# Solid angles of a tetrahedron

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?

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).

For $\epsilon=0.4$, the solid angles at $C$ and $D$ in Dmitri's example are more than twice the solid angle at $A$: $24.5^\circ$ vs. $12.1^\circ$. (And the area of $\triangle BCD = 0.41 > 0.32 = \triangle ABC$.) Nice!

For random tetrahedra with vertices in the unit cube, in $100,000$ trials I get $8,355$ counterexamples including (after rounding) $\lbrace (0.57, 0.28, 0.95), (0.15, 0.54, 0.87), (0.45, 0.97, 0.68), (0.96, 0.38, 0.24) \rbrace$.

The reason I looked for a counterexample is that I don't think the statement is plausible. As you move around in the space of tetrahedra, the largest face changes. Also, the largest solid angle changes. There seems to be no general reason for these changes to happen at exactly the same place.

Here is a construction based on the observation above. Let $A = (0,0,0)$, $B=(5,1,0)$, $C=(5,-2,0)$. These are the vertices of a triangle in the $xy$-plane. Let $D(t) = (0,0,t)$.

For small $t \gt 0$, $AD$ is small, so $\triangle DAB$ and $\triangle DAC$ have small areas. $\triangle DBC$ orthogonally projects onto $\triangle ABC$, so $\triangle DBC$ is slightly larger than $\triangle ABC$. When $t$ is large, $\triangle DAC$ becomes larger than $\triangle DBC$. The transition occurs at $t = \frac{3\sqrt{5}}{2} = 3.3541.$ The solid angle at $A$ is constant, while the solid angle at $B$ increases and eventually becomes larger. The transition occurs at $t = \frac{69+3\sqrt{29}}{20} = 4.2578$. When $3.3541 \lt t \lt 4.2578$, the largest solid angle is at $A$, while the side with the largest area is $\triangle DAC$.

• Douglas, it seems clear that there are counterexamples. But I am not sure which part of your second paragraph fails to hold when you replace "tetrahedra" with "triangles". Also, 8335 out of 100k seems rather small... – Vidit Nanda Mar 19 '13 at 0:36
• In a triangle with two equal angles, the sides opposite these are equal. This is because of symmetry. But a tetrahedron with two equal solid angles need not have any symmetry. – Robert Israel Mar 19 '13 at 1:13
• @Vel Nias: I don't think $8\%$ counterexamples is small. My guess was that there would be closer to $1\%$ counterexamples. – Douglas Zare Mar 19 '13 at 2:33

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).