Abstract
A geometric \(k\)configuration is a collection of points and straight lines in the plane so that \(k\) points lie on each line and \(k\) lines pass through this point. We introduce a new construction method for constructing \(k\)configurations with nontrivial dihedral or chiral (i.e., purely rotational) symmetry, for any \(k \ge 3\); the configurations produced have \(2^{k2}\) symmetry classes of points and lines. The construction method produces the only known infinite class of symmetric geometric 7configurations, the second known infinite class of symmetric geometric 6configurations, and the only known 6configurations with chiral symmetry.
Similar content being viewed by others
Avoid common mistakes on your manuscript.
1 Introduction
A \(k\)configuration, also known as a \((n_{k})\) geometric configuration, is a collection of \(n\) points and \(n\) straight lines, typically in the Euclidean or projective plane, so that each point lies on \(k\) lines and each line passes through \(k\) points. Constructing new configurations by using geometric symmetries has proven to be a highly productive technique; see, e.g., [3, 7, 9, 10, 12]. However, most of the previous work on configurations has focused on constructions that produce 3 and 4configurations; relatively little work has been done on producing or classifying \(k\)configurations for \(k > 4\).
The configurations in this paper were initially found when one of the authors was trying to reconstruct the single known geometric (\(20_{4}\)) configuration, which was found by Branko Grünbaum a few years ago. The first published diagram of the configuration appeared in 2008, in [11]; that diagram is reproduced as our Fig.1. In that article, he very briefly described his construction technique, saying “What I first managed to do was to construct a \((20_{4})\) configuration of pseudolines, by merging two \((10_{3})\) configurations. Looking at the result, I wondered whether one could straighten the pseudolines. As it turned out, this is possible...”
The construction outlined in this paper is a significantly different construction from the one used by Grünbaum. In fact, his configuration is the first in an infinite family of configurations. Not only is the configuration generalizable to other dihedrally symmetric 4configurations (which Grünbaum had noticed very soon after the previous discovery (private communication, and also discussed in [12, Chap. 3])), but the construction algorithm described in this paper also extends to produce chirally symmetric 4configurations, and to \(k\)configurations for arbitrary \(k\). The construction technique, with several workedout examples, will be described in the next section, and subsequent sections will prove the validity of the technique.
In particular, this new construction technique produces the first known 6configurations with chiral symmetry (the infinite family of 6configurations constructed in [4, 5] have dihedral symmetry) and the first known infinite class of symmetric 7configurations at all.
2 The Configuration Construction Algorithm
Throughout this paper, the notation \(\mathcal{C }_{O}(Q)\) refers to a circle centered at point \(O\) and passing through point \(Q\). The notation \(\mathcal{C }(P,Q,R)\) refers to the circumcircle of points \(P\), \(Q\), and \(R\). The line between points \(P\) and \(Q\) is denoted by either \(P \vee Q\) or \(\overline{PQ}\), and the intersection of lines \(\ell \) and \(m\) is denoted \(\ell \wedge m\).
Definition 1
Given a regular convex \(m\)gon of points cyclically labelled as \(w_{i}\), a line of span \(s\) is any line \(w_{i}\vee w_{i+s}\).
Definition 2
Given a regular convex \(m\)gon of points centered at \(O\), cyclically labelled as \(w_{i}\), the circle of span \(s\) through \(w_{j}\) is the circumcircle \(\mathcal{C }(w_{j}, O, w_{js})\); note that the line \(w_{js}\vee w_{j}\) is a span \(s\) line.
Definition 3
Given a regular convex \(m\)gon of points centered at \(\mathcal{O }\), cyclically labelled as \(w_{i}\), and integers \(a\), \(b\) with \( 1 \le a, b \le \lfloor \tfrac{m}{2}\rfloor \), an integer \(d\) is welldefined if the circle \(\mathcal{C }(w_{d}, \mathcal{O }, w_{db})\) intersects the line \(w_{0} \vee w_{a}\) in at least one point.
The algorithm for the construction of a configuration with symbol \(\mathcal{A }(m; a,b; d_{1}, d_{2}, \ldots , d_{k2})\) is iterative. The parameters \(m,a,b\) and \(k\) are all positive integers, with \(1 \le a,b< \tfrac{m}{2}, k\ge 3; d_1,\ldots ,d_{k2}\) are integers between \(0\) and \(m1\), modulo \(m\). A number of nontrivial constraints are required to ensure the incidence structure produced by the algorithm is a configuration. A detailed discussion of parameters occurs in Sect. 6. The configurations \(\mathcal{A }(m; a,b; d)\) were discussed by Grünbaum in [12, Sect. 2.7] as a class of chiral astral \(3\)configurations; that is, as 3configurations with chiral geometric symmetries and precisely two symmetry classes of points and lines. In that reference, they were denoted \(m\#(b,c; d)\).
2.1 The Construction Algorithm
To construct \(\mathcal{A }(m; a,b; d_{1}, \ldots , d_{k2})\) where each \(d_{j}\) is a welldefined parameter, let \(N = k2\), and define \(S = \{1, 2, \ldots , N\}\) to be the index set associated with \(\{d_{1}, \ldots , d_{N}\}\). For each subset \(\sigma \subseteq S\), we will construct a symmetry class of points and a symmetry class of lines indexed by \(\sigma \). All indices are taken modulo \(m\).

Step 0: Construct a regular convex \(m\)gon with center \(\mathcal{O }\), and label the points cyclically as
$$\begin{aligned} (v)_{0}, (v)_{1}, \ldots ,(v)_{m1}. \end{aligned}$$Here, we will interpret \((v)_{i}\) as \((v_{\emptyset })_{i}\), but for notational convenience, we suppress the \(\emptyset \) subscript. Then construct span \(a\) lines \((L)_{i} = (v)_{i} \vee (v)_{i+a}\). (Again, by \((L)_{i}\), we mean \((L_{\emptyset })_{i}\).)

Step 1: For each \(d_{j}\), construct the span \(b\) circle \(c_{j}:=\mathcal{C }((v)_{d_{j}}, \mathcal{O }, (v)_{d_{j}b})\). For each \(c_{j}\), let \((v_j)_0\) be a point of intersection of \(c_{j}\) and \((L_{0})_{0}\); since the parameter \(d_{j}\) is welldefined, such an intersection point must exist. Again, we interpret \(v_j\) as \(v_{\{j\}}\) but suppress the brackets for convenience. Define \((v_j)_i\) to be the rotation of \((v_j)_0\) by \(\tfrac{2\pi i}{m}\) about \(\mathcal{O }\). Then construct span \(b\) lines \((L_{j})_{i} = (v_{j})_{i} \vee (v_{j})_{i+b}.\)

Step \(n\), for \(1<n\le k2\): Let \(\sigma \) be any subset of \(S\) of size \(n\). For any \(q\) in \(\sigma \), define \(\sigma _{q}\) to be the set \(\sigma \setminus \{q\}\). Construct vertex \((v_{\sigma })_{0}\) to be the common intersection of all lines \((L_{\sigma _{q}})_{0}\) for each \(q \in \sigma .\) (It remains to show that this vertex \((v_{\sigma })_{0}\) is welldefined.) Define \((v_{\sigma })_{i}\) to be the rotation of \((v_{\sigma })_{0}\) by \(\tfrac{2 \pi i}{m}\) about \(\mathcal{O }\). Next, define \((L_{\sigma })_{i} := (v_{\sigma })_{i} \vee (v_{\sigma })_{i+\delta _{n}}\), where
$$\begin{aligned} \delta _{n} = \Big \{\begin{array}{ll} a &{} \text{ if } n \text{ is } \text{ even };\\ b &{} \text{ if } n \text{ is } \text{ odd }. \end{array} \end{aligned}$$
Definition 4
A welldefined parameter \(d_{j}\) is valid if the circle \(c_{j}\) intersects line \((L)_{0}\) at a point which does not lie on any \((L)_{t}\) other than \((L)_{0}\).
In Sect. 4, we show that the points \((v_{\sigma })_{i}\) are welldefined and discuss properties of the lines \((L_{\sigma })_{i}\). In Sect. 5, we show that, given valid parameters, the algorithm produces a \(k\)configuration with \(2^{k2}\) symmetry classes of points and lines. A discussion of parameters follows in Sect. 6.
Note that for each set of discrete parameters, there may be multiple configurations corresponding to the same symbol, depending on which intersection of \(c_{j}\) with \((L)_{0}\) is chosen; we indicated which intersection should be used by the use of primes (or not) on the \(d_{j}\). Intuitively, \(d_{j}^{\prime }\) indicates that we choose the leftmost intersection, while \(d_{j}\) means that we choose the rightmost intersection. More formally, if we parameterize \((L)_{0}\) as \((L)_{0} = (1t) (v)_{0} + t (v)_{a}\) and \(c_{j}\) intersects \((L)_{0}\) at two distinct points associated with parameter values \(t = \lambda _{j}, \lambda _{j}^{\prime }\) with \(\lambda _{j} < \lambda _{j}^{\prime }\), then the primed notation \(d_{j}^{\prime }\) indicates that we choose the intersection point \((v_{1})_{0} = (1\lambda _{j}^{\prime }) (v)_{0} + \lambda _{j}^{\prime }(v)_{a}\), while the unprimed notation \(d_{j}\) indicates that we choose \((v_{1})_{0} = (1\lambda _{j}) (v)_{0} + \lambda _{j}(v)_{a}\). For example, Fig. 5 shows the configuration \(\mathcal{A }(9;4,2;1,3^{\prime })\), which is not geometrically congruent to the configuration \(\mathcal{A }(9; 4,2; 1^{\prime }, 3^{\prime })\), although the two configurations are combinatorially isomorphic; the construction of configuration \(\mathcal{A }(9; 4,2; 1^{\prime }, 3^{\prime })\) is described carefully in Example 2 and shown in Fig. 4.
2.2 Three Examples
Here, we outline the steps in constructing the 3configuration \(\mathcal{A }(5;2,2;1^{\prime })\), the 4configuration \(\mathcal{A }(9; 4,2; 1^{\prime }, 3^{\prime })\), and the 5configuration \(\mathcal{A }(10;3,3;1,2,4)\).
Example 1
The 3configuration \(\mathcal{A }(5; 2,2; 1^{\prime })\).
In this configuration, \(m = 5\), \(a = b = 2\), and \(d_{1} = 1^{\prime }\); the configuration is shown in Fig. 2. To construct the configuration, begin with the vertices of a regular 5gon centered at \(\mathcal{O }\), labelled consecutively as \((v)_{i}\), and construct lines of span \(2\) with respect to these points: that is, construct the lines \((L)_{i}:= (v)_{i} \wedge (v)_{i+2}\).
Secondly, since \(d_{1} = 1\), construct the circle \(\mathcal{C }_1\) passing through \((v)_{1}\), \(\mathcal{O }\), and \((v)_{12} = (v)_{1} = (v)_{4}\). This circle intersects line \((L)_{0}\) in two places. Figure 2 shows both possibilities. Since the symbol \(d_{1}\) is primed, choose the leftmost intersection point, which is labelled as \((v_1)_0.\) Next, the points \((v_1)_i\) are constructed as the images under rotation by \(\tfrac{2 \pi i}{m}\) of \((v_1)_0.\) Adding lines \((L_1)_i=(v_1)_i \vee (v_1)_{i+2}\) of span 2 completes the construction.
Notice that in this case, the two configurations \(\mathcal{A }(5; 2,2; 1)\) and \(\mathcal{A }(5; 2,2; 1^{\prime })\) are geometrically similar (as well as combinatorially isomorphic), although the labels and colors are different.
Example 2
The 4configuration \(\mathcal{A }(9; 4,2; 1^{\prime }, 3^{\prime })\).
Here, since \(N = 2\), we will be constructing a 4configuration; the various steps in the construction are illustrated in Fig. 3, and the final configuration is shown in Fig. 4. Note that \(m =9, a = 4, b = 2, d_{1} = 1^{\prime }\), and \(d_{2} = 3^{\prime }\).
Since \(m = 9\), begin by constructing the vertices \((v)_{i}\) of a regular convex \(9\)gon, colored black. The vertex \((v)_{0}\) is shown larger and black in Fig. 3a. Next, since \(a = 4\), construct lines \((L)_{i}\) of span \(4\) with respect to these vertices. The line \((L)_{0}\) is shown thick and black in Fig. 3b.
Now construct two circumcircles of span \(b = 2\) passing through vertices \((v)_{1}\) (since \(d_{1} = 1\)) and \((v)_{3}\) (since \(d_{2} = 3\)), respectively. That is, construct the circles \(c_{1} = \mathcal{C }((v)_{1}, \mathcal{O }, (v)_{1})\) (Fig. 3c) and \(c_{2} = \mathcal{C }((v)_{3}, \mathcal{O }, (v)_{1})\) (Fig. 3d). To construct the next two classes of points, we look to see if the \(d_{i}\) are primed. Both are, so we will choose the leftmost intersection of circle \(c_{j}\) with line \((L)_{0}\) to define as \((v_{j})_{0}\). The points \((v_{1})_{0}\) and \((v_{2})_{0}\) are colored blue and red, respectively, and both are shown larger. The points \((v_{j})_{i}\) are constructed as the images under rotation by \(\tfrac{2 \pi i}{m}\) of \((v_{j})_{0}\) (Fig. 3c, d).
Now, we construct the lines \((L_{1})_{i}\) of span \(b = 2\) with respect to the points \((v_{1})_{i}\) and the lines \((L_{2})_{i}\) of span \(2\) with respect to the points \((v_{2})_{i}\) (Fig. 3e).
Next, we construct the point \((v_{12})_{0}\) (shown green in Fig. 3f) as the intersection of lines \((L_{1})_{0}\) and \((L_{2})_{0}\) and then construct the rotated images \((v_{12})_{i}\). When the line \((L_{12})_{0}\) of span \(a = 4\) is constructed, it passes through points \((v_{1})_{d_{2}}\) and \((v_{2})_{d_{1}}\). (This is not obvious and will be proved in Sect. 4.) The resulting incidence structure has four lines passing through each point and four points lying on each line, so it is a 4configuration, shown in Fig. 4. It has ninefold rotational symmetry. Note this configuration is not geometrically congruent to \(\mathcal A (9;4,2;1,3^{\prime })\), shown in Fig. 5, although it is combinatorially isomorphic.
Example 3
The 5configuration \(\mathcal{A }(10; 3,3; 1,2,4)\)
In this example, we construct the 5configuration \(\mathcal{A }(10; 3,3; 1,2,4)\); intermediate steps in the construction are shown in Fig. 6 and the completed configuration is shown in Fig. 7. As before, this configuration begins with the construction of the vertices of a regular \(m\)gon centered at \(\mathcal{O }\), where in this case, \(m = 10\). Since \(a = 3\), we construct all lines of span 3. Here, \(b = 3\) as well, and \(d_{1} = 1\), \(d_{2} = 2\), \(d_{3} = 4\), so we construct three circumcircles, each of span 3, beginning at the points \((v)_{1}\), \((v)_{2}\), \((v)_{4}\) respectively (shown in dark blue, medium blue, and light blue in Fig. 6a). Since none of the \(d_{j}\) are primed, we choose the rightmost intersection of each circle with line \((L)_{0}\), again using dark, medium and light blue to construct the points \((v_{1})_{0}\), \((v_{2})_{0}\), \((v_{3})_{0}\), respectively (also shown in Fig. 6a).
Next, the points \((v_i)_j\) are constructed along with lines of span \(b\) with respect to each of the \((v_{j})_{i}\). In Fig. 6b, the lines \((L_{1})_{0}\), \((L_{2})_{0}\), \((L_{3})_{0}\) are shown in dark, medium, and light blue, respectively. Also in Fig. 6b, we construct the point \((v_{12})_{0}\) (red) as the intersection of lines \((L_{1})_{0}\) and \((L_{2})_{0}\), the point \((v_{13})_{0}\) (orange) as the intersection of lines \((L_{1})_{0}\) and \((L_{3})_{0}\), and the point \((v_{23})_{0}\) (yellow) as the intersection of lines \((L_{2})_{0}\) and \((L_{3})_{0}\).
In Fig. 6c, we construct the rotated images \((L_{j})_{i}\) and \((v_{jk})_{i}\) by rotating appropriate elements by \(\tfrac{2 \pi i}{m}\) about \(\mathcal{O }\). Notice that the points \((v_{23})\) do not lie on lines \((L_{1})\), although they are very close.
In the next step of the construction, shown in Fig. 6d, we construct the lines \((L_{12})_{0}\), \((L_{13})_{0}\) and \((L_{23})_{0}\), where, for example, \((L_{12})_{0}\) is constructed to pass through \((v_{12})_{0}\) and \((v_{12})_{3}\), all shown in red. Notice that \((L_{12})_{0}\) passes through a dark blue point, which is in fact \((v_{1})_{2}\), and a medium blue point, \((v_{2})_{1}\), and recall that \(d_{1} = 1\) and \(d_{2} = 2\). That is, \((L_{12})_{0}\) passes through \((v_{1})_{d_{2}}\) and \((v_{2})_{d_{1}}\); similar relationships can be verified for \((L_{13})_{0}\) (orange) and \((L_{23})_{0}\) (yellow). Furthermore, notice that the three lines \((L_{12})_{0}\), \((L_{13})_{0}\), and \((L_{23})_{0}\) all intersect in a single point, which we will call \((v_{123})_{0}\), indicated in green.
In Fig. 7, we show the completed configuration, formed by first constructing the rotated images of the lines \((L_{12})_{0}\), \((L_{23})_{0}\) and \((L_{13})_{0}\) and the point \((v_{123})_{0}\). Finally, we construct one more symmetry class of lines, called \((L_{123})_{i}\), which are lines that pass through \((v_{123})_{i}\) and \((v_{123})_{i+3}\); notice that line \((L_{123})_{0}\) (green) passes through \((v_{12})_{4} = (v_{12})_{d_{3}}\) (red), \((v_{13})_{2} = (v_{13})_{d_{2}}\) (orange), and \((v_{23})_{1} = (v_{23})_{d_{1}}\) (yellow). There are a number of near misses, indicated by pink circles, where symmetry classes of points are almost, but not quite, incident with various symmetry classes of lines.
3 Preliminary Geometric Lemmas
To prove the validity of the construction, we will need several elementary geometric lemmas. The first two are wellknown results from elementary Euclidean Geometry (see, e.g., [1]). Recall that a cyclic quadrilateral is a quadrilateral whose vertices are concyclic—that is, they all lie on a circle.
Lemma 1
A quadrilateral \(ABCD\) is cyclic if and only if one pair of opposite angles is supplementary.
Lemma 2
(Cyclic Quad Lemma) Quadrilateral \(ABCD\) is cyclic if and only if \(\angle ABD = \angle ACD\).
We will also need the following:
Lemma 3
(Equal Spans Lemma) Let \(\mathcal{C }, \mathcal D \) be two concentric circles with center \(O\), and let \(u, u^{\prime }\) be on \(\mathcal{C }\). Let \(\angle uOu^{\prime } = \alpha \). Let \(R\) be any point on \(\mathcal{D }\) that does not lie on the line \(O \vee u,\) and construct \(R^{\prime }\) so that \(\angle ROR^{\prime }=\alpha \). Finally let \(Q\) be the intersection of \(\overline{uu^{\prime }}\) and \(\overline{RR^{\prime }}\). Then the four points \(u, R, Q, O\) are concyclic.
Proof
There are several cases to consider. First \(\mathcal{C }\) or \(\mathcal{D }\) may have larger radius. Second, either point \(R\) or \(R^{\prime }\) may be within the sector determined by \(uOu^{\prime }\) or both may be outside it. Finally, the point \(Q\) may occur on the line one side or the other of segment \(R \vee R^{\prime }\) or it may occur between points \(R\) and \(R^{\prime }\). (Similarly, for line \(u \vee u^{\prime }\).) All of these cases have effectively the same proof: an application of Lemma 1. We will give a detailed proof in one case, shown in Fig. 8, where \(\mathcal{D }\) is the smaller circle, \(R\) and \(R^{\prime }\) are both outside sector \(uOu^{\prime }\), and the point \(Q\) occurs past \(R\) and \(u^{\prime }\).
By construction, \(\triangle uOu^{\prime }\) and \(\triangle ROR^{\prime }\) are both isosceles with apex angle \(\alpha \), so the two triangles are similar; hence, their base angles are equal (shown blue doublemarked). Note that angle \(\angle ORQ\) (shown green triplemarked) is supplementary to \(\angle R^{\prime }RO\), since \(R^{\prime }, R, Q\) are collinear. Therefore, since all the base angles are equal, \( \angle ORQ\) is supplementary to \(\angle u^{\prime }uO\), which equals \(\angle QuO\) because \(Q, u^{\prime }, u\) are collinear. It follows that \(O, u, Q,R\) form a cyclic quadrilateral and hence are concyclic. \(\square \)
Grünbaum used a specialized case of the forward direction of the following lemma in [12, Sect. 2.8], in his discussion of a geometric construction for chirally symmetric astral 3configurations. In [6], one author [LWB] discusses ways to construct a number of infinite families of symmetric 3configurations using a modified version of the forward direction of the lemma.
Lemma 4
(Configuration Construction Lemma) Let \(R\) and \(R^{\prime }\) be two points on circle \(C_{O}(R)\) centered at \(O\), and let \(\angle ROR^{\prime } = \alpha \). If \(Q\) is any point on the circumcircle \( \mathcal{C }(R,O,R^{\prime })\) not on line \(O \vee R,\) and \(Q^{\prime }\) is on \(C_{O}(Q)\) so that \(\angle QOQ^{\prime } = \alpha \), then \(\overline{QQ^{\prime }}\) passes through \(R^{\prime }\). Conversely, if \(Q\) and \(Q^{\prime }\) are any two points on a circle centered at \(O\) so that \(\angle QOQ^{\prime } = \alpha \) and \(Q, R^{\prime }, Q^{\prime }\) are all collinear, then \(Q\) lies on the circumcircle \(\mathcal{C }(R,O,R^{\prime })\).
Proof
\([\implies ]\) Observe that \(C_O(R)\) and \(C_O(Q)\) are two concentric circles such that \(\angle ROR^{\prime } = \angle QOQ^{\prime }\) and \(Q\) does not lie on line \(O \vee R.\) Thus, the Equal Spans Lemma implies that the point \(P=\overline{QQ^{\prime }} \wedge \overline{RR^{\prime }}\) is concyclic with \(O,R,\) and \(Q.\) Since we assumed \(O,R\) and \(Q\) lie on \(C_O(R),\) the point \(P\) must also lie on \(C_O(R).\) Now, \(P\) lies on \(C_O(R)\) and on the line \(\overline{RR^{\prime }}\) which forces \(P=R^{\prime }.\) Thus, line \(\overline{QQ^{\prime }}\) passes through \(R^{\prime }.\)
\([\Longleftarrow ]\) Again, circles \(C_O(R)\) and \(C_O(Q)\) are concentric and \(\angle ROR^{\prime } = \angle QOQ^{\prime }.\) Further, we assume \(Q,\) \(Q^{\prime }\) and \(R^{\prime }\) are collinear. Hence, \(\overline{QQ^{\prime }} \wedge \overline{RR^{\prime }} = R^{\prime }.\) Thus, the Equal Spans Lemma implies that \(O,\) \(R\), \(Q\), and \(R^{\prime }\) are concyclic. Thus, \(Q\) must lie on \(C_O(R).\) \(\square \)
Lemma 5
(CircumcircleSpan Lemma) If points \(v_{1}, v_{2}, \ldots , v_{n}, O\) lie on a single circle \(\mathcal{C }\), and for each \(i\), \(v_{i}^{\prime }\) is constructed so that \(Ov_{i} = Ov_{i}^{\prime }\) and \(\angle v_{i}O v_{i}^{\prime } = \alpha \) for some fixed angle \(\alpha \), then the lines \(v_{i}\vee v_{i}^{\prime }\) all intersect in a single point. Moreover, this point lies on \(\mathcal{C }\).
Proof
By induction on \(n\). For each \(j\), define \(L_{j} = v_{j}\vee v_{j}^{\prime }\). Define \(Z\) as the intersection of \(L_{1}\) with \(\mathcal{C }\) other than \(v_{1}\) (see Fig. 9).
Base case: Clearly, \(L_{1}\) passes through \(Z\).
Induction hypothesis: Suppose that for all \(j\) with \(1 \le j \le n1\) that \(L_{j}\) passes through \(Z\). We need to show that \(L_{n}\) also passes through \(Z\).
By construction, \(v_{1}\) and \(v_{1}^{\prime }\) lie on a circle centered at \(O\), \(v_{n}\) and \(v_{n}^{\prime }\) lie on a different circle centered at \(O\), and \(\angle v_1 O V_1^{\prime } = \angle v_n O v_n^{\prime }.\) Thus, if \(Q\) is the intersection of \(L_{n}\) and \(L_{1}\), then the Equal Spans Lemma implies that \(v_{1}\), \(v_{n}\), \(O\) and \(Q\) are concyclic. But \(v_{1}, v_{n}\) and \(O\) already lie on \(\mathcal{C }\), and the other intersection of \(L_{1}\) and \(\mathcal{C }\) is \(Z\). Hence \(Z = Q\). \(\square \)
4 The Validity of the Construction
Theorem 6
For any \(k \ge 3\), given a positive integer \(m \ge 5\) and a choice of integers \(1 \le a, b < \tfrac{m}{2}\) and \(k2\) valid discrete parameters \(d_{1}, d_{2}, \ldots , d_{k2},\) it is possible to construct a symmetric \(k\)configuration \(\mathcal{A }(m; a,b; d_{1}, \ldots , d_{k2})\), with \(m(2^{k2})\) points and lines, either \(C_{m}\) or \(D_{m}\) symmetry, and \(2^{k2}\) symmetry classes of oints and lines.
To prove this, we will show the following results: (1) For each \(\sigma \subseteq S\), the vertex \((v_{\sigma })_{0}\) is welldefined; (2) for each \(\sigma \subseteq S,\) the line \((L_{\sigma })_{0}\) passes through the points \((v_{\sigma _{q}})_{d_{q}}\) for each \(q \in \sigma \) where \(\sigma _q=\sigma \setminus \{q\}\); (3) every point \((v_{\sigma })_{0}\) has \(k\) lines passing through it; (4) every line \((L_{\sigma })_{0}\) has \(k\) points lying on it; (5) the construction method produces a total of \(2^{k2}\) symmetry classes of points and lines; (6) the resulting configuration has the desired symmetry group (and thus the desired total number of points and lines).
Theorem 7
For each \(\sigma \subseteq S\), the vertex \((v_{\sigma })_{0}\) is welldefined: that is, it lies on the common intersection point of all lines \((L_{\sigma _{q}})_{0}\), for each \(q \in \sigma \). In addition, the set of points \(\{ (v_{\sigma _{q}})_{0}: q \in \sigma \} \cup \{\mathcal{O }, (v_{\sigma })_{0}\}\) are all concyclic, on a circle named \(\mathcal{C }_{\sigma }\).
Proof
By induction on \(n\), the size of the set \(\sigma \).

Base cases:

\(n = 0\): In this case, the vertex \((v_{\emptyset })_{0} = (v)_{0}\) is welldefined as the vertex of a regular convex \(m\)gon. In addition, there are infinitely many circles passing through \((v)_{0}\) and the origin.

\(n = 1\): Let \(\sigma =\{j\}=j\), for ease of notation. Each vertex \((v_{1})_{0}\), \((v_{2})_{0}\), ..., \((v_{N})_{0}\) is constructed to lie on the line \((L)_{0}\), so these vertices are well defined. In this case, the set \(\{ (v_{\sigma _{q}})_{0}: q \in \sigma \} \cup \{\mathcal{O }, (v_{\sigma })_{0}\}=\{(v)_0,\mathcal{O },(v_j)_0\}\) which is clearly concyclic. Note that the circle \(\mathcal{C }_{\{j\}}\) defined here is not the same as the circle \(c_{j}\) used to determine the points \((v_{j})_{0}\) in the construction algorithm.

\(n = 2\): Choose any subset \(\sigma = \{x, y\}\) of \(S\). In the construction algorithm, we defined \((v_{xy})_{0}\) to be the intersection of lines \((L_{x})_{0}\) and \((L_{y})_{0}\).

Since by construction lines \((L_{x})_{0}\) and \((L_{y})_{0}\) are both lines of span \(b\) passing through points \((v_{x})_{0}\) and \((v_{y})_{0}\), respectively, by the Equal Spans Lemma, the points \((v_{x})_{0}\), \((v_{y})_{0}\), \(\mathcal{O }\) and the intersection of \((L_{x})_{0}\) and \((L_{y})_{0}\), which we called \((v_{xy})_{0}\), are all concyclic, on a circle we define to be \(\mathcal{C }_{xy}\).


Induction Hypothesis: For any \(j\) with \(0 \le j \le n1\), let \(\psi \) be any subset of \(S\) with \(\psi  = j\), and assume that \((v_{\psi })_{0}\) is defined to be the common intersection of lines \((L_{\psi _{q}})_{0}\) for all \(q \in \psi \). In addition, assume that the set of points \(\{ (v_{\psi _{q}})_0: q \in \psi \} \cup \{\mathcal{O }, (v_{\psi })_0\}\) are all concyclic, on the circle named \(\mathcal{C }_{\psi }\).
We need to show that for any \(2 < n \le N\) and for any \(\sigma \subset S\) with \(\sigma  = n\), that \((v_{\sigma })_{0}\) lies on the common intersection of the lines \((L_{\sigma _{q}})_{0}\) for all \(q \in \sigma \). It suffices to show that for any distinct \(p, q, r\) in \(\sigma \), that the lines \((L_{\sigma _{p}})_{0}\), \((L_{\sigma _{q}})_{0}\) and \((L_{\sigma _{r}})_{0}\) intersect in a single point. To do this, we will first show that the points \((v_{\sigma _{r}})_0, (v_{\sigma _{p}})_0, (v_{\sigma _{q}})_0, \mathcal{O }\) are concyclic.
Let \(\varphi = \sigma \setminus \{p, q, r\}\), and let \(\varphi x := \varphi \cup \{x\}\), for \(x \subseteq \{p, q, r\}\). Note that \(\varphi pqr = \sigma \) and \(\varphi pq = \sigma _{r}\), \(\varphi pr = \sigma _{q}\), \(\varphi qr = \sigma _{p}\).
In the remainder of the proof, we will suppress the “0” subscript to simplify notation; that is, \(v_{\varphi q}\) indicates the point \((v_{\varphi q})_{0}\), and \(L_{\varphi q}\) indicates the line \((L_{\varphi q})_{0}\), etc.
By the Construction Algorithm and the induction hypothesis, since \(\varphi  = n3\), we have constructed (see Figure 10) line \(L_{\varphi }\) (black), which contains vertices \(v_{\varphi p}\), \(v_{\varphi q}\), and \(v_{\varphi r}\) (blue, green, cyan, respectively). Also by the algorithm, we have constructed lines \(L_{\varphi p}\) (blue), \(L_{\varphi q}\) (green), and \(L_{\varphi r}\) (cyan), and by the induction hypothesis, these intersect pairwise in points \(v_{\varphi pq}\) (red), \(v_{\varphi pr}\) (magenta), \(v_{\varphi qr}\) (orange). Moreover, the points \(v_{\varphi p}, v_{\varphi q}, v_{\varphi pq}, \mathcal{O }\) are concyclic on circle \(\mathcal{C }_{\varphi pq}\) (red), and similarly for \(\mathcal{C }_{\varphi pr}\) (magenta), \(\mathcal{C }_{\varphi qr}\) (orange).
Considering the segment \(\mathcal{O }v_{\varphi p}\) and circle \(\mathcal{C }_{\varphi pr}\) (magenta), using the Cyclic Quad Lemma we conclude that \(\angle \mathcal{O }v_{\varphi r} v_{\varphi p} = \angle \mathcal{O }v_{\varphi p r} v_{\varphi p}\) (gray triplemarked angles). Using the same segment and circle \(\mathcal{C }_{\varphi pq}\) (red), we conclude that \(\angle \mathcal{O }v_{\varphi q} v_{\varphi p} = \angle \mathcal{O }v_{\varphi pq} v_{\varphi p}\) (blue doublemarked angles). Since \(v_{\varphi pq}, v_{\varphi pr}, v_{\varphi p}\) all lie on line \(L_{\varphi p}\), and \(v_{\varphi q}, v_{\varphi r}, v_{\varphi p}\) all lie on line \(L_{\varphi }\), \(\angle v_{\varphi pq} v_{\varphi pr} \mathcal{O } = \angle v_{\varphi q} v_{\varphi r} \mathcal{O }\) (red singlemarked angles), since they are both supplementary to equal angles.
Therefore, \(\triangle \mathcal{O }v_{\varphi pq} v_{\varphi pr} \sim \triangle \mathcal{O }v_{\varphi q} v_{\varphi r}\) (gray triangles), since they have two pairs of equal angles. Hence the third angles in each triangle are equal (bottom green quadruplemarked angles); that is, \(\angle v_{\varphi r} \mathcal{O } v_{\varphi q} = \angle v_{\varphi pr} \mathcal{O } v_{\varphi pq}\).
However, using segment \(v_{\varphi r} v_{\varphi q}\) and circle \(\mathcal{C }_{\varphi qr}\) (orange), \(\angle v_{\varphi r} v_{\varphi qr} v_{\varphi q} = \angle v_{\varphi r} \mathcal{O } v_{\varphi q}\) (green quadruplemarked angles). But \(\angle v_{\varphi r} v_{\varphi qr} v_{\varphi q}\) is determined by lines \(L_{\varphi q}\) and \(L_{\varphi r}\), so \(\angle v_{\varphi r} v_{\varphi qr} v_{\varphi q} = \angle v_{\varphi pq} v_{\varphi qr} v_{\varphi pr}\). Thus, we can conclude that
Hence by the Cyclic Quad Lemma, we conclude that \(v_{\varphi pq} = v_{\sigma _{r}}, v_{\varphi qr} = v_{\sigma _{p}}, v_{\varphi pr} = v_{\sigma _{q}}, \mathcal{O }\) are concyclic, on a circle (green dashed) which we shall name \(\mathcal{C }_{ \varphi pqr} = \mathcal{C }_{\sigma }\).
Therefore, if we construct lines \(L_{\varphi pq} = L_{\sigma _{r}}\), \(L_{\varphi pr} = L_{\sigma _{q}}\), \(L_{\varphi qr} = L_{\sigma _{p}}\) of span \(\delta _{n1}\) (with respect to \(\mathcal{O }\)) through vertices \(v_{\varphi pq} = v_{\sigma _{r}}\), \(v_{\varphi pr} = v_{\sigma _{q}}\), \(v_{\varphi qr} = v_{\sigma _{p}}\), respectively, by the CircumcircleSpan Lemma, these three lines will intersect at a single point \(v_{\varphi pqr} = v_{\sigma }\) which also lies on \(\mathcal{C }_{\sigma }\). \(\square \)
Lemma 8
Let \(\sigma \) be a nonempty subset of \(S\) with \(\sigma  = n\). Then the circle \(\mathcal{C }_\sigma \) contains \((v_\sigma )_{({\delta }_{n1})}.\) That is, \(\mathcal{C }_\sigma \) is a circle of span \(\delta _{n1}\) with respect to the class of points \(v_\sigma \).
Proof
Let \(\sigma \) be a nonempty subset of \(S\) of cardinality \(n.\) For ease of reference, let \(\delta =\delta _{n1}.\) Recall from Theorem 7 that \(\mathcal{C }_\sigma \) contains the points \((v_\sigma )_0\), \(\mathcal{O }\), and \((v_{\sigma _q})_0\) for every \(q \in \sigma .\)
Let \(q\) be an arbitrary element in \(\sigma \) and consider the concentric circles \(\mathcal{C }_\mathcal{O }( (v_\sigma )_0)\) and \(\mathcal{C }_\mathcal{O }( (v_{\sigma _q})_0),\) and points \((v_{\sigma _q})_0,\) \((v_{\sigma _q})_\delta ,\) \((v_{\sigma })_{\delta },\) and \((v_{\sigma })_0\). Clearly, \(\angle (v_{\sigma _q})_0\mathcal{O }(v_{\sigma _q})_\delta = \angle (v_{\sigma })_{\delta }\mathcal{O }(v_{\sigma })_0\). Thus, by the Equal Spans Lemma, the point of intersection \(Q = \overline{(v_{\sigma _q})_0(v_{\sigma _q})_\delta } \wedge \overline{(v_{\sigma })_{\delta }(v_{\sigma })_0}\) is concyclic with the points \((v_{\sigma _q})_0, \mathcal{O },\) and \((v_\sigma )_{\delta }.\) Call this circle \(\mathcal{C }^{\prime }.\)
Note that by definition of \((L_{\sigma _q})_0,\) the line \((L_{\sigma _q})_0 := \overline{(v_{\sigma _q})_0(v_{\sigma _q})_\delta }\) and contains the point \((v_\sigma )_0\), by Theorem 7. Hence, \(Q=(v_\sigma )_0.\) But now circle \(\mathcal{C }^{\prime }\) contains the three points: \((v_\sigma )_0\), \(\mathcal{O }\), and \((v_{\sigma _q})_0\) and so \(\mathcal{C }^{\prime }=\mathcal C _\sigma .\) Thus, \(\mathcal{C }_\sigma \) includes the point \((v_{\sigma })_{\delta }\), which was what we wanted to show. \(\square \)
Theorem 9
For each \(\sigma \subseteq S\) with \(\sigma  = n\), the line \((L_{\sigma })_{0}\) passes through the points \((v_{\sigma _{q}})_{d_{q}}\) for each \(q \in \sigma \) (as well as through the two points \((v_{\sigma })_{0}\) and \((v_{\sigma })_{\delta _{n}}\)).
Proof
Again, we proceed by induction on the size of \(\sigma \).

Base cases:

\(n = 0\): If \(n = 0\), then \(\sigma = \emptyset \), and by construction, line \((L)_{0}\) passes through \((v)_{0}\) and \((v)_{a}\), and there are no other vertices to consider.

\(n = 1\): The line \((L_{j})_{0}\) is defined to pass through \((v_{j})_{0}\) and \((v_{j})_{b}\). We need to show that it also passes through \((v)_{d_{j}}\). But this follows immediately from the Circumcircle Construction Lemma: line \((L_{j})_{0}\) is a span \(b\) line, and by construction, point \((v_{j})_{0}\) lies on the circumcircle \(c_{j} := \mathcal{C }((v)_{d_j}, \mathcal{O }, (v)_{d_{j}b})\), so \((L_{j})_{0}\) also passes through \( (v)_{d_{j}}\).


Induction Hypothesis: Suppose that for any \(\psi \subseteq S\) with \(\psi  =r \) for any \(0 \le r\le n1\) that line \((L_{\psi })_{0}:= (v_{\psi })_{0} \vee (v_{\psi })_{\delta _{r}}\) passes through each point \((v_{\psi _{q}})_{d_{q}}\) for all \(q \in \psi \).
We need to show for any \(\sigma \subseteq S\) with \(\sigma  = n\) that \((L_{\sigma })_{0}\) passes through the points \((v_{\sigma _{p}})_{d_{p}}\) for each \(p \in \sigma \).
Choose any \(p, q \in \sigma \). Let \(\sigma _{pq} = \sigma \setminus \{p,q\} = (\sigma _{q})_{p} = (\sigma _{p})_{q}\) and let \(\delta =\delta _{n} = \delta _{n2}.\) Note that by construction, \((v_{\sigma _{p}})_{0}\) and \((v_{\sigma _{q}})_{0}\) both lie on \((L_{\sigma _{pq}})_{0}\), which is of span \(\delta \). Construct \((L_{\sigma _{q}})_{0}\) passing through \((v_{\sigma _{q}})_{0}\). By the induction hypothesis applied to \(\sigma _{q}\) which has size \(n1\), we conclude that since \(\sigma _{pq} \cup \{p\} = \sigma _{q},\) \((v_{(\sigma _{q})_{p}})_{d_{p}} = (v_{\sigma _{pq}})_{d_{p}}\) lies on \((L_{\sigma _{q}})_{0}\). See Fig. 11 to follow along in the diagram.
Now let \(C = \mathcal{C }((v_{\sigma _{p}})_{d_{p}}, \mathcal{O }, (v_{\sigma _{p}})_{d_{p}  \delta })\) (shown red and thick). If we rotate \(C\) by angle \(\tfrac{2 \pi d_{p}}{m}\) about \(\mathcal{O }\), then we get the circle \(C^{\prime } =\mathcal{C }((v_{\sigma _{p}})_{0}, \mathcal{O }, (v_{\sigma _{p}})_{\delta })\) (red and dotted). Recall that circle \(\mathcal{C }_{\sigma _{p}}\) contains \((v_{(\sigma _{p})_{q}})_{0}\), along with \((v_{\sigma _{p}})_{0}\) and \(\mathcal{O }\), and by Lemma 8, it also contains \((v_{\sigma _{p}})_{\delta _{n2}} = (v_{\sigma _{p}})_{\delta }\), since \(\sigma _{p} = n1\). Therefore, \(C^{\prime } = \mathcal{C }_{\sigma _{p}}\).
Because \(\mathcal{C }_{\sigma _{p}}\) contains \((v_{\sigma _{pq}})_{0}\), it follows that \(C\) contains \((v_{\sigma _{pq}})_{d_{p}}\). Define \(Z\) to be the intersection of \((L_{\sigma _{q}})_{0}\) and \(C\) different from \((v_{\sigma _{pq}})_{d_{p}}\).
By construction the gray triangles \(\triangle (v_{\sigma _{p}})_{d_{p}} \mathcal{O } (v_{\sigma _{pq}})_{d_{p}}\) and \(\triangle (v_{\sigma _{p}})_{0}\mathcal{O }(v_{\sigma _{pq}})_{0}\) are congruent. Hence \(\angle \mathcal{O }(v_{\sigma _{p}})_{d_{p}} (v_{\sigma _{pq}})_{d_{p}} = \angle \mathcal{O }(v_{\sigma _{p}})_{0} (v_{\sigma _{pq}})_{0} = s\) (shown as pink triplemarked angles). However, using segment \(\mathcal{O }(v_{\sigma _{pq}})_{d_{p}}\), circle \(C\), and the Cyclic Quad lemma, we conclude that \(\angle \mathcal{O }(v_{\sigma _{p}})_{d_{p}} (v_{\sigma _{pq}})_{d_{p}} = \angle \mathcal{O } Z (v_{\sigma _{pq}})_{d_{p}}\) (also marked \(s\)).
But \(\angle \mathcal{O } Z (v_{\sigma _{pq}})_{d_{p}} = \angle \mathcal{O } Z (v_{\sigma _{q}})_{0}\), since \(Z\), \((v_{\sigma _{pq}})_{d_{p}}\), and \((v_{\sigma _{q}})_{0}\) are collinear on \((L_{\sigma _{q}})_{0}\) (dark blue). Thus, we conclude that \(\angle \mathcal{O } Z (v_{\sigma _{pq}})_{d_{p}} = \angle \mathcal{O } Z (v_{\sigma _{q}})_{0}\), so using Cyclic Quad one more time, with the segment \(\mathcal{O } (v_{\sigma _{q}})_{0}\) subtending the angles (shown blue and dotted), we conclude that \(Z\) is on \(\mathcal{C }((v_{\sigma _{q}})_{0}, \mathcal{O }, (v_{\sigma _{p}})_{0})\) (shown green and dashed). But this circle is precisely \(\mathcal{C }_{\sigma } \), and from the definition of \(\mathcal{C }_{\sigma }\), the point \((v_{\sigma })_{0}\), which also lies on \((L_{\sigma _{q}})_{0}\), lies on \(\mathcal{C }_{\sigma }\). That is, the line \((L_{\sigma _{q}})_{0}\) intersects \(\mathcal{C }_{\sigma }\) in two points, \(Z\) and \((v_{\sigma })_{0}\), which are both distinct from \((v_{\sigma _{q}})_{0}\) (which is also on \((L_{\sigma _{q}})_{0}\) and \(\mathcal{C }_{\sigma }\)), so it follows that \(Z = (v_{\sigma })_{0}\). Let \(Z^{\prime } = (v_{\sigma })_{\delta }\); then the line \(ZZ^{\prime }\) is the line \((L_{\sigma })_{0}\).
Now apply the Configuration Construction Lemma using the points \((v_{\sigma _{p}})_{d_p}\) and \((v_{\sigma _{p}})_{d_{p}  \delta }\), which are both on \(C\), along with \(Z = (v_{\sigma })_{0}\), which is now also on \(C\), to conclude that the line \(ZZ^{\prime } = (L_{\sigma })_{0}\) passes through \((v_{\sigma _{p}})_{d_{p}}\), which is what we needed to show. \(\square \)
5 Is It a Configuration?
In this section, we will show that, assuming the use of valid parameters, the incidence structure produced by the Construction Algorithm is actually a \(k\)configuration, and count the number of symmetry classes of points and lines, completing the proof of Theorem 6.
Choose any \(\sigma \subseteq S\) (where \(S = k2\)), and suppose \(\sigma  = n\). Consider \((v_{\sigma })_{0}\). By symmetry, it suffices to show that \((v_{\sigma })_{0}\) has \(k\) lines passing through it.
Note that if \(\sigma  = 0\), then \(\sigma = \emptyset \), and \((v_{\emptyset })_{0}\) lies only on the two lines \((L_{\emptyset })_{0}\) and \((L_{\emptyset })_{a}\) and the \((k2)\) subsequently defined lines \((L_{\{p\}})_{d_p}\) for \(p\in S.\) If \(\sigma  = k2\), then \(\sigma = S\) and \((v_{S})_{0}\) lies only on the \(k2\) previously defined lines \((L_{S_{q}})_{0}\) for \(q \in S\) and the two lines \((L_{S})_{0}\) and \((L_{S})_{\delta _{k2}}\).
In the other cases, by construction, \((v_{\sigma })_{0}\) lies on the \(n\) previously defined lines \((L_{\sigma _{q}})_{0}\) for each \(q \in \sigma \), the pair of lines \((L_{\sigma })_{0}\) and \((L_{\sigma })_{\delta _{n}}\) (for \(\delta _{n} = a\) or \(b\) depending on the parity of \(n\)), and the \((k2)  n\) subsequently defined lines \((L_{\sigma _{p}})_{d_p}\) for each \(p \notin \sigma \), for a total of \(n + 2 + (k2)  n = k\) lines passing through each point.
Similarly, consider \((L_{\sigma })_{0}\). By symmetry, it suffices to show that \((L_{\sigma })_{0}\) has \(k\) points lying on it. If \(\sigma  = 0\), then \((L_{\emptyset })_{0}\) contains the two points \((v_{\emptyset })_{0}\) and \((v_{\emptyset })_{a}\) and the \((k2)\) subsequently defined points \((v_{\{p\}})_{0}\) for \(p \in S.\) If \(\sigma  = k2\), then \((L_{S})_{0}\) only passes through the \(k2\) previously defined vertices \((v_{S_{q}})_{d_{q}}\) for \(q \in S\) and the two points \((v_{S})_{0}\) and \((v_{S})_{\delta _{k2}}\). In the general case, by construction, \((L_{\sigma })_{0}\) contains \(n\) previously defined vertices \((v_{\sigma _{q}})_{d_{q}}\) for each \(q \in \sigma \), the pair of points \((v_{\sigma })_{0}\) and \((v_{\sigma })_{\delta _{n}}\), and the \((k2)  n\) subsequently defined vertices \((v_{\sigma p})_{0}\) for each \(p \notin \sigma \), for a total of \(n + 2 + (k2)  n = k\) points lying on each line.
In total, we have constructed \(\genfrac(){0.0pt}{}{k2}{\sigma }\) symmetry classes of points and lines for each \(\sigma \subset \{1, 2, \ldots , k2\}\), for a total of \(2^{k2}\) symmetry classes of points and lines. Each symmetry class contains \(m\) elements for a total of \(m2^{k2}\) points and lines.
By construction, the configurations that have been constructed have \(m\)fold rotational symmetry. Given appropriate choices of discrete parameters, the configurations, such as the configuration shown in Fig. 1, may in fact have dihedral symmetry and fewer symmetry classes; classifying which parameters yield configurations with dihedral symmetry is beyond the scope of this paper and will be addressed in a future work.
This completes the proof of Theorem 6.
6 Valid Parameters and Further Directions
Theorem 6 shows that if valid parameters are chosen for \(\mathcal{A }(m; a,b; d_{1}, \ldots , d_{k2})\), then following the construction techniques outlined in Sect. 2, a \(k\)configuration is produced. However, the theorem is silent as to which parameters are valid. Unfortunately, it turns out that completely characterizing precisely which parameters are valid is fairly complex and beyond the scope of this paper. In this section, we will identify large sets of valid parameters. A more complete characterization of valid discrete parameters will be discussed in a subsequent work.
In this work we present the following theorems.
Theorem 10
In the configuration \(\mathcal{A }(m; a,b; d)\), if \(d = 0, a, b, a+b\), then the construction method produces a degenerate configuration: either a \((2,4)\) or \((4,2)\)configuration.
Proof
Consider circumcircle \(\mathcal{C } = \mathcal C (v_{d}, \mathcal{O }, v_{db})\). By construction, it is a span \(b\) circle.
Suppose that \(d = b\); then \(\mathcal{C }\) intersects \((L)_{0}\) at \((v)_{bb} = (v)_{0}\) and at some other point. If we call the second intersection point \(w_{0}\) and use this as the point \((v_{1})_{0}\) in the construction algorithm, the algorithm has us construct a second line \((L_{1})_{0}\) which passes through \((v_{1})_{0}\), \((v_{1})_{b}\) and \((v)_{d} = (v)_{b}\) in this case (since \(d = b\)). On the other hand, the line \((L)_{b}\) passes through the points \((v)_{b}, (v)_{b+a}\), and \((v_{1})_{b}\). That is, both lines \((L_{1})_{0}\) and \((L)_{b}\) pass through the two points \((v)_{b}\) and \((v_{1})_{b}\), so they are the same line. Hence the result of the construction is an incidence structure in which every point has two lines passing through it. Namely, \((v)_{i}\) has lines \((L)_{i}\) and \((L)_{ia}\) through it, and \((v_{1})_{i}\) has lines \((L)_{i+b}\) and \((L)_{i+bb} = (L)_{i}\) passing through it. Each line contains four points. Namely, \((L)_{i}\) contains points \((v)_{i}\), \((v)_{i+a}\), \((v_{1})_{i}\) and \((v_{1})_{i+b}\). Thus, the incidence structure is a \((4,2)\)configuration.
On the other hand, if we let \((v_{1})_{0} = (v)_{0}\) (choosing the other possible intersection of \(\mathcal{C }\) with line \((L)_{0}\)), then the line \((L_{1})_{0}\) is distinct from \((L)_{0}\) (as long as \(a \ne b\)), and each line has only two points on it. Namely, \((L)_{i}\) contains \((v)_{i}\) and \((v)_{i+a}\), while \((L_{1})_{i}\) contains \((v)_{i}\) and \((v)_{i+b}\). Each point has four lines passing through it. Namely, \((v)_{i}\) contains lines \((L)_{i}\), \((L)_{ia}\), \((L_{1})_{i}\) and \((L_{1})_{ib}\). Thus the algorithm produces a \((2,4)\)configuration.
In the other possible cases for \(d\), if \(d = 0\) then \(\mathcal{C }\) passes through \((v)_{d}= (v)_{0}\), which is on \((L)_{0}\); if \(d = a\) then \(\mathcal{C }\) passes through \((v)_{d}= (v)_{a}\), which is on \((L)_{0}\); and if \(d = a+b\) then \(\mathcal{C }\) passes through \((v)_{db}= (v)_{a}\), which is on \((L)_{0}\). In each case, arguing as above, we achieve a \((4,2)\)configuration by choosing the “other” point on \((L)_{0}\) as our point \((v_{1})_{0}\), while choosing the point \((v)_{a}\) or \((v)_{0}\) as the point \((v_{1})_{0}\) yields a \((2,4)\)configuration. \(\square \)
Lemma 11
Given a circle \(\mathcal{C }\) centered at \(\mathcal{O }\) and four points on \(\mathcal{C }\) labelled \(v_0, v_a, v_d,\) and \(v_{db},\) let \(L_0=v_0\vee v_a\) and \(\mathcal{C }_d=\mathcal{C }(\mathcal{O }, v_d,v_{db}).\) Let \(m_a\) be the midpoint of chord \(v_0v_a\) and let \(w_0\) and \(w^{\prime }_0\) be points of intersection between \(L_0\) and \(\mathcal{C }_d\). Let \(2\alpha = \angle v_0\mathcal{O }v_a,\) \(2\beta =\angle v_d\mathcal{O }v_{db}\), \(\tau _1=\angle m_a\mathcal{O }w_0,\) \(\tau _2=\angle m_a\mathcal{O }w^{\prime }_0,\) and \(2\delta = \angle v_d\mathcal{O }v_0.\) Then

(i)
\(2\delta =\alpha +\beta \tau _1\tau _2\);

(ii)
\(\cos (\alpha ) \cos (\beta )=\cos (\tau _1)\cos (\tau _2).\)
Proof
Construct \(\mathcal{C }, v_0, v_a, v_d,v_{db},L_0,\mathcal{C }_d,m_a,w_0,w^{\prime }_0,\tau _1, \tau _2,\) and \(\delta \) as described (see Fig. 12). Construct point \(w_{b}\) by rotating \(w_0\) clockwise by \(\angle v_d\mathcal{O }v_{db} = 2\beta \) and construct \(m_b,\) the midpoint of segment \(w_0w_{b}.\) From the Configuration Construction Lemma, line \(L_1=w_0 \vee w_{b}\) also intersects \(v_{db}.\) Observe that \(v_{db}\) and \(w^{\prime }_0\) both lie on \(\mathcal{C }_d\) and both subtend segment \(\mathcal{O }w_0.\) Thus, \(\angle \mathcal{O }v_{db}w_0 \; = \; \angle \mathcal{O }w^{\prime }_0w_0\) and right triangles \(\triangle \mathcal{O }m_bv_{db}\) and \(\triangle \mathcal{O }m_aw^{\prime }_{0}\) are congruent. Thus \(\angle w_{b}\mathcal{O }v_{db}=\tau _2\beta \) and by rotational symmetry, \(\angle w_0\mathcal{O }v_d=\tau _2\beta .\) Finally, \(2\delta =\angle v_d \mathcal{O }v_0=(\angle m_a\mathcal{O }v_0\angle m_a \mathcal{O }w_0)(\angle w_0\mathcal{O }v_d)=(\alpha \tau _1)  (\tau _2\beta )= \alpha + \beta  \tau _{1}  \tau _{2}.\)
For part (ii), we observe that by elementary righttriangle trigonometry, \(\cos \alpha ={\mathcal{O }m_a}/{\mathcal{O }v_a}\), since \(m_{a}\) was constructed so that \(\mathcal{O }m_{a} \!\perp \! v_{0}v_{a}\). Thus, \(\mathcal{O }m_a\!=\!({\cos \alpha })({\mathcal{O }v_a}).\) Now, triangle \(\triangle \mathcal{O }m_aw_0\) is also a right triangle, with angle \(\angle m_{a}Ow_{0} = \tau _{1}\), adjacent side \(\mathcal{O }m_{a}\), and hypotenuse \(\mathcal{O }w_0.\) Thus,
or, equivalently,
Similarly, since \(\mathcal{O }m_{b} \perp w_{0}v_{db}\), triangle \(\triangle m_b\mathcal{O }v_{db}\) is also a right triangle. From part (i), angle \(\angle m_{b}\mathcal{O }v_{db} = \tau _2\). Thus, \(\cos ( \tau _2) ={\mathcal{O }m_b}/{\mathcal{O }v_{db}}\) or equivalently \(\mathcal{O }m_b=(\cos \tau _2)(\mathcal{O }v_{db}).\) Now triangle \(\triangle \mathcal{O }m_bw_{b}\) is also a right triangle, with angle \(\angle m_b\mathcal{O }w_{b}=\beta \), adjacent side \(\mathcal{O }m_b\), and hypotenuse \(\mathcal{O }w_{b}\). Thus,
or, equivalently,
Since triangle \(\triangle w_0\mathcal{O }w_{b}\) is isosceles by construction, \(\mathcal{O }w_0=\mathcal{O }w_{b}.\) Hence,
Since both \(v_a\) and \(v_{db}\) lie on \(\mathcal{C },\) we know \(\mathcal{O }v_a=\mathcal{O }v_{db},\) so
and part (ii) is proved. \(\square \)
As a consequence of Lemma 11, if any four of the angles \(\alpha ,\beta , \tau _1, \tau _2\) or \(\delta \) are multiples of a particular angle, the fifth must be also. In particular, in the configuration construction algorithm, \(\alpha , \beta ,\) and \(\delta \) are always multiples of \(2\pi /m.\) Furthermore, for some positive integer \(r < m/2\), \(\tau _1=r\pi /m\) if and only if \(w_0=L_0 \wedge L_{r}\), where \(L_{r}\) is the rotation of \(L_0\) by \(2r\pi /m.\) Lemma 11 implies that if a circumcircle intersects \(L_0\) in two points, either both of those points of intersection correspond to intersections of the form \(L_0 \wedge L_r\) for some \(r\) or neither do. That is:
Theorem 12
(Extra Incidences) In a configuration \(\mathcal{A }(m; a, b; d)\), circumcircle \(c = \mathcal{C }(v_{d}, \mathcal{O }, v_{db})\) intersects line \(L_{0}\) at the point of intersection of \(L_{0}\) and \(L_{t}\) precisely when either

(i)
\(d \in \{ 0, a, b, a+b \}\) and the configuration produced is degenerate; or

(ii)
the discrete parameters \(a, b, d, t\) satisfy
$$\begin{aligned} \cos \left( a \frac{\pi }{m}\right) \cos \left( b\frac{\pi }{m} \right) = \cos \left( t \frac{\pi }{m} \right) \cos \left( (2dab+t)\frac{\pi }{m}\right) . \end{aligned}$$(1)
Proof
Suppose \(c = \mathcal{C }(v_{d}, \mathcal{O }, v_{db})\) intersects \((L)_{0}\) at two points, \(w_{0}\) and \(w_{0}^{\prime }\). We may apply Lemma 11 to the geometric construction, where
to conclude that if \(\tau _{1}\) and \(\tau _{2}\) are defined as in the proof of Lemma 11 that \(\tau _{1}+ \tau _{2}\) is also an integer multiple of \(\pi /m\), since
If \(w_{0}\) lies at the intersection of \(L_{0}\) and \(L_{t}\), then \(\tau _{1} = \tfrac{ t \pi }{m}\), so it follows that \(\tau _{2}\) must also be an integer multiple of \(\tfrac{\pi }{m}\), say \(\tau _{2} = \tfrac{ t_{2}\pi }{m}\). Thus, if a circumcircle intersects \(L_0\) in two points, either both of those points of intersection correspond to intersections of the form \(L_0 \wedge {L}_{r}\) for some \(r\) or neither do, and this happens precisely when
We showed previously that \(2\delta = \alpha + \beta  \tau _{1}  \tau _{2}\), or in this case,
so \(2d = a+b t  t_{2}\). Solving for \(t_{2}\), we see that \(t_{2} = 2d  a  b+t\), and substituting into Eq. (2) yields Eq. (1).
If \(d = 0, a, b, a+b\), then by the proof techniques in Theorem 10, we can show that \(t = b, b, b, b\) respectively, and Eq. (1) is trivially satisfied. For example, if \(d = a\), then Eq. (1) becomes
However, Lemma 10 showed that if \(d = 0, a, b, a+b\), then the configurations produced are degenerate. \(\square \)
Equation (2) is a familiar equation in the study of configurations: it is the hard axiom to fulfill in the construction of a certain class of highly symmetric 4configurations, the astral 4configurations; see [2, 10] and [12, Sect. 3.6]. These astral 4configurations have symbol \(m\#(a,b; c,d)\) and the parameters are required to satisfy certain axioms. Myerson in [13] determined all rational solutions to
and elementary trigonometry converts these solutions into integer solutions to (2). One interesting result is that all solutions require \(6 \mid m\). A consequence of Theorem 12 is that the circumcircle construction outlined in this paper provides a novel construction for producing astral 4configurations. For example, applying the construction to \(\mathcal{A }(12; 4,4; 1)\), where \(d_{1} = 1\) is a welldefined but not valid parameter choice, produces a \((24_{4})\) astral 4configuration, rather than the expected 3configuration, because if \(m = 12, a = b = 4, t = 1, t_{2} = 5\), then it is known (and easy to show) that (2) is satisfied, and \(2d = a+bt_{1}t_{2} = 8  6 = 2\), so \(d = 1\). That is:
Corollary 13
Given parameters \(m, a, b, d, t\) that satisfy (1), the configuration \(\mathcal{A }(m; a,b; d)\) is the astral 4configuration \(m\#(a, t; b, 2dab+t)\).
In addition, if we apply the construction algorithm using the symbol \(\mathcal{A }(12; 4,4; 1,3, 5)\), which would be expected to produce a 5configuration, because of the extra incidences caused by the fact that \(d_{1} = 1\) a \((96_{6})\) configuration is generated which has only 8 symmetry classes of points and lines. This is now the smallest known 6configuration, shown in Fig. 13.
Lemma 14
Given \(m\), \(a\), \(b\), if \(d \in \{1, 2, \ldots , a+b1\} \setminus \{a,b\}\), then circle \(c = \mathcal{C }((v)_{d}, \mathcal{O }, (v)_{db})\) intersects \((L)_{0}\) in two distinct points; in particular, if \(d \in \{1, 2, \ldots , a+b1\} \setminus \{a,b\}\) then \(d\) is a elldefined parameter.
Proof
Under these conditions, at least one of \((v)_{d}\) and \((v)_{db}\) is separated from \(\mathcal{O }\) by \((L)_{0}\): specifically, if \(d\) is strictly between \(0\) and \(a\), then \((L)_{0}\) separates \(\mathcal{O }\) and \((v)_{d}\), while if \(d\) is strictly between \(a\) and \(a+b\), \((v)_{db}\) is separated from \(\mathcal{O }\) by \((L)_{0}\). In either case, the circle \(c\) intersects \((L)_{0}\) twice. \(\square \)
For certain values of \(m, a, b\) there are other welldefined parameters \(d\); these will be discussed in a subsequent work.
Theorem 15
If \(m \ge k+1\) when \(k\) is even or if \(m \ge k+2\) for \(k\) odd, and if \(6 \not \mid m\), then it is possible to construct a \(k\)configuration with \(2^{k2}\) symmetry classes of points and lines, by choosing \(a = b = \lfloor \tfrac{m1}{2} \rfloor \) and any subset of size \(k2\) from the set \(S = \{ 1, 2, \ldots , a1, a+1, \ldots , 2a1\}\) to be the sequence \(d_{1}, \ldots , d_{k2}\) in the configuration \(\mathcal{A }(m a,b; d_{1}, \ldots , d_{k2})\).
Proof
Suppose \(k\) is even. Then
so \(a \ge \left\lfloor {k}/{2}\right\rfloor = {k}/{2}\). Hence \(S = 2a2 \ge k2\).
On the other hand, if \(k\) is odd, then \(a = \lfloor {(m1)}/{2}\rfloor \) and \(m \ge k+2\), so \(a \ge \lfloor {(k+1)}/{2}\rfloor \); therefore
In either case, it is possible to choose a subset of size \(k2\) from \(S\) to serve as a sequence of \(d_{j}\) in the construction algorithm. By Lemma 14, each of these \(d_{j}\) is welldefined. So it is possible to implement the construction algorithm.
It remains to show that each \(d_j\) is valid; that is, there are no additional intersections. By Theorem 12, the only way that a circle \(c_{j}\) can intersect a point that is the point of intersection of \((L)_{0}\) and \((L)_{t}\) for some \(t\) is either when \(d_{j} = 0, a, b, a+b\), which we have disallowed, or when \(a, b, d, t\) satisfy (1). However, solutions to (1), in the form of (2), have been completely characterized by Meyerson in [13] and all require that \(6 \mid m\), so if \(6\not \mid m\), there are no extra intersections. \(\square \)
We conjecture that if \(k\) is even, the configuration
is minimal for this construction, while if \(k\) is odd, the configuration
is minimal (although any other subset of size \(k2\) from
would also work), unless there are extra incidences. Note that the configuration of this type for \(k = 4\) is the configuration \(\mathcal{A }(5; 2,2; 1,3)\), a \((204)\) configuration. The configuration \(\mathcal{A }(7; 3,3; 1,2,4)\) produces a \((56_{5})\) configuration, but the configuration \(\mathcal{A }(12; 4,4; 1,3)\), which has extra incidences, produces a \((48_{5})\) configuration (also discussed in [8]). The configuration \(\mathcal{A }(7;3,3; 1,2,4,5)\) is a \((112)_{6}\) configuration, slightly larger than the previously known recordholder, a \((110_{6})\) configuration constructed in [4]; both configurations have dihedral symmetry. The configuration \(\mathcal{A }(9; 4,4; 1,2,3,5,6)\) produces a chiral \((9 \cdot 2^{5}_{7}) = (288_{7})\) configuration, the smallest known. Unfortunately, neither \(\mathcal{A }(7;3,3; 1,2,4,5)\) nor \(\mathcal{A }(9; 4,4; 1,2,3,5,6)\) can be drawn intelligibly at a scale appropriate to this paper, because many of the point classes cluster too close to the center of the configuration; for example, if the radius of the points \((v)\) is set to 1, then the configuration \(\mathcal{A }(7;3,3; 1,2,4,5)\) has point classes with smallest radius approximately \(0.0629012\) and largest radius approximately \(15.8979\) (and the next largest radius is \(4.32402\)).
References
AltshillerCourt, N.: College Geometry: An Introduction to the Modern Geometry of the Triangle and the Circle. Dover Publications, New York (2007)
Berman, L.W.: A characterization of astral \((n_4)\) configurations. Discrete Comput. Geom. 26(4), 603–612 (2001)
Berman, L.W.: Movable \((n_4)\) configurations. Electron. J. Combin. 13(1). Research Paper 104, 30 pp (2006)
Berman, L.W.: Constructing highly incident configurations. Discrete Comput. Geom. 46(3), 447–470 (2011)
Berman, L.W.: Erratum to: Constructing highly incident configurations. Discrete Comput. Geom. 46(3), 471 (2011)
Berman, L.W.: Geometric constructions for 3configurations with nontrivial geometric symmetry (in review)
Berman, L.W., Grünbaum, B.: Deletion constructions of symmetric 4configurations. Part I. Contrib. Discret. Math. 5(1), 18–33 (2010)
Berman, L.W., Laura, Ng.: Constructing 5configurations with chiral symmetry. Electron. J. Combin. 17(1). Research Paper 2, 14 pp (2010)
Boben, M., Pisanski, T.: Polycyclic configurations. Eur. J. Combin. 24(4), 431–457 (2003)
Grünbaum, B.: Astral \((n_4)\) configurations. Geombinatorics 9(3), 127–134 (2000)
Grünbaum, B.: Musings on an example of Danzer’s. Eur. J. Combin. 29(8), 1910–1918 (2008)
Grünbaum, B.: Configurations of Points and Lines. Graduate Studies in Mathematics, vol. 103. American Mathematical Society, Providence, RI (2009)
Myerson, G.: Rational products of sines of rational angles. Aequationes Math. 45(1), 70–82 (1993)
Acknowledgments
The first author thanks Marston Condor, Barry Monson, Egon Schulte, and Asia Ivic Weiss for inviting her to participate in the 2011 Workshop on Symmetry in Graphs, Maps and Polytopes at the Fields Institute, where the \((96_{6})\) configuration was found.
Author information
Authors and Affiliations
Corresponding author
Rights and permissions
About this article
Cite this article
Berman, L.W., Faudree, J.R. Highly Incident Configurations with Chiral Symmetry. Discrete Comput Geom 49, 671–694 (2013). https://doi.org/10.1007/s0045401394940
Received:
Accepted:
Published:
Issue Date:
DOI: https://doi.org/10.1007/s0045401394940