Problem
GEO-B3-M01-P023 Contacts of a Chain
Circles \(R_1\) and \(R_2\) are tangent at \(A\). Circles \(S_1,\ldots,S_n\) are tangent to both \(R_1,R_2\), and \(S_i\) is tangent to \(S_{i+1}\) at \(T_i\). Prove that points \(T_1,\ldots,T_{n-1}\) lie on one circle through \(A\), or on one line.
Hint 1. Invert centered at \(A\).
Hint 2. Circles \(R_1,R_2\) become parallel lines.
E. Full solution. After inversion centered at \(A\), circles \(R_1\) and \(R_2\) become parallel lines. Each circle \(S_i\), tangent to both, maps to a circle tangent to these two parallel lines. Hence all images \(S_i^*\) have the same radius, and their centers lie on the midline between the parallel lines. If neighboring equal circles are tangent, their tangency points lie on the same midline. Thus points \(T_i^*\) are collinear. The inverse image of a line not passing through the inversion center is a circle through \(A\); if the line passes through \(A\), it remains a line. Therefore the original points \(T_i\) lie on one circle through \(A\), or on one line.
This is direct preparation for porisms of chains of circles.