Maximum Regions Vees Can Divide a Circle

The Circle Division by Lines problem (link) asks into how many regions, at most, one can divide a circle (or: the plane) with $$n$$ chords (or: lines).

I am wondering about a similar question, but for which the dividing curves are not chords/lines, but rather $$\mathsf{V}$$-shaped curves, which can be oriented in any direction. Stated succinctly:

What is the maximum number of pieces in which it is possible to divide a circle for a given number of $$\mathsf{V}$$-shaped cuts, where the cuts can be oriented in any direction?

RE: Initial Comments I am fine with the assumption that the vertex must be inside the circle and the angle should be fixed across all $$\mathsf{V}$$-cuts. But, I am open to suggestions as others see fit!

RE: Solution In fact, the formula provided works irrespective of the vertex angle. In retrospect, Euler's Formula is a wise way to solve the problem and, with a quick check, one finds that plugging in $$n=1$$ and $$n=2$$, respectively, yields $$2(1)^2 - 1 + 1 = \fbox{2}$$ and $$2(2)^2 - 2 + 1 = \fbox{7}$$, as desired.

For example, when there are a total of $$2$$ $$\mathsf{V}$$-cuts, I believe the maximum is $$7$$ regions:

What is the maximum number of regions for $$n \in \mathbb{Z}^+$$ total $$\mathsf{V}$$-cuts?

• $n=1 \implies R=2$ and $n=2 \implies R=7$. If we have enough terms, maybe we can observe a recurrence relationship? – Mohammad Zuhair Khan May 14 '19 at 15:52
• @MohammadZuhairKhan Approaching this with small cases and looking for a pattern seems like a potentially wise heuristic to me. That is certainly how I broach the circle-chord problem when teaching it. But, I am not sure how to assert with confidence in the case of e.g. $n=4$ that the maximum is achieved. (Maybe there is a way to come up with a meaningful upper bound?) In other words: I'm not quite sure how to generate terms! – Benjamin Dickman May 14 '19 at 15:57
• You can produce more regions if the cuts are allowed to touch the edge or have the tip of the $V$ outside of the circle. – Vasya May 14 '19 at 15:57
• Does the vertex of the V have to be inside the circle? If not, you can have $3$ regions for $n=1$ and $9$ for $n=2$. – Robert Israel May 14 '19 at 15:58
• @RobertIsrael The version with the vertex inside the circle and the angle fixed might be the easiest to broach, first, but I am open to modifications. Do you have a suggestion either way? – Benjamin Dickman May 14 '19 at 16:02

We consider a simple case when vertices of $$\vee$$-cuts must be inside the circle, but their angles can be the same or differ (it turns out, that this doesn’t change the answer). A configuration of $$\vee$$-cuts can be naturally considered as a planar connected (multi)graph. Let $$0\le V_{ij}\le 4$$ be the number of intersection points between $$i$$-th and $$j$$-th $$\vee$$’s. Put $$S=\frac 12\sum_{i\ne j} V_{ij}\le 2n(n-1).$$ It is easy to see that the graph has $$V=2n+\tfrac 12\sum_{i\ne j} V_{ij}=2n+S$$ vertices and $$E=2n+\sum_{i}\left(1+\sum_{j\ne i}V_{ij}\right)=3n+2S$$ edges. By Euler’s formula, the graph $$G$$ has
$$F=1+E-V=1+3n+2S-2n-S=1+n+S\le 2n^2-n+1$$
inner faces and the equality is attained iff each $$V_{ij}$$ equals $$4$$.
• I think you meant $E= 2n+\sum_i\left(1+\sum_{j\neq i}V_{ij}\right)$? Also, your wording kinda hints at it, but configurations in which $V_{ij}$ is $4$ for all pairs $i,j$ can be built, so that bound is tight. – N.Bach May 18 '19 at 8:06