Carefully analyzing this eventually leads to the correct solution (maybe– it really is tricky!)

]]>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?

]]>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.

]]>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.

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