Krdytkyu

Let $A_1, A_2, A_3, A'_1, A'_2, A'_3$ be the vertices of a (not-necessarily convex) octahedron; here $X'$ is the vertex not on an edge with $X$. Suppose that the four non-adjacent triangular faces $A_1 A_2 A_3$, $A_1 A'_2 A'_3$,$A'_1 A_2 A'_3$,$A'_1 A'_2 A_3$ are equilateral with sides of lengths $x_0, x_1, x_2, x_3$.

I am interested in knowing the conditions on $x_0,x_1,x_2,x_3$ that must be satisfied for such a configuration to exist; and how unique the configuration is. This is a preliminary step to understanding this Question on Octahedra
and the answer by @Blue.

Clearly it is necessary that $x_i>0$ for each $i$; and by using the triangle inequality on the other four faces, that $x_i+x_j>x_k$ for every 3-set ${i,j,k}subset {0,1,2,3}$.

Question 1: are these conditions sufficient for existence? I think so, but have only a hand-waving argument and would like something better.

It seems to me that if such a configuration exists for some $x_0,x_1,x_2,x_3$ then in general if we fix $A_2, A_3, A'_1, A'_2, A'_3$ there are two points $A_1$ which satisfy the right conditions, giving a convex octahedron and a non-convex one. Again I can only wave my hands, and my intuition may be wrong. So,

Question 2: what can be proved about uniqueness?

Lastly, to get some feeling for which $(x_0,x_1, x_2,x_3)$ yield an octahedron, I would like to understand their geometry.

Question 3: Describe geometrically the (subset of) the solid bounded by

begin{align*}
x_0+x_1+x_2+x_3&=1;&\
x_i&=0, &i=0,dots,3; \
x_i+x_j&=x_k &textrm{ for every 3-set } {i,j,k}subset {0,1,2,3}
end{align*}

which yield solutions.

This is not an answer, but just a conjecture about Question 1.

Consider your octahedron "unfolded" on a plane, and in particular its equilateral face $ABC$ of side $x_0$ and its three neighbouring (not equilateral) faces $ABE$, $BCD$, $ACF$ (see diagrams below). We can imagine to "fold" these neighbouring faces (all on the same side of the equilateral face) until $DE=DB$, $DF=CF$ and $FE=AE$. We have then three equations for three unknown parameters (the positions of points $D$, $E$, $F$ along the three circles where they are constrained to lie).

But if you try that with GeoGebra, it turns out that the construction succeeds only if, in the unfolded planar graph, $DEge DB$, $DFge CF$ and $FEge AE$. Thus the first diagram shows a constructible octahedron, while the second one is not constructible, because $DE<DB$. I haven't a proof for that (yet) and would like to know other opinions on this conjecture.

A short comment on Question 3: I think the first condition can always be met by a suitable rescaling, the second one gives a single point and the third one (as written) implies the second one. But maybe I just don't understand what you are asking.