Three circles through P: checking the converse of the radical-axis argument
midnaMain·(edited)
I can see why a second common point should force the three opposite-vertex diagonals to concur. In the converse, how do I know the second intersection is real and distinct from P rather than a tangency? I'd like to settle that before calling the proof complete.
Problem:
Let be a cyclic and convex hexagon. A point is called *admissible* if it does not lie on the circumcircle of or on any of the lines , and . An admissible point is called *fantastic* if , and intersect in exactly two points. Prove that if there exists a fantastic point, then all admissible points are fantastic.
Source: Baltic Way 2023 Shortlist · MNET-01J8
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