My thinking is thus: First Assume the wolves and hare are on the same line of longitude. The hare at the equator and equidistant from the wolves
However the hare moves the wolves can always stay at the same longitude as the hare by moving along their own line of latitude. Because the hare is between the wolves their latitude circles will be smaller than the hares. This gives them a free longitudinal component of motion they can use to close with the hare. Although the hare can move out of its equidistant point it will always be caught.. eventually
Now to consider out initial assumption. We can partially dispense with the equidistant requirement as the hare is able to move closer to one wolf or the other from its starting point without escaping or increasing its distance from either wolf.
We can also define our longitudinal axis as the plane of W-H-W. as long as the animals are on a plane they can be considered on a line of longitude
So, can the wolves move to position the hare between them and on a plane regardless of the hares movement? well if the wolves each move to a pole they form a plane with the rabbit wherever it is. The rabbit may not be in the same hemisphere, trapping one wolve behind the 'hump' of the planet. but this wolf is still able to maintain its longitudinal match with the hare and the nearer wolf can chase the hare towards the equator until they are all in the same hemisphere
NOW! max time to capture.
The best strategy from the hare is to move such that the wolves come closer at the slowest rate. This is achieved by having the maximum longitudinal speed.
Lets consider the case where the wolves are at the poles and the hare at the equator.
The hare runs along the equator and the wolves match its longitude but slowly approach. They cant run directly to the closest intersection point because the hare will reverse direction and escape. As they get closer and closer to the hare their courses become more and more parallel. But we are asked to consider them as points. Luckily they don't need to eat
1h smell radius. much harder. have to think
OK so, The wolves can detect the hare when they are 1h away from it. If they all happen to be on a plane as before they can catch it. this is the case where 2 wolves are 2hrs apart and the hare is directly between them.
So 12 wolves (24h/2) can start at one pole and, equally spaced, move to the other pole eventually trapping the hare
Say we have 1 less wolf, the ring of wolves almost reaches the equator and then MUST leave a gap. but can the rabbit squeeze though an infinitesimal gap? the wolves move at speed s same as the hare, the overlapped region of smell expands at sqrt(1-s^2)... no thats on a flat plane, for a sphere its.. arccos(cos(1)/cos(s)) ? which is always less than s. Ergo the hare can escape from between any two closing spherical caps
OK, Considering the strategy for 3 wolves.
Assuming the same best case (for the hare) starting point with wolves on poles and hare on equator but with one spare wolf at one pole. The 2 wolves move as before but the spare wolf moves to an arbitrary point on the equator to try and cut off the rabbit.
In this case the hare is still caught at infinity ( i think the same infinity, have to do the maths) as it can approach the third wolf but switch direction immediately before capture. As long as its speed remains constant the approach time of the original wolves is the same and the third wolf cannot catch it.
However what if the original two wolves vary their strategy, dropping back to prevent the reverse direction tactic? tbc..