Curiosity: the rodef and nirdaf are each a single point in a closed circle. A single point can’t carry a gun (since a single point has no dimension, and hence no mass). Alternatively, shooting the nirdaf would violate the given assumption that both move with the same maximum speed.
Here’s a hint for how to visualize the problem:
Let’s assume for a moment that we have a closed square with vertices at (0,0),(0,1), (1,0) and (1,1). This is a simple unit square. Now, assume the rodef begins at one corner of the square, say (1,1), and the nirdaf begins at the opposite corner (0,0). So the nirdaf is sqrt(2) units away from the rodef.
Now, as a conjecture, it would be a reasonable assumption that the rodef and nirdaf would have opposite strategies; namely, the rodef wants to minimize distance between the rodef and nirdaf while the nirdaf wants to maximize distance between the rodef and nirdaf.
In this example, the nirdaf actually has no incentive to move whatsoever. This is because (0,0) is as far as the nirdaf can get from (1,1) where the rodef is. So the nirdaf is maximizing distance from the rodef by remaining at (0,0). The rodef’s optimal strategy from (1,1) is to move in a straight line towards (0,0) because the shortest distance between two points is a straight line (extra credit to whoever proves that the shortest distance between two points is a straight line).
Since the nirdaf will never have an incentive to move given the rodef’s strategy (moving away from (0,0) at any time will shorten the distance with the rodef), and the rodef continually gets closer, then it is clear to see that the rodef will eventually “corner the nirdaf” and catch the nirdaf at (0,0).
So in this contrived example, the rodef catches the nirdaf regardless of what the nirdaf tries to do. The riddle, though, is what happens in a circle?
Another hint:
The optimal strategy for the rodef to minimize distance to the nirdaf is to move along the straight line connecting the rodef’s location on the circle with the nirdaf’s location on the circle.