us consider the condition that three points $A,B,C$ exist between two parallel lines $L_3$ and $L_4$ whose slope is $\tan(\theta+\frac{\pi}{2})\ (0\lt \theta\lt\frac{\pi}{2})$ where the distance between $L_3$ and $L_4$ is $d$.
We may suppose that $L_3$ passes through $A$, so we may write
$$L_1 : y=\tan\left(\theta+\frac{\pi}{2}\right)x,\qquad L_2 : y=\tan\left(\theta+\frac{\pi}{2}\right)x+\frac{d}{\sin\theta}$$
The condition that three points $A,B,C$ exist between $L_3$ and $L_4$ is 
$$0\le \tan\left(\theta+\frac{\pi}{2}\right)\times 5+\frac{d}{\sin\theta}\quad \text{and}\quad 12\le\tan\left(\theta+\frac{\pi}{2}\right)\times 5+\frac{d}{\sin\theta},$$
i.e.
$$\cos\theta\le\frac d5\quad\text{and}\quad 12\sin\theta+5\cos\theta\le d\tag2$$

Finally, let us prove that if there exists a pair of $(d,\theta)$ satisfying both $(1)$ and $(2)$, then $d\ge \frac{144}{\sqrt{193}}$.
We have
$$\small\begin{align}&(1)(2)\\\\&\implies\cos\theta\le\frac{d}{12}\quad\text{and}\quad \sqrt{1-\cos^2\theta}\le\frac{d-5\cos\theta}{12}
\\\\&\implies \cos\theta\le\frac{d}{12}\quad\text{and}\quad 1-\cos^2\theta\le \frac{d^2-10d\cos\theta+25\cos^2\theta}{144}
\\\\&\implies \cos\theta\le\frac{d}{12}\quad\text{and}\quad 169\cos^2\theta-10d\cos\theta+\color{blue}{d^2-144}\ge 0
\\\\&\implies \frac{5d+12\sqrt{169-d^2}}{169}\le \cos\theta\quad\text{and}\quad \cos\theta\le\frac{d}{12}
\\\\&\left(\because\ \text{we already know that $d=\frac{144}{\sqrt{193}}\approx 10.4$ is possible, so we may suppose that $\color{blue}{d^2-144}\lt 0$}\right)
\end{align}$$
In order for such $\theta$ to exist, we have to have 
$$\begin{align}&\frac{5d+12\sqrt{169-d^2}}{169}\le \frac{d}{12}
\\\\&\implies\sqrt{169-d^2}\le\frac{109}{144}d
\\\\&\implies 169-d^2\ge 0\quad\text{and}\quad 169-d^2\le \left(\frac{109}{144}d\right)^2
\\\\&\implies 0\lt d\le 13\quad\text{and}\quad d\ge\frac{144}{\sqrt{193}}
\\\\&\implies \frac{144}{\sqrt{193}}\le d\le 13\qquad\blacksquare\end{align}$$
