Geometry with Trigonometry

(Marvins-Underground-K-12) #1

224 Vector and complex-number methods Ch. 11


Then

p^24 +q^24 =p^24 +(

(p 1 − 1 )
q 1

)^2 p^24 ,

p^24 +q^24
q 4

=


p^24
q 4

(


1 +(


p 1 − 1
q 1

)^2


)


=p 4.

−q 1
p 1 − 1

(


1 +(


p 1 − 1
q 1

)^2


)


=


2 p 1 − 1
p 1

1


1 +(p^1 q− 11 )^2

−q 1
p 1 − 1

(


1 +(


p 1 − 1
q 1

)^2


)


=−


2 p 1 − 1
p 1

q 1
p 1 − 1

.


By (11.6.4) the circumcentreZ( 16 iii)of[Z 4 ,Z 8 ,Z 18 ]where

z 4 =z 8 +(p 4 +q 4 ı)(z 18 −z 8 )

is given by


z( 16 iii)=z 8 +^12

[


1 +


p^24 +q^24
q 4

ı−

p 4
q 4

ı

]


(z 18 −z 8 )

=z 8 +^12


⎣ 1 +


p^24 ( 1 +(p^1 −^1 )

2
q^21 )
q 4

ı+

q 1
p 1 − 1

ı


⎦(z 18 −z 8 )

=z 8 +^12

[


1 +p 4

−q 1
p 1 − 1

(


1 +


(p 1 − 1 )^2
q^21

)


ı+

q 1
p 1 − 1

ı

]


(z 18 −z 8 )

=z 8 +^12

[


1 −


2 p 1 − 1
p 1

q 1
p 1 − 1

ı+

q 1
p 1 − 1

ı+

q 1
p 1 − 1

ı

]


(z 18 −z 8 )

=z 8 +^12

[


1 +


q 1
p 1 − 1

(



2 p 1 − 1
p 1

+ 1


)


ı

]


(z 24 −z 8 )

=z 8 +^12 [ 1 −

q 1
p 1

ı](z 18 −z 8 ).

Continuing from this we have
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