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?


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

Discussion

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Copyright © 1996-2018 Alexander Bogomolny

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.


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

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.)

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Copyright © 1996-2018 Alexander Bogomolny
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