This problem was the most technical one out of all the five problems. While the equilbrium angle could be found
relatively easily, the oscillation period required calculation of few integrals, and taking few derivatives which were not too difficult in principle, but required patience and accuracy. Although such calculations can be considered to be the most boring part of physics, it is also an inevitable part of it.
Without doubt, the best solution was made by Navneel Singhal. This is because he was able to avoid calculation of the integrals by noticing that these integrals give actually the moment of inertia of the triangle. To see this, let us use coordinate system with the origin at the centre of mass, with horizontal xaxis and vertical yaxis. As hinted , we can assume the velocity of the centre of mass to be negligibly small when calculating the kinetic energy of the system. Using the first hint, one can easily deduce that v_{y} must be a linear function of y: v_{y}=ay+b. Since the centre of mass is almost at rest, averaging v_{y} over the entire water mass should give zero (when neglecting quadratically small terms). Average of y gives the coordinate of the centre of mass, which is zero; hence, b=0 so that v_{y}=ay. The condition of incompressibility of water can be expressed in terms of derivatives as dv_{x}/dx+dv_{y}/dy=0, hence v_{x}=ax. So, the double kinetic energy of a water element is 2dK=a^{2}(x^{2}+y^{2})dm, hence K=a^{2}I/2, where I is the moment of inertia with respect to the centre of mass. Navneel's solution, however, is not overhelmingly better than other solutions. Therefore he gets only half of the bonus points, and the rest is divided between the other four welldocumented solutions: Thomas Bergamaschi; Dylan Toh; Tóbiás Marozsák; Muhammad Farhan Husain. And here are the results. Number of fully correct solutions: 8.

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