# Tiling the plane with finitely many congruent pieces

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.

Write $$A_i=T_i(A)$$ for $$A=A_1$$, where each $$T_i$$ is a rigid motion. For each $$i$$, we have $$|A_i\cap B(r)|=|T_i(A\cap T_i^{-1}(B(r))|=|A\cap T_i^{-1}(B(r))|$$. The symmetric difference between this set and $$A\cap B(r)$$ is contained in the symmetric difference of $$T_i^{-1}(B(r))$$ and $$B(r)$$, whose measure is $$O(r)$$. Hence the measures of $$A_i\cap B(r)$$ are all equal up to $$O(r)$$. Since the sum of their measures is $$B(r)$$, this implies $$|A_i\cap B(r)|=B(r)/n+O(r)$$ for each $$i$$.