A round cake has icing on the top but not the bottom. Cut out a piece in the usual shape
(a sector of a circle with vertex at the center), remove it, turn it upside down, and replace
it in the cake to restore roundness. Do the same with the next piece; i.e., take a piece
with the same vertex angle, but rotated counterclockwise from the first one so that one
boundary edge coincides with a boundary edge of the first piece. Remove it, turn it
upside-down, and replace it. Keep doing this in a counterclockwise direction. The figure
shows the situation after two steps when the central angle is 90°. If θ is the central angle
of the pieces, let f (θ) be the number of steps needed so that, under repeated cutting-andflipping just described, all icing returns to the top of the cake, with f (θ) set to ∞ if this
never happens. For example, f(90°) = 8.
Test your intuition: What is f(181°)?
The original source of this problem was Problem 188.8.131.52 in the 1968 Moscow Math Olympiad [L, p. 90] in a slightly different form. The variant given here is in [Winkler].
[L] 60-odd years of Moscow Mathematical Olympiads, G. Leites editor, G. Halperin, A Tolpygo, P Grozman A Shapovalov, V Prasolov, A Fomenko
[W] Mathematical Mind-Benders by Peter Winkler, published by AK Peters, Wellesley, Mass., 2007.