Article: CJB/2012/232
The Coconic Hexagon with Main Diagonals Concurrent
Christopher Bradley
Figure 1
Almost complete diagram without labels
1
6c
5b
2f
tF
tB
5
2
6
tC
1e
To 5G
1G
tE
To 6c
1
bf
3a
26
tA
15
6b
ac
2a
1f
P
3
A
U 5a
ae
F
To ad
13
1d
6e
B
1c
tD
Q
3b
bd
C
2d
24
4G 4b
E
R D
4c 3e
ce
4a
46
T
G
4f
S 5d
To 36
df
35
6d
2e
2G
To 3G
5c
3f
tC
To Z
14
be
Figure 2
25
A slightly condensed diagram with labels
cf
1. Discussion
ABCDEF is a hexagon inscribed in a conic and is such that its main diagonals AD, BE, CF are
concurrent at a point G. Sides AB, BC, ... are labelled 1, 2, ... . Points such as 14 are then the
intersection of AB and DE. Tangents are drawn at A, B, ..., so that the tangents at A and B meet
at P. Point Q is the intersection of the tangents at B and C and so on. The tangents at A, B, … are
also labelled tA, tB,… and points such as ac are thus the intersections of tA and tC. It is proved that
PQRSTU is a conic and that the six points ac, bf, ae, df, ce, bd lie on a conic. It is akso proved
that 13, 26, 15, 46, 35, 24 lie on a conic.
Points 14, 25, 36 is the polar line of G with respect to conic ABCDEF and the polar line of
PQRSTU, consisting of ad, be, cf, unsurprisingly coincides with it.
2
A point such as 5d is the intersection of line EF and the tangent at D and there are 24 such points
and we prove they lie on three 6 point conics, namely 6c 2f 5b 2e 5c 3f and 2a 1e 4a 5d 4b 1d
and 1f 1c 2d 4c 4f 5a. The other 6 points lie on the two lines 3 and 6.
Finally a point such as 3G means the intersection of line 3 with the appropriate diagonal BGE. It
is finally shown that 1G, 3G, 5G are collinear as are 2G, 4G, 6G and that they meet at a point Z
on the polar line.
2. Setting the Scene
We use homogeneous projective co-ordinates with ABC the triangle of reference and G the unit
point. We take the conic in which the hexagon is inscribed to have equation
fyz + gzx + hxy = 0.
(2.1)
The equation of AG is y = z and this meets the conic (1.1) at the point 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).
The tangent to conic (1.1) at the point (p, q, r) has equation
(gr + hq)x + (fr + hp)y + (fq + gp)z = 0.
(2.2)
Giving p, q, r suitable values we find the equations of the tangents to be
tA = UP : hy + gz = 0,
tB = PQ : fz + hx = 0,
tC = QR: gx + fy = 0,
tD = RS : (g + h)2x + fgy + hfz = 0,
tE = ST : fgx + (h + f)2y + ghz = 0,
tF = TU : hfx + ghy + (f + g)2z = 0.
(2.3)
(2.4)
(2.5)
(2.6)
(2.7)
(2.8)
The intersection points of the tangents have co-ordinates:
P(f, g, – h), Q(– f, g, h), R(– f, g, 2g + h), S(– f((h + f – g), – g(g + h – f), 2fg + h(f + g + h)),
T(– 2gh –f(f + g + h), g(f + g – h), h(h + f – g)), U(f + 2g, g, – h).
(2.9)
3. The conic PQRSTU
We may now show that P, Q, R, S, T, U lie on a conic with equation
ux2 + vy2 + wz2 + 2lyz + 2mzx + 2nxy = 0,
where
u = 2fh2(g + h), v = 2fh(f2 + f(g + 2h) + h(g + h)), w = 2f2h(f + g),
3
(3.1)
l = f(f2(g – h) + f(g2 – h2) + gh(g + h)), m = f2(g2 + gh + 2h2) + fg(g + h)2 + g2h(g + h),
n = h(f2(g – h) + f(g2 – h2) + gh(g + h)).
(3.2)
4. Points of intersection of tangents
The points of intersection of tangents other than P, Q, ... have co-ordinates:
ac: (f, – g, h), ad: (f(g – h), – g(g + h), h(g + h)), ae: (f + 2h, – g, h), bd: (– f, g + 2h, h),
be: (– f(h + f), g(f – h), h(h + f)), bf: (– f, – 2f – g, h), ce: (f, – g, 2f + h),
cf: (f(f + g), – g(f + g), h(g – f)), df: (– f(f + g – h), f(g + 2h) + g(g + h), – h(g + h – f)).
(4.1)
The line ad be cf, the polar of G with respect to conic PQRSTU has the rather satisfying equation
(g + h)x + (h + f)y + (f + g)z = 0.
(4.2)
It may now be shown that the points ac, ae, bd, bf, ce, df lie on a conic with an equation of the
form (3.1), where
u = 2fgh(g + h), v = – 2fgh(h + f), w = 2fgh(f + g), l = – f(f2(g + h) + f(g + h)2 + gh(g – h)),
m = – g(f(f + g + h)(g + 3h) + gh(g + h)), n = h(f2(g – h) – (fg + gh + hf)(g + h)).
(4.3)
5. Equations of the sides of ABCDEF and their points of intersection
The equation of 1 = AB is z = 0 and the equation of 2 = BC is x = 0. The equation of CD is
(g + h)x + fy = 0 and the equation of 4 = DE is (g + h)x + (h + f)y – hz = 0.
The equation of 5 = EF is fx – (h + f)y – (f + g)z = 0 and that of 6 = FA is hy + (f + g)z = 0
(5.1)
Various points of intersection have co-ordinates as follows:
14: (– (h + f), g + h, 0), 25(0, – (f + g), h + f), 36(f(f + g), – (f + g)(g + h), h(g + h)).
(5.2)
It may be seen that these three points also lie on the common polar line with equation (4.2).
Other points of intersection have co-ordinates:
13: (– f, g + h, 0), 15: (h + f, f, 0), 24: (0, h, h + f), 26(0, – (f + g), h),
35: (f(f + g), – (f + g)(g + h), f(f + g + h) + h(g + h)),
46: ( f(f + g + h) + h(g + h), – (f + g)(g + h), h(g + h)).
(5.3)
It may now be shown that the points 13, 15, 24, 26, 35, 46 lie on a conic with equation of the
form (3.1), with
u = 2hf(g + h), v = – 2hf(h + f), w = 2hf(f + g), l = – f(f(f + g + h) + h(g – h)),
m = – f(f + g + h)(g + 2h) – gh(g + h), n = h(f(f – g – h) – h(g + h)).
(5.4)
4
6. The 24 points of intersection of the sides and the tangents
We give the co-ordinates of these points in order of the sides AB, BC, ... .
1c: (– f, g, 0); 1d: (– fg,(g + h)2, 0); 1e: (– (h + f)2, fg, 0); 1f: (– g, f, 0); 2d: (0, – h, g);
2e: (0, – gh, (h + f)2); 2a: (0, – g, h); 3e: (fg, – g(g + h), f (f + 2g + 2h) + h(g + h));
3f: (– f(f + g)2, (f + g)2(g + h), h(f2 – g(g + h)), 3a: (fg, – g(g + h), h(g + h)); 3b: (– f, g + h, h);
4f: (f2 + fg + gh, – f(g + h) – g2, h(g – f)); 4a: (fg + h(g + h), – g(g + h), h(g + h));
4b: ( – f(h + f), f(g + h) + h2, h(h + f)); 4c: (f, – g, f – g); 5a: (h – g, – g, h);
5b: (– f(h + f), – f2 – hf – gh, h(h + f)); 5c: (f(f + g), – g(f + g). f2 + fg + gh);
5d: (f(g – h), – fh – g(g + h), fg + h(g + h)); 6b: (f, f + g, – h); 6c: (f(f + g), – g(f + g), gh);
6d: (f(fg + g2 – h2), – (f + g)(g + h)2, h(g + h)2); 6e; (f2 + f(g + 2h) + h(2g + h), – g((f + g), gh).
(6.1) – (6.24)
7. The additional six point Conics
Conic 6c 2f 5b 2e 5c 3f
This has an equation of the form (3.1), where
u = 2gh(f(f + g + 2h) + (g + h)2),
v = 2gh(h + f)2,
w = 2gh(f + g)2,
l = f4 + 2f3(g + h) + f2(g2 + 4gh + h2) + 2fgh(g + h) + 2g2h2,
m = g(f3 + f2(2g + h) + f(g + h)2 + g2h),
n = h(f3 + f2(g + 2h) + f(2g2 + gh + h2) + g2h).
(7.1)
(7.2)
(7.3)
(7.4)
(7.5)
(7.6)
Conic 2a 1e 4a 5d 4b 1d
This has an equation of the form (3.1), where
u = 2fgh(g + h)2,
v = 2fgh(h + f)2,
w = 2fgh(f2 + f(g + 2h) + g2 + gh + h2),
l = f(f2(g2 + h2) + fh(2g2 + gh + 2h2) + h2(2g2 + gh + h2)),
m = g(f2(g2 + gh + h2) + fh(g + h)2 + h2(g + h)2)),
n = h(f2(2g2 + 2gh + h2) + 2fh(g + h)2 + h2(g + h)2)).
(7.7)
(7.8)
(7.9)
(7.10)
(7.11)
(7.12)
Conic 1f 1c 2d 4c 4f 5a
This has an equation of the form (3.1), where
u = 2fgh,
v = 2fgh,
w = 2hf(f + g – h),
l = f(fg + g2 – gh + h2),
m = – f2h + f(g2 + h2) + g3,
(7.13)
(7.14)
(7.15)
(7.16)
(7.17)
5
n = h(f2 + g2).
(7.18)
The remaining 6 points do not lie on a non-degenerate conic as they lie three by three on lines 3
and 6.
8. Two additional three point lines and where they meet
Lines AD, BE CF have equations y = z, z = x, x = y respectively. Points have co-ordinates
1G = CF^AB: (1, 1, 0), 2G = AD^BC: (0, 1, 1), 3G = BE^CD: (f, – (g + h), f),
4G = CF^DE: (h, h, f + g + 2h), 5G= AD^EF: (2f + g + h, f, f),6G – BE^FA: (h, – (f + g), h).
(8.1)
It may now be shown that 1G, 3G, 5G lie on the line with equation
f(x – y) – (f + g + h)z = 0,
(8.2)
And also that 2G, 4G, 6G lie on the line with equation
(f + g + h)x + h(y – z) = 0/
(8.3)
These two lines meet at the point Z with co-ordinates
Z(h(2f + g + h), – f2 – f(2g + h) – (g + h)2, f(f + g + 2h)).
(8.4)
Also Z lies on the polar line (4.2). David Monk has pointed out that these are Pascal lines so
presumably Z is a Steiner point.
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