How can an arbitrary quadrilateral be cut into pieces with two cuts so that a parallelogram can be formed from the pieces?
I. Solution: (Fig. 31.) The quadrilateral must be cut in such a way that the new vertices replace the old ones, with opposite vertices having equal angles.

![](https://cdn.mathpix.com/cropped/2024_05_02_ab09a3b16f5bc6e4c253g-1.jpg?height=553&width=704&top_left_y=304&top_left_x=770)

At the intersection of two straight cuts, two pairs of equal angles are formed among the four angles created, and the sum of the original angles of the quadrilateral is $360^{\circ}$. Thus, we can hope for a solution where the original four vertices are brought together. To get a quadrilateral again, it is necessary that the sides of the angles fit together in equal segments. This can be achieved by having the cutting lines pass through the midpoints of the sides.

Thus, we arrive at the following solution: Cut the quadrilateral along the midlines (lines connecting the midpoints of opposite sides); then sequentially flip the resulting pieces around the midpoints of the sides to the adjacent side, so that the two halves of the side overlap. (The already flipped pieces are further flipped together with the next one.) This way, the four vertices will coincide, and since the sum of the angles in them is $360^{\circ}$, no part will be left over, and the pieces will not overlap. At the points where the rotation occurred, a vertex might form, but this does not happen because the two angles that come together are the same as they were before, just with different sides fitting together, and thus they continue to form a straight angle. The resulting figure is therefore a quadrilateral. The angles at the opposite vertices are the vertex angles formed by the cuts. Thus, we have indeed assembled a parallelogram.

Solved by: Bognár J., Csernók L., Czibere T., and Nagy F., Gehér L., Gősy S., Kővári T., Párkány M., Réthy Eszter, Róna P., Szépfalussy P., Tarnóczi T., Vígh Magda, Vörös M.

Partially: Boda I., Ungár Veronika.

Remarks: The sides of the parallelogram are twice the length of one segment of the midlines, so it follows from our result the already known theorem (Problem 73, p. 30) that the midlines of a quadrilateral bisect each other.

Ed.

Thus, we constructed a parallelogram from the quadrilateral, whose sides are the midlines, and the angle between them is the angle formed by the midlines. The area of the new quadrilateral is naturally equal to the area of the original, so if we denote the lengths of the midlines by $e, f$, and the angle between them by $\omega$, we get the following method for calculating the area of any quadrilateral:

$$
t=e \cdot f \cdot \sin \omega
$$

Péter Róna (Budapest Evangelical Gymnasium VII. grade)

II. Solution: Several solved the problem by combining folding. Connect the midpoints of adjacent sides. Fold the quadrilateral so that the two opposite sides of the smaller quadrilateral (parallelogram) lie on the same line. (They do not necessarily overlap). The first cut should be along this line. Two triangles fall off, leaving a hexagon. Keeping the two triangles together, slide them so that their vertices coincide. Then cut them in the direction of the diagonal of the original quadrilateral. The resulting four triangles can be used to complete the hexagon into a parallelogram as shown in Fig. 31b (the lines of the cuts are drawn thickly).

![](https://cdn.mathpix.com/cropped/2024_05_02_ab09a3b16f5bc6e4c253g-2.jpg?height=581&width=461&top_left_y=154&top_left_x=862)

Solved by: Szabó E., Szarvas F.

A similar solution was found by: Gacsányi S.

Remark: The sides of the resulting parallelogram are parallel to the diagonals, and one is equal in length to the corresponding diagonal, while the other is half as long as the corresponding diagonal. Thus, if the lengths of the diagonals are denoted by $g, h$, and the angle between them by $\varphi$, the area can also be calculated as follows: $t=\frac{g \cdot h}{2} \sin \varphi$.