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

hexagon divided into 24 equilateral triangles

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.

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

enter image description here

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


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


1 Answer 1


You can convince yourself that:

  1. The two points at the acute angles of a parallelogram uniquely determine that parallelogram;
  2. Any two points that don't lie on a common line in the grid can be the two points at the acute angles of some parallelogram.

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.)

  • $\begingroup$ Thank you for a great answer. Exactly what I was looking for! $\endgroup$
    – Bysshed
    Jun 12, 2023 at 0:20
  • $\begingroup$ Unless you count degenerate, 0-area parallelograms, in which case you can remove the restriction that the points not be on the same line, and then it's just 171. $\endgroup$ Jun 12, 2023 at 14:53

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