Mark T. wrote:
If you use pencil and paper, sure two horizontal lines divide the thing into three pieces that are the same. But y'all have been talking, open sets, closed sets, point sets, etc. Down to that level of detail I don't see how to trivially make equal thirds. Actually, I don't see how to do it at all. If the trivial way is to make two equally spaced cuts, the middle third shares two sides, the other two share one. How are the points on the cuts assigned?
Jim P. replied: << There's no way to do the dissection if you play the game that way (and if the pieces have polygonal interiors and boundaries composed of open, half-open, and closed intervals). Each region of this sort has an "Euler measure" (aka combinatorial Euler characteristic), which you can compute by decomposing it into points, open intervals, and open 2-cells, and then computing V-E+F, where V counts the points, E counts the open intervals, and F counts the open 2-cells. This quantity is always a whole number; it doesn't depend on how you decompose a region, and it's invariant under congruence. . . .
I seem to recall once showing that *none* of the 5 Platonic solids (thought of as polyhedral surfaces) could be used for an exact down-to-the-last-point partition into congruent connected pieces by using its face interiors as disjoint open sets and then adjoining part of each's boundary. Using radial projection, these can all be considered as being about partitions of the usual sphere. In fact I'm aware of only aware of ONE way to partition the sphere into conguruent connected pieces each of which is an open set with an interval of its boundary adjoined. (Or infinitely many ways if the boundary is allowed to be a finie union of intervals.) (Apologies if this has come up in this venue before.) --Dan