Is there a two dimensional shape (living in a plane) that can be divided into $2$ and $3$ but not $6$ equal parts of same size and shape? This question is a simpler take on this puzzling.SE question.

If such a shape exists, the $3$ parts can't be symmetric and the $2$ parts can't be divisible into $3$ equal parts.

I'm not sure how to formalize this, but I feel that group theory could have a result about this.

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    $\begingroup$ How are you defining (or the original OP) part? Because under a loose definition of part, the answer is obviously no. $\endgroup$ Dec 29, 2021 at 14:18
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    $\begingroup$ @DonThousand In the same way as when talking about a dissection. How is the answer "obviously no under a loose definition"? $\endgroup$
    – Eod J.
    Dec 29, 2021 at 14:35
  • $\begingroup$ @Eod.J Under the dissection definition, it is not obvious, I agree, but if we let a part simply be a measurable set, the fact that a figure can be divided into any $n$ equal parts is a simple consequence of Fubini's theorem. $\endgroup$ Dec 29, 2021 at 14:36
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    $\begingroup$ I think to speak about groups we need to define "figure" and "division" more precisely. Naive definition would be "split figure as a set into two parts s.t. there is distance-preserving bijection between any two parts", but then even splitting square into two parts becomes non-trivial. We can require parts to be measurable and allow bijection ignore sets of measure 0, but it also complicates things. $\endgroup$
    – mihaild
    Dec 30, 2021 at 10:10
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    $\begingroup$ @mihaild It is not nearly so complicated as you make it out to be. If you really want to formalize it in a measure-theoretic sense we can say that the shape is the union of Euclidean transformations of a single shape, which transformations intersect pairwise in (at most) a finite union of piecewise $C^\infty$ curves, but this is one of those cases where the 'intuitive' definition of dissection has managed to serve just fine for centuries — we talk about tessellations of the plane all the time without being terribly concerned about the boundary chunks. $\endgroup$ Jan 1, 2022 at 14:06

2 Answers 2


Showing that a figure can't be dissected into a certain number of connected* parts is exceedingly hard (I don't actually know of any negative results of this form beyond the case of $2$ pieces, and even then it could be quite difficult). So it is quite unlikely that you will get any answers which can rigorously demonstrate they have no decomposition into $6$ pieces. (You might hope that one's intuition suffices, but this seems not to be true, unless your intuition thinks it plausible that you could cut an equilateral triangle into 5 congruent pieces!)

That said, there are some shapes which seem like promising candidates, and for which I haven't found any dissections into $6$ pieces (counterexamples welcome in the comments). As inspired by this followup question, we might want to reframe the problem: take an asymmetric shape, put three copies of it together to get a symmetric result, and hope that the union does not have some unrelated division into six congruent parts.

The term for this at least in the world of polyforms is an oddity with 3 pieces (plus the restriction that the base piece be asymmetric), which turns up examples like the following polyomino constructions:

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At the very least, it seems likely that any cases like these will have $6$-fold decompositions in ways unrelated to their $2$-fold and $3$-fold dissections; to the extent such things occur, I think it is reasonable to call them essentially a "coincidence", and for them to suggest that no regularity in the structure of tilings is forcing any result of this form.

Expanding beyond polyominoes, we could try similar strategies on other kinds of polyforms. With polyiamonds we have some examples from the same website, though fewer:

enter image description here enter image description here

Polyhexes offer us some more possibilities, though of course they can be interpreted as polyiamonds with six times as many cells too. (I checked the last of the ones shown here, and it is not the union of any six $15$-iamonds when interpreted as a polyiamond.)

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(Note that the first and second examples here give an infinite family of oddities by extending the lengths of all three "legs" of the shape.)

Exploring the more exotic polyforms, we note that polyocts can be used in place of polyominoes for all of the cases using a square grid - this puts additional restrictions on how any $6$-fold dissection might work, and rules out e.g. trying to scale up the polyomino to a larger grid and find a dissection there. Here's an example to show what I mean, but of course this can be generalized to any polyomino example:

enter image description here

In addition, the polyct oddity page gives us one genuinely novel example, shown below.

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Finally, we find several very weird examples within the polybirds:

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(Note that the last example here can be modified by adding arbitrary decorations to the sides of the squares, so long as they possess full $K_4$ (not $D_4$) symmetry.)

Again, I have no proof that any of these shapes don't have some miraculous decomposition into six pieces, but there seems no particular reason to expect them to.

We can actually make all of the above examples even harder to split into $6$ pieces, if we give up on being simply-connected (which many violate anyway), by extending the same idea as with the polyocts. The idea is to cut "random" dihedrally-symmetric holes into the components of each polyfrom, as though cutting out paper snowflakes:

enter image description here

This doesn't affect the two-fold or three-fold decompositions, but it severely restricts any attempt to do clever things with a six-fold decomposition, and essentially forces such a decomposition to respect the polyform boundaries pretty closely.

*Allowing pieces to be disconnected will invalidate e.g. the naive polyomino solutions, but the snowflake-cutting technique still seems to present a challenge for any counterexamples as far as I can tell. I've focused on the connected case implicitly here since it seems the more natural one to me.


Not an answer, but a long comment.

Below is a somewhat arbitrary construction which:


  • Is infinite and unbounded, and has no obvious ways to be made bounded;
  • Is a discrete set of points, and thus has zero area;
  • I don't know how to prove that it cannot be divided into 6 equal parts (although I will argue that this is likely true);


  • Provides evidence that OP's question does not have any trivial obstructions;
  • Might lead to an answer to OP's question by the hands of the people smarter than me.

Here's the construction.

Start by taking the following triangular subdivision of the plane. triangles

Then, arbitrarily number the smaller triangles with some integer values such that each neighboring pair of triangles' numbers differ by exactly 1. This is one of uncountably infinitely many ways to do this. enter image description here

We next form the 3-dimensional set $$S=\left\{(x,y,z)\in\mathbb{R}^3:\begin{matrix}\text{$(x,y)$ is a vertex of a small triangle}\\\text{and $z$ is that triangle's number}\end{matrix}\right\}$$ So $S$ is a union of vertices of triangles, each projecting onto a small triangle but also vertically displaced by that triangle's corresponding altitude. $S$ should not contain line segments or surfaces, only a few many points are allowed.

We pick some linear transformation $f:\mathbb{R}^3\to\mathbb{R}^2$ such that when represented as a $3\times 2$ matrix, all possible entries of the matrix are algebraically independent over $\mathbb{Q}(S)$ (abuse of notation: we adjoin $\mathbb{Q}$ by all coodinates of points of $S$). In effect this gives a "sufficiently random" projection that maps $S$ back to the 2-dimensional plane. Specifically, the function $f$ bijectively maps between the sets $S$ and $f(S)\subseteq\mathbb{R}^2$ while still being linear.

  • Claim: $f(S)$ divides into two equal parts. This is because $S$ comes in pairs of points $\vec{u},\vec{v}\in S$ sharing the same $x,y$ coordinates, differing only in the $z$ coordinate and by exactly $1$. So we divide $S$ into the bottom points and the top points which differ by a single translation by $(0,0,1)$. And under the linear mapping $f$, $f(S)$ is divided into two parts differing by a single translation by $f(0,0,1)$.

  • Claim: $f(S)$ divides into three equal parts. This is because $S$ also comes in triplets of points $\vec{u},\vec{v},\vec{w}\in S$ sharing the same $z$ coordinate and forming one translate of a small triangle originally in the plane. We similarly divide $S$ by the three possible positions of a vertex from a fixed small triangle, and under the linear map $f$, $f(S)$ is divided into three equal parts that differ by translations in the plane only.

  • Conjecture: $f(S)$ cannot divide into six equal parts. This is where we had to pick $f$ to be sufficiently random when we did so. We also need $S$ to be sufficiently random as well, because otherwise if, for example, all gray triangles receive altitude 0 and all black triangles receive altitude 1, then $S$ partitions into sextuplets that are vertices of triangular prisms. When we made totally random choices whenever we could have done so, I don't see any obvious way how $S$ (and $f(S)$) could be partitioned into six equal parts.

Thus $f(S)$ would be a mock candidate construction for a shape that the OP asked for.

  • $\begingroup$ When one connects pairs of points in $S$ that differ by a translation which showed up in the construction of the 2-division or the 3-division, the graph structure on $S$ resembles the Cayley graph of the free product $\mathbb{Z}_2*\mathbb{Z}_3$. $\endgroup$
    – Edward H
    Feb 26 at 8:16
  • $\begingroup$ Of course only after posting this did I realize the OP had asked for simply connected shapes. $\endgroup$
    – Edward H
    Feb 26 at 8:22

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