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Cut the cake and flip it too


Go to solution Solved by Prof. Templeton,

Question

So it's your birthday! wink.gif
And the round thing with candles on it is from Carvel's.

You blow out the candles in one try, and your wish comes true!
A personalized BrainDen puzzle - which, unfortunately,
you must answer correctly before eating the sweet cold stuff.

It's French vanilla ice cream, covered with dark Swiss chocolate icing.

A protractor is provided, which is set at random to an angle X.
You must cut a piece at that angle, turn the piece upside down and replace it.
Because it's made of ice cream, it heals itself [re-freezes].

You then cut a 2nd piece, at angle X beginning where the 1st piece ended, and turn it upside down.
Beginning where the 2nd piece ends, a 3rd piece at angle X is cut and flipped. And so on.
Each successive piece begins at angle [n-1]X and ends at angle nX, [shown] leaving a path of vanilla ice cream in its wake.

 

post-1048-1232208680.gif

 

Eventually you will work your way around the cake and begin
cutting into the part of the cake that has been flipped upside down.
When that is flipped, of course, the chocolate icing comes back on top.

Here's the puzzle. If the cut-and-flip-at-angle-X process is continued,
will there ever come a point where the cake is restored to its original state?

Answer correctly, and we all have cake. biggrin.gif
Otherwise, it's wait until next year. sad.gif

 

Since

X was chosen at random, the probability that it is rational is zero.
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Yes with a maximum of 720 times around because then your 'll have x*720 which is a multiple of the twice degrees of the cake thus making you flip the enter cake

And being diabetic you guys enjoy the cake w/o me ^^

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I say no.

If your angle is truly an irrational number (unable to be expressed as a fraction or decimal), and you have to multiply it by ONLY rational & whole numbers (number of cuts) to find a multiple of 360 (degrees, for starting point), you won't be able to succeed.

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Since X was chosen at random, the probability that it is rational is zero.

Also, note to neida, it really is an ice cream cake, not a waxing crescent moon. ;)

well I'm pretty sure that...

irrational number * rational number=irrational number so NO it would not be possible to restore tha cake to its original condition. you would need an irrational number of cuts to restore the cake, which is obviously impossible.

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the edges trade places. This becomes important after you go all the way around the cake and have a slice that is part vanilla and part chocolate on top. You can't simply change the vanilla portion to chocolate and leave it in the same sector, it trades spots with the sector of chocolate that becomes vanilla. Then you go around the cake again and come to that one sliver that's different. Your last cut will be on the trailing edge of the sliver that's different, so you won't be cutting the sliver into smaller pieces. Your next cut has the different sector on one side, so when it's flipped, it's now on the other side, and is still the same size. So essentially, you're chasing a sliver of cake the size of the remainder after you divide 360 by X around the whole cake infinitely, you will never return the cake to its original state unless there is zero remainder after you take 360/X.

This is making me hungry. :D

edit: grammar

Edited by lazboy
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edit: grammar

the edges trade places. This becomes important after you go all the way around the cake and have a slice that is part vanilla and part chocolate on top. You can't simply change the vanilla portion to chocolate and leave it in the same sector, it trades spots with the sector of chocolate that becomes vanilla. Then you go around the cake again and come to that one sliver that's different. Your last cut will be on the trailing edge of the sliver that's different, so you won't be cutting the sliver into smaller pieces. Your next cut has the different sector on one side, so when it's flipped, it's now on the other side, and is still the same size. So essentially, you're chasing a sliver of cake the size of the remainder after you divide 360 by X around the whole cake infinitely, you will never return the cake to its original state unless there is zero remainder after you take 360/X.

This is making me hungry. :D

Not only will you never be able to make a cut in precisely the same place as your first cut, the chocolate coating thing makes it extra-impossible

;)
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No one eats cake, so far.

Except Izzy and Guyver, but I'm keeping it frozen until they give more convincing justification.

And Izzy is right, I wouldn't have posted it if the answer had been "No".

Spoiler for lazboy::

There are 3, maybe 4, steps to the proof.

You've made the first necessary observation - without this observation, the answer would clearly be No.

Can you discover the next step?

Spoiler for Nudge, if you want one:

I can't think of a gentle enough nudge that doesn't more of less hand it to you.

Maybe think about this question, along the lines of my sig B))

Could the answer be Yes, even if the cut boundaries never repeat?

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I say no.

If your angle is truly an irrational number (unable to be expressed as a fraction or decimal), and you have to multiply it by ONLY rational & whole numbers (number of cuts) to find a multiple of 360 (degrees, for starting point), you won't be able to succeed.

Can a random number be irrational?

Or can you cut a slice of cake in an irrational angle?

After a round if you are not at start point (360/x is not an integer), you will begin to cut beginning from a new point but again end at a newer point. Thus you will never end at your last starting point. So only if 360/x is integer, it is possible, though I'm not sure and bnv.s problems are never expected to be so simple.

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[1] Can a random number be irrational?

[2] Or can you cut a slice of cake in an irrational angle?

  1. Between any two rational numbers, no matter how nearly equal they are,
    there is an infinity of irrational numbers. Given that the irrationals "outnumber"
    [a poor choice of word, since they in fact are not countable at all; the rationals are countable]
    the rationals so greatly, a real number chosen at random has zero probability of being rational.

    Zero probability does not mean impossible.


  2. Well, it's a mathematical thing. Any physical cut will disturb the pie.
    But you can mathematically "cut" the continuum of real angles at an irrational value, yes.

    It's precisely these "cuts" that are used to define the real numbers.
    [Google "Dedekind cut" or just look here.]
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  1. Between any two rational numbers, no matter how nearly equal they are,
    there is an infinity of irrational numbers. Given that the irrationals "outnumber"
    [a poor choice of word, since they in fact are not countable at all; the rationals are countable]
    the rationals so greatly, a real number chosen at random has zero probability of being rational.

    Zero probability does not mean impossible.

But isn't there also an infinite number of rational numbers between any two rational numbers, thus balancing out the probabilities?

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But isn't there also an infinite number of rational numbers between any two rational numbers, thus balancing out the probabilities?

Both statements are true.

But you can arrange the rationals into a square array

[number the columns and rows with integers and the elements of the array are their quotient]

and "count" them, following an expanding diagonal path through the array.

So there is a 1-1 correspondence between the natural numbers 1, 2, 3 ... and the rationals p/q.

You cannot so arrange the reals.

Between any two entries of the discrete array of rationals there is an infinity of reals.

You can't squeeze more rationals into the table - they're all there.

We say that countable infinite sets [like the natural and the rational numbers] have a certain cardinality [Aleph0]

The cardinality of the reals is a higher cardinality.

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Just so you guys know, I've had ice cream cake twice since the posting of this riddle. :P

Which is pretty awesome, considering my annual cake-intake isn't that high.

Well, even if the angle is an irrational number, does it matter? Even if the angle happened to be pi (no, cake :P), irrational number or not, you'd still be able to cut at that exact angle each time.

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Bravo! B))

Can you generalize for other ranges of x?

First: As you know, in software programming, requesting a random number will always give an integer. My fault was this. Actually a random number is as writing a random number of digits on a paper and then randomly place a decimal point at anywhere. So the possibility of this number of being a rational is nearly equal to zero, as you said. I wrote this for any others those knows random numbers incorrectly, as me.

If that is wrong please correct me.

Second: In profs answer, I couldn't get why the colors of regions, as I indicated below as A and B, are different ? What have I missed?

post-10474-1232441198.jpg
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You don't need to generate a truly irrational number to explore the general solution.

There are really only two cases that differ: X = 360/N or not.

In the first case, say X=90o, the Nth cut gets you back to the beginning with everything white.

Then the 2Nth cut turns it all brown again.

In the second case, say X=91o, the Nth cut overlaps into the white area.

The amazing result is the boundaries between white and brown areas don't move!

This is true whether X is rational [90o] or irrational [say 91.333....o].

Eventually the colored areas all turn brown again because there are only

a finite number of color combinations they can have.

The 2nd cut starts in brown region and further, into the white region.

The part that is brown turns white [area B], and the white part turns brown [area A].

Then that piece is flipped over, moving A backwards and B forward.

Now the really unexpected result is that the third cut [lower left drawing] goes exactly

to the end of the A region. It becomes white and gets flipped back to the "3:00" position,

setting up the 4th cut, which again reached exactly to the end of that region

to turn everything brown.

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The 2nd cut starts in brown region and further, into the white region.

The part that is brown turns white [area B], and the white part turns brown [area A].

Then that piece is flipped over, moving A backwards and B forward.

Now the really unexpected result is that the third cut [lower left drawing] goes exactly

to the end of the A region. It becomes white and gets flipped back to the "3:00" position,

setting up the 4th cut, which again reached exactly to the end of that region

to turn everything brown.

Thanks for explanation, I worked only on turning upside down, and missed to move pieces foreward and backward by flipping. An amazing question!

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Let X = Y + Z; where Y = 360o/n and Z < Y.

For example, if X = 91o, then Y = 90o, and Z = 1o.

In this case n, which is the least integer that is greater than or equal to 360o/X, is 4.

What can we say about C, the number of cuts required to restore the cake?

That is, relate C to n.

As an example, PT's value of X was 180o < X < 360o; that is, Y = 180o [n = 2] and 0o < Z < 180o.

He found, for n = 2 that C = 4.

For starters, solve the next two cases: determine C, after first locating and counting the light/dark areas.

  1. n=3 [say X=121o]
  2. n=4 [say X=91o].
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and came up with 12 moves for X=135° (which would be n=3, Y=120, and Z=15 using your nomenclature). I assume this is the answer for all 120<X<180 then? I've been trying to come up with a spreadsheet that I can input any X value and return a solution, but haven't had much luck yet.

I don't know how I got messed up when I tried to visualize it in my head for my last post (and incorrectly stated you'd be chasing the remainder around the cake forever), but once I actually drew out each step, I got it to work. Perhaps this X=Y+Z would come in handy...

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and came up with 12 moves for X=135° (which would be n=3, Y=120, and Z=15 using your nomenclature). I assume this is the answer for all 120<X<180 then? I've been trying to come up with a spreadsheet that I can input any X value and return a solution, but haven't had much luck yet.

I don't know how I got messed up when I tried to visualize it in my head for my last post (and incorrectly stated you'd be chasing the remainder around the cake forever), but once I actually drew out each step, I got it to work. Perhaps this X=Y+Z would come in handy...

Here's a neat trick.

Rotate the entire cake by -X before each cut.

If you do that, a smaller number of color boundaries are created.

The areas break out into two groups:

  1. n large areas [with 1 and n abutting at 0o] and
  2. n-1 small areas in between.
Large areas are X-nZ; small areas are nZ.

n[X-nZ] + [n-1][nZ] = nX -n2Z + n2Z - nZ = n[X-Z] = nY = 360.

So a proof that finite number of steps does it is there are only 22n-1 color combinations.

So it has to cycle after no more than that many cuts.

But it's better than that: it seems the large areas flip over and back in 2n moves,

while the small areas flip over and back [being one fewer] in 2[n-1] moves.

So the least common multiple 2n[n-1] moves would do both.

For n=2 that's 4 PT's case

For n=3 that's 12

For n=4 that's 24 - which means I'm half through at 12 cuts.

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Here's the puzzle. If the cut-and-flip-at-angle-X process is continued,

will there ever come a point where the cake is restored to its original

state?

No, the cake's original state was un-cut.

Edited by Llam4
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