With two Heron-like formulas for tetrahedral volume in Euclidean space, it makes sense to investigate what happens in non-Euclidean —specifically, hyperbolic— space. After all, a number of hedronometric formulas (for instance, the Laws of Cosines) have counterparts in both geometries.

Unfortunately, the search for a hyperbolic hedronometric formula for volume is hindered by the fact that the most-accessible general formula (due to Derevnin and Mednykh) for hyperbolic volume is a thorny integral parameterized by the dihedral angles \(A,B,C,D,E,F\) (with \(A, B, C\) meeting at a vertex and opposing, respectively, \(D, E, F\)):

$$V = -\frac{1}{4}\int_{\theta_0-\phi}^{\theta_0+\phi}\log \frac{\cos\frac{A+B+C+\theta}{2} \cos\frac{A+E+F+\theta}{2} \cos\frac{D+B+F+\theta}{2} \cos\frac{D+E+C+\theta}{2}}{\sin\frac{A+D+B+E+\theta}{2} \sin\frac{A+D+C+F+\theta}{2} \sin\frac{B+E+C+F+\theta}{2} \sin\frac{\theta}{2}} d\theta $$

where \(\theta_0\) and \(\phi\) themselves are given by elaborate expressions in the dihedral angles.

The good news is that, because the integral is parameterized by dihedral angles, and because dihedral angles are related neatly to face and pseudoface areas by a Law of Cosines

$$\cos\frac{Y}{2} \cos\frac{Z}{2} + \sin\frac{Y}{2} \sin\frac{Z}{2} \cos A = \cos \frac{H}{2} = \cos\frac{W}{2} \cos\frac{X}{2} + \sin\frac{W}{2}\sin\frac{X}{2} \cos D$$

with dihedrals \(A, D\) between respective face pairs \(\{Y,Z\}\) and \(\{W,X\}\), we’re *pretty close* to Pseudo-Heronic formula: we just have to solve for the angles from the areas, and then plug the solutions in as the integral parameters. This turns out to be a symbolic nightmare (for the most part), so that expressing the Derevnin-Mednykh formula in terms of \(W, X, Y, Z, H, J, K\) seems pretty hopeless; but, we can, at least, do the solving and plugging-in *numerically* for tetrahedra we handle on a case-by-case basis. Of course, that amounts to a *strategy*, not a *formula*, for computing volume, but at this point, that’s the best I know how to do.

The **Open Question**:

What’s better than the best I know how to do?

Interestingly, there’s an aspect of tetrahedral volume analysis that has a very nice hedronometric counterpart. The determinant of (angular) Gram matrix of a tetrahedron,

$$G := \left[ \begin{array}{cccc}1 & -\cos A & -\cos B & -\cos F \\ – \cos A & 1 & -\cos C & -\cos E \\ -\cos B & -\cos C & 1 & -\cos D \\ -\cos F & -\cos E & -\cos D & 1 \end{array}\right]$$

is a fundamental quantity in any geometry. (Fun fact: in Euclidean, hyperbolic, and spherical space, the determinant’s value is respectively zero, non-positive, or non-negative.) The quantity actually figures into the Derevnin-Mednykh formula. Hedronometrically, *it factors!* Specifically, in hyperbolic space:

$$-4 \overline{W_2}^2\overline{X_2}^2\overline{Y_2}^2\overline{Z_2}^2\cdot \det G = \begin{array}{c}\left( \ddot{H_2} + \ddot{J_2}\ddot{K_2} – \ddot{W_2}\ddot{X_2} – \ddot{Y_2} \ddot{Z_2}\right) \\ \cdot\left( \ddot{J_2} + \ddot{K_2}\ddot{H_2} – \ddot{W_2}\ddot{Y_2} – \ddot{Z_2} \ddot{X_2}\right) \\ \cdot \left( \ddot{K_2} + \ddot{H_2}\ddot{J_2} – \ddot{W_2}\ddot{Z_2} – \ddot{X_2} \ddot{Y_2}\right) \end{array}$$

Here, I’m using my “Morse code” shorthand for sine and cosine:

$$\overline{W_2} := \sin\frac{W}{2} \qquad \ddot{W_2} := \cos\frac{W}{2}$$

Even better, one can show that, for a *perfect* tetrahedron (where opposite edges are perpendicular), the three factors on the right are nicely *proportional* in a way that defines a symmetric value, \(M\),

$$\begin{align}

\ddot{H_2} \left( \ddot{H_2} + \ddot{J_2}\ddot{K_2} – \ddot{W_2}\ddot{X_2} – \ddot{Y_2} \ddot{Z_2}\right) &= \ddot{J_2} \left( \ddot{J_2} + \ddot{K_2}\ddot{H_2} – \ddot{W_2}\ddot{Y_2} – \ddot{Z_2} \ddot{X_2}\right) \\

&= \ddot{K_2}\left( \ddot{K_2} + \ddot{H_2}\ddot{J_2} – \ddot{W_2}\ddot{Z_2} – \ddot{X_2} \ddot{Y_2}\right) \\

&=: \ddot{H_2}\ddot{J_2}\ddot{K_2} + M

\end{align}

$$ so that we have

$$-4 \overline{W_2}^2\overline{X_2}^2\overline{Y_2}^2\overline{Z_2}^2\cdot \ddot{H_2}\ddot{J_2}\ddot{K_2} \cdot \det G = \left( \ddot{H_2}\ddot{J_2}\ddot{K_2} + M \right)^3$$

(In theory, a perfect tetrahedron is completely determined by its face areas, so we “could” eliminate references to \(H, J, K\) in this formula. However, the result here isn’t at all pretty.)

This factorization may (or may not) speak to a hedronometric dynamic locked deep within the Derevnin-Mednykh volume integral, but I suspect that the analysis requires more expertise in differential geometry than I possess at this time.

A complete discussion of all of this stuff —including an example deriving a hedronometric volume integral tailored to very specific types of tetrahedra— is in my note from 2009: “Heron-like Strategies for Non-Euclidean Tetrahedral Volume”.