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I am an electrical engineer system. I live in Viet Nam. I am not a Mathematician. I construct and found a problem as follows:

Let a cubic, and five conics $(C_1)$, $(C_2)$, $(C_3)$, $(C_4)$, $(C_5)$. The Cubic and a conic $(C_i)$ have six common points. Conic $(C_i)$ and conic $(C_{i+1}$ and the cubic have three common points for $i=1,2,3,4$. Let $P_1, P_2, P_3$ lie on $(C_1)$ and the cubic but don't lie on $(C_2)$ and $P_4, P_5, P_6$ lie on $(C_5)$ and cubic but don't lie on $(C_4)$. Then $P_1, P_2, P_3, P_4, P_5, P_6$ lie on a conic. I need a proof? Could you help me give your proof?

The cubic is three lines in this figure

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Here is a slightly related question of mine from a few months ago, that you might find interesting: mathoverflow.net/q/208292/27742 – Gabriel Nivasch Feb 29 at 8:01
    
Thank to You very much. Today I think about the problem about and found a conjecture at here: mathoverflow.net/questions/232430/a-problem-of-four-curves I hope that you like, and I need a answer. @GabrielNivasch – Oai Thanh Đào Feb 29 at 15:34
up vote 19 down vote accepted

Recall the quartic version of the Cayley-Bacharach theorem:

Theorem: Consider two quartics in general position, which intersect in $16$ points (by Bezout's theorem). Then if a third quartic passes through $13$ of those points, it also passes through the other $3$.

This can be used to establish a simpler result:

Lemma: Let $\Gamma$ be a cubic, and let $T_1, T_2, T_3, T_4$ be triples of points on $\Gamma$. Suppose that for each $i \in \{1, 2, 3\}$, the six points $T_i \sqcup T_{i+1}$ lie on a conic $C_i$. Then the six points $T_1 \sqcup T_4$ also lie on a conic.

Proof: We have an obvious set of $12$ points, and wish to extend it to a set of $16$ points so that Cayley-Bacharach is applicable. We let $P$ be the fourth intersection point of $C_1$ and $C_2$ (i.e. the one that isn't in $T_2$), and $Q$ be the fourth intersection point of $C_2$ and $C_3$. Now draw the line $l$ through $P$ and $Q$; it intersects $C_1$ again at a point $R$, and $C_3$ again at a point $S$.

Then $T_1 \sqcup T_2 \sqcup T_3 \sqcup T_4 \sqcup \{P, Q, R, S \}$ is a set of $16$ points. There are two quartics passing through it, namely $C_1 \cup C_3$ and $\Gamma \cup \ell$. Note that $C_2$ passes through $8$ of these $16$ points; let $C_4$ be the conic passing through any $5$ of the other $8$ points. Then $C_2 \cup C_4$ is a quartic passing through $13$ of the points, so passes through the other $3$ by Cayley-Bacharach. The result follows.

Corollary: Let $\Gamma$ be a cubic, and let $T_1, T_2, \dots, T_{2k}$ be triples of points on $\Gamma$. Suppose that for each $i \in \{1, 2, \dots, 2k - 1\}$, the six points $T_i \sqcup T_{i+1}$ lie on a conic $C_i$. Then the six points $T_1 \sqcup T_{2k}$ also lie on a conic.

Proof: Repeated application of previous result.

Your theorem is the case $k = 3$.

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