A parallelogram $ABCD$ is given as in the figure. A point $E$ moves along side $AB$ and a point $G$ moves along side $CD$ such that segment $EG$ is parallel to $AD$. When the intersection $F$ of segments $EG$ and $BD$ was at one-fifth of the diagonal $BD$ (closer to point $D$), the area of the shaded part of the parallelogram was $1 \mathrm{~cm}^{2}$ larger than when it was at two-fifths (again closer to $D$). Determine the area of parallelogram $ABCD$.

(E. Patáková)

![](https://cdn.mathpix.com/cropped/2024_04_17_b0839f028217e6476178g-1.jpg?height=394&width=669&top_left_y=725&top_left_x=702)
Let the length of side $AB$ be denoted by $a$ and the height of the parallelogram $ABCD$ to this side be denoted by $v$; the area of the parallelogram is equal to $av$. Triangles $DGF$ and $BEF$ are similar by the $uu$ theorem (angles $DFG$ and $BFE$ are vertical, angles $FEB$ and $FGD$ are alternate). Further, let $v_1$ be the height of triangle $DGF$ to side $DG$ and $v_2$ be the height of triangle $BEF$ to side $BE$.

a) Point $F$ is at one fifth of the diagonal $BD$. The similarity ratio of triangles $DGF$ and $BEF$ in this case is $1:4$. For the corresponding heights of these triangles, we have $v_1 + v_2 = v$. From this, it follows that $v_1 = \frac{1}{5} v$ and $v_2 = \frac{4}{5} v$. Note that $|DG| = |AE|$, and thus for the lengths of the corresponding sides of these triangles, we have $|DG| + |BE| = a$. Therefore, $|DG| = \frac{1}{5} a$ and $|BE| = \frac{4}{5} a$, and the areas of the shaded triangles are:

$$
\begin{aligned}
& S_{DGF} = \frac{1}{2} \left( \frac{1}{5} a \cdot \frac{1}{5} v \right) = \frac{1}{50} av, \\
& S_{BEF} = \frac{1}{2} \left( \frac{4}{5} a \cdot \frac{4}{5} v \right) = \frac{16}{50} av.
\end{aligned}
$$

The shaded part of the parallelogram in this case has an area of

$$
S_{DGF} + S_{BEF} = \frac{17}{50} av.
$$

b) Point $F$ is at two fifths of the diagonal $BD$. The similarity ratio of triangles $DGF$ and $BEF$ in this case is $2:3$. Similarly to the previous paragraph, we derive that $S_{DGF} = \frac{1}{2} \left( \frac{2}{5} a \cdot \frac{2}{5} v \right) = \frac{4}{50} av$ and $S_{BEF} = \frac{1}{2} \left( \frac{3}{5} a \cdot \frac{3}{5} v \right) = \frac{9}{50} av$. The shaded part of the parallelogram in this case has an area of

$$
S_{DGF} + S_{BEF} = \frac{13}{50} av.
$$

From the problem statement, we know that the difference $\frac{17}{50} av - \frac{13}{50} av = \frac{2}{25} av$ is exactly $1 \text{ cm}^2$. From this, it follows that $av = \frac{25}{2} \text{ cm}^2$. The area of the parallelogram $ABCD$ is $12.5 \text{ cm}^2$.

Another solution. Divide segments $AB$ and $BC$ into fifths. Parallels to these segments drawn through the resulting points will divide the parallelogram $ABCD$ into 25 identical small parallelograms.

If we also draw parallels to the diagonal $BD$, the parallelogram $ABCD$ will be divided into 50 identical small triangles, see the figure.

![](https://cdn.mathpix.com/cropped/2024_04_17_b0839f028217e6476178g-2.jpg?height=388&width=668&top_left_y=340&top_left_x=700)

If point $F$ is at one fifth of the diagonal $BD$, then the shaded part of the parallelogram consists of 17 small triangles. If point $F$ is at two fifths of the diagonal $BD$, the shaded part of the parallelogram in this case consists of 13 small triangles.
![](https://cdn.mathpix.com/cropped/2024_04_17_b0839f028217e6476178g-2.jpg?height=390&width=1280&top_left_y=984&top_left_x=388)

The difference of $1 \text{ cm}^2$ corresponds to the area of 4 small triangles. The area of the parallelogram $ABCD$ is therefore equal to $50 \cdot \frac{1}{4} = \frac{25}{2} = 12.5 \text{ cm}^2$.

Evaluation. 1 point for identifying a property that can be used to compare the areas of the shaded triangles (e.g., similarity, division into triangles, etc.); 3 points for expressing a relationship from which the exact difference in areas in the given situations can be determined (e.g., an equation derived from the similarity of triangles, determining how many small triangles the shaded parts differ by, etc.); 2 points for calculating the area of the parallelogram $ABCD$.