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Article: CJB/2011/133
Conics in a Semi-Regular Hexagon
Christopher Bradley
Abstract: Hexagon AFBDCE with ABC an equilateral triangle and AD, BE, CF concurrent at
the centroid P of ABC is inscribed in a conic, and as such is defined as a semi-regular hexagon.
It is proved that the six circumcentres of triangles AFP, FBP, BDP, DCP, CPE, EPA are co-conic
as are their six orthocentres. It is also proved that the intersections of contiguous Euler lines are
co-conic. Cabri indicates that the six in-centres and six nine-point centres are also co-conic.
A
61
H1
F
O1
I1
O6
H6
I6
12
E
56
O2
H5
I5
I2
P
O5
H2
23
B
H3
I3
I4
O3
O4
45
H4
34
D
Fig. 1
A semi-regular hexagon and its conics
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C
1. The co-ordinates of D, E, F
We use areal co-ordinates with equilateral triangle ABC as triangle of reference. This means we
may set a = b = c = 1 in known formulae. We suppose the conic in which the hexagon AFBDCE
is inscribed has equation
fyz + gzx + hxy = 0.
(1.1)
The centroid P has co-ordinates (1, 1, 1) so the equation of AP is y = z. Line AP meets the conic
again at D with co-ordinates D (– f, g + h, g + h). Similarly E and F have co-ordinates E (h + f,
– g, h + f) and F(f + g, f + g, – h).
2. The circumcircles of the six triangles AFP, FBP, BDP, DCP, CEP, EAP
Circles with a = b = c = 1 have equations of the form
yz + zx + xy + (ux + vy + wz)(x + y + z) = 0.
(2.1)
With known co-ordinates for their vertices it is straightforward to calculate u, v, w for each of
the six circles. These are:
AFP: u = 0, v = – (f + g – h)/(2f + 2g – h), w = – (f + g)/(2f + 2g – h).
(2.2)
FBP: u = – (f + g – h)/(2f + 2g – h), v = 0, w = – (f + g)/(2f + 2g – h).
(2.3)
BDP: u = – (g + h)/(2g + 2h – f), v = 0, w = – (g + h – f)/(2g + 2h – f).
(2.4)
DCP: u = – (g + h)/(2g + 2h – f), v = – (g + h – f)/(2g + 2h – f), w = 0.
(2.5)
CEP: u = – (h + f – g)/(2h + 2f – g), v = – (h + f)/(2h + 2f – g), w = 0.
(2.6)
EAP: u = 0, v = – (h + f)/(2h + 2f – g), w = – (h + f – g)/(2h + 2f – g).
(2.7)
3. Centres O1 – O6 of the six circles
With a = b = c = 1, the centre of a circle with equation of the form (2.1) has co-ordinates (x, y,
z), where
x = 2u – v – w + 1, y = 2v – w – u + 1, z = 2w – u – v + 1.
(3.1)
So for the six circles in Section 2 we have:
AFP: x = 2, y = (f + g + h)/(2f + 2g – h), z = (f + g – 2h)/(2f + 2g – h).
FBP: x = (f + g + h))/(2f + 2g – h), y = 2, z = (f + g – 2h)/(2f + 2g – h).
BDP: x = (g + h – 2f)/(2g + 2h – f), y = 2, z = (f + g + h)/(2g + 2h – f).
DCP: x = (g + h – 2f)/(2g + 2h – f), (f + g + h)/(2g + 2h – f), z = 2.
CEP: x = (f + g + h)/(2h + 2f – g), y = (h + f – 2g)/(2h + 2f – g), z = 2.
EAP: x = 2, y = (h + f – 2g)/(2h + 2f – g), z = (f + g + h)/(2h + 2f – g).
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(3.2)
(3.3)
(3.4)
(3.5)
(3.6)
(3.7)
4. The Conic of Circumcentres
The equation of a conic is of the form
ux2 + vy2 + wz2 + 2lyz + 2mzx + 2nxy = 0.
(4.1)
From the co-ordinates in Section 3 we obtain the equation of the conic passing through O1 – O6.
Its values are
u = 2(5f3 – 6f2g – 6f2h – 9fg2 – 9fgh – 9fh2 + 2g3 – 3g2h – 3gh2 + 2h3),
(4.2)
with v and w to be obtained from Equation (4.2) by cyclic change of f, g, h.
And
l = 4f3 + 24f2g + 24f2h + 9fg2 + 18fgh + 9fh2 – 11g3 + 12g2h + 12gh2 – 11h3,
(4.3)
with m and n to be obtained from Equation (4.3) by cyclic change of f, g, h.
5. Modification of a known formula
When a = b = c = 1 then the line perpendicular to px + qy + rz = 0 through the point with coordinates (l, m, n) is
The (m(p + q – 2r) – n(r + p – 2q))x + (n(q + r – 2p) – l(p + q – 2r)y +
(l(r + p – 2q) – m(q + r – 2p)z = 0.
(5.1)
6. The orthocentres H1– H6 of the six triangles
Equation (5.1) must be used repeatedly.
The equation of AF is
hy + (f + g)z = 0.
The equation of the line through P perpendicular to AF is
(f + g – h)x – y(f + g) + hz = 0.
(6.1)
(6.2)
The equation of the line AP is
y – z = 0.
(6.3)
The equation of the line through F perpendicular to AP is
(f + g – h)x – (f + g)y – (f + g)z = 0.
(6.4)
Solving equations (6.2) and (6.4) we find the co-ordinates of H1 are H1((f + g), (f + g – h), 0).
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The equation of BF is
hx + (f + g)z = 0.
The equation of the line through P perpendicular to BF is
(f + g)x – (f + g – h)y – hz = 0.
(6.5)
(6.6)
The equation of BP is
x – z = 0.
(6.7)
The equation of the line through F perpendicular to BP is
(f + g)x – (f + g – h)y + (f + g)z = 0.
(6.8)
Solving equations (6.6) and (6.8) we find the co-ordinates of H2 are H2((f + g – h), (f + g), 0).
Similarly the co-ordinates of H3 – H6 are
H3(0, g + h, g + h – f), H4(0, g + h – f, g + h), H5(h + f – g, 0, h + f), H6(h + f, 0, h + f – g).
7. The Conic of Orthocentres
Using the co-ordinates obtained in Section 6 we find the equation of the conic of orthocentres to
be of the form (4.1) with
u = 2(f + g)(g + h)(h + f)(f + g – h)(g + h – f)(h + f – g), with v = w = u and
(7.1)
2
2
l = – (f + g)(f + g – h)(h + f – g)(h + f)(f – 2f(g + h) + 2(g + h) ),
(7.2)
where m, n may be obtained from l by cyclic change of f, g, h.
8. The Equations of the Six Euler Lines
From the co-ordinates of the circumcentres and orthocentres we may obtain the equations of the
Euler lines. The equation of O1H1 is
(f + g – h)x – (f + g)y – (3f + 3g – h)z = 0,
(8.1)
and the equation of O2H2 is
(f + g)x – (f + g – h)y + (3f + 3g – h)z = 0.
Their point of intersection 12 has co-ordinates 12(3f + 3g – h, 3f + 3g – h, – h).
(8.2)
(8.3)
The points of intersection 34, 56 have co-ordinates that may be obtained from those of 12 by
cyclic change of x, y, z and f, g, h.
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The equation of O3H3 is
(3g + 3h – f)x – (g + h – f)y + (g + h)z = 0.
The point of interaction of O2H2 and O3H3 is the point 23, which has co-ordinates
23(3f2 + f(g – 3h) – 2g(g + h), 3f2 – f(5g + 9h) – (g + h)(8g – 3h),
– (f(2g + 3h) + (g + h)(2g – 3h))
(8.4)
(8.5)
The points of intersection 45, 61 have co-ordinates that may be obtained from those of 23 by
cyclic change of x, y, z and f, g, h.
9. The Euler Conic through 12, 23, 34, 45, 56, 61
From the co-ordinates derived in Section 8 we may obtain the equation of the Euler Conic which
is of the form (4.1) with
u = 2(f – 3(g + h))((2f2 + f(g + h) – (g – h)2),
(9.1)
where v and w may be obtained from u by cyclic change of f, g, h, and
l = 2f3 + 14f2(g + h) + f(7g2 + 16gh + 7h2) – 5g3 + 9g2h + 9gh2 – 5h3,
(9.2)
and where m, n may be obtained from l by cyclic change of f, g, h.
10. Further Comments
Cabri indicates that the six in-centres are also co-conic and also the six nine-point centres, but
not other main triangle centres such as the centroids or symmedian points. It may also be
observed that the hexagon formed by the contiguous points of intersection of the Euler lines has
opposite sides parallel.
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BRISTOL BS8 2SP.
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