# Symmetric Circle Configurations from Regular Skeletal Polyhedra
## 5 E-Constructed Point-Circle Configurations
### Chiral Polyhedra and the E-Construction  
Chiral polyhedra do not lend themselves to point-circle configurations via the E-construction as the coplanarity condition fails in each case. This is remarkable, as chiral polyhedra are almost regular, failing regularity just by a factor of 2. In a way, this seems to indicate that the highly symmetric circle configurations arising from regular skeletal polyhedra are somewhat exceptional among E-constructed configurations. The chiral polyhedra belong to six two-parameter families of polyhedra, each of which also includes exactly two regular polyhedra parameterized in the same way as the chiral polyhedra. In three families, one regular polyhedron is finite; in the other three, both regular polyhedra are infinite. Thus, altogether, the six families contain nine infinite regular polyhedra, as mentioned in the previous section. Checking coplanarity for these six families is not entirely straightforward as the two parameters are integral or real. By way of example, we explain the procedure for the family of polyhedra \(\{P(a,b)\}_{a,b}\) of type \(\{6,6\}\), with integral parameters \(a,b\), relatively prime, in which the chiral polyhedra have skew hexagonal faces and vertex figures ([18] (Section 5)). The symmetry group of a polyhedron \(Q:=P(a,b)\) is generated by  
\[ \begin{array}{l}s_{1}\colon(x_{1},x_{2},x_{3})\mapsto(-x_{2},x_{3},x_{1})+(0,-b,-a),\\ s_{2}\colon(x_{1},x_{2},x_{3})\mapsto(-x_{3},x_{1},x_{2}).\end{array} \]  
Figure 14: Two views of the circle configurations for \(\{4,6\,|\,4\}\). The figure on the right shows how a \(2\times 2\times 2\) block of cubes of the underlying cubical tessellation contributes to the circle configuration: six of the eight cubes contribute two circles per cube, and the other two do not contribute a circle. Here, \(s_{1}\) is a rotatory reflection (a rotation about an axis followed by a reflection in a plane perpendicular to the axis) that cyclically permutes the vertices of the hexagonal base face \(f\). The generator \(s_{2}\) is a rotatory reflection that, when applied to the vertex \(u\) in the base edge \(e\) distinct from the base vertex \(v\), produces the hexagonal vertex figure at \(v\); in the present case, \(v=o\). The symmetry \(t:=s_{1}s_{2}\) is given by  
\[t\colon(x_{1},x_{2},x_{3})\mapsto(-x_{1},x_{2},-x_{3})+(a,0,b)\]  
and is a half-turn whose rotation axis passes through the midpoint of \(e\) and is perpendicular to \(e\). The vertices, edges, and faces of the polyhedron are the images of \(v\), \(e\), and \(f\) under the group \(G(Q)\). Note that \(e=\{o,u\}\) and \(u=t(o)=(a,0,b)\). The vertices of the vertex figure at \(o\) are given by \(s_{2}^{i}(u)\) with \(j=0,\ldots,5\); of these six vertices, only \(s_{2}(u)=(-b,-a,0)\) and \(s_{2}^{-1}(u)=(0,-b,-a)=s_{1}(o)\) are points in the block \(N(e)\). The two remaining points in \(N(e)\) are the images of the latter two under \(t\), namely, \((a+b,-a,b)\) and \((a,-b,a+b)\), respectively. Thus  
\[N(e)=\{(-b,-a,0),(0,-b,-a),(a+b,-a,b),(a,-b,a+b)\}.\]  
For a point \(x=(x_{1},x_{2},x_{3})\) in \(\mathbb{E}^{3}\) define the vector \(\hat{x}\) in \(\mathbb{E}^{4}\) by \(\hat{x}:=(x_{1},x_{2},x_{3},1)\). Then it is clear that four points \(v_{1},\ldots,v_{4}\) in \(\mathbb{E}^{3}\) are affinely dependent (that is, coplanar) if and only if the respective vectors \(\hat{v}_{1},\ldots,\hat{v}_{4}\) are linearly dependent in \(\mathbb{E}^{4}\) (that is, have vanishing determinant). For the four points in \(N(e)\) the determinant is given by  
\[\det \begin{pmatrix}-b&0&a+b&a\\ -a&-b&-a&-b\\ 0&-a&b&a+b\\ 1&1&1&1\end{pmatrix} =2a^{3}+2a^{2}b-2 ab^{2}-2b^{3}=2(a+b)^{2}(a-b)\,.\]  
Thus, the determinant is \(0\) if and only if \(a=\pm b\). It follows that \(P(a,b)\) satisfies the coplanarity condition if and only if \(a=\pm b\). These parameter values characterize precisely the two regular polyhedra in the infinite family of polyhedra \(\{P(a,b)\}_{a,b}\), namely \(\{6,6\mid 3\}\) and \(\{6,6\}_{4}\).