Slovenia — National Math Olympiad

Denote the sides of the parallelogram by $a$ and $b$. We may assume that $a \ge b$. The small triangles which share at least one point with the sides of the parallelogram are marked in the figure. We can find their number by subtracting $2$ triangles in the corners and the number of the triangles inside the white parallelogram from the number of all triangles in the enlarged parallelogram.

When $b \le 2$ there is no white parallelogram. If $b = 1$, then the number of the small triangles is equal to the number of the triangles inside a parallelogram of size $(a + 2) \times 3$, minus $2$. There are $2 \cdot 3(a + 2)$ of the triangles in the parallelogram, so $6(a + 2) - 2 = 46$. This implies $a = 6$.

If $b = 2$, then there are $2 \cdot (a + 2) \cdot 4 - 2 = 46$ small triangles and this implies $a = 8$.

Let $b \ge 3$. The enlarged parallelogram contains $2(a + 2)(b + 2)$ small triangles and the white parallelogram contains $2(a - 2)(b - 2)$ small triangles. We get
$$2(a + 2)(b + 2) - 2(a - 2)(b - 2) - 2 = 46,$$
and this implies $a + b = 6$. Since $a \ge b \ge 3$, this is only possible when $a = b = 3$.

Let us find the areas of these parallelograms. Since the acute angle in these parallelograms measures $60^\circ$, the areas are equal to $\frac{\sqrt{3}ab}{2}$.

When $a = 6$, $b = 1$ we get $3\sqrt{3}$.

When $a = b = 3$ we get $\frac{9\sqrt{3}}{2}$.

When $a = 4$, $b = 2$ we get $4\sqrt{3}$.