This conjecture is a generalization of [A theorem for cubic-A generalization of Carnot theorem, in MSE question](https://math.stackexchange.com/questions/1440544/a-theorem-for-cubic-a-generalization-of-carnot-theorem). I'm an electrical engineer, I am not a mathematician. I don't know how to prove this result.


Let $A\_1A\_2....A\_n$ be a n-gons, Let $m$ points $B\_{ij}$ lie on $A\_iA\_{i+1}$ for j=1, 2,...., m. Then $m.n$ points $B\_{ij}$ for $i=1, 2, 3,...., n$ and $j=1, 2,...., m$ lie on a curve of degree $m$ if only if:


$$\prod\_{i=1}^n \prod\_{j=1}^m \frac{\overline{B\_{ij}A\_i}}{\overline{B\_{ij}A\_{i+1}}}=1$$


Note that: $A\_{m+1}=A\_1$



A conjecture associated with n-gon cut curve of degree m
I believe you meant to type "curve of degree m". Let's prove the "only if" direction first. If you can prove it for triangles then you can triangulate your $n$-gon and multiply together the expressions for each triangle. So let's assume $n=3$.


Now, pick coordinates $(t\_0,t\_1,t\_2)$ on $\mathbb P^2$ so that the lines $t\_i=0$ correspond to the three lines of your triangle, and let $at\_0+bt\_1+bt\_2$ be a generic line that doesn't pass through any of your points. Consider the three functions $$f\_i=\frac{t\_i}{at\_0+bt\_1+ct\_2}$$
on your degree $m$ curve. [Weil reciprocity](https://en.wikipedia.org/wiki/Weil_reciprocity_law) then tells you that $\prod\_{P} (f\_i,f\_j)\_P=1$ so you can write $$\prod\_P (f\_0,f\_1)\_P(f\_1,f\_2)\_P(f\_2,f\_0)\_P=1$$
and by the definition of the Weil symbol this ends up being exactly the product $\prod\_{i=1}^n \prod\_{j=1}^m \frac{\overline{B\_{ij}A\_i}}{\overline{B\_{ij}A\_{i+1}}}$ in your statement. I learned this proof [here](http://www.maths.tcd.ie/report_series/tcdmath/tcdm1303.pdf). Note that the result is true even if you let points coincide and count them with multiplicities. 


To got the other way, omit one of the points and pick a degree $m$ curve through the remaining $3m-1$, by the calculation above the remaining point will be uniquely determined.


