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?
Best Answer
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 $4$.