What is the radius of a regular right-angled dodecahedron in $\mathbb{H}^3$?

Question

What is the distance from the center to a vertex of a regular right-angled dodecahedron in $\mathbb{H}^3$? A right-angled dodecahedron is a regular dodecahedron with all dihedral angles equal to $\frac{\pi}{2}$; e.g., see Example 5 in Chapter 13 of Ratcliffe's Foundations of Hyperbolic Manifolds. The video Not Knot features some true perspective views of regular right-angled dodecahedra.

Answer 1

The edges of the dodecahedron are the sides of the dodecahedron's faces. These are regular right-angled pentagons which sit in their own copies of $\mathbb{H}^2$. A right-angled pentagon consists of $5$ isosceles triangles having angles $\frac{2\pi}{5}$, $\frac{\pi}{4}$, $\frac{\pi}{4}$. Theorem 3.5.4. (The Second Law of Cosines) of Ratcliffe's book says, "If $\alpha$, $\beta$, $\gamma$ are the angles of a hyperbolic triangle and $a$, $b$, $c$ are the lengths of the opposite sides, then $$\cosh c = \frac{\cos \alpha \cos \beta + \cos \gamma}{\sin \alpha \sin \beta}."$$ From this we determine that the length of a side of a right-angled pentagon, and so that of an edge of the right-angled dodecahedron, is $\mathrm{acosh}(1 + 2 \cos(\frac{2 \pi}{5}))$.

Now we can apply the hyperbolic law of sines to determine the distance $R$ from the center of the dodecahedron to a vertex. I.e. Theorem 3.5.2. (Law of Sines) of Ratcliffe's book says, "If $\alpha$, $\beta$, $\gamma$ are the angles of a hyperbolic triangle and $a$, $b$, $c$ are the lengths of the opposite sides, then $$\frac{\sinh a}{\sin \alpha} = \frac{\sinh b}{\sin \beta} = \frac{\sinh c}{\sin \gamma}."$$ To apply this we just need to know the angle subtended by an edge of the right-angled dodecahedron, which is the same angle as in the Euclidean case of the angle subtended by an edge of a regular dodecahedron. This MSE post calculates this to be $\mathrm{acos}\frac{\sqrt{5}}{3}$. Applying the hyperbolic law of sines to the right triangle formed by bisecting the isosceles triangle with one vertex at the center of the dodecahedron and base side equal to an edge, we find (with near miraculous simplification) $$R = \mathrm{asinh} \frac{\sinh(\frac{1}{2} \mathrm{acosh}(1 + 2 \cos(\frac{2\pi}{5})))}{\sin(\frac{1}{2} \mathrm{acosh}(\frac{\sqrt{5}}{3}))} = \mathrm{asinh}(\frac{1}{2}\sqrt{3(1 + \sqrt{5})}) \approx 1.22645687.$$