Hermit
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Yes, that's exactly what I was thinking. If it's easier for the spider to return to the bisector along a line perpendicular to the bisector, then it might also be easier to crawl from the bisector along the same perpendicular. 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.
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 see the answer as v/sqrt(3)+o(v). In that case, the ant has no winning strategy.
Yes! The spider's speed must be greater than the ant's speed divided by the square root of three!!! u>v/31/2
This is the condition under which it can move its "three-pointed star of death" towards the ant. In the case of equality, it can only hold the ant in the corner, preventing it from escaping, but it also cannot bring the plane in which it crawls (the plane in which the three-pointed star is located) closer to the vertex of the corner in which the ant is trapped! But if it's greater, then, without letting the ant out of the corner, it can bring the plane closer to it!
Thank you all for your interest in the puzzle!
Go ahead, Ment, pose the next one!
Всё не так плохо как Вы думаете. Всё намного хуже!