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Pascal Lines: Steiner and Kirkman Theorems6 distinct points can be reordered to form 60 distinct hexagons. If the points are concyclic or lie on a conic, then, according to Pascal's theorem, they define a Pascal line. Concerning the latter, Steiner and Kirkman proved two remarkable theorems: Theorem (J. Steiner)The Pascal lines of the hexagons ABCDEF, ADEBCF, and ADCFEB are concurrent. Theorem (T. P. Kirkman)The Pascal lines of the hexagons ABFDCE, AEFBDC, and ABDFEC are concurrent.
Copyright © 1996-2009 Alexander Bogomolny
ExplanationFirst, let's introduce a convenient notation. Below, lAB will denote the left-hand side an equation LemmaLet points A, B, C, and D lie on the circle given by a second degree equation
where l and m are some real numbers. ProofLet X be a point concyclic with, but different from, A, B, C, or D. None of lAB(X), lCD(X), lBC(X), or lAD(X) equals 0. It is then possible to find l1 and m1 such that
Define f1 = l1lABlCD + m1lBClAD. What curve is described by the second degree equation Clearly, f1(A) = f1(B) = f1(C) = f1(D) = f1(X) = 0. Since five points define a second degree equation up to a multiplicative constant, f1 = af, for a constant a.
First of all we shall use Lemma to prove Pascal's theorem. The proofs of Steiner's and Kirkman's theorems will follow. Pascal's TheoremIf the points A, B, C, D, E, and F are concyclic, then the three points of intersection of lines AB and DE, BC and EF, and CD and FA are colinear. ProofThe quadrilaterals ABCD, AFED, and BEFC are inscribed in a circle whose equation is taken to be
From (1) and (2),
Let X be the point of intersection of AB and ED: Similarly, the point of intersection of CD and AF lies on the same line
Proof of Steiner's TheoremFrom (2) and (3), as in the proof of Pascal's theorem, we obtain that the points of intersection of lines AF and BE, ED and CF, AD and BC lie on the line
meet at a point. Indeed, if the first two lines intersect at Y, then
so that m1m2lAD(Y) = m1m3lBC(Y) = m2m3lEF(Y), from which we conclude that
provided m2 is not 0. However, at least one of the six coefficients is bound to be nonzero. It is then possible to rearrange the argument so as to end up with dividing by one which is not 0. Proof of Kirkman's TheoremThe proof of Pascal's theorem starts with three quadrilaterals ABCD, AFED, and BEFC. If instead, we chose the quadrilaterals ABFE, ABDC, and CDFE, we would have obtained the statement of Kirkman's theorem. Remark
References
 Pascal and Brianchon Theorems
Copyright © 1996-2009 Alexander Bogomolny |
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