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Puzzle from http://www2.maths.bris.ac.uk/~majwm/compendium/cakeslice.php

A piece of angle $x$ is cut from a cake, which is purple on top and yellow underneath, and turned upside down. Then another piece of angle $x$ is cut from where the last cut is made, and the cuts carry on going clockwise around the cake. Why will the cake become all purple on top after finitely many moves, whether or not $x$ is rational?

Also, will the cake still return to purple after finitely many moves if the cake has 3 or more layers and instead of the pieces being turned upside down the layers are cyclically permuted? Or if the angles of consecutive cuts cycle through some finite sequence $x_1...x_k$?

Angela Pretorius
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    The layers are not just permuted, they are flipped, which is quite essential. – WimC May 20 '13 at 11:06
  • There's a discussion and demonstration at http://demonstrations.wolfram.com/TheCakeIcingPuzzle/ and a pointer to a solution in Peter Winkler's book, Mathematical Mind-Benders. – Gerry Myerson May 20 '13 at 11:08

1 Answers1

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It should have been obvious that merely permuting the pieces wouldn't work.

If a line between a yellow and a purple section was reached on the cut that ended at angle $nx$, then if the cut ending at angle $mx$ is the next piece cut that contains that line, the line will be flipped to the position that will be reached on the $m+(m-1)-n \ th$ cut. The way all the lines get flipped over and turned into lines that will be reached later forces the cake to return to purple eventually, and the same kind of thing happens for when there are two angles $x_1,x_2$.

When there are three angles this argument does not work because when the cut that reaches $(n+1)x_1+nx_2+nx_3$ goes over a line at $(m+1)x_1+(m+1)x_2+mx_3$ it will flip it to $(2n-m)x_1+(2n-m-1)x_2+(2n-m)x_3$ which is not a cut that is guaranteed to be reached later.

Angela Pretorius
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