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.
$H(3)=1,H(6)=12...H(20)=966$
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?