Gorgonzola has explained how to carve a dodecahedron out of a cube. In that approach there is a dodecahedron sitting inside a cube. I'm going to show a different construction of the (regular) dodecahedron which starts from a cube and constructs a dodecahedron on the outside.

Begin with a cube with edges of length the the golden ratio, which I'll denote by t. Recall that

t2 - t - 1 =0 (*)
Why does the golden ratio come into this? Well if you think about a regular pentagon with edge length 1. Then the length of a diagonal is t. A diagonal of the pentagon is a straight line joining any two non-adjacent vertices. Here's a picture (with slightly odd scaling)
1
____
1 /    \  1
/______\
t
\    /
1  \  / 1
\/

The dodecahdron construction is quite simple. On each square face of the cube we put up a tent. The tent has four sides two of them are isosceles triangles and two of them are trapezoids. Here's a picture (not to scale) of the types of tent side indicating the edge lengths.

/\         _1__
1/  \1     1/    \1
/____\     /______\
t           t
Looking down on a face of the cube from above the tent looks like this
______
|\      /|
| \____/ |
| /    \ |
|/______\|
When we put up a tent on a face of the cube adjacent to the this one we do it so that the ridge of the new tent is perpendicular to the ridge of our exisiting tent. Here's a picture (to a slightly weird scale) of two adjacent faces of the cube indicating this.
______
|\      /|
| \____/ |  <--- cube face
| /    \ |
|/______\|
|        |
| \    / |
|  \  /  |
|   \/   |
|   |    |  <--- cube face
|   |    |
|   /\   |
|  /  \  |
| /    \ |
|________|
Notice that at the edge of the cube where the two tents meet we have a pentagon whose edges have length 1. It's clear now that we have indeed constructed a regular dodecahedron with a cube inscribed, except for one small detail. The pentagon
____
/    \
/______\

\    /
\  /
\/
has a fold line (along the edge of the cube) and we really should ensure that that the pentagon is not actually folded. In other words we have to check that the trapezoid and the isosceles triangle are in the same plane. Let a be the angle that the trapezoid makes with the cube face and let b be the angle that the isosceles triangle makes with the cube face. We must show that a+b=pi.

We need some more notation. Let x be the height of the tent and let y be the height of the isosceles triangle. So we have the triangles:

/|           /|           /|\
y / | x        / | x      1 / | \ 1
/  |         /  |         / y|  \
/b__|        /a__|        /___|___\

(t-1)/2       t/2          t/2 t/2
By the Pythagorean theorem we have
y2 + (t/2)2 = 1 and y2 = x2 + ((t-1)/2)2
If we eliminate y and use (*) we can deduce from these equations that t=1/2. It's now easy to compute that tan a = 1/t and tan b = 1/(t-1). So using (*) again we see that tan a.tan b = 1. It follows that a+b = pi, as required.

Thus we have constructed a dodecahedron with an inscribed cube. Clearly there is some choice here, there are five cubes inscribed in the dodecahedron.

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