This problem was aimed, first and foremost, to demonstrate that in some cases, geometrical approach to relativistic kinematics using xict diagram is much more efficient than algebraic approach based on Lorentz transform. More or less as expected, none of the participants were able to figure out the geometric approach without hints. Even with the hints, only one contesant was able to work out fully geometrical solution, and five more found semigeometrical solutions in which case they relied on algebraic results.
The best solution award goes to the only fullygeometric solution: Prathyush Poduval. As hinted, the trajectory of the spaceship in xictdiagram is a circle; Prathyush derives this fact from the finding that in proper time τ, the tangent of the xicttrajectory rotates at a constant angular speed. Alternatively, one could have expressed the curvature radius of the trajectory as the ratio of an infinitesimal arc length, i.e. an increment of the interval ds=d(icτ), and the corresponding rotation angle of the tangent dα=dgτ/ic; the result R=c^{2}/g is independent of time, so the curve is a circle. One should also notice that due to equality ds=d(icτ), the length of a xicttrajectory is given by the corresponding proper time, s=icτ. What is left to do is to apply the theorem of inscribed angles; more detailed explanations of this last step can be found in other solutions, see below. For a reference, here is his bruteforce solution, which is also among the most economically written bruteforce solutions. Halfgeometric solutions making use of the inscribed angles theorem together with equality s=icτ (earning bonus factor 1.1). Balázs Németh  a detailed and nicely written solution; Navneel Singhal; Peter Elek; Thomas Bergamaschi  another welldocumented solution; Alkın Kaz. While the bruteforce approach is mathematically longer, it is still useful to have a look on some of such solutions. I have selected the bestdocumented solutions from the first week. Tóbiás Marozsák; Balázs Németh; Dylan Toh; Satoshi Yoshida; Elvinas Ribinskas; Finally, here is a selection of correct answers (assuming c=1 and g=1)  these might look different, but are equivalent, nevertheless. And here are the results. Number of fully correct solutions: 53. Names in italic correspond to unofficial participants (they get their deserved speed bonus, but do not advance the count for the next speed bonuses

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