This is a painfully easy problem, yet the answer continues to escape me. I am seeking a general formula that can be employed to determine the number of diagonals of a regular polygon that are parallel to at least one of its sides. A quadrilateral has no diagonals that meet this criterion, a pentagon has 5, a hexagon has 3, an octagon has 8, etc. I know that the general forumla for the number of diagonals $N_d$ is
$N_d=\frac{n(n-3)}{2}$,
but is there a similar formulation for parallel diagonals? Any help here would be much appreciated. Also, finding the number of diagonals that are not parallel to any of the polygon's sides would be helpful as well…
Best Answer
If $n$ is odd, all of the diagonals are parallel to one side.
If $n$ is even, then it is just a little harder. Label the vertices $0,1,2,\dots,n-1$. A diagonal is a pair of these points which are more than $1$ apart, and it is parallel to an edge if their difference is odd.
It's easier to pick diagonals which are not parallel - because you can pick any $2$ nodes that are even labeled, or any two nodes that are odd-labeled.
This means there are $2\binom{n/2}2 = \frac{n(n-2)}{4}$ diagonals which are not parallel, so $\frac{n(n-3)}{2}-\frac{n(n-2)}{4} = \frac{n(n-4)}{4}$ are parallel, when $n$ is even.