TITLE: Octahedra with four equilateral faces
QUESTION [5 upvotes]: 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.

REPLY [0 votes]: Without loss of generality, let me assume the lengths are scaled so the shortest is 1. I'll label the four lengths $(1, a, b, c)$ withere $1 \leq a \leq b \leq c$.
Then the condition $c < a + 1$ is sufficient to ensure all the triangle inequalities are met.
Approaching it a different way from Intelligenti pauca, I came to a similar obstruction for $(1, 1, 1.8, 1.8)$. If you consider the wedge formed by two equilateral triangles of side length 1.8 meeting at a point, joined by two isosceles triangles with side lengths 1.8, 1.8, and 1, the base is a 1 by 1.8 rectangle; the two equilateral triangles of side length 1 cannot reach across it (since the height of the triangle is $\sqrt{3}/2 \approx 0.87$).
This also fails Intelligenti pauca's condition, which we can rephrase by saying that when we unfold the three non-equilateral triangles adjacent to the equilateral triangle of side length 1, the angles in the gaps must be greater than $60^\circ$.
Thus the sums of the angles in the two non-equilateral triangles at a vertex must be less than $240^\circ$.
Let me crudely edit Intelligenti pauca's picture:

The marked angle must be greater than $60^\circ$, so $\beta+ \gamma$ must be less than $240^\circ$.
If $a = 1$ and $b = c = 1.8$ as in my example, then the angles $\beta$ and $\gamma$ are about $128.3^\circ$, so we see the criterion fails.
To apply this more generally, the law of cosines tells us the angle $\beta$ is $\cos^{-1}(\frac{1+a^2-b^2}{2a})$ and $\gamma$ is $\cos^{-1}(\frac{1+a^2-c^2}{2a})$.
Since $\gamma \geq \beta$, if $\beta$ is more than $120^\circ$ we already know the situation is impossible. Hence
$$
\frac{1+a^2-b^2}{2a} > \frac{-1}{2}, \text{ or } \  a^2 + a - b^2 > -1.
$$
This fails for $1^2 + 1 - 1.8^2 = -1.24$ in my example, but not for
$1.1^2 + 1.1 - 1.6^2 = -0.25$ in Intelligenti pauca's example.
If $\gamma$ is more than $120^\circ$, we can't automatically say if the total angle is too big without knowing $\beta$.
Using the idenity $\cos^{-1}(A) + \cos^{-1}(B) = \cos^{-1}(AB - \sqrt{(1 - A^2)(1-B^2)})$,
we get the condition $AB - \sqrt{(1 - A^2)(1-B^2)} < \frac{-1}{2}$ where $A = \frac{1+a^2-b^2}{2a}$ and $B = \frac{1+a^2-c^2}{2a}$.
We can rearrange that condition to
$$
A^2 + AB + B^2 < \frac{3}{4}.
$$
In terms of the triangle side lengths $a$, $b$, and $c$, that expands to the truly horrendous expression
$$
3 a^4 + 3a^2 + b^4 + b^2c^2 +c^4 + 3 < 3a^2b^2 + 3a^2c^2 + 3b^4 + 3c^2.
$$
When combined with $a + 1 > c \geq b \geq a \geq 1$, this is a necessary and sufficient condition for the four lengths to appear as sides of equilateral triangles on an octahedron.
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