\section*{Problem 3 - V11023}

The quadrilaterals \(V_{1}, V_{2}, V_{3}\) agree in the diagonals \(e\) and \(f\). In \(V_{1}\), these diagonals intersect at an angle of \(30^{\circ}\), in \(V_{2}\) at \(45^{\circ}\), and in \(V_{3}\) at \(60^{\circ}\).

How do the areas of the three quadrilaterals compare?
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![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1170.jpg?height=356&width=1378&top_left_y=684&top_left_x=341)

Each of the quadrilaterals is divided into four sub-triangles by the diagonals. This results in the diagonal segments \(e_{1}, e_{2}, f_{1}, f_{2}\) with \(e=e_{1}+e_{2}\) and \(f=f_{1}+f_{2}\). With the intersection angle \(\alpha\) of the diagonals, the area of the quadrilateral is then given by

\[
\begin{aligned}
F & =\frac{1}{2} e_{1} \cdot f_{1} \sin \alpha+\frac{1}{2} e_{2} \cdot f_{2} \sin \alpha+\frac{1}{2} e_{1} \cdot f_{2} \sin \left(180^{\circ}-\alpha\right)+\frac{1}{2} e_{2} \cdot f_{1} \sin \left(180^{\circ}-\alpha\right) \\
& =\frac{1}{2} \sin \alpha \cdot\left(e_{1} f_{1}+e_{2} f_{2}+e_{1} f_{2}+e_{2} f_{1}\right) \\
& =\frac{1}{2} \sin \alpha \cdot\left(e_{1}+e_{2}\right)\left(f_{1}+f_{2}\right)=\frac{1}{2} \sin \alpha \cdot e \cdot f
\end{aligned}
\]

Thus, the ratios of the areas are

\[
\begin{array}{r}
F_{1}: F_{2}=\sin 30^{\circ}: \sin 45^{\circ}=\frac{1}{2}: \frac{1}{2} \sqrt{2}=1: \sqrt{2} \\
F_{2}: F_{3}=\sin 45^{\circ}: \sin 60^{\circ}=\frac{1}{2} \sqrt{2}: \frac{1}{2} \sqrt{3}=\sqrt{2}: \sqrt{3}
\end{array}
\]

The areas of the three quadrilaterals are in the ratio \(1: \sqrt{2}: \sqrt{3}\).