Geometrical Paradoxes

Two polygons in the plane are congruent by dissection if one of them can be decomposed into finitely many polygonal pieces that can be rearranged using isometries (and ignoring boundaries) to form the other polygon.

It is clear that polygons that are congruent by dissection have the same area. A simple proof can be given by efficiently making use of the fact that congruence by dissection is an equivalence relation.

Let's prove that two polygons of same area are always congruent by dissection.

We do this first for a triangle. Any triangle is congruent by dissection to a rectangle:

Any rectangle, in turn, can be transformed to a square. The animation shows how it is done if rectangle's length is at most 4 times its width:

If the rectangle length is more than 4 times its width, it can be transformed to a rectangle of suitable proportions:

Finally, any number of squares can be transformed into one square iteratively:

Thus, any polygon is congruent by dissection to a square of the same size.

This means congruence by dissection is equivalence relation. By unobvious reasons, congruence by dissection is strongly connected to another equivalence relation, equidecomposability.

Suppose $G$ acts on $X$ and $A, B \in X$. Then $A$ and $B$ are $G$- equidecomposable (sometimes called finitely $G$-equidecomposable or piecewise $G$-congruent) if $A$ and $B$ can each be partitioned into the same finite number of respectively $G$-congruent pieces.

Theorem: If $D$ is a countable subset of $\mathbb{S}^2$, then $\mathbb{S}^2$ and $\mathbb{S}^2 \\ D$ are $SO_3 (\mathbb{R})$-equidecomposable (using two pieces).

This leads to the Banach–Tarski Paradox itself:

The sphere $\mathbb{S}2$ is $SO_3 (\mathbb{R})$-paradoxical, as is any sphere centered at the origin. Moreover, any solid ball in $\mathbb{R}^3$ is $G_3$-paradoxical and $\mathbb{R}^3$ itself is paradoxical.