The angle "alpha" through which the 4 cubes must rotate in order to define a regular Dodecahedron is given by the equation:
where "p" is the Goldern ratio
alpha approximately equals 44.47751219 degrees.
One half of this value is approximately 22.2387561 degrees.
To derive angle alpha, I considered 3 vertices of the 120 Polyhedron.
These vectors all have the same magnitude given by
mag = mag(v3) = sqrt(p^4 + p^4 + p^4) = sqrt(3)p^2
The dot product v2 and v1:
v1.v2 = mag(v1)mag(v2)cos(theta) = p^4
(sqrt(3)p^2)(sqrt(3)p^2)cos(theta) = p^4
cos(theta) = 1/3
The projection of v2 onto v1:
A = mag(v2)cos(theta) = (sqrt(3)p^2)/3 = p^2 / sqrt(3)
The vector A defined to be in the direction of v1 and of
length A can be defined as
A = (0, p/3, (p^3/3))
Then the distance from A to v2 is
d = 2 sqrt(2/3) p^2
The distance from v2 to v3 is
mag(v2 - v1) = mag(p^2 - p^3, -p^2, p^2 - p)
which reduces to mag = 2p.
So we have a triangle with 2 sides equal to
d = 2 sqrt(2/3) p^2
and one side equal to
mag = 2p
With this information, we find the angle alpha to be
2 sqrt(2/3) p^2 sin(alpha / 2) = p
Solving for sin(alpha/2) we get