Unusual Cake Slicing

[PaulT]

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 cake-ness of the puzzle is purely aesthetic; we're not looking for lateral-thinking 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 third-hand or so. Supposedly there is a solution (I have my doubts). But don't blame me if it's impossible.

[quintopia]

I'm going to assume the axiom of choice and that the poison is at a single point-mass. 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.

[PaulT]

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 similar-slices-with-the-centre-in-one-of-them 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.

[Puck]

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.

[Hix]

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 counter-example 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.

[Langlier]

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*

[quintopia]

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 counter-example 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."

[Darth Eru]

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.

[quintopia]

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. \exists X \in S s.t. the origin O \in X and O \not\in S\setminus X, and s.t. \forall X_1, X_2 \in S, X_1 \cong X_2, and s.t. \bigcup_{X \in S}X=D.

[Alan Hicks]

Is it a pound cake?

[jaap]

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.

[Darth Eru]

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.

[NathanielJ]

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).

[DaMullet]

edit: Oop, shit.

Spoiler:

Cut it in half with a swirl, like a yin-yang.

[crzftx]

Cut it in half with a swirl, like a yin-yang.

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.

[jaap]

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 re-read 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 triangle-ish left-overs you've been able to ignore that.

[troyp]

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.

[quintopia]

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.

[troyp]

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 doubled-up 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 double-post, so I did. Never even noticed that little button before)]

[Notch]

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.

[PaulT]

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 piece-of-pie 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 third-hand along a line of mathematicians, so they were probably quite precise with their language.

[fyjham]

Unless you allow for throwing out off-cuts 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

[quintopia]

It's also quite possible if we use toroidal geometry, in which case, I believe crzftx's answer is a solution.

[jaap]

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).

[Puck]

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 pie-slice 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.

[quintopia]

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.

[troyp]

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:

WLOG, consider the unit circle.
Make any number of cuts. Each cut divides a piece into 2 new pieces.
defn: "net convexity" C = (length of convex perimeter of the piece) - (length of concanve perimeter of the piece)
Any length of cut which is convex wrt one piece is convex wrt the other, and vice versa.
Using this fact, by induction on the number of pieces,
sum (over all pieces) C = 2pi
let n= total number of pieces, e= total number of pieces with a part of the circle in its perimeter (edge piece)
All pieces are the same shape, so for each piece, C=2pi/n

Now consider an edge piece. It contains a portion of the unit circle of 2pi/e length, which is convex.
Suppose it contains another convex region in its border.
This region resulted from a cut within the circle, so there must be a corresponding concave region on the border of an adjacent piece.
Since all pieces are the same shape, the corresponding concave region must also be on the piece's own border.
The same argument applies swapping convex and concave regions of border.
Therefore, any internal region of convex border is balanced by concave border, and vice versa.
Thus, for an edge piece, the net convexity results purely from the region of border it inherits from the unit circle and C=2pi/e
Hence e=n. Every piece is an edge piece.

Assertion: The centre O of the unit circle lies on the border of every edge piece.
Proof: [I'm stuck on this last step. It's easy if I can assume that the "edge region" of an edge piece corresponds to the edge region of each of the others, but technically I can't.]

[jestingrabbit]

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:

WLOG, consider the unit circle.
Make any number of cuts. Each cut divides a piece into 2 new pieces.
defn: "net convexity" C = (length of convex perimeter of the piece) - (length of concanve perimeter of the piece)
Any length of cut which is convex wrt one piece is convex wrt the other, and vice versa.
Using this fact, by induction on the number of pieces,
sum (over all pieces) C = 2pi
let n= total number of pieces, e= total number of pieces with a part of the circle in its perimeter (edge piece)
All pieces are the same shape, so for each piece, C=2pi/n

Now consider an edge piece. It contains a portion of the unit circle of 2pi/e length, which is convex.
Suppose it contains another convex region in its border.
This region resulted from a cut within the circle, so there must be a corresponding concave region on the border of an adjacent piece.
Since all pieces are the same shape, the corresponding concave region must also be on the piece's own border.
The same argument applies swapping convex and concave regions of border.
Therefore, any internal region of convex border is balanced by concave border, and vice versa.
Thus, for an edge piece, the net convexity results purely from the region of border it inherits from the unit circle and C=2pi/e
Hence e=n. Every piece is an edge piece.

Assertion: The centre O of the unit circle lies on the border of every edge piece.
Proof: [I'm stuck on this last step. It's easy if I can assume that the "edge region" of an edge piece corresponds to the edge region of each of the others, but technically I can't.]

There are problems with this other than what you're stuck on.

Spoiler:

1. I think you've disproved the existence of Hix's example, but I know Hix's example exists. Contradiction.

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.

[jaap]

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!)

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/n-2pi = (e-n)*2pi/n, to be compensated for by the n-e interior pieces.

Sorry to be such a spoilsport. I don't even have anything constructive to offer instead.

[Random832]

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 pie-slice style cuts, but off-center so that the poison is in one slice. It should be close enough that no-one 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 non-identical pieces.

[Godskalken]

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.

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/n-2pi = (e-n)*2pi/n, to be compensated for by the n-e interior pieces.

Uhm, if that's correct you just showed that e-n=n-e, 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.

[jaap]

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/n-2pi = (e-n)*2pi/n, to be compensated for by the n-e interior pieces.

Uhm, if that's correct you just showed that e-n=n-e, 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.

[Random832]

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 counter-example 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:

hix.PNG (4.49 KiB) Viewed 7804 times

[jestingrabbit]

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.

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/n-2pi = (e-n)*2pi/n, to be compensated for by the n-e interior pieces.

Uhm, if that's correct you just showed that e-n=n-e, 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.

[Random832]

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.

[troyp]

@ 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).

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...

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:

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]

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?)

[PaulT]

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'.

[TheSwaminator]

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:

Circle raidus = 1
n = # of pieces
Total area = pi
Area of each piece = pi/n

You can only have three types of edges.
Straight edge
Curved edge based on radius 1.
Generic (any other type of edge, such as sinusoidal) (There can be an infinite # of generic edges, and each would be treated the same exact way that I’m going to treat generic edges in my examples.)

s = straight edge length per piece
c = convex curved edge degrees per piece (angle is proportional to length)
k = concave curved edge degrees per piece (angle is proportional to length)
g1 = length of one side of Generic edge per piece (assuming that the edge is different from either side)
g2 = length of the other side of Generic edge per piece

g1 = g2, range 0-> infinity
s can be any #, 0 to infinity (Let’s not debate whether or not it could be infinity, please?)

Ls = length of all the straight cut divisions = s*n/2
Lg = length of all generic edge divisions = g1*n = g2*n = (g1 + g2)*n/2 [To make it look like Ls]
Lc = degrees of all the curved cut divisions {added by cutter} = n*k

Original amount of convexness to be filled, 360 degrees of radius 1 arcs.

n*c = n*k + 360
c-k = 360/n

What’s really important here is Lc.
As Lc (or n*k) increases, n*c increases by the same amount, but not proportionally.
When Lc = 0, there will be no inside pieces, because all pieces have 360/n convex degrees, and the only place to put that edge is against the outer rim of the cake.
Only when Lc>0 can you have inside pieces (duh, I know).
I think that we're all certain that you must have inside pieces to have a legitimate answer, so Lc >0.

n*c = n*k + 360
c = k + 360/n
c/k = 1 + 360/(n*k)
c/k = (Lc + 360)/Lc

If there exists an answer, having n pieces and a total length of Lc in curved cuts, then each piece must have an area of pi/n and the ratio of convex to concave edges on each piece must be 1 + 360/Lc, which is always greater than 1, but which decreases as pieces are divided using curved edges.

I now want to bring up another factor. Many types of pieces can be broken up.

Spoiler:

If someone can find a solution with similar pieces of different sizes, if the piece shape allows for breaking up of the piece into similar, smaller pieces, and if all of the original piece size ratios can be broken up, then that incorrect solution can be solved by the breaking up of pieces. Notice Random832’s picture (slightly unpdated).

hix.PNG (4.93 KiB) Viewed 6840 times

He has triangle like pieces being split with a straight edge to form two pieces. Note that those smallest pieces cannot be broken anymore (I think?). Go back to the triangle. That triangle can be made into a diamond by joining each two consecutive triangles together, forming 3 diamond shapes.

Notice that the 3 diamonds have ¾ convex and that the 6 triangles have 2/3 convex.
Notice again that the split triangles have added straight edge, but the same ratio of convex to concave.
The more pieces, the less (relatively) convex, assuming that you split the pieces with curved edges.

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.

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 non-postulate) 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 cake-eating 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!

[++$_]

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.

If the polygons are not required to be regular, it's quite possible. Wolfram has some hexagons, and even a pentagon, here.