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 $4\sqrt{3}$.