Problem
GEO-B3-M02-P016 Antipodal Projectivity
B. New Original Problem. A circle \(\omega\), a point \(M\in\omega\), and a line \(l\) not passing through \(M\) are given. For \(X\in l\), the line \(MX\) meets \(\omega\) again at \(Y\). Let \(Y'\) be the point of the circle antipodal to \(Y\). The line \(MY'\) meets \(l\) at \(X'\). Prove that the map \(X\mapsto X'\) is a projective transformation of the line \(l\).
C. Hint 1. Break the map into three steps.
D. Hint 2. The half-turn of the circle also gives a projective transformation after parametrising the circle by a line.
E. Full Solution.
The map consists of three steps: projection from \(l\) to the circle with center \(M\), the antipodal map on the circle, and projection from the circle back to \(l\) with center \(M\).
The first and third steps are projective after identifying the circle with a line. The antipodal map is the restriction of a plane motion preserving the circle, and therefore in the same parametrisation it also gives a projectivity.
The composition of three projective maps is a projective transformation of \(l\). Hence \(X\mapsto X'\) is projective.
The task is useful before problems about fixed points of such compositions.