A curious relation between angles and lengths of edges of a tetrahedron Consider a Euclidean tetrahedron with lengths of edges
$$
l_{12}, l_{13}, l_{14}, l_{23}, l_{24}, l_{34}
$$ 
and dihedral angles 
$$
\alpha_{12}, \alpha_{13}, \alpha_{14}, 
\alpha_{23}, \alpha_{24}, \alpha_{34}.
$$ 
Consider solid angles
\begin{split}
 &\Omega_1=\alpha_{12}+\alpha_{13}+\alpha_{14}-\pi \\
 &\Omega_2=\alpha_{12}+\alpha_{23}+\alpha_{24}-\pi \\
 &\Omega_3=\alpha_{13}+\alpha_{23}+\alpha_{34}-\pi \\
 &\Omega_4=\alpha_{14}+\alpha_{24}+\alpha_{34}-\pi \\
\end{split}
and perimeters of faces
\begin{split}
 &P_1=l_{23}+l_{34}+l_{24} \\
 &P_2=l_{13}+l_{14}+l_{34} \\
 &P_3=l_{12}+l_{14}+l_{24} \\
 &P_4=l_{12}+l_{23}+l_{13}. \\
\end{split}
Then the following cross-ratios are equal to each other:
$$
[e^{i\Omega_1}, e^{i\Omega_2}, e^{i\Omega_3}, e^{i\Omega_4}]=[P_1, P_2, P_3, P_4].
$$
Question: Is it known? I have found a proof of this statement (see here), but it involves quite tricky algebraic geometry. It will be very interesting to me to see a more elementary approach. 
Addition:
Similar statements hold in spherical and hyperbolic geometry.
For a spherical tetrahedron 
$$
[e^{i\Omega_1}, e^{i\Omega_2}, e^{i\Omega_3}, e^{i\Omega_4}]=[e^{iP_1}, e^{iP_2}, e^{iP_3}, e^{iP_4}].
$$
For a hyperbolic tetrahedron 
$$
[e^{i\Omega_1}, e^{i\Omega_2}, e^{i\Omega_3}, e^{i\Omega_4}]=[e^{P_1}, e^{P_2}, e^{P_3}, e^{P_4}].
$$
Addition 2:
One can prove a more general statement, which I formulate in the spherical case (but it is true in other geometries after appropriate modifications). There exists a $PSL_2(\mathbb{C})-$ transformation, sending eight numbers
$$
1, e^{i\Omega_1}, e^{i\Omega_2}, e^{i\Omega_3}, e^{i\Omega_4}, e^{i(\alpha_{12}+\alpha_{23}+\alpha_{34}+\alpha_{14})}, e^{i(\alpha_{12}+\alpha_{24}+\alpha_{34}+\alpha_{13})},
e^{i(\alpha_{13}+\alpha_{23}+\alpha_{24}+\alpha_{14})}
$$
to
$$
1, e^{iP_1}, e^{iP_2}, e^{iP_3}, e^{iP_4}, e^{i(l_{12}+l_{23}+l_{34}+l_{14})}, e^{i(l_{12}+l_{24}+l_{34}+l_{13})},
e^{i(l_{13}+l_{23}+l_{24}+l_{14})}.
$$
I know neither an elementary proof of this statement nor any interpretation of the coefficients of the $PSL_2(\mathbb{C})-$ transformation.
 A: Euclidean case
Using the formula for the tan of the half solid angle that Robin Houston quotes, and expressing everything in terms of edge lengths by using the cosine law to convert the dot products, I end up with the following linear relationship:
$$\frac{1}{P_i} = \alpha_E \cot\left(\frac{\Omega_i}{2}\right)+\beta_E$$
where:
$$\alpha_E = \frac{12 V}{P_1 P_2 P_3 P_4}$$
$V$ is the volume of the tetrahedron:
$$V = \frac{1}{12\sqrt{2}}\sqrt{\left|
\begin{array}{ccccc}
 0 & l_{12}^2 & l_{14}^2 & l_{13}^2 & 1 \\
 l_{12}^2 & 0 & l_{24}^2 & l_{23}^2 & 1 \\
 l_{14}^2 & l_{24}^2 & 0 & l_{34}^2 & 1 \\
 l_{13}^2 & l_{23}^2 & l_{34}^2 & 0 & 1 \\
 1 & 1 & 1 & 1 & 0
\end{array}
\right|}$$
and:
$$\beta_E = \frac{\left(\Sigma_{i \lt j} l_{ij}P_i P_j\right) - 2 \left(\Sigma l_{ij} l_{pq} l_{st} \right)}{P_1 P_2 P_3 P_4}$$
where the second summation is over the 16 choices of distinct triples of edges that do not all share a common vertex.  There are $\binom{6}{3}=20$ triples of distinct edges; if we omit the four triples that are incident on each of the four vertices, we obtain the 16 that appear in that sum.
Another expression for $\beta_E$ is:
$$\beta_E = \frac{\text{Permanent}\left(
\begin{array}{ccccc}
 0 & l_{12} & l_{14} & l_{13} & 1 \\
 l_{12} & 0 & l_{24} & l_{23} & 1 \\
 l_{14} & l_{24} & 0 & l_{34} & 1 \\
 l_{13} & l_{23} & l_{34} & 0 & 1 \\
 1 & 1 & 1 & 1 & 0
\end{array}
\right)
}{2 P_1 P_2 P_3 P_4}$$
Spherical case
In the spherical case, starting from a tetrahedron with specified edge lengths, a similar relationship is given by:
$$\cot\left(\frac{P_i}{2}\right) = \alpha_S \cot\left(\frac{\Omega_i}{2}\right) + \beta_S$$
where:
$$\alpha_S = \frac{\sqrt{\gamma_S}}{4 \sin\left(\frac{P_1}{2}\right)\sin\left(\frac{P_2}{2}\right)\sin\left(\frac{P_3}{2}\right)\sin\left(\frac{P_4}{2}\right)}$$
and:
$$\gamma_S =
\left|
\begin{array}{cccc}
 1 & \cos \left(l_{12}\right) & \cos \left(l_{13}\right) & \cos \left(l_{14}\right) \\
 \cos \left(l_{12}\right) & 1 & \cos \left(l_{23}\right) & \cos \left(l_{24}\right) \\
 \cos \left(l_{13}\right) & \cos \left(l_{23}\right) & 1 & \cos \left(l_{34}\right) \\
 \cos \left(l_{14}\right) & \cos \left(l_{24}\right) & \cos \left(l_{34}\right) & 1 \\
\end{array}
\right|$$
A series expansion of $\gamma_S$ to sixth order in the edge lengths yields the expected relationship:
$$\sqrt{\gamma_s} \approx 6 V_E$$
where $V_E$ is the volume computed from the same edge lengths using Cayley's Euclidean formula.  (Note that there is a factor of 2 difference between the slopes of the two lines in the limit of small edge lengths, because we are using the cotan of the half perimeter on the $y$ axis in the spherical case.)
$$\beta_S = \frac{\Sigma_i \sin\left(Q_i\right) - \sin\left(\Sigma_{j\lt k} l_{jk}\right) - \sin\left(l_{12}+l_{34}\right) - \sin\left(l_{13}+l_{24}\right) - \sin\left(l_{14}+l_{23}\right)}{8 \sin\left(\frac{P_1}{2}\right)\sin\left(\frac{P_2}{2}\right)\sin\left(\frac{P_3}{2}\right)\sin\left(\frac{P_4}{2}\right)}$$
where:
$Q_i$ is the sum of the edge lengths for the edges incident on vertex $i$.
A: This is not an answer to the question, but an experimental observation that suggests a sharper conjecture: it’s only written as an answer because I’d like to flesh it out a bit more than there’s room for in a comment.
The image of $e^{i\Omega}$ under the Möbius transformation $z\mapsto\frac{i-zi}{1+z}$ is $\tan(\frac\Omega2)$, so an equivalent statement of Rudenko’s theorem is that 
$$
[\tan(\frac{\Omega_1}2), \tan(\frac{\Omega_2}2), \tan(\frac{\Omega_3}2), \tan(\frac{\Omega_4}2)]=[P_1, P_2, P_3, P_4].
$$
Experimentally, something stronger seems to be true: the four points


*

*$(1/\tan(\frac{\Omega_1}2), 1/P_1)$

*$(1/\tan(\frac{\Omega_2}2), 1/P_2)$

*$(1/\tan(\frac{\Omega_3}2), 1/P_3)$

*$(1/\tan(\frac{\Omega_4}2), 1/P_4)$
are collinear.
It’s fairly easy to check this numerically for particular examples, because $\tan(\frac{\Omega}2)$ has a nice expression that can be computed easily from coordinates:
$$
\tan \left( \frac{\Omega}{2} \right) =
  \frac{\left|\vec a\cdot(\vec b\times\vec c)\right|}{abc + \left(\vec a \cdot \vec b\right)c + \left(\vec a \cdot \vec c\right)b + \left(\vec b \cdot \vec c\right)a}
$$
where $\vec a$, $\vec b$ and $\vec c$ are the vectors from the vertex in question to the other three vertices, and $a$, $b$ and $c$ are the lengths of these vectors.
Since the expression in the numerator is six times the volume of the tetrahedron, it is the same at all four vertices, so to check collinearity it suffices to compute the denominator.
I haven’t played with the spherical or hyperbolic cases at all.
A: Here one can find an algebra-geometric proof of the trigonometric relations. The main point is that given a (say, spherical) tetrahedron $T$ one can construct a rational elliptic surface $X_T$ with the property that surfaces corresponding to a tetrahedron and its dual are isomorphic. The cross-ratios in the question are some invariants of $X_T$ computed in two different ways. These invariants are analogs of Cayley invariants of cubic surfaces: cross-ratios of four points in which four lines intersect the fifths. 
