# Number of parallelograms in a hexagon of equilateral triangles

The figure shows a regular hexagon divided into 24 congruent equilateral triangles. How many parallelograms are there in the figure?

How would you solve this problem? Is there a generalisation for larger hexagons?

My attempt:

I counted 87 parallelograms. My method was to count the number of parallelograms in one orientation and then multiply this answer by three. The image below is meant to illustrate the three orientations (red, green and blue). The number on each vertex is the number of blue parallelograms such that the vertex is at the top left.

$$87 = 3 (1+2+3+4+2+3+4+3+4+3)$$

This method seems difficult to generalise and I was hoping to see some other solutions. (I am also not certain the answer is correct)

Reference:

This question is an investigation that appears on page 3 of the solutions to the 2022 UKMT senior challenge solutions

So we can start with $$\binom{19}{2} = 19 \cdot 9 = 171$$ pairs of points, and subtract the $$3 \left[ \binom 32 + \binom 42 + \binom 52 + \binom 42 + \binom 32\right] = 3 \cdot 28 = 84$$ pairs of points on a common line. We are left with $$87$$, the number of parallelograms.
The general formula for a hexagon with a side of length $$n$$ can be found in the same way; after a bunch of summations, we get $$\frac{9 n^4 + 4 n^3 - n}{2}$$, which is OEIS sequence A047786. The comments in the OEIS sequence don't really explain what it's for, but the cited paper Arrays and brooks gives it on p. 30 as the formula for $$b_{n2}$$, defined exactly as the number of ways to place two points on the hexagon that are not on a common grid line. (The paper studies $$b_{nk}$$, the number of ways to place $$k$$ points, for which it does not find a closed form.)