This is a problem I have had for a while. For a triangle, the side opposite the largest angle has the largest length (and similarly for smallest angle). For a tetrahedron, the question is whether the face opposite the largest solid angle (trihedral angle – the area of the spherical triangle with angles equal to the dihedral angles incident on one vertex) has the largest area?
I can prove that this is true for tetrahedra coming from sphere packings (each vertex $v_i$ has a weight $r_i$ and the edge lengths are equal to $r_i+r_j$, etc.) but this excludes degenerations of a tetrahedron such as the one going to a box with two diagonals.
I suspect this is true but have no proof. Perhaps someone else does?
Best Answer
I think this statement does not hold. The following tetrahedron should give a counterexample:
$A=(-0.5, 0, 0)$, $B=(1,0,0)$, $C=(0,\varepsilon^2, \varepsilon)$, $D=(0,\varepsilon^2, -\varepsilon)$, $1>>\varepsilon>0$.
Here $BCD$ has largest area, but the spherical angles at $C$ and $D$ should be significantly larger than the one at $A$. (I have done the calculation only approximatively, but I believe it is true).