B1. Connect the midpoint $\mathrm{M}$ of PS with the midpoint $\mathrm{N}$ of $\mathrm{AB}$. Triangle MPN is a 30-60-90 triangle, where $M N=2$ and thus $P N=\frac{2}{\sqrt{3}}$

Conclusion: $P B=2+\frac{2}{\sqrt{3}}$. The question asks for twice the area of an equilateral triangle with side $2+\frac{2}{3} \sqrt{3}$. The half base is thus $1+\frac{1}{3} \sqrt{3}$ and the height

$\sqrt{3}+1$. The required area is thus

![](https://cdn.mathpix.com/cropped/2024_04_17_774e242ba19fa9593e58g-2.jpg?height=446&width=499&top_left_y=1019&top_left_x=1244)
$2 \times\left(1+\frac{1}{3} \sqrt{3}\right) \times(1+\sqrt{3})=4+\frac{8}{3} \sqrt{3}$.
B1. $A B C D$ is a square with side 4.

$P$ lies on $A B, S$ on $C D$ and $Q$ and $R$ on $P S$ such that

1) $A P=C S$ and
2) the triangles $P B R$ and $S D Q$ are both equilateral.

What is the total area of these two triangles?

![](https://cdn.mathpix.com/cropped/2024_04_17_6ccf422d4ecbc0dab674g-2.jpg?height=325&width=345&top_left_y=240&top_left_x=1438)