How many unique circles can we draw on vertices of a $n$-sided regular polygon? To draw a circle, pick two distinct vertices. One is the center of the circle, and the other determines the radius.
Let $a(n)$ be the solution.
The maximal number of such circles we can draw on distinct pairs of $k$ arbitrary points gives an upper bound and should be equal to $2!\binom{k}{2}=k(k-1)$. That is, we know:
$$ a(n)\le n(n-1) $$
We can start counting $a(n)=0,2,3,8,10,18,21,32,36,50,\dots$ for $n=1,2,3,\dots$
But I'm not sure how to find a closed form for the sequence.
I searched OEIS and it appears it corresponds to a sequence with following closed form:
$$a(n)=n\left\lfloor \frac{n}{2}\right\rfloor$$
Can we prove this?
Best Answer
Consider drawing circles centered at point $P$ to all other points. If we pick some point $Q$ such that there are $k$ points between $P,Q$ when walking along the polygon clockwise, then this is the same as picking $Q'$ such that there are $k$ points between $P,Q'$ when walking along the polygon anticlockwise.
If $n$ is odd, this means that exactly half circles drawn from some point $P$ will be duplicates.
If $n$ is even, we have a point $Q^{*}$ which which is exactly opposite our point $P$ and does not have a corresponding "duplicate pair" point. In other words, if $n$ is even we have one less pair of duplicates but one more circle, per point.
Hence, we have:
$$\begin{align} a(n)&=\begin{cases} \left(\frac12(n-1)\right)n,&n\text{ is odd}\\ \left(\frac12(n-2)+1\right)n,&n\text{ is even} \end{cases} \end{align}$$
Which can be simplified to:
$$ a(n)=n\left\lfloor \frac{n}{2} \right\rfloor $$
Confirming the observations.