Problem Link.
An equilateral triangle with integer side length $n≥3$ is divided into $n^2$ equilateral triangles with side length 1 as shown in the diagram in the above link. The vertices of these triangles constitute a triangular lattice with $\frac{(n+1)(n+2)}{2}$ lattice points.

Let $H(n)$ be the number of all regular hexagons that can be found by connecting 6 of these points.
I can't understand why $H(6)=12$. I can only count $11$ regular hexagons.
What I do is, whenever $n=3m$, it is the first triangle with a regular hexagon of size $m$. And on each successive $n$, the number by which hexagons of size $x$ increase, increments.
This also follows that for $n=6=3.2$, it will have only $1$ regular hexagon of size $2$, and number of hexagons of size $1=10$.
$\therefore H(6) =11 \ne 12$. Where am I wrong?


You get one hexagon of side length $\sqrt3$ by joining the midpoint of the big triangle's base to the upper vertex of the second and fifth triangles on that base. Then go straight up to the midpoints of the sides of the big triangle. Then complete the hexagon. Remember, you're allowed to join any six points of the lattice that produce a regular hexagon. Does that help?

  • $\begingroup$ :| I am also working on it. And I thought a hexagon of even length will give rise to other such hexagon like you described. Still my answer is incorrect. $\endgroup$ – maverick Apr 2 '17 at 15:52
  • $\begingroup$ @maverick--After verifying that H(n) = 966 for n = 20, and working out H(n) for n = 4 to 19, I've derived three general expressions for H(n). They're identical in their first three terms, but differ in the remaining term(s) depending on whether n = 0, 1, or 2 (mod 3). I'm not sure how I'd go about doing the summation for n = 3 to 12345 when each expression is valid only for one in three values of n. Are you still working on it? Is the answer published yet? Good problem. $\endgroup$ – Edward Porcella Apr 6 '17 at 23:38
  • $\begingroup$ No i haven't solved this yet. Can you elaborate on your approach? $\endgroup$ – maverick Apr 7 '17 at 7:52
  • $\begingroup$ @maverick--Besides integer sides > 1, and the diagonal of the rhombus noted above, diagonals of other parallelograms become sides of possible hexagons. E.g. the diagonal of a pl-gram with sides 2,1 is the side of a possible hexagon beginning with n= 9. (Putting a hexagon into each corner of the big triangle, it's easy to count how many fit by noting the size of the triangle formed by their centers. Note that if the diagonal is not of a rhombus, two hexagons fit around each center.) The totals for n = 3-20 are enough to infer the three expressions I mentioned. Does this make sense so far? $\endgroup$ – Edward Porcella Apr 7 '17 at 16:14

Observe that 6 is the smallest side length which can contain hexagon of side length 2, similar to side length 3 which will contain hexagon of side 1.

In this hexagon for side length 2, connect all the midpoints of sides of the hexagon, this will form another regular hexagon.

Considering hexagon of side length $x$, you can choose $x-1$ points from all the sides to form $x-1$ extra regular hexagons in addition to 1 which we already have. This hint is enough for the problem.


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