The following question came up at a conference and a solution took a while to find.

Puzzle. Find a way of cutting a pizza into finitely many congruent pieces such that at least one piece of pizza has no crust on it.

We can make this more concrete,

Let $D$ be the unit disc in the plane $\mathbb{R}^2$. Find a finite set of subsets of $D$, $\mathcal{A}=\{A_i\subset D\}_{i=0}^n$, such that

  • for each $i$, $A_i$ is simply connected and equal to the closure of its interior
  • for each $i, j$ with $i\neq j$, $\operatorname{int}(A_i)\cap \operatorname{int}(A_j)=\emptyset$
  • $\bigcup\mathcal{A}=D$
  • for each $i,j$, $A_i=t(A_j)$ where $t$ is a (possibly orientation reversing) rigid transformation of the plane
  • for some $i$, $\lambda(A_i\cap\partial D)=0$ where $\lambda$ is the Lebesgue measure on the boundary circle.

Note that we require only that $\lambda(A_i\cap\partial D)=0$ and not that $A_i\cap\partial D=\emptyset$. I know of a solution but am interested in what kinds of solutions other people can find, and so I welcome the attempt.

Dan Rust
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  • Don't you want $\lambda(\partial A_i\cap\partial D)=0$ where $\lambda$ is the measure on the boundary of the circle (you should have all the dimensions the same else the lower dimensional piece has zero measure anyway)? – Mark Bennet Sep 01 '13 at 18:56
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    @MarkBennet: I think Daniel is considering the unit circle to be the boundary of the unit disk. – robjohn Sep 01 '13 at 18:59
  • Well $A_i\cap\partial D$ is a subset of the boundary circle. – Dan Rust Sep 01 '13 at 19:00
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    @DanielRust I was reading the measure as the measure on the disk. I agree it is OK as is. – Mark Bennet Sep 01 '13 at 19:01
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    Related on mathoverflow: http://mathoverflow.net/questions/17313/is-it-possible-to-dissect-a-disk-into-congruent-pieces-so-that-a-neighborhood-o/17451#17451 – Daniel R Sep 01 '13 at 19:48
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    I'd imagine the solution involves some multiple of pie. – Joel B Sep 01 '13 at 22:33
  • D-D-D-DUCPLICATE !!!!11 http://math.stackexchange.com/questions/71959/decomposing-a-circle-into-similar-pieces/71977 – Abdulh Khazzak Gustav ElFakiri Sep 02 '13 at 00:18
  • @AbdulhKhazzakGustavElFakiri The question is different and identical to the MO link Daniel R mentioned above. – Dan Rust Sep 02 '13 at 00:20
  • So the whole circle is pizza, and the crust is the outside line, of width zero? – Warren P Sep 02 '13 at 02:01
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    I wonder how the mathematical formulation is "more concrete" than the physical description of the problem! – Ari Brodsky Sep 02 '13 at 03:03
  • In the mathematical formulation, you don't say anywhere that the subsets $A_i$ should be pairwise disjoint (or even almost pairwise disjoint). That would seem to be essential; otherwise the problem might be trivial. – Ari Brodsky Sep 02 '13 at 03:05
  • @AriBrodsky You're entirely correct! I'll add that condition. – Dan Rust Sep 02 '13 at 03:06
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    But is there a way to use a pizza cutter (or other kitchen utensils) to cut a pizza into one of these arrangements? I would enjoy seeing a photo of a real pizza cut this way... – Hammerite Sep 02 '13 at 03:16
  • For the record, only @WarrenP has asked if the pizza is circular, everyone else has assumed the condition. And everyone knows pies are square. – Anthony Sep 02 '13 at 05:05
  • @Anthony: pies are round, cornbread are square... – robjohn Sep 02 '13 at 05:51
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    Every answer thus far involves the pieces touching the crust ( 1 point ). Is there an answer that doesn't require touching the crust? – krikara Sep 03 '13 at 07:41
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    What if the restaurant had only lasagnas? – VividD Apr 17 '15 at 07:13
  • A [recent preprint](http://arxiv.org/abs/1512.03794) seems to suggest there are an uncountable set of solutions to this problem. **neat!!** – Dan Rust Dec 18 '15 at 11:32
  • @robjohn I ask this to you because you are awesome :) What is "int"? And why do we need to have int$A_i \cap$ int$A_j = \emptyset$? – Laplacian Dec 25 '20 at 09:45
  • @robjohn P.s. Merry Christmas!! – Laplacian Dec 25 '20 at 09:52
  • @Turing: $\operatorname{int}(A_i)\cap \operatorname{int}(A_j)=\emptyset$ means that the interiors of the $A_i$ are disjoint. Merry Christmas! – robjohn Dec 25 '20 at 23:14
  • This is the most complex description I have ever seen. – Dmitry Kamenetsky Oct 11 '21 at 22:28

3 Answers3


Here is another with 12 pieces, but all pieces have the same orientation:

$\hspace{32mm}$enter image description here

Using this idea, the pizza can be divided into $6n$ equal pieces with the same orientation for any $n$. However, to have some pieces with no crust, we need $n\gt1$. Above is $n=2$, here is $n=3$:

$\hspace{32mm}$enter image description here

To cut a pizza like this, a blade shaped like, and as long as one sixth of, the circumference of the pizza would be most useful, since all of the cuts are this size and shape.

Here is Mathematica code that will generate these sliced pizzas for any $n$:

Pizza[n_] := 
 Module[{g, arcs = {Thickness[1.3/400], Circle[{0, 0}, 1]}}, 
  For[i = 0, i < 6, For[j = 0, j < n, AppendTo[arcs,
     Rotate[Rotate[Circle[{-1, 0}, 1, {0, Pi/3}],
       j Pi/3/n, {-1/2, Sqrt[3]/2}], i Pi/3, {0, 0}]]; ++j]; ++i]; 
  Show[Graphics[arcs], ImageSize -> 400, 
   PlotRange -> 1.01 {{-1, 1}, {-1, 1}}]]


I thought of the construction of a regular hexagon: you draw a circle with a compass, and then mark arcs on the circle whose chords are the radius of the circle. Due to the properties of equilateral triangles, each arc is exactly $1/6$ of the circumference of the circle, and the chords of those arcs form a regular hexagon. At each vertex of the hexagon, the compass will span to the next vertex (by construction) and to the center of the circle (again, by construction).

Connecting each vertex to the center with arcs centered at the previous vertex, we get the circle tiled by $6$ curvy triangles with congruent sides; two convex sides and one concave side. The centers of the convex sides are the opposite vertices of the curvy triangle. Since the chords of the curved sides have a length $1$ radius, we can trace out the interior convex sides with a congruent arc rotating about the opposite vertex.

$\hspace{32mm}$enter image description here

Since we can sweep out these $6$ triangles with these congruent arcs, we can split up the curvy triangles into any number of congruent pieces with these arcs.

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Here is one solution in $12$ pieces.

enter image description here

Robert Israel
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  • This is the solution I had in mind as well. There are also (a family of) solutions where all pieces have the same orientation. – Dan Rust Sep 01 '13 at 19:02
  • +1: I was working on a drawing of this. Back to the drawing board :-) – robjohn Sep 01 '13 at 19:03
  • @DanielRust One begins by putting a circular arc of equal curvature to the circle from the six obvious points on the boundary to the six interior points at which a straight line meets an arc (delete the straight lines. Then the congruent pieces can be subdivided? – Mark Bennet Sep 01 '13 at 19:56
  • Similar to the drawing here: http://mathoverflow.net/a/17451/12357 – Joel Reyes Noche Sep 02 '13 at 04:20
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    Six of the pieces here only have crust on one corner, but didn't the problem say absolutely none? – Hakanai Sep 02 '13 at 05:29
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    @Trejkaz: The requirement is only that the measure of crust included in some piece $A_i$, i.e. $\lambda(A_i \cap \partial D)$, be 0. A single point of idealized zero-thickness crust is OK since that has measure 0. – The_Sympathizer Sep 02 '13 at 05:43
  • Here's an SVG of this solution with each slice a different color to show how not only how this works but how neat it is to have zorro at your birthday party. http://jsfiddle.net/crazytonyi/8dHYP/1/ – Anthony Sep 02 '13 at 09:50
  • Hmmm, yummy Escher-type pizza... – Matemáticos Chibchas Sep 02 '13 at 21:48
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    Someone has to ask this: What if we really require $A_i\cap \partial D = \emptyset$ for at least one $i$, that is at least one piece, including its boundary, must lie entirely in the interior of the disk? – Jeppe Stig Nielsen Sep 02 '13 at 21:48
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    @JeppeStigNielsen: that appears to be an open problem – Robert Israel Sep 02 '13 at 22:22
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    @mike4ty4 It's fair enough to have ideals, but someone who is thinking more about the pizza than the semantics will probably consider a 0-width crust to be less than ideal. Illustrating the danger of stating mathematical problems in real-world terms which people might relate with. :D – Hakanai Sep 03 '13 at 06:45
  • True. But isn't it easy to accomodate a nonzero-width crust simply by modifying the construction so the straight lines terminate higher up the curved ones, so as to not include any crust? – The_Sympathizer Sep 04 '13 at 02:45
  • @mike4ty4: I don't see how you could do that and keep the pieces all congruent. – Robert Israel Sep 04 '13 at 04:06
  • @Robert Israel: What about this: http://img850.imageshack.us/img850/2949/lj0s.png ? (The dotted circle is the inner border of the thick crust) Looks pretty congruent to me, at least as best as I could draw. – The_Sympathizer Sep 04 '13 at 06:09
  • @mike4ty4: Do you think a region with four corners can be congruent to a region with three corners? – Robert Israel Sep 04 '13 at 07:09
  • Umm... Oh! Now I see... I didn't read it carefully enough -- we need EVERY region congruent to EVERY other. Never mind! – The_Sympathizer Sep 04 '13 at 07:46
  • So I wonder: can you do it with thick crust? :) – The_Sympathizer Sep 04 '13 at 07:48
  • @RobertIsrael I used one of your brilliant images in [this answer](http://math.stackexchange.com/questions/890706/what-is-the-oldest-open-problem-in-geometry/895525#895525) but I'm mindful that you might not approve me using it. If you would like me to remove the image I'm more than happy to respect that. – Dan Rust Aug 12 '14 at 20:39

If this violates the parameters in a clear way, consider this a teaching opportunity. Would this count:

enter image description here

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    How are these pieces "congruent"? – nbubis Sep 03 '13 at 22:37
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    This does violate the conditions, as we need all pieces to be congruent. That is, every piece must be identical (shape, size, etc.) except for orientation. Good try though--it was what popped in my head first before I noticed the "congruent" requirement. – apnorton Sep 03 '13 at 22:38
  • So the center violates it then? Would that mean at least one slice needs to meet the crust? – Anthony Sep 03 '13 at 22:45
  • unfortunately the boundary of each slice has a different curvature (which violates the congruency condition). – Dan Rust Sep 03 '13 at 22:48
  • What if I split it in quarters with two more slices, then it would just be the middle 4 that were off, right? – Anthony Sep 03 '13 at 23:54
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    @Anthony - Congruent means that you can overlay the shapes one on top of the other. rings of different sizes do not have this property. – nbubis Sep 05 '13 at 19:52
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    Anthony, you're confusing congruence with similarity. – dfeuer Sep 09 '13 at 16:20