Pascal Lines: Steiner and Kirkman Theorems

6 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.

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Explanation

Explanation

First, let's introduce a convenient notation. Below, l_AB will denote the left-hand side an equation l_AB = 0 of the straight line joining points A and B. (l_AB is obviously defined up to a constant factor.) Thus, l_AB(A) = l_AB(B) = 0.

Lemma

Let points A, B, C, and D lie on the circle given by a second degree equation f = 0. Then

f = ll_ABl_CD + ml_BCl_AD,

where l and m are some real numbers.

Proof

Let X be a point concyclic with, but different from, A, B, C, or D. None of l_AB(X), l_CD(X), l_BC(X), or l_AD(X) equals 0. It is then possible to find l₁ and m₁ such that

l₁l_AB(X)l_CD(X) + m₁l_BC(X)l_AD(X) = 0.

Define f₁ = l₁l_ABl_CD + m₁l_BCl_AD. What curve is described by the second degree equation f₁ = 0?

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 Theorem

If 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.

Proof

The quadrilaterals ABCD, AFED, and BEFC are inscribed in a circle whose equation is taken to be f = 0. f, therefore, may be represented in either one of the three ways:

(1)	f = l₁l_ABl_CD + m₁l_BCl_AD,
(2)	f = l₂l_AFl_ED + m₂l_ADl_EF,
(3)	f = l₃l_BEl_CF + m₃l_BCl_EF.

From (1) and (2),

f = l₁l_ABl_CD - l₂l_AFl_ED = (m₁l_BC - m₂l_EF)l_AD.

Let X be the point of intersection of AB and ED: l_AB(X) = 0 and l_ED(X) = 0. Then both l_ABl_CD and l_AFl_ED vanish at X, while l_AD does not. It therefore follows that the function m₁l_BC - m₂l_EF vanishes at X. In other words, X lies on m₁l_BC = m₂l_EF.

Similarly, the point of intersection of CD and AF lies on the same line m₁l_BC = m₂l_EF, and, for, the intersection of BC and EF that is obvious.

Proof of Steiner's Theorem

From (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 m₂l_AD = m₃l_BC. Similarly, it follows from (1) and (3) that the point of intersection of AB and CF, CD and BE, AD and EF all lie on the line m₁l_AD = m₃l_EF. It is easy to check that the three lines at hand

m₁l_BC = m₂l_EF, m₂l_AD = m₃l_BC and m₁l_AD = m₃l_EF

meet at a point. Indeed, if the first two lines intersect at Y, then

m₁l_BC(Y) = m₂l_EF(Y) and m₂l_AD(Y) = m₃l_BC(Y),

so that m₁m₂l_AD(Y) = m₁m₃l_BC(Y) = m₂m₃l_EF(Y), from which we conclude that

m₁l_AD(Y) = m₃l_EF(Y),

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 Theorem

The 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

However the symmetry between Steiner's and Kirkman's theorems is begging. For example, it can be shown that each of the 60 Pascal lines that correspond to six points on a circle belong to exactly one Steiner triple, while each belongs to exactly three Kirkman's triples. So there are 20 Steiner points and 60 Kirkman's points.
There are even more elegant proofs of the two theorems.

References

V. V. Prasolov, Essays on Numbers and Figures, AMS, 2000
G. Salmon, Treatise on Conic Sections, Chelsea Pub, 6e, 1960

Pascal and Brianchon Theorems

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