The Eleven-Slice Pizza

Is there a way to slice a pizza into eleven equal-area slices using only four cuts?

During a late-night Slack chat a last week, the subject somehow wandered onto eleven-slice pizzas, and the above question came up. I immediately demonstrated that if we don’t worry about area, it’s easy to get eleven slices with four cuts:


So is it possible to precisely place these four cuts so that the pieces are equal in area?

Since the area of a unit circle is \pi, and there are eleven slices, each slice needs to have an area of one eleventh of \pi.

A_A = A_B = \ldots = A_J = A_K = \cfrac{\pi}{11}

Determining the Central Angles

Let’s consider the area on one side of one cut – in this case, the area made up of slices A, B, E, and I:


Because ABEI has four slices, we know that its area would need to be \frac{4}{11} \pi. It’s also a circular segment cut off by a chord, so we can use the circular segment area formula (all calculations will use radians):

A_{ABEI} = \cfrac{R^2}{2} (\alpha - \sin \alpha)

\frac{4}{11} \pi = \cfrac{1^2}{2} (\alpha - \sin \alpha)

\frac{4}{11} \pi = \cfrac{1}{2} (\alpha - \sin \alpha)

Using Wolfram Alpha to solve for \alpha:

\alpha \approx 2.706389 \text{ rad}

Now, note that segment ACFJ is symmetrical to ABEI, so it has the same central angle \alpha.


That gives us the central angle for two of the chords, which indirectly defines their position relative to the center of the circle. We can find the other two central angles in a similar manner:


Segment EGIJK consists of five slices, so its area is \frac{5}{11} \pi. We’ll call its central angle \beta.


Segment FHIJK also has five slices, so it has the same \frac{5}{11} \pi area and same central angle \beta as segment EGIJK. Once again using the circular segment area formula:

\frac{5}{11} \pi = \cfrac{1}{2} (\beta- \sin \beta)

\beta \approx 2.998549 \text{ rad}

So we now know the central angle (and thus position relative to the center) of all four chords. However, we don’t yet know the chords’ positions relative to each other.

Determining the Chords’ Relative Positions

Let’s now consider the overlap between segments ABEI and ACFJ:


The overlap between these two segments is slice A, which needs to have an area of \frac{1}{11} \pi. We would like to determine the central angle \gamma, which will give us the position of these segments’ chords relative to each other.

The area of slice A is equal to the sum of the area of circular segment RS and the area of \triangle RST.

The area of circular segment RS is once again determined by:

A_{RS} = \frac{1}{2}(\gamma - \sin \gamma)

To get the area of \triangle RST, however, we’ll have to do some trigonometry to find angles \angle SRT, \angle RST, \angle RTS, and the length of side \overline{RS}.

Because \triangle PRS is an isosceles triangle, we know its two base angles are the same. Since the radian angles of a triangle add up to \pi:

\angle PRS = \frac{1}{2} (\pi - \gamma)

Since the chords formed by central angles \alpha also define isosceles triangles:

\angle PRT = \frac{1}{2} (\pi - \alpha)

Solving for all three angles of \triangle RST:

\angle SRT = \angle PRS - \angle PRT

\angle SRT = \frac{1}{2} (\pi - \gamma) - \frac{1}{2} (\pi - \alpha)

\angle SRT = \frac{1}{2} \pi - \frac{1}{2} \gamma - \frac{1}{2} \pi + \frac{1}{2} \alpha

\boldsymbol{\angle SRT} = \frac{1}{2} (\alpha - \gamma)

\boldsymbol{\angle RST} = \angle SRT = \frac{1}{2} (\alpha - \gamma)

\angle RTS = \pi - 2 \angle SRT

\angle RTS = \pi - (2)(\frac{1}{2})(\alpha - \gamma)

\angle RTS = \pi - (\alpha - \gamma)

\boldsymbol{\angle RTS} = \pi + \gamma - \alpha

Solving for the length of \overline{RS} using the Pythagorean Theorem, and knowing that the hypotenuse is equal to the radius of 1:

\sin(\frac{\gamma}{2}) = \cfrac{\frac{\overline{RS}}{2}}{1}

\overline{RS} = 2 \sin(\frac{\gamma}{2})

From the three angles and one side, we can then solve for the area of \triangle RST:

A_{\triangle RST} = \cfrac{\overline{RS}^2 \sin(\angle SRT) \sin(\angle RST)}{2 \sin(\angle RTS)}

A_{\triangle RST} = \cfrac{(2 \sin(\frac{\gamma}{2}))^2 \sin(\frac{1}{2} (\alpha - \gamma)) \sin(\frac{1}{2} (\alpha - \gamma))}{2 \sin(\pi + \gamma - \alpha)}

A_{\triangle RST} = \cfrac{(2 \sin(\frac{\gamma}{2}))^2 \sin^2(\frac{1}{2} (\alpha - \gamma))}{2 \sin(\pi + \gamma - \alpha)}

Now, since we know that the area of the circular segment plus the area of the triangle need to add up to \frac{1}{11} \pi, we can solve for \gamma:

A_A = A_{RS} + A_{\triangle RST}

\frac{1}{11} \pi = \frac{1}{2}(\gamma - \sin \gamma) + \cfrac{(2 \sin(\frac{\gamma}{2}))^2 \sin^2(\frac{1}{2} (\alpha - \gamma))}{2 \sin(\pi + \gamma - \alpha)}

\frac{1}{11} \pi = \frac{1}{2}(\gamma - \sin \gamma) + \cfrac{(2 \sin(\frac{\gamma}{2}))^2 \sin^2(\frac{1}{2} (2.706389- \gamma))}{2 \sin(\pi + \gamma - 2.706389)}

\gamma \approx 0.870954 \text{ rad}

Now we know the central angle of each of the first two chords (\alpha), the central angle of the overlap between them (\gamma), and the central angle of each of the second two chords (\beta). We just need to find the central angle of the second overlap (which we’ll call \delta) to completely define the position of all four chords.


 Looking at this layout, it’s the same as what we just did, with three exceptions:

  • The overlap area (IJK) is three slices instead of one, so the area of the overlap is \frac{3}{11} \pi instead of \frac{1}{11} \pi
  • The central angle for each of the two chords is \beta \approx 2.998549 \text{ rad} instead of \alpha \approx 2.706389 \text{ rad}
  • We’re solving for \delta instead of \gamma

So given that, we can simply reuse the final A_A formula above, making the appropriate substitutions:

A_{IJK} = A_{RS} + A_{\triangle RST}

\frac{3}{11} \pi = \frac{1}{2}(\delta- \sin \delta) + \cfrac{(2 \sin(\frac{\delta}{2}))^2 \sin^2(\frac{1}{2} (\beta - \delta))}{2 \sin(\pi + \delta - \beta)}

\frac{3}{11} \pi = \frac{1}{2}(\delta- \sin \delta) + \cfrac{(2 \sin(\frac{\delta}{2}))^2 \sin^2(\frac{1}{2} (2.998549- \delta))}{2 \sin(\pi + \delta- 2.998549)}

\delta \approx 1.849541 \text{ rad}

This means we now know the central angles of all four chords and the central angle of the overlap between each pair of same-length chords:

\alpha \approx 2.706389 \text{ rad}
\beta \approx 2.998549 \text{ rad}
\gamma \approx 0.870954 \text{ rad}
\delta \approx 1.849541 \text{ rad}

Since the layout needs to be symmetrical for the areas to be equal, we know that the intersection of each pair of same-length chords must rest on the same diameter of the circle. From all of that, we know the exact position all four chords must be in to make all of the following true:

A_{A} = \frac{1}{11} \pi
A_{IJK} = \frac{3}{11} \pi
A_{ABEI} = A_{ACFJ} = \frac{4}{11} \pi
A_{EGIJK} = A_{FHIJK} = \frac{5}{11} \pi

So if we create the cuts with the above \alpha\beta\gamma, and \delta angles, and the remaining slices each have an area of \frac{1}{11} \pi, then we’ve solved the problem. On the other hand, if we create the cuts and any of the remaining slices don’t have an area of \frac{1}{11} \pi, then we know that it’s impossible to get eleven equal slices with four cuts, since any change of position of the chords would make one or more of the areas we listed above incorrect.

Drawing the Calculated Cuts

Let’s then draw the cuts using the calculated actual positions. Starting with areas ABEI and ACFJ:


So far, it’s looking good; BEI and CFJ both appear to be three times larger than A, which is what we would want. Now let’s draw the final two cuts:



So even though the sets of slices we calculated are an appropriate portion of the pizza, the remaining slices clearly are not even. As stated above, there’s no way we can move the cuts without making the sections we calculated the wrong size. Therefore, we’ve demonstrated that while we can cut a pizza into eleven slices using only four cuts, it’s not possible to cut the pizza so that those eleven slices have equal area.


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