About the relativistic wagon simultaneity problem I'm getting utterly confused with this problem. The situation is basically to know what events to take in one frame in order to see them synchronized in another. I sketched the situation as this:

There are two wagons A and B of rest length $\ell$ moving with speed $-v$ and $v$ as seen by an observer C. In the wagon B there's a lamp that blinks regularly.
Suppose (this is actually the original problem's metaphor) this is a relativistic dance contest where C is the judge and the observers at A and B are a dancing couple. What I want to know is when should the observers at A and B should dance after one blink so that the judge C sees their dance synchronized.
What I'm trying is to see the events of their individual reaction of the observers at the frame of the lamp and then transform them to the frame of the jury. So, letting $(t_B,x_B,0,0)$ be the event when the dancer at B reacts and $(t_A,x_A,0,0)$ the event when the dancer at A reacts as seen from the lamp's frame, in the jury's frame,
$$t^\prime_B=\gamma_v(t_B-v\frac{x_B}{c^2})\\
t^\prime_A=\gamma_{-v}(t_A+v\frac{x_A}{c^2})$$
which must be equal in order for C to see this events as simultaneous, then (as $\gamma_v=\gamma_{-v}$)
$$t_B-v\frac{x_B}{c^2}=t_A+v\frac{x_A}{c^2}$$
as seen from the lamp, $x_B=\ell$ always, so
$$t_A=t_B-v\frac{\ell}{c^2}-v\frac{x_A}{c^2}$$
and from here I think I don't really know how to proceed; $x_A$ should be at least equal to $-\ell$, so I'm letting $x_A=-x_\alpha-\ell$ for some $x_\alpha>0$ which I hope to be able to determine, hence
$$t_A=t_B-v\frac{x_\alpha}{c^2}$$ If I let $x_\alpha=0$, how come $t_A^\prime=t_B^\prime$ implies $t_A=t_B$? What should I do with $x_\alpha$? I'm also thinking that letting $t_B=\frac{\ell}{c}$ would be nice since this is the time that takes light to reach the observer in B, then
$$t_A=\frac{\ell}{c}-v\frac{x_\alpha}{c^2}$$ but then again I don't know what to do with $x_\alpha$. I also thought of taking three reference frames, two for the wagons and one for C, but I did not get anywhere. I also thought that velocity addition should get in the way or time dilation, but I cant see where. So how should this problem be handled and using what concepts?
 A: I do not think the example you used is good to illustrate the relativity of simultaneity. It has the additional complication of using 3 frames of reference, when two should be enough. I could not follow your derivation, but here is a much simpler one and easier to understand. 
Imagine two observers, one  midway inside a speeding traincar and another observer standing on a platform as the train moves past.
A flash of light is given off at the center of the traincar just as the two observers pass each other. The observer on board the train sees the front and back of the traincar at fixed distances from the source of light and as such, according to this observer, the light will reach the front and back of the traincar at the same time.
The observer standing on the platform, on the other hand, sees the rear of the traincar moving (catching up) toward the point at which the flash was given off and the front of the traincar moving away from it. As the speed of light is finite and the same in all directions for all observers, the light headed for the back of the train will have less distance to cover than the light headed for the front. Thus, the flashes of light will strike the ends of the traincar at different times.The explanation is best understood using two graphs: 
1) from the point of view of the person in the train, the light reaches both ends simultaneously
 e 
2) From the point of view of the person on the platform, the light reaches the back of the train first  

