A lamington is a piece of cake in the shape of a rectangular prism (rectangular cuboid). Each surface is coated with chocolate icing and coconut. I want to share a lamington with two other friends. The icing is the best bit. The icing on corners is particularly good, because of the high surface/volume ratio.

It's possible to cut three pieces with the same volume and same surface area of icing. But can I divide a lamington into three congruent pieces - where each piece is the same shape (modulo reflection or rotation), and has the same surfaces coated with icing.

BTW: I'm not after an "envy free" algorithm that lets different people cut and choose.

  • $\begingroup$ I can't find it - maybe it was not here, but on MathOverflow - but there was a question on cutting a rectangle into $n$ congruent pieces, and for various small primes values of $n$ (like 3) there was no solution other than $n-1$ parallel cuts. What I'm suggesting is that even ignoring the icing it may not be that easy to cut your prism into three congruent pieces in any but the trivial way. $\endgroup$ – Gerry Myerson Jul 28 '11 at 5:35
  • $\begingroup$ Found it, math.stackexchange.com/questions/50085/… which also has a link to a discussion on MO. $\endgroup$ – Gerry Myerson Jul 28 '11 at 5:54

There's a solution if the "pieces" don't have to be contiguous: Cut out cubes from each of the corners, with side length half the smallest side length of the lamington. Each of the remaining rectangular prisms can simply be cut into three congruent pieces with congruent icing. Distribute $3n$ of the cubes, and divide the $8-3n$ remaining cubes into three pieces along their $3$-fold symmetry axis.

  • $\begingroup$ Slick, very slick. $\endgroup$ – Brian M. Scott Jul 28 '11 at 6:49
  • $\begingroup$ Yes, great answer. Thanks Joriki! I wonder if there's an answer that will yield contiguous pieces... $\endgroup$ – John Jul 30 '11 at 5:35

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