How many non-congruent triangles can be formed from $n$ equally spaced vertices on a circle?

Question

How many non-congruent triangles can be formed from $n$ equally spaced vertices on a circle? I tried using the stars and bars method but it isn't working for me. Partitioning $n$ into $3$ parts looks promising though.

Answer 1

A very quick answer, which may be incorrect. It is $\,\,[\dfrac{n-1}{2}]$

where [ ] denotes integer part!! It gives correct number from $n=3$ to $n=10$.

Assuming of course that congruent means equal!!

Answer 2

Number of partitions of $n$ into $3$ parts is exactly the answer. The length of a side of a triangle is determined by the number of vertices that lie between the two vertices at its end. But instead of vertices it would be easier to make sense of it in terms of arcs. So $n$ vertices will partition the circle into $n$ arcs of equal length and any two edges joining any two vertices will be same length if and only if the number of arcs joining those vertices are same. Now since the total number of arcs is $n$, the lengths of edges of any triangle are determined completely by the partitions of $n$.

Take simple examples to understand it better

$n = 3$
Only partition $1+1+1$ and only one triangle.

$n = 4$
$1+1+2$
The only possible triangle is one with $2$ edges joining adjacent vertices and last edge joining the remaining open ends(obviously $2$ arcs in between).

$n = 5$
This is not easy to visualise. Earlier I thought there were $4$ of them but then I checked partitions which are only $2$, namely $1+1+3$ and $1+2+2$ and then I recounted and found there are indeed only $2$ of them.

I am not sure how would a rigorous proof of this will go but it is certainly easy to understand with few trys.