Geometry with Trigonometry

Sec. 11.6 Mobile coordinates 217

11.6.8 Centroids of similar triangles erected on the sides of a triangle....

Given an arbitrary triangle[Z 1 ,Z 2 ,Z 3 ], if we consider pointsZ 4 ,Z 5 andZ 6 defined by

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

z 6 =z 1 +(p 1 +q 1 ı)(z 2 −z 1 ),

for some non-zero real numbersp 1 andq 1 ,thenwe have triangles[Z 2 ,Z 3 ,Z 4 ],
[Z 3 ,Z 1 ,Z 5 ],[Z 1 ,Z 2 ,Z 6 ]erected on the sides[Z 2 ,Z 3 ],[Z 3 ,Z 1 ]and[Z 1 ,Z 2 ], respectively,
which are similar to each other and have the same orientation as each other.
We denote the centroid of[Z 1 ,Z 2 ,Z 3 ]byZ 7 so thatz 7 =^13 (z 1 +z 2 +z 3 )and the
centroid of[Z 4 ,Z 5 ,Z 6 ]byZ′ 7 so that

z′ 7 =^13 [z 2 +(p 1 +q 1 ı)(z 3 −z 2 )+z 3 +(p 1 +q 1 ı)(z 1 −z 3 )+z 1 +(p 1 +q 1 ı)(z 2 −z 1 ),

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

=^13 (z 1 +z 2 +z 3 ).

HenceZ 7 ′is the same point asZ 7.

First we denote the centroid of[Z 2 ,Z 3 ,Z 4 ]byZ( 7 i)and then

z( 7 i)=^13 [z 2 +z 3 +z 4 ]=^13 [z 2 +z 3 +z 2 +(p 1 +q 1 ı)(z 3 −z 2 )].

Secondly we denote the centroid ofZ 3 ,Z 1 ,Z 5 ]byZ 7 (ii)and then

z( 7 ii)=^13 [z 3 +z 1 +z 3 +(p 1 +q 1 ı)(z 1 −z 3 )].

Finally we denote the centroid of[Z 1 ,Z 2 ,Z 6 ]byZ( 7 iii)and then

z( 7 iii)=^13 [z 1 +z 2 +z 1 +(p 1 +q 1 ı)(z 2 −z 1 )].

Then we have that

z( 7 i)+z( 7 ii)+z( 7 iii)

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

=z 1 +z 2 +z 3.

It follows that the centroid of the triangle[Z 7 (i),Z( 7 ii),Z 7 (iii)]is the same point as the
centroidZ 7 of the original triangle[Z 1 ,Z 2 ,Z 3 ].