Problem
GEO-B1-M05-P032 An Operation with a Perpendicular Bisector
Three points are marked on the plane. Then the following operation is repeated: choose already marked points \(A,B,C\) and mark point \(D\), the reflection of \(A\) across the perpendicular bisector of \(BC\). Prove that if after several operations three distinct marked points become collinear, then the three initial points were collinear.
C. Hint 1. Argue by contradiction: assume the initial points were not collinear.
D. Hint 2. Then the initial points lie on a common circle. Check that the operation preserves this circle.
E. Full solution.
Assume the three initial points were not collinear. Then they lie on a unique circle \(\omega\).
We prove by induction that every newly marked point also lies on \(\omega\). Suppose chosen points \(A,B,C\) already lie on \(\omega\), and \(D\) is obtained by reflecting \(A\) across the perpendicular bisector of \(BC\).
The perpendicular bisector of chord \(BC\) passes through the center of circle \(\omega\). Hence \(\omega\) is symmetric with respect to this line. Therefore the reflection of point \(A\), which lies on \(\omega\), also lies on \(\omega\). Thus \(D\in\omega\).
So all marked points lie on one circle. But a line can meet a circle in at most two points, unless it coincides with the circle, which is impossible. Therefore three distinct marked points could not become collinear. This is a contradiction.
Hence, if three collinear points do appear, the three initial points must have been collinear.
A. Source analysis. Main objects: a perpendicular bisector, reflection, and a common circle. The obvious approach is to track coordinates, but the hidden observation is that the operation preserves the circle through the initial points. Number of key ideas: 2.
F. Difficulty justification. This is Level 6: a regional-style problem based on a circle invariant under reflection.
G. Check. This is not a one-step exercise: one must choose an invariant and prove it is preserved at each step.