Geometry with Trigonometry

Sec. 11.6 Mobile coordinates 211

By cyclic rotation we can write down the other two coefficients and so have

Z 16 =

1

2

cosα

sinβsinγ

Z 1 +

1

2

cosβ

sinγsinα

Z 2 +

1

2

cosγ

sinαsinβ

Z 3.

That the sum of the coefficients here is equal to 1 follows from the identity

sin2α+sin2β+sin2γ=4sinαsinβsinγ,

for the angles of a triangle. For

sin2αF+sin2βF+sin2γF

=sin(αF+βF)cos(αF−βF)+2sinγFcosγF

=2sinγF[cos(αF−βF)+cosγF]

=2sinγF[cos(αF−βF)−cos(αF+βF)]

=2sinγF.2sinαFsinβF.

11.6.5 Other distinguished points for a triangle ...............

For the centroidZ 7 of[Z 1 ,Z 2 ,Z 3 ]we have by 11.5.3 thatz 7 =^13 (z 1 +z 2 +z 3 )and so

z 7 −z 2 =^13 (z 1 +z 2 +z 3 )−z 2 =^13 (z 1 −z 2 )+^13 (z 3 −z 2 )

=^13 (p 1 +q 1 ı)(z 3 −z 2 )+^13 (z 3 −z 2 )

=^13 (p 1 + 1 +q 1 ı)(z 3 −z 2 ). (11.6.6)

This givesthe complex coordinate of the centroid.
We next turn to the orthocentre of this triangle. PointsZon the line throughZ 1
perpendicular toZ 2 Z 3 have complex coordinates of the form

z=z 2 +(p 1 +qı)(z 3 −z 2 )=z 3 +(p 1 − 1 +qı)(z 3 −z 2 )

=z 3 +

p 1 − 1 +qı

p 1 +q 1 ı

(z 1 −z 2 ).

ForZ 3 Zto be also perpendicular toZ 1 Z 2 we also need the coefficient ofz 1 −z 2 here
to be purely imaginary. Thus we need

(p 1 − 1 )p 1 +qq 1 =0i.e.q=−

(p 1 − 1 )p 1

q 1

,

and so obtain

z 11 =z 2 +p 1

(

1 −

p 1 − 1

q 1

ı

)

(z 3 −z 2 ), (11.6.7)

asthe complex coordinate of the orthocentre.
It takes more of an effort to deal with the incentre. The point with complex co-
ordinatez 2 +|Z 21 ,Z 3 |(z 3 −z 2 )lies on the half-line[Z 2 ,Z 3 at a unit distance fromZ 2.