@article {MATHEDUC.06499626,
author = {Fisher, J. Chris and Schr\"oder, Eberhard M. and Stevens, Jan},
title = {Circle incidence theorems.},
year = {2015},
journal = {Forum Geometricorum},
volume = {15},
issn = {1534-1178},
pages = {211-228},
publisher = {Florida Atlantic University, Department of Mathematical Sciences, Boca Raton, FL},
abstract = {Let us consider $n$ points $A_1, A_2, \dots, A_n$ $(n\ge 5)$ in the plane, such that no three are collinear and the lines $l_i=\langle A_{i-1},A_{i+1}\rangle$ and $l_{i+1}=\langle A_i,A_{i+2}\rangle$ intersect in a point $B_{i,i+1}$ (indices are considered modulo $n$). Let $c_{i,i+1}$ be the circle passing through the points $A_i$, $B_{i,i+1}$ and $A_{i+1}$, and let $g_i$ be the radical axis of the circles $c_{i-1,i}$ and $c_{i,i+1}$. Larry Hoehn discovered a concurrence theorem about pentagons, which says that the five radical axes $g_1, g_2, \dots, g_5$ lie in a pencil (they are concurrent or parallel). For $n\ge 6$, the radical axes $g_1, g_2, \dots, g_n$ do not lie in a pencil in general. In this paper, the authors show that in the case $n=6$ they lie in a pencil if and only if the six points $B_{i,i+1}$ lie on a conic. Also, the theorem about pentagons is generalized to $n$-gons. The main theorem states that, if the lines $g_1, g_2,\dots, g_{n-3}$ lie in a pencil, then the remaining three radical axes $g_{n-2}$, $g_{n-1}$ and $g_n$ lie in the same pencil. The proof is given in the affine plane over an arbitrary field with weaker assumptions.},
reviewer = {Hiroshi Okumura (Osaka)},
msc2010 = {G45xx (G75xx)},
identifier = {2016c.00746},
}