18 Apr
2006
18 Apr
'06
11:43 p.m.
Franklin wrote: << 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 . . .. . . .
How do you see this last assertion (without assuming convexity)? --Dan << -----Original Message----- From: Michael Kleber <michael.kleber@gmail.com> . . . 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.