In hyperbolic geometry, exactly how big is a dodecahedron composed entirely of right angles?

Question

Specifically, I need to know the distance from the center to the vertices, and the distance to the faces.

Answer 1

One way to calculate this is by considering suitable triangles inside the dodecahedron and using hyperbolic trigonometry. This is almost the same as you could do in euclidean geometry, only you start with some triangle with fixed angles which fixes its lengths in hyperbolic geometry. It is a valid approach, however this probably leads to very ugly calculations so I'll only outline which triangles you consider. And I really hope someone finds a clever way to calculate this.

There is a triangle formed by the center $A$ of a dodecahedron, the center $B$ of a face and the center $C$ of an edge of the face. This triangle has one 90°-angle at $C$ and one 45°-angle at $B$. If we can calculate the distance $\ell$ between the two vertices of these angles then we can also calculate the other lengths in the triangle and one of these is the distance between the center and the center of a face.

To calculate the missing length $\ell$, consider the triangle that you obtain by connecting in one of the faces the center with a vertex and with the center of an adjacent edge. Call the vertex $D$ and take the center of the edge to be $C$ as before. This gives you a triangle with a 90°-angle at the center of the edge, a 45°-angle at the vertex and a 72°-angle at the center of the face. So you can calculate all the lengths in this triangle.

Now if you also want to calculate the distance from the center to the dodecahedron to the vertices, you consider the riangles $ABC$ and $ACD$ to get from $AB$ via $AC$ to $AD$.