# Dividing square into two congruent parts.

Recently I came acrooss that Putnam problem https://prase.cz/kalva/putnam/psoln/psol6412.html. I've been thiniking about other figures which cannot be divided into two congruent parts. Here's my proof for line segment

Suppose indirectly that it is possible to divide a line segment $$\overline {AB}$$ into two congruent parts $$\overline {AB} = X \sqcup Y$$. Consider the bijection of $$f: X \rightarrow Y$$ which is an isometric involution and its inverse$$g: Y \rightarrow X$$. Let $$O$$ be the center of $$\overline {AB}$$ segment. Let us assume without loss of generality that point $$O \in X$$. So its reflection $$O '$$ belongs to the set $$Y$$. Now consider the set of $$T$$ points in segment $$\overline {AB}$$ separated by $$ka$$ from point $$O$$, where $$a = | OO '|$$, and $$k \in \mathbb {Z} _ {\geqslant 0}$$. Every element of this set has reflection in this set. This is because the distance between each of these points and its reflection is a multiple of $$a$$, so this reflection belongs to $$T$$. However, since $$O$$ is a $$\overline {AB}$$ middle, then on both sides of $$AB$$ there are the same number of points from the set $$T$$, so the set $$T$$ has an odd power. Given that the involution has no fixed points, so the set $$T$$ have an even power - a contradiction.

I want to prove the same for the square, but I'm struggling with that. I would prefer "easy" solution like that from putnam and mine, but any help will be greatly appreciated.

I know that you asked for an easy solution, but there is a general result that no compact set with odd Euler characteristic can be divided into two congruent parts. In particular, anything homeomorphic to a line segment or disc (like the square) has Euler characteristic $$1$$, and so this cannot be done.

This is because, if your set $$X$$ can be divided into congruent $$Y_1, Y_2$$, then there exists an involution $$f\colon X\rightarrow X$$ such that $$f(Y_1) = Y_2, f(Y_2) = Y_1$$, so that $$f$$ cannot have any fixed points.

But see the discussion in this post: Does every involution have a fixed point?, which shows that any involution of a compact space with odd Euler characteristic has a fixed point.

By the way, this explains why removing the centre from the disc allows it to be partitioned into congruent pieces, since a punctured disc has Euler characteristic $$0$$.