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CHASLES' THEOREM

By Hubert Shutrick

1. Pencils of Lines

Let MATH and MATH where $f_{s} = 0$ and $f_{t} = 0$ are two different lines in the plane. Each pair $s,t$ such that $(s,t) \neq (0,0)$ defines a line

  MATH

and these lines form a pencil of lines with base $f_{s},f_{t}$. For $r \neq 0$ the parameters $(rs,rt)$ give the same line as $(s,t)$ and so the parameters are homogeneous. If the lines $f_{s} = 0$ and $f_{t} = 0$ have a common point, then any line in the pencil also goes through the point. Conversely, if $f = 0$ is any line through the common point, then the columns of the matrix


  MATH

are linearly dependent using the common point, and, since the first two rows are independent, the last row must be a linear combination of them. Hence, $f = 0$ is in the pencil. A similar argument shows that, if $f_{s} = 0$ and $f_{t} = 0$ are parallel and $f = 0$ is also parallel, then it is in the pencil: the pencil consists of all lines parallel with the base lines.

2. Change of Base

If


  MATH

give two distinct lines of the pencil, then $f_{u},f_{v}$ can be a new base.


  MATH

The transition from $(u,v)$ to $(s,t)$ is given by the matrix formula


  MATH

It is important to note that the base is determined by $f_{s}, f_{t}$ but not by the lines $f_{s} = 0$ and $f_{t} = 0$ because, if $s_{u}\neq t_{v}$, the new base MATH has the same base lines with parameters $(0,1),(1,0)$ but the other lines get new parameters. To define the base geometrically, a third line, say MATH is required and the above change of base gives it parameters $(1,1)$.

3. Cross-ratio

In terms of the homogeneous parameters for the pencil, the convenient definition of cross-ratio is


  MATH

It agrees with the one parameter version if the parameter is replaced by $s/t$.


Its importance in this context is that it is depends only on the four lines and not on the choice of homogeneous parameters. This becomes obvious using that the determinant of the product of two matrices is the product of the determinants:


  MATH

If three different points are given along with the cross-ratio $k$ of them with a forth point, then the forth point is uniquely determined. To prove this, choose the base so that the points have parameters MATH. By the symmetry properties of cross-ratio they can be assumed to occur in the above order.


  MATH

4. Projective Plane

The projective plane is obtained from the ordinary Euclidean plane by adjoining an extra point for each pencil of parallel lines. It is the point `at infinity' where the lines of the pencil meet. To accommodate the new points it is convenient to introduce homogeneous coordinates. This is done by replacing the usual $x,y$ coordinates by $x/z,y/z$. Each MATH represents a point of the plane but, if $r \neq 0$, then $(rx,ry,rz)$ represents the same point so the coordinates are homogeneous.


A line in the plane is given by an equation $ax+by+cz = 0$, where not all of the coefficients are zero. In particular, $z=0$ gives the line at infinity and parallel lines


  MATH

meet at $(b,-a,0)$ as required.


The line joining points MATH and MATH can be described by homogeneous parameters $s,t$ since


  MATH

is on the line for each $(s,t) \neq (0,0)$. If MATH is a point not on the line, then the equation of the line joining it to the point with parameters $s,t$ is given by


  MATH

This can be rewritten


  MATH

The parameters for the line are the parameters for the pencil of lines through MATH. Therefore the cross-ratio of four points on a line is the same as the cross-ratio of the lines joining a point to them implying that cross-ratio is invariant under projection.

5. Change of Coordinates

The coordinate vectors MATH form a base for the coordinate system in the sense that $(x,y,z)= xX+yY+zZ $. If the points defined by


  MATH

are not collinear, then they can form a new base:


  MATH

In matrix form


  MATH

The matrix is invertible since the points are not collinear.


Note that the points defined by $X,Y,Z$ are not sufficient to determine the base because


  MATH

defines a new base with the same base points if $x_{x},y_{y},z_{z}$ are not all equal. To define a base geometrically, an extra point is needed, say, MATH which should not be on a line joining two of the other three points. The change of base above then gives this point coordinates $(1,1,1)$.

6. Chasles' Theorem

The theorem states that, for four points on a non-degenerate conic, the cross-ratio of the pencil of four lines from a fifth point of the conic to the given points does not depend on the choice of the fifth point.


To prove this choose a coordinate system such that $X$ and $Y$ are points of the conic, $Z$ is the intersection of the tangents from $X$ and $Y$ and the $(1,1,1)$ point is on the conic. The transformation of coordinates is linear and homogeneous so the conic will still be of the form


  MATH

Since the line $y=0$ is tangent at $(1,0,0)$, $a = g = 0$ to give the double root $z = 0$. Similarly, $b = f = 0$. Inserting $(1,1,1)$ reduces the equation to


  MATH

Consider the pencil of lines through $X$ with base $y,-z$. The line with parameters $s,t$ is $sy=tz$ and it meets the conic again at the point $(s^2,t^2,st)$. Thus, $s,t$ also give a homogeneous parametrisation of the conic. Now, consider the pencil through $Y$ with base $z,-x$. The line $sz=tx$ also meets the conic again at $(s^2,t^2,st)$: it gives the same parametrisation of the conic. It follows that the cross-ratios of pencils from $X$ and $Y$ to four points of the conic must be the same. Since $X$ and $Y$ were arbitrary points of the conic, the theorem is true.


(Chasles' theorem leads to an elegant projective proof of Pascal's Theorem. Also, another proof of Chasles' Theorem appears elsewhere.)

Copyright © 1996-2008 Alexander Bogomolny

28777977Page copy protected against web site content infringement by Copyscape



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