Space diagonals of a dodecahedron I have been studying on platonic solids for a while and figuring out properties of dodecahedron. A dodecahedron with sidelength $a$ has $60$ surface diagonals and $100$ space diagonals, $10$ being long, $30$ being medium, and $60$ being short. I have figured out length of long diagonal is $a\sqrt{1+\phi^{4}}$ and medium diagonal is $a\phi^{2}$ but I couldn't find the short diagonal.
 A: You actually are most of the way to getting the short space diagonals. Let A be any vertex of the regular dodecahedron and B be the opposite vertex. If C is any other vertex then $\triangle ABC$ is a right triangle with right angle at C because all vertices lie in a sphere with diameter $\overline{AB}$. If we choose C so that $\overline{AC}$ is a face diagonal whose length is $a\phi$ (diagonals of a regular pentagon with side length $a$), and if we use $a\sqrt{1+\phi^4}$ as the length of the hypotenuse, we are left with $a\sqrt{1-\phi^2+\phi^4}$ as the length of the short space diagonals.
If we plug in $\phi^2=1+\phi$ and $\phi^4=2+3\phi$ (shout-out to Fibonacci) we discover that in fact the short space diagonals measure $a\phi\sqrt2$ and the long ones simplify to $a\phi\sqrt3$, the ratios seeming as if we were dealing with squares and cubes rather than pentagons and dodecahedra. What gives?
The face diagonals of a regular dodecahedron have the property that if you select one from each of the twelve faces in the proper manner, you define the edges of a cube (see the drawing in heropup's answer). Then your short space diagonals of the dodecahedron are face diagonals of the cubes, each edge of which measures $a\phi$, and the long space diagonals of the dodecahedron are space diagonals of the cubes.
A: A regular dodecahedron can be placed in $\mathbb R^3$ such that the vertices are $$(\pm 1, 0, \pm \phi^2),$$ their circular permutations $(0, \pm \phi^2, \pm 1), (\pm \phi^2, \pm 1, 0)$, and the inscribed cube $(\pm \phi, \pm \phi, \pm \phi)$, where $\phi = (1 + \sqrt{5})/2$ is the golden ratio.  The circumradius is therefore $\phi \sqrt{3}$, and the dodecahedral edge length is simply $2$.  Then the face diagonals have length equal to the edge length of the cube, which is $2\phi$, the shortest interior diagonal between two vertices will be the distance between, for instance, $(-1, 0, \phi^2)$ and $(\phi^2, 1, 0)$, which is $2 \sqrt{3 + \sqrt{5}}$.  Alternatively, it can be computed as simply $\sqrt{2}$ times the length of the cube edge, i.e. $2 \sqrt{2} \phi$.  It is straightforward to compute all of the interior diagonal lengths and normalize these by the edge length, which I leave as an exercise.
I don't have time to create my own diagram, so I pulled one from the Internet.  This construction of the regular dodecahedron dates back to at least to the time of Euclid, as it appears in the Elements.
The ratio of the edge length of the cube to the edge length of the dodeachedron is $\phi$.

