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:

enter image description here

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

  • $\begingroup$ $n=1 \implies R=2$ and $n=2 \implies R=7$. If we have enough terms, maybe we can observe a recurrence relationship? $\endgroup$ – Mohammad Zuhair Khan May 14 '19 at 15:52
  • $\begingroup$ @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! $\endgroup$ – Benjamin Dickman May 14 '19 at 15:57
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    $\begingroup$ 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. $\endgroup$ – Vasya May 14 '19 at 15:57
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    $\begingroup$ 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 $n=2$. $\endgroup$ – Robert Israel May 14 '19 at 15:58
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    $\begingroup$ @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? $\endgroup$ – 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 $4$.

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    $\begingroup$ 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. $\endgroup$ – N.Bach May 18 '19 at 8:06
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    $\begingroup$ @N.Bach Yes, you are right. Thanks. $\endgroup$ – Alex Ravsky May 18 '19 at 9:29
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    $\begingroup$ Euler's Formula is a great idea in retrospect; this is nice, thank you! $\endgroup$ – Benjamin Dickman May 18 '19 at 16:00

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