# Minimal pizza cutting

Given a circle, we want to divide it into $$n$$ connected equally sized pieces. In such a way that the total length of the cutting is minimal. What can we say about the solution for each $$n$$. Are they unique (up to some symmetry). Do all cuttings arise from the intersection of three straight segments with angle 120º?

These are conjectured to be the best solutions for $$n \in \{2, 3, 4, 7\}$$

Case $$n= 2$$ goes as follows: For simplicity the radius is $$1$$. If there is only one or no points on the circumference, then we know the minimal curve is the circumference, which has length $$\frac{2\pi}{\sqrt{2}}$$ bigger than 2 and we are done. Hence, we have at least two points in the circumference. If they are in the same diameter, the shortest curve is the line, so the cutting has at least length 2.

Now, pick a diameter parallel to $$\overline{AC}$$, which leaves both points on the same side. Since the curve connecting both needs to have area $$\pi/2$$ in needs to contain at least one point on the other side of the diameter.

Now, $$\overline{AB}$$ is bounded below by a straight line, same with $$\overline{BC}$$. And the sum of those two is smaller than $$\overline{AO} + \overline{OC}$$ where $$O$$ is the center of the circle.

For $$n\in\{3, 4, 7\}$$ I don't have a proof and they might not be minimal.

As for the conditions on the cuttings, probably something like picewise $$C^2$$ is enough.

• Why skipping $n=5,6$? Commented Aug 6, 2020 at 19:52
• @WhatsUp I have no guess of what they might look like. Case 7 is specially symmetric Commented Aug 6, 2020 at 20:20
• Your solution for $n=7$ generalizes to having a regular $(n-1)$-gon in the center instead of a hexagon. Might these be solutions for $n=5,6$? And have you already checken whether having a triangle in the center is worse than your proposed solution for $n=4$? Except for the case $n=7$, these solutions do not have $120^\circ$ angles. Commented Aug 6, 2020 at 20:40
• Cox and Flikkema have some candidate solutions from small n here: doi.org/10.37236/317, and the reference therein will lead you to some literature on the existence and regularity of solutions to this problem (not sure about uniqueness though). In general the arcs of constant curvature and meet in threes at 120º, but not necessarily straight. Commented Aug 6, 2020 at 21:29
• Another condition (that I believe rules out your n=4 candidate) is that the arcs meet the boundary perpendicularly. Commented Aug 6, 2020 at 21:36

Headley, Francis, and Simon Cox. "Least-perimeter partition of the disc into $$N$$ regions of two different areas." arXiv:1901.00319 (2019). arXiv abstract.
Their first figure illustrates Yoav's points that the arcs "meet in threes at $$120^\circ$$, but not necessarily straight," and that "the arcs meet the boundary perpendicularly":