[Math] What are the rules for a Tetartoid pentagon

Question

The tetartoid (also tetragonal pentagonal dodecahedron, pentagon-tritetrahedron, and tetrahedric pentagon dodecahedron) is a dodecahedron with chiral tetrahedral symmetry. It has twelve identical pentagonal faces. It has edges of the form a b b c c. The pentagon itself can be completely determined by the sides b&c and the angles bb bc cc. The last side and other two angles are then determined. But that doesn't not ensure the tetartoid can be built.

What are the rules for a tetartoid pentagon?

As mentioned below, I 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.

I've put together a Mathematica Demonstration for this, The Tetartoid.

Answer 1

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)))