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AlfredFlaßhaar15-Moonstone
15-Moonstone
July 23, 2023
Solved

Cut the cake

  • July 23, 2023
  • 1 reply
  • 3421 views

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!

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

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

 

1 reply

W
Werner_E25-Diamond I
25-Diamond I
July 23, 2023

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 ...

 

A
AlfredFlaßhaar15-MoonstoneAuthor
15-Moonstone
July 25, 2023

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.

A
AlfredFlaßhaar15-MoonstoneAuthor
15-Moonstone
July 25, 2023

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.

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