# Intersections of circles drawn on vertices of regular polygons

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 $$a(n)$$?

• 4 sets of 3 rings in the 4, 5 sets of 4 rings in the 5, etc.
– user645636
Jan 24 '20 at 15:27
• each pair of annuli have 8 intersection points encompassing the two intersections.
– user645636
Jan 24 '20 at 15:53
• Quick note: Someone made an OEIS sequence for this- oeis.org/A331702. It's attributed to this exact MSE question. Aug 31 '21 at 22:56

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 $$a(n) \approx 0.089 \, n^{4.14}$$

$$a(n) \leq n + \frac{n^3}2 + \frac{n^2}6 (n-1)(n-2) = n + \frac{n^2}3 + \frac{n^4}6$$

$$\hspace{1cm}$$