Let γ1,…,γn be a regular polygon in R2=C. A second polygon ˜γ1,…,˜γn is said to be a Darboux transform of γ if three conditions are satisfied. The first two conditions are:
1) The distance between corresponding points is a fixed number ℓ. This means for all j we have
|˜γj−γj|=ℓ.
2) Corresponding edges have the same length. This means
|˜γj+1−˜γj|=|γj+1−γj|.
These two conditions are trivially met by a polygon ˜γ that is just a translate of γ, i.e. ˜γj=γj+a for some fixed a∈R2. In this case for all j the four points γj,γj+1,˜γj,˜γj+1 form a parallelogram. To eliminate this trivial solution we add a third condition:
3) Given the points γj,γj+1 and ˜γj the next point ˜γj+1 is chosen to be different from ˜γj+(γj+1−γj), if possible.
Here the word “if possible” refer to the case where γj,γj+1,˜γj lie on a straight line. In this case the parallelogram solution is the only one that complies with the first two conditions.
We denote the edge lengths by
vj=|γj+1−γj|
and introduce unit vectors Tj and Sj such that
γj+1−γj=vjTj˜γj−γj=ℓSj.
Now it is easy to see that the three conditions above amount to saying that Sj+1 is obtained from Sj (the parallelogram solution) by reflecting it in that diagonal of the parallelogramm that does not contain γj.
The direction of this diagonal is given by the vector ℓSj–vjTj. Reflection in a vector a∈C is the map
z↦aˉaˉz.
Using this we obtain
Sj+1=ℓSj–vjTjℓˉSj–vjˉTjˉSj=ℓSj−vjTjℓ−vjˉTjSj.
Obviously the four points γj,γj+1,˜γj,˜γj+1 lie on a circle. One might formulate this by saying that the polygons γ and ˜γ are “enveloped” by a sequence of circles in such a way that the arclength on the two envelopes is in correspondence.