There are clearly lots of choices for what restrictions you put on the pieces. Personally, I'd be happy to learn anything: a proof that the obvious trisection is the only one possible with polygonal pieces would be great, as would an abstract, AC-dependent construction that created three congruent point-sets whose union is the square. (Yes, Harold, you're right that the former would satisfy my son more than the latter, but I'm less picky.) Franklin Adams-Watters wrote

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[...]

and Dan Asimov asked for justification. Talking with my colleague Will Brockman this morning, we came up with the same result, if you assume that the pieces can't pass through each other -- three disjoint connected open sets whose closures' union is the square, I suppose. Here's a line of reasoning. --------- There must be some piece which touch at least two corners (since 4>3). (By "touches", of course, I mean the corner is in the closure of the piece.) First, it can't touch two opposite corners. Let the square's vertices be ABCD, and suppose a piece touched A and C. Then every piece must touch two opposite corners (only points of separation sqrt(2)), and no piece can touch B and D (since it can't pass through the one touching A and C). That leaves all three pieces touching A and C, which is impossible: there are only four automorphisms of the plane taking the line segment AC to itself, and we can't use two pieces that are reflections of one another in the line BD, because those pieces would intersect wherever either one touched BD. So assume some piece touches A and B; we want to show it touches the entire line segment AB. If not, consider the closure of the piece, along with the line segment AB: this is now non-simply-connected. The hole in this thing must be filled in, and it would have to be filled in with a congruent piece -- which therefore also has a hole the same size. That's clearly impossible: if you fill a hole of size S using a piece with a hole of size S, then the non-hope part of the piece has size zero, while we need it to have measure 1/3. --------- That last paragraph is, I suppose, a general argument that if you assume pieces can't interpenetrate, then connected implies simply connected. So okay, we have one piece running along a whole edge of the square. Where to next? We've used the fact that the diagonals are the same length and are the only diameters. That's also a property of rectangles, and note that the 2x3 rectangle *does* have a nontrivial congruent trisection into three dominos. So we'll need to a property that is exclusively squarish. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.