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I need to prove the following theorem

Let the hexagon $ABCDEF$ be inscribed in the nondegenerate conic $q=V(f)$. Assume that $A,B,C,D,E,F$ are distinct. Let $P=\overline{FA}\cap \overline{CD}, Q=\overline{AB}\cap \overline{DE}$ and $R=\overline{BC}\cap \overline{EF}$. Prove that $P, Q$ and $R$ are collinear.

I need to do it in two steps, in the first step we take $G\in q$ to be any other point and then I need to show that there is a homogeneous cubic $c$ so that $V(c)$ vanishes at $G$ and at $A,B,C,D,E,F,P,Q,R$.

Then I need to show that $V(c)$ is the union of $q$ and a line $l$ and that $l$ goes through $P,Q$ and $R$.

I am not sure how to construct the required cubic $c$.

Jimmy R
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1 Answers1

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The most direct route to prove this (as mentioned below in the comments) is using Bezout's theorem, as seen here: http://en.wikipedia.org/wiki/Pascal%27s_theorem#Proof_using_B.C3.A9zout.27s_theorem

Another way of proving Pascal's Theorem is using the Cayley-Bacharach Theorem. In fact the proof of the Cayley-Bacharach theorem uses Bezout's theorem a few times.

Partition your set of 6 lines into two sets of three, call them $X$ and $Y$, such that $X\cap Y$ is exactly the 9 points $A,B,C,D,E,F,P,Q,R$. $X$ and $Y$ are each degenerate cubics. Now let us define $Z$ as the union of the conic $q$ and the line passing through $P$ and $Q$. $Z$ is also a conic, and it passes through 8 of the 9 intersection points of $X$ and $Y$, so by the Cayley-Bacharch Theorem, it must contain the $9^{th}$ point (which is $R$ in this case). This means $R$ has to be on the line passing through $P$ and $Q$, that is, the three points are collinear.

Sergio Da Silva
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