Triple intersection criteria:               Torsten Sillke
-----------------------------

In the following the complex conjugate is indicated by a prime.

Lemma 2 [1 p 56 + 58]

   Let a, b, c three complex numbers
   then the following conditions are equivalent:

   (1)  a, b, c are collinear

   (2)  (b - a)/(c - a) is real

   (3)  (b - a)/(c - a) = (b' - a')/(c' - a')

   (4)  |  a   a'  1  |
        |  b   b'  1  |  =  0
        |  c   c'  1  |


Lemma 2.9.2  [1 p 103]

   Suppose t1, t2, t3, t4 are points on the unit circle.
   Then (the extensions of) the chords joining the points
   t1, t2 and t3, t4 meet at z with

	  t1 + t2 - t3 - t4
     z' = -----------------
            t1 t2 - t3 t4

   Proof: We know the equations of the lines passing through 
   the points t1 and t2, and that of t3 and t4 are

     z + t1 t2 z' = t1 + t2,   z + t3 t4 z' = t3 + t4

   respectively. This follows from Lemma 2 and the 
   factorization (t1 + t2)(t1' - t2') = t2 t1' - t1 t2'.
   Hence the intersection of the two lines gives the 
   requested result.


Corollary

   Suppose t1, t2, t3, t4, t5, t6 are points on the unit circle.
   Draw the diaogonals for (t1, t2), (t3, t4), and (t5, t6).
   Then the following conditions are equivalent:

   (1)  (the extensions of) the three chords meet at one point,
	or the three diagonals are parallel.
   (2)  (t1 - t4)(t3 - t6)(t5 - t2) = (t1 - t6)(t3 - t2)(t5 - t4)
   (3)  |t1 - t4||t3 - t6||t5 - t2| = |t1 - t6||t3 - t2||t5 - t4|
   
   Proof:
   (1) <=> (2)
   Case: there are at most two chords parallel.
     Assume (t1, t2) is not parallel to (t3, t4) and (t5, t6).
     Then t1 t2 - t3 t4 != 0 and t1 t2 - t5 t6 != 0.

         triple-diagonal intersection

         t1 + t2 - t3 - t4     t1 + t2 - t5 - t6
     <=> -----------------  =  -----------------
	   t1 t2 - t3 t4         t1 t2 - t5 t6

     <=> (t1 + t2 - t3 - t4)(t1 t2 - t5 t6) = (t1 + t2 - t5 - t6)(t1 t2 - t3 t4)

     <=> (t1 - t4)(t3 - t6)(t5 - t2) = (t1 - t6)(t3 - t2)(t5 - t4)

   Case: all chords are parallel.
     Then t1 t2 - t3 t4  =  t1 t2 - t5 t6  =  t3 t4 - t5 t6  =  0.

     0 = (t1 + t2 - t3 - t4)(t1 t2 - t5 t6) = (t1 + t2 - t5 - t6)(t1 t2 - t3 t4)

     =>  (t1 - t4)(t3 - t6)(t5 - t2) = (t1 - t6)(t3 - t2)(t5 - t4)

   (2)  => (3)
     trivial

   (3)  => (1)
     ugly


References:

[1] Liang-shin Hahn;
   Complex Numbers and Geometry,
   MAA 1994 (Spectrum Series), ISBN 0-88385-510-0

[2] A. I. Markuschewitsch;
   Komplexe Zahlen und konforme Abbildungen,
   4. Aufl., Berlin 1973

[3] I. M. Jaglom;
   Komplexe Zahlen und ihre Anwendungen in der Geometrie,
   Moskau, 1969

[4] Herbert Pieper;
   Die komplexen Zahlen - Theorie - Praxis - Geschichte,
   Deutscher Verlag der Wissenschaften, Berlin, 1988, 2. Aufl.