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Any straight line through the center point will create 2 identical halves. Any 2 straight lines through the center point at 90 degrees to each other will creat
by vanishing 15y ago
Any straight line through the center point will create 2 identical halves.
Any 2 straight lines through the center point at 90 degrees to each other will create 4 identical pieces.
In both of the above cases each line segment from the center to the edge can be distorted in any way which does not intersect the edge of the square or any of the other lines and the distortion can be rotated 180 degrees in the first case or 90 degrees 3 times in the second to create new identical shapes.
The only special case seems to be using 4 lines to divide the square into 8 pieces which is, I think, the only configuration of 8 pieces.
I think that's every possible solution.
I'm weak at math, but I think that means there are an infinite number of configurations for each of an infinite number of configurations for both of the first 2 solutions, and then there's that 1 extra solution. So does that mean there are uncountably infinite solutions? I'm not sure how you would apply the diagonal method to this.
- ColinWright 15y agoYou've found two infinite families and one sporadic solution. There is another sporadic solution, and the 8 piece solution you've found is not, in fact a sporadic. And yes, the infinities are uncountable.
- vanishing 15y agoAh, yes. You can also rotate the 8-fold solution by 22.5 degrees for another solution.
- ColinWright 15y agoHmm. Don't think so.
- vanishing 15y agoRight. As I said, I'm weak in math.
- ColinWright 15y agoYou can't be that weak at what I call math, although your experiences with math education might be unhappy ones. You have found more solutions than I first found - that can't be bad. Half my life is spent helping people discover that they're good at "proper math" even when they think they're bad at "school math."
- pozorvlak 15y agopuzzles Oh, right - you can vary the shape of the diagonal lines, provided each "arm" has rotational symmetry about its mid-point. OK, so we've got three infinite families and the trivial solution (only one piece - is that your sporadic solution?). I think that might be all: each piece can contain 4, 2, 1 or 1/2 of the original square's corners, since (lacuna) all pieces must contain the same number of corners and further subdividing the corners (into 1/3s, say) would mean some pieces don't touch the centre (another lacuna).
- ColinWright 15y agoOK, that's now the set of solutions I've got. You've also gone some way to showing them to be complete. More to do, though. And now do it for an equilateral triangle.
- ColinWright 15y agoOK - I've now seen a "solution" with 16 "pieces." My head hurts.