TITLE: Dissecting a square
QUESTION [6 upvotes]: Edited - some comments may now be out-of-date.
I thought I had a complete set of solutions to this:
Cut a square into identical pieces so
that they all touch the center point.

It became clear after some discussions that I was very, very wrong.
There are infinite families of solutions, and a sporadic.  So I have
two questions:

What do you think is a complete set of solutions?
What techniques and approaches can I use to prove that the ones
I have are all there are?

Hope that's clearer. Thanks.

REPLY [2 votes]: The number of solutions is the maximal possible, namely $2^{2^{\aleph_0}}$. But we need to be specific about some definitions. To begin, what does it mean to cut into identical pieces? There seems to be agreement on this point that it means to partition the square into finitely many pieces each of which can be rotated into the other. The other definition I will assume is that "touching the center point" means having that point in the closure of each set. We also need to assume that we actually have a partition of the square minus the centre point, because otherwise there is no solution except for the square itself because the centre point can belong to only one member of the partition.  
Given all of this, let $[(p^0_\xi,p^1_\xi)]_{\xi\in 2^{\aleph_0}}$ enumerate all pairs of points in the square symmetric about the central point. For any function $f:2^{\aleph_0}\to 2$ let $X(f)$
 be the set of all $p_\xi^{f(\xi)}$ and $Y(f)$ the complement. This is a partition of the square into two pieces as desired, and there are $2^{2^{\aleph_0}}$ such partitions.