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Every dissection I can think of that cuts a square into two contiguous congruent shapes seems to be rotationally symmetric. (Allowing disjoint shapes allows for dissections that aren't.) Is there a simple proof or counterexample?

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    $\begingroup$ I suspect you mean it has inversion symmetry? "Rotationally symmetric" would mean that it's invariant under arbitrary rotations about the centre. $\endgroup$
    – joriki
    Jan 3, 2023 at 13:15

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Yes, any such dissection must have the two pieces related to each other by a $180^\circ$ rotation about the center of the square.

Suppose we have cut a unit square into a red piece and a green piece, hereafter $R$ and $G$. Consider the restrictions of $R$ and $G$ to the perimeter of the square, which we'll denote $R_p$ and $G_p$. The requirement that $R$ and $G$ be contiguous forces $R_p$ and $G_p$ to be connected segments of the boundary, so there are just two points on the boundary of the square where we transition from red to green and vice verse.

Now, let's consider the longer of these two pieces of boundary, WLOG $R_p$. Since it is of length at least $2$, it contains one full side, say $AB$. It will then also contain some pieces of $BC$ and $DA$ whose length totals at least $1$.

I claim that a shape congruent to $R_p$ can only fit in the square when it's placed to align with an axis. Why is this? Consider, for a given skew angle, how much length we can pack into the square at that angle:

enter image description here

You can do a bit of algebra and check that the sum of the blue lengths is always less than $1$ for any orientation of the red unit segment in the diagram. This means that for any way we tilt $R_p$ that isn't axis-aligned, there won't be room to fit it in the square.

What does this mean? It means that if $f$ is the isometry of the plane such that $f(R) = B$, $f(R_p)$ carries us to some axis-aligned image of $R_p$ within $B$. In particular, $f(\overline{AB})$ must be either parallel to $\overline{AB}$ or parallel to $\overline{BC}$.

  • Case 1: $f(\overline{AB})$ is parallel to $\overline{BC}$. This means that there is a point $p\in \overline{AB}$ contained in both $\overline{AB}$ (red) and $f(\overline{AB})$ (green). The only way this can happen while preserving contiguity is if $p=A$ or $p=B$ (without loss of generality we'll assume the former). This means that $R_p$ is exactly equal to the union of the line segments $\overline{AB}$ and $\overline{BC}$, and $f$ is exactly the reflection across the line $\overline{AC}$. But this forces the dissection to be the one which cuts the square into two congruent triangles, which is rotationally symmetric.

  • Case 2: $f(\overline{AB})$ is parallel to $AB$. Since $R$ is entirely on one side of the line segment $\overline{AB}$, $G$ will be entirely on one side of the line segment $f(\overline{AB})$. This means that either (2.1) $f(AB)=CD$ or (2.2) $R$ and $G$ are each rectangles.

    • (2.1) In the former case, $f$ is either a reflection across the line joining the midpoints of $BC$ and $AD$, or $f$ is a 180 degree rotation about the center. The latter is exactly what we wanted to show, and the former forces our dissection to be two congruent rectangles side by side, which can also be viewed as a dissection with rotational symmetry.
    • (2.2) The only way for this to happen is if $R$ and $G$ are of the rectangular form described above, which can be viewed as a 180 degree rotation.
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