A Solution to Crux Mathematicorum Problem 5109
Crux Mathematicorum Problem 5109, followed by a synthetic fixed-point argument and a projective generalization.
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1 classical · 1 solution
Crux Mathematicorum Problem 5109, followed by a synthetic fixed-point argument and a projective generalization.
Two short proofs from quadratic equations
Subtracting circle equations produces the radical axes; adding normalized equations of two parabolas with perpendicular axes produces a circle through their four intersections.