Question: Suppose I put an ant in a tiny racecar on every face of a soccer ball. Each ant then drives around the edges of her face counterclockwise. The goal is to prove that two of the ants will eventually collide (provided they aren't allowed to stand still or go arbitrarily slow).

My brother told me about this result, but I can't quite seem to prove it. Instead of a soccer ball, we should be able to use any connected graph on a sphere (provided that there are no vertices of valence 1). We may as well assume there are no vertices of valence 2 either, since you can always just fuse the two edges.

I (and some people I've talked to) have come up with a number of observations and approaches:

* Notice that if two ants are ever on the same edge, then they will crash, so the problem is discrete. You can just keep track of which edge each ant is on, and let the ants move one at a time. Then the goal is to show that there is no way for the ants to move without crashing unless some ant only moves a finite number of times.
* You can assume all the faces are triangles. If there is a face with more than three edges, then you can triangulate it and make the ants on the triangles move in such a way that it looks exactly the same "from the outside". If there is a 1-gon, it's easy to show the ants will crash. If there is a 2-gon, it's easy to show that you can turn it into an edge without changing whether or not there is a crash.
* One approach is to induct on the number of faces. If there is a counterexample, I feel like you should either be able to fuse two adjacent faces or shrink one face to a point to get a smaller counterexample, but I can't get either of these approaches to work.
* If you have a counterexample on a graph, I think you get a counterexample on the dual graph. Have the dual ant be on the next edge along which a (non-dual) ant will pass through the given vertex.
* It feels like there might be a very slick solution using the hairy ball theorem.

Answer: 
I want to use the hairy ball theorem in the following way. Assume for simplicity that we start at time 0 and each ant returns to its original location at time 1. Suppose no two ants ever meet; then by compactness there is ε > 0 such that no two ants are pass through the same point at times closer than ε. Let's deform each face slightly inward near its vertices so that it becomes a smooth embedding of a disc into S^2. Also assume that ants move along the boundary with positive speed (so that time defines an homeomorphism from S^1 = [0, 1] / {0, 1} to the boundary of this disc).

For each face define a vector field on the boundary by p -> Rt(p)v(p), where v(p) is the unit tangent vector to the disc at p, t(p) is the time that the ant is at point p, and Ra is rotation (clockwise around the outward vector) by 2πa radians. This vector field extends over the disc. Extend it to the rest of the sphere by making it drop off to 0 rapidly outside the disc.

The sum of all these vector fields is nonzero everywhere except possibly near the vertices, because along any edge, the angle between the vectors coming from the two faces is bounded below in terms of ε. If one could control what happens near the vertices, this would contradict the hairy ball theorem.
Say there are n faces. The phase space on which the ants are moving are on is T^n, and the ants movement is described by a curve on T^n that has speed bounded from below in any direction. Hence, it suffices to show that if you cut T^n along the collision loci, you reduce the dimension of the H^1. This is a purely combinatiorial claim, which (it seems - I did not check all the details) you can prove by induction.