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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.

Thanks

JosephK
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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
  • The lower bound $2\pi$ follows immediately from the relationship with the sum of the vertex angles given here. Now you just have to show that the sum increases as you move a point towards the opposite face. – joriki Feb 21 '13 at 10:54
  • Can you explain how did you get $2\pi$ from that sum please? – JosephK Feb 21 '13 at 11:10
  • Well, the sum of the vertex angles is positive. – joriki Feb 21 '13 at 11:16
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    @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

1 Answers1

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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.

achille hui
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