Solution to Exercise 129.3

Problem:

IN Exer 129.3 of Norman Wildwerger’s Math Foundations video: MF129: The projective line, circles, and a proof of the CQQ theorem we are asked to show that the quadrea of a triangle (necessarily ‘cyclic’) can be obtained from the signed area formula:

½ (x₁y₂ – x₂y₁ + x₂y₃ – x₃y₂ + x₃y₁ – x₁y₃)

Using the substitutions for a rational point lying on the unit circle centred at the origin :

[x(u,t)_i , y(u,t)_i] = [u_i² – t_i²/ u_i²+ t_i² , 2u_it_i / u_i²+ t_i²]

we must show the same formula becomes:

2 (u₁t₂-u₂t₁)(u₂t₃-u₃t₂)(u₃t₁-u₁t₃)/(u₁²+ t₁²)(u₂²+ t₂²)(u₃²+ t₃²)

 

Solution:

BEGIN by factoring a single ‘y‘ (say y₁ ) in the area formula, make substitutions and place over a common denominator:

½ ( x₃ – x₂ )y₁

   =  (½) 2u₁t₁((u₃²- t₃²)(u₂²+ t₂²)  –  (u₂²- t₂²)(u₃²+ t₃²))/(u₁²+t₁²)(u₂²+t₂²)(u₃²+t₃²)

   =          u₁t₁((u₃²- t₃²)(u₂²+ t₂²)  –  (u₂²- t₂²)(u₃²+ t₃²))/(u₁²+t₁²)(u₂²+t₂²)(u₃²+t₃²)

EXPAND the numerator and collect like terms:

               u₁t₁((u₃² –  t₃²)(u₂² +  t₂²)  –  (u₂² –  t₂²)(u₃² +  t₃²))/(u₁² + …

   =         u₁t₁(u₃²u₂²– t₃²t₂² + u₃²t₂²- t₃²u₂²)-(u₂²u₃²– t₂²t₃² + u₂²t₃²- t₂²u₃²)/(u₁² + …

   =        2u₁t₁(u₃²t₂² – t₃²u₂²)/(u₁² + t₁²)(u₂² + t₂²)(u₃² + t₃²)

WE notice a difference of squares in the numerator, but rather than factoring, instead distribute allowing the collection and permuting of terms:

      2u₁t₁(u₃²t₂² – t₃²u₂²)

   =        2u₁t₁(u₃²t₂²) –  2u₁t₁(u₂²t₃²)

   =        2 (u₁t₁)(u₃t₂)(u₃t₂) –  2(u₁t₁)(u₂t₃)(u₂t₃)

By commutativity (of multiplication) we amend the two distinct factors to beome three:

   =        2(u₁t₂)(u₃t₁)(u₃t₂) –  2(u₁t₃)(u₂t₁)(u₂t₃)

[in simpler notation, we write ‘(u₁t₂)’ as ‘(12)’ – signifying a ‘u before t‘ convention ]

CONTINUING, we have shown:

½ ( x₃ – x₂ )y₁

   =       2 ((u₁t₂)(u₃t_i)(u₃t₂) –  (u₁t₃)(u₂t_i)(u₂t₃))/(u₁² + …

   =       2 ((12)(31)(32) – (13)(21)(23))/…

AND thus:

½ (x₁y₂ – x₂y₁ + x₂y₃ – x₃y₂ + x₃y₁ – x₁y₃)

   =       2 ((12)(31)(32) – (13)(21)(23)) +

  ((23)(12)(13) – (21)(32)(31)) +

((31)(23)(21) – (32)(13)(12))/…

NOTE how ‘(32)’ and ‘(23)’ each appear beside two matching pairs in the second and third lines:

[ ‘(12)(13)’ = ‘(13)(12)’ & ‘(21)(31)’ = ‘(31)(21)’ ]

but beside two unmatched pairs in the first line:

[ (12)(31)(32)  –  (13)(21)(23) ]

However adding the following terms, as the second line:

+ (13)(21)(32*) –  (12)(31)(23*)

will give a ‘balanced’ expression (minus the adjustment term) of eight ‘triple’ terms which (by the binomial theorem) can then be factorised.

FROM  the ‘enlarged’ expression:

        2 ((12)(31)(32) – (13)(21)(23)) +

(13)(21)(32*) – (12)(31)(23*) +

  ((23)(12)(13) – (21)(32)(31)) +

((31)(23)(21) – (32)(13)(12))/…

–  [(13)(21)(32) – (12)(31)(23)]/…

We note that algebraically, since ‘like’  terms in corresponding positions still commute, the adjustment terms can be ignored. That is:

[(13)(21)(32*) – (12)(31)(23*)]/… =  0

THUS, for the area of a triangle, we have:

½ (x₁y₂ – x₂y₁ + x₂y₃ – x₃y₂ + x₃y₁ – x₁y₃)

=      2 (((23) – (32))((12)(13) – (12)(31)) –  ((23) – (32))((21)(13) – (21)(31)))/…

=      2 ( (23) – (32) ) ( (12)(13) – (12)(31) –  (21)(13) – (21)(31) )/…

=      2 ( (23) – (32) ) ( (12)( (13) – (31) ) –  (21)( (13) – (31)) )/…

=      2 ( (23) – (32) ) ( (12) –  (21) )( (13) – (31) )/…

=      2 ( (12) – (21) ) ( (23) –  (32) )( (13) – (31) )/…

=     – 2 (u₁t₂-u₂t₁)(u₂t₃-u₃t₂)(u₃t₁-u₁t₃)/(u₁²+ t₁²)(u₂²+ t₂²)(u₃²+ t₃²)

WHICH is (almost!) what we required to show since there is a ‘negative’ sign upon inspection of the terms in the original product.