# Dissecting a square into congruent pieces that all touch the centre

Edited for clarity:

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?

I hope that's clearer. Thanks.

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– Qiaochu Yuan May 26 '11 at 13:44

What kind of cutting is allowed? There is also a set of solutions starting from dividing the square into eighths with 45-45-90 triangles (assuming reflection is allowed). You can then change the straight lines into curves as described by Phonon. Maybe not in the spirit of the question is to consider all the rays from the center to the square. Partition them into four sets making sure that each set of four rays at 90 degree angles go into distinct sets. $24^{\mathbf{c}}$ solutions!

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I think it's best if the pieces are connected, perhaps even pathwise connected. Certainly that's what I originally intended. More discussion over on mathoverflow as commented and lunk above. – Colin D Wright May 26 '11 at 14:18
@Colin: is "lunk" the past participle of "to link"? – Ross Millikan Jun 9 '11 at 21:52

Infinite set 1. One line through the center at any angle.
Infinite set 2. Two perpendicular lines through the center at any angle.
Sporadic set 1. 8 45-45-90 triangles.

More interesting is what identical shapes can make a square. Non-right triangles with angles $(\pi/8,\pi/4,5\pi/8)$, $(\pi/4,\pi/3,5\pi/12)$, and $(\pi/12,\pi/4,2\pi/3)$ will do it, as shown by Laczkovich. There are also rectifiable polyominoes, with pages by Reid, Clarke, and Friedman.

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There's an infinite number of solutions. The general form involves a big + with its two lines crossing at the center of the square and perpendicular to each other. When you rotate it at any angle, it will cut the square into four equal parts. The shapes of the four pieces will also be the same provided we can rotate them. Because any angle works, there are infinitely many solutions.

Update: For that matter, because the four lines have the same distance from the center to a side of the square, they could be any identical curves going from the center outward at right angles to each other (this is probably better defined in terms of reflectional and rotational transformations than angles, but I assume the naive understanding of a right angle). As long as these curves always reach a side of a square and don't cross, you've got equal division. This also describes just one family, but it's broader I think.

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That's one infinite family. However, I have learned that there are other infinite families, and the one you describe isn't as general as it can be. This is what prompted the question. I originally thought the answer you give here is complete, but it isn't. – Colin D Wright May 26 '11 at 13:28
I see. It wasn't completely clear what you meant from the question. – Phonon May 26 '11 at 13:30
I've now edited the question in an attempt to make it clearer. – Colin D Wright May 26 '11 at 13:39
You can also take any of the solutions in this family and perturb it by adding arbitrary bumps and wiggles to one of the arms of the +, and then adding identical bumps and wiggles to the other three arms. The solution space is thus very large indeed. – MJD Apr 12 '12 at 5:20