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I'd like to ask if someone can help me out with this problem. I have to determine what is the lower and upper bound for sum (the largest and smallest sum I can get) of dihedral angles in arbitrary Tetrahedron and prove that. I'm ok with hint for proof, but I'd be grateful for lower and upper bound and reason for that.


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Possible candidates for the bounds might be a degenerate oblate tetrahedron with sum $3\pi$, a degenerate prolate tetrahedron with sum $2.5\pi$ and a regular tetrahedron with sum $6\arccos\frac13\approx2.35\pi$, so the first and last of those might be the bounds. I don't know how to prove that though. – joriki Feb 21 '13 at 10:07
This code finds that the bounds are $2\pi$ and $3\pi$, but again I don't know how to prove that. – joriki Feb 21 '13 at 10:40
Thanks for that! I hope someone will help me out with that proof:) – JosephK Feb 21 '13 at 10:50
Ah, there are actually two different kinds of degenerate flattened tetrahedrons. If a point approaches the opposite face, three angles tend to $\pi$ and three to $0$, whereas if two edges approach each other, their two angles tend to $\pi$ and the other four tend to $0$. So those are the boundary cases. I would think that you might be able to prove that by showing that the sum always increases/decreases as you deform the tetrahedron into one of those shapes. – joriki Feb 21 '13 at 10:51
@joriki using the formula you are given, if one can show the sum of four inner solid angles of a tetrahedron is bounded by $2\pi$, one can also deduce the upper bound of sum of dihedral angles is $3\pi$. – achille hui Feb 21 '13 at 12:06
up vote 4 down vote accepted

Lemma: Sum of the 4 internal solid angles of a tetrahedron is bounded above by $2\pi$.

Start with a non-degenerate tetrahedron $\langle p_1p_2p_3p_4 \rangle$. Let $p = p_i$ be one its vertices and $\vec{n} \in S^2$ be any unit vector. Aside from a set of measure zero in choosing $\vec{n}$, the projection of $p_j, j = 1\ldots4$ onto a plane orthogonal to $\vec{n}$ are in general positions (i.e. no 3 points are collinear). When the images of the vertices are in general positions, a necessary condition for either $\vec{n}$ or $-\vec{n}$ belong to the inner solid angle at $p$ is $p$'s image lies in the interior of the triangle formed by the images of other 3 vertices. So aside from a set of exception of measure zero, the unit vectors in the 4 inner solid angles are "disjoint". When one view tetrahedron $\langle p_1p_2p_3p_4 \rangle$ as the convex hull of its vertices, the vertices are extremal points. This in turn implies for any unit vector, $\vec{n}$ and $-\vec{n}$ cannot belong to the inner solid angle of $p$ at the same time.

From this we can conclude (up to a set of exception of measure zero), at most half of the unit vectors belongs to the 4 inner solid angles of a tetrahedron. The almost disjointness of the inner solid angles then forces their sum to be at most $2\pi$.

Back to original problem

Let $\Omega_p$ be the internal solid angle and $\phi_{p,i}, i = 1\ldots 3$ be the three dihedral angles at vertex $p$. The wiki page mentioned by @joriki tell us:

$$\Omega_p = \sum_{i=1}^3 \phi_{p,i} - \pi$$

Notice each $\Omega_p \ge 0$ and we have shown $\sum_{p}\Omega_{p} \le 2\pi$. We get:

$$\begin{align} & 0 \le \sum_p \sum_{i=1}^3 \phi_{p,i} - 4\pi \le 2\pi\\ \implies & 2\pi \le \frac12 \sum_p \sum_{i=1}^3 \phi_{p,i} \le 3\pi \end{align}$$

When we sum the dihedral angles over $p$ and $i$, every dihedral angles with be counted twice. This means the expression $\frac12 \sum_p \sum_{i=1}^3 \phi_{p,i}$ above is nothing but the sum of the 6 dihedral angles of a tetrahedron.

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I was given very similar problem in school, really thanks for this. +1 – Noturab Feb 21 '13 at 15:04

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