I have a cross in the form of one square with four identical squares surrounding it

   _
 _|_|_
|_|_|_|
  |_|

Cut it into the fewest number of pieces possible such that, when rearranged the pieces form two smaller crosses of identical size.

(In reply to re: 4 pieces minus epsilon by Sir Percivale)

It was meant to show that at points A and B, the pieces don't go all the way to the corners, but are instead slightly rounded.  This make only four pieces, but they don't form two exact crosses.  They would only form two exact crosses if the corners were not rounded, but then there would be 8 pieces as my original solution.

This 4 piece solution of mine is not really legitamate, in other words.  It only works in the limit as the roundedness of the corners decreases, hence the epsilon reference from calculus.

Posted by Jer on 2006-03-08 12:33:47