Unusual symmetries of the Cayley-Menger determinant for the volume of tetrahedra Suppose you have a tetrahedron $T$ in Euclidean space with edge lengths $\ell_{01}$, $\ell_{02}$, $\ell_{03}$, $\ell_{12}$, $\ell_{13}$, and $\ell_{23}$. Now consider the tetrahedron $T'$ with edge lengths
$$\begin{aligned}
\ell'_{02} &= \ell_{02} &
  \ell'_{13} &= \ell_{13}\\
\ell'_{01} &= s-\ell_{01} &
  \ell'_{12} &= s-\ell_{12}\\
\ell'_{23} &= s-\ell_{23}&
  \ell'_{03} &= s-\ell_{03}
\end{aligned}
$$
where $s = (\ell_{01} + \ell_{12} + \ell_{23} + \ell_{03})/2$.
If the edge lengths of $T'$ are positive and satisfy the triangle inequality, then the volume of $T'$ equals the volume of $T$. In particular, if $T$ is a flat tetrahedron in $\mathbb{R}^2$, then $T'$ is as well. This is easily verified by plugging the values $\ell'_{ij}$ above into the Cayley-Menger determinant.
In fact, it's possible to show that the linear symmetries of $\mathbb{R}^6$ that preserve the Cayley-Menger determinant form the Weyl group $D_6$, of order $2^5 * 6! = 23040$. This is a factor of $15$ times larger than the natural geometric symmetries obtained by permuting the vertices of the tetrahedron and negating the coordinates.
The transformations don't always take Euclidean tetrahedra to Euclidean tetrahedra, but they do sometimes. For instance, if you start with an equilateral tetrahedron $T$ with all side lengths equal to $1$, then $T'$ is also an equilateral tetrahedron. Thus if $T$ is a generic Euclidean tetrahedron close to equilateral, $T'$ will also be one, and $T$ and $T'$ will not be related by a Euclidean symmetry.
I can't be the first person to observe this. (In fact, I vaguely recall hearing about this in the context of quantum groups and the Jones polynomial.) What's the history? How to best understand these transformations (without expanding out the determinant)? Are $T$ and $T'$ scissors congruent? Etc.
 A: It may be worth noting that for some choices of six lengths that there exist as many as 30 inequivalent tetrahedra with these six lengths. 
A: These are the so-called Regge symmetries, described by T. Regge in a 1970-ish paper. For a bit on it, with references, see the paper
Philip P. Boalch, MR 2342290 Regge and Okamoto symmetries, Comm. Math. Phys. 276 (2007), no. 1, 117--130.\
A: The following comment in the question intrigued me:

In fact, it's possible to show that the linear symmetries of
$\mathbb{R}^6$ that preserve the Cayley-Menger determinant form the
Weyl group $D_6$, of order $2^5 * 6! = 23040$.

In particular, this suggests that it should be possible to perform a change of basis such that the Cayley–Menger determinant is a symmetric polynomial in the six variables, which is also invariant under even numbers of sign changes.
Some experimentation confirmed that the following variables work in the sense that the geometric and Regge symmetries all preserve the 12-element set $\{ \pm u, \pm v, \pm w, \pm x, \pm y, \pm z \}$:

*

*$u = \ell_{01} + \ell_{23}$;

*$v = \ell_{02} + \ell_{13}$;

*$w = \ell_{03} + \ell_{12}$;

*$x = \ell_{01} - \ell_{23}$;

*$y = \ell_{02} - \ell_{13}$;

*$z = \ell_{03} - \ell_{12}$;

More succinctly, the variables are the three sums of opposite edges and the three differences of opposite edges. The Cayley–Menger determinant can then be shown to be equal to the following 57-term polynomial:
$$ uvwxyz - \dfrac{1}{16} \sum_\text{sym}^{6} u^6 + \dfrac{1}{16} \sum_\text{sym}^{30} u^4 v^2 - \dfrac{1}{8} \sum_\text{sym}^{20} u^2 v^2 w^2 $$
where the ‘symmetric sums’ are over all distinct monomials obtained by permuting the six variables (and the number above the sum is the number of terms). This is visibly invariant under the Weyl group $D_6$ generated by arbitrary permutations and even numbers of sign changes.
A: To address the scissors congruence question at the end of the post: the Regge symmetries produce tetrahedra which are scissors congruent. This is proved in Section 6 (Theorem 9 and Corollary 10) of 


*

*J. Roberts. Classical $6j$-symbols and the tetrahedron. Geom. & Top. 3 (1999), pp. 21-66. (link to paper on arXiv)
The argument is indirect, proving that the Regge symmetries preserve volume and Dehn invariant, then deducing the statement on scissors congruence from Sydler's theorem.
The corresponding statement for hyperbolic tetrahedra was proved (by construction) in 


*

*Y. Mohanty. The Regge symmetry is a scissors congruence in hyperbolic space. Alg. Geom. Top. 3 (2003), 1-31. (link to paper on arXiv)
A: Let me give a geometric interpretation for the case of tetrahedra of volume zero. The statement becomes as follows:
Given four positive numbers $a,b,c,d$ that satisfy the quadrangle inequalities, denote
$$
s=\frac{a+b+c+d}2, \quad a'=s-a \text{ etc.}
$$
Take a quadrilateral with side lengths $a,b,c,d$ (in this cyclic order). If its diagonals have lengths $x,y$, then there exists a quadrilateral with side lengths $a',b',c',d'$ and the same diagonal lengths.
Again, this can be proved by a direct computation, but Arseniy Akopyan  noticed that this is equivalent to the Ivory lemma: diagonals in a curvilinear quadrilateral formed by four confocal conics are equal, see the left half of the picture.

Indeed, with two opposite vertices of an $(a,b,c,d)$-quadrilateral as the foci, draw two ellipses and two hyperbolas through the other two vertices. This gives a curvilinear quadrilateral whose diagonal is also a diagonal of our $(a,b,c,d)$-quadrilateral, see the right half of the picture. The relations between $a,b,c,d$ and $a',b',c',d'$:
$$
a+b=c'+d', c-d=d'-c' \text{ etc.}
$$
imply that the other two vertices of this curvilinear quadrilateral form together with the foci an $(a',b',c',d')$-quadrilateral.
As a side remark: I studied configuration spaces of planar quadrilaterals in this preprint, but didn't know about Regge symmetries. Thanks to Dylan and Igor for bringing this up!
There is an analog of the Ivory lemma in dimension three, but there is no word about volumes in it. Still, it would be interesting to look at the corresponding configuration of confocal surfaces (one of the families degenerates).
Also note that the Ivory lemma holds in the hyperbolic and spherical geometry as well.
