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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.
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.
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 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.
Darth Eru wrote:I think it's impossible, and here's my reasoning:
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.
Will wrote:Andrew Jackson was all kinds of badass.
The Mighty Thesaurus wrote:HACKS ARE STING OUR SYLLES AND SING THEM TO TERRISTS!
DaMullet wrote:Cut it in half with a swirl, like a yin-yang.
crzftx wrote:The shapes must have an even number of sides so that half can be "in"s and half can be "out"s.
Darth Eru wrote:I never actually called it a proof.
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.
quintopia wrote:It's also quite possible if we use toroidal geometry, in which case, I believe crzftx's answer is a solution.
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.
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).
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:
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
troyp wrote:Now consider an edge piece. It contains a portion of the unit circle of 2pi/e length, which is convex.
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.
jestingrabbit wrote:1. I think you've disproved the existence of Hix's example, but I know Hix's example exists. Contradiction.
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/n-2pi = (e-n)*2pi/n, to be compensated for by the n-e interior pieces.
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/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.
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 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.
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.
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/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.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
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.
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.
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.
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?)
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'.
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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