Suppose the nasty foreign runner runs along the curve $y=x^2$, so their position at time $t$ is $(t,t^2)$. Of course this supposes that the runner can run infinitely fast, but that's not the point!

Meanwhile the foreign runner is at $(\frac{1}{4}\tan \theta ,\frac{1}{4}{\tan }^2\theta )$.

Suppose we identify the plane with $C$, so the inner runner's path is specified by the continuously differentiable function $\gamma :[0,T]\to C$.

Then if the two runners are always level, the position of the second runner at time $t$ is $\gamma (t)-i{e}^{i\theta (t)}$ where $\theta (t)=$ argument of $\gamma \text{'}(t)$.

So the velocity (a 2-vector identified with a complex number) of the second runner at time $t$ is

$\gamma \text{'}(t)+\theta \text{'}(t)e^{i\theta (t)}$.

These two components are now parallel, since $\theta (t)$ and hence $\theta \text{'}(t)$ is real. (We probably need to assume $\theta (t)$ is increasing here.)

So thespeed is just $|\gamma \text{'}(t)|+|\theta \text{'}(t)|$ and the integral of this along the curve is the length of the first runner's pathplus the change in $\theta (t)$.

Oh dear. Now I'd bettergo and look for the mistake.

We can get away with assuming that $\gamma$ is once continuously differentiable everywhere and twice differentiable except at finitely many points. This certainly covers all designs I've seen for stadium tracks.

We also need to assume that at points where it reverses its curvature the radius of curvature is at least 1, or the second runner ends up running backwards, which is rather unrealistic.

I think it may be possible to get away with an arbitrary once continuously differentiable curve $\gamma$ by approximating it with a sequence of curves ${\gamma }_n$, each of which is twice continuously differentiable, and such that ${\gamma }_n\to \gamma$ and ${\gamma }_n\text{'}\to \gamma \text{'}$ (both uniformly). Neither of these alone implies the other, which is why George's piecewise-linear approximation argument doesn't quite work as far as I can see (it's at least not obvious to me that the second should occur in this case). I'm fairly sure that such a sequence always exists.