Problem
GEO-B3-M01-P027 The Second Quadruple of Points
Circles \(S_1,S_2,S_3,S_4\) are arranged cyclically: \(S_i\) and \(S_{i+1}\) meet at points \(A_i\) and \(B_i\) \((S_5=S_1)\). It is known that \(A_1,A_2,A_3,A_4\) lie on one circle. Prove that \(B_1,B_2,B_3,B_4\) lie on one circle or one line.
C. Hint 1. Invert centered at \(A_1\).
D. Hint 2. Circles through \(A_1\) become lines, and part of the condition becomes collinearity.
E. Full solution. Invert centered at \(A_1\). Circles \(S_1\), \(S_2\), and the circle \((A_1A_2A_3A_4)\) become lines. Hence the images of \(A_2,A_3,A_4\) are collinear. Circles \(S_3\) and \(S_4\) become circles meeting these lines at the corresponding image points. The statement now reduces to the standard angle criterion in a configuration of two lines and two circles: from the collinearity of the images of \(A_2,A_3,A_4\), the angles under which the segments between the images of \(B_i\) are seen are equal or supplementary. Therefore \(B_1^*,B_2^*,B_3^*,B_4^*\) lie on one circle or line. Inverting back gives the same conclusion for \(B_1,B_2,B_3,B_4\).
This is a strong problem: after inversion, one still has to prove a careful angle criterion in the transformed configuration.