Bugs walking in a plane

The counterexample I thought I had here doesn't work.

Here is a proof. Since none of the bugs are moving in the same direction, any pair of lines determined by the velocity vectors intersect. Let $C$ denote the convex hull of these intersection points. Since after waiting a sufficiently long period of time the bugs will be arbitrarily far away from $C$, if we "zoom out" far enough $C$ will become arbitrarily small with respect to the convex hull of the location of the bugs. It follows that we can assume that $C$ is arbitrarily small to begin with.

We now claim that the bugs eventually form a convex polygon in which the angle at each vertex is strictly less than $\pi$. To do this it suffices to examine a configuration of three bugs $a, b, c$ in consecutive counterclockwise order. Pick a coordinate system in which the centroid of $C$ is the origin and $b$ travels in the positive $y$-direction (hence $a$ travels to the right and $c$ travels to the left). Then it is easy to see that regardless of where $a, b, c$ initially begin along their routes, $b$ will eventually have $y$-coordinate greater than either $a$ or $c$, so angle $abc$ will eventually be strictly less than $\pi$.

It follows that by waiting sufficiently long the bugs will always form a convex polygon. In fact, the bugs are approximating the convex polygon whose vertices are the unit velocity vectors of the bugs.

Assuming no bugs get squashed in the process:

At $\lim_{t\to\infty}$ when $t$ is time, the bugs' beginning points shrink to $\frac{P}{t} = 0$ as observed when zoomed out. This means the final position of each bug lies on $\sqrt{r^2 + (vt)^2}$, or on the edge of a huge circle. Thus, you can see that all the bugs make a convex polygon (or $N$-gon) since a circle can be thought of as a regular (and convex) $\infty$-gon.

Of course this isn't the most vigorous proof in the world, but it's written in the style that most humans can understand.