What are the rules for a Tetartoid pentagon?

As explained in this page a tetartoid can be built starting from a tetrahedron, as follows. Take a tetrahedron of unit edges and vertices $V_1$, $V_2$, $V_3$ and $V_4$. Upon every edge $V_iV_j$ construct two points $P_{ij}$ and $P_{ji}$ having a fixed distance $s$ from $V_i$ and $V_j$ respectively ($0\le s\le1/2$). Join the center $C_{ijk}$ of every face $V_iV_jV_k$ (vertices taken counterclockwise) with $P_{ij}$, $P_{jk}$ and $P_{ki}$, so that every face is divided into three quadrilaterals (12 of them in the whole tetrahedron). But notice that every quadrilateral $C_{ijk}P_{ij}V_jP_{jk}$ can be also seen as a degenerate pentagon $C_{ijk}P_{ij}P_{ji}V_jP_{jk}$, because vertex $P_{ji}$ lies between $P_{ij}$ and $V_j$.

Let now $O$ be the tetrahedron center and consider the six planes $OV_iV_j$ passing through the center and each edge. From point $P_{ij}$ draw the perpendicular line to $OV_iV_j$ and take on it a point $Q_{ij}$ such that $P_{ij}Q_{ij}=t$, where $t$ is another fixed parameter. Do the same for $P_{ji}$ but choose point $Q_{ji}$ so that it lies on the opposite side of the edge $V_iV_j$ with respect to $Q_{ij}$. The orientation of the three points $Q_{ij}$ sitting around the same vertex must be the same for every vertex.

Every pentagon $C_{ijk}P_{ij}P_{ji}V_jP_{jk}$ can be then transformed into a (non-degenerate) pentagon $C'_{ijk}Q_{ij}Q_{ji}V'_jQ_{jk}$, where $C'_{ijk}$ and $V'_j$ are the intersections of lines $OC_{ijk}$ and $OV_j$ respectively with plane $Q_{ij}Q_{ji}Q_{jk}$. These twelve pentagons are precisely the faces of a tetartoid, which is parameterized by $s$ and $t$.

So, one can compute the lengths of the sides and the angles of the faces as a function of $s$ and $t$ and that should be the answer to your question. I've found the explicit expressions with a symbolic computation software: they are quite complex, but if needed I can write them here.

ADDENDA.

Here's a GeoGebra file I've made, showing a tetartoid generated from a tetrahedron: http://tube.geogebra.org/material/show/id/1481233 .

And here are the values of $a^2$, $b^2$, $c^2$, $\cos\beta$ and $\cos\gamma$ as a function of parameters $s$ and $t$, where $\beta=\angle bb$ and $\gamma=\angle cc$. I chose to present $a$ instead of $\angle bc$ because it has a much simpler expression.

b^2 = ((1 + 3 (-1 + s) s + (-1 + t) t) (3 (1 - 2 s)^2 s^2 + 
     2 (1 - 2 s) s t + (3 + 4 s (-3 + 4 s)) t^2 - 4 t^3 + 
     4 t^4))/(-3 s + 6 s^2 + t (-1 + 2 t))^2

c^2 = ((s^2 + 3 t^2) (-4 s^3 + 4 s^4 + 2 s (1 - 6 t) t + 
     t^2 (5 + 12 (-1 + t) t) + 
     s^2 (1 + 4 t (-1 + 4 t))))/(s (-1 + 2 s) - 3 t + 6 t^2)^2

a^2 = 1 - 4 s + 4 s^2 + 4 t^2;

cos beta = (-3 (1 - 2 s)^2 s^2 + 
   2 s (-1 + 2 s) t + (5 + 4 s (-3 + 2 s)) t^2 - 4 t^3 + 4 t^4)/(
  6 (1 - 2 s)^2 s^2 + 4 (1 - 2 s) s t + 2 (3 + 4 s (-3 + 4 s)) t^2 - 
   8 t^3 + 8 t^4)

cos gamma = ((s - t) (s + 4 s^3 + 4 s^2 (-1 + t) + 12 s t^2 + 
     t (-1 + 12 (-1 + t) t)))/(-8 s^3 + 8 s^4 + 4 s (1 - 6 t) t + 
   2 t^2 (5 + 12 (-1 + t) t) + s^2 (2 + 8 t (-1 + 4 t)))

Managed to solve this. Choose numbers $0\le a\le b\le c$. Calculate
$n = a^2 c - b c^2 ,$
$d_1 = a^2 - a b + b^2 + a c - 2 b c ,$
$d_2 = a^2 + a b + b^2 - a c - 2 b c .$

Then, if $n \times d_1 \times d_2 \ne 0$,

$$((a, b, c),(-a, -b, c),(-n, -n, n)/d_1, (-c, -a, b),(-n, n, n)/d_2)$$

gives a tetartoid pentagon which can be multiplied by the tetrahedral group to give the full figure. For example, (4,8,20) generates

$$((4, 8, 20),(-4, -8, 20),(-15, -15, 15),(-20, -4, 8),(-10, 10, 10))$$