Talk:Dissection problem
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honeycombs
[edit]“ | ... It is not true, however, that any polyhedron has a dissection into any other polyhedron of the same volume. This process is possible, however, for any two honecombs (such as cube) in three dimension and any two zonohedra of equal volume (in any dimension). | ” |
What does it mean to dissect a honecomb (such as cube) into a honecomb? —Tamfang (talk) 04:32, 5 December 2011 (UTC)
- I think what it means is that the funamental domains of periodic honeycombs can be dissected into each other. E.g. a cube is the fundamental domain for one honeycomb, the truncated octahedron is the fundamental domain for a different honeycomb, so there is a dissection from a cube to a truncated octahedron. —David Eppstein (talk) 04:55, 5 December 2011 (UTC)
Monsky's theorem
[edit]It can't be true that "Most polygons cannot be equidissected". Any polygon can be disscected into any other polygon, e.g. into a 3x1 rtectangle. This can be disscected uinto 3 equal squares, each of which can be dissected into 4 pieces which can form an equilateral triangle. HuPi (talk) 19:42, 3 May 2023 (UTC)
- Equidissection does not allow the reassembly of pieces, only partition. —David Eppstein (talk) 20:55, 3 May 2023 (UTC)
- Thanks. I should have read the linked article on equidissection.
- I've been trying to determine how many pieces are needed to dissect two squares to be reassembled as one square. Five pieces are always sufficient, and if the ratio of the sides of the two squares is x >= 1, then four pieces are sufficient for x = 1 or 2k(k+1)/(2k+1) for positive integral k. Do you know a more complete answer? regards HuPi (talk) 17:26, 14 May 2023 (UTC)
- There's an extended discussion of exactly this question in Chapter 7 ("First Steps") of Greg Frederickson's book Dissections: Plane and Fancy. —David Eppstein (talk) 18:59, 14 May 2023 (UTC)
- I've been trying to determine how many pieces are needed to dissect two squares to be reassembled as one square. Five pieces are always sufficient, and if the ratio of the sides of the two squares is x >= 1, then four pieces are sufficient for x = 1 or 2k(k+1)/(2k+1) for positive integral k. Do you know a more complete answer? regards HuPi (talk) 17:26, 14 May 2023 (UTC)