[Math] Volume of an Irregular Octahedron from edge lengths

Question

Does anyone know how to calculate the volume of an irregular octahedron from the lengths of the edges?

The octahedron has triangular faces, but the only information are the edge lengths.

Alternatively, how might I calculate the length of a line drawn ( the yellow dashed line) between two opposing vertices? If this known it can be split into 4 irregular tetrahedra.

In the Image the different edge lengths are coour coded, i.e all green edges are length x, blue edges are length y…

Answer 1

Let $w$, $x$, $y$, $z$ be the edge lengths, and let $d$ be a body diagonal. Then $s:=d^2$ satisfies this quartic (barring typos):

$$\begin{align} 0 &= s^4 (-w + x + y + z) (w - x + y + z) (w + x - y + z)(w + x + y - z) \\[4pt] &- 12 s^3 (w x + y z)(w y + x z)(wz+x y) \\[4pt] &+2 s^2 \left(\begin{array}{c} \sigma_1^4 \sigma_2^2 - 4 \sigma_1^2 \sigma_2^3 - 2 \sigma_1^5 \sigma_3 + 8 \sigma_1^3 \sigma_2 \sigma_3 \\ + 8 \sigma_1 \sigma_2^2 \sigma_3 - 10 \sigma_1^2 \sigma_3^2 - 12 \sigma_2 \sigma_3^2 + 8 \sigma_1^4 \sigma_4 \\ - 44 \sigma_1^2 \sigma_2 \sigma_4 + 32 \sigma_2^2 \sigma_4 + 48 \sigma_1 \sigma_3 \sigma_4 - 32 \sigma_4^2 \end{array}\right) \\[4pt] &-4 s (w x + y z)(w y + zx)(wz+xy) \left( \begin{array}{c} 2 \sigma_1^4 - 8 \sigma_1^2 \sigma_2 + 3 \sigma_2^2 + 10 \sigma_1 \sigma_3 - 8 \sigma_4 \end{array}\right) \\[4pt] &-\left(\begin{array}{c} \sigma_1^4 \sigma_2^4 - 4 \sigma_1^2 \sigma_2^5 - 4 \sigma_1^5 \sigma_2^2 \sigma_3 + 16 \sigma_1^3 \sigma_2^3 \sigma_3 \\ + 8 \sigma_1 \sigma_2^4 \sigma_3 + 8 \sigma_1^4 \sigma_2 \sigma_3^2 - 64 \sigma_1^2 \sigma_2^2 \sigma_3^2 + 64 \sigma_1 \sigma_2 \sigma_3^3 \\ - 12 \sigma_3^4 + 8 \sigma_1^6 \sigma_2 \sigma_4 - 48 \sigma_1^4 \sigma_2^2 \sigma_4 + 64 \sigma_1^2 \sigma_2^3 \sigma_4 \\ - 16 \sigma_2^4 \sigma_4 + 64 \sigma_1^3 \sigma_2 \sigma_3 \sigma_4 - 64 \sigma_1 \sigma_2^2 \sigma_3 \sigma_4 - 24 \sigma_1^2 \sigma_3^2 \sigma_4 \\ - 32 \sigma_2 \sigma_3^2 \sigma_4 - 12 \sigma_1^4 \sigma_4^2 - 32 \sigma_1^2 \sigma_2 \sigma_4^2 + 64 \sigma_2^2 \sigma_4^2 \end{array}\right) \end{align} \tag{$\star$}$$ where the $\sigma_i$ are the symmetric polynomials $$\sigma_1 := w + x + y + z \qquad \sigma_2 := w x + w y + w z + x y + x z + y z$$ $$\sigma_3 := w x y + w x z + w y z + x y z \qquad \sigma_4 := w x y z$$

As a sanity check, when $w=x=y=z$ (ie, the octahedron is regular), equation $(\star)$ reduces to $$16 s w^4 (s-2 w^2)^3 = 0$$ with root $s = 2 w^2$ corresponding to $d = w \sqrt{2}$, as expected.

The discriminant of $(\star)$ is

$$\begin{align}\Delta = 4096&\cdot(w^2 - x^2)^2 (w^2 - y^2)^2 (w^2-z^2)^2 (x^2 - y^2)^2 (x^2-z^2)^2 (y^2 - z^2)^2 \\ &\cdot (w^2 - x^2 + y^2 - z^2 - w y - z x )^2 (w^2 - x^2 + y^2 - z^2 + w y + z x )^2 \\ &\cdot (w^2 + x^2 - y^2 - z^2 - w x - y z )^2 (w^2 + x^2 - y^2 - z^2 + w x + y z )^2 \\ &\cdot (w^2 - x^2 - y^2 + z^2 - w z - x y )^2 (w^2 - x^2 - y^2 + z^2 + w z + x y )^2 \end{align}$$

If $\Delta = 0$ (that is, when at least two edge-lengths match, or at least one of those other factors vanishes), then the polynomial has a multiple root. Otherwise, the polynomial has either four real, or else four non-real, roots; Descartes' Rule of Signs, and/or or the techniques described here could potentially help refine our understanding of the nature of the roots, but the coefficients are sufficiently complicated that a general sign analysis is difficult.

In the all-real-roots case, due to symbolic symmetry, the roots must correspond to the (squares of the) three "body diagonals" of the octahedron, and a spare. (Perhaps the last root is negative, and cannot be the square of a diagonal.)

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As for volume ... The complicated nature of the diagonal length doesn't inspire a great deal of hope for a nice answer. Nevertheless, I'm still crunching some symbols in Mathematica, and will update this answer later.

... well, it's later ...

So far, finding a polynomial involving volume and side-lengths has proven difficult. Intervening formulas are simply enormous (with $(\star)$ being svelte by comparison). With no final result to show, I'll give some formulas that may help others.

We can think of the solid as four tetrahedra joined along the yellow diagonal, $d$. Writing $V_w$ for the volume of the tetrahedron with an equilateral $w$-colored face, the Cayley-Menger determinant tells us that

$$\begin{align} 144\,V_w^2 &= w^2 \left(\; -w^4 - y^4 - z^4 - d^4 + w^2 y^2 + w^2 z^2 + w^2 d^2 + y^2 z^2 + y^2 d^2 + z^2 d^2\;\right) \\ 144\,V_x^2 &= x^2 \left(\; - x^4 - y^4 - z^4 - d^4 + x^2 y^2 + x^2 z^2 + x^2 d^2 + y^2 z^2 + y^2 d^2 + z^2 d^2 \;\right) \\ 144\,V_y^2 &= y^2 \left(\; -y^4-w^2-x^4-d^4 + y^2 w^2 + y^2 x^2 + y^2 d^2 + w^2 x^2 + w^2 d^2 + x^2 d^2 \;\right) \\ 144\,V_z^2 &= z^2 \left(\; -z^4-w^2-x^4-d^4 + z^2 w^2 + z^2 x^2 + z^2 d^2 + w^2 x^2 + w^2 d^2 + x^2 d^2 \;\right) \\ \end{align}$$ where I have taken the $w$- and $x$-colored equilaterals to share the vertex at one end of $d$, and the $y$- and $z$-colored equilaterals to share the vertex at the other end of $d$.

The octahedral volume ($V$) is, of course, the sum of these tetrahedral volumes, but the square roots prevent combination. Through a process of repeated squarings (or iterations of Mathematica's Resultant function), one gets a polynomial with $V$ and even powers of the $V_{-}$s:

$$\begin{align} 0 &= V^{16} \\ &- 8 V^{14} \sigma_1 \\ &+ 4 V^{12}\left(7\sigma_1^2-4\sigma_2\right) \\ &- 8 V^{10} \left(7 \sigma_1^3 - 12 \sigma_1 \sigma_2 + 16 \sigma_3 \right) \\ &+2 V^8 \left(35 \sigma_1^4 - 120 \sigma_1^2 \sigma_2 + 48 \sigma_2^2 + 256 \sigma_1 \sigma_3 - 1088 \sigma_4\right) \\ &-8 V^6 \left(7 \sigma_1^5 - 40 \sigma_1^3 \sigma_2 + 48 \sigma_1 \sigma_2^2 + 96 \sigma_1^2 \sigma_3 - 128 \sigma_2 \sigma_3 - 320 \sigma_1 \sigma_4\right) \\ &+4 V^4 \left(\begin{array}{c} 7 \sigma_1^6 - 60 \sigma_1^4 \sigma_2 + 144 \sigma_1^2 \sigma_2^2 - 64 \sigma_2^3 + 128 \sigma_1^3 \sigma_3 \\ - 512 \sigma_1 \sigma_2 \sigma_3 + 1024 \sigma_3^2 + 320 \sigma_1^2 \sigma_4 - 1792 \sigma_2 \sigma_4 \end{array}\right) \\ &-8 V^2 \left(\begin{array}{c}\sigma_1^7 - 12 \sigma_1^5 \sigma_2 + 48 \sigma_1^3 \sigma_2^2 - 64 \sigma_1 \sigma_2^3 + 16 \sigma_1^4 \sigma_3 \\ - 128 \sigma_1^2 \sigma_2 \sigma_3 + 256 \sigma_2^2 \sigma_3 + 192 \sigma_1^3 \sigma_4 - 768 \sigma_1 \sigma_2 \sigma_4 + 1024 \sigma_3 \sigma_4\end{array}\right) \\ &+\left(\sigma_1^4 - 8 \sigma_1^2 \sigma_2 + 16 \sigma_2^2 - 64 \sigma_4\right)^2 \end{align} \tag{$\star\star$}$$

Here, the $\sigma_i$ are also symmetric polynomials, but in the squares of tetrahedral volumes:

$$\sigma_1 := V_w^2 + V_x^2 + V_y^2 + V_z^2 \qquad \sigma_2 := V_w^2 V_x^2 + V_w^2 V_y^2 + V_w^2 V_z^2 + V_x^2 V_y^2 + V_x^2 V_z^2 + V_y^2 V_z^2$$ $$\sigma_3 := V_w^2 V_x^2 V_y^2 + V_w^2 V_x^2 V_z^2 + V_w^2 V_y^2 V_z^2 + V_x^2 V_y^2 V_z^2 \qquad \sigma_4 := V_w^2 V_x^2 V_y^2 V_z^2$$

That's all well and good, I suppose, but recall that $(\star\star)$ involves the diagonal length $d$. A full solution to the problem posed would require eliminating $d$ from $(\star\star)$, using $(\star)$. Although it's just a matter of hitting the equations once more with Resultant, the operation on polynomials this degree and size is quite computationally expensive. I've let my laptop chug away at the problem for a good while without completion; I may try again later.