Square $ABCD$ has sides of length $6$. Points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{CD}$, respectively, with $AP=3$ and $DQ=\frac{18}5$. A path begins along the segment from $P$ to $Q$ and continues by reflecting against the sides of $ABCD$ (with congruent incoming and outgoing angles). If the path hits a vertex of the square, it terminates there; otherwise it continues forever. At which vertex does the path terminate? (Answer with $A$, $B$, $C$, $D$, or say that the path continues forever.)

1. Unfold the reflections: the path becomes a straight line in the plane tiled by $6\times6$ squares, starting at $P=(0,3)$ (with $A=(0,0)$, $B=(6,0)$, $C=(6,6)$, $D=(0,6)$) through $Q=\left(\frac{18}5,6\right)$. The direction is $\left(\frac{18}5,3\right)\parallel(6,5)$.
2. The line is $(6t,\ 3+5t)$. It meets a corner $(6i,6j)$ of the tiling when $t=i$ and $3+5i=6j$.
3. $i=1,2$ give $8,13$, not multiples of $6$; $i=3$ gives $18=6\cdot3$. So the first corner is $(18,18)$ with $i=3$, $j=3$ both odd.
4. Odd $i$ means $x=6$ and odd $j$ means $y=6$, so the path ends at the vertex $(6,6)=\boxed{C}$.