# 5 - On the Intersection of Curves  pp. 25-27

By Arthur Cayley
• By Arthur Cayley

• Publisher: Cambridge University Press

Online Publication Date:October 2010

Original Publication Year:1889

Online ISBN:9780511703676

Paperback ISBN:9781108004930

• Chapter DOI: http://dx.doi.org/10.1017/CBO9780511703676.006

Subjects: History of mathematical texts

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The following theorem is quoted in a note of Chasles' Aperçu Historique &c., Memoires de Bruxelles, tom. xi. p. 149, where M. Chasles employs it in the demonstration of Pascal's theorem: “If a curve of the third order pass through, eight of the points of intersection of two curves of the third order, it passes through the ninth point of intersection.” The application in question is so elegant, that it deserves to be generally known. Consider a hexagon inscribed in a conic section. The aggregate of three alternate sides may be looked upon as forming a curve of the third order, and that of the remaining sides, a second curve of the same order. These two intersect in nine points, viz. the six angular points of the hexagon, and the three points which are the intersections of pairs of opposite sides. Suppose a curve of the third order passing through eight of these points, viz. the aggregate of the conic section passing through the angular points of the hexagon, and of the line joining two of the three intersections of pairs of opposite sides. This passes through the ninth point, by the theorem of Chasles, i.e. the three intersections of pairs of opposite sides lie in the same straight line, (since obviously the third intersection does not lie in the conic section); which is Pascal's theorem.

The demonstration of the above property of curves of the third order is one of extreme simplicity. Let U = 0, V = 0, be the equations of two curves of the third order, the curve of the same order which passes through eight of their points of intersection (which may be considered as eight perfectly arbitrary points), and a ninth arbitrary point, will be perfectly determinate.

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