If the edges of a regular pentagon are 1, then the diagonals (any vertex to any vertex) is going to be phi, the golden mean. What's golden about it? Self similarity, early fractal: the shorter part to the longer part is the same ratio as the longer part to the whole. You can solve it algebraically, with the quadratic equation. Maybe play some Gilbert and Sullivan for background music, if you're a major general wannabee.
So where does the regular pentagon fit in? As the 12 faces of the pentagonal dodecahedron, for one. In alchemical literature, this was often Universe itself, with the other four Platonics serving to model fire, water, air and earth, nevermind in what order for now. The idea was these elements get shaken and stirred, resulting in all the compounds we take for granted. This was a primitive theory, but prefigured what we'd later teacher as the Periodic Table of Elements (expanded to 92, plus the transuranics).
The pentagonal dodeca has a dual, the icosahedron, that's also Platonic. In general the 5 Platonics work like this, under the operation of "make my dual":
P. Dodeca <--> Icosa Octa <--> Cube Tetra <--> Tetra
In other words, the tetrahedron is dual to itself. We may choose to define an implicit inside-outing between the tetrahedron and its dual, such that the expansion outward to the "bigger" volumes happens within left- and right-handed schema, which we might label Positive and Negative respectively (or vice versa).
In one of the dynamic models, the icosahedra occur near the winged extremes, around a twist point (modeled by the cuboctahedron in Fuller's jitterbug projections), after which the pendulum swings back towards zero again (the tetrahedron).
In cartoon form we might show a tiger, leaping back and forth through a hoop (an inside-outing point), perching successively on two stands, on the left and on the right. I've had this one on file for some years, shared it with Applewhite and Ken Snelson as I recall, and on Synergetics-L.
left .......... right tiger --->x<--- tiger icosa --------- icosa
Anyway, all dynamism and circus acts aside, you've got the two tetrahedra nesting in the cube as face diagonals (stella octangula), with the latter's edges bisecting the octahedron's, to make 12 criss-crosses of short and long diagonals (cube edge and octa edge respectively). These 12 crosses define the facets of another dodecahedron, the rhombic one, so fascinating to Kepler, a space-filler.
These rhombic dodecahedra define the voronoi cells of CCP spheres, i.e. balls closest packed in a tetrahedral stack, as with cannon balls or oranges in the supermarket.
[icosahedron phase fits here, dual to pentagonal dodeca]
20 = Cuboctahedron (dual to rhombic dodeca, 12-around-1)
Of course CCP is closely related to HCP, same density (~0.71) and to an infinite number of same-density Barlow packings. You can achieve these through lamination, e.g. of triangles atop traiangles (squares atop squares just nets you the CCP as you don't have the two choices -- more on this buried in geometry-research, in my dialog with John Conway (not to be confused with Damian Conway, at least as brilliant, but in a different way)).