# The Sydler π/4 polyhedron

## Getting a 3d print

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

**We follow the steps by Jean-Pierre Sydler to construct a polyhedron that is scissors congruent to a cube and where all dihedral angles are right except for one which is 45 degrees.**

This polyhedron is used in Sydler's proof of Sydler's Theorem which is related to Hilbert's third problem and states that every polyhedron is determined by its volume and Dehn invariant up to scissors congruence.

**References:** J.-P. Sydler, "Conditions necessaires et suffisantes pour l'equivalence des polyedres de lâ€™espace euclidien a trois dimensions", Commentarii mathematici Helvetici (1965/66) Volume: 40, page 43-80
## 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.
## 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.
## Step 3

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

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 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.
## 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.
## The result

## Simplified version

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