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.

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    $\begingroup$ @MarkBennet: I think Daniel is considering the unit circle to be the boundary of the unit disk. $\endgroup$
    – robjohn
    Commented Sep 1, 2013 at 18:59
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    $\begingroup$ Related on mathoverflow: mathoverflow.net/questions/17313/… $\endgroup$
    – Daniel R
    Commented Sep 1, 2013 at 19:48
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    $\begingroup$ I'd imagine the solution involves some multiple of pie. $\endgroup$
    – Joel B
    Commented Sep 1, 2013 at 22:33
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    $\begingroup$ I wonder how the mathematical formulation is "more concrete" than the physical description of the problem! $\endgroup$ Commented Sep 2, 2013 at 3:03
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    $\begingroup$ Every answer thus far involves the pieces touching the crust ( 1 point ). Is there an answer that doesn't require touching the crust? $\endgroup$
    – krikara
    Commented Sep 3, 2013 at 7:41

3 Answers 3


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.

  • $\begingroup$ Is it hard to find a general form for all possible solutions? $\endgroup$ Commented Sep 2, 2013 at 3:18
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    $\begingroup$ @PratyushSarkar: it's likely to be harder to prove that you have all possible solutions than to find what turn out to be all possible solutions. Are there solutions other than mine and robjohn's (and their reflections and rotations?). $\endgroup$ Commented Sep 2, 2013 at 3:38
  • $\begingroup$ Curious, what made you think of this solution? $\endgroup$
    – hadsed
    Commented Sep 5, 2013 at 16:06
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    $\begingroup$ @hadsed: I have added a section on the motivation. $\endgroup$
    – robjohn
    Commented Sep 5, 2013 at 19:49
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    $\begingroup$ It moves! It moves! $\endgroup$
    – Pedro
    Commented Sep 5, 2013 at 21:55

Here is one solution in $12$ pieces.

enter image description here

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    $\begingroup$ Six of the pieces here only have crust on one corner, but didn't the problem say absolutely none? $\endgroup$
    – Hakanai
    Commented Sep 2, 2013 at 5:29
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    $\begingroup$ @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. $\endgroup$ Commented Sep 2, 2013 at 5:43
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    $\begingroup$ 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? $\endgroup$ Commented Sep 2, 2013 at 21:48
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    $\begingroup$ @JeppeStigNielsen: that appears to be an open problem $\endgroup$ Commented Sep 2, 2013 at 22:22
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    $\begingroup$ @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 $\endgroup$
    – Hakanai
    Commented Sep 3, 2013 at 6:45

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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    $\begingroup$ How are these pieces "congruent"? $\endgroup$ Commented Sep 3, 2013 at 22:37
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    $\begingroup$ 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. $\endgroup$
    – apnorton
    Commented Sep 3, 2013 at 22:38
  • $\begingroup$ So the center violates it then? Would that mean at least one slice needs to meet the crust? $\endgroup$
    – Anthony
    Commented Sep 3, 2013 at 22:45
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    $\begingroup$ @Anthony - Congruent means that you can overlay the shapes one on top of the other. rings of different sizes do not have this property. $\endgroup$ Commented Sep 5, 2013 at 19:52
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    $\begingroup$ Anthony, you're confusing congruence with similarity. $\endgroup$
    – dfeuer
    Commented Sep 9, 2013 at 16:20

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