B3. Due to the symmetry of the figure in the line $P R$, we can limit ourselves to calculating the length of $P Q R$, half of the path, when determining the length of $P Q R S P$. $P Q F=R Q A$ Reflect $A B C D E F$ in the line $A F$, see figure.

$$
R^{\prime} Q A=R Q A \text {, so } P Q F=R^{\prime} Q A \text {. }
$$

Thus, $Q$ lies on $P R^{\prime}$. Therefore, the length of the path $P Q R$ is equal to the length of $P R^{\prime}$. Let $T$ be the midpoint of $E^{\prime} F . P T=F D=4 \sqrt{3}$, because triangle $A F D$ is a $30^{\circ} 60^{\circ} 90^{\circ}$ triangle with $A F=4, A D=8, F D=4 \sqrt{3}$.

$$
\begin{aligned}
& T R^{\prime}=\frac{A F+E^{\prime} B^{\prime}}{2}=6, \text { so } \\
& P R^{\prime}=\sqrt{6^{2}+(4 \sqrt{3})^{2}}=\sqrt{84}=2 \sqrt{21} \text { Therefore } \\
& P Q R S P=4 \sqrt{21} .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_774e242ba19fa9593e58g-3.jpg?height=582&width=734&top_left_y=240&top_left_x=1089)
B3. On a billiard table in the shape of a regular hexagon $A B C D E F$ with side 4, a ball is struck from $P$ towards $Q$. Calculate the length of the path $P Q R S P$, neglecting the dimensions of the ball. $P$ is the midpoint of $D E$ and $R$ is the midpoint of $A B, Q$ lies on $A F$ and $S$ lies on $B C$.