12

Not a definite answer but I narrowed it down to 2 possibilities. It's either In this case or the second possible answer is Because


8

For two spiders, one of the spiders has to be at least as fast as the ant in order for the spiders to catch the ant. There are 3 situations: Both spiders are slower than the ant: The ant can avoid forever At least one spider is as fast as the ant: The spiders can catch the ant At least one spider is faster than the ant: The spiders can catch the ant ...


8

An upper bound for the 3 spider case:


6

I feel like 1/2+ε is enough for the 3 spider case. The spiders can start (or take their sweet time to get) in this configuration: And then the two slow spiders can follow the red arrows to ensure the ant will be trapped, while the "fast" spider keeps watch on its edge. Unless I overlooked something, this seems to work, but doesn't prove there isn'...


6

Let's focus on the A few observations: Conversely: This already resolves C as To avoid tedious edge cases I'll assume that the enclosure is a rectangle with one side completely open. B A


4

For my analysis, I will assume that all we get to specify is the sum of the speeds of the spiders and that the sum we specify must work for any possible assignment of that total speed to the spiders where each spider must be given a non-zero speed. Four Spiders It wasn't asked, but first lets take a look at the case of 4 spiders. With 4, any positive sum ...


3

Here is a strategy giving a better uppper-bound for the 3 spider case. And there is a strategy for two spiders, one with speed exactly 1 and one with speed ε.


1

A couple of words are confusing: By "hover" I assume you mean it's moving (generally means to stay in place) By "retain" I assume you mean some kind of average speed needed. Clearly our speed will accelerate and not be maintained as a constant. So, maybe:


Only top voted, non community-wiki answers of a minimum length are eligible