TITLE: Solid angles of a tetrahedron
QUESTION [13 upvotes]: 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?

REPLY [5 votes]: One can show that  the face opposite the largest solid angle has the largest perimeter. This follows directly from here (see the  answer of Greg Egan with an explicit linear relation).