Consider a regular convex polygon with $n$ sides. I know that the number of lines connecting any two points of the $n$-gon is given by $n \choose 2$, and hence the number of diagonals in the $n$-gon is given by ${n\choose{2}} – n$.
However I wish to find the number of diagonals such that there are $k$ vertices between the two vertices that are connected by the diagonal.
I denote the number of such diagonals by $D_n(k)$.
The maximum number of vertices in between any two diagonals is just $n-3$, but I have noticed if we put $k$ in between $\lfloor{\frac{n}{2}}\rfloor – 1$ and $n-2$, we start to recount the same diagonals that are already counted for some lower $k$. For example when we count for a pentagon, $k=1$ and $k=2$ just give the same diagonals.
To be more precise, I think that $D_n(k) = D_n(n-2-k)$. So I think it makes sense to restrict $k$ such that $0< k < \lfloor{\frac{n}{2}}\rfloor$.
For example:
$$
D_4(1) = 2 \\
D_5(1) = 5 \\
D_6(1) = 6 \\
D_6(2) = 3 \\
D_7(1) = 7 \\
D_7(2) = 7
$$ and so on. What I have noticed is that for $n>4$ and $k=1$, $D_n(k) = n$, but I am not sure if that is always the case. Is there some closed form for $D_n(k)$ or just some method to work it out?
Thanks to user @OscarLanzi, I have worked out the for odd $n$, the number of diagonals for all possible $k$ is $n$, whereas for even $n$, it is $n$ for $k \in [1, \frac{n}{2} – 2]$ and $\frac{n}{2}$ for $k=\frac{n}{2} -1$.
Best Answer
Make a table with eught columns and five rows. The columns represent $n$ values from $3$ to $10$, the rows represent $k$ values from $0$ to $4$. Fill in the table with the appropriate diagonal counts, leaving areas blank where $k\ge n/2$.
What relationship do you notice between the diagonal counts $n$ and $k$ in most of the table? Can you see why certain numbers along the "edge" of the entries are less than that?
Draw out the polygons for some combinations of $n$ and $k$ to see what is happening geometrically.