1 Introduction

The classical version of Miquel’s Pentagon theorem on the Riemann sphere can be formulated as follows:

Theorem 1

Let \(h_1,\ldots ,h_5\) be five different Möbius circles which intersect each other at a point I and such that any three of them only meet in I. Then, for \(i\in \{1,\ldots ,5\}\), \(h_{i-1}\) and \(h_{i+1}\) meet in I and a second point \(q_i\), and \(h_{i-2}\) and \(h_{i+2}\) meet in I and a second point \(s_i\) (indices read cyclically). Let \(k_i\) be the Möbius circle through \(s_i, q_{i-1}, q_{i+1}\). Then, for \(i \in \{1,\ldots ,5\}\), \(k_{i-1}\) and \(k_{i+1}\) meet in \(q_i\) and a second point \(p_i\), and the points \(p_1,\ldots ,p_5\) all lie on one common Möbius circle c.

The situation is shown in Fig. 1. Miquel’s original proof can be found in [4, Théorème III]. It is based on classical angle theorems. A computer assisted algebraic proof which uses null bracket algebra has been published in [3]. A simple algebraic proof based on the cross ratio has been discussed in [1].

The assumption that the Möbius circles \(h_i\) intersect (not touch) each other in I implies that the points \(q_i\) and \(s_i\) are different from I. In addition, since we assume that any three of the circles \(h_i\) only meet in I, we have that the 10 points \(q_i, s_i\) are pairwise distinct. These assumptions can be relaxed if one is interested in degenerate cases of the configuration. Using the Möbius transformation \(z\mapsto 1/(z-I)\) we may always assume that \(I=\infty \). In this case, the Möbius circles \(h_i\) are lines in the complex plane.

Fig. 1
figure 1

The classical Pentagon theorem

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The idea is now to replace the circles \(k_i\) through \(s_i,q_{i-1},q_{i+1}\) by circles \(k_i\) through \(s_i,q_{i-2},q_{i+2}\). This variant has apparently not yet been treated in the literature. Surprisingly, this configuration shows numerous incidences, even significantly more than the classical Pentagon Theorem 1.

Theorem 2

Let \(h_1,\ldots ,h_5\) be five different Möbius circles which intersect each other at a point I and such that any three of them only meet in I. Then, for \(i\in \{1,\ldots ,5\}\), \(h_{i-1}\) and \(h_{i+1}\) meet in I and a second point \(q_i\), and \(h_{i-2}\) and \(h_{i+2}\) meet in I and a second point \(s_i\) (indices read cyclically). Let \(k_i\) be the Möbius circle through \(s_i, q_{i-2}, q_{i+2}\), and let \(p_{ij}, r_{ij}\) be the points of intersection of \(k_i\) and \(k_j\). Then, for \(i\in \{1,\ldots ,5\}\), the following quadruples of points are concyclic:

  • \(p_{i+1,i-2},r_{i+1,i-2}, p_{i+2,i-1}, r_{i+2,i-1}\) lie on a circle \(c_i\) (see Fig. 2).

  • \(q_i,p_{i-1,i+1},r_{i-1,i+1},p_{i+2,i-2}\) lie on a circle \(d_i\) (see Fig. 3).

  • \(q_{i-1},q_{i+1}, p_{i+1,i+2},p_{i-2,i-1}\) lie on a circle \(e_i\) (see Fig. 4).

It does not matter which of the two points of intersection of the circles \(k_i\) and \(k_j\) is denoted by \(p_{ij}\) or \(r_{ij}\). However, let us agree that the intersection of \(k_{i-2}\) and \(k_{i+2}\) is \(q_i=r_{i+2,i-2}\) while the second point of intersection of \(k_{i-2}\) and \(k_{i+2}\) will be denoted \(p_{i+2,i-2}\). Note that \(k_{i-2}\) and \(k_{i+2}\) always have two points of intersection, \(q_i=r_{i+2,i-2}\) and \(p_{i+2,i-2}\). However, whether \(k_i\) and \(k_{i+2}\) have common points depends on the position of the points \(q_i\).

A geometric proof of Theorem 2 would certainly be possible. On the other hand, a direct approach using coordinates in order to compute the various points leads to horrible and really long expressions. Here, we propose a proof based on calculations with complex numbers and the theory of the power of a point with respect to a circle.

Interestingly, not only the Miquel Pentagon Theorem has this new relative in Theorem 2, but recently also a new variant of Morley’s Five Circles Theorem was discovered (see [2]).

Fig. 2
figure 2

First sibling of the Pentagon theorem

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Fig. 3
figure 3

Second sibling of the Pentagon theorem

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Fig. 4
figure 4

Third sibling of the Pentagon theorem

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It turns out that Theorem 2 follows easily from another incidence result, the mother of the siblings in Fig. 5, which we formulate now.

Theorem 3

In the configuration of Theorem 2 the 10 Möbius circles though \(I,p_{ij},r_{ij}\), \(i\ne j\in \{1,\ldots ,5\}\), are either concurrent in a point \(J\ne I\), or they touch each other in the point I.

Fig. 5
figure 5

The mother of the siblings

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2 Siblings of the Pentagon Theorem: The proof

In this section, we carry out the computations in the complex plane \(\mathbb {C}\). In particular, z denotes a complex variable, and \({\bar{z}}\) is its complex conjugate. The equation of a line through two different points p, q is given by

$$\begin{aligned} (p-z)({\bar{q}}-{\bar{z}})= ({\bar{p}}-{\bar{z}})(q-z). \end{aligned}$$

Indeed \(z=p\) and \(z=q\) are solutions of this equation, and expanded it has the form

$$\begin{aligned} {\bar{a}}z+a{\bar{z}}=c, \quad a\in \mathbb {C}\setminus \{0\},c\in \mathbb {R}, \end{aligned}$$

of a line.

Similarly, the equation of a circle through three different points p, q, r (which do not lie on a line) is given by

$$\begin{aligned} (p-q)(r-z)({\bar{p}}-{\bar{z}})({\bar{r}}-{\bar{q}}) = ({\bar{p}}-{\bar{q}})({\bar{r}}-{\bar{z}})(p-z)(r-q) \end{aligned}$$

since \(z=p,z=q,z=r\) are solutions of this equation, and expanded it has the form

$$\begin{aligned} (z-c)({\bar{z}}-{\bar{c}})=r,\quad c\in \mathbb {C},r\in \mathbb {R}_+, \end{aligned}$$

of a circle.

Observe that the group of Möbius transformations

$$\begin{aligned} z\mapsto \frac{az+b}{cz+d}, \quad \det \begin{pmatrix}a&{}b\\ c&{}d\end{pmatrix}\ne 0,\quad a,b,c,d\in \mathbb {C}, \end{aligned}$$

(with the usual convention \(1/0=\infty , 1/\infty =0\)) is sharply 3-transitive on the set of points and maps blocks (i.e. circles or lines) to blocks. The Möbius transformation \(z\mapsto 1/(z-I)\) maps the point I to the point \(\infty \). Hence we may assume without loss of generality that I is the point \(\infty \).

We first compute the points \(s_i\). The blocks \(h_1, \ldots , h_5\) are lines of the form

$$\begin{aligned} h_i: (q_{i-1}-z)({\bar{q}}_{i+1}-{\bar{z}}) =({\bar{q}}_{i-1}-{\bar{z}})(q_{i+1}-z). \end{aligned}$$

The point \(s_i\ne \infty \) is the intersection of the lines \(h_{i-2}\) and \(h_{i+2}\). Solving the corresponding linear system of the two equations yields

$$\begin{aligned} s_i= \frac{ (q_{i-2}-q_{i+1}) (q_{i+2} \bar{q}_{i-1}-q_{i-1} \bar{q}_{i+2}) - (q_{i-1}-q_{i+2}) (q_{i+1} \bar{q}_{i-2}-q_{i-2} \bar{q}_{i+1}) }{ (q_{i+2}-q_{i-1}) (\bar{q}_{i-2}-\bar{q}_{i+1}) - (q_{i-2}-q_{i+1}) (\bar{q}_{i+2} - \bar{q}_{i-1})}.\nonumber \\ \end{aligned}$$
(1)

The equation of the block \(k_i\) through the points \(s_i,q_{i-2},q_{i+2}\) is then given by

$$\begin{aligned} (s_i - q_{i+2})(q_{i-2} - z)({\bar{s}}_i - {\bar{z}})({\bar{q}}_{i-2} - {\bar{q}}_{i+2}) \\ = ({\bar{s}}_i - {\bar{q}}_{i+2})({\bar{q}}_{i-2} - {\bar{z}})(s_i - z)(q_{i-2} - q_{i+2}). \end{aligned}$$

By expanding the products we can bring this equation into the standard form

$$\begin{aligned} k_i\,:\,(z-c_i)({\bar{z}}-{\bar{c}}_i) = r_i \end{aligned}$$
(2)

with

$$\begin{aligned} c_i= & {} \frac{q_{i-2}q_{i+2} (\bar{q}_{i-2}-\bar{q}_{i+2})+ q_{i+2}s_i (\bar{q}_{i+2}-\bar{s}_i)+s_iq_{i-2} (\bar{s}_i-\bar{q}_{i-2})}{\bar{q}_{i-2} (q_{i+2}-s_i)+\bar{q}_{i+2} (s_i-q_{i-2})+\bar{s}_i(q_{i-2}-q_{i+2}) },\\ r_i= & {} -\frac{ (q_{i-2}-q_{i+2}) (\bar{q}_{i-2}-\bar{q}_{i+2}) (q_{i+2}-s_i) (\bar{q}_{i+2}-\bar{s}_i) (s_i-q_{i-2}) (\bar{s}_i-\bar{q}_{i-2}) }{\bigl (q_{i-2} (\bar{q}_{i+2}-\bar{s}_i)+q_{i+2} (\bar{s}_i-\bar{q}_{i-2})+(s_i (\bar{q}_{i-2}-\bar{q}_{i+2})\bigr )^2}. \end{aligned}$$

Using (1) in these formulas we obtain

$$\begin{aligned} c_i= & {} \frac{q_{i-2} q_{i-1} ({\bar{q}}_{i+1}-{\bar{q}}_{i-2})+q_{i+2} (q_{i+1} ({\bar{q}}_{i+2}-{\bar{q}}_{i-1})+q_{i-2} ({\bar{q}}_{i-2}+{\bar{q}}_{i-1}-{\bar{q}}_{i+1}-{\bar{q}}_{i+2}))}{q_{i-1} ({\bar{q}}_{i+1}-{\bar{q}}_{i-2})+q_{i+2} ({\bar{q}}_{i-2}-{\bar{q}}_{i+1})+(q_{i+1}-q_{i-2}) ({\bar{q}}_{i+2}-{\bar{q}}_{i-1})},\\ r_i= & {} - \frac{(q_{i+2}-q_{i-2}) ({\bar{q}}_{i+2}-{\bar{q}}_{i-2}) (q_{i-2}-q_{i+1}) ({\bar{q}}_{i-2}-{\bar{q}}_{i+1}) (q_{i+2}-q_{i-1}) ({\bar{q}}_{i+2}-{\bar{q}}_{i-1}) }{\bigl (q_{i+2} ({\bar{q}}_{i+1}-{\bar{q}}_{i-2})+ q_{i-1} ({\bar{q}}_{i-2}-{\bar{q}}_{i+1})-(q_{i+1}-q_{i-2}) ({\bar{q}}_{i+2}-{\bar{q}}_{i-1})\bigr )^2}. \end{aligned}$$

The equation of the radical axis \(a_{ij}\) of the circles \(k_i\) and \(k_j\) results by eliminating \(z{\bar{z}}\) from their respective equations (2). We obtain

$$\begin{aligned} a_{ij}\,:\, z({\bar{c}}_j-{\bar{c}}_i) + {\bar{z}}(c_j-c_i) = c_j{\bar{c}}_j -c_i{\bar{c}}_i +r_i-r_j. \end{aligned}$$
(3)

One finds that the intersection of any two of these radical axes is the point

$$\begin{aligned} J=\frac{\sum \nolimits _{i=1}^5 {\bar{q}}_i(q_{i-1}q_{i-2}-q_{i+1}q_{i+2})}{ \sum \nolimits _{i=1}^5 {\bar{q}}_i(q_{i-1}+q_{i-2}-q_{i+1}-q_{i+2})}. \end{aligned}$$
(4)

This can also be checked by substituting for z the given expression (4) for J in the equation (3) of the radical axis \(a_{ij}\). If the denominator in (4) is different from 0, the radical axes meet in \(J\ne I\), otherwise they are parallel. This completes the proof of Theorem 3.

An immediate consequence of Theorem 3 is the following.

Proposition 4

The point J has the same power with respect to all circles \(k_i\).

Indeed this follows directly from the fact that the radical axis of any two circles \(k_i,k_j\) passes through J.

The proof of Theorem 2 follows now directly from Proposition 4. Namely, consider four arbitrary points p, q, u, v among the points \(p_{ij}, r_{ij}\) such that p, q belong to one of the radical axes, and u, v belong to another radical axis. Then p, q, u, v are concyclic. This gives, apart from the circles \(k_i\), exactly the circles \(c_i,d_i\) and \(e_i\) in Theorem 2. And as a final conclusion we get the following.

Proposition 5

The radical axes of each two of the circles \(k_i,c_i,d_i,e_i\) either all meet in J or are all parallel.