Non-empty, disjoint subsets of $\Bbb R^2,$ both isometric to their union? This question was posed to me by a friend who tells me the answer is yes, but I cannot see why. 
Does there exist two nonempty, disjoint sets $A, B \subset \mathbb{R}^2$ such that $A$ and $B$ are both isometric to their union?
I cannot for the life of me come up with a way of constructing both sets - I've noticed that the measure of both has to be $0$, but that's about it.
 A: Let the group with presentation $\langle x,y \mid x^2=e \rangle$ (that is, the free product $C_2 * \mathbb Z$) act on the plane by mapping $x$ to a rotation by $180^\circ$ about $(1,0)$ and $y$ to a rotation about the origin by some angle $\theta$ whose cosine is transcendental.
(Originally I claimed that this action is faithful, but actually it isn't -- the group element $y^{-2}xyxyxy^{-2}xyxyx$ has the identity action no matter what $\theta$ is. Fortunately less can do for this particular purpose).
Let $K$ be the set of group elements that can be written on the form
$$ y^{n_m}xy^{n_{m-1}}x\cdots xy^{n_1} x y^{n_0} $$
where $m\ge 0$, and each $n_j\ge 0$ except that $n_0$ may be negative.
This set is not a subgroup, but it has two useful properties:


*

*$K$ is the disjoint union of $\{yk\mid k\in K\}$ and $\{xyk\mid k\in K\}$.

*Different elements of $K$ map the point $p_0 = (3,0)$ to different points. (This can be seen by going to the complex plane where $y$ is multiplication by $e^{i\theta}$ and $x$ is the map $z\mapsto 2-z$. Then each element of $K$ maps $3$ to a different Laurent polynomial in $e^{i\theta}$ with integer coefficients, and those all have distinct values because $e^{i\theta}$ is transcendental. Phew!)
Now set
$$ A = \{ykp_0 \mid k\in K\} \qquad B = \{xykp_0 \mid k\in K \} $$
Then $A$, $B$, and $A\cup B$ are related by rigid motions of the plane, namely:
$$ y^{-1}A = (xy)^{-1}B = \{kp_0 \mid k\in K \} = A \cup B $$
so they are isometric.

This construction is inspired by the initial step of the proof for the Banach-Tarski paradox. It doesn't need the axiom of choice because it doesn't need to select a representative of each orbit, because it is not required that $A\cup B$ is an entire pre-existing shape.
It is natural to ask how $A$ looks -- but unfortunately it can't really be seen: it is dense in $\mathbb R^2$.
