Geometry with Trigonometry

218 Vector and complex-number methods Ch. 11

Z 1

Z 2

Z 3

Z 4

Z 5

Z 6

Figure 11.15. Similar triangles on sides of triangle.

Z (^1)

Z 2

Z 3

Z 4

Z 5

Z 6

11.6.9 Circumcentres of similar triangles on sides of triangle .....

In a more complicated fashion than in the last subsection, for an arbitrary triangle
[Z 1 ,Z 2 ,Z 3 ]suppose that we take pointsZ 4 ,Z 5 andZ 6 so that

z 4 =z 2 +(p 1 +q 1 ı)(z 3 −z 2 ),z 3 =z 1 +(p 1 +q 1 ı)(z 5 −z 1 ),

z 2 =z 6 +(p 1 +q 1 ı)(z 1 −z 6 ), (11.6.10)

so thatwe have similar triangles once again on the sides of the original trian-
gle but now in the correspondences(Z 2 ,Z 3 ,Z 4 )→(Z 1 ,Z 5 ,Z 3 )→(Z 6 ,Z 1 ,Z 2 ).We

letZ( 16 i),Z 16 (ii),Z 16 (iii)be the circumcentres of these three similar triangles, so that by
(1.6.4)we have

z( 16 i)−z 2 =

1

2

(

1 +

p^21 −p 1 +q^21

q 1

ı

)

(z 3 −z 2 ), (11.6.11)

z( 16 ii)−z 1 =

1

2

(

1 +

p^21 −p 1 +q^21

q 1

ı

)

(z 5 −z 1 ), (11.6.12)

z( 16 iii)−z 6 =

1

2

(

1 +

p^21 −p 1 +q^21

q 1

ı

)

(z 1 −z 6 ). (11.6.13)

But by (11.6.9) we havez 5 −z 1 =p 1 +^1 q 1 ı(z 3 −z 1 )andz 2 −z 1 =( 1 −p 1 −q 1 ı)(z 6 −z 1 ),

so thatz 6 −z 1 = 1 −p 11 −q 1 ı(z 2 −z 1 ).