[spoiler]This method does not guarantee catching the student within a particular time but does guarantee catching them eventually. I assume that the psychic student knows where the professors are but not where they will go next. The three professors start on the top vertex and each takes a different edge so they end up on the three vertices on the base. The student must now be on one of the three edges on the base. I can’t see how to cover all three edges but I can cover two of them. This gives a 2/3 chance of catching the student whatever he does and the professors just repeat the process until they get lucky which they must do eventually. It requires some coordination between the professors which may be a problem. At random one of the professors moves along one of the base edges. There is a 1/3 chance that he finds the student. This professor, and the other professor on the same vertex, now walk to the third professor. There is a 1/3 chance that they catch the student. But now we are back at the starting position with all three professors on the same vertex. So the professors can execute the same strategy again, each time they have a 2/3 chance of catching the student. Assuming it takes 1 hour to cover an edge we can calculate how long we expect them to take, E = 1/3 x 2 hours + 1/3 x 3 hours + 1/3 x (3 + E) hours which solves to give E = 4 hours.[/spoiler]
]]>Maybe not though. I think that puzzle may be impossible.
Perhaps it’s blind professors and a *slower* student…
]]>spiders don’t know where the spider is (so we have to make sure the fly cannot pass us)
spiders may meet and bypass each other
We start off with all spiders in one corner. Two spiders traverse the same egde, the second spider of those two traverses another adjacent edge. (see diagram below)
.

. _ .
These two edges and three corners cannot have a fly (or else it has been eaten). So the middle one now traverses the diagonal edge. This causes three edges to be free of the fly.
. .
 /
. _ .
Now we sacrifice an edge (either the left of bottom one). So one spider on the outside edge goes back to the bottom left and goes up the diagonal.
. .
 /
.
Then it connects the two upper vertices by traversing it’s edge, forming a triangle.
. _ .
 /
.
The only thing left is for all three spiders to simultaneously move into the invisited edges and cornering the fly in the last corner.
[/spoiler]