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Given any integer $n$, any rectangular region or any sector of a disc (including the full disk as a boundary case) can be cut into $n$ mutually congruent pieces - by equally spaced parallel lines and lines radiating from a point at equal angular spacing respectively.

Intuitively, this property generalizes to some deformations of the rectangle which continue to allow partition into $n$ congruent pieces by mutually parallel and equally spaced polylines or curves for any $n$ and for some deformations of a sector which can be congruent partitioned by mutually congruent curves radiating from a single point. Among non-convex regions, the portion between two concentric arcs of different radii and equal angle measure has this property.

Question: Are there any other classes of planar regions which can be cut into $n$ mutually congruent and connected regions for any $n$? The answer seems negative. But is there a proof?

Note 1: It seems that in non-Euclidean geometry, the sector (including the disk as a limit) is the only class of planar figures that allows partition into n mutually congruent pieces for all $n$.

Note 2: In 3D, one readily has parallelopipeds and suitable slices of regions with axial symmetry (sphere, torus, cone...) which can obviously be cut into n mutually congruent 3D regions for any $n$.

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  • $\begingroup$ Somewhat related: math.stackexchange.com/questions/1838617/… and also math.stackexchange.com/questions/8288/… $\endgroup$ Commented Jan 13, 2021 at 12:05
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    $\begingroup$ I'm confused by your comment that "there seem to be no non-convex planar region at all with such a property": as you mentioned, sectors of discs have this property, and some sectors of discs are not convex. $\endgroup$ Commented Sep 4, 2022 at 21:48
  • $\begingroup$ Thanks for the correction. Yes, the portion between two concentric arcs with equal angle measure is not convex and it has the property we need! Edited the question incorporating your observation. $\endgroup$ Commented Sep 5, 2022 at 4:32
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    $\begingroup$ Is there a proof that a regular pentagon can’t be cut into three congruent pieces? It would seem silly to ask the more complicated question if there’s no proof for the simplest non-rectangular subcase. $\endgroup$
    – user44143
    Commented Sep 16, 2022 at 4:22
  • $\begingroup$ There indeed are planar regions that cannot be cut into n mutually congruent pieces for some values of n and some regions which cannot be cut into n congruent pieces for any n>1.The present question only asks whether there are classes of shapes other than those it (the question) mentions with the property of allowing partition into n congruent connected pieces for all values of n. $\endgroup$ Commented Sep 17, 2022 at 6:33

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The planar regions that can be cut into $n$ congruent pieces (for all $n$) seem to be almost exactly those of the form

$$a\le r\le b\\ 0\le\theta\le c$$

in polar coordinates. This includes these cases:

  • $a=0, b<\infty, c=2\pi$ — a disk
  • $a=0, b<\infty, c<2\pi$ — a circular sector (pizza-slice)
  • $a=0, b=\infty, c=2\pi$ — the entire plane
  • $a=0, b=\infty, c<2\pi$ — an infinite pizza-slice
  • $a>0, b<\infty, c=2\pi$ — an annulus (difference of two concentric disks)
  • $a>0, b<\infty, c<2\pi$ — an annular sector (the polar analogue of a rectangle)
  • $a>0, b=\infty, c=2\pi$ — everything in the plane except a disk
  • $a>0, b=\infty, c<2\pi$ — an infinite pizza-slice with a bite taken out

But there are also some special cases intuitively involving a point at infinity but which I'm not sure how to describe rigorously. I'll handwave here by pretending that $\frac{\infty}{2}$ is a thing.

  • $a=\frac{\infty}{2}, b<\infty, c<2\pi$ — a rectangle
  • $a=\frac{\infty}{2}, b=\infty, c<2\pi$ — an "infinitely tall rectangle," the region between two vertical lines chopped off with a horizontal chop

Two cases are even weirder:

  • $a=\frac{\infty}{2}, b<\infty, c=2\pi$ — an "infinitely wide rectangle," the region between two horizontal lines. This can be cut into $n$ congruent pieces, but only by slicing it into horizontal slices, not vertical slices as you might expect from the analogy

  • $a=\frac{\infty}{2}, b=\infty, c=2\pi$ — the half-plane. This can't be cut into $n$ congruent pieces by either horizontal or vertical slices; but coincidentally it happens to define the exact same region as the "infinite pizza-slice" case above ($a=0, b=\infty, c=\pi$).

And then there's another wrinkle: each case can be deformed, as you pointed out. A rectangle can become a parallelogram or even a chevron (dissectable into $n$ thin chevrons bundled together) or zigzag. A circular sector can have its two straight sides zigzaggified in the same way. For example, if you take the Reuleaux triangle and invert ("pop in") one of its three sides to make a sort of Starfleet emblem... well, that's a valid region, I think.

All this is to say that your conjecture seems basically plausible; but there are so many handwavey bits here — especially what kinds of "zigzaggifications" are allowed — that I'm not sure my answer is adding much value. ;) Also, I was surprised by the failures of my polar-coordinates system in the $a=\frac{\infty}{2}, c=2\pi$ cases: it produced dissectable regions but they were dissectable only in ways different from what the analogy predicted (horizontal slices instead of vertical; radial slices instead of parallel). That suggests that there may be other ways to dissect things we haven't predicted at all, yet.

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