# Does this hexagon theorem have a name?

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 $$(L)$$

• 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 $y$. Jun 1, 2022 at 15:08