Stan Wagon, TYI 23: Ladies and Gentlemen: The Answer

TYI 32, kindly offered by Stan Wagon asked

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; If the central angle
of the pieces is 181°, what is the number of steps needed so that, under repeated cutting-and flipping, all icing returns to the top of the cake?

Here are your answers:

And the correct answer is:


This entry was posted in Combinatorics, Geometry, Test your intuition and tagged , . Bookmark the permalink.

8 Responses to Stan Wagon, TYI 23: Ladies and Gentlemen: The Answer

  1. Stan Wagon says:

    The diagram is a little exaggerated, given for 195 degrees, not 181; but it is all the same.

    The fact that the right answer was guessed 9% of the time is actually pretty good. In my experience, everyone with mathematics knowledge gets this wrong by guessing. Note that 6 radians is also a nice one to try. It is irrational, so many might thing infinity is the answer. But f(theta) = 4 for any angle between pi and 2 pi. Thanks to Gil Kalai for posting this. The general formula is:

    f(A) = 2 n (n+1) where n = Floor[2 Pi / A], except the easy case where 2 Pi / A is an integer, in which case it is 2 n.

    • Gil Kalai says:

      Stan, What about g(A)?

      • Dan Carmon says:

        If instead of rotating the knife you rotate the cake, then you would have had g(A) = f(A) in all the above cases. When the cake is fixed and the knife is rotating, it follows that after every f(A) moves, the cake has simply rotated by f(A)*A. If A is not a rational multiple of 2 \pi, the cake would never return to the original position; and if A = 2\pi q with q rational, then g(A) would be the lcm of f(A) and the denominator of q.

        For example, for A = 181 degrees, after every 4 iterations the entire cake is rotated by 4 degrees (as seen in the diagram), and would return to the original position after g(181) = 360 iterations.

      • Stan Wagon says:

        If theta is in radians, then the cake will return to its original state iff theta is rational in pi, say (p/q) pi. Suppose 0 < p/q < 2. The easy case is 2 pi / theta is an integer n; then 2n steps. Otherwise 2 k n (n+1) steps where n is Floor[2 pi / n] and k is q / gcd(q, n (n+1)).

        The ideas Carmon posted are all correct for this problem.

      • Stan Wagon says:

        The formula I just posted is due to Dan Velleman. I am starting work on a problem book with him, as a followup to my Which Way Did the Bicycle Go? book from 22 years ago.

  2. Alex Kritchevsky says:

    Ah, I think a lot of people, including myself, didn’t realize the implications of actually flipping the piece over, rather than just inverting it ‘in place’. As soon as I realized I was leaving that out the answer became obvious.

    • Stan Wagon says:

      A key point is the word “turn” in the problem. The piece is removed and TURNED: it is rotated. It is not a reflection, which is of course not a rigid motion. Most hear the words upside-down and think reflection — but we have to use a rigid motion.

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