Collision Course On The Plaza

The key is to exploit the symmetry within the problem. In fact, if we take a snapshot of the three cats at any time $t_1$ between $T_0$ and $T_0+\Delta t$ , they will be forming an equilateral triangle relative to each other, which shrinks and counterclockwise as $t_1$ approaches $T_0+\Delta t$ . We can then draw this diagram:

Cat $B$ 's velocity vector can be separated into two component vectors, one of magnitude $5\sqrt{3}$ m/s pointing perpendicular from Cat $A$ 's velocity vector, and one of magnitude $5$ m/s pointing 'towards' Cat $A$ 's velocity vector. As Cat $A$ 's velocity vector has a magnitude of $10$ m/s, the combined effect is as if Cat $A$ were running towards Cat $B$ at a velocity of $15$ m/s. The question asks for $\Delta t$ : as Cat $A$ has velocity $15$ m/s, the cats will collide in $\frac{270}{15}=18$ seconds.

Note that it was okay to calculate the time for only Cat $A$ , as all three cats collide at the same time.