# 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, lAB will denote the left-hand side an equation lAB = 0 of the straight line joining points A and B. (lAB is obviously defined up to a constant factor.) Thus, lAB(A) = lAB(B) = 0.

### Lemma

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

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 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 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 = l1lABlCD + m1lBClAD, (2) f = l2lAFlED + m2lADlEF, (3) f = l3lBElCF + m3lBClEF.

From (1) and (2),

f = l1lABlCD - l2lAFlED = (m1lBC - m2lEF)lAD.

Let X be the point of intersection of AB and ED: lAB(X) = 0 and lED(X) = 0. Then both lABlCD and lAFlED vanish at X, while lAD does not. It therefore follows that the function m1lBC - m2lEF vanishes at X. In other words, X lies on m1lBC = m2lEF.

Similarly, the point of intersection of CD and AF lies on the same line m1lBC = m2lEF, 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 m2lAD = m3lBC. 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 m1lAD = m3lEF. It is easy to check that the three lines at hand

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

m1lBC(Y) = m2lEF(Y) and m2lAD(Y) = m3lBC(Y),

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 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

1. 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.

2. There are even more elegant proofs of the two theorems.

### References

1. V. V. Prasolov, Essays on Numbers and Figures, AMS, 2000
2. G. Salmon, Treatise on Conic Sections, Chelsea Pub, 6e, 1960 ### Pascal and Brianchon Theorems 