232 Vector and complex-number methods Ch. 11
and this is a non-zero multiple of
∣∣
∣∣
∣∣
∣∣c^2 +a^2 −b^2 (c^2 +a^2 −b^2 )(a^2 +b^2 −c^2 ) 1
c^2 +a^2 −b^2 + 2 a^2 −[(c+a)^2 −b^2 ][(c−a)^2 −b^2 ] 3
a(c+a−b) −a(c+a−b)[(c−a)^2 −b^2 ] 1∣∣
∣∣
∣∣
∣∣
,
the value of which is 4(ca^5 −c^3 a^3 −ba^5 +a^3 b^3 +c^3 ba^2 −b^3 ca^2 ). This factorizes as
4 a^2 (b−c)(c−a)(a−b)(a+b+c)and sothe incentre lies on the Euler line if and
only if the triangle is isosceles.
11.7.4 Miquel’s theorem, 1838 .......................
Z 1
Z 2
Z 3
Z 4
Z 5
Z 6
Figure 11.20(a). Miquel’s theorem.
Z 1
Z 2
Z 3
Z 4
Z 5
Z 6
Figure 11.20(b). Miquel’s theorem.Let Z 1 ,Z 2 ,Z 3 be non-collinear points, and Z 4 ∈Z 2 Z 3 ,Z 5 ∈Z 3 Z 1 ,Z 6 ∈Z 1 Z 2 be distinct
from Z 1 ,Z 2 and Z 3 .LetC 1 ,C 2 ,C 3 be the circumcircles of[Z 1 ,Z 5 ,Z 6 ],[Z 2 ,Z 6 ,Z 4 ],
[Z 3 ,Z 4 ,Z 5 ], respectively. ThenC 1 ,C 2 ,C 3 have a point in common.
Proof. Suppose that these circles have centres the pointsW 1 ,W 2 ,W 3 , respectively.
We first assume thatC 2 andC 3 meet at a second pointZ 7 =Z 4 .IfZ 7 =Z 6 the result
is trivially true, so we may exclude that case. AsZ 1 ,Z 2 ,Z 6 are collinear, we havez 1 −
z 6 =−ν(z 2 −z 6 ),for some non-zeroνinR.AsZ 2 ,Z 4 ,Z 6 ,Z 7 are concyclic, by 10.9.3
we havezz 72 −−zz 66 =ρzz^27 −−zz^44 , for some non-zeroρinR.AsZ 2 ,Z 3 ,Z 4 are collinear, we have
z 2 −z 4 =−λ(z 3 −z 4 ), for some non-zeroλinR.AsZ 3 ,Z 4 ,Z 5 ,Z 7 are concyclic,
we havezz 73 −−zz^44 =σzz 73 −−zz 55 , for some non-zeroσinR. On combining these we have
z 1 −z 6
z 7 −z 6 =νρλσ
z 3 −z 5
z 7 −z 5. It follows by 10.9.3 thatZ^1 ,Z^5 ,Z^6 ,Z^7 are concyclic.
We suppose secondly thatC 2 andC 3 have a common tangent atZ 2. It is convenient
to suppose thatz 1 =z 2 +(p 1 +ıq 1 )(z 3 −z 2 )andw 2 =z 2 +(p+ıq)(z 3 −z 2 ).Then
the footZ 7 of the perpendicular fromW 2 toZ 2 Z 3 has complex coordinatez 7 =z 2 +
p(z 3 −z 2 ), and hencez 4 =z 2 + 2 p(z 3 −z 2 ). Then the mid-pointZ 8 ofZ 3 andZ 4 has
complex coordinatez 8 =z 2 +(p+^12 )(z 3 −z 2 ). It follows that for the centreW 3 ofC 3