Dividing square into two congruent parts.

Question

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.

Answer 1

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 $0$.