# Problem 2 – Hint 5

An extended summary of the previous hints: the solution can be constructed by using/showing/noticing the following things. Finding the fastest lap time by a fixed-sized triangle is the same as finding the largest triangle by a fixed lap time. Fastest lap (largest triangle) means that the boy is running always on the verge of slipping. Hence the speed in the hodograph plane is constant so that a fixed running time corresponds to a fixed curve length. With a fixed curve length in the hodograph plane, maximizing the displacement in the direction of an $x$-axis (in $x-y$-plane) is the same as maximizing the average $x$-component of the velocity in the hodograph plane (averaged over the length of the curve). If we consider running between two neighbouring vertices of the triangle, the initial and final directions of the velocity vector can be deduced from symmetry considerations, hence the Hint 4 can be used.

Please submit the solution to this problem via e-mail to physcs.cup@gmail.com. There will be no more hints for the Problem 2. The next update of the results will be published at 13:00 GMT, 6th February 2022.