Heron’s formula provides the area, \(A\), of a triangle from the lengths, \(a, b, c\), of its edges:
$$A = s (s-a)(s-b)(s-c) \qquad \text{where}\qquad s := \frac{1}{2}\left(a+b+c\right)$$
The Cayley-Menger determinant generalizes this formula and can provide the “content” of an any-dimensional simplex from the lengths of its edges, but I want something hedronometric.
We cannot expect to compute the volume, \(V\), of a tetrahedron from the areas, \(W, X, Y, Z\), of its faces. A tetrahedron has six degrees of freedom —for instance, you can leave any five edges as-is and still adjust the sixth— so that its nature cannot be captured by just four parameters. For a hedronometric “Heron’s formula” to exist, then, we have two avenues to explore:
- Introduce more faces
- Handle fewer tetrahedra
The “introduce more faces” route may sound ridiculous —a tetrahedron only has four faces!— but a hedronometrist knows that a tetrahedron has three pseudofaces, with areas \(H, J, K\), that can join the fun. This ostensibly gives us fully seven degrees of freedom, although the Sum of Squares identity imposes a dependency that reduces this to the proper six:
$$W^2 + X^2 + Y^2 + Z^2 = H^2 + J^2 + K^2$$
Lo and behold! We can derive a face-based volume formula,
$$\begin{align}81 V^4 &= 2 W^2 X^2 Y^2 + 2 W^2 Y^2 Z^2 + 2 W^2 Z^2 X^2 + 2 X^2 Y^2 Z^2 + H^2 J^2 K^2 \\[0.5em] &- H^2 \left( W^2 X^2 + Y^2 Z^2 \right) – J^2 \left( W^2 Y^2 + Z^2 X^2 \right) – K^2 \left( W^2 Z^2 + X^2 Y^2 \right)\end{align}$$
which I call “Pseudo-Heron”. The form of the right-hand side suggests that it’s the determinant of some Cayley-Menger-like matrix (and I have other reasons to suspect Cayley-Menger-like ties), but I have so far been unable to construct an appropriate matrix.
An Open Question here:
What is the Cayley-Menger-like matrix whose determinant gives the Pseudo-Heron volume formula?
(By the way: I won’t go into it here, but this formula bears some resemblance to the Third Law of Cosines for non-Euclidean tetrahedra.)
The other route to a hedronometric Heron’s formula —“handle fewer tetrahedra”— raises lots of possibilities, since there are lots of families of tetrahedra admitting just four degrees of freedom.
For instance, a “right-corner” tetrahedron —with three edges meeting at a right angles at a vertex opposite “hypotenuse-face” \(W\)— has pseudoface areas computable from face areas
$$H^2 = X^2 + Y^2 \qquad J^2 = Y^2 + Z^2 \qquad K^2 = Z^2 + X^2$$
so that the Pseudo-Heron formula collapses to a single term:
$$9 V^2 = 2 X Y Z$$
Nice. Now, we could present all sorts of contrived families with face-only volume formulas, but we want something fairly general. What’s a good choice?
Early on in my hedronometric studies, I fixated on a class of tetrahedron with a special property: opposite edges determined orthogonal vectors. The property simplified a number of equations and generally made things nicer, so I called such tetrahedra “perfect”. (The mathematical literature has the less-honorific term “orthogonal”.) As it turns out, perfect tetrahedra have an interesting conceptual tie to Heron’s original formula.
(Too-)Briefly: Heron’s formula for (non-obtuse) triangle area is equivalent to the Pythagorean Theorem for right corner tetrahedra; when you prop-up a triangle in 3-space, with one vertex on each coordinate axis (you can only do this with non-obtuse triangles), then that triangle is the hypotenuse-face of a tetrahedron with a right corner at the origin. We can express the lengths of that tetrahedron’s three perpendicular edges, and the areas of its leg-faces, in terms of the lengths of the edges about the original triangle, and then use the relation \(W^2 = X^2 + Y^2 + Z^2\) to get Heron’s formula (or vice-versa). Easy.
The point is that Heron’s area formula is an application of the tetrahedral Pythagorean Theorem when we can prop-up a triangle against the coordinate axes in 3-space. It’s perhaps not-unreasonable, then, to seek a Heron’s volume formula for tetrahedra we can prop-up against the coordinate axes in 4-space. In 4-space, any pair of coordinate axes —and thus, the plane containing them— is perpendicular to the remaining pair —and the plane containing them. A proppable tetrahedron has opposing edges in such perpendicular planes: the edges determine orthogonal vectors, so a proppable tetrahedron is perfect!
Perfect tetrahedra admit only four degrees of freedom: the original six are reduced by the conditions that opposite edges are perpendicular. (If two pairs are perpendicular, then the last pair must be as well. That’s why we lose only two degrees of freedom from perfection.) As such, their volume can be determined by the areas of their faces. Unfortunately, the resulting “Heron’s formula” for that volume is nowhere-near as clean at Heron’s triangle formula (or the Pseudo-Heron tetrahedron formula); at best, we can relate volume and face areas in an enormous polynomial … essentially a quartic in \(V^3\).
The “Heron Quartic” is too complicated to give here, so I’ll just defer discussion to my note (from 2005) covering all this stuff: “Heron-like Results for Tetrahedral Volume”.
The Open Question(s) here:
What is the best way to express the Heron Quartic, and what is it trying to tell us?