Unit circle is divided into $n$ equal pieces, what is the least value of the perimeters of the $n$ parts?

Solution 1:

The isoperimetric inequality gives the lower bound: $$L_n\geq 2\pi\sqrt{n},$$ and I believe we can achieve: $$L_n\leq C\cdot\sqrt{n}$$ through the following strategy: choose some integers $m_1,\ldots,m_k$ such that: $$0\leq m_1\leq m_2\leq\ldots\leq m_k,\qquad m_1+\ldots+m_k = n$$ and consider concentric regular polygons $\Gamma_1,\ldots,\Gamma_k$, with center in the center of the circle, such that $\Gamma_i$ has $m_i$ vertices and area $\frac{m_1+\ldots+m_i}{n}\pi$, then split the "polygonal annulus" between $\Gamma_i$ and $\Gamma_{i+1}$, where $\Gamma_{n+1}$ is the original circle, into $m_i$ equal parts with a set of $m_i$ segments having minimum perimeter. If the sequence $m_1,\ldots,m_k$ is properly chosen (close to be an arithmetic sequence, in order that all the pieces are "almost equal" one each other), we stay quite close to the lower bound for $L_n$. I still do not have a rigorous proof ot his, but the conjecture $$L_n = \Theta(\sqrt{n})$$ is very likely to hold in my opinion. Just to make my argument visually clear, here there is a likely almost-minimal split with $n=10$, with $m_1=0,m_2=3,m_3=7$:

In order to get the "reversed isoperimetric inequality" we just need to show that we can take all the pieces "close enough to be a circle". It is quite trivial that we can have all the pieces with at most six sides (including the curvilinear one for the pieces on the boundary of the circle) and convex, we just need to prove that for any piece, there exists a circumscribed rectangle whose side ratio (as the ratio between the diameter and the "orthogonal width") is less a certain constant $K$, in order to prove that $$L_n\leq 2\pi K'\sqrt{n}.$$ The problem now boils down to two components: the arithmetic one, that is the choice of the best sequence $\{m_i\}$, and the geometric one, i.e. the hunt for the $K$ and $K'$ constants.

Lemma $1$: The diameter of any piece belonging to the annulus $A_i$ between $\Gamma_i$ and $\Gamma_{i+1}$ is bounded by $\frac{1}{m_i}$ times the length of the circumcircle of $\Gamma_{i+1}$, hence by $$\frac{1}{m_i}\sqrt{\frac{m_1+\ldots+m_{i+1}}{n}\cdot\frac{2\pi}{n\sin\frac{2\pi}{m_i}}}.$$ Moreover, for any piece belonging to $A_i$, the "orthogonal width" is bounded below by the difference of the circumradii of $\Gamma_i$ and $\Gamma_{i+1}$, hence $K$ depends only on the sequence $\{m_i\}.$

Lemma $2$: For any convex shape with area $A$ and "Kakutani ratio" $\leq K$, the perimeter is bounded by: $$ 2\left(\sqrt{K}+\frac{1}{\sqrt{K}}\right)\sqrt{A}.$$

Assuming that for any $n$ we can achieve the realistic constant $K=3$ through Lemma 1, we have:

$$ L_n \leq 3\pi\sqrt{n}.$$

This bound is worse than the trivial $2\pi+2n$-bound for any $n\leq 15$, but it is asymptotically optimal. The interesting thing is that $K$ may be big only for pieces that lie near to the center of the circle, so, essentially, only for $m_1$ pieces. This leads to a stronger conjecture:

$$ L_n \leq 2\pi\sqrt{n} + K''. $$

However, this very last line may incredibly difficult to prove, if not even wrong. If my algorithm leads to the optimal solution, the case $n=9$ (in which we have a trisected hexagon in the center, with $K\approx\sqrt{3}$) gives that $K''$ must be at least $2.7$.

Solution 2:

DISCLAIMER: This is not an answer! It just provides visual and numerical aspects of the problem in an illustrative way.

I made dynamic diagrams producing configurations similar to the one the OP gave in the question. My method was simply to assume that each annulus would have a width of approximately $\sqrt{\pi/n}$ which is the side length of $n$ identical squares having the same area as the circle. Then I adjusted the width by simply rounding to match a whole number of segments of area $\pi/n$ into each annulus. The result can be seen here:

It would be much harder to program a more fine tuned approach. I chose thise circular and in that quite regular construction due to its simplicity. Yet it seems pretty good already!

If anyone care to play around with it more, they can visit my interactive file that goes up to $n=700$ in this manner. Note how the actual perimeter of these configurations seem to follow a curve close to the lower bound of $2\pi\sqrt n$ given by Jack D'Aurizio in his answer. Here is the interactive file to play with:

http://www.geogebratube.org/student/m138801

Solution 3:

For $n>2$, let $a=\sqrt{\frac{2\pi}{3n\sqrt 3}} $, so that the area $ a^2\frac32\sqrt 3$ of a regular with side length $a$ is $\frac \pi n$. Overlay the unit circle with a corresponding hexagonal lattice. We may assume wlog. that no lattice point is on the circle. Then there are $O(\sqrt n)$ hexagons intersecting the circle. More precisely, these intersecting hexagons are within the annulus with inner radius $1-2a$ and outer radius $1+2a$, hence their combined area is $<\pi(1+2a)^2-\pi(1-2a)^2=8\pi a$ and their count $m$ is $$m<8na=\sqrt{\frac{2\pi}{3\sqrt 3}}\cdot \sqrt n.$$ At most four of the edges of such a hexagon are fully inside the unit circle, hence the total length $p$ of these edges is $$p\le 4am<32na^2 =\frac{64\pi}{3\sqrt 3}.$$ (Admittedly, the constant factor in tis estimate is very crude). The $n'\ge n-m$ hexagons that are fully inside the unit circle contribute $6an'$ to the perimeter sum. We partition the remaining area into $n-n'\le m$ parts with $(n-n')$ radial lines of lenth $<2a$, hence these $n-n'$ parts contribute at most $2\pi+p+2(n-n')\cdot 2a$ to the perimeter sum. We conclude that $$\begin{align}L_n &\le 6an'+2\pi +p+4a(n-n')\\ &\le 6an -2a(n-n') +p+2\pi\\ &\le 6an +p+2\pi\\ &= 2\sqrt{{2\pi\sqrt 3}}\cdot \sqrt n+\frac{64\pi}{3\sqrt 3}+2\pi.\end{align}$$ Numerically, $$\begin{align}2\sqrt{2\pi\sqrt 3}&\approx 6.597817 \end{align}$$ and remarkably, this is just about $1.05$ times $2\pi$.