It was good to meet you today (26 August 2015)!  I hope the following
will be a do-able hw project for the "4-day weekend" that will help
you get familiar with Platonic polyhedra and configurations of points
on the unit 2-sphere S^2 in R^3.  [Follow-up 02 September 2015: please
write up what you can to turn in at class on September 09!]

PROJECT 0.  Explore the geometry of the tetrahedron (self-dual, where
"duality" exchanges vertices for faces), cube, octahedron (dual to
cube), dodecahedron (which you guys built - and 4-colored - today!)
and icosahedron (dual to dodecahedron) scaled so that their vertices
are configurations of points on the unit 2-sphere S^2 in R^3
(i.e. collections of unit vectors).  In particular, try working out
the spherical distances (i.e. angles), or the cosine of those angles
(i.e. dot products), for distinct pairs of points in each such
configuration.  (The dodecahedron and icosahedron may be very tricky,
so don't worry if you get stuck....*)

			*	*	*

For the tetrahedron with one vertex at (0, 0, 1) we found the others
must be of the form (*, *, -1/3) and so the cosine of any pair (why
"any" and not "some"?) is -1/3.

If you use another picture of the tetrahedron, with vertices at the 4
"red" corners of the cube 

(1, 1, 1), (1, -1, -1), (-1, 1, -1), (-1, -1, 1)

(alternating with the 4 "blue" corners - recall these are dual to the
faces of the "red/blue" octahedron we also built today), and again
scale the cube vertices to make them unit vectors, you should get the
same result: cosine (or dot product) -1/3.

Of course, this means that for the cube itself some of the cosines
will also be -1/3, but there will also other cosines for the cube.

From this, we could in principle work out the Coulomb (or other
reciprocal power-law) energy of these configurations explicitly - all
will be critical configurations, but not all are minima - please
ponder which are minima and which aren't?

[*You can think of the dodecahedron as a cube with six "hip roofs"
erected on its faces: the "roof ridges" on opposite cube faces are
parallel, and exactly one choice for the location of each ridge makes
the resulting pentagonal face regular.

The wikipedia web-page

https://en.wikipedia.org/wiki/Dodecahedron 

gives an explicit formulation for this "hip-roofs-on-a-cube" picture
of the regular pentagonal dodedecadron, in fact a "movie" of all the
"pyritohedra" (irregular dodecahedra with "pyrite-crystal symmetry")
interpolating between a cube (each face decorated with extra edge and
vertices to make degenerate pentagonal dodedecadron) and a rhombic
dodecahedron (I've edited the wikipedia text slightly to jibe with my
terminology):

The coordinates of the eight vertices of the original cube are:

(±1, ±1, ±1)

The coordinates of the 12 vertices of the ridges are:

(0, ±(1 + h), ±(1 − h^2)) 

(±(1 + h), ±(1 − h^2), 0) 

(±(1 − h^2), 0, ±(1 + h)) 

where h is the height of the hip roof above the faces of the cube.

[We found that the determinant of the 3x3 matrix whose rows were the
vectors u, v, w gotten by subtracting (1, 0, 1) from three nearby
vertices ((1 − h^2), 0, (1 + h)), ((1 + h), -(1 − h^2), 0) and 
((1 +h), (1 − h^2), 0) must vanish, and thus these seemingly curious 
choices of coordinates means the pentagonal faces are not "bent" along
the cube edges; in other words, for all choices of h in the open interval
(0,1), the resulting polyhedron really has 12 faces, which are non-
regular (generally, but see below for the special value of h which makes 
them regular) pentagons; the (generic) polyhedron in this family has 
"pyritohedral" symmetry (the symmetry of the cube decorated with the 3 
pairs of extra edges, slightly smaller than the cube symmetry group).]

When h = 1, the six ridges degenerate to points and a rhombic
dodecahedron is formed.  When h = 0, the ridges are absorbed in the
faces of the cube, and the pyritohedron reduces to a cube.  When h =
(√5 − 1)/2, the inverse of the golden ratio

	      	  	φ := (√5 + 1)/2, 

the result is a regular dodecahedron.

(Of course, remember to rescale at the end to make these 20 vertices
into unit vectors!)

			*	*	*

And for the icosahedron (and its relatives) here's a nice way to
work out the coordinates of its 12 vertices:

Consider the vectors defined by all the possible cyclic permutations
and sign-flips of coordinates of the form (2, 1, 0).  These coordinates
represent the truncated octahedron with alternated vertices deleted.

This construction is called a snub tetrahedron in its regular
icosahedron form, generated by the same operations carried out
starting with the vector (φ, 1, 0), where φ is the golden ratio

	      	  	φ := (√5 + 1)/2

- see https://en.wikipedia.org/wiki/Icosahedron and also John Baez
(September 11, 2011) http://math.ucr.edu/home/baez/golden.html  ]