Pascal: Necessary and Sufficient

The applet below illustrates the following construction. Given three points P, Q, R and a circle C(O). Choose a point A on the circle and extend AP to meet the circle in B'. Extend B'Q to meet the circle in C. Extend CR to meet the circle in A'. Extend A'P to meet the circle in B. Extend BQ to meet the circle in C' and, finally, extend C'R to meet the circle in, say, X. Under what conditions will X coincide with A?

Discussion

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The question is surely related to Pascal's theorem: from Pascal's theorem it follows that if X = A then necessarily P, Q, and R are collinear. We want to prove that the converse is also true. If P, Q, R are collinear then, for every choice of A, X = A.

So, assume P, Q, R are collinear but X ≠ A. Let R' be the intersection of AC' and A'C. Then, by Pascal's theorem, P, Q, and R' are collinear. A contradiction comes from the fact that both R and R' lie on A'C and also on PQ. Thus they can't be different. But if R = R' then both AC' and XC' pass through that point, meaning that they define one and the same line. This completes the proof.

(This argument is a very slight modification of the one from the Diameters and Chords article.)

Pascal and Brianchon Theorems

• Pascal's Theorem
• Pascal in Ellipse
• Pascal's Theorem, Homogeneous Coordinates
• Projective Proof of Pascal's Theorem
• Pascal Lines: Steiner and Kirkman Theorems
• Brianchon's theorem
• Brianchon in Ellipse
• The Mirror Property of Altitudes via Pascal's Hexagram
• Pappus' Theorem
• Pencils of Cubics
• Three Tangents, Three Chords in Ellipse
• MacLaurin's Construction of Conics
• Pascal in a Cyclic Quadrilateral
• Parallel Chords
• Parallel Chords in Ellipse
• Construction of Conics from Pascal's Theorem
• Pascal: Necessary and Sufficient
• Diameters and Chords
• Chasing Angles in Pascal's Hexagon

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