Imagine the task as solved. The result - the parallelogram is centrally symmetric to the intersection point $P$ of the diagonals. Reflect the given $A B C \triangle$ over $P$. The parallelogram remains unchanged, as the reflection maps it onto itself; the sides of the reflected triangle $A^{\prime} B^{\prime} C^{\prime} \triangle$ necessarily pass through the vertices of the parallelogram (Figure 1), because the sides of the original $A B C \triangle$ also pass through these - corresponding - points in the reflection.

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Figure 1

Therefore, the vertices of the sought parallelogram can only be those points where the sides of the original and reflected triangles intersect.

At most 6 such intersection points $(1,2,3,4,5,6)$ can arise, i.e., 3 pairs of reflected points $(1,4 ; 2,5 ; 3,6)$, which indicate 3 possible diagonals for the sought parallelogram. Out of these 3 diagonals, we can choose two in three different ways. Each pair of diagonals defines a different parallelogram that meets the conditions of the problem.

Thus, the number of solutions is $3(1245 ; 3461 ; 5623)$, as long as the original triangle $A B C \triangle$ and the reflected triangle $A^{\prime} B^{\prime} C^{\prime} \triangle$ have 6 intersection points, meaning that the reflection of any side of the original triangle over $P$ intersects the other two sides. This is necessary and sufficient if, for each side, the distance of $P$ from the side is less than half the height corresponding to that side, or in other words, the point $P$ is inside the triangle defined by the midlines of $A B C \triangle$.

If $P$ lies on a midline (Figure 2), then two parallelograms coincide (in the figure, $1245 \equiv 3461$), and the third (5623) degenerates into a line segment $\left(B^{\prime} B\right)$. In this case, there is only one solution.

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Figure 2

If $P$ is outside the triangle defined by the midlines (Figure 3), then only two intersection points $(1,4)$, or one diagonal (14), arise.

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In this case, there is no solution.

Kengyel Vilma (Bp. I., Szilágyi E. lg. II. o. t.)