4. A billiard table is in the shape of a regular hexagon $A B C D E F$. A ball is struck from the midpoint $P$ of $A B$, hits a point $Q$ on $B C$, and then successively strikes the sides $C D, D E, E F, F A$, finally hitting a point on $A B$. Let $\angle B P Q=\theta$, find the range of values for $\theta$. Hint: Use the principle that the angle of incidence equals the angle of reflection.


Answer
As shown in the figure, construct a regular hexagon $A B C D E F$ successively
symmetric about $B C, C D, D E, E F, F A$
Draw $A M, B N$ perpendicular to the original line $A B$ through $A, B$, with the feet of the perpendiculars at $M, N$.
Assuming the side length of the regular hexagon is 2, then $|A M|=$
$|B N|=3 \sqrt{3},|P M|=8,|P N|=10$.
Since $\tan \angle E P M=\frac{2 \sqrt{3}}{5}>\tan \angle A P M=\frac{3 \sqrt{3}}{8}, \tan \angle F P M=\frac{2 \sqrt{3}}{7}<\tan \angle A P M$
$=\frac{3 \sqrt{3}}{10}$, then $\angle B P N \leqslant \theta \leqslant \angle A P M \Rightarrow \frac{3 \sqrt{3}}{10} \leqslant \tan \theta \leqslant \frac{3 \sqrt{3}}{8}$.
Therefore, the range of $\theta$ is $\left[\arctan \frac{3 \sqrt{3}}{10}, \arctan \frac{3 \sqrt{3}}{8}\right]$.