Quiz

Uran is right, I think. Trap the ant in the corner of a tetrahedron (this is possible, as the ant needs to change edges. If the ant only crawls along one edge, the task of catching it becomes much simpler). Accordingly, the spider at that moment should be on the bisector to the corner, at the vertex where the ant is. If the ant moves, the spider should move along the perpendicular to the ant's edge, and it should have enough speed to cover this segment while the ant crawls to its base. At the same time, the ant may try to play with the spider, making continuous false sorties, without intending to finally escape from the trap. This will essentially ensure its safety with the minimum speed of the spider (it will not be able to get closer), so the spider's speed should be even infinitesimally greater.
The perpendicular to the edge from the bisector of the tetrahedron... After long calculations, I realized that it is apparently sqrt(2) times smaller than the distance to its base from the vertex. Then, if the ant's speed is v, the spider's speed is
v/sqrt(2) + o(v)