Cut the cake

[AlfredFlaßhaar]

A cylindrical cake (torte) has a chocolate icing. A knife is used to gradually cut out pieces of cake with the angle alpha and then flipped over and reinserted into the cake so that the icing is on the bottom. We keep doing that - after one circuit, the cuts can end up in pieces that have already been cut. Of course, the glaze moves back up, ...
If e.g. alpha = Pi/2 (90°), then after two turns the glaze is completely on top again.
Question: For which angle alpha is the glaze completely up again after a finite number of steps?

Friendly puzzle greetings!

[Werner_E]

Guess we can assume alpha = p/q*2 pi with p and q being positive integers, p<=q and gcd(p,q)=1.
Interesting case is p<>1..

Or would you also use non-rational multiples of 2 pi? Guess this would mean an infinite number of turns and cuts.

What exactly is the question?

When you ask
"For which angle alpha is the glaze completely up again after a finite number of steps?"

you already gave one possible answer with 90° and two turns 😉

With my addition above any angle can be used and the number of turns until the whole cake is in its initial state (despite of the cuts , we have to assume a self-healing cake) would be finite for all angles.

EDIT: Thinking about it for a while I guess that the angle actually could also be a irrational multiple of 2pi, too and the number of turns would be finite for any angle chosen. At first glance I missed the fact that the colors of an already 2-color piece do not only change but are also switches in position.

I guess its easier if we think about cutting at the same position all the time and assume the cake to be turned for the angle.
But I must confess that its too hot here for further thinking and digging deeper into this riddle to come up with a solid proof.

Being a Mathcad (and Prime) forum coming up with a Mathcad 15 animation would be appropriate but I am not going to go for this in the near future.

So the questions of interest are

1) Give a proof that for any real angle alpha the number of turns needed to arrive at the initial state is finite

2) Give a formula in terms of alpha for the number of turns needed to return to the initial state.
3) Use Mathcad to create an animation of the process for any number. You may limit the number of turns shown in the animation.

EDIT: Well, I have to restrict that I now a t least think that it may be possible that also for irrational multiples of 2 pi the original state returns in finite time. I am not quite sure that this will always be the case - it's just a gut feeling - not to mention that I don't know how it would be proved or how the aforementioned formula could be derived 😉

Thats work for someone else ...

[AlfredFlaßhaar]

It is not difficult to show that rational parts of pi give solutions. But according to Weyl's theorem about the uniform distribution mod 1, it can be assumed that there is no solution for irrational parts. At least I didn't find any. There is a nice book by Hlawka about the equal distribution mod 1 😉 .
But you're right that the task here is out of place for Mathcad. Only I found the task published anonymously on the Internet simply attractive and too good to forget. At my age I finally have time to deal with interesting trivia. So I beg your indulgence.

[AlfredFlaßhaar]

p. s.
Presumably there are solutions for parts of 2*pi as quadratic irrational numbers. This is probably due to the periodicity of the continued fraction expansion, while rational numbers only have finite expansions that then lead to the solution of the problem.

[Werner_E]

Actually any angle (no matter if rational or irrational, either in terms of pi or in terms of degrees) will turn the cake to its initial state after a finite number of cuts & turnovers. This is very amazing and actually counter intuitive.

First of all, I thought I had a simple proof that it does not work with irrational multiples of 2 pi. But there I had made the mental mistake to swap only the colors of the pie piece and have not mirrored the areas, as it happens when turning the piece.

Without mirroring the areas, the last cut would have to be made at a point where a previous cut was made, and this contradicts the assumption of an irrational multiple of 2 pi.

I suspect that you are making the same mistake in assuming that you can use Weyl's theorem to show that it is not possible for irrational angles to recover the pie in finite time.

By the way, a nice visualization can be found here.

https://jsfiddle.net/9fxgknkf/
Angles can be chosen in whole degrees only via a selection list. But irrational numbers cannot be represented in the computer anyway.

We should probably refrain from animating in Mathcad for arbitrary angles, because even whole degrees can have very long periods, which would go beyond the capabilities of Mathcad animation with only 999 frames (and would also be boring to look at)

[AlfredFlaßhaar]

As I understood the task, something wrong was assumed of me. By modeling it as a clock face starting at 6 o'clock and continuing anti-clockwise, an indexed sequence of points equidistant in an arc is then generated on the periphery. If the process is to be terminated after a finite number of steps, a point generated earlier must be hit and the process returns the desired result after repetition. This gives a simple linear Diophantine equation, which immediately implies the need for the rational part of 2*pi. Under this assumption, Weyl's theorem excludes irrational parts.

How is the existence of irrational parts proved? Do you know the origin of the cake task? What did I assume wrong?

[Werner_E · Accepted solution]

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If the process is to be terminated after a finite number of steps, a point generated earlier must be hit

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No, thats exactly the same error I made first, too.  I had assumed that only the color(s) of the piece of cake would change, but of course the whole piece (the colors) must be mirrored additionally and thats the reason why the cake can be restored after a finite number of cuts without making the last cut at position which already was cutted earlier.

I must confess that after realizing my error, I had not looked into the problem very deeply. But somewhere in the back of my mind it was nevertheless always present and after some time I had the feeling that I knew the task from somewhere. At first I thought of Martin Gardener's books, but I somehow had the feeling that it must be younger.
To cut a long story short - this brain teaser can be found in one of Peter Winkler's beautiful books, "Mathematical Mind Benders" from 2007,  The book was also translated into German under the title "Mehr mathematische Rätsel für Liebhaber".

And as luck would have it, the relevant pages are currently fully included in Google's sample reading. (Chapter 9 - "Wirkliche Herausforderungen")
Have fun:  https://books.google.at/books?id=e74hBAAAQBAJ&pg=PA165&hl=de&source=gbs_toc_r&cad=3#v=onepage&q&f=false

[AlfredFlaßhaar]

Das Buch besorge ich mir 🙂 . Herzlichen Dank und freundliche Grüße,

Alfred Flaßhaar

[AlfredFlaßhaar]

If you are still interested in the solution, here is my short one
View of the poodle core (hereinafter I refer to it as a section).
Overall process on the piece of cake: cutting, lifting, tilting, inserting,
Rotate cake by angle z):

The goal is---> glaze completely back up, no matter what
Central angle x is cut, cake is self-healing in the cut

After a finite number of cuts, there are only finitely many (for each
Pattern limited to the top) x and z pieces as a pattern on the cake,
that permute

Remaining piece with central angle z satisfies k*x+z=2*pi, therefore k*x<2*pi<(k+1)*pi

k is calculated from the floor function floor(2*pi/x) and from this z

floor doesn't care whether the argument is rational or irrational

Each cut has exactly one predecessor and successor, starting pattern
(glaze above) is one of the finite number of possibilities (permutations
of the x and z pieces) and must therefore be repeated in the cycle.

Overall, the task would have a better presentation of the solution
earned in the book by P. Winkler. The actual meaning of the task is in
well hidden from the original formulation and does not indicate
that an algorithm (namely rotating the cake by z) is sought,
which pushes the glaze back up.

[AlfredFlaßhaar]

Correction:

... Remaining piece with central angle z satisfies k*x+z=2*pi, therefore k*x<2*pi<(k+1)*x ...