## A Hedronometric Theorem of Menger

In 1928, Karl Menger outlined necessary and sufficient conditions for a set of edge lengths to determine an actual, non-degenerate, tetrahedron. The conditions amount to dead-simple sanity checks that the consequent face areas and volume have to be positive real numbers.

In the short note “A Hedronometric Theorem of Menger”, I derive (as the title suggests) a hedronometric counterpart of Menger’s result, giving necessary and sufficient conditions for a set of face and pseudoface areas to determine a tetrahedron. It’s really about time I did.

The note came about when I observed a “Heronic duality” that I had never noticed (nor even suspected) before. Writing “$$[\bullet]$$” for what I’ll call the Heronic product,

$$[x,y,z] := \left(x+y+z\right)\left(-x+y+z\right)\left(x-y+z\right)\left(x+y-z\right)$$

we know that the area, $$A$$, of a triangle with edge lengths $$x, y, z$$ is given by

$$16 A^2 = [x,y,z]$$

That’s Heron’s formula (hence, the term “Heronic product”), and it gives us areas from edge lengths.

As it turns out, Heron products give us tetrahedral edge lengths from face and pseudoface areas (and volume); for instance, if $$a$$ is the edge between faces $$Y$$ and $$Z$$, and thus also a diagonal of pseudoface $$H$$, then

$$9V^2 a^2 = \left[H,Y,Z\right]$$

Going back the other way, we can even express the areas of tetrahedral pseudofaces as Heronic products:

$$16 H^2 = \left[a,d,j\right] \qquad \text{where} \qquad j^2 = a^2-b^2+c^2+d^2-e^2+f^2$$

The Heronic product is quite the workhorse!

This note answers the an Open Question from my “Heron-like Results …” post, where I wrote that I had “other reasons to suspect Cayley-Menger-like ties” to the Pseudo-Heron volume formula. There, I wondered what matrix had that formula as its determinant; in the note, that matrix is $$\Delta^\star$$. (There may be a better one, but that’s the best I have right now.)

Update. I’ve added a Corollary that gives conditions for just six faces —three standard, three pseudo— to determine a tetrahedron.

Posted 24 November, 2012 by in Hedronometry, Open Question