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Consider three points $A, B$ and $C$ in the projective plane [the white points in the picture below] not all on one line. Next, choose a point $O$ [the blue point] and draw the lines that connect $O$ with $A, B$ and $C$ respectively. We will refer to these lines as the "blue lines". Then, we repeat this procedure with another point $O'$ in the plane [the red point in the picture]; the corresponding lines will be called the "red lines".

We then define the points of intersection of the blue line through A and the red line through B, the blue line through B and the red line through C, the blue line through C and the red line through A, and connect these intersection points [the purple points] via lines with $C, A$ and $B$ respectively. Prove that the latter three lines [the white lines in the picture] are concurrent.


I would like to obtain a proof of the statement above by constructing suitable (degenerate) cubics and/or conics, and applying classical theorems.

Thanks

enter image description here

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  • $\begingroup$ Could you add a picture clarifying the two plans you refer to and the various points? $\endgroup$ – Moti Nov 24 '18 at 15:32
  • $\begingroup$ Where is O? In what plan is O'? If intersected all are in same plan - right? $\endgroup$ – Moti Nov 24 '18 at 15:33
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    $\begingroup$ Seems like a variant of Desargues’ theorem. $\endgroup$ – amd Nov 24 '18 at 22:40
  • $\begingroup$ @Moti What is not clear about my question? The points O and O' are arbitrarily chosen in the plane. I will add a picture later of a possible configuration. $\endgroup$ – pdm Nov 24 '18 at 23:30
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    $\begingroup$ @amd: Pappos, not Desargues. Desargues is a $(10_3,10_3)$ configuration: 10 points each incident with 3 lines, and 10 lines each incident with 3 points. Pappos is $(9_3,9_3)$. Since these two are kind of the fundamental theorems in projective incidence geometry, most things boil down to one or the other, and counting helps distinguishing them. As an interesting fact aside, note that Pappos implies Desargues but not vice versa. Not relevant here. $\endgroup$ – MvG Nov 25 '18 at 16:25
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Inspect the combinatorics of your configuration. You have $9$ points, each of which is incident with $3$ lines of the configuration. Conversely you have $9$ lines, each incident with $3$ points. So the whole configuration is a $(9_3,9_3)$ configuration. There is just one such configuration which corresponds to a projective incidence theorem, namely Pappos' theorem (or Pappus', if you prefer that transliteration). Take a close look and you should be able to identify that configuration.

If you want to, you can view Pappos' theorem as a special case of Pascal's theorem, where the conic degenerates to a pair of lines. Or you might consider triplets of lines as a degenerate cubic and then treat Pappos' theorem as a special case of the Cayley-Bacharach theorem.

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The first step is to use the projective feature of the plane in the following way -

enter image description here

Let me know if this get you closer to the solution you looked for

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