Pascal's Theorem - Proof Using Cubic Curves

Proof Using Cubic Curves

Pascal's theorem has a short proof using the Cayley–Bacharach theorem that given any 8 points in general position, there is a unique ninth point such that all cubics through the first 8 also pass through the ninth point. In particular if 2 general cubics intersect in 8 points then any other cubic through the same 8 points meets the ninth point of intersection of the first two cubics. Pascal's theorem follows by taking the 8 points as the 6 points on the hexagon and two of the points (say, M and N in the figure) on the would-be Pascal line, and the ninth point as the third point (P in the figure). The first two cubics are two sets of 3 lines through the 6 points on the hexagon (for instance, the set AB, CD, EF, and the set BC, DE, FA), and the third cubic is the union of the conic and the line MN. Here the "ninth intersection" P cannot lie on the conic by genericity, and hence it lies on MN.

The Cayley–Bacharach theorem is also used to prove that the group operation on cubic elliptic curves is associative. The same group operation can be applied on a cone if we choose a point E on the cone and a line MP in the plane. The sum of A and B is obtained by first finding the intersection point of line AB with MP, which is M. Next A and B add up to the second intersection point of the cone with line EM, which is D. Thus if Q is the second intersection point of the cone with line EN, then

Thus the group operation is associative. On the other hand, Pascal's theorem follows from the above associativity formula, and thus from the associativity of the group operation of elliptic curves by way of continuity.