Article: CJB/153/2011
More properties of the Incentre
Christopher Bradley
Abstract: In a triangle ABC, with incentre I, the midpoints of AI, BI, CI are located.
Perpendicular bisectors of AD, BE, CF are drawn. These meet the interiors of the sides of ABC
at six points that are shown to lie on an ellipse. As the figure develops to include the excentres
several other circles and conics are located.
Z
a2
b1
C'
A
32
N
Q P
O
2
B
a3
D
W
31`
23
E
M
1 V
I
U
13
Y
3
F
21`
12
L
A'
b3
c2
X
Fig. 1
1
C
B' c1
1. Introduction
Given triangle ABC the incentre I is located and also the circumcentre O. Points D, E, F are the
midpoints of AI, BI, CI respectively.
EF meets side AB at 31 and side CA at 21. Points 12 and 32 lie on side FD where it intersects
BC and AB respectively. Points 23 and 13 lie on side DE where it intersects CA and BC
respectively. It is proved that thee six points lie on a conic.
Lines 31 13 and 12 21 meet at a point A' on AI and points B' and C' are similarly defined. It is
shown that the circle A'B'C' has its centre at a point P lying on IO.
The points A', E, F, 13, 12 lie on a conic passing through points 2 on BI and 3 on CI. The points
B', F, D, 23, 21 lie on a conic passing through 3 on CI and a point 1 on AI. The points C', D, E,
32, 31 lie on a conic passing through the previously defined points 1 on AI and 2 on BI.
The lines through A, B, C perpendicular to AI, BI, CI form the triangle XYZ of excentres. It is
shown that circle XYZ has centre Q also lying on IO.
The line B'C' meets ZX at the point b1 and the point XY at the c1. Points c2, a2 lie on C'A'
where it intersects XY and YZ respectively. Points a3 and b3 lie on A'B' where it intersects YZ
and ZX respectively. It is shown these six points of intersection lie on a conic.
2. Triangle DEF
Point D is the midpoint of AU and so has co-ordinates D(2a + b + c, b, c). Similarly E and F
have co-ordinates E(a, 2b + c + a, c), F(a, b, 2c + a + b).
It follows that EF has equation
(a + 2(b + c))x – a(y + z) = 0.
(2.1)
(b + 2(c + a))y – b(z + x) = 0,
(2.2)
(c + 2(a + b))z – c(x + y) = 0.
(2.3)
Similarly FD, DE have equations
and
3. The six points 13, 12, 21, 23, 32, 31 and the conic on which they lie
DE meets BC at 13 with co-ordinates (0, 2a + 2b + c, c).
2
FD meets BC at 12 with co-ordinates (0, b, 2c + 2a + b).
EF meets CA at 21 with co-ordinates (a, 0, 2b + 2c + a).
DE meets CA at 23 with co-ordinates (2a + 2b + c, 0, c).
FD meets AB at 32 with co-ordinates (2c + 2a + b, b, 0).
EF meets AB at 31 with co-ordinates (a, 2b + 2c + a, 0).
It may now be shown that these six points lie on a conic with an equation of the form
ux2 + vy2 + wz2 + 2fyz + 2gzx + 2hxy = 0,
(3.1)
where
u = bc(2b + 2c + a), v = ca(2c + 2a + b), w = ab(2a + 2b + c),
f = – a(2a2 + b2 + c2 + 3bc + 3ca + 3ab), g = – b(2b2 + c2 + a2 + 3ca + 3ab + 3bc),
h = – c(2c2 + a2 + b2 + 3ab + 3bc + 3ca).
(3.2)
The line AI meets EF at U with co-ordinates U(a(b + c), b(a + 2b + 2c), c(a + 2b + 2c)) and the
conic with equation (3.2) at L with co-ordinates (x, y, z), where
x = 2a2 + 3b2 + 3c2 + 6bc + 6ca + 6ab – (a + b + c)√(8a2 + 9b2 + 9c2 + 18bc + 20ca + 20ab),
y = b(a + 2b + 2c), z = c(a + 2b + 2c).
(3.3)
Points V, M on BI and points W, N on CI have co-ordinates that may be obtained from those of
U, L by cyclic change of x, y, z and a, b, c.
4. Points A', B', C' and circle A'B'C' with centre P on IO
The line 23 32 has equation
– bcx + c(2a + b + 2c)y + b(2a + 2b + c)z = 0.
(4.1)
The line 31 13 has equation
c(a + 2b + 2c)x – cay + a(2a + 2b + c)z = 0.
(4.2)
These lines meet at the point C' with co-ordinates
(x, y, z) = (a(2a + 2b + c), b(2a + 2b + c), – c(a + b + 2c)).
(4.3)
Points A', B' have co-ordinates that may be written down from those of C' by cyclic change of x,
y, z and a, b, c. Note that it is immediate that A' lies on AI etc.
Circle A'B'C' has an equation of the form
a2yz + b2zx + c2xy + (fx + gy + hz)(x + y + z) = 0,
where
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(4.4)
f = bc(a + 2b + 2c)/(4(a + b + c)), g = ca(2a + b + 2c)/(4(a + b + c)),
h = ab(2a + 2b + c)/(4(a + b + c)).
(4.5)
The x- co-ordinate of the centre of a circle in the form (4.4) is
x = – (a4 – a2(b2 + c2 + 2f – g – h) + (b2 – c2)(g – h)).
(4.6)
It follows that P, the centre of circle A'B'C', has x co-ordinate
x = – (a/2)(2a3 + a2(b + c) – 2a(b2 + bc + c2) – (b2 + c2)(b – c)).
(4.7)
It may now be checked that P lies on IO with equation
bc(b – c)(b + c – a)x + ca(c – a)(c + a – b)y + ab(a – b)(a + b – c)z = 0.
(4.8)
5. The conic A' 13 12 E F and the point 2 (and others by cyclic change)
It may now be checked that the five points A', 13, 12, E, F lie on a conic with equation of the
form (3.1), where
u = bc(a + 2b + 2c)(3a2 + 2b2 + 2c2 + 4bc + 6ca + 6ab),
v = a2c(a + 2b + 2c)((2a + b + 2c), w = a2b(a + 2b + 2c)(2a + 2b + c),
f = – a2(a + 2b + 2c)(2a2 + b2 + c2 + 3bc + 3ca + 3ab),
g = ab(a3 + 3a2b + a(b + c)(2b – 3c) – c(b + c)2),
h = ac(a3 + 3a2c + a(b + c)(2c – 3b) – b(b + c)2).
(5.1)
The conics B' 21 23 F D and C' 32 31 D E now have equations that may be written down by
cyclic change of u, v, w, and a, b, c, and f, g, h.
The conic A' 13 12 E F meets the line BI at the point 2 with co-ordinates (x, y, z), where
x = a(a + 2b + 2c)(2a + b + 2c),
y = b(5a2 + 2b2 + 4c2 + 6bc + 9ca + 8ab),
(5.2)
z = c(a + 2b + 2c)(2a + b + 2c).
Co-ordinates of points 3, 1 may be obtained from (5.2) by cyclic change of x, y, z and a, b, c.
6. Points X, Y, Z and circle XYZ, centre Q on IO
The points X, Y, Z have co-ordinates X(– a, b, c), Y(a, – b, c), Z(a, b, – c) and it is now easy to
find the equation of circle XYZ. It has the form of Equation (4.4) with f = bc, g = ca, h = ab. The
x- co-ordinate of its centre Q (using Equation (4.6)) is
x = – a(a3 + a2(b + c) – a(b + c)2 – (b2 – c2)(b – c)).
(6.1)
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the y- and z- co-ordinates follow from Equation (6.1) by cyclic change of a, b, c. It may now be
checked that Q lies on IO, with Equation (4.8).
7. Points a2, a3, b3, b1, c1, c2 and the conic on which they lie
The line YZ has equation cy + bz = 0. The line C'A' has equation
c(a + 2b + 2c)x – cay + a(2a + 2b + c)z = 0.
(7.1)
These two lines meet at the point a2, which therefore has co-ordinates (x, y, z), where
x = a(2a + 3b + c), y = b(a + 2b + 2c), z = – c(a + 2b + 2c).
(7.2)
The line A'B' has equation
b(a + 2b + 2c)x + a(2a + b + 2c)y – abz = 0.
(7.3)
The intersection of YZ and A'B' is the point a3, which therefore ha co-ordinates (x, y, z), where
x = a(2a + b + 3c), y = – b(a + 2b + 2c), z = c(a + 2b + 2c).
(7.4)
The co-ordinates of b3 and c1 follow from those of a2 by cyclic change of x, y, z and a, b, c and
the co-ordinates of b1 and c2 follow from those of a3 by cyclic change of x, y, z and a, b, c.
It is now possible to work out the equation of the conic through a2, a3, b3, b1, c1, c2 and it has
the form of equation (3.1) with
u = b2c2(a + 2b + 2c),
(7.5)
with v, w following from u by cyclic change of a, b, c and
f = a2bc(6a2 + 4b2 + 4c2 + 9bc + 10ca + 10ab),
with g, h following from f by cyclic change of a, b, c.
Flat 4,
Terrill Court,
12-14, Apsley Road,
BRISTOL BS8 2SP.
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(7.6)
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