DH. Ice Cream Cake
(I have no idea why cakes are so popular in math puzzles, but here is another conundrum)
Peter Winkler gives us one more puzzle from his book Mathematical Mind Benders and tells us a little bit about why good puzzles are like good jokes.
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notovny said,
January 17, 2008 at 7:16 pm
This might be a bit of a crazy take on it. Haven’t read the book, so no clue if this is the “real” answer.
I also don’t know if posting an answer in the comments is frowned upon, so perhaps a bit of spoiler space?
Okay.
Assume the cake is a real object rather than an abstraction. As a result, it’s made up of molecules, and the molecules are made up of atoms.
You’re not going to cut atoms with a knife. Not unless you’re swinging it ludicrously hard.
As a result, regardless of what angle you pick, each slice is going to contain an integer number of atoms. Assume you’re exact enough that the same number of atoms is between each slice. (this is probably going to assume that the cake is perfectly circular and cylindrically symmetrical) If that’s the case the slice you’re picking and flipping is a rational fraction of the cake.
Assume this rational fraction is p/q , in lowest terms. On cut q+1, you’ll be putting the knife exactly in the same place it was on the first cut. Cut q+2, you’ll flip the same section of the cake that you flipped the first time. Continue around and you’ll be flipping back all the same sections you did before, until cut 2q+1, which flips the last section of the cake, and puts the knife back in its starting position, with all the icing back on top.
As mentioned this may take a ridiculously long time. Particularly if q = the number of atoms in the cake. But it is a finite number of steps.
strauss said,
January 19, 2008 at 4:20 pm
Hi, ice cream is, as every chemist knows, a mathematically homogenous substance, directly derived from the very fabric of the universe.
Or at least the ice cream in this cake is! This is the right argument for a rational angle of slice, but the amazing aspect of this puzzle is that it works just as well for irrational angles as well.
jlundell said,
January 19, 2008 at 8:32 pm
Clearly (for some definition of “clearly”), the answer must have to do with slice-flipping. If the first cut is at 0, then by the time we come all around the cake and cut a slice that includes the cut at 0, when we flip that slice, that first cut is no longer at 0. Presumably it all works out (along with getting the top back on top), but I haven’t figured out a notation to make the calculation at all straightforward.
eron said,
March 24, 2008 at 12:01 pm
Hey hi, first of all, congratulations for your program from germany (although I’m spanish)!
I’ve been quite puzzled with this puzzle for some days now because it looks rather easy to prove that the ice cream can’t be put back on top, unless I’m not getting it right…
I assume the following:
One starts cutting at an angle of A=0 (wherever that is) and the cutting angle is X. Of course if X is some rational number N/M times 2*pi it’s easy to see that after N*M slices one gets back to the starting point A=0.
The important point is this. No matter what the angle X is, if the cake actually gets all the ice cream back on top, this has to happen at the initial point X. I think this is also easy to see.
If that is the case, then one has done N slices after which one has managed to complete M turns around the cake, yielding:
N * X = M * 2*pi
which doesn’t work for all X.
What am I getting wrong?
strauss said,
March 24, 2008 at 12:06 pm
This is a really strange problem—it can take a truly astronomical number of flips before things work out. The key is the word “flip” When you have an irrational angle, when you come back around, a chunk of cake you’re cutting into will be upside down. When you cut this and flip it again, two molecules of cake that were next to each other at first might be moved apart…
Carefully analyzing this eventually leads to the correct solution (maybe– it really is tricky!)
Shawn said,
March 31, 2014 at 2:17 pm
I should probably be practicing differential equations before my class tomorrow instead of analyzing a puzzle and posting my results on a post that hasn’t had comments in over six years, but I don’t often do what I should. I haven’t read the solution in the book, but I have thought about this problem for many hours. Here goes: Let us imagine the ice cream cake is a right circular cylinder. If the volume of the cake is the volume bounded from below by the plane z = 0, from above by the plane z = 1, and constrained by the inequality x^2 + y^2 <= 1, we see that its projection on the x,y plane is the unit circle. Let us make our first cut of the ice cream cake such that our angle x (which I will refer to as ? to avoid confusion with the Cartesian coordinates) is an angle in standard position (i.e., the initial side will start at the line x=y=0 and go through the point (1,0,0) and the terminal side will be ? radians counterclockwise on the x,y plane). There is not a straightforward interpretation for what to do for |?| > 2pi nor for imaginary values of ?. Also, negative ? values can be simply thought of as its negation in the other direction, which for the purposes of this problem will yield symmetric and similar results. Therefore, I shall only concern myself with angles weakly between 0 and 2pi. For ? = 0, the problem is trivial. The terminal side and initial side of the angle will be the same, so there will be no cake to flip. After any number of flips, the icing will be (“back”) on top of the cake. For an angle of 2pi, the terminal side and initial side will again be the same, but this time the entire cake is the slice. After any even number of flips, the icing will be all on the top. As Peter Winkler said, an integer divisor of 2pi will also result in all the icing returning to the top, as during the second revolution, the slices made will be the exact slices as the previous revolution, and thus all slices will be flipped. The analysis gets slightly more complicated, however, for other values of ?. One thing that I have found that is important to keep in mind, and which I imagine Chaim was referring to above, is that [spoiler] when you flip a slice that is part icing on the bottom and part icing on the top, not only will the icing change from top to bottom and bottom to top, but also because the entire slice is rotated, the icing is also reflected across the line given by the terminal side of the angle, so to speak. For example, if you flip a slice whose top is 2/3 icing in a contiguous region contacting the terminal angle, the result will a slice whose top has 2/3 not icing, but touching the initial side of the angle. If you don’t understand what we mean, try drawing a slice of cake with icing the way we described and actually flip it.[/spoiler]