# Symmetric Circle Configurations from Regular Skeletal Polyhedra
## 6 V-Constructed Point-Circle Configurations
If \(s(v)=v\), we are done since \(s\) fixes the base vertex; in particular, this settles the case when \(v=z\). Let \(v\
eq z\), and suppose that \(s(v)=\vartheta\). Then, \(\vartheta\) is a vertex of \(Q\) since \(s\in G(Q)\). Also, as \(s\) permutes the \(q\) points of \(C\) and the vertex figures of \(Q\) are congruent planar regular \(q\)-gons, the vertices \(v\) and \(\vartheta\) must share the same vertex figure. By the vertex-transitivity of \(G(Q)\), this \"bipyramidal structure\" must occur at every circle of the configuration, including the circle \(C^{\prime}\) associated with \(v^{\prime}\). As \(s\) fixes \(v^{\prime}\), it maps edge \(e=\{v^{\prime},v\}\) of \(Q\) to an edge, namely \(\{v^{\prime},\vartheta\}\), and both edges connect \(v^{\prime}\) to points in \(C^{\prime}\). Next, since \(s\) also fixes \(z\), it must leave the line \(l^{\prime}\) through \(v^{\prime},z\) pointwise fixed and induce a plane isometry on the plane \(H^{\prime}\) perpendicular to \(l^{\prime}\) at \(z\) that maps \(v\) to \(\vartheta\). This plane isometry has an invariant line, namely \(H\cap H^{\prime}\), and thus must be a half-turn about a point or a reflection in the invariant line; in the latter case, the reflection line must be the perpendicular bisector of \(v\) and \(\widehat{v}\) in \(H^{\prime}\). It follows that \(s\) itself must be a half-turn about \(l^{\prime}\) or the reflection in the plane \(H\), the latter being the perpendicular bisector of \(v\) and \(\vartheta\) in \(\mathbb{E}^{3}\). It follows that \(H\) and \(H^{\prime}\) are perpendicular, that \(H^{\prime}\) contains \(C^{\prime}\), and that \(z\) is the center of both \(C\) and \(C^{\prime}\). In particular, since \(v\) and \(\vartheta\) are opposite vertices in \(C^{\prime}\), the number \(q\) must be even. Thus \(q=4\) or \(q=6\). We show that the case \(q=4\) leads to the two exceptional cases mentioned in the theorem and that the case \(q=6\) can be eliminated. Suppose \(q=4\). Then the two points of \(C\cap C^{\prime}\) are vertices of \(Q\) and thus contribute points to \(C\) and \(C^{\prime}\); they also are connected to \(v^{\prime}\) by edges of \(Q\). It follows that \(Q\) must contain the entire 1-skeleton of the octahedron spanned by the six points of \(C\) and \(C^{\prime}\); in fact, as these six vertices already account for the full vertex valency, \(4\) (=\(q\)), the 1-skeleton of \(Q\) must coincide with the 1-skeleton of the octahedron. This situation can only occur if \(Q\) is an octahedron or the Petrie dual of the octahedron. In the case of the octahedron \(\{3,4\}\) and its Petrie dual \(\{6,4\}_{3}\), the above isometry \(s\) either is the reflection \(r_{C}\) in the plane \(H\) through \(C\) or the half-turn about \(l^{\prime}\); but the latter is just \(r_{C}r_{2}\). Note that both \(r_{C}\) and \(r_{C}r_{2}\) belong to \(G(Q)\), and both are possible choices for \(s\) for the octahedron or its Petrie dual. In either case, \(r_{C}s\) stabilizes the point-circle pair \((v^{\prime},C)\) and fixes the base vertex \(v\) and base edge \(e\). It follows that \(r_{C}s\in\langle r_{2}\rangle\). Thus the stabilizer of \((v^{\prime},C)\) in \(G(Q)\) is \(\langle r_{2},r_{C}\rangle\), which is isomorphic to \(C_{2}\times C_{2}\) as \(r_{2}\) and \(r_{C}\) commute. What remains is to reject the possibility that \(q=6\). No finite regular polyhedron has 6-valent vertices. Planar polyhedra and blends of planar polyhedra with a line segment cannot have bipyramidal structure of the above kind at the circles of the corresponding V-constructed point-circle configuration. Blends of planar polyhedra with an apeirogon all have skew vertex figures and can be discounted. This leaves the pure polyhedra of which only two have planar hexagonal vertex figures, namely \(\{6,6\}_{4}\) and its Petrie dual \(\{4,6\}_{6}\). In both cases, each vertex is the center of its vertex figure, and thus, we are in the case \(v=z\) already discussed above.