Geometry with Trigonometry

220 Vector and complex-number methods Ch. 11

By (11.6.4) the circumcentreZ′ 16 of the triangle[Z 4 ,Z 5 ,Z 6 ]is given by

z′ 16 =z 5 +^12

[

1 +

p^21 +q^21

q 1

ı−

p 1

q 1

ı

]

(z 6 −z 5 ), (11.6.15)

and if we relate this to the original triangle it becomes

z′ 16 =z 2 +^12 ( 1 +p 1 +q 1 )ı(z 3 −z 2 )+^12

[

1 +−

p 1

q 1

+

p^21 +q^21

q 1

ı

]

(−^12 )(z 3 −z 2 )

=z 2 +^12

[

1 +p 1 +q 1 ı−^12 +^12 p^1

q 1

−p

2

1 +q

2

1

2 q 1

ı

]

(z 3 −z 2 )

=z 2 +^12

[

1

2 +p^1 +q^1 ı+

1

2

p 1

q 1

ı−

p^21 +q^21

2 q 1

ı

]

(z 3 −z 2 ). (11.6.16)

We now find another point on this circle with centre (11.6.16). The footZ 8 of the
perpendicular fromZ 1 ontoZ 2 Z 3 has complex coordinatez 8 =z 2 +p 1 (z 3 −z 2 ).We
seek the circumcentre of the triangle[Z 4 ,Z 8 ,Z 6 ]. For the moment let us take

z 4 =z 8 +(p 2 +q 2 ı)(z 6 −z 8 ) (11.6.17)

from which

z 4

=( 1 −p 2 −q 2 ı)z 8 +(p 2 +q 2 ı)z 6

=( 1 −p 2 −q 2 ı)[z 2 +p 1 (z 3 −z 2 )]+ (p 2 +q 2 ı)[z 2 +^12 (p 1 +q 1 ı)(z 3 −z 2 )]

=( 1 −p 2 −q 2 ı+p 2 +q 2 ı)z 2 +[p 1 ( 1 −p 2 −q 2 ı)+(p 2 +q 2 ı)^12 (p 1 +q 1 ı)](z 3 −z 2 )

=z 2 +

{

p 1 ( 1 −p 2 )+^12 (p 2 p 1 −q 2 q 1 )+[−p 1 q 2 +^12 (p 2 q 1 +p 1 q 2 )]ı]

}

(z 3 −z 2 )

=z 2 +^12 (z 3 −z 2 ).

Thus
p 1 ( 1 −p 2 )+^12 (p 2 p 1 −q 2 q 1 )+[−p 1 q 2 +^12 (p 2 q 1 +p 1 q 2 )]ı=^12 ,

from which

1

2 (p^2 q^1 −p^1 q^2 )=^0 , p^1 (^1 −p^2 )+

1

2 (p^2 p^1 −q^2 q^1 )=

1

2.

Asp 1 =0wehaveq 2 =pp^21 q 1 .As2p 1 − 2 p 1 p 2 +p 2 p 1 −q 2 q 1 =1wehave
2 p 1 −p 1 p 2 = 1 +q 2 q 1. Hence 2p 1 −p 1 p 2 = 1 +pp^21 q^21 and so

2 p 1 − 1 =p 2 (

q^21

p 1

+p 1 ), p 2 =

( 2 p 1 − 1 )p 1

p^21 +q^21

,

q 2 =

q 1

p 1

( 2 p 1 − 1 )p 1

p^21 +q^21

=

( 2 p 1 − 1 )q 1

p^21 +q^21

.