Dichoral angle in 4D platonic solid from schlafli symbol

Question

Is there a way to find the dichoral angle between two cells of a 4D platonic solids solely from its schlafli symbol ?

I'm thinking of a trigonometric identity similar to the one for 3D platonic solids on this page: https://en.wikipedia.org/wiki/Platonic_solid $$ \sin{\theta\over 2} = \frac{\cos\left(\frac{\pi}{q}\right)}{\sin\left(\frac{\pi}{p}\right)} $$

I have found a table which gives these angles for the 4D platonic solids on this page : http://answers.google.com/answers/threadview/id/497054.html but nowhere have I found such a formula.

Answer 1

I get this formula:

$$\sin^2 \frac{\theta\{p,q,r\}}{2} = \frac{-\sin^2\dfrac{\pi}{p}\;\cos^2\dfrac{\pi}{r}}{\cos\left(\dfrac{\pi}{p}+\dfrac{\pi}{q}\right)\;\cos\left(\dfrac{\pi}{p}-\dfrac{\pi}{q}\right)} \tag{$\star$}$$ where $\theta\{p,q,r\}$ is the dichoral angle for the $\{p,q,r\}$ polychoron.

(Note: That negative sign cancels with $\cos(\pi/p+\pi/q)$, which is negative for all $(p,q)$ pairs under consideration.)

The formula is consistent with the values given in the extensive Google Answers entry by "mathtalk-ga".

$$\begin{array}{c|c|c|c} \text{polychoron} & \{p,q,r\} & \sin^2(\theta/2) & \theta \;\text{(á la mathtalk)} \\ \hline 4\text{-simplex} & \{3,3,3\} & 3/8 & \operatorname{arccos}\frac14 = 75.5\ldots^\circ \\ 4\text{-cube} & \{4,3,3\} & 1/2 & \phantom{2}\pi/2 = \phantom{1}90^\circ\\ 4\text{-orthoplex} & \{3,3,4\} & 3/4 & 2\pi/3 = 120^\circ\\ 24\text{-cell} & \{3,4,3\} & 3/4 & 2\pi/3 = 120^\circ \\ 120\text{-cell} & \{5,3,3\} &\frac18(5+\sqrt{5}) & 4\pi/5 = 144^\circ\\ 600\text{-cell} & \{3,3,5\} & \frac3{16}(3+\sqrt{5}) & 2 \operatorname{arctan}\left((2+\sqrt{5})\sqrt{3}\right) = 164.4\ldots^\circ \end{array}$$

Deriving the formula was a straightforward (albeit tedious and unenlightening) matter of establishing four-dimensional vectors $A$, $B$, $C$, $D$, subject to these constraints: $$|A| = |B| =|C|=|D| $$ $$\begin{align} \angle AOB = \angle COB = \angle DOB &= \theta\{p\}\phantom{,q} \quad(=\text{angle of $p$-gon}) \\ \angle ABC = \angle ABD &= \theta\{q\}\phantom{p,\,} \quad(=\text{angle of $q$-gon}) \\ \text{dihedral}\;\angle\;\text{of planes}\;ABC\;\text{and}\;ABD &= \theta\{q,r\} \quad(=\text{angle of $\{q,r\}$-hedron (see OP)}) \end{align}$$

Then I expressed $\cos\theta\{p,q,r\}$ in terms of the normals to hyperplanes $OABP$ and $OABQ$. A few passes with Mathematica's Resultant feature eliminated all the free variables, leaving only an expression in various $\theta$s, which I manipulated into $(\star)$. $\square$