Using only a compass, draw all possible circles on the vertices of a regular $n$-sided polygon.
(That is, in every ordered pair of vertices one is the center, and their distance is the radius.)
How many intersections are there?
Let $a(n)$ be the intersection count for given $n\in\mathbb N$.
First three terms $a(1),a(2),a(3)=0,2,6$ are simple. The next three terms are:
Notice that the circle set (given by a $n$-sided polygon) can be split into $n$ symmetric regions.
Let $A_n$ count intersections inside one of the $n$ regions. Let $\delta_n\in\{0,1\}$ compensate for when there is one extra central intersection. This implies we can write every term as:
$$a(n)=nA_n+\delta_n$$
The first $20$ terms should be (constructed in GeoGebra):
$$\begin{array}{} a(1) &= \space\space0 &= \space\space1\cdot0 \\ a(2) &= \space\space2 &= \space\space2\cdot1 \\ a(3) &= \space\space6 &= \space\space3\cdot2 \\ a(4) &= 40 &= \space\space4\cdot10 \\ a(5) &= 55 &= \space\space5\cdot11 \\ a(6) &= 145&= \space\space6\cdot24 + 1 \\ a(7) &= 238&= \space\space7\cdot34 \\ a(8) &= 584&= \space\space8\cdot73 \\ a(9) &= 612&= \space\space9\cdot68 \\ a(10) &= 1350&= 10\cdot135 \\ a(11) &= 1804&= 11\cdot164 \\ a(12) &= 2401&= 12\cdot200+1 \\ a(13) &= 3523&= 13\cdot271 \\ a(14) &= 5180&= 14\cdot370 \\ a(15) &= 6150&= 15\cdot410 \\ a(16) &= 9312&= 16\cdot582 \\ a(17) &= 11101&= 17\cdot653 \\ a(18) &= 13645&= 18\cdot758+1 \\ a(19) &= 17746&= 19\cdot934 \\ a(20) &= 22300&= 20\cdot1115 \\ \dots \end{array}$$
We can notice that it seems $\delta_n=1$ if and only if $n$ is a multiple of $6$.
Are there any other patterns? Is it possible to find a closed form for $a(n)$?
My attempt:
WLOG, Let $V_1,V_2,\dots,V_n$ be vertices of a regular $n$-sided polygon with circumradius $1$.
We can take $V_i=(x_i,y_i)=(\cos(\frac{2i\pi}{n}),\sin(\frac{2i\pi}{n})),i=1,2,\dots,n$.
The $k$-th diagonal from some vertex $V$ of the polygon will have length $2\sin(\frac{k\pi}{n})$.
At each vertex $V$, we will have $c=1,2,\dots,\left\lfloor\frac{n}{2}\right\rfloor$ circles$^{[1]}$ with radii $r_c=2\sin(\frac{c\pi}{n})$.
$$ (i,c)\text{-Circle}\dots\space\space \left(x-\cos\frac{2i\pi}{n}\right)^2+\left(y-\sin\frac{2i\pi}{n}\right)^2=\left(2\sin\frac{c\pi}{n}\right)^2 $$
Is it possible to derive a closed form for the number of intersections from this?
I believe I managed to solve a simpler problem:
"At each vertex $V$, consider only one circle of radius $r$."
Then the number of such intersections $I(n,r)$ should be:
$$ I(n,r)=\begin{cases} (n-1)n, & r \gt 1\\ (n-1)n - n\left\lfloor\frac{n}{2}\right\rfloor+1, & r=1\\ (n-2)n, & \sin(\frac{\left(\frac{n}{2}-1\right)\pi}{n})\lt r\lt\sin(\frac{\pi}{2})=1\\ (n-3)n, & r = \sin(\frac{\left(\frac{n}{2}-1\right)\pi}{n})\\ \dots & \dots \\ (2k)n, & \sin(\frac{k\pi}{n}) \lt r \lt \sin(\frac{(k+1)\pi}{n})\\ (2k-1)n, & r = \sin(\frac{k\pi}{n})\\ \dots & \dots \\ 4n, & \sin(\frac{2\pi}{n}) \lt r \lt \sin(\frac{3\pi}{n})\\ 3n, & r = \sin(\frac{2\pi}{n})\\ 2n, & \sin(\frac{\pi}{n}) \lt r \lt \sin(\frac{2\pi}{n})\\ 1n, & r = \sin(\frac{\pi}{n})\\ 0, & r \lt \sin(\frac{\pi}{n}) \end{cases} $$
This solves the problem of intersections for any $r\in\mathbb R_{+}$ but for only one layer of circles.
In the original problem, we have $\left\lfloor\frac{n}{2}\right\rfloor$ layers of circles with different radii on each layer. The radii of circles between layers have specific ratios (determined by $n$): radii are diagonals of the regular $n$-sided polygon.
My idea was to use $I(n,r_c),c=1,2,\dots,\left\lfloor\frac{n}{2}\right\rfloor$ to get to $a(n)$. But, I get lost when trying to add and subtract all of the unique and duplicate intersections.
How can we solve the original problem and find $a(n)$?
We can find an upper bound by considering the case $n \to \infty :$ There are $\left \lfloor \frac{n}2 \right \rfloor \leq \frac{n}2$ radii. Let the distance between the points be $1$, every circle with radius $k$ has two unique intersections with $1+2(k-1)$ circles with the same radius so in total
$$2n \sum_{k=1}^{n/2} 1+2(k-1) = \frac{n^3}2$$
Every circle with radius $k$ has four unique intersections with $1+2(k-1)$ circles for every bigger radius:
$$4n \sum_{k=1}^{n/2} \bigl( 1+2(k-1) \bigr) \left( \frac{n}2 - k \right) = \frac{n^2}6 (n-1)(n-2)$$
In order to not overcount the starting points we consider them seperately. The upper bound is shown below as well as the fit $a(n) \approx 0.089 \, n^{4.14}$
$$a(n) \leq n + \frac{n^3}2 + \frac{n^2}6 (n-1)(n-2) = n + \frac{n^2}3 + \frac{n^4}6$$
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