Quiz

Yes, that's exactly what I was thinking. If it's more convenient for the spider to return to the bisector along a line perpendicular to the bisector, then it might also be more convenient to crawl along that same perpendicular line from the bisector. In other words, let the spider intercept the ant along a line perpendicular to its bisector, specifically the one that intersects the edge of the tetrahedron. Then it will need a speed of v/sqrt(3) to intercept it.
But here's the catch: the ant can constantly return to the vertex and change edges. It won't be able to escape the trap that way, but the spider will also have to return to the same point, and it won't be able to get close to the vertex. So, in the end, I think the answer is v/sqrt(3) + o(v). In that case, the ant has no winning strategy.