Question below.
Some background:
Take an isosceles obtuse triangle of the form
with $\alpha = \frac{n-1}{n}\pi$ for some $n \geq 3$ ($\beta=\frac{\pi}{2n}$)
If you look at the class of convex polyhedra such that each face is congruent to this triangle, then the vertices are of the following types:
- $\alpha^2\beta^2$
- $\beta^4$, $\beta^6$, …, $\beta^{4n-2}$
- $\alpha\beta^{2n}$
A type indicates which face angles go around it.
Lets look at what happens if such a polyhedron has a vertex of type $\beta^m$ ($m=4,6, …, 4n-2$):
The vertices colored red, only allow for the type $\alpha^2\beta^2$ and so they can be filled in. The next picture shows this:
where lines that are marked similarly should be pasted together. This can be done in all vertices colored red, obtaining
This fixes thus a unique abstract polyhedron $P_m$.
Abstract here also fixes the face angles and the edge lengths (up to scale).
This shape (if it is geometrically realizable) somewhat resembles a trapezohedron where the faces are bent on the long diagonal.
Which of the $P_m$ is geometrically realizable as a convex polyhedron? (we do not allow edges with dihedral angle $\pi$) And how would you prove this?
After trying out with some paper models, it seems that the only polyhedron that is geometrically realizable is $P_4$.
I can prove that $P_6$ is never geometrically realizable as a convex polyhedron $Q$:
Any net of $Q$ can be folded to a parallelepiped with kite faces (angles $\alpha$, $2\beta$, $\alpha$, $2\beta$). By Alexandrov's uniqueness theorem, this means that $Q$ is a parallelepiped; and thus has dihedral angles $\pi$ (two triangles joined together with their long edge form a face of this parallelepiped)
At first I thought I could extend this proof to $m > 6$ by showing that the net folds to a trapezohedron, but proving this is much harder than in case of the parallelepiped; because now we have to show that there exists a geometrically realizable trapezohedron with all faces congruent to a kite with angles $\alpha, 2\beta, \alpha, 2\beta$. Moreover, it might be that this is not the case; which means we have to find another way to prove this.
Question: How to prove that $P_m$ is not geometrically realizable as a convex polyhedron for $m > 6$? How to prove that $P_4$ is geometrically realizable as a convex polyhedron? Which general techniques exist for proving similar results?
I would prefer answers that do not use coordinate calculations.
Results that I think could help:
Lemma 2a p 162 of Convex Polyhedra
If two convex polyhedral angles, distinct from dihedral angles
and possibly degenerate, have corresponding planar angles of equal measure
while not all of their dihedral angles are equal, then there are at least four
sign changes in the differences between the corresponding dihedral angles as
we go around the vertices.
This can be used to find minimal and maximal dihedral angles.
For example, in a $\alpha^2\beta^2$ vertex, if we set the dihedral angle between the two $\beta$ to $\pi$; this is still realizable. This angle should decrease. Therefore the opposite angle has to decrease as well; which shows that the dihedral angle between the two $\alpha$ reaches its maximum when the dihedral angle between the two $\beta$ is $\pi$. This dihedral angle $A$ satisfies the following formula:
$\cos(A) = \frac{\cos(2\beta) – \cos^2(\alpha)}{\sin^2(\alpha)} = \frac{-\cos(\alpha) – \cos^2(\alpha)}{\sin^2(\alpha)}$
This question is part of a search to classify the convex polyhedra with congruent isosceles triangular faces (the faces do not need to be transitive).
Best Answer
I finally got around to an answer to my question. Let us start with the following observation:
Triangles come in pairs and by looking at these pairs differently, we get a different abstract polyhedron (which is in some sense dual). However, both abstract polyhedron have the same metric and according to the Alexandrov's uniqueness theorem, at most one can be geometrically realizable. Now, it's not too difficult to see that by flipping each pair of triangles likes this, we get that $P_m$ (for $m \geq 6$) is transformed to the skeleton of a gyroelongated $\frac{m}{2}$-gonal bipyramid.
Now we prove that any gyroelongated $n$-gonal bipyramid with congruent isosceles triangular faces is always geometrically realizable for $n > 3$. Such a gyroelongated bipyramid has $n$ rotational symmetry. It consists of two apex vertices and two rings of 'middle' vertices. These two rings form an antiprism. Basically, the figure is a an antiprism with two pyramids pasted on the regular $n$-gons of the antiprism.
By scaling, we can always have that the vertices on the two rings are exactly distance $1$ away from the rotational symmetry axis of the figure.
We need to check that this polyhedron does not have any dihedral angles larger than $\pi$. The dihedral angles between triangular faces of the antiprism are always less than $\pi$, and so are the dihedral angles between the triangular faces of the pyramid. The dihedral angle we need to check is the one between the antiprism and the pyramid (the red edge in the picture above). Note that the purple edges and the symmetry axis lie in a single plane. Therefore, we have the following situation (line through $A$ and $E$ is symmetry axis):
Now, we have the following
We know that $\varphi$ and $\psi$ are in between $0$ and $\frac{\pi}{2}$. Therefore, $\varphi > \psi$ iff $\cos(\varphi) < \cos(\psi)$. Now $\cos(\varphi) = \frac{1-\cos\left(\frac{\pi}{n}\right)}{x}$ and $\cos(\psi) = \frac{\cos\left(\frac{\pi}{n}\right)}{x}$ So $\varphi > \psi$ iff $1-\cos\left(\frac{\pi}{n}\right) < \cos\left(\frac{\pi}{n}\right)$ iff $\cos\left(\frac{\pi}{n}\right) > \frac{1}{2} = \cos\left(\frac{\pi}{3}\right)$. Which is equivalent to $\frac{\pi}{n} < \frac{\pi}{3}$ or $n > 3$. Therefore, the gyroelongated $n$-gonal bipyramid is always realizable for $n > 3$ independent on the angles of the isosceles faces. Therefore by the Alexandrov's uniqueness theorem, $P_m$ is not geometrically realizable for $m > 6$. Note that $\varphi = \psi$ iff $n=3$ and thus a gyroelongated $3$-gonal bipyramid is actually a parallelepiped, hence $P_6$ is not geometrically realizable.
The only figure that is left is $P_4$. We prove that it is always geometrically realizable. Take the following doubly covered polygon where only the red edges are not 'glued'.
If you would make a paper model of this, you would get a 'hole' with four border edges at the top. It is not difficult to see that the left part of the figure has a single degree of freedom (it is a single vertex of degree 4). We can make it so that $F$ moves out of the plane, vertex $A$ and $C$ move to the right and vertex $E$ stays stationary. Since it is symmetric, the right side does not restrict this movement. Then by an argument akin to the intermediate value theorem, there is a point where $A$ and $B$ coincide. By a symmetry argument, it lies in the plane through the point $E$, $F$ and $F$ of the other side. Note that $C$ and $D$ do not lie in this plane. Therefore, it forms a polyhedron with the same abstract structure as $P_4$.
The dihedral angles at $AC$, $BD$, $CE$, and $DE$ start at $0$ and increase because the point $F$ moves out of the plane. They cannot become more than or equal to $\pi$ because that would mean that $C$ and $D$ would lie in the plane of $E$, $F$ and the other $F$. The dihedral angles at $CF$, $DF$, and $EF$ start at $\pi$ and decrease due to the same reason. Similarly, they cannot become $0$. The same argument also shows that the dihedral angle of the red lines is strictly convex.
So this shows that $P_4$ is realizable as long as $\alpha^2\beta^2$ and $\beta^4$ are strictly convex vertex figures (even for non-obtuse triangles). We have that $\alpha^2\beta^2$ is strictly convex when $2\pi - 2\alpha - 2\beta > 0$ iff $\pi - \alpha > 0$, which is always the case. Similarly, $\beta^4$ is always strictly convex because $2\pi - 4\beta > 0$ since $\beta < \frac{\pi}{2}$. So for every isosceles triangle, $P_4$ is geometrically realizable as a strictly convex polyhedron.
Now onto other classes of polyhedra with congruent isosceles triangular faces!