# Regular polygon compass circles

How many unique circles can we draw on vertices of a \(n\)-sided regular polygon? To draw a circle, pick two distinct vertices. One of the vertices is the center of the circle. The distance between the center and the other vertex is the radius. Two circles are distinct if they have different centers, different radii, or both. No degenerate circles, having zero radius, are allowed.

Let \(a(n)\) be the solution. The number of pairs of distinct points we can select from \(k\) points is \(k(k - 1),\) as there are \(k\) ways of picking the first point, and \(k - 1\) ways of picking the second. This gives the upper bound

$$a(n) \le n(n - 1)$$Playing with regular \(n\)-gons for small \(n,\) it's not too difficult to find \(a(n) = 3, 8 ,10, 18.\) After doing so, it seems that the difficulty is determining the number of distinct circles that will be centered at each vertex. It also seems that circles come in pairs. That is, if a line of symmetry is draw through some vertex, then its two adjacent points determine the same circle. Generalizing this idea a bit further: Consider drawing circles centered at point \(P\) to all other points. If we pick some point \(Q\) such that there are \(k\) points between \(P\) and \(Q\) when walking along the polygon clockwise, then this is the same as picking \(Q'\) such that there are \(k\) points between \(P\) and \(Q'\) 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 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

$$\begin{align} a(n) = \begin{cases} \left(\dfrac{n - 1}{2}\right)n & n\text{ is odd} \\[1em] \left(\dfrac{n - 2}{2} + 1\right)n & n\text{ is even} \end{cases} \end{align}$$This can be simplified to

$$\begin{align} a(n) = \begin{cases} \left(\dfrac{n - 1}{2}\right)n & n\text{ is odd} \\[1em] \left(\dfrac{n}{2}\right)n & n\text{ is even} \end{cases} \end{align}$$From which it's not to difficult to see that

$$a(n) = \left\lfloor \dfrac{n}{2} \right\rfloor n$$Here's the OEIS page for the sequence \(a(n).\)