Problem 698

In the triangular grid, there are three sets of parallel lines, denoted as $W_{a}$, $W_{b}$, and $W_{c}$, parallel to $BC$, $CA$, and $AB$ respectively. Let $BC = CA = AB = n$. Extend $AB$ to point $B'$ and $AC$ to point $C'$ such that $BB' = CC' = 1$, and connect $B'C'$. Extend the diagonal grid lines to intersect with $B'C'$, forming a $\triangle AB'C'$ with a regular triangular grid of size $n+1$. Denote the line $B'C'$ as $l$, and the points on $l$ from left to right as $A_{1} = B'$, $A_{2}$, ..., $A_{n+2} = C'$.

Notice that in Figure 1, each parallelogram is formed by two sets of parallel lines.

First, consider the parallelograms formed by $W_{b}$ and $W_{c}$ (referred to as A-type). Take any A-type parallelogram $DEFG$ (as shown in Figure 1), where $DE, GF \in W_{b}$ and $EF, DG \in W_{c}$. Extend $DG$, $EF$, $GF$, and $DE$ to intersect with line $l$ at points $D'$, $E'$, $F'$, and $G'$ respectively. It is easy to see that there is a one-to-one correspondence between the parallelogram $\square DEFG$ and the ordered quadruple of points $(D', E', F', G')$ on line $l$. Therefore, the number of A-type parallelograms $f_{a}(n) = \mathrm{C}_{n+2}^{4}$.

Similarly, the number of B-type parallelograms (formed by $W_{c}$ and $W_{a}$) and C-type parallelograms (formed by $W_{a}$ and $W_{c}$) is the same, i.e., $f_{b}(n) = f_{c}(n) = \mathrm{C}_{n+2}^{4}$.
Therefore, $f(n) = 3 \mathrm{C}_{n+2}^{4}$.