A3. A rectangle with sides of length 1 and $\sqrt{3}$ is folded over a diagonal.

Calculate the area of the common part, the shaded triangle in the figure.

![](https://cdn.mathpix.com/cropped/2024_04_17_2582a2b087f6953c1edag-1.jpg?height=189&width=248&top_left_y=2033&top_left_x=1578)
A3 $\mathrm{AC}=2$. The triangle that is folded is half of an equilateral triangle. $\angle C A B=30^{\circ}$. Due to the folding, $\angle C A E=30^{\circ}$ also holds. Thus, $\angle E A D=30^{\circ}$. Triangles $A B C$ and $A D E$ are similar, so $D E: D A=B C: B A$. From this, it follows that $D E=\frac{1}{3} \sqrt{3}$. Therefore, $E C=\frac{2}{3} \sqrt{3}$ and the desired area is

![](https://cdn.mathpix.com/cropped/2024_04_17_7fa3f9281a1c56596c60g-1.jpg?height=334&width=503&top_left_y=1523&top_left_x=1256)
thus equal to $\frac{1}{2} \times 1 \times \frac{2}{3} \sqrt{3}=\frac{1}{3} \sqrt{3}$.