There are some interesting possibilities for dissecting a cube into three congruent pieces. Select a triagonal, and split the cube in half with a plane perp-bisecting the triagonal. The two pieces have regular triangle symmetry, and each can be cut into thirds using 120deg wedges with the triagonal as the spine. The starting wedge angular position is arbitrary, so the two halves of the cube can have different cutting positions. Wedges from the two halves can be glued for a rather peculiar looking congruent trisection. I believe one of Gardner's columns was about dissections, and said the square-into-5-congruent- pieces problem had only the trivial solution, and that a proof existed. Rich
Rich wrote:
There are some interesting possibilities for dissecting a cube into three congruent pieces. Select a triagonal, and split the cube in half with a plane perp-bisecting the triagonal.[...]
This 3-fold axis of symmetry means you can get away with almost anything -- make any 2d surface in 3-space whose boundary is a line and which doesn't intersect itself when rotated by a multiple of 120 degrees, and cut the cube according to that.
I believe one of Gardner's columns was about dissections, and said the square-into-5-congruent- pieces problem had only the trivial solution, and that a proof existed.
I just found this last night (thanks to the MAA's wonderful all-Gardner-columns-on-a-CD set!). It took me a while, because it's *not* in his column on dissections into congruent pieces. It's one of his "Nine Problems" cols, reproduced as chapter 15 in The Unexpected Hanging. He shows two nontrivial congruent-4-sections to mislead the reader, then asks for a congruent-5-section of a square. The answer shows the obvious one, with a little ha-ha-you- took-a-long-time-to-think-of-this-didn't-you, and states that it's unique without even any mention of the notion of a proof. Obviously it'd be nicest to prove there's no other trisection in a way which works for 5, and as many other numbers as possible, as well. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
I see now that Torsten Sillke's problems collection includes the page http://www.mathematik.uni-bielefeld.de/~sillke/PROBLEMS/dissection which includes the citation to the aforementioned Gardner column (that would have saved me a bunch of time...) as well as the following: --------- Question A: =========== For which k does there exist a dissection of the n-dimensional hypercube into k conguent connected parts which are not boxes? In other words: For which k do there exist subsets A_1, ..., A_n of the unit hypercube s.t. i) all A_i open, connected ii) the A_i are pairwise disjoined and congruent iii) the closure of the union of the A_i is the hypercube If 1<k<=n, there is such a dissection. If in addition k is a prime, there are infinitely many dissections. --------- I'm not sure why the "k is prime" is there -- it seems to me that the n-dim hypercube is acted on by a symmetry of order k for all k<=n -- just cycle the first k coordinates -- and any fundamental domain for that action, of which there are plenty if k>2, is the requested dissection. I'm just recycling the argument for trisection of the cube in my last message, of course. Torsten's page also includes, further down, a reference to the following note, whose abstract (but not full text, I think) I can get to on line: --------- 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 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 --------- So you can't do it with a polyomino. --Michael Kleber
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Schroeppel, Richard