I've never seen this problem before, but I'm sure the answer is no. Since there are four corners and only 3 pieces, one of the pieces must have two of the corners. It must then have the edge connecting them (opposite corners won't work at all). The other pieces must have a congruent segment. One can do this with the opposite edge, but the other must have this edge somewhere else. This is the one step I don't quite see - we need to show that this edge is parallel to the originally identified edge. It can't be perpendicular to it, but conceivably it could be at a diagonal. Once this is established, the edge divides the square into 2 parts, which must be in the ratio 2 to 1 for a solution to be possible; and then we have one rectangular piece and only one way to get the other two. (This is assuming that each piece must be contiguous. I don't see any solution with disconnected pieces, but there's a lot more "wiggle room".) Franklin T. Adams-Watters -----Original Message----- From: Michael Kleber <michael.kleber@gmail.com> Surely everyone other than me already knows this, but when my 7-year-old son asked me this over dinner tonight, I couldn't give him a satisfying answer. Is it possible to dissect a square into 3 congruent pieces in any way other than the trivial one? Surely the answer is no, but I can't think of a proof. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen. _______________________________________________ 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