Geometry with Trigonometry

Sec. 11.6 Mobile coordinates 219

Also, by (11.6.12)

z( 16 iii)+(p 1 +q 1 ı)(z( 16 ii)−z( 16 iii))

=( 1 −p 1 −q 1 ı)z( 16 iii)+(p 1 +q 1 ı)z( 16 ii)

=( 1 −p 1 −q 1 ı)(z 16 (iii)−z 1 )+(p 1 +q 1 ı)(z( 16 ii)−z 1 )+( 1 −q 1 ı+z 1 ı)z 1

=( 1 −p 1 −q 1 ı)(z( 16 iii)−z 1 )+(p 1 +q 1 ı)(z( 16 ii)−z 1 )+z 1

=^12

[

1 −

p^21 −p 1 +q^21

q 1

ı

]

(z 2 −z 1 )+^12

[

1 +

p^21 −p 1 +q^21

q 1

ı

]

(z 3 −z 1 )+z 1

=^12 (z 2 +z 3 )+^12

[

p^21 −p 1 +q^21

q 1

ı(z 3 −z 1 )−

p^21 −p 1 +q^21

q 1

ı(z 2 −z 1 )

]

=^12 (z 2 +z 3 )+^12

p^21 −p 1 +q^21

q 1

ı(z 3 −z 1 −(z 2 −z 1 ))

=^12 (z 2 +z 3 )+^12

p^21 −p 1 +q^21

q 1

ı(z 3 −z 2 )

=z( 16 i).

It follows thatthe triangle[Z 16 (i),Z 16 (ii),Z( 16 iii)]is also similar to the similar triangles

above, in the correspondence(Z 2 ,Z 3 ,Z 4 )→(Z( 16 iii),Z 16 (ii),Z( 16 i)).
In the particular case whenp 1 = 1 / 2 ,q 1 =

√

3 /2, the similar triangles are all
equilateral triangles, that is all three sides have equal lengths. In this case the last
result is known asNapoleon’s theorem. It is easier to prove than the more general
case, as for it we can work just with centroids.

11.6.10 Triangle with vertices the mid-points of sides of given triangle

For a triangle[Z 1 ,Z 2 ,Z 3 ]letZ 4 ,Z 5 ,Z 6 be the mid-points of the the sides[Z 2 ,Z 3 ],
[Z 3 ,Z 1 ],[Z 1 ,Z 2 ], respectively. Suppose thatz 1 =z 2 +(p 1 +q 1 ı)(z 3 −z 2 ).Now

z 4 =^12 (z 2 +z 3 )=z 2 +^12 (z 3 −z 2 ),

z 5 =^12 (z 3 +z 1 )=z 2 +^12 (z 3 −z 2 )+^12 [z 2 +(p 1 +q 1 ı(z 3 −z 2 )]

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

z 6 =^12 (z 1 +z 2 )=^12 [z 2 +(p 1 +q 1 ı(z 3 −z 2 )+z 2 ]=z 2 +^12 (p 1 +q 1 ı)(z 3 −z 2 ).

Then

z 6 −z 5 =^12 (p 1 +q 1 ı− 1 −p 1 −q 1 ı)(z 3 −z 2 )=−^12 (z 3 −z 2 ),

z 4 −z 5 =z 2 +^12 (z 3 −z 2 )−z 2 −^12 ( 1 +p 1 +q 1 ı)(z 3 −z 2 )

=−^12 (p 1 +q 1 ı)(z 3 −z 2 )=(p 1 +q 1 ı)(z 6 −z 5 ). (11.6.14)

Thus the triangle[Z 4 ,Z 5 ,Z 6 ]is similar to the original triangle[Z 1 ,Z 2 ,Z 3 ].