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?



Solid angles of a tetrahedron
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).


