Geometry with Trigonometry

(Marvins-Underground-K-12) #1

Sec. 11.7 Some well-known theorems 231


11.7.3 The incentre on the Euler line of a triangle ...........


We suppose that we have the mobile coordinatesz 1 −z 2 =(p 1 +q 1 ı)(z 3 −z 2 ),where
p 1 andq 1 are real numbers andq 1 =0. Thenz 1 −z 3 =(p 1 − 1 +q 1 ı)(z 3 −z 2 ),andas
in 11.6.3 we have


z 1 −z 2
z 3 −z 2

=p 1 +q 1 ı=c
a

cisβ, z^2 −z^3
z 1 −z 3

=^1 −p^1 +q^1 ı
( 1 −p 1 )^2 +q^21

=a
b

cisγ,

z 3 −z 1
z 2 −z 1

=


p 1 − 1 +q 1 ı
p 1 +q 1 ı

=


b
c

cisα,

where we are using our standard notation. Thenp^21 +q^21 =c


2
a^2 ,(^1 −p^1 )

(^2) +q 2
1 =
b^2
a^2.
We recall that the orthocentre, centroid and incentre have mobile coordinatesp 1 +
p 1 ( 1 −p 1 )
q 1 ı,
p 1 + 1
3 +
q 1
3 ı,
p 1 +



p^21 +q^21 +q 1 ı
1 +


p^21 +q^21


( 1 −p 1 )^2 +q^21 respectively.
Now 1− 2 p 1 =b

(^2) −c 2
a^2 ,p^1 =
c^2 +a^2 −b^2
2 a^2 ,^1 −p^1 =
a^2 +b^2 −c^2
2 a^2 ,p^1 +^1 =
c^2 +a^2 −b^2 + 2 a^2
2 a^2.
Moreover
q^21 =
c^2
a^2
−p^21 =
c^2
a^2



(


c^2 +a^2 −b^2
2 a^2

) 2


=−


(c^2 +a^2 −b^2 )^2 − 4 c^2 a^2
4 a^4

=−

[(c+a)^2 −b^2 ][(c−a)^2 −b^2 ]
4 a^4

,


while


p 1 +


p^21 +q^21 =

c^2 +a^2 −b^2
2 a^2

+


c
a

=


(c+a)^2 −b^2
2 a^2

,


1 +



p^21 +q^21 +


( 1 −p 1 )^2 +q^21 = 1 +

c
a

+


b
a

=


a+b+c
a

,


so that


p 1 +


p^21 +q^21

1 +


p^21 +q^21 +


( 1 −p 1 )^2 +q^21

=


(c+a)^2 −b^2
2 a(a+b+c)

=


c+a−b
2 a

,


q^21
1 +


p^21 +q^21 +


( 1 −p 1 )^2 +q^21

=−


1


4 a^3

[(c+a)^2 −b^2 ][(c−a)^2 −b^2 ]
a+b+c

=−


1


4 a^3

(c+a−b)[(c−a)^2 −b^2 ].

The determinant for collinearity, on multiplying the middle column byq 1 ,is


∣∣
∣∣
∣∣
∣∣

c^2 +a^2 −b^2
2 a^2

(c^2 +a^2 −b^2 )(a^2 +b^2 −c^2 )
4 a^41
c^2 +a^2 −b^2 + 2 a^2
6 a^2 −

[(c+a)^2 −b^2 ][(c−a)^2 −b^2 ]
12 a^41
c+a−b
2 a −

(c+a−b)[(c−a)^2 −b^2 ]
4 a^31

∣∣


∣∣


∣∣


∣∣



,

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