αpolyhedra
Introduction
Definition: An αpolyhedron 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 nontrivial αpolyhedron, 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 αpolyhedra where α is not commensurable with 360°. Combining Sydler's ideas with Jessen [Theorem 6; Jes68], one expects:
Conjecture: An αpolyhedron 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
 Henry Segerman has made a video about the Sydler π/4polyhedron.
 Shortly after the video was released, Robin Houston found a much simpler polyhedron with the same property, a 45°dodecahedron constructed as follows:
 Consider the hexahedron SABCT in [Figure 2; Syd65] with α=45°.
 Take two copies of SABCT, scale them by half and join them along the triangle ABC. Call this polyhedron P_{0}. It is a arccos(1/3)≈109.5°octahedron.
 Take a third copy of SABCT and subtract P_{0} in such a way that the result is a 135°decahedron. Call this polyhedron P_{1}.
 Take a cuboid of size √2×√2×1 aligned with six faces of P_{1} and subtract P_{1} from it. The result is a 45°dodecahedron.
 A arccos(1/3)≈109.5°octahedron fell out of the construction of the 45°dodecahedron.
 A 135°decahedron fell out of the construction of the 45°dodecahedron.
 Robin Houston constructed a 15°polyhedron. He has several versions of 15°polyhedra. Here is one:
 Start with a prism with angles 15°, 75° and 90°. We now focus on the 75° edge. We want to add polyhedra to it to make it a multiple of 90°. We assume it has length 2.
 Add the polyhedron with two nonright angles 15° and 75° along two edges that line up and have length 1 each.
 Take the polyhedron in [Figure 1; Syd53] with α=15°. Besides the 15° and 75° angle, it has some 45° angles.
 Add prisms with angle 45°, 45° and 90° to make room where necessary.
 Cap off the 45° angles with a 135°decahedra.
 Our 75° edge now has turned into three colinear edges of length one each with angles 75°+15°=90°, 75°+75°=150° and 75°. Similarly to the previous step, we add a polyhedron with two nonright angles 30° and 60°.
 Our edge now has turned into three colinear edges with angles 90°, 150°+30°=180° and 75°+60°=135°. Thus, adding a 45°polyhedron fixes it.
 Robin Houston constructed a 60°polyhedron.
 Start with the characteristic orthoscheme of the cube which is a tetrahedron with nonright angles 45°, 60° and 45°.
 Cap off the 45° dihedral angles with a 135°decahedron.

Robin Houston constructed a simpler 15°polyhedron (annotated version).
 Take a prism with angles 15°, 75° and 90°.
 Add a 60°polyhedron and a 135°dodecahedron along the 75°edge to make a 75°+60°+135°=270° angle.
 Here are some more pointers to construct more:

We have already seen examples from two interesting families of polyhedra given by Sydler:
 The hexahedron SABCT in [Figure 2; Syd65] has two nonright dihedral angles 180°α at BC and β at AS related by (3 + cos 2α)(1  cos β) = 4. As we have already seen, it seems very versatile. We can, for example, pick α=60° and stick it on the 60°polyhedron to obtain a βpolyhedron with β=arccos(3/5)≈126.9°. Adding prisms to make room, we can apply this trick to other polyhedra.
 [Figure 1; Syd53] shows a polyhedron with dihedral angles α at edge CC'' and 90°α at edge C'C'' and all other dihedral angles being multiples of 45°. The edges CC'' and C'C'' line up and have the same length. The 45°angles can be capped off, for example, in the case of α=30°, resulting in this polyhedron.

Some other families of polyhedra given by Sydler:
 Assume we have a 180°/npolyhedron. Extrude the polygon [Figure 7; Syd53] to obtain a polyhedron and cap it off with copies of the 180°/npolyhedron to obtain a k·180°/npolyhedron. In that paper, Sydler shows how to transform a polyhedron with all dihedral angles being rational (that is a rational multiple of 360°) to one where all dihedral angles are multiples of 45°. Capping of 45° angles, the arguments can also be used to construct any αpolyhedron with α rational.
 In [Syd52], Sydler considers the polyhedron RSTABCD. It can be used to "move" nonrational dihedral angles around.
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 scissorscongruent 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 πζ (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.
Step 2
After this step, we still have two extra angles (call them β and πβ), but they are along edges (labeled 2 and 3) in the same plane.
Step 3
Before we can continue, we need to add a box (which is scissorscongruent to a cube).
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).
Step 5
Before we continue, we need to add some other boxes and prisms (all of which are scissorscongruent 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.
Step 6
To get rid of the two extra angles β and πβ (labeled 2 and 3), we remove a prism over a polygon with angles β, πβ, and right angles. The polygon must be carefully chosen such that the prism is entirely contained in the polyhedron.
The result
Simplified version
With some variations to the above steps, I was able to make a slightly simplier polyhedron with the same property.
Bibliography
 [Jes68] B. Jessen, The algebra of polytopes and the Dehn–Sydler theorem, Math. Scand. 22 (1968), 241–256.
 [Syd44] J.P. Sydler, Sur la décomposition des polyèdres, Commentarii mathematici Helvetici 16 (1943/1944), 266273.
 [Syd52] J.P. Sydler, Sur les conditions nécessaires pour l'équivalence des polyèdres euclidiens, Elemente der Mathematik 7 (1952), 3, 4972.
 [Syd53] J.P. Sydler, Sur l'équivalence des polyèdres à dièdres rationnels, Elemente der Mathematik 8 (1953), 4, 7579.
 [Syd59] J.P. Sydler, Sur quelques polyèdres équivalents obtenus par un procédé en chaînes, Elemente der Mathematik 14 (1959), 5, 100109.
 [Syd65] J.P. Sydler, Conditions necessaires et suffisantes pour l'equivalence des polyedres de l’espace euclidien a trois dimensions, Commentarii mathematici Helvetici 40 (1965/66), 4380.