α-polyhedra

Introduction

Definition: An α-polyhedron is a polyhedron where all dihedral angles are right (that is 90° or 270°) except for one angle which is α.

It is easy to find such a polyhedron for α=90°, namely, just take the cube. But for other angles (not a multiple of 90°), I didn't even believe such a polyhedron could exists...
Until I read that Sydler [Syd65] claims there is a 45°-polyhedron. Following his instructions, I explicitly build the first non-trivial α-polyhedron, see below.
It seems very likely that the ideas in Sydler's papers [Syd44, Syd52, Syd53, Syd59, Syd65] should yield a polyhedron for 45°, 30°, 60°, 22.5°, ... ‐ that is for any rational multiple of 360°.
There are even some α-polyhedra where α is not commensurable with 360°. Combining Sydler's ideas with Jessen [Theorem 6; Jes68], one expects:

Conjecture: An α-polyhedron exists if and only if sin(α) is algebraic (that α is the root of a polynomial with integral coefficients).

Sydler was motivated to (abstractly) resolve Hilbert's third problem about scissors congruence and the Dehn invariant. But I want to actually see these polyhedra!

Update

The Sydler π/4 polyhedron

Getting a 3d print

Order a 3d print on Shapeways here or download the obj or stl file.

Step 1

We start with a prism over an isosceles right triangle. This prism is scissors-congruent to a cube but has two π/4 dihedral angles (labeled 1 and 2). After this step, we are left with only one edge with angle π/4. However, we introduced two angles ζ and π-ζ (labeled 3, respectively, 4).
From now on, we always call the angles of a polyhedron which are not right or π/4 the extra angles.

Sydler π/4 polyhedron: Step 1 by Matthias Goerner on Sketchfab

Step 2

After this step, we still have two extra angles (call them β and π-β), but they are along edges (labeled 2 and 3) in the same plane.

Sydler π/4 polyhedron: Step 2 by Matthias Goerner on Sketchfab

Step 3

Before we can continue, we need to add a box (which is scissors-congruent to a cube).

Sydler π/4 polyhedron: Step 3 by Matthias Goerner on Sketchfab

Step 4

Consider a polyhedron with 6 triangles and label one pair of opposite edges by 2 and 3. We can find a such a polyhedron Q (green) such that the angle at the edge labeled 2 is π - α and all unlabeled edges have right angles. We call the angle at edge 3 β. Also see [Figure 2; Syd65].
If we remove Q from one place and add it along the edge with angle α, we achieve that the two extra angles (labeled 3 and 4) of the resulting polyhedron are now along edges parallel as line segments (in the sense that they span a rectangle and not just a parallelogram).

Sydler π/4 polyhedron: Step 4 by Matthias Goerner on Sketchfab

Step 5

Before we continue, we need to add some other boxes and prisms (all of which are scissors-congruent to cubes). Some of them are not strictly necessary but provide breathing room when performing the next step so we can get a polyhedron more ameanable to 3d printing.

Sydler pi/4 polyhedron: Step 5 by Matthias Goerner on Sketchfab

Step 6

To get rid of the two extra angles β and π-β (labeled 2 and 3), we remove a prism over a polygon with angles β, π-β, and right angles. The polygon must be carefully chosen such that the prism is entirely contained in the polyhedron.

Sydler pi/4 polyhedron: Step 6 by Matthias Goerner on Sketchfab

The result

Sydler pi/4 polyhedron by Matthias Goerner on Sketchfab

Simplified version

With some variations to the above steps, I was able to make a slightly simplier polyhedron with the same property.

Sydler pi/4 polyhedron: Version 2 (simplified) by Matthias Goerner on Sketchfab

Bibliography