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Let $P$ be a polygon with $180^\circ$ rotational symmetry. Let $O$ be the center of $P$ and suppose $P$ is dissected into congruent polygons $A$ and $B$. Must the $180^\circ$ rotation around $O$ send $A$ to $B$?

This conundrum appears to have stumped everyone I know, and we don't know if the answer is "yes" or "no".

If the problem is too difficult, the case in which $P,A$, and $B$ are polyominos (polygons formed by joining unit squares along their sides) is also of interest.

RavenclawPrefect
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greenturtle3141
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  • The question in title and the body don't match. Anyway, there is no way to partition a polygon into two disjoint congruent polygon -- think about which subpolygon the point $O$ belongs to. – user10354138 Oct 15 '23 at 04:57
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    "Partition" should be interpreted in a layman's sense; ignore boundaries. – greenturtle3141 Oct 15 '23 at 05:02
  • Substituting "partition" with "dissection" to make this clearer. – greenturtle3141 Oct 15 '23 at 05:14
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    Depends on what you admit as a polygon. For example, I can write a square as the union of two (self-intersecting) quadrilaterals. – user10354138 Oct 15 '23 at 05:22
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    @user10354138 The question is obviously about non-self-intersecting polygons. – lily Oct 15 '23 at 05:42
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    I think the close votes for this question are misplaced. This question is obviously not a homework problem, and natural/elegant enough to motivate itself. Adding "context" here would not improve the question. – RavenclawPrefect Oct 17 '23 at 00:39
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    I have a proof too long to fit here that for grid-aligned congruent polyominos $A,B$ to disjointly union to a polyomino $P$ $180^\circ$ rotationally symmetric about point $O$, if $A,B$ are not already $180^\circ$ rotated images about $O$ of each other, then there must be a point $O'\ne O$ such that $A,B$ are $90^\circ$ rotated images of each other about $O'$. Having found an infinite parametrized family of not-necessarily-simply-connected such pairs $(A,B)$ for fixed $O,O'$, I suspect examples where $A,B$ are connected exist, but only findable by some long computer search. – Edward H Nov 02 '23 at 04:23
  • @EdwardH Can you at least post such an example, which will show that the answer to this question is a "no"? (at least if we relax it to not-necessarily-simply-connected) – Calvin Lin Nov 04 '23 at 15:36
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    @CalvinLin As you wish: Note that when we relax to disconnected polyominos, even something simpler like 1x1 squares centered at {(-1,0),(1,0)} and {(0,-1),(0,1)} work. In the particular case I mentioned, something like 1x1 squares centered at {(-1,-3),(-1,-2),(-1,0),(0,-2),(1,1),(1,2),(2,2),(3,-2)} and {(-2,1),(-2,2),(-1,1),(0,-1),(2,-1),(2,0),(2,3),(3,-1)} work. – Edward H Nov 04 '23 at 23:51

2 Answers2

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Since no-one has tried an answer yet here goes...

It's to do with whether it's possible for one polygon to be rotated to make the other e.g. by 90 degrees, but not 180 degrees.

In diagram 1) region A is the join of the two polygons

enter image description here

If we have polygon 1 rotated to make polygon 2, we have region B. Since the entire polygon has rotational symmetry order 2, we have region C as a copy of region A and B...diagram 2)

then we get region D, since polygon 2 is a rotation of 1, then region E due to the rotational symmetry of the entire polygon etc... leading to diagram 3

This would not be allowed, since there are gaps along the join. It could be argued however that the join could be designed to eliminate any gaps or overlaps by making the join rotationally symmetric, order 2 like this

enter image description here

But this makes the two polygons rotations of each other with a 180 degree rotation. So if this reasoning is correct, it doesn't seem possible for the second polygon to be a rotation of the first by 90 degrees.

Extension: The same kind of argument seems to apply for other rotations and part rotations, an inward pointing region at A is reproduced at B, by a 36 degree rotation, then C due to the 180 rotational symmetry of the whole shape, then D by a 36 degree rotation etc...

enter image description here

Starting with A, we end up with A and J at the join, requiring the join to be rotationally symmetric (order 2) making the red polygon a rotation, by 180 degrees of the first, so far it's looking as though the answer to your question is 'yes'.

John Hunter
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  • What if the two congruent polygons are mirror images of each other, not just rotations? – Jaap Scherphuis Nov 06 '23 at 09:07
  • @JaapScherphuis Yes, the post just shows that if we try to have the second polygon as a 90 degree rotation of the first, then we end up with polygons with rotational symmetry of order 4 (and that includes the 180 degree rotation symmetry), so if polygons exist that are not 180 degree rotations of each other, there must be a reflection involved – John Hunter Nov 06 '23 at 09:17
  • @Edward H labels A, B,C etc... are different parts (or regions, set of connected edges) )of polygons 1 and 2. The final diagram is a new pair of polygons, separate to the first 3 diagrams labelled 1),2),3). Points P and Q on the final diagram are points on the ends of the join of the two polygons. Hope it makes sense... – John Hunter Nov 06 '23 at 12:32
  • @JaapScherphuis The mirror version has a simple proof for polygons in general: First you show by a barycentric argument the center of symmetry $O$ lands on the line $\ell$ of reflection. Then by connectedness the two tiles lie on different sides of $\ell$. It follows next that the two tiles are rotationally symmetric across $O$. Glide reflections and translations go into a same proof for general polygons too (but different from the reflections proof). Somehow the only "hard" case is rotations about a point. – Edward H Nov 07 '23 at 00:22
  • @Edward H so if the two polygons exist, maybe one is rotated by e.g. 120 degrees to get the other, (or any angle that doesn't give the 180 degree rotational symmetry), maybe with a reflection too. Perhaps each being made up of joined equilateral triangles?? – John Hunter Nov 07 '23 at 08:45
  • Yes, $120^\circ$ rotations also may or may not be possible. Polyamonds and polyhexes may or may not be possible. The general case of rotations about a point is beyond my understanding so far. However reflections are already ruled out, as well as anything other than a pure rotation. – Edward H Nov 07 '23 at 11:33
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Here's an incomplete proof. An equivalent question is whether a polygon $P$ with $180^\circ$ rotational symmetry about $O$ can be divided into two congruent polygons such that the dividing curve $D$ between the polygons does not have $180^\circ$ symmetry about $O$.

First, we can rule out $D$ having $180^\circ$ symmetry about a point $O'\neq O$. This is because, in such a case, we can shift $D$ to have it's center at $O$ (perhaps extending $D$ in a symmetric way if necessary), which will result in a new division of $P$ into $A'$ and $B'$. By the $180^\circ$ symmetry of $P$ and the new shifted $D$ about $O$, $A'$ and $B'$ must have the same area. However, this shifting must've increased the area of one of $A,B$ (and correspondingly decreased the other), meaning $A,B$ had different areas. But this contradicts them being equivalent.

So now suppose we have a $D$ that is not symmetric about any point. $D$ then looks different whether you are approaching $D$ from the $A$ side or the $B$ side. Let's call the two versions "innie" and "outie" respectively.

Now, each of $A$, $B$, $P$ may have several copies of $D$ along their respective boundaries, either as innies or outies. Suppose for the moment that the various innies and outies cannot overlap except possibly at their endpoints. Then the number of innies of $P$ and outies of $P$ must each be finite, and moreover must each be even, by rotational symmetry. The total number of innies between $A$ and $B$ is therefore odd, and the total number of outies odd as well, since the boundary $D$ contributes one more innie to $A$ and outie to $B$ that were not included in the boundary of $P$. But this is impossible: since $A$ and $B$ are equivalent, they have the same number of innies (and same number of outies) and therefore the total number of innies and outies between them must each be even.

It remains to handle the case where the copies of the dividing curve $D$ found along the boundaries can intersect each other. There are a couple potential issues. The first is that the number of copies along the boundaries may be infinite. This can only occur, however, if $D$ is a line segment (in which case a longer line segment somewhere on the boundary would contain infinitely many copies of $D$). But line segments have rotational symmetry, which we've already ruled out.

The bigger problem is that you may have a copy $D$' of $D$ along the boundary of $A$ or $B$, such that part of $D'$ overlaps $D$, and part does not. Then $D'$ is counted in the count of innies or outies of $A$, but not in the count of innies or outies of $P$. I suspect you can handle this by breaking $D$,$D'$ into the segments of overlap and non-overlap and doing a similar counting argument, but I haven't found a concise proof of this yet.

AAA
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