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If I cut a perfectly circular pizza through its center 6 times at 30° angles, I get 12 pieces of equal area.

If I don't have to cut through the center, I can cut in a grid shape to divide the pizza into 16 pieces, but only the middle four will be of equal area.

Theoretically I could cut the pizza into a total of 22 pieces using 6 cuts, but more often than not none of them would be of equal area.

For 6 cuts, what is the largest number of pieces of equal size I can obtain (using only straight cuts that go through the entire pizza)? What about $n$ cuts?


And since people are posting answers to this effect, no, you are not allowed to fold the pizza or rearrange slices. This is not a lateral thinking puzzle.

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  • $\begingroup$ And we are only talking about straight line cuts? $\endgroup$ – skv Nov 7 '14 at 3:09
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    $\begingroup$ May one obtain extra pieces that are a different area but don't count? Or does every piece obtained have to have the same area? $\endgroup$ – xnor Nov 7 '14 at 6:41
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    $\begingroup$ You can have extra pieces that don't count, yes. $\endgroup$ – user88 Nov 7 '14 at 6:42
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    $\begingroup$ I think I can get 16 for n=6, but it's hard to prove or draw. It use an intermediate value theorem argument to show that three shapes can be made to have equal areas by tuning two parameters. $\endgroup$ – xnor Nov 7 '14 at 9:45
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    $\begingroup$ Your comment about left-over pieces being allowed should really be in the question body, not just a comment. Most similar puzzles I have seen do not allow left over pieces. $\endgroup$ – DeveloperInDevelopment Nov 8 '14 at 7:23
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UPDATE: I've added a large $N$ solution for multiples of 3 that slightly betters OP's solution at $3\times (\frac N 3 - 1)^2 + 9$, see end of this post.

Just to show that @humn is not the only one capable of wasting eyewatering amounts of pizza here are

15

tiny but equal pieces of pizza made using 6 cuts.

enter image description here
Due to symmetries there are only tree kinds of pieces; equalizing those costs 2 degrees of freedom which we can afford: Let $P$ be the point in the upper center where the blue and orange triangles meet. Then we can adjust the distance of P to the center and the angle between the lines meeting at $P$.

$N = 3n$ solution:

Example $N=12$, 36 slices: enter image description here

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  • $\begingroup$ Beautiful 3-fold symmetry, and only 4 discarded pieces! $\endgroup$ – humn Nov 2 '20 at 18:45
  • $\begingroup$ @humn, actually, it's seven. There are very small discarded bits at the outer tips of the three quadrilaterals. $\endgroup$ – Paul Panzer Nov 2 '20 at 19:24
  • $\begingroup$ Those teensy scraps are exquisitely subtle, @Paul Panzer, and just make me appreciate your dissection all the more as every cut crosses every other cut to produce the maximum possible 22 total pieces $\endgroup$ – humn Nov 3 '20 at 2:48
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☆   Revised to introduce a general solution of limited value, to include a solution for 7 cuts, and to acknowledge a better 6-cut solution   ☆

What about $n$ cuts?

Hats off to Paul Panzer’s tasty general solution, which produces more equally sized pieces than puzzle poser Joe Z.’s general sample solution for  $n = 6k > 6$  cuts.   The parabolic result-curves of both those general solutions, however, begin slower than a linearly growing general solution that produces $4(n{-}3)$ equally sized pieces from $n = 2k \ge 6$  cuts.   This linearly growing general solution is an improvement only in the narrow domain of  $6 \,{<}\, n \,{<}\, 12$  cuts.

None of the general solutions so far cover the case of $n \,{=}\, 7$ cuts. Here is a solution with rectangular symmetry whose number of equal pieces can easily be matched by Paul Panzer’s specific solution for 6 cuts merely by slicing a single segment of pizza along an appropriately measured chord.


         7 cuts, 16 equally sized pieces

  • Triangles E through L are congruent with area a b / 2.

  • Suitably adjusted values for a and b can make the areas of pieces A and B also a b / 2, and thus of pieces C, D, M, N, O and P as well.   This is because increasing the value of a decreases the sum of areas A+B while increasing the value of b increases the ratio of areas B/A.


Original post follows (slightly edited), which led to the solutions above.

For 6 cuts, what is the largest number of pieces of equal size I can obtain (using only straight cuts that go through the entire pizza)?

This is not the most possible with $n \,{=}\, 6$ cuts but here are 14 equally sized pieces, A through N, out of 20 total pieces. This has 1 equal piece fewer than Paul Panzer’s beautiful 3-fold symmetric solution and 2 equal pieces more than both the  $2n \,{=}\, 12$  “naively” sliced sectors of the puzzle statement and the  $ \displaystyle \raise1ex\strut \small \big( {\raise-.4ex n \over 2} \kern.05em{-}\kern.1em 1 \big) \!\!\; \raise1.8ex{\scriptsize 2} \normalsize \! + 8 = 12 $   equal pieces produced by the general algorithm in puzzle poser Joe Z.’s sample solution.

  • The dimensions shown for triangles E and G clearly make both of their areas a b, and thus of pieces F, H, I and J as well.

  • Suitably adjusted values for a and b can make the areas of pieces A and B also a b, and thus of pieces C, D, K, L, M and N as well.   This is because increasing the value of a decreases the sum of areas A+B while increasing the value of b increases the ratio of areas B/A.

A series of experiments led to the above dissection . . .

. . . along with some later experiments.

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If you can fold the pizza, with one cut you could ideally create infinite slices. Just fold the pizza N times along the symmetry axis and cut it in the middle.

What you obtain is 2(N-1) triangular slices folded in half, and ready for eating.

5 folds, 1 cut (I know what you're thinking, dirty mind):

enter image description here

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  • $\begingroup$ In my mind it works (I got for induction up to 3 folds), I need to try with a piece of paper though :) $\endgroup$ – clabacchio Nov 7 '14 at 12:53
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    $\begingroup$ yes please experiment with the paper up to at least 8 folds. $\endgroup$ – Kenshin Nov 7 '14 at 13:06
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    $\begingroup$ @Mew I see what you did there ;/ $\endgroup$ – clabacchio Nov 7 '14 at 13:08
  • $\begingroup$ "'if' you can fold the pizza" I've updated the question statement to clarify that you can't. $\endgroup$ – user88 Nov 7 '14 at 16:50
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I'm aware of an algorithm for $(\frac{n}{2}-1)^2 + 8$ pieces given an even $n \ge 6$ cuts as follows:

Inscribe a square $S$ inside the circle. Let $d$ be some distance from the edge of the square, and cut $\frac{n}{2}$ times, equally spaced between $d$ away from one edge to $d$ away from the opposite edge, both horizontally and vertically (for a total of $n$ cuts).

By the Intermediate Value Theorem, there exists some value $d$ for which the 8 edge pieces adjacent to the corners and the inside pieces all have equal area, which is a total of $(\frac{n}{2}-1)^2 + 8$ pieces.

This overtakes the naive solution of $2n$ at $n = 8$. It gives $17$ pieces for $8$ cuts, $24$ pieces for $10$ cuts, $33$ pieces for $12$ cuts, etc.

There may be a solution better than this one, though. Anyone else willing to give it a shot?

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I can only create 64 slices.

Step 1. Cut the pizza from North to South (2 slices)

Step 2. Cut the pizza from East to West (4 slicles)

Step 3. Lay the 4 slices on top of each other pointing East and cut from East to West (8 slices)

Step 4. Lay the 8 slices again on top of each other and slice (16 slices)

Step 5. Repeat 32 slices

Step 6. Repeat again 64 slices

The maximum number of slices is thus given by 2^n, where n is the number of cuts.

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  • $\begingroup$ Since maximum possible number of slices is given as; $$\frac{n*(n+1)}{2}+1=22$$ which can be obtained by intersection of 6 lines, putting slices on top of each other is invalid. $\endgroup$ – shyos Nov 7 '14 at 11:43
  • $\begingroup$ @shyos, clearly your formula doesn't take into account my method of stacking pizza slices on top of one another before cutting. Where in the question does it state it is invalid? $\endgroup$ – Kenshin Nov 7 '14 at 11:44
  • $\begingroup$ as OP states,"Theoretically I could cut the pizza into a total of 22 pieces using 6 cuts, but more often than not none of them would be of equal area." $\endgroup$ – shyos Nov 7 '14 at 11:50
  • $\begingroup$ @shyos, he's talking theoretically, but practically you can cut it into 64 equal slices. $\endgroup$ – Kenshin Nov 7 '14 at 11:52
  • $\begingroup$ He uses 'theoretically', since it is the maximum possible slices for n=6. But those slices wont have equal area. So slice number should be less than < 22 $\endgroup$ – shyos Nov 7 '14 at 11:55
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With a pizza, I guess the answer is logically 12, like you said. A taller cylinder is a different scenario, as you could cut through the middle height-wise, but this is not viable with a pizza. You could technically cut a cylinder into n discs of equal size.

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  • $\begingroup$ We're assuming that the pizza is flat, not a cylinder. $\endgroup$ – user88 Nov 7 '14 at 6:57
  • $\begingroup$ Also, I know of a way to get 17 pieces for 8 cuts. The answer isn't always $2n$. $\endgroup$ – user88 Nov 7 '14 at 6:58
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Cut two equilateral triangles? This should give you both the same size pieces on the outside, as well as same size pieces on the inside.

enter image description here

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  • $\begingroup$ That's two groups of 6 pieces, though, which only counts as 6 because you can only consider one group at a time. $\endgroup$ – user88 Nov 7 '14 at 5:17

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