Wikipedia:Reference desk/Archives/Mathematics/2012 August 13

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August 13

Hyperball cutting fractions.

While a pizza can be cut into 7 pieces with three cuts, I don't believe that it can be cut evenly into sevenths. Is there a way to figure out how good it can be? i.e. for cutting B^n with m cuts, there must be a single piece which makes up at least X percent of the original B^n...Naraht (talk) 20:08, 13 August 2012 (UTC)

Why do you think it can't? I would guess that it can — each of the three lines has three degrees of freedom (two for position of midpoint, one for slope), making nine altogether, and you need just six to balance seven proportions. That doesn't prove you can do it, but you do at least have enough knobs; you just might not be able to turn one of them far enough. --Trovatore (talk) 22:31, 13 August 2012 (UTC)

Oh, wait, once you know the position of the midpoint you know the slope. Still, you have six degrees of freedom, which should be enough. --Trovatore (talk) 22:36, 13 August 2012 (UTC)

You've no real freedom, you need to have 3/7 of the pizza on one side of each cut so all three are the same distance from the centre. You then need 1/7 in the area between each pair of cuts and you'd need that to happen when they were 120° to each other for iI don't have much of a feeling for whether it should be possible to cut the pizza into equal slices t all to join up right and it doesn't. Dmcq (talk) 22:59, 13 August 2012 (UTC)

OK, sounds right. --Trovatore (talk) 00:02, 14 August 2012 (UTC)

Who says the cental portion needs to be an equilateral triangle? It just needs to have area 1/7. -- SGBailey (talk) 06:24, 14 August 2012 (UTC)

Do the cuts have to be straight lines? Double sharp (talk) 07:14, 14 August 2012 (UTC)

Yes, also with your name you might be interested in the double edged version of the mezzaluna ;-) In the interests of completeness I gues I should also point to the possibly more familiar pizza cutter. Dmcq (talk) 08:11, 14 August 2012 (UTC)

By the way for the case of a torus (or doughnut though I think the jam or cream gives problems when you squash them - you need a very sharp knife and some bits are very small) you can dissect them into 13 pieces with three cuts without moving the pieces, see[1]. Of course if you can move the pieces you can cut a pizza into eight equal pieces with three cuts. More interestingly it is then also possible to divide it into seven equal pieces though stacking pieces on top of each other may cause them to stick and get messy. Dmcq (talk) 19:10, 14 August 2012 (UTC)

If I had to guess, I would say that it is not possible to get equal cuts, based on the case with threefold symmetry (all angles 60 and 120 degrees). It should be fairly easy to determine the best case scenario under this ansatz, but it's not a calculation that I would wish to do myself. I've also not really thought about how to prove this ansatz (if true). For the general case, I would consider devising a numerical scheme. It's probably easiest to minimize (numerically) the sum of the squares of the areas, so I would start with that and see if there is any obvious structure to the solution. Sławomir Biały (talk) 19:44, 14 August 2012 (UTC)

Specific case

OK, let's drop this back to the specific two dimensional 3 cut-seven slice situation. There are 3 types of slice, the center slice, the three slices that border the center slice at a side and the three slices that border the center slice at a corner. It seems to be that the only chance for 7 equal slices is to have the three slices at 120° rotation from each other giving an equilateral triangle in the center. The only degree of freedom then is how far from the center these lines are. If the lines are at the center, *with the center slice shrunk to nothing* then the "corner" slice is equal to the "side slice", however as the cut moves toward the edge, the corner slices lose area more quickly than the side slices (for example if moved out only 10%, the side piece has lost the 10% "tip", but the corner slices have lost area to the center slice *and* to both side pieces that it borders. So with that original assumption, seven equal pieces is impossible. The question is how close to equal can you get...

With a symmetrical arrangement where each cut is a distance of r from the center, the areas of each kind of slice are:

3 Sqrt[3] r^2
1/4 (Sqrt[3] + 2 \[Pi] - 10 Sqrt[3] r^2 + 6 r Sqrt[1 - r^2] + r Sqrt[1 + 2 r (r + Sqrt[3 - 3 r^2])] - 
    Sqrt[3 - 3 r^2] Sqrt[1 + 2 r (r + Sqrt[3 - 3 r^2])] + 4 ArcSin[1/2 (r - Sqrt[3 - 3 r^2])])
1/8 (-Sqrt[3] + 10 Sqrt[3] r^2 - 10 r Sqrt[1 - r^2] - r Sqrt[1 + 2 r^2 + 2 Sqrt[3] r Sqrt[1 - r^2]] + 
    Sqrt[3] Sqrt[-(-1 + r^2) (1 + 2 r^2 + 2 Sqrt[3] r Sqrt[1 - r^2])] - 4 ArcSin[r] - 4 ArcSin[r/2 - 1/2 Sqrt[3 - 3 r^2]])

The smallest slice is largest when r=0.196315..., when there are 4 slices with 6.3744% of the area each and 3 with 24.8341%. -- Meni Rosenfeld (talk) 09:08, 15 August 2012 (UTC)

Let's fix the two dimensional case. A chord corresponding to a cut has to divide the (unit) circle in two pieces $A$ and $B$ whose areas are in proportion $(3:4)$, thus of area $3\pi /7$ and $4\pi /7$. This immediately fixes the distance of the chord from the center (we need not solve the equation, which is I guess $x-\sin x=6\pi /7$ for the angle $x$ under which the segment is seen). That is, the three cutting chords are tangent to the same inner concentric circle. If we then consider the piece $A$, and note that the two triangular pieces of the three into which it is sub-divided have equal areas, the symmetry of the configuration immediately follows, hence the impossibility as already observed above by Sławomir Biały. --pma 10:50, 15 August 2012 (UTC)([edit]: I note only now Meni's answer, on these lines)

The above argument seems to generalize partially for n-dimensional Euclidean balls; at least, the cutting hyperplanes must be tangent to the same inner concentric sphere. I don't see now if this yelds to the symmetry as before, though. What about the 3 dimensional case, that is, the Euclidean 3-ball to be cut into 15 pieces of equal volume by 4 planes? [edit] Actually any two planar cuts must divide the volume in fixed proportions -namely, (3:4:4:4) which fixes their angle, because they are tangent to a given sphere. So the 3-dimensional case has to be symmetric too (that is, the central piece is a regular tetrahedron centered at the origin). From the symmetry it should easily follow that the equal subdivision is impossible, as before.--pma 11:03, 15 August 2012 (UTC)

Impossibility of equipartitions by hyperplanes (the $n$-dimensional case, $n\geq 2$ ) So, even if the Euclidean $n$-ball can be divided by $n$ orthogonal hyperplanes into $2^{n}$ pieces of equal $n$-dimensional volume, a partition with $n+1$ hyperplanes never produces $2^{n+1}-1$ pieces of equal volume. The argument is a plain generalization of the $2$ and $3$ dimensional case above. A partition of the $n$-ball by $n+1$ hyperplanes in generic position consists of $2^{n+1}-1$ pieces, one for each face of the skeleton of a symplex, which is the only interior piece, and the only polyhedral one. Each hyperplane divides the ball into two halves, made by $2^{n}-1$ and $2^{n}$ pieces of the partition (the latter containing the central symplex). Therefore, assuming the pieces have equal volume, the volumes of the two halves cut by the hyperplane are uniquely determined. Therefore the distance from the hyperplane to the origin is also determined (the volume depends strictly monotonically from the distance). Thus the hyperplanes are all tangent to a sphere of given radius. Then, we consider any pair of hyperplanes and notice that, again, they divide the ball into four pieces that must contain a fixed number of pieces, namely $(2^{n-1}-1,2^{n-1},2^{n-1},2^{n-1})$ (any two hyperplanes share some $(n-2)$-face of the inner symplex). This implies again that the angle between the two hyperplanes is uniquely determined (again by a monotonicity dependence of the volume from the angle, given that the hyperplanes are tangent to the same sphere). This proves that the inner symplex is regular, and centered at the origin. But such a partition can't be made of pieces of equal volume: for instance, the piece $A$ corresponding to a face, strictly contains the symmetric copy of the piece $B$ corresponding to the opposite vertex (that is $-B\subset A$ ). --pma 12:58, 15 August 2012 (UTC)

Meni, how the heck did you get those equations?Naraht (talk) 01:47, 16 August 2012 (UTC)

Using the silicon half of my brain. Or my silicon overlord. Or some other expression that signifies the fact that I don't really know how to do calculations anymore, I just know how to use Mathematica. In particular the main part of the code is

cond[B_] := Apply[And, Table[(-1)^B[[i]] ({x, y}.{Cos[2 \[Pi] i/3], Sin[2 \[Pi] i/3]} - r) <= 0, {i, 3}]]
Integrate[Boole[x^2 + y^2 <= 1 && cond[{1, 1, 0}]], {x, -1, 1}, {y, -1, 1}, Assumptions -> {0 < r < 1/2}]

And then just altering the integral expression for the different kinds of slices, plotting a graph and numerically solving equality of the relevant slices. -- Meni Rosenfeld (talk) 14:04, 16 August 2012 (UTC)