= Solution
The <world lines> in the train frame are $x_A=L/2$, $x_B=-L/2$ and $x_C=-vt$. In the station frame, <length contraction> gives
$$
x'_A=vt'+\frac{L}{2\gamma},\qquad x'_B=vt'-\frac{L}{2\gamma},\qquad x'_C=0.
$$
The emission event is $O=(0,0)$ in both frames. The signal <world lines> in $S'$ are $x'=ut'$ and $x'=-ut'$; their intersections with the train endpoints define the reception events $R_A,R_B$. Their velocities in $S$, from the <relativistic velocity-addition formula>, are $(u-v)/(1-uv/c^2)$ and $-(u+v)/(1+uv/c^2)$.
The two platform-alignment events occur at $(x',t')=(\pm L/(2\gamma),0)$ in $S'$. The <Lorentz transformation> puts them at $(x,t)=(\pm L/2,\mp vL/(2c^2))$ in $S$, on the tilted simultaneity line $t=-vx/c^2$. The <spacetime diagrams> show all three observers, both signals, their receptions, and these distinct endpoint-alignment events:
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ib/paper-4-train-signals.png]
{title=Train and station spacetime diagrams with signal receptions and simultaneous platform alignments}
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