If we inscribe a regular dodecahedron in the same cube we have 12 vertices at (±1, 0, ±u) and its even permutations, along with these eight: (±v, ±v, ±v).
We must use the fact that the distance of these two classes of vertex are equidistant from the origin:
3v2 = 1 + u2.
As above we must also equate two typical edge lengths: from (1, 0, −u) to (1, 0, u) and from (1, 0, u) to (v, v, v).
4u2 = (1−v)2 + (u−v)2 + v2
4u2 = 1 − 2v + v2 + u2 − 2uv + v2 + v2
3u2 − 3v2 + 2uv + 2v − 1 = 0
Slightly messy:
One face has vertex set {(1, −t, 0), (1, t, 0), (t, 0, 1)}.
The center of that face is ((t+2)/3, 0, ⅓).
Lets skip the ⅓'s and scale down later.
We have a face center at (t+2, 0, 1).
A neighboring face has vertex set {(1, −t, 0), (1, t, 0), (t, 0, −1)}.
The corresponding (scaled up) face center is (t+2, 0, −1).
The line between these two icosahedron face centers is an edge of the dual dodecahedron.
We want this edge to have a coordinate value 1.
We scale by
1/(t+2) = 1/((((√5) − 1)/2)+2) = 2/((√5) + 3)
= 2((√5) − 3)/(((√5) − 3)((√5) + 3))
= 2((√5) − 3)/(−4)
= (3 − √5)/2 = 0.3819660112501051 .
With this scaling the dodecahedron vertex at (t+2, 0, 1) moves to (1, 0, (3 − √5)/2) and we learn that u = (3 − √5)/2.
Another neighbor of the first face has vertices at {(0, 1, t), (1, t, 0), (t, 0, 1)}.
The scaled up center of that face is at (t+1, t+1, t+1).
We see that
v = (t+1)/(t+2) = ((√5) + 1)/((√5) + 3)
=((√5) − 3)((√5) + 1)/((√5) − 3)((√5) + 3))
= (2 − 2√5)/(−4) = ((√5) − 1)/2 = 0.6180339887498949 .
Our large dodecahedron has 12 vertices at (±1, ±(t+2), 0) including even permutations, and 8 more at (±(t+1), ±(t+1), ±(t+1)). The edges are all 2 long.
Scaled down by 1/(t+2) = (3 − √5)/2 we have 12 vertices at (±(3 − √5)/2, ±1, 0) and 8 more at (±t, ±t, ±t) where t=((√5) − 1)/2. The edges are all 3 − √5 long. (u = (3 − √5)/2 and v = t = ((√5) − 1)/2.)