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January 23

Dissections into equal parts

This StackExchange question asks whether there exists a figure which can be divided into 2 equal parts, and into 3 equal parts, but not 6 equal parts (with "equality" being defined in terms of rigid transformations.) This seems to be a rather complicated question, as finding dissections into equal parts (or proving they can't exist) is adjacent to highly nontrivial problems like finding whether equidissections of shapes can exist. As such, I was wondering whether there are any explicit constructions, methods of disproof, or relevant literature for this type of problem. GalacticShoe (talk) 05:17, 23 January 2022 (UTC)[reply]

Note that the StackExchange question is about two-dimensional shapes. It is conceivable that there are only examples in higher dimensions (and conversely). --Lambiam 11:22, 23 January 2022 (UTC)[reply]

The dissection into two congruent parts, let's call them "left" and "right", and that into three congruent parts, say "upper", "middle" and "lower", can be combined to give a dissection into six parts, "upper-left", ..., "lower-right". These are not known to be necessarily congruent. The three ✽-left parts form a dissection of the left part, and likewise for the right. There is an automorphism witnessing the left–right congruence; applying it to the three ✽-right parts also results in a dissection of the left part. I think that I have proved that if these two dissections of the left part into three coincide (possibly in a different permutation, where e.g. the middle-right part is sent to the lower-left part), all six parts are congruent. --Lambiam 11:37, 24 January 2022 (UTC)[reply]

Hey Lambiam, thanks for the response! The case where the two dissections coincide is particularly interesting and I'm going to try to look more into it, also considering the case where there the two dissections differ. Once again, thanks for taking the time to help! GalacticShoe (talk) 17:56, 29 January 2022 (UTC)[reply]

If the two dissections do not coincide, their combination induces a further dissection of the left part into potentially up to nine parts. Likewise for the right part. The automorphisms witnessing the upper–middle–lower congruences can also be combined with each other and with the left–right automorphism; repeating this generates dissections into more and more parts. I think I can prove that in a tame topological space like Euclidean space at some point this subdissection process comes to an end with a dissection into a finite number of jigsaw pieces such that any of the automorphisms maps whole pieces to whole pieces. The original proof that this implies congruence for six pieces does not work here, because it involves considering all possible pairings one by one, which is not possible for an unknown number of pieces. --Lambiam 21:30, 29 January 2022 (UTC)[reply]

Wait, what about a regular hexagon, or circle? Think of a 6-slice pizza. That's 6 congruent parts, or 3 congruent parts each consisting of 2 slices joined together, or 2 congruent parts each of 3 joined slices. 2601:648:8202:350:0:0:0:C115 (talk) 05:03, 28 January 2022 (UTC)[reply]

The question asks for "2 or 3 but not 6". —Tamfang (talk) 06:32, 28 January 2022 (UTC)[reply]