# Moscow Seven Sisters

Fix $$n$$ points in the plane in generic position, i.e. no three of them on the same line, etc. The number of lines joining two of them is $${n \choose 2}$$. The number of regions in which $$\ell$$ lines cut the plane is $$\leq \ell (\ell+1)/2+1$$. By moving around in the plane, we observe the $$n$$ points in different circular orderings. Since there are $$(n-1)!$$ such orderings, it is clear that for $$n \geq 7$$ we cannot observe the points in all distinct orderings, while for $$n \leq 4$$ this is possible. What happens for $$n=5$$ and 6?

• This is a very important missing piece of information, without which the question is unanswerable. Who are Moscow seven sisters and how are they related to this problem? – Batominovski Oct 19 '18 at 21:14
• @Batominovski For $n=7$, you get seven points, which you can picture as seven buildings in the city of Moscow, Wikipedia is your friend. – jj_p Oct 19 '18 at 21:38
• @jj_p: Requiring a reader to have to perform a web search to understand the context of a question is poor authorship. In this case, it would be most helpful for you to include your above comment as part of the question itself (not everyone reads comments), providing a specific link to whatever Wikipedia entry you believe is relevant. – Blue Oct 20 '18 at 8:56
• @Blue That is just a curiosity, totally irrelevant to understand the math question.. – jj_p Oct 20 '18 at 17:49

Here is just a small correction, the maximum number $$F_n$$ of possible regions created by straight lines passing through a given set of $$n$$ points is $$F_n:=\begin{cases}1&\text{if }n\in\{0,1\}\,,\\3\binom{n}{4}+3\binom{n}{2}-n+1&\text{if }n=2,3,4,\ldots\end{cases}\,.$$ It is easy to show that $$F_n<(n-1)!\text{ for }n=6,7,8,\ldots\,.$$
Thus, it remains to verify whether all cyclic permutations of the $$n$$ points can be seen on the plane when $$n=5$$. After some trials, I believe that the answer for $$n=5$$ is that there does not exist such a point configuration. However, I have no proof.
• Your $F_n$ is wrong, already for $n=4$. The correct one is the one I wrote. So $n=5$ and 6 cannot be discarded that way. – jj_p Oct 20 '18 at 17:46
• Show me a configuration with $4$ points such the lines connecting two of them create more than $18$ regions. You probbly used the formula for $\displaystyle\binom{n}{2}$ lines in general position (where you can indeed create $\displaystyle\frac12\,\binom{n}{2}\,\Biggr(\binom{n}{2}+1\Biggl)+1$ regions), but these lines are not in general position. The $n$ points are, and each of these points will be the intersection of $n-1$ lines. – Batominovski Oct 20 '18 at 17:49