10
$\begingroup$

I have a puzzle to construct a polyhedron where the measure of every dihedral angle is $\frac\pi4$ radians or prove that none can exist. The polyhedron may self-intersect however it must be finite, bounded, and dyadic as well as have planar faces.

Here's a quick glossary of terms as I use them:

  • Polyhedron: A polytope in Euclidean 3-space.
  • Finite: There are only a finite number of elements in the polyhedron.
  • Bounded: There exists a sphere of finite radius which encloses every vertex of the polyhedron.
  • Dyadic: Each edge is incident on exactly two faces.
  • Planar faces: For every face all vertices of that face lie on some plane.

These are all pretty standard when talking about polyhedra.

As an example, if we were to imagine a 2-dimensional version of this problem, a regular octagram (Schläfli symbol $\{8/3\}$) would be a solution. A square bowtie would be another solution (illustration). Both are polygons with all interior angles measuring $\frac\pi4$ radians.

Of course a polyhedron is tougher since there is more freedom to arrange things in 3 space.

$\endgroup$
15
  • 1
    $\begingroup$ @OscarLanzi Yes, the interlinked squares have angles of π/2, so not that one. But I'll make it clear in the question. $\endgroup$ Jul 7, 2022 at 16:16
  • 1
    $\begingroup$ In my hint I was referring to embedded polyhedra, I do not know about immersed ones, I do not even know what an "interior angle" would mean in the setting of non-embedded polyhedra. $\endgroup$ Jul 9, 2022 at 13:39
  • 1
    $\begingroup$ @MoisheKohan 1. What specifically is unclear to you about the definition? The links are not intended to provide any part of the definition. That bullet point is conveying that we are talking about objects in 3 space, the other bullet points are providing the specifics of what we mean by polyhedron. i.e. nothing exotic or skew. 2. I've just removed the "interior" language entirely. It does change the problem a bit, but not in a way that effects the result. Defining it completely would just be a distraction. $\endgroup$ Jul 12, 2022 at 19:33
  • 1
    $\begingroup$ Arguably "interior" still persists in the problem description. How do define "every element of the polyhedron" for the purposes of testing boundedness? For example, which points are inside the bowtie when some of the angles are apparently open-ended (for otherwise we would get an angle of $\pi/2$ where two sides intersect. I sort of see what you are trying to define, but details like that may affect an attempt at proving non-existence. $\endgroup$ Jul 12, 2022 at 20:15
  • 1
    $\begingroup$ @JyrkiLahtonen This is a good point. In this case since the number of elements must be finite what I really mean is that the vertices must be bounded rather than elements. $\endgroup$ Jul 12, 2022 at 20:20

0

You must log in to answer this question.

Browse other questions tagged .