Note: Hyperbolic Dodecahedra Ball Model

This is a computation of the hyperplanes that bound a regular dodecahedron with dihedral angle $\theta$ in the Poincare ball model, useful if you want to draw an extrinsic picture of such a dodecahedron.

Describing Faces Given Dihedral Angle

Geodesics in the ball model are either planes through the origin or spheres orthogonal to the bounding unit 2-sphere. We will center our dodecahedron at the origin of the ball so all bounding planes are segments of spheres, which can be represented by giving their (euclidean) center $c\in\mathbb{R}^3$ and their (euclidean) radius $r>0$. But as all these spheres must be orthogonal to the boundary of the ball model, these degrees of freedom depend on one another. We can make this relationship concrete by noticing that two circles of radii $r_1,r_2$ whose centers are distance $d$ apart intersect orthogonally if $r_1^2+r_2^2=d^2$, by the pythagorean theorem.

PICTURE

Thus, a circle centered at $c\in\mathbb{R}^2$ of radius $r$ is a hyperbolic geodesic if $$r^2+1=|c|^2$$ Thus, we can encode a geodesic plane by a unit vector $\hat{u}\in\mathbb{S}^2$ and a radius $r$, from which the center can be recovered as $c=\sqrt{r^2+1}\hat{u}$. For a hyperbolic dodecahedron, the unit vectors corresponding to face centers are parameterized by the vertices of a regular euclidean icosahedron. Connecting these two vertices to the center yields an icoseles triangle, where if the distance to the verticies is 1, the remaining edge length is $2/\sqrt{2+\phi}$.

PICTURE

To work with dodecahedra of dihedral angle $\theta$, we will need a generalization of this relationship for circles making angle $\theta$. This follows from the analogous diagram, replacing the pythagorean theorem with the Law of Cosines. One subtlety - the triangle involved does not itself have angle $\theta$, as that is the angle of tangents to the circles at their intersection, and the triangle is made of radii. A quick diagram reveals however that this new angle $\chi$ is just $\pi-\theta$:

DIAGRAM

Applying the law of cosines yields $r_1^2+r_2^2-2r_1r_2\cos\chi=d^2$, and as $\cos\chi=-\cos\theta$ this becomes

$$r_1^2+r_2^2+2r_1r_2\cos\theta = d^2$$

Now for a regular dodecahedron, the radii for all geodesic sides are equal, and this simplifies to $2r^2+2r^2\cos\theta = d^2$, or

$$2r^2(1+\cos\theta)=d^2$$

With $|c|$ as the distance of the centers from the ball model origin, we can draw a familiar triangle using two adjacent face centers: a triangle from the icosahedron. And we know its trigonometry: its base $d$ is $2/\sqrt{2+\phi}$ times the two equal heights. Finally, we already know $|c|^2=r^2+1$ (each geodesic plane is orthogonal to the boundary):

THREE PICTURES OF THESE THREE CASES

$$d^2=2r^2(1+\cos\theta)$$ $$d=\frac{2}{\sqrt{2+\phi}}|c|$$ $$|c|^2=1+r^2$$

Putting these three facts together lets us get a single equation for the missing radius $r$:

$$\frac{4}{2+\phi}(1+r^2)=2r^2(1+\cos\theta)$$

Solving for radius yields

$$r^2=\frac{\frac{2}{2+\phi}}{1+\cos\phi-\frac{2}{2+\phi}}$$

Simplifying a bit

$$r=\frac{1}{\sqrt{(2+\phi)\left(\frac{1+\cos\theta}{2}\right)-1}}$$

Using the half angle identity gives an alternative description

$$r=\frac{1}{\sqrt{(2+\phi)\cos^2\frac\theta 2-1}}$$

However this is often easier to work with and remember if we instead compute $1/r^2$:

$$\frac{1}{r^2}=(2+\phi)\cos^2\frac\theta 2 -1$$

Specific Dodecahedra

Plugging in different dihedral angles gives different dodecahedra, much like in our intrinsic computations.

Right-Angled

If $\theta=\pi/2$ then $\cos\theta = 0$ and

$$\begin{align} \frac{1}{r^2} &=(2+\phi)\frac{1+\cos\theta}{2}-1\
&=(2+\phi)\frac{1+0}{2}-1\
&=\frac{2+\phi}{2}-1\
&=\frac\phi 2 \end{align}$$

$$\implies r=\sqrt{\frac{2}{\phi}}$$

Seifert-Weber

Here the dihedral angle is $\theta=\cos\frac{2\pi}{5}$. Using the half angle identity it will prove useful to know $\cos\frac\pi 5=\frac\phi 2$. Plugging this in,

$$\begin{align} \frac{1}{r^2} &=(2+\phi)\cos^2\frac\theta 2-1\
&=(2+\phi)\cos^2\frac\pi 5-1\
&=(2+\phi)\frac{\phi}{2}-1 \end{align}$$

Using the arithmetic of the golden ratio, this simplifies as

$$(2+\phi)\frac{\phi}{2}-1=\frac{2\phi+\phi^2}{2}-1=\frac{3\phi+1}{2}-1=\frac{3\phi-1}{2}$$

Thus the radius is

$$r=\sqrt{\frac{2}{3\phi-1}}$$

Generalizing to Other Polytopes

The same reasoning generalizes directly to other regular polytopes (say, finding coxeter cubes, etc). Let $s$ be the distance between two adjacent face centers of a polytope (that is, the side length of its dual if the vertices are on the unit sphere): in the case above we had $s=\frac{2}{\sqrt{2+\phi}}$. Then our fundamental equation for radius remains unchanged:

$f^2(1+r^2)=2r^2(1+\cos\theta)$$

And we can solve this for radius, by multiplying out and collecting terms with $r$:

$$r^2=\frac{f^2}{2(1+\cos\theta)-f^2}$$

Or potentially simpler, using the half angle identity $\cos\frac\theta 2 = \sqrt{\frac{1+\cos\theta}{2}}$

$$r^2=\frac{f^2}{4\cos\frac\theta 2-f^2}=\frac{1}{\frac{4}{f^2}\cos^2\frac\theta 2-1}$$

$$\implies r = \frac{1}{\sqrt{\left(\frac{2}{f}\cos\frac\theta 2\right)^2-1}}$$