Consider a convex polyhedron $\mathcal{P}$ in $\mathbb{E}^3$ with $n$ edges. By convex polyhedron, I mean the convex hull of a finite set of points in $\mathbb{E}^3$ whose affine hull is all of $\mathbb{E}^3$. And by edge, I mean a line segment along which precisely 2 distinct faces meet at an angle less than $\pi$. Let $\theta_\mathrm{max}(\mathcal{P})$ be the maximal dihedral angle of $\mathcal{P}$. It seems intuitive to me that, as $n$ grows, $\theta_\mathrm{max}(\mathcal{P})$ must approach $\pi$, regardless of which polyhedron I choose. To make this more precise, I am suggesting that
$$\inf\{\theta_\mathrm{max}(\mathcal{P})~|~\mathcal{P}\text{ a convex polyhedron with }n\text{ edges}\}\longrightarrow\pi\quad\text{as}\quad n\to\infty.$$
Is this statement true? If so, how does one prove it? Please reference any literature you might have used in an answer.
Edit 25/06/20 – I believe this problem is interesting because if we ask the question about vertices and their conical angles instead, we can show that the maximum always approaches $2\pi$ by just using Gauss-Bonnet. This is in some way an analogous statement question one dimension higher.
Best Answer
No, it's not true.
Take a regular $n$-gon prism, and cut off the vertices, with planes sloped at some angle $\beta$, so that the original edges are reduced to points. (You can ensure that the horizontal and vertical edges disappear at the same time by adjusting the height of the prism. But we don't really need such precision; we only need to get rid of the vertical edges, which are the ones with a large dihedral angle.) The result is a rectified prism.
The square (rather, rhombus) faces have normal vectors
$$a_k=\left(\cos\frac{(2k+1)\pi}{n},\;\sin\frac{(2k+1)\pi}{n},\;0\right),$$
the triangular faces have normal vectors
$$b^\pm_k=\left(\sin\beta\cos\frac{2k\pi}{n},\;\sin\beta\sin\frac{2k\pi}{n},\;\pm\cos\beta\right),$$
and the $n$-gon faces have normal vectors
$$c^\pm=(0,0,\pm1).$$
Consider these as points on the unit sphere (see Gauss map). The exterior (or supplementary) dihedral angle at an edge is just the spherical distance between the two faces' normal vectors. The angle deficit at a vertex is the area of the spherical polygon formed by the surrounding faces' normal vectors. In fact, the Gauss map gives another realization of the dual polyhedron, as a tiling of the sphere. So your question is whether a spherical tiling with a large number of vertices (or edges, or tiles) must have a short edge.
This particular polyhedron has only two types of edge. The exterior dihedral angles are given by
$$\cos\theta_{ab}=a_k\cdot b_k^\pm=a_k\cdot b_{k+1}^\pm=\sin\beta\cos\frac\pi n$$
(between a square and a triangle), and
$$\cos\theta_{bc}=b_k^\pm\cdot c^\pm=\cos\beta$$
(between a triangle and an $n$-gon). Clearly, as $n\to\infty$, neither $\theta_{ab}$ nor $\theta_{bc}\to0$.