Walking around Santa Cruz, track around the soccer field I was recently walking around the track at UCSC, and I noticed that the track didn't always curve inward. Sometimes it curved the other way. Compare this (A convex track): http://www.irelandaerialphotography.com/dr_f2_3838.jpg to this (A concave track): http://www.motoyard.com/trackday/images/track-chuckwalla-l.gif . In the latter, there are portions of the track where the outer edge ( the edge bordering the infinite face, if this was a planar graph) is longer than the inner edge (the edge bordering the inner field). In a convex track, the outer edge is always longer than the inner edge. But is it possible to construct a (concave) track such that the inner edge is the same length as the outer edge? If not, how does one prove that it is impossible? The track must have a width greater than zero. (If the width is zero, the inner and outer edge will be one and the same, and the question is meaningless.) As an extension, is it possible to construct a track such that all paths along it are of the same length? A path is defined as the set of all points equidistant from the outside edge all the way around the track and equidistant from the inside edge all the way around the track, though these two distances do not have to be equal.
 A: Here's an answer that only works only on a technicality. I think the formulation of the question should rule out this answer. Draw a wiggly closed path, then draw a less wiggly closed path around it which has the same length. It should look something like this:
   __________
 /            \
|  /\/\/\/\/\_ |
|  \_/\/\/\/\/ |
 \____________/

The inside and outside boundaries have the same length. Moreover, every "path" has the same length simply because there are no paths:

A path is defined as the set of all points equidistant from the outside edge all the way around the track and equidistant from the inside edge all the way around the track, though these two distances do not have to be equal.

A: Let $s\mapsto(x(s), y(s))$, $0\leq s\leq L(\Gamma)$, be a simply closed  curve $\Gamma$ parametrized by arc length. Then the parallel curve $\Gamma_\epsilon$ at distance $\epsilon$ from $\Gamma$ has the parametric representation $s\mapsto (u(s),v(s))$ with  $$u(s)=x(s)-\epsilon\dot y(s),\quad v(s)=y(s)+\epsilon \dot x(s).$$
Here $\epsilon$ may have either sign; I omit the discussion of this point. It follows from Frenet's formula that the line element of $\Gamma_\epsilon$ is given by $d\sigma=(1-\epsilon\kappa)ds$, where  $\kappa$ denotes the curvature of $\Gamma$ and we assume  $\epsilon\kappa(s)<1$ for all 
$s$. Integrating we obtain
$$L(\Gamma_\epsilon)=L(\Gamma)-\epsilon\int_0^{L(\Gamma)}\kappa(s)ds = L(\Gamma)-2\pi \epsilon,$$
the latter equation following from the fact that the total curvature of a Jordan curve in the plane is $2\pi$ (up to sign).
A: This may be enough to get you started.
In building a model railroad track, to make a complete circle, I need 12 curved pieces.
Assuming I want a more elaborate layout, but want to keep things on the same level and
allow no branches or crossovers ( i.e. I have only straight and curved pieces and no
risers to lift track off my flat building surface), I need many curved pieces.  If
I start at one place, go counterclockwise around the track, and count pieces that
curve left and pieces that curve right, I get 12 more pieces that curve left than curve
right after I complete one loop and return to my starting place.
Is your scenario similar to building railroad track?
Gerhard "Ask Me About System Design" Paseman, 2010.08.15
A: One possible formulation is to consider a smooth closed curve $\Gamma$ and to define the track as the Minkowski sum of $\Gamma$ and the circle of radius $r$. If $\Gamma$ is nice enough, so that the curvature at any point doesn't exceed $1/r$, then the outer boundary has length equal to the length of the inner boundary plus $4\pi r$. (That is, the perimeter of the Minkowski sum of two regions is equal to the sum of the individual perimeters in most nice cases.)
A: If you afford to use a bridge or a crossing, a figure eight track can at the same time: be (mostly in case of a bridge) horizontal, and have the same length for its two edges.
A: Allow self-intersections.  Take the symmetric figure eight.
That does the trick. And it has the bonus that each half is convex. 
This is what Kloeckner might have been trying to say. 
(  As a kid, we used to drive model
slot cars on these. I imagine you can find such tracks raced by real cars. )
A: I think what you're coming up against is the simplest version of the Gauss–Bonnet theorem. 
Say we have a reference runner running around the inside track.  A runner next to her will have to speed up or slow down to keep pace, depending on "how much the track is turning", a parameter known as the 'curvature' of a plane curve.  You can get the total distance of the nearby lane by integrating the curvature over the entire track (this is always an integer multiple of $2\pi$ — see Total curvature) and multiplying by the width of the lane.  So if your track doesn't intersect itself then you get total curvature $2\pi$.   In short, you can never get away from this phenomenon, and in fact the shape of the track doesn't matter at all. 
In 3 dimensions you can fix this of course, by running on the surface of a cylinder, or more interestingly, you can electrically charge your runners and shoot them in a tokamak fusion reactor - then you will finally get a fair race.
