Counting Hexagons

An equilateral triangle with integer side length $n \ge 3$ is divided into $n^2$ equilateral triangles with side length 1 as shown in the diagram below.
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.

For example, $H(3)=1$, $H(6)=12$ and $H(20)=966$.
Find $\displaystyle \sum_{n=3}^{12345} H(n)$.

This is a highly complex combinatorics problem, often seen in math competitions or advanced coursework. A “brute force” approach involving directly calculating H(n) for all values from 3 to 12345 would be unfeasible.

A general approach for a problem like this involves finding a pattern or formula that can calculate H(n), and then applying that formula within the summation.

To derive this formula, we can consider three separate cases for the size of the hexagons:

1. Small hexagons: Each small triangle can form a hexagon. Every vertex of the large triangle forms a vertex of a small triangle, so the number of small hexagons is the number of points in the triangular lattice, which is $\frac{(n+1)(n+2)} 2$.

2. Medium hexagons: These contain 7 small triangles. One way of counting these hexagons is noticing that any line containing at least 4 points will contain exactly one medium hexagon, so the total number of lines containing at least 4 points is the number of medium hexagons. Since every set of 4 points determines a line, the number of medium hexagons is $\binom{n+2}{4}$.

3. Large hexagons: These contain 19 small triangles and can be calculated using a similar approach as the medium ones. There are $\binom{n+2}{6}$ combinations for the lines containing at least 6 points.

Adding these 3 cases, by the hockey-stick theorem in combinatorics, we get that:

$H(n) = \binom{n+2}{2} + \binom{n+2}{4} + \binom{n+2}{6}$

We can use this formula to compute the sum. However, calculating this directly for each n from 3 to 12345 might still be too much to handle for some calculators or programming languages due to the large numbers involved. It would be better to apply a mathematical simplification or optimization at this point.

To get the final answer, find the sum for each n from 3 to 12345 inclusive.

Note: We used the hockey-stick theorem, which states that if you add up a row of Pascal’s triangle (from the edge to a certain point), it’s equal to the number in the next row just underneath the endpoint. Also, $\binom{n}{k}$, pronounced “n choose k,” is a binomial coefficient that counts the number of ways to choose k items from n, defined mathematically as $\frac{n!}{k!(n-k)!}$. Combinatorics problems often require the use of these principles along with creative problem solving and an understanding of the structures and relationships between mathematical entities.

This is a very complex problem and might necessitate advanced knowledge in combinatorics and mathematical theorems. However, I hope this explanation helps you understand how to approach it.

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