If I have an equidissection of a square into various polygons, and I want to map each point on the square to a point on the unit circle such that it each piece (which is not necessarily still a polygon) still has the same area as each other piece.

Generally, I've tried using the square of side length $2$ centered at $0, 0$ and I tried to map $r, \theta$ to $\frac{r}{r_{max}}, \theta$ where $r_{max}$ is the largest $r$ such that the point $r, \theta$ fits inside the square, but respective area is in this case not preserved unless the $\theta$ values of each point are constant, meaning the area is $0$.

Is there some sort of way I can use the Schwarz-Christoffel mapping backwards to map a square to the infinite half-plane and then reapply it to turn that half-plane into a polygon of side-length $\infty$ (i.e. a circle)? And is there any way that would likely preserve respective area?

Is there another mapping I should try?

  • $\begingroup$ I woulnd't like to discourage you, but am very pessimistic at the attempt you do: preserving areas can be done in 3D (example Mollweide projection) but in 2D, as long as you have one or two "shapes", you will succed to map them onto "shapes" with the same area, but with many shapes that have to be mapped into the unit disk... $\endgroup$ – Jean Marie May 10 '16 at 16:05
  • $\begingroup$ @JeanMarie I was trying to keep $\theta$ constant and find a function $f(r)$ that acts from $[0, r_{max}]$ to $[0, 1]$. Do you have any tips? $\endgroup$ – Carl Schildkraut May 10 '16 at 18:06

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