In a parallelogram $ABCD$ with an area of 2 square units, a line parallel to $AD$ intersects the perimeter of the parallelogram at points $P$ and $R$, and a line parallel to $AB$ intersects the perimeter at points $S$ and $Q$, as shown in the figure. What is the sum of the areas of triangles $AQR$, $BRS$, $DPQ$, and $CSP$?

![](https://cdn.mathpix.com/cropped/2024_05_02_8be36c8c0ab680fed1dfg-1.jpg?height=274&width=399&top_left_y=302&top_left_x=863)
I. solution. Let the areas of the four small parallelograms be $t_{a}, t_{b}, t_{c}, t_{d}$, and the area of the large parallelogram be $T$.

Then, taking into account the "falling" parts, the areas of the triangles can be written as follows:

$$
\begin{array}{ll}
T_{A Q R}=T-\left(\frac{t_{c}}{2}+\frac{t_{a}+t_{b}}{2}+\frac{t_{a}+t_{d}}{2}\right) ; & T_{B R S}=T-\left(\frac{t_{d}}{2}+\frac{t_{a}+t_{b}}{2}+\frac{t_{b}+t_{c}}{2}\right) \\
T_{C S P}=T-\left(\frac{t_{a}}{2}+\frac{t_{c}+t_{d}}{2}+\frac{t_{b}+t_{c}}{2}\right) ; & T_{D P Q}=T-\left(\frac{t_{b}}{2}+\frac{t_{a}+t_{d}}{2}+\frac{t_{c}+t_{d}}{2}\right)
\end{array}
$$

The sum of the areas of the four triangles:

$$
T_{\text {total }}=4 T-\frac{5\left(t_{a}+t_{b}+t_{c}+t_{d}\right)}{2}
$$

We know that $T=t_{a}+t_{b}+t_{c}+t_{d}$. Based on this, $T_{\text {total }}=4 T-\frac{5 T}{2}=\frac{3}{2} T=\frac{3}{2} \cdot 2=3$.

Therefore, the sum of the areas of the four triangles is 3 units.

II. solution. Let's create a diagram according to the example.

Let the sought area sum be $t$. Denote the height of the parallelogram by $m$, and its two parts by $m_{1}$ and $m_{2}$ as shown in the diagram. Let's write down the area of each triangle:

$$
\begin{aligned}
& T_{A Q R}=2-\frac{D R \cdot m+A B \cdot m_{1}+R C \cdot m_{2}}{2} \\
& T_{B R S}=2-\frac{R C \cdot m+A B \cdot m_{1}+D R \cdot m_{2}}{2} \\
& T_{D P Q}=2-\frac{A P \cdot m+P B \cdot m_{1}+D C \cdot m_{2}}{2} \\
& T_{C S P}=2-\frac{P B \cdot m+A P \cdot m_{1}+D C \cdot m_{2}}{2}
\end{aligned}
$$

Adding the areas:

$$
\begin{aligned}
t & =8-\frac{D R \cdot m+A B \cdot m_{1}+R C \cdot m_{2}}{2}-\frac{R C \cdot m+A B \cdot m_{1}+D R \cdot m_{2}}{2}- \\
& -\frac{A P \cdot m+P B \cdot m_{1}+D C \cdot m_{2}}{2}-\frac{P B \cdot m+A P \cdot m_{1}+D C \cdot m_{2}}{2}
\end{aligned}
$$

Further simplifying:

$$
\begin{aligned}
2 t & =16-m \cdot(D R+R C+A P+P B)- \\
& -m_{1} \cdot(A B+A B+P B+A P)-m_{2} \cdot(R C+D R+D C+D C) \\
2 t & =16-2 m \cdot A B-3 m_{1} \cdot A B-3 m_{2} \cdot A B=16-A B\left(2 m+3\left(m_{1}+m_{2}\right)\right)
\end{aligned}
$$

Since $A B\left(2 m+3\left(m_{1}+m_{2}\right)\right)=A B \cdot 5 m=10$, then $2 t=6$, and thus the sum of the areas of the triangles $A Q R, B R S, D Q R$ and CSP is 3 units.

Kiss-Tóth Christian (Budapest, Árpád Gymnasium, 9th grade)