Interesting. I hadn't looked this up before. But it seems from here that:
There's a general way to convert from ground to wheel and vice versa. If you have a wheel in polar co-ordinates $(\rho,\theta)$ and ground in cartesian co-ordinates $(x,y)$, you can convert from one to the other:
Wheel $\rightarrow$ Ground: $y = \rho$ and $x=\int\rho d\theta$.
Ground $\rightarrow$ Wheel: $\rho = y$ and $\theta=\int\frac{1}{y}dx$
This gives the general idea, and it's little more than a math exercise after this.
- Driving a triangle:
Wheel is $\rho=1/\cos(\theta)$ for $\pi/3\leq\theta\leq\pi/3$ (for a suitably scaled triangle). So you can parametricize the curve with $y = 1/\cos(\theta)$ and $x = \ln(\sec(\theta)+\tan(\theta))$. I don't know if that's a special curve. The square is a repeating inverted catenary. This one looks similar. Obviously the angle at the bottom should meet at $\pi/3$ instead of $\pi/4$, but it's qualitatively similar. It's here.

- Driving on triangles:
Your ground is $y=x$ (+ reflections and periodicity). So you get $\rho=e^\theta$. Do that for $0\leq\theta\leq\pi/4$ and then reflect and repeat. You'll end up with a four-pointed star, something like this:

- Inverted cycloids:
I'll have to look up the formula for a cycloid, but it will be something like a cardioid. Edit. Oh, yeah, so the math works out neatly. Cycloid is $x=(\theta-\sin(\theta))$ and $y=(1-\cos(\theta))$ (actually parametricized by $t$, but I'm cheating a bit because I know how it will work out). So the wheel is $\rho=(1-cos(\theta))$ and $\theta=\int\frac{1}{1-\cos(\theta)}dx = \theta$. Justifying my cheat. So it ends up as $\rho=(1-cos(\theta))$:
