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When I read Cox's Theorem, and Clifford's Circle Theorem and Miquel six circles theorem, I found the conjecture as folowing. And I checked the conjecture by the Geogebra sofware, the conjecture is right. I am looking for a proof of the conjecture.

Conjecture: Let four points $P_1, P_2, P_3, P_4$ on a plane $\sigma$. Let six planes $\sigma_{12}$, $\sigma_{13}$, $\sigma_{14}$, $\sigma_{23}$, $\sigma_{24}$ such that the plane $\sigma_{ij}$ through two points $P_i$ and $P_j$. Let $P_{ijk}$ be a common points of three planes $\sigma_{ij}, \sigma_{jk}, \sigma_{ki}$. Then show that four points $P_{123}, P_{124}, P_{134}, P_{234}$ lie on a plane.

See also Cox's Theorem

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closed as unclear what you're asking by Ramiro de la Vega, Dima Pasechnik, Alex Degtyarev, Stefan Kohl, Per Alexandersson Mar 29 '16 at 21:12

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    $\begingroup$ Unless I misread something, your statement is simply the dual of Cox's theorem as stated on the MathWorld page you provided: so it follows from it by projective duality (or by dualizing the proof). This is not a research-level question. $\endgroup$ – Gro-Tsen Mar 29 '16 at 21:05
  • $\begingroup$ I am not sure that I am right. But I get this ideas from Miquel six circles theorem. The Cox's theorem similarly the Clifford circles theorem. The Clifford circles theorem and Miquel six circles theorem is not same. I will think about your comment in one day and reply again. @Gro-Tsen . Thanks to You. $\endgroup$ – Oai Thanh Đào Mar 30 '16 at 0:55
  • $\begingroup$ Dear Mister @Gro-Tsen , the conjecture is not dual of the Cox's theorem. $\endgroup$ – Oai Thanh Đào Mar 30 '16 at 7:29
  • $\begingroup$ The dual of the statement contained in the first half of the MathWorld article "Cox's Theorem" is: «Let $P_1,\ldots,P_4$ be four points in general position on a plane $σ$ and let $σ_{ij}$ be a plane on the line $P_i P_j$. Let $P_{ijk}$ denote the intersection of $σ_{ij},σ_{ik},σ_{jk}$. Then the four points $P_{234},P_{134},P_{124},P_{123}$ are all on one plane $σ_{1234}$.» In what way is this different from the "conjecture" you wrote? Even the notation is the same! $\endgroup$ – Gro-Tsen Mar 30 '16 at 12:11
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    $\begingroup$ The dual of any theorem of incidence projective geometry is also a theorem, this doesn't need to be proved anew for each theorem. Read any good textbook on projective geometry (in short, take a nondegenerate quadric and use the polarity w.r.t. this quadric to reduce the theorem to its dual). Alternatively, take a purely incidence proof of Cox's theorem and dualize the proof. Again, this is well-known projective geometry, not research: please ask on math.stackexchange.com for more explanations on projective duality if you need, but not here. $\endgroup$ – Gro-Tsen Mar 30 '16 at 13:39