I think the example below requires pieces with a circular boundary, that can't be polygonized. Take a square, and remove a smaller interior square, oriented at an angle of atan(1/2). Also remove a small notch along one side, and toward a corner. The goal is to dissect this into the same larger square, but with the smaller square hole + notch rotated to make the orientation parallel to the larger square. The minimal piece-count solution (2) is to just cut a circular piece that includes the small square, and rotate it. --------------- | | | ---- | | < | | | | | | | ---- | The goal polygon. | | | | --------------- Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com on behalf of franktaw@netscape.net Sent: Thu 4/20/2006 9:19 AM To: math-fun@mailman.xmission.com Subject: Re: [math-fun] Square thirds All the concern about open sets, etc., is because when you get down to the point-set level, the obvious division *isn't* three identical pieces. Basically, the problem can be restated as dividing the square into polygons, such that the polygons cover the square, and intersect only on their edges. Slightly more generally, one can look for simply connected open regions that are the interior of their closure, such that their pairwise intersections is empty, and the union of their closures is the entire square. I think it's true that if a polygon has a dissection in the latter form, it has one into polygons, but I don't know if that has been proved. My intuition is that you can take any curved edges and approximate them by piecewise linear edges, getting a polygon; but I don't see how to make that rigorous. Franklin T. Adams-Watters -----Original Message----- From: Torgerson, Mark D mdtorge@sandia.gov I'm still stuck on the trivial way of thirding the square. 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? Mark -----Original Message----- From: math-fun-bounces+mdtorge=sandia.gov@mailman.xmission.com [mailto:math-fun-bounces+mdtorge=sandia.gov@mailman.xmission.com] On Behalf Of Fred lunnon Sent: Wednesday, April 19, 2006 2:38 PM To: dasimov@earthlink.net; math-fun Subject: Re: [math-fun] Square thirds On 4/19/06, dasimov@earthlink.net <dasimov@earthlink.net> wrote:

A RELATED PROBLEM that some people could've predicted I'd ask is:

In what ways can the square torus be decomposed into n congruent pieces for any positive n (other than the obvious n parallel annuli decomposition [whose slopes, btw, can be any fixed rational number]) ?

--Dan

Well, hexagonal and triangular tilings as well, obviously: including the famous dissection into 7 hexagons, each adjacent at an edge to all the others. Fred Lunnon _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ___________________________________________________ Try the New Netscape Mail Today! Virtually Spam-Free | More Storage | Import Your Contact List http://mail.netscape.com _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun