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So this answer uses a trick, which some may consider a loophole, in fact it uses two tricks. So I'm going to explain how it works.
If the OP wants to patch this approach out of the challenge, I will delete my answer.
The exact definition of a polyhedron varies from person to person. Some circumstances require that a polyhedron be finite and not self intersect, this gives 5 regular polyhedra, some allow self intersection but still require finiteness, this gives 9 regular polyhedra, some times Euclidean tilings are considered as well giving 12.
However in the beginning of the 20th century Petrie and Coxeter discovered 3 regular polyhedra that hadn't been considered before. These were infinite polyhedra with flat non-intersecting faces, however they didn't have insides, so they hadn't been considered before. Pictured below is a section of the mucube, one of the three:
This discovery lead to a new definition of polytope (and by extension a new definition of polyhedron). This definition was pretty much just based on vertices and edges. The higher dimensional elements no longer needed to have exact interiors. This means you can have the above polyhedron, which divides space into two symmetric portions, but also you can have polyhedra whose faces aren't planar.
For example, classically a polygon needs to be "flat" meaning all of its vertices lie in a plane. However skew polygons can have vertices which lie in higher dimensional space. For example here's a 4-dimensional regular polygon:
Image by Plasmath see this page for liscensing
(We will come back to this particular polygon later)
The same can also be done with polyhedra. You can have polyhedra whose vertices span 4 or greater dimensions.
Lots of great work would be done with this definition. I'd especially like to point out the work of McMullen and Schulte. They classified all the regular polyhedra in 3-space, which has been summarized in this excellent youtube video:
jan Misali, there are 48 regular polyhedra
This definition is the one I am using in this answer.
Now the challenge says:
Your task is to output the vertex coordinates of a regular dodecahedron. The size, orientation, and position of the dodecahedron are up to you, as long as it is regular.
So I just need to choose a regular dodecahedron. The dodecahedron I chose is the "Simplicial embedded dodecahedron". In general you can take any regular polyhedron with n vertices and create an embedding of it on a (n-1)-simplex. This embedding is still a regular polyhedron with full flag transitivity, but its vertices are that of the (n-1)-simplex.
So to output the vertices of the simplicial embedded dodecahedron, I need to output the vertices of the 19-simplex. At first glance I've just made my problem much harder. The vertices of the 19-simplex centered at the origin are nasty, way worse than the vertices of the convex regular dodecahedron.
However here's where the second trick comes in. The vertices of a regular n-simplex are actually really simple if you express them in n+1 dimensions. For example you can give the vertices of the triangle as:
This is a regular triangle. And likewise the vertices of the 19 simplex can be given as all permutations of \$(1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)\$
So in summary, I am giving the vertices for a 19 dimensional dodecahedron in 20 dimensional space.
I can't exactly render this in any way that is helpful. But its faces I can say its faces are the regular 4D pentagon shown earlier. If you want to try to imagine 12 of those in 20 dimensional space.