Consider a triangle $ABC$ and a point $P$ inside it. Construct a parallelogram inscribed in the triangle such that its diagonals intersect at point $P$. (We only consider this parallelogram to be inscribed if all four of its vertices lie on the perimeter of the triangle, meaning no vertex lies on the extension of any side of the triangle.)
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.

![](https://cdn.mathpix.com/cropped/2024_05_02_54186d9f73015c6bd80ag-1.jpg?height=572&width=653&top_left_y=343&top_left_x=714)

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.

![](https://cdn.mathpix.com/cropped/2024_05_02_54186d9f73015c6bd80ag-1.jpg?height=566&width=740&top_left_y=1514&top_left_x=671)

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.

![](https://cdn.mathpix.com/cropped/2024_05_02_54186d9f73015c6bd80ag-2.jpg?height=654&width=808&top_left_y=123&top_left_x=645)

In this case, there is no solution.

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