Counting tetrahedrons inside a dodecahedron

Question

The regular dodecahedron has $20$ vertices, and $12$ faces which are (congruent) regular pentagons; three pentagons meet at a vertex.

Fixing a vertex $v$ of dodecahedron, we count the other vertices, whose minimum distance from $v$ is $3$. One can see the Schlegel diagram of the dodecahedron here.

There are exactly $6$ vertices at distance $3$ from $v$: call them $x,x',y,y',z,z'$.

With figure below (Schlegel diagram), we can see that we can partition these six vertices into two sets, such that vertices in each set are at distance $3$ from each other.

Thus, from the vertex $v$, we produced $2$ tetrahedrons, with one vertex common.

But, I was unable to see from this construction, why do we get only $10$ regular tetrahedrons inside regular dodecahedron? Can anybody explain it?

(Note: I know some other way - going via cubes, and then tetrahedrons; but, here I am trying to understand it from the Schlegel diagram, for better understanding of the diagram and explanation.)

why this question came? There are $20$ vertices in Dodecahedron, and tetrahedron has $4$ vertices, so if we want to get $10$ tetrahedrons, I was unable to see how the repetition of vertices is coming in tetrahedrons while distributing vertices of dodecahedron.

Answer 1

For any vertex $v$, we find $2$ regular tetrahedrons and therefore we get a total number of $20\cdot 2=40$ tetrahedrons. But in this way, each tetrahedron is counted $4$ times (recall that a tetrahedron has 4 vertices). Hence there are just $40/4=10$ different regular tetrahedrons.

Answer 2

The only regular tetrahedra in the dodecahedron occur for choices of vertices in the Schlegel diagram for which the distance between any two tetrahedral vertices is $3$. It is not difficult (using a vertex coloring argument) to show, for example, if $v$ is chosen as a tetrahedral vertex (WLOG), then the only other candidate vertices are $(x, x', y, y', z, z')$. But since the tetrahedral graph is $3$-regular, these six choices must correspond to at most two valid tetrahedra, since these must be partitioned into two triplets.

So if $v$ admits exactly two tetrahedra, and there are $20$ dodecahedral vertices, then there are $20 \times 2 / 4 = 10$ distinct tetrahedra: $20$ by permuting dodecahedral vertices, $2$ for two distinct tetrahedra per vertex, then divided by $4$ because each tetrahedron has four vertices.