# A conjecture like Cayley–Bacharach theorem

Let six points $A, A', B, B', C, C'$ lie on a conic and a cubic. Let a conic through $B, B', C, C'$ and meets the cubic again at $A_1, A_2$. Let a conic through $C, C', A, A'$ and meets the cubic again at $B_1, B_2$. Let a conic through $A, A', B, B'$ and meets the cubic again at $C_1, C_2$. Then six points $A_1,A_2, B_1, B_2, C_1, C_2$ lie on a conic. In this case if we let $A_1, B_1, C_1$ are collinear then $A_2, B_2, C_2$ are collinear.

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Special case 1, The cubis are three lines. Let six points $A, A', B, B', C, C'$ lie on a conic. Let a conic through $B, B', C, C'$ and meets line $AA'$ at $A_1, A_2$ Let a conic through $C, C', A, A'$ and meets line $BB'$ at $B_1, B_2$ Let a conic through $A, A', B, B'$ and meets line $CC'$ at $C_1, C_2$. Then six points $A_1,A_2, B_1, B_2, C_1, C_2$ lie on a conic. In this case if $A_1, B_1, C_1$ are collinear then $A_2, B_2, C_2$ are collinear.

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Special case 2, A conic and a cubic: Let a conic meets a cubic at six points $A, B, C, A', B', C'$. Let $AB', A'B$ cut the cubic at $C_1, C_2$ respectively; Let $BC', B'C$ cut the cubic again at $A_1, A_2$; Let $AC', A'C$ cut the cubic at $B_2, B_1$ then $A_1, B_1, C_1$ are collinear and $A_2, B_2, C_2$ are collinear.

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Special case 3, The conics are lines: Let two lines meet a cubic at $A, B, C$; and $A', B', C'$ respectively. Let $AB', A'B$ cut the cubic at $C_1, C_2$ respectively; Let $BC', B'C$ cut the cubic again at $A_1, A_2$; Let $AC', A'C$ cut the cubic at $B_2, B_1$ then $A_1, B_1, C_1$ are collinear and $A_2, B_2, C_2$ are collinear.

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Special case 4, The cubic are a line and a conic: Let the cubic is a conic ($\gamma$) through four points $A, A', B, B'$ and line $CC'$ on the plane). Let points $A, B, C, A', B', C'$ lie on another conic. Let $AB', A'B$ meet $CC'$ at $C_1, C_2$; let $AC', A'C, BC', B'C$ meet conic $\gamma$ at $B_2, B_1, A_1, A_2$. Then six points $A_1, A_2, B_1, B_2, C_1, C_2$ lie on a conic.

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• I checked the conjecture by software, I see this result is true. But I don't have a solution – Oai Thanh Đào Sep 5 '15 at 9:42

If the cubic is smooth and given the usual group law with an inflexion as identity, the condition that given six points $P_1,\ldots,P_6$ on the cubic they also lie on a conic is $P_1+\cdots+P_6=0$. In your case $$A+A'+B+B'+C+C'=0$$ $$A_1+A_2+B+B'+C+C'=0$$ $$A+A'+B_1+B_2+C+C'=0$$ $$A+A'+B+B'+C_1+C_2=0$$ Adding the last three equations and using the first, gives $A_1+A_2+B_1+B_2+C_1+C_2=0$, which is what you want. The general case probably follows by degenerating the cubic.
• Let two curves intersections at $\frac{d^2+3d}{2}+\frac{l^2+3l}{2}+2$ points. Let $\frac{d^2+3d}{2}+1$ points lie on a curve of degree $d$, then $\frac{l^2+3l}{2}+1$ extant points lie on a curve of degree $l$. – Oai Thanh Đào Sep 9 '15 at 10:12
• Dear Professor José @FelipeVoloch , I propose a conjecture as follows: A Conjecture: Let two curves intersect at $\frac{d^2+3d}{2}+\frac{l^2+3l}{2}+2$ points. Let the $\frac{d^2+3d}{2}+1$ points lie on a curve of degree $d$, then the $\frac{l^2+3l}{2}+1$ extant points lie on a curve of degree $l$. I am sorry if the conjecture is wrong. – Oai Thanh Đào Sep 9 '15 at 10:23