Problem
GEO-B3-M01-P015 Tangency Points in a Segment
In a circular segment with chord \(AB\), two circles are inscribed, each tangent to chord \(AB\) and to the arc of the segment. They meet at points \(M\) and \(N\). Prove that line \(MN\) passes through the fixed point of the arc equidistant from \(A\) and \(B\).
Hint 1. Choose the midpoint of arc \(AB\) as the inversion center.
Hint 2. This inversion sends chord \(AB\) to arc \(AB\).
E. Full solution. Let \(P\) be the point of the arc equidistant from \(A\) and \(B\). Take the inversion centered at \(P\) with power \(PA^2\). It swaps chord \(AB\) and arc \(AB\) of the segment: points \(A\) and \(B\) remain fixed, while the circle through \(P,A,B\) maps to line \(AB\). Circles inscribed in the segment map to circles of the same type. For two such circles, their common chord and its image must pass through the inversion center, because both circles are tangent to the two mutually inverse boundaries. Therefore their common chord \(MN\) passes through \(P\).
This is already an olympiad move: one must choose a special point of the arc, not an arbitrary center.