# 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.

What if applet does not run? |

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

## Explanation

First, let's introduce a convenient notation. Below, *l*_{AB} will denote the left-hand side an equation *l*_{AB} = 0*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

*f* = l*l*_{AB}*l*_{CD} + m*l*_{BC}*l*_{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 l_{1} and m_{1} such that

l_{1}*l*_{AB}(X)*l*_{CD}(X) + m_{1}*l*_{BC}(X)*l*_{AD}(X) = 0.

Define *f*_{1} = l_{1}*l*_{AB}*l*_{CD} + m_{1}*l*_{BC}*l*_{AD}. What curve is described by the second degree equation *f*_{1} = 0?

Clearly, *f*_{1}(A) = *f*_{1}(B) = *f*_{1}(C) = *f*_{1}(D) = *f*_{1}(X) = 0. Since five points define a second degree equation up to a multiplicative constant, *f*_{1} = a*f*, 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.

(1) | f = l_{1}l_{AB}l_{CD} + m_{1}l_{BC}l_{AD}, |

(2) | f = l_{2}l_{AF}l_{ED} + m_{2}l_{AD}l_{EF}, |

(3) | f = l_{3}l_{BE}l_{CF} + m_{3}l_{BC}l_{EF}. |

From (1) and (2),

*f* = l_{1}*l*_{AB}*l*_{CD} - l_{2}*l*_{AF}*l*_{ED} = (m_{1}*l*_{BC} - m_{2}*l*_{EF})*l*_{AD}.

Let X be the point of intersection of AB and ED: *l*_{AB}(X) = 0*l*_{ED}(X) = 0.*l*_{AB}*l*_{CD} and *l*_{AF}*l*_{ED} vanish at X, while *l*_{AD} does not. It therefore follows that the function _{1}*l*_{BC} - m_{2}*l*_{EF}_{1}*l*_{BC} = m_{2}*l*_{EF}.

Similarly, the point of intersection of CD and AF lies on the same line _{1}*l*_{BC} = m_{2}*l*_{EF},

### 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 _{2}*l*_{AD} = m_{3}*l*_{BC}._{1}*l*_{AD} = m_{3}*l*_{EF}.

m_{1}*l*_{BC} = m_{2}*l*_{EF}, m_{2}*l*_{AD} = m_{3}*l*_{BC} and m_{1}*l*_{AD} = m_{3}*l*_{EF}

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

m_{1}*l*_{BC}(Y) = m_{2}*l*_{EF}(Y) and m_{2}*l*_{AD}(Y) = m_{3}*l*_{BC}(Y),

so that m_{1}m_{2}*l*_{AD}(Y) = m_{1}m_{3}*l*_{BC}(Y) = m_{2}m_{3}*l*_{EF}(Y), from which we conclude that

m_{1}*l*_{AD}(Y) = m_{3}*l*_{EF}(Y),

provided m_{2} 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

- Pascal's Theorem
- Pascal in Ellipse
- Pascal's Theorem, Homogeneous Coordinates
- Projective Proof of Pascal's Theorem
- Pascal Lines: Steiner and Kirkman Theorems
- Brianchon's theorem
- Brianchon in Ellipse
- The Mirror Property of Altitudes via Pascal's Hexagram
- Pappus' Theorem
- Pencils of Cubics
- Three Tangents, Three Chords in Ellipse
- MacLaurin's Construction of Conics
- Pascal in a Cyclic Quadrilateral
- Parallel Chords
- Parallel Chords in Ellipse
- Construction of Conics from Pascal's Theorem
- Pascal: Necessary and Sufficient
- Diameters and Chords
- Chasing Angles in Pascal's Hexagon
- Two Triangles Inscribed in a Conic
- Two Triangles Inscribed in a Conic - with Solution
- Two Pascals Merge into One
- Surprise: Right Angle in Circle

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

69824498