Pascal's theorem: Difference between revisions

Revision as of 21:00, 13 February 2022

Revision as of 17:36, 16 June 2022

Line 1:

{{short description|Theorem on the collinearity of three points generated from a hexagon inscribed on a conic}}

[[File:Pascaltheoremgenericwithlabels.svg|thumb|250px|Pascal line {{math|''GHK''}} of self-crossing hexagon {{math|''ABCDEF''}} inscribed in ellipse. Opposite sides of hexagon have the same color.]]

[[File:THPascal.svg|thumb|250px|right|The intersections of the extended opposite sides of simple [[cyclic polygon|cyclic]] hexagon {{math|''ABCDEF''}} (right) lie on the Pascal line MNP (left).]]

[[Image:Pascal'sTheoremLetteredColored.PNG|thumb|250px|Self-crossing hexagon {{math|''ABCDEF''}}, inscribed in a circle. Its sides are extended so that pairs of opposite sides intersect on Pascal's line. Each pair of extended opposite sides has its own color: one red, one yellow, one blue. Pascal's line is shown in white.]]

Line 54 ⟶ 53:

==Proof using cubic curves==

[[File:THPascal.svg|thumb|250px|right|The intersections of the extended opposite sides of simple [[cyclic polygon|cyclic]] hexagon {{math|''ABCDEF''}} (right) lie on the Pascal line MNP (left).]]

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, {{math|''M''}} and {{math|''N''}} in the figure) on the would-be Pascal line, and the ninth point as the third point ({{math|''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 {{math|''AB, CD, EF''}}, and the set {{math|''BC, DE, FA''}}), and the third cubic is the union of the conic and the line {{math|''MN''}}. Here the "ninth intersection" {{math|''P''}} cannot lie on the conic by genericity, and hence it lies on {{math|''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 coneconic if we choose a point {{math|''E''}} on the coneconic and a line {{math|''MP''}} in the plane. The sum of {{math|''A''}} and {{math|''B''}} is obtained by first finding the intersection point of line {{math|''AB''}} with {{math|''MP''}}, which is {{math|''M''}}. Next {{math|''A''}} and {{math|''B''}} add up to the second intersection point of the coneconic with line {{math|''EM''}}, which is {{math|''D''}}. Thus if {{math|''Q''}} is the second intersection point of the coneconic with line {{math|''EN''}}, then

:<math>(A + B) + C = D + C = Q = A + F = A + (B + C)</math>