Problem
GEO-B3-M01-P025 Fixed Circle of Contacts
Two circles \(R_1\) and \(R_2\) are tangent at \(A\). Circles \(S_1,\ldots,S_n\) are tangent to both \(R_1\) and \(R_2\), with \(S_i\) tangent to \(S_{i+1}\) at \(T_i\). Also \(S_n\) is tangent to \(S_1\). Prove that points \(T_1,\ldots,T_n\) lie on one circle passing through \(A\).
C. Hint 1. The inversion center should be the tangency point \(A\).
D. Hint 2. After inversion, \(R_1\) and \(R_2\) become parallel lines, and all \(S_i\) become equal circles.
E. Full solution. Invert centered at \(A\). Since \(R_1\) and \(R_2\) are tangent at the inversion center, their images are two parallel lines. Each circle \(S_i\), tangent to both given circles, maps to a circle tangent to the two parallel lines. Hence all \(S_i^*\) have the same radius, and their centers lie on the midline between the parallel lines. If two equal circles are tangent, their tangency point lies on the line of centers; therefore all points \(T_i^*\) lie on that midline. The inverse image of this line is a circle passing through the inversion center \(A\). Hence the original points \(T_i\) lie on one circle through \(A\).
The final-level difficulty lies in choosing the center: without inversion, the statement about many tangency points looks like a strong theorem about a chain of circles.