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Question : Do you know this property of a hexagon?

Consider the configuration: Six points $A_1$, $A_2$, $A_3$, $A_4$, $A_5$, $A_6$ in a plane and let six points $B_i \in A_iA_{i+1}$ for $i=1, 2,\dots, 6$ and $7\equiv 1$.

  1. Let six points $B_1$, $\cdots$, $B_6$ are collinear and $$\frac{\overline{B_1A_1}}{\overline{B_1A_2}}. \frac{\overline{B_2A_2}}{\overline{B_2A_3}}. \frac{\overline{B_3A_3}}{\overline{B_3A_4}}.\frac{\overline{B_4A_4}}{\overline{B_4A_5}}. \frac{\overline{B_5A_5}}{\overline{B_5A_6}}. \frac{\overline{B_6A_6}}{\overline{B_6A_1}}=1$$ then six points $A_1$, $A_2$, $A_3$, $A_4$, $A_5$, $A_6$ lie on a conic.
  2. Let six points $A_1$, $A_2$, $A_3$, $A_4$, $A_5$, $A_6$ lie on a conic and five points $B_1$, $\cdots$, $B_5$ lie on line $(L)$ and $$\frac{\overline{B_1A_1}}{\overline{B_1A_2}}. \frac{\overline{B_2A_2}}{\overline{B_2A_3}}. \frac{\overline{B_3A_3}}{\overline{B_3A_4}}.\frac{\overline{B_4A_4}}{\overline{B_4A_5}}. \frac{\overline{B_5A_5}}{\overline{B_5A_6}}. \frac{\overline{B_6A_6}}{\overline{B_6A_1}}=1$$ then $B_6$ also lie on $(L)$

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    $\begingroup$ "Can I call it Viet Nam theorem?" seems like a strange question. If it's known, then you probably can't name it; and, if it isn't known, then you can call it anything you like, as long as you can convince other people to call it that. (But I have no idea if it's known.) $\endgroup$
    – LSpice
    Commented Jun 1, 2022 at 14:52
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    $\begingroup$ This is probably not what you want so only a comment. There is a simple analytic method to prove and extend this result: you may assume that the $A$ points are $(0,0)$, $(1,0)$, $(0,1)$, $(a,b)$, $(c,d)$, $(x,y)$, You can then compute the $B$´s and hence the area of the corresponding hexagon (using wedge products). It is then easy to calculate that the latter´s vanishing (or even constancy) is a quadratic equation in $x$ and $y$. $\endgroup$
    – memorial
    Commented Jun 1, 2022 at 15:08
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    $\begingroup$ Some comments: (1) The standard name that authors should give their theorems would be something like "Theorem 1". It's up to the rest of the community to decide if they want to name it otherwise. (2) This appears to be a theorem in projective geometry; these ratios are really cross-ratios between the secants of the conic and the intersections with two lines, the line displayed and the line at infinity, but by projective invariance of cross-ratio one could in fact take two arbitrary lines. (c) It may also be that one can replace the conic and two lines with a more general quartic curve. $\endgroup$
    – Terry Tao
    Commented Jun 1, 2022 at 17:24
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    $\begingroup$ ... indeed, now that I think about it a bit more, this result is presumably reflecting some property of the Jacobian of a quartic curve. Perhaps someone more expert in algebraic geometry than I can figure out the actual connection. $\endgroup$
    – Terry Tao
    Commented Jun 1, 2022 at 17:27
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    $\begingroup$ Thank Professors @TerryTao, LSpice and Memorial for your comment. $\endgroup$
    – Đào Thanh Oai
    Commented Jun 1, 2022 at 23:36

1 Answer 1

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There is a similar equation in Lazare Carnot's "Essai sur la théorie des transversales". However, Carnot's equation seems to hold for an arbitrary n-gon, which is a bit strange.

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