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Edited - some comments may now be out-of-date.

I thought I had a complete set of solutions to this:

Cut a square into identical pieces so
that they all touch the center point.

It became clear after some discussions that I was very, very wrong.

There are infinite families of solutions, and a sporadic. So I have two questions:

  1. What do you think is a complete set of solutions?

  2. What techniques and approaches can I use to prove that the ones I have are all there are?

Hope that's clearer. Thanks.

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What exactly is your question? How can we help you to show that your answer is "complete", when we don't know your answer, let alone your proof? –  Abel Stolz May 26 '11 at 13:15
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It seems to me that there are infinitely many solutions. Voting to close. –  Steven Gubkin May 26 '11 at 13:17
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@Colin: that question may be clear, but that is not the question you asked. The question you asked is "did I miss any solutions, and how can I prove that my answer is complete," and neither of these questions is possible to answer without knowing what your answer is. –  Qiaochu Yuan May 26 '11 at 13:21
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@Colin: also, cross-posting on math.SE (math.stackexchange.com/questions/41499/…) is discouraged. Generally we want people to decide on one site, and if it turns out not to be appropriate (or doesn't get any answers on math.SE) then ask on the other site. –  Qiaochu Yuan May 26 '11 at 13:43
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Too many or's. Choose something. Meanwhile, I'll make my choice. I do not want to turn it into a topology problem, so I'll not assume the pieces connected. I do not want it to become some amenability question either, so I suggest to assume that each set is a closure of its interior with boundary of zero measure. Under these assumptions, look at artofproblemsolving.com/Forum/viewtopic.php?f=48&t=408538 Is it one of the solutions you knew? –  fedja May 26 '11 at 20:26

3 Answers 3

up vote 10 down vote accepted

You can take any of a variety of paths from the center to the edges alt text.

There are various ways to draw a a path from one vertex of a small square to the opposite one and get two symmetric pieces (although one can't just use horizontal and vertical segments). Then 4 of those little squares yield a dissection into 8 pieces.

Later addition Under more relaxed conditions, here is an example with 32 pieces (and a suggestion that maybe there is no limit) . If you look at it the right way you might be able to convince yourself that each piece is path-wise connected. The 4 colored pieces are congruent by rotation around the center of the area shown (which is a quarter of the full square). Using a reflection will give 4 more pieces for a total of eight. These eight pieces (so far) fit together to fill in the square shown and each touches each corner. Put together 4 copies of this (32 pieces total) to get the full square.

enter image description here

As it is, each piece in the full square touches one corner and the center. I have not totally convinced myself, but it seems that it should be possible to divide each piece into 4 by quadrisecting each acute angle and then recolor in such a way as to have all 32 pieces each touching all 4 corners. If so, then 4 copies of that figure could be arranged and give a square partitioned into 132 pieces all touching its center. If that is correct then there should be no limit.

comments

1) If you want respect on this site then stop hinting that you have a perhaps complete classification with three families plus one sporadic (I don't mind that but some people here do). Describe them carefully enough that people can decide if they have others that they can think of. Is the sporadic case the entire square? I gather that you think every solution uses 1,2,4 or 8 pieces. I imagine that is true but the proof of that alone would be a good start and might not be that easy (see comment 3 below.)

2) Your description would probably make clear what you mean by piece, but the most general definition commonly seen (although I have now used a looser definition above) might be something like : "a closed topological disk in the plane with boundary a simple closed curve." You wish to find a finite set of such pieces, all congruent (reflection allowed), disjoint interiors, union the square and the center on the boundary of each piece (ignoring the one piece case...). Tedious, but worth saying anyway. For reference below: a polyomino is such a tile made of unit squares meeting edge to edge.

3) It is fun and challenging to find a missed example for claims of the sort: "this is all tilings." It can be surprisingly difficult (compared to the "obviousness" of the result) and rather tedious (to my taste) to prove that there are not any exceptions. Here is a 7 page paper discussing when you can split a polygon into 2 congruent shapes:

Splitting a Polygon into Two Congruent Pieces Kimmo Eriksson The American Mathematical Monthly Vol. 103, No. 5 (May, 1996), pp. 393-400 It might be relevant for this problem and at least is an example of how to prove such things.

I can believe that the only way to partition a rectangle into 3 congruent pieces is if the pieces are themselves rectangles. Here is an 8 page proof:

Samuel J. Maltby Trisecting a rectangle Journal of Combinatorial Theory, Series A Volume 66, Issue 1, April 1994, Pages 40-52

It cites the following classic (to those who follow these matters) 6 page note proving a long conjectured result (again the methods in the paper could be relevant):

Polyominoes of order 3 do not exist I. N. Stewart and A. Wormstein Journal of Combinatorial Theory, Series A Volume 61, Issue 1, September 1992, Pages 130-136

Abstract: The order of a polyomino is the minimum number of congruent copies that can tile a rectangle. It is an open question whether any polyomino can have an odd order greater than one. Klarner has conjectured that no polyomino of order three exists. We prove Klarner's conjecture by showing that if three congruent copies of a polyomino tile a rectangle then the polyomino itself is rectangular. The proof uses simple observations about the topology of a hypothetical tiling, and symmetry arguments play a key role.

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2  
To fill out: Take a path from a point on the boundary to the centre. Rotate it by a quarter turn, half turn, three quarters about the centre of the square. A condition is required so that these paths do not cross each other, but if they don't they divide the square into four congruent pieces. For eight piece solutions, divide into four by horizontal and vertical straight lines through the centre. Join the centre to the edge with a path having twofold rotational symmetry around the midpoint of the diagonal from centre to vertex and lying within the quarter. Replicate in the other quarters. –  Mark Bennet May 26 '11 at 17:19
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@Colin D Wright: This is not a discussion site. This is a question and answer site. Say what you know if you want me to spend time thinking about it. –  Douglas Zare May 26 '11 at 19:31
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@Colin I forgive you and I don't disagree! I'm just telling you what I've seen. I'd still like to know what your 3 families and sporadic case are. –  Aaron Meyerowitz May 26 '11 at 20:23
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My infinite families are as follows. Take any non self-intersecting path from the center to the border such that a 180 degree rotation does not intersect. That divides the square into two. Similarly for a 90 degree rotation, which divides the square into 4. Now divide into four squares, and for each small square, divide diagonally with a curve that has 180 degree symmetry. That divides the square into 8. The sporadic is the trivial case - one piece. But fedja has just demonstrated a 16 piece dissection, so now I know I know nothing. –  Colin D Wright May 26 '11 at 20:56
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@Aaron: This is a cool answer. I urge you to rephrase the beginning of comment (1), however, even though the OP seems not to have been offended. –  Daniel Litt May 29 '11 at 7:44

I think there are very few such solutions. The pieces must be identical, and they must touch the center. Consider the segment joining the center with one of the vertices. Then all small figures (in which you split the square) must contain a segment of this length, and there are only four such segments. Any such segment belongs to at most two small figures, and we find that there are at most $8$ small figures. From here on it is easy to see that the possible splits are:

  • the square itself
  • the square cut by a diagonal
  • the square cut by two diagonals
  • the square cut by parallel lines through the center
  • the square cut by parallel lines through the center and by its diagonals
  • the square cut by a line through the center
  • the square cut by two orthogonal lines through its center.
  • the square cut by any smooth curve symmetric by its center.
  • the square cut by any smooth curve symmetric by its center, and the rotate of this curve by $\pi/2$.

There are indeed many solutions. Sorry for my initial remark. I think that essentially the square can be dissected in 2,4 or 8 parts. The 8 parts is unique. The 2 parts cutting must be symmetric by its center, and the 4 parts cutting must be made such that is invariant by a $\pi/2$ rotation.

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That's what I originally thought, but there are many, many more. Hence asking about how to prove the completeness of the enumeration. –  Colin D Wright May 26 '11 at 13:26
    
Present one of your many more solutions, to see where my answer is wrong. –  Beni Bogosel May 26 '11 at 13:29
    
@Beni: Do you mean, literally, "must contain a segment of this length"? –  Joseph O'Rourke May 26 '11 at 13:35
    
Draw a semicircle with one end at the centre and the other touching the side of the square in the middle. Rotate 180 degrees. Now you have a sort of Ying-Yang in a square. –  Colin D Wright May 26 '11 at 13:36
    
You're getting closer to the solutions I have. I have three infinite families and a sporadic. The challenge is to show that these are all, and the question is what techniques people might suggest for approaching this. –  Colin D Wright May 26 '11 at 13:47

The number of solutions is the maximal possible, namely $2^{2^{\aleph_0}}$. But we need to be specific about some definitions. To begin, what does it mean to cut into identical pieces? There seems to be agreement on this point that it means to partition the square into finitely many pieces each of which can be rotated into the other. The other definition I will assume is that "touching the center point" means having that point in the closure of each set. We also need to assume that we actually have a partition of the square minus the centre point, because otherwise there is no solution except for the square itself because the centre point can belong to only one member of the partition.

Given all of this, let $[(p^0_\xi,p^1_\xi)]_{\xi\in 2^{\aleph_0}}$ enumerate all pairs of points in the square symmetric about the central point. For any function $f:2^{\aleph_0}\to 2$ let $X(f)$ be the set of all $p_\xi^{f(\xi)}$ and $Y(f)$ the complement. This is a partition of the square into two pieces as desired, and there are $2^{2^{\aleph_0}}$ such partitions.

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Yes, but it's not all the solutions. Agreed that more solutions won't increase the cardinality, but not including all the solutions seems "sub-optimal." What if you additionally require the "pieces" to be connected (in some sense)? –  Colin D Wright May 26 '11 at 18:48
    
If you ask for all partitions into two pieces, the sets $X(f)$ and $Y(f)$ I describe do give you all solutions. If you ask for partitions into $n$ pieces then a similar argument, using $n^\text{th}$-roots of unity, also yields a description of all partitions. If you ask for connected pieces, that is a different question. –  Juris Steprans May 26 '11 at 19:51
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Even connected is not enough though. For example let the square in question be centred at the origin and have width 2. For any subset $X\subseteq (0,1]$ consider the partition consisting of the open upper half of the square together with $X$ and $[-1,0) \setminus -X$. This again yields many connected partitions. So one should probably ask for partitions into open connected sets and define "partition" to mean you partition all but a closed nowhere dense subset of the square (so that you can ignore the boundaries). –  Juris Steprans May 26 '11 at 20:06
    
Or define two partitions to be equivalent if they agree on all but a nowhere dense set, which is probably the same, but I'd need to work hard to think about that clearly enough. Very helpful - thank you. –  Colin D Wright May 26 '11 at 20:22

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