Suppose $A_1,\dots,A_n$ are subsets of the plane that are all related by rigid motions such that $|(A_1 \cup \dots \cup A_n)^c| = 0$ and $|A_i \cap A_j| = 0$ for all $1 \leq i < j \leq n$, where $|S|$ denotes the Lebesgue measure of $S$.

Must each $A_i$ have the property that $|A_i \cap B(r)|/|B(r)| \rightarrow 1/n$ as $r \rightarrow \infty$, where $B(r)$ is the disk of radius $r$ centered at 0?

The answer is clearly “yes” if the $A_i$’s are all obtained from one another by rotation about a point (e.g., consider the Fatou sets for Newton’s algorithm applied to the polynomial $z^3 - 1$). Maybe this is the only way to tile the plane by $n$ congruent pieces, but I don’t see why it should be true.

Suppose $A_1,\dots,A_n$ are measurable subsets of the plane that are all related by rigid motions such that $|(A_1 \cup \dots \cup A_n)^c| = 0$ and $|A_i \cap A_j| = 0$ for all $1 \leq i < j \leq n$, where $|S|$ denotes the Lebesgue measure of $S$.

Must each $A_i$ have the property that $|A_i \cap B(r)|/|B(r)| \rightarrow 1/n$ as $r \rightarrow \infty$, where $B(r)$ is the disk of radius $r$ centered at 0?

The answer is clearly “yes” if the $A_i$’s are all obtained from one another by rotation about a point (e.g., consider the Fatou sets for Newton’s algorithm applied to the polynomial $z^3 - 1$). Maybe this is the only way to tile the plane by $n$ congruent pieces, but I don’t see why it should be true.