I am trying to describe a Mobius band in the shape of a triangle like this:
parametrically in terms of its $x$, $y$, and $z$ functions. Is this even possible?
I know a basic mobius strip can be represented parametrically by
$$x=\left[R+s\cos\left(\frac 12t\right)\right]\cos t$$ $$y=\left[R+s\cos\left(\frac 12t\right)\right]\sin t$$ $$z=s\sin\left(\frac 12t\right)$$
but I would imagine the triangle shape would effect the equations quiet a bit..
any pointers in the right direction would be greatly appreciated.
In order to make the Moebius band $M$ smooth at the creases in your figure we have to take care that the tangent planes are vertical there. We begin by producing an S-shaped strip $S$ making up a third of the total surface. This strip will project vertically onto a trapezoid with two sides parallel to the $y$-axis and having width $2h$ in the $x$-direction. Choose $0<h<a$, and $c>0$, and put $$S:\quad(t,u)\mapsto {\bf r}(t,u):=\left(a+u,\ \sqrt{3}\left(a-{u\over3}\right)\sin{t\over2}, \ c\sin t\right)\qquad(-\pi\leq t\leq\pi, \ -h\leq u\leq h)\ .$$ The factor $\sin{t\over2}$ has derivative $0$ at $t=\pm\pi$ and will produce vertical tangent planes along the crease lines. The following figure shows this strip for the choices $a=4$, $h=1.5$, $c=1.5$:
The total band $M$ is the union of three congruent such strips, whereby the other two are obtained from $S$ through rotations of $\pm120^\circ$ around the $z$-axis: