Unusual Cake Slicing
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Unusual Cake Slicing
Here's the puzzle:
You have a circular cake. There is a spot of poison in the exact centre of the cake. You must cut the cake up into some number of pieces (greater than 1) so that each piece is the same size and shape (rotations and reflections fine) and so that the poison falls solely in one slice.
I should clarify that the cakeness of the puzzle is purely aesthetic; we're not looking for lateralthinking answers (slicing the cake in the horizontal plane or whatever). If you prefer, you have to divide a circle into n pieces (n>1) so that the pieces are all the same size and shape and the central point is in the interior of one of the pieces.
Now, I heard this puzzle thirdhand or so. Supposedly there is a solution (I have my doubts). But don't blame me if it's impossible.
You have a circular cake. There is a spot of poison in the exact centre of the cake. You must cut the cake up into some number of pieces (greater than 1) so that each piece is the same size and shape (rotations and reflections fine) and so that the poison falls solely in one slice.
I should clarify that the cakeness of the puzzle is purely aesthetic; we're not looking for lateralthinking answers (slicing the cake in the horizontal plane or whatever). If you prefer, you have to divide a circle into n pieces (n>1) so that the pieces are all the same size and shape and the central point is in the interior of one of the pieces.
Now, I heard this puzzle thirdhand or so. Supposedly there is a solution (I have my doubts). But don't blame me if it's impossible.
Re: Unusual Cake Slicing
I'm going to assume the axiom of choice and that the poison is at a single pointmass. I divide the cake into many immeasurable subsets and translate them to reassemble them into a square. Then I cut the square into 4 smaller squares, only one of which contains the point. If the point must lie on the boundary, I cut it into four triangles.
On a more serious note, I don't believe it's possible (assuming we must cut into a countable number of slices), considering all lines of symmetry of the disk pass through the center (and the center is the origin of rotational symmetry as well) it seems that the center point must be on the boundary of any slices of a set of congruent slices.
On a more serious note, I don't believe it's possible (assuming we must cut into a countable number of slices), considering all lines of symmetry of the disk pass through the center (and the center is the origin of rotational symmetry as well) it seems that the center point must be on the boundary of any slices of a set of congruent slices.
Re: Unusual Cake Slicing
I knew there was a condition I forgot to add... finitely many slices.
I don't think it seems likely there's a solution either, but neither can I see that it's obviously impossible. An equilateral triangular cake has all its lines of symmetry passing through the centre, which is also the centre of rotational symmetry. But you can slice it into 4 similarsliceswiththecentreinoneofthem fairly trivially.
Edit: Addendum: Also, the solution that supposedly exists does not employ any kind of fractal slices. You should be able to do it to a real cake (I guess the cake might have to be very large to be able to cut with the required precision though). If a fractal solution does exist I'd still like to know, mind.
I don't think it seems likely there's a solution either, but neither can I see that it's obviously impossible. An equilateral triangular cake has all its lines of symmetry passing through the centre, which is also the centre of rotational symmetry. But you can slice it into 4 similarsliceswiththecentreinoneofthem fairly trivially.
Edit: Addendum: Also, the solution that supposedly exists does not employ any kind of fractal slices. You should be able to do it to a real cake (I guess the cake might have to be very large to be able to cut with the required precision though). If a fractal solution does exist I'd still like to know, mind.
Re: Unusual Cake Slicing
Well, I'm going to try to prove that it's impossible, and this should either work, or help us figure out where to look for a solution, at least.
Basically, the idea I'm working with is that any piece you cut either is or is not an edge piece  i.e. it includes part of the edge of the original cake. All pieces cannot be edge pieces  it is impossible to isolate the center with a set of identical pieces if they are all connected to the edge. (I'm not quite sure how to prove that lemma, but I'm pretty confident that it's true.) Therefore, we must be cutting one piece out of the middle of the cake. That piece must have one of its edges be identical to a segment of the original cake's edge  that is, somewhere in the cake we must have a cut that is the same shape as the edge of the cake.
That's where my brain locks up.
Basically, the idea I'm working with is that any piece you cut either is or is not an edge piece  i.e. it includes part of the edge of the original cake. All pieces cannot be edge pieces  it is impossible to isolate the center with a set of identical pieces if they are all connected to the edge. (I'm not quite sure how to prove that lemma, but I'm pretty confident that it's true.) Therefore, we must be cutting one piece out of the middle of the cake. That piece must have one of its edges be identical to a segment of the original cake's edge  that is, somewhere in the cake we must have a cut that is the same shape as the edge of the cake.
That's where my brain locks up.
22/7 wrote:If I could have an alternate horn that would yell "If you use your turn signal, I'll let you in" loud enough to hear inside another car, I would pay nearly any amount of money for it.
Re: Unusual Cake Slicing
quintopia wrote:On a more serious note, I don't believe it's possible (assuming we must cut into a countable number of slices), considering all lines of symmetry of the disk pass through the center (and the center is the origin of rotational symmetry as well) it seems that the center point must be on the boundary of any slices of a set of congruent slices.
This doesn't solve the original problem, but I have a counterexample to "the center point must be on the boundary of any slices of a set of congruent slices."
Set a compass's radius to the radius of the cake, and put one end of the compass on the circumference of the cake. Trace a 60 degree arc from the center of the cake to another point on the circumference (which will be 60 degrees around the circumference from the first circumference point). Trace out 5 more arcs this way, and the cake will be divided into 6 congruent "equiarcular" "triangles". All 6 of them touch the cake center, of course, but note that each has a line of symmetry that does not pass through the cake center. So cut out the 6 "triangles" and then cut each along its line of symmetry to get 12 congruent pieces, only 6 of which touch the cake center.
Edit / Double Post
My best idea for solving the problem goes something like this: Imagine a regular hexagon surrounded by 6 congruent regular hexagons. Taken together, they form a shape which is an approximation to a circle. Now pretend it really is a circle.
Re: Unusual Cake Slicing
i'm making the assumption that this cake is perfectly circular with no frosting and is flat on top and bottom. the single point of the cake is in the dead center.
I'd then cut the cake horizontally into an odd number of pieces > 1
the poison would be in the center of my horizontally cut pieces
*edit  just noticed that my solution was mentioned in the original problem... doh*
I'd then cut the cake horizontally into an odd number of pieces > 1
the poison would be in the center of my horizontally cut pieces
*edit  just noticed that my solution was mentioned in the original problem... doh*
Re: Unusual Cake Slicing
Hix wrote:quintopia wrote:On a more serious note, I don't believe it's possible (assuming we must cut into a countable number of slices), considering all lines of symmetry of the disk pass through the center (and the center is the origin of rotational symmetry as well) it seems that the center point must be on the boundary of any slices of a set of congruent slices.
This doesn't solve the original problem, but I have a counterexample to "the center point must be on the boundary of any slices of a set of congruent slices."
Set a compass's radius to the radius of the cake, and put one end of the compass on the circumference of the cake. Trace a 60 degree arc from the center of the cake to another point on the circumference (which will be 60 degrees around the circumference from the first circumference point). Trace out 5 more arcs this way, and the cake will be divided into 6 congruent "equiarcular" "triangles". All 6 of them touch the cake center, of course, but note that each has a line of symmetry that does not pass through the cake center. So cut out the 6 "triangles" and then cut each along its line of symmetry to get 12 congruent pieces, only 6 of which touch the cake center.
I meant that it must be on the boundary of some (>=1) slices (which implies it touches at least two), whereas the problem states the point must be in the interior. In other words, when I said "any slices" I was using "slice" to mean "series of cuts."
Re: Unusual Cake Slicing
I think it's impossible, and here's my reasoning:
Assume there is a solution.
As Puck said, there are either edge slices or inner slices. Considering an edge slice, it has as one side an arc, with a radius equal to the radius of the cake. Since the solution requires > 1 slices, the arc cannot be a full circle, so there must be at least two corners (and thus at least two sides) on an edge slice. That's lemma 1
Lemma 2 is that there must be an inner slice. If all the slices in the solution were edges, there would be no way for them to not pass a side through the centre while retaining equal size and shape. So if there is a solution there must be at least one inner slice.
Since the inner slice must be the same shape as an edge slice, the inner slice must (by Lemma 1) have an arc with the same radius as the cake and at least two corners.
Now is where the intuition comes in:
Looking at an edge piece, the back of it (the edge(s) that aren't on the outside of the cake) must fit into an overall pattern. You can either design the back so that it will fit another back, or so that it will fit the outer arc (possibly some combination of both, but that won't really affect my proof) Here's the problem: if you design it to fit the arc, then the pattern will keep on going until you reach the other side of the cake (maybe not directly across). At that point, what do you do? You can't fit the back to the back, so the cake won't be circular. If you design the back to fit another back, then what happens next? You have the outer arc of the cake, and then an inner piece with an arc facing inward. Other possibilities include cutting up concentric rings into pieces, but then the center can't be the same shape. Any way you look at it that strictly follows the definition of the solution will lead you into a problem where there is a leftover slice that isn't the same size or shape or you can't form a circle. I can't prove it, but I'm pretty sure.
There is one possible solution, but it requires interpreting the rules a bit. Technically, concentric rings could be made so they all had the same volumes, and they are still in a circular shape. In that way, you could cut the cake into pieces of the same "size" and "shape". But I don't think that's the intended solution.
Assume there is a solution.
As Puck said, there are either edge slices or inner slices. Considering an edge slice, it has as one side an arc, with a radius equal to the radius of the cake. Since the solution requires > 1 slices, the arc cannot be a full circle, so there must be at least two corners (and thus at least two sides) on an edge slice. That's lemma 1
Lemma 2 is that there must be an inner slice. If all the slices in the solution were edges, there would be no way for them to not pass a side through the centre while retaining equal size and shape. So if there is a solution there must be at least one inner slice.
Since the inner slice must be the same shape as an edge slice, the inner slice must (by Lemma 1) have an arc with the same radius as the cake and at least two corners.
Now is where the intuition comes in:
Looking at an edge piece, the back of it (the edge(s) that aren't on the outside of the cake) must fit into an overall pattern. You can either design the back so that it will fit another back, or so that it will fit the outer arc (possibly some combination of both, but that won't really affect my proof) Here's the problem: if you design it to fit the arc, then the pattern will keep on going until you reach the other side of the cake (maybe not directly across). At that point, what do you do? You can't fit the back to the back, so the cake won't be circular. If you design the back to fit another back, then what happens next? You have the outer arc of the cake, and then an inner piece with an arc facing inward. Other possibilities include cutting up concentric rings into pieces, but then the center can't be the same shape. Any way you look at it that strictly follows the definition of the solution will lead you into a problem where there is a leftover slice that isn't the same size or shape or you can't form a circle. I can't prove it, but I'm pretty sure.
There is one possible solution, but it requires interpreting the rules a bit. Technically, concentric rings could be made so they all had the same volumes, and they are still in a circular shape. In that way, you could cut the cake into pieces of the same "size" and "shape". But I don't think that's the intended solution.
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Re: Unusual Cake Slicing
I didn't understand that proof. I couldn't even find the proofs of the lemmas. What do you mean by "sides" in the first lemma?
Let me redefine the problem in more mathematical terms:
Let D be the unit disk. Find a finite set S of cardinality >1 of closed connected subsets of D that only overlap at their boundaries, s.t. [imath]\exists X \in S[/imath] s.t. the origin [imath]O \in X[/imath] and [imath]O \not\in S\setminus X[/imath], and s.t. [imath]\forall X_1, X_2 \in S, X_1 \cong X_2,[/imath] and s.t. [imath]\bigcup_{X \in S}X=D[/imath].
Let me redefine the problem in more mathematical terms:
Let D be the unit disk. Find a finite set S of cardinality >1 of closed connected subsets of D that only overlap at their boundaries, s.t. [imath]\exists X \in S[/imath] s.t. the origin [imath]O \in X[/imath] and [imath]O \not\in S\setminus X[/imath], and s.t. [imath]\forall X_1, X_2 \in S, X_1 \cong X_2,[/imath] and s.t. [imath]\bigcup_{X \in S}X=D[/imath].

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Re: Unusual Cake Slicing
Is it a pound cake?
Re: Unusual Cake Slicing
Darth Eru wrote:I think it's impossible, and here's my reasoning:
It seems reasonable. When you cut the cake up, each cut makes the same length of convex as of concave piece sides. You start with a surplus of convex side from the outer edge of the cake. All pieces are supposed to be identical, so each must also have a more convex boundary than concave.
While it is clear that such a piece cannot tile the whole plane, there is no obvious reason why it couldn't form a disc just like Hix's example, even if there have to be internal pieces.
Re: Unusual Cake Slicing
I never actually called it a proof. My exact words were "here's my reasoning", and later on I even stated that I couldn't actually prove my result. The lemmas are basically assumptions that I made from trying to find possible counterexamples and failing. They aren't proven, but I also doubt you can disprove them, since I'm pretty sure they're true.
As for sides, I thought it was a pretty well known geometric axiom that any two dimensional figure has at least one side. If a figure has at least two corners (do I have to explain what that is too?), then it logically has at least two sides.
As for understanding the rest of it, sometimes I find it difficult to put the scenarios I'm imagining into coherent language, especially if it's a visual problem I'm trying to discuss (such as this). So sorry if it doesn't make sense.
BTW, putting the problem into set notation, while somewhat laudable in and of itself, would be more useful if you then used it to help solve the problem, rather than leaving it as evidence that you're smarter than the person who was trying to actually solve it, albeit ineffectively.
As for sides, I thought it was a pretty well known geometric axiom that any two dimensional figure has at least one side. If a figure has at least two corners (do I have to explain what that is too?), then it logically has at least two sides.
As for understanding the rest of it, sometimes I find it difficult to put the scenarios I'm imagining into coherent language, especially if it's a visual problem I'm trying to discuss (such as this). So sorry if it doesn't make sense.
BTW, putting the problem into set notation, while somewhat laudable in and of itself, would be more useful if you then used it to help solve the problem, rather than leaving it as evidence that you're smarter than the person who was trying to actually solve it, albeit ineffectively.
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 NathanielJ
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Re: Unusual Cake Slicing
Darth Eru wrote:As for sides, I thought it was a pretty well known geometric axiom that any two dimensional figure has at least one side. If a figure has at least two corners (do I have to explain what that is too?), then it logically has at least two sides.
The problem is, you can easily cut it in such a way that each piece has just one corner and just some pieces have just one side (eg. have one of the "corners" be rounded in so that you just have a sharper corner on the bordering piece).
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Re: Unusual Cake Slicing
edit: Oop, shit.
Spoiler:
Will wrote:Andrew Jackson was all kinds of badass.
The Mighty Thesaurus wrote:HACKS ARE STING OUR SYLLES AND SING THEM TO TERRISTS!
Re: Unusual Cake Slicing
DaMullet wrote:Cut it in half with a swirl, like a yinyang.
no need for spoiler, that won't work. The only way the yin yang cut won't go through the direct center is if the sides aren't the same
I have a guess. If you can find something, like the hexagon below, with an even number of sides, that can repeat it's exact shape (like the way four equilateral triangles make a bigger one), you'll have your solution. The shapes must have an even number of sides so that half can be "in"s and half can be "out"s. Does there exist such a shape?
EDIT: I guess I didn't think about it as hard as I thought. There must be a central piece and there can be no outer edge made up by more than one piece for my "solution" to work. It also seems that the building block shapes needn't make up the same shape. If you could make a regular polygon out of a bunch of other polygons rotated around a central piece, it should work.
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Re: Unusual Cake Slicing
crzftx wrote:The shapes must have an even number of sides so that half can be "in"s and half can be "out"s.
Why? Hix's example doesn't have that. His example only fails at the central point not being interior.
Also reread my previous post. If you want to divide up the whole disc into identical pieces, they must have more 'out' than 'in' (in terms of total perimeter length rather than number of sides). Because your first attempt doesn't deal with the triangleish leftovers you've been able to ignore that.
Re: Unusual Cake Slicing
Just came across this and found it interesting. I think it's impossible, but I'm way too tired to even try to prove it.
I like jaap's observation about the amount of convex and concave perimeter, but I can't think how to use it. I had another thought though: can someone prove that a solution would have to be radially symmetric? That sounds like something you might be able to prove and it would be simple from there.
I like jaap's observation about the amount of convex and concave perimeter, but I can't think how to use it. I had another thought though: can someone prove that a solution would have to be radially symmetric? That sounds like something you might be able to prove and it would be simple from there.
Re: Unusual Cake Slicing
Darth Eru wrote:I never actually called it a proof.
My bad. I thought it was an outline of a proof, and was trying to get some clarification. Didn't mean to sound like an asshole or anything. I would have proved it if I could, but after phrasing it in set theoretic terms, no proof idea was forthcoming, so I left it at that. I do think my rephrasing added something that wasn't mentioned explicitly before (namely, that the pieces are closed sets overlapping at their boundaries) so it wasn't a complete waste of time. That little piece of info would be useful if we could prove that the center point must be on a piece's boundary.
Re: Unusual Cake Slicing
My suggestion re: rotational symmetry was a bit redundant, as it was already mentioned as early as, er, the first response. You'll have to forgive my inability to read last night  my cognitive functions were failing. Anyway, I still think this is the way to go.
@PaulT: an equilateral triangle has symmetry on rotation through 2pi/3, but a circle has symmetry under any rotation. Notice the obvious solution for the triangle still obeys the symmetry (not sure if this has to be true for any solution).
[Edit: copied text of doubledup post into this one]
Wait...I'm liking jaap's idea again.
Think of a general piece, and then the edge pieces
Each piece has 2pi*r/n "extra" concavity, where n=no. pieces
Each edge piece has 2pi*r/e of the circumference, where e=no. edge pieces. Any other concavity an edge piece has is cancelled out by some convexity. So n=e. Every piece is an edge piece  The only way to cut is through the centre and you can't avoid the poison.
That works doesn't it? (seems too simple...)
[edit: belatedly discovered I could delete my own doublepost, so I did. Never even noticed that little button before)]
@PaulT: an equilateral triangle has symmetry on rotation through 2pi/3, but a circle has symmetry under any rotation. Notice the obvious solution for the triangle still obeys the symmetry (not sure if this has to be true for any solution).
[Edit: copied text of doubledup post into this one]
Wait...I'm liking jaap's idea again.
Think of a general piece, and then the edge pieces
Each piece has 2pi*r/n "extra" concavity, where n=no. pieces
Each edge piece has 2pi*r/e of the circumference, where e=no. edge pieces. Any other concavity an edge piece has is cancelled out by some convexity. So n=e. Every piece is an edge piece  The only way to cut is through the centre and you can't avoid the poison.
That works doesn't it? (seems too simple...)
[edit: belatedly discovered I could delete my own doublepost, so I did. Never even noticed that little button before)]
Last edited by troyp on Mon Sep 22, 2008 3:22 am UTC, edited 3 times in total.
Re: Unusual Cake Slicing
The original question seems strange. If it should be possible to do with a real cake, it seems to me it's supposed to be about cutting a cake fairly without including the center in any piece, not having the center piece have the same shape as the other pieces. In which case the question is trivial.
My guess is that the question got confused somewhere along the second party.
My guess is that the question got confused somewhere along the second party.
Re: Unusual Cake Slicing
Notch wrote:The original question seems strange. If it should be possible to do with a real cake, it seems to me it's supposed to be about cutting a cake fairly without including the center in any piece, not having the center piece have the same shape as the other pieces. In which case the question is trivial.
My guess is that the question got confused somewhere along the second party.
No, people are doing the right question. Maybe having poison in the centre is a bit misleading. All that's required is that the poison is on one specific piece of cake, not spread across many as it would be with, say, normal pieceofpie slices. If it helps, I like to think that I'll be handing the pieces around and I want to poison someone without them realising, so their piece has to look like everyone else's.
Edit: I guess you're actually suggesting that the puzzle got deformed before it reached me. Obviously I can't know, but it was thirdhand along a line of mathematicians, so they were probably quite precise with their language.
Re: Unusual Cake Slicing
Unless you allow for throwing out offcuts at the edges (Which makes it so trivial I assume it's against the rules, but if your mate is annoying technically it wasn't excluded ) or let us cut it thin enough that the edges are for all intents and purposes straight (And you said it must have a finite number of cuts) I just don't think it's possible.
You're basically looking for a shape which can tessellate a circle without any edge being on the center... and I just don't think that it's possible to do it without symmetry. I can't prove it mathematically, but it just seems logical...
Now if you want to poison a couple of people while still having safe slices for yourself, I could hook you up, but that's more a reinterpretation than an answer
You're basically looking for a shape which can tessellate a circle without any edge being on the center... and I just don't think that it's possible to do it without symmetry. I can't prove it mathematically, but it just seems logical...
Now if you want to poison a couple of people while still having safe slices for yourself, I could hook you up, but that's more a reinterpretation than an answer
Re: Unusual Cake Slicing
It's also quite possible if we use toroidal geometry, in which case, I believe crzftx's answer is a solution.
Re: Unusual Cake Slicing
quintopia wrote:It's also quite possible if we use toroidal geometry, in which case, I believe crzftx's answer is a solution.
A donut has no poisoned center.
And if you're going to use topology, you might as well start with a square instead of a disc, which has the easy solution of cutting it into 3 equal rectangles.
There are easy answers for the equilateral triangle and for the square. Are there any other regular polygons for which the problem can be solved? I doubt even that is possible, though the hexagon seems the likeliest candidate (and then possibly the octagon or dodecagon).
Re: Unusual Cake Slicing
I think the point that Notch was making was that if this were a real cake, no one would be eating the poisoned piece, so no one would care what size or shape it was, just so the rest of the pieces are identical. (In this case, the solution is easy  cut a circle of arbitrarily small size around the poisoned center, then pieslice the rest of the cake into however many pieces you desire.)
By reasoning like Darth Eru's, I remain convinced that the puzzle as originally stated is impossible, but I'm unsure how to prove it.
By reasoning like Darth Eru's, I remain convinced that the puzzle as originally stated is impossible, but I'm unsure how to prove it.
22/7 wrote:If I could have an alternate horn that would yell "If you use your turn signal, I'll let you in" loud enough to hear inside another car, I would pay nearly any amount of money for it.
Re: Unusual Cake Slicing
jaap wrote:There are easy answers for the equilateral triangle and for the square. Are there any other regular polygons for which the problem can be solved? I doubt even that is possible, though the hexagon seems the likeliest candidate (and then possibly the octagon or dodecagon).
It's possible with any triangle and most quadrilaterals in fact. We should be asking "are there any other convex polygons for which the problem can be solved?"
The answer is yes: an irregular hexagon made out of 10 equilateral triangles works.
Re: Unusual Cake Slicing
If anyone's interested, I wrote up most of a (rough, informal) proof of the impossibility of the original problem, based on the line of reasoning first brought up by Darth Eru. I got stuck at the end at a step I thought would be easy (corresponding to DE's lemma 2). I think the rest is right, although I couldn't guarantee it. [spoilered for brevity]
Spoiler:
 jestingrabbit
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Re: Unusual Cake Slicing
troyp wrote:If anyone's interested, I wrote up most of a (rough, informal) proof of the impossibility of the original problem, based on the line of reasoning first brought up by Darth Eru. I got stuck at the end at a step I thought would be easy (corresponding to DE's lemma 2). I think the rest is right, although I couldn't guarantee it. [spoilered for brevity]Spoiler:
There are problems with this other than what you're stuck on.
Spoiler:
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Re: Unusual Cake Slicing
troyp wrote:Now consider an edge piece. It contains a portion of the unit circle of 2pi/e length, which is convex.
That does not follow. The pieces could have two or more convex sections of different lengths. Some edge pieces might use one of the convex sections on the outside, others might use another. On average they use 2pi/e for the outside, but that doesn't mean they all use the same amount.
(Ninja'd!)
troyp wrote:This region resulted from a cut within the circle, so there must be a corresponding concave region on the border of an adjacent piece.
The corresponding concave part need not be all on one piece  the convex section might lie adjacent to two or more other pieces, and being identical, they might even be using the same concave section.
As far as I can see, all we know is that each piece has C_{piece} = 2pi/n. The edge pieces together have C_{edges} = e*2pi/n. This leaves an inner border of e*2pi/n2pi = (en)*2pi/n, to be compensated for by the ne interior pieces.
Sorry to be such a spoilsport. I don't even have anything constructive to offer instead.
Re: Unusual Cake Slicing
Based on "I like to think that I'll be handing the pieces around and I want to poison someone without them realising, so their piece has to look like everyone else's." a practical solution would be: cut ordinary pieslice style cuts, but offcenter so that the poison is in one slice. It should be close enough that noone will notice, and if the guy who gets the poison does notice he isn't going to say anything (after all, he's getting the bigger slice)
I think the problem as actually described (all pieces have to be identical) is impossible, unless you allow for discarding nonidentical pieces.
I think the problem as actually described (all pieces have to be identical) is impossible, unless you allow for discarding nonidentical pieces.
 Godskalken
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Re: Unusual Cake Slicing
jestingrabbit wrote:1. I think you've disproved the existence of Hix's example, but I know Hix's example exists. Contradiction.
I don't think you've understood the proof. Hix's example is certainly possible under those conditions.
jaap wrote:each piece has C_piece = 2pi/n. The edge pieces together have C_edges = e*2pi/n. This leaves an inner border of e*2pi/n2pi = (en)*2pi/n, to be compensated for by the ne interior pieces.
Uhm, if that's correct you just showed that en=ne, or, n=e.
My guess is that the original puzzle was actually a lateral thinking thingy, and that the "cut horizontally" solution is the intended one.
Re: Unusual Cake Slicing
Godskalken wrote:jaap wrote:each piece has C_piece = 2pi/n. The edge pieces together have C_edges = e*2pi/n. This leaves an inner border of e*2pi/n2pi = (en)*2pi/n, to be compensated for by the ne interior pieces.
Uhm, if that's correct you just showed that en=ne, or, n=e.
No, the sign flips because concave fits against convex. The surplus of convex sides on the inner pieces is compensated for by the inside of the ring of edge pieces having a surplus of concave sides.
Re: Unusual Cake Slicing
Hix wrote:quintopia wrote:On a more serious note, I don't believe it's possible (assuming we must cut into a countable number of slices), considering all lines of symmetry of the disk pass through the center (and the center is the origin of rotational symmetry as well) it seems that the center point must be on the boundary of any slices of a set of congruent slices.
This doesn't solve the original problem, but I have a counterexample to "the center point must be on the boundary of any slices of a set of congruent slices."
Set a compass's radius to the radius of the cake, and put one end of the compass on the circumference of the cake. Trace a 60 degree arc from the center of the cake to another point on the circumference (which will be 60 degrees around the circumference from the first circumference point). Trace out 5 more arcs this way, and the cake will be divided into 6 congruent "equiarcular" "triangles". All 6 of them touch the cake center, of course, but note that each has a line of symmetry that does not pass through the cake center. So cut out the 6 "triangles" and then cut each along its line of symmetry to get 12 congruent pieces, only 6 of which touch the cake center.
It took me a while to get this, here's a visual aid for anyone who doesn't quite get what you're talking about:
 jestingrabbit
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Re: Unusual Cake Slicing
Godskalken wrote:jestingrabbit wrote:1. I think you've disproved the existence of Hix's example, but I know Hix's example exists. Contradiction.
I don't think you've understood the proof. Hix's example is certainly possible under those conditions.
The proof claims quite directly that there are no internal pieces. Hix's example has internal pieces. Contradiction.
Godskalken wrote:jaap wrote:each piece has C_piece = 2pi/n. The edge pieces together have C_edges = e*2pi/n. This leaves an inner border of e*2pi/n2pi = (en)*2pi/n, to be compensated for by the ne interior pieces.
Uhm, if that's correct you just showed that en=ne, or, n=e.
My guess is that the original puzzle was actually a lateral thinking thingy, and that the "cut horizontally" solution is the intended one.
Given that it arose in mathematics, or at least with mathematicians, I tend to think that we've got the question as it was initially put. Not every question has an easy answer.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Re: Unusual Cake Slicing
I've come to the conclusion that the problem as stated is impossible. PaulT: If you in fact have a solution that you believe to be the "right" answer, please post it in spoiler tags or PM me so that I can explain how it does not actually satisfy the conditions of the problem.
Re: Unusual Cake Slicing
@ jaap: that's okay, killing flawed proofs is constructive
I don't think your second objection is valid, though. You can cut the sections wherever necessary. I was only using the lengths and for any small section there's exactly one region on either side (more than 2 only meet at a point which has zero length).
I agree with your first objection (see below).
Well, Hix's example certainly demolishes the unproved assertion at the end (I didn't get this example when I first read it, I had to go back and draw it  it's a neat construction). I guess it's *technically* consistent with my result since all hix's slices have at least a point on the edge :/
Anyway, of more concern is this...
(sigh) This is what got me when I tried to prove the last part  the assumption that all pieces had the corresponding piece on the edge. I was sure I'd used this assumption elsewhere, but when I went back to fix it/add a note, I couldn't find it. I half convinced myself the proof didn't require it after all.
ah well, I'll see if I can salvage anything from it later.
edit:
(My suspicion was the same as Notch's: that it was originally a simple problem with constant areas, not shapes.)
So you think there may be a solution? Interesting. I'll admit I'm less convinced than I was before that there's not (though still pretty convinced).
I don't think your second objection is valid, though. You can cut the sections wherever necessary. I was only using the lengths and for any small section there's exactly one region on either side (more than 2 only meet at a point which has zero length).
I agree with your first objection (see below).
Godskalken wrote:jestingrabbit wrote:1. I think you've disproved the existence of Hix's example, but I know Hix's example exists. Contradiction.
I don't think you've understood the proof. Hix's example is certainly possible under those conditions.
Well, Hix's example certainly demolishes the unproved assertion at the end (I didn't get this example when I first read it, I had to go back and draw it  it's a neat construction). I guess it's *technically* consistent with my result since all hix's slices have at least a point on the edge :/
Anyway, of more concern is this...
jesting rabbit wrote:2. What about edge pieces that use different parts of the edge to make the circle? ie my shape has two convex sides, S1 and S2. Some of the edge pieces have S1 on the circle, some have S2.
(sigh) This is what got me when I tried to prove the last part  the assumption that all pieces had the corresponding piece on the edge. I was sure I'd used this assumption elsewhere, but when I went back to fix it/add a note, I couldn't find it. I half convinced myself the proof didn't require it after all.
ah well, I'll see if I can salvage anything from it later.
edit:
jesting rabbit wrote:Given that it arose in mathematics, or at least with mathematicians, I tend to think that we've got the question as it was initially put. Not every question has an easy answer.
(My suspicion was the same as Notch's: that it was originally a simple problem with constant areas, not shapes.)
So you think there may be a solution? Interesting. I'll admit I'm less convinced than I was before that there's not (though still pretty convinced).
 MartianInvader
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Re: Unusual Cake Slicing
Very interesting one. A question for clarification: Are we assuming that all the pieces must be connected? (i.e., if I divvied it up into 5 squares and 5 circles, could I claim 1 piece is a square + a circle?)
Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!
Re: Unusual Cake Slicing
MartianInvader wrote:Very interesting one. A question for clarification: Are we assuming that all the pieces must be connected? (i.e., if I divvied it up into 5 squares and 5 circles, could I claim 1 piece is a square + a circle?)
I'd say all the pieces have to be 'in one piece'.

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Re: Unusual Cake Slicing
PaulT wrote:MartianInvader wrote:Very interesting one. A question for clarification: Are we assuming that all the pieces must be connected? (i.e., if I divvied it up into 5 squares and 5 circles, could I claim 1 piece is a square + a circle?)
I'd say all the pieces have to be 'in one piece'.
Otherwise, you could use two pieces of different size and shape. The only extra constraint would be that they would have to be aligned (a pretty large constraint). Even with "unconnected" pieces, I don't think that there is a solution.
Now, I've done some math dealing with the ratio of convex to concave edges in the answer (if it exists).
I'm thinking that if we go through the math of all possible answers, it might lead to a proof that there are no answers that satisfy all requirements.
Here is my work:
Spoiler:
I now want to bring up another factor. Many types of pieces can be broken up.
Spoiler:
If anyone really, really wants to explore this topic, read Escher’s Regular Division of the Plane, or this book written by Doris Schattschneider, M.C. Escher: Visions of Symmetry. I heard her speak at the opening of a temporary exhibition of Escher’s work in my hometown, and she would provide us with the means to prove or disprove the possibility of an answer. Just as Euclidean geometry and spherical geometry have rules, I bet that Euclidean division of the plane has some postulate (or nonpostulate) rules that would be of some benefit to us.
By the way, if this puzzle can be solved, Escher would have done it a long time ago, and he didn't. (Look at the middle point)
http://www.math.sunysb.edu/~scott/mat11 ... cleIII.jpg
Once we eventually figure out how to answer this problem, someone has to devise a problem in which you must make sure that a specific person eats the piece with the poison in it. (I’m thinking something having people with different arm lengths going at the cake and each taking a piece, or something having you make sure that the # of cakeeating people invited = the number of pieces. Then you must make sure that the man eats last, and assuming that people take outer pieces first, he’ll get poisoned.)
YAY! Finally!
Re: Unusual Cake Slicing
If the polygons are not required to be regular, it's quite possible. Wolfram has some hexagons, and even a pentagon, here.Equilateral triangles and squares can always be broken up into geometically similar shapes of the same type. Equilateral triangles and squares can be broken up into N pieces where N is the square of a prime number. (1,4,9,16,25) Interestingly, I can find no ways to break up any polygons with more than 4 sides into mini, similar copies of itself.
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