Problem
GEO-B3-M02-P018 Pascal in Reverse
B. New Original Problem. Points \(A,B,C,D,E,F\) lie on one conic. Let \(P=AB\cap DE\) and \(Q=BC\cap EF\). The line \(PQ\) meets \(CD\) at \(R\). Prove that \(A,F,R\) are collinear.
C. Hint 1. Apply Pascal to the hexagon \(ABCDEF\).
D. Hint 2. The third Pascal point must already lie on the line \(PQ\).
E. Full Solution.
By Pascal's theorem for the hexagon \(ABCDEF\), the points
\[ P=AB\cap DE,\quad Q=BC\cap EF,\quad R_0=CD\cap FA \]
are collinear. The first two are \(P\) and \(Q\), so the third point \(R_0\) lies on the line \(PQ\).
But by definition \(R\) is the intersection of \(PQ\) with \(CD\). The point \(R_0\) also lies on \(PQ\) and on \(CD\). Hence \(R=R_0\), and therefore \(R\in FA\). Thus \(A,F,R\) are collinear.
A good task showing that Pascal can be used not only to prove collinearity, but also to recover a missing line.