Cross ratios and harmonic conjugate elements




Dividing ratio

Take P1 and P2 as two different real points on a line a, and take point P on a different from P2. From the theory about homogeneous coordinates we know that
 
PP1
----- = (P1,P2,P) = the dividing ratio of point P relative to (P1,P2)
PP2
Remark: If (P1,P2,P) < 0, point P is between P1 and P2.

In a cartesian coordinate system we take P(x,y); P1(x1,y1); P2(x2,y2). Then we know :

 
             x1 - k x2
        x = ------------   and
              1 - k

             y1 - k y2
        y =  -------------
               1 - k

Solving these equations for k, we find
 
             x - x1
        k = --------      and
             x - x2

             y - y1
        k = ---------
             y - y2

We also know that the ideal point of P1P2 has a dividing ratio = 1.

Cross Ratio

Take four different, regular points A,B,C,D on a line.
By definition we have :
 
        cross ratio of the ordered points A,B,C,D
           (A,B,C)
        = ---------
           (A,B,D)
We denote the cross ratio of the ordered points A,B,C,D as (A,B,C,D).

Cross Ratio and coordinates

Take A(x1,y1),B(x2,y2),C(x3,y3),D(x4,y4) on a line.
 
              (A,B,C)    CA     DA      x3 - x1     x4 - x1
(A,B,C,D) =  -------- = ---- : ---- =  --------- : ----------
              (A,B,D)    CB     DB      x3 - x2     x4 - x2

                                        y3 - y1     y4 - y1
                                    =   -------- : ----------
                                        y3 - y2     y4 - y2

Points in harmonic range

If (A,B,C,D) = -1 , we say that
A,B,C,D are in harmonic range or
A,B,C,D is a harmonic system of points or
C and D are harmonic conjugate points relative to A and B.
D is the harmonic conjugate point to C relative to A and B.

Coordinates of points in harmonic range

 
        (A,B,C,D) = -1
<=>
        x3 - x1     x4 - x1
        -------- : -------- = -1
        x3 - x2     x4 - x2
<=>
        x3 - x1      x4 - x1
        -------- = - -------
        x3 - x2      x4 - x2
<=>
        (x3 - x1)(x4 - x2) = - (x4 - x1)(x3 - x2)
<=>
        ...
<=>
        2(x1 x2 + x3 x4) = (x1 + x2)(x3 + x4)
Similarly, we find
 
        (A,B,C,D) = -1
<=>
        2(y1 y2 + y3 y4) = (y1 + y2)(y3 + y4)

Properties

Harmonic points in special coordinate systems

Cross Ratio and homogeneous coordinates

In an arbitrary coordinate system we have A(x1,y1,z1), B(x2,y2,z2). Since C and D are on line AB, there is a h and h' such that
 
        C(x1 + h x2, y1 + h y2, z1 + h z2)
        D(x1 + h' x2, y1 + h' y2, z1 + h' z2)
Then:
        (A,B,C) =

        x1 + h x2    x1
        --------- - ----
        z1 + h z2    z1                 h z2
        ---------------- = ... = ... = ------
        x1 + h x2    x2                 z1
        --------- - ---
        z1 + h z2    z2

Similarly
                  - h' z2
        (A,B,D) = --------
                    z1
And from these results

                     h
        (A,B,C,D) = ---
                     h'

Extending the notion of cross ratio to projective level.

Say A,B,C,D are four different collinear points in the projective plane. Then
 
 A(x1,y1,z1), B(x2,y2,z2),
 C(x1 + h x2, y1 + h y2, z1 + h z2), D(x1 + h' x2, y1 + h' y2, z1 + h' z2)
Now BY DEFINITION :
 
                     h
        (A,B,C,D) = ---
                     h'
and all previous properties hold.
It can be proved that cross ratio is a projective invariant.
With previous definition :
 
        (A,B,C,D) = -1 <=>  h' = - h <=> h + h' = 0

Special harmonic points in the affine plane

Choose A(x1,y1,1) and B(x2,y2,1) different and regular points.
 
        C(x1 + h x2, y1 + h y2, 1 + h)
        D(x1 + h' x2, y1 + h' y2, 1 + h')

        If h = -1 and h' = 1 then (A,B,C,D) = -1
but then we have
        C(x1 - x2, y1 - y2, 0)  and D(x1 + x2, y1 + y2, 2)
<=>
                                        x1 + x2  y1 + y2
        C(x1 - x2, y1 - y2, 0)  and D( --------,---------, 1)
                                           2        2
<=>
        C is the ideal point of AB and D is the midpoint of [AB]
Conclusion:
The midpoint of the segment [AB] is the harmonic conjugate point to the ideal point of the line AB relative to A and B.

An ordered quartet of lines

Take an ordered quartet of concurrent lines (a,b,c,d) in the projective plane.
 
        a: u1 x + v1 y + w1 z = 0
        b: u2 x + v2 y + w2 z = 0
        c: (u1 + h u2)x + (v1 + h v2)y + (w1 + h w2)z = 0
        d: (u1 + h'u2)x + (v1 + h'v2)y + (w1 + h'w2)z = 0
A line e intersects these lines respectively in A,B,C and D.
 
        e: uo x + vo y + wo z = 0
then
        A has coordinates (x1,y1,z1) =
           | vo   wo |     | uo   wo |   | uo   vo |
        (  | v1   w1 | , - | u1   w1 | , | u1   v1 | )

        B has coordinates (x2,y2,z2) =
           | vo   wo |     | uo   wo |   | uo   vo |
        (  | v2   w2 | , - | u2   w2 | , | u2   v2 | )

        C has coordinates
    | vo         wo  |    | uo         wo  |   | uo          vo  |
 (  | v1+hv2   w1+hw2|, - | u1+hu2   w1+hw2| ,  | u1+hu2   v1+hv2 | )

    | vo   wo |    | vo   wo |     | uo   wo |   | uo   wo |
 =( | v1   w1 | + h| v2   w2 | , - | u1   w1 |- h| u2   w2 | ,

                                    | uo   vo |    | uo   vo |
                                    | u1   v1 |+ h | u2   v2 | )

 = (x1 + h x2, y1 + h y2, s1 + h z2)

and here the same h appears as in the line c!

Similarly we find for D (x1 + h'x2, y1 + h'y2, s1 + h'z2)
and here the same h' appears as in the line d
From this, we deduce an important corollary : Take an ordered quartet of concurrent lines (a,b,c,d) in the projective plane. The cross ratio (A,B,C,D) = k is independent of the choice of the line e! So, we can define

Cross ratio of an ordered quartet of lines

From previous conclusion we can state: The cross ratio (a,b,c,d) is by definition the cross ratio of the intersection points of the ordered quartet of lines with an arbitrary line.

Harmonic quartet of lines

The ordered quartet of lines is harmonic if and only if (a,b,c,d) = -1. we say that
a,b,c,d are in harmonic range or
a,b,c,d is a harmonic system of lines or
c and d are harmonic conjugate lines relative to a and b.
d is the harmonic conjugate line to c relative to a and b.

Orthogonality and harmonic quartet of lines

If in a quartet of lines a,b,c, and d, the line c is orthogonal to d, then
 
        (a,b,c,d) = -1  <=>     b and c are the bisectors of a and b
The proof is left as an exercise.

Construction of a harmonic quartet

We start from the figure

We'll prove that (a,b,c,d) = -1 and (A,B,C,D) = -1

 
(A,B,C,D) = (a,b,c,d)

                intersection with line A'D'

        = (A',B',C',D')

        = (S'A',S'B',S'C',S'D') =

                intersection with line AD

        = (B,A,C,D)

Thus,  k = (A,B,C,D) = (B,A,C,D)

                CA     DA                    CB     DB
Now (A,B,C,D) = --- : ----   and (B,A,C,D) = --- : ----
                CB     DB                    CA     DA

So,          1
        k = ---  <=>    k2  = 1 <=> k = -1 or k = 1
             k
But a cross ratio = 1 is impossible for a quartet of points.
So, (A,B,C,D) = -1 and (a,b,c,d) = -1

Corollary




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