= Solution
Use the spatial <internal-wave ray tracing> model, with $dx/dz=\pm16z$, and reflect the rays so that the <frequency> and tangential <wave number> are preserved. At a vertical wall the horizontal energy direction reverses; at a horizontal wall the vertical energy direction reverses. On the sloping wall the reflection depends on <internal-wave slope criticality>. The three-reflection orbit below encounters that wall at a height above $3/16$, so the wall reverses the horizontal energy direction there.
First take the packet to leave the bottom upward and to the left, with $0<x_0<6$. The four segments of the loop are
$$
\begin{aligned}
x&=x_0-8z^2 &&\text{to the left wall},\\
x&=8z^2-x_0 &&\text{to the top},\\
x&=16-x_0-8z^2 &&\text{to the sloping wall},\\
x&=6+3z_s+8(z^2-z_s^2) &&\text{back to the bottom}.
\end{aligned}
$$
The left-wall reflection is at $z=\sqrt{x_0/8}$ and the top reflection at $x=8-x_0$. The slope intersection solves $8z_s^2+3z_s+x_0-10=0$, giving the <internal-wave ray return map>
$$
\boxed{z_s=\frac{\sqrt{329-32x_0}-3}{16},\qquad
R(x_0)=6+3z_s-8z_s^2=x_0-4+6z_s.}
$$
Thus the next formal bottom reflection is at $(R(x_0),0)$. On $0<x_0<6$, the slope heights are between approximately $0.544$ and $0.946$, consistent with the assumed reflection type.
An <internal-wave attractor> is a stable periodic ray, so solve $R(x_*)=x_*$. This gives $z_s=2/3$, and therefore
$$
\boxed{x_*=40/9.}
$$
The successive vertices are
$$
\boxed{(40/9,0)\longrightarrow(0,\sqrt5/3)\longrightarrow(32/9,1)\longrightarrow(8,2/3)\longrightarrow(40/9,0).}
$$
This is \b[clockwise circulation] in the $(x,z)$ plane. Differentiating the <internal-wave ray return map> shows
$$
R'(x_0)=\frac{16z_s-3}{16z_s+3},\qquad \boxed{R'(x_*)=23/41<1.}
$$
Neighbouring bottom intersections converge to the periodic orbit, geometrically focusing the energy beam.
For completeness, the opposite upward launch direction has the reversed three-reflection itinerary: slope, top, left. In the interval where this itinerary is valid, put
$$
z_r=\frac{3+\sqrt{201-32x_0}}{16},\qquad
\boxed{R_{\rm rev}(x_0)=x_0+4-6z_r.}
$$
The three-reflection reverse map is $R^{-1}$, with domain $R(0)<x_0<R(6)$. Its fixed point is the same, but $R_{\rm rev}'(x_*)=41/23>1$. \b[Counterclockwise packets defocus from that orbit.] Except for the exactly periodic reverse ray, they eventually leave this itinerary. In the full geometric reflection construction, omitted-side or additional-slope reflections redirect generic rays into the clockwise attracting branch. Tangency and exactly critical reflections are singular exceptions requiring finite-wavelength or dissipative treatment.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-74-wave-attractor.png]
{title=Clockwise spatial internal-wave attractor and convergence of neighbouring ray loops in the trapezoidal basin}
The bottom reflection is a formal spatial-ray idealization: part (b) shows that the <WKB approximation> itself fails near $z=0$, so the diagram does not imply arrival and reflection there in finite WKB travel time. In a real fluid, <viscous attenuation of an internal-wave beam>, finite <wavelength>, <internal-wave breaking> and ensuing <turbulence> and mixing limit the beam width and energy density. Unlimited focusing is a property of the ideal geometric model, not a prediction for the resolved physical wave.
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