Shortest path #1

Consider a regular hexahedron like in the image on the left.

It's clear that $ABCD$ is a regular tetrahedron, as all edges of it are of the same length.

Then we can see that the minimum length of $\overline{PQ}$ would be the height of the hexahedron, as it literally cannot be shorter. (Look at the top and the bottom faces of the hexahedron.)

So, $P$ and $Q$ would be arranged like above, as midpoints of $\overline{AB}$ and $\overline{CD}.$

Since the edge length of the hexahedron is $\dfrac{6}{\sqrt{2}}=3\sqrt{2},$ (look at the top or the bottom face of the hexahedron again.)

the height of the hexahedron is also $m=3\sqrt{2}.$

$\therefore~m^2=\left(3\sqrt{2}\right)^2=\boxed{18}.$

Let's place the tetrahedron in the $xyz$ space. By simple trigonometric considerations, the vertices came to be $B(-3,0,0)$ , $C(3,0,0)$ , $D(0,6\frac{\sqrt{3}}{2},0)$ and $A(0,\sqrt{3},\sqrt{24})$ . Let's now consider the vector $\overline{DC}$ : its direction is given by the line $y=3\sqrt{3}-\sqrt{3}x$ (that could be written as a parametric curve $(h,3\sqrt{3}-\sqrt{3}h,0)$ ). So, $P$ is, as a locus of points,

$\displaystyle Q=\{(x,y,z)\in \mathbb{R^3} \hspace{3pt} | \hspace{3pt} x=h,\hspace{3pt} y=3\sqrt{3}-\sqrt{3}h,\hspace{3pt} z=0, \hspace{10pt} h\in \mathbb{R}, \hspace{3pt} 0 \leq h \leq 3 \}$

With the same procedure we find $P$ :

$\displaystyle P=\{(x,y,z)\in \mathbb{R^3} \hspace{3pt} | \hspace{3pt} x=3t-3,\hspace{3pt} y=\sqrt{3}t,\hspace{3pt} z=\sqrt{24}t, \hspace{10pt} t\in \mathbb{R}, \hspace{3pt} 0 \leq t \leq 1 \}$

Now we can define $||\overline{PQ}||=d(t,h)$ as:

$\displaystyle d(t,h)=\sqrt{(3t-3-h)^2+(\sqrt{3}t+\sqrt{3}h-3\sqrt{3})^2+24t^2}$

Let's solve the minimization problem:

$\displaystyle \nabla d(t,h)=\overline{0} \hspace{5pt} \Longrightarrow \hspace{5pt} \begin{cases} \displaystyle \frac{\partial d(t,h)}{\partial t}=0 \\[3pt] \displaystyle \frac{\partial d(t,h)}{\partial h}=0 \end{cases} \hspace{5pt} \Longrightarrow \hspace{5pt} \begin{cases} \displaystyle -3(3t-3-h)-\sqrt{3}(\sqrt{3}t+\sqrt{3}h-3\sqrt{3})-24t^2=0 \\[3pt] \displaystyle 3(3t-3-h)-\sqrt{3}(\sqrt{3}t+\sqrt{3}h-3\sqrt{3})=0 \end{cases} \hspace{5pt} \Longrightarrow \hspace{5pt} \begin{cases} \displaystyle t=\frac{1}{2} \\[3pt] \displaystyle h=\frac{3}{2} \end{cases}$

Eventually,

$\displaystyle d\left(\frac{1}{2}, \frac{3}{2}\right)=3\sqrt{2}=m, \hspace{10pt} m^2=18$ .