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Article: CJB/2011/162
When 24 Points form three 8-point Conics and 12 Concurrent Lines
Christopher Bradley
Abstract: A construction is described involving two chords of a conic that results in 24 points
that form three 8-point Conics and 12 Concurrent Lines.
twuv
atdv
13
To 45
L1
58
L3
17
14
18
dtbv
W
A
48
15
awcu
78
avdt
D
38
47
P
12
T
ctdv
To 16
25
C
67
36
26
23
56
34
68
37
U
B
L6
V
46
To 28
avct
24
27
btcv
L8
budw
L2
tuvw
L4
35
btdv
auwc
L7
L5
Fig. 1
57
Three 8-point conics and
12 concurrent lines
1. Introduction
In a given conic Σ two chords AC and BD are drawn intersecting at a point P. The tangents at A,
B, C, D are labelled L1, L2, L3, L4 respectively. Lines Li, Lj intersect at a point labelled ij. It
follows that points 13 and 24 lie on the polar p of P with respect to Σ. Line 12 34 meets Σ again
at T and V and line 14 23 meets Σ again at W and U. It is proved that these two lines both pass
through P. This result and some others in this article may be proved by Brianchon’s theorem.
1
28
(Private communication with G C Smith, University of Bath). The tangents at T, U, V, W are
labelled L5, L6, L7, L8 respectively. It follows that points 57 and 68 both lie on p. Using the same
definition of points as before we may now define points 12, 14, 15, 16, 17, 18 (all on L1), 23, 25,
26, 27, 28 (all on L2), 34, 35, 36, 37, 38 (all on L3), 45, 46, 47, 48 (all on L4), 56, 58 (both on L5),
67 and 78. The points thus labelled other than those on p are the 24 points of the title. The
following are the main results proved in this article (i) Points 12, 58, 14, 78, 34, 67, 23, 56 lie on
a conic; (ii) points 35, 27, 46, 38, 17, 45, 28, 16 lie on a conic; (iii) points 15, 18, 48, 47, 37, 36,
26, 25 lie on a conic; (iv) lines 15 37, 25 47, 26 48, 67 58, 12 34, 17 35, 56 78, 23 14, 38 16, 28
46, 27 45, and 18, 36 are concurrent at P; (v) line UW contains 4 more important points and p
contains 8 more important points. These features are all shown in Fig. 1. In addition to the three
8-point conics listed Cabri II plus indicates that there are two more 8-point conics A, B, U, W,
56, 27, 17, 58 and C, D, U, W, 78, 45, 35, 67 and numerous 6-point conics.
Proofs are presented in the projective plane using homogeneous projective co-ordinates.
2. The conic Σ, chords AC, BD and the tangents L1 – L4 and their intersections
We suppose the conic Σ has equation y2 = zx, with A the point A(1, 0, 0), B the point B(1, 1, 1),
C the point C(0, 0, 1) and the point P on AC having co-ordinates P(s, 0, t).
Later for convenience we shall set s = (2uv) 2 and t = (u2 – v2)2 in order that s, t, and s + t have
rational square roots.
The equation of BP is
tx + (s – t)y – sz = 0.
(2.1)
This meets the conic Σ at the point D with co-ordinates D(s2, – st, t2). The tangent to the conic at
a general point (1, q, q2) (with parameter q) has equation
2qy = q2x + z.
(2.2)
The equations of the tangents at A, B, C, D may now be determined.
L1 Tangent at A: q = 0,
z = 0.
L2 Tangent at B: q = 1,
2y = x + z.
L3 Tangent at C: q infinite,
x = 0.
L4 Tangent at D: q = – t/s
t2x + 2sty + s2z = 0.
(2.3)
(2.4)
(2.5)
(2.6)
The co-ordinates of the six intersection of these lines one with another may now be obtained.
12: x = 2, y = 1, z = 0.
13: x = 0, y = 1, z = 0.
14: x = – 2s, y = t, z = 0.
2
23: x = 0, y = 1, z = 2.
24: x = – 2s, y = t – s, z = 2t.
34: x = 0, y = – s, z = 2t.
The points 13 and 24 determine the polar line p, which therefore has equation
tx + sz = 0.
(2.7)
3. The second quadrilateral TUVW and lines L5 – L8
The equation of the line 12 34 is
tx – 2ty – sz = 0.
(3.1)
This line meets the conic at points we label T and V whose co-ordinates are
T: x = 4u4, y = 2u2 (u2 –v2), z = (u2 – v2)2,
V: 4v4, y = 2v2 (v2 – u2), z = (u2 – v2)2.
Here the values of s, t in terms of u, v (section 2) have been used.
The equation of the line 14 23 is
tx + 2sy – sz = 0.
(3.2)
This line meets the conic at points we label U and W whose co-ordinates are
U: x = 4u2v2, y = 2uv (u + v) 2, z = (u + v) 4,
W: x = 4u2v2, y = – 2uv (u – v) 2, z = (u – v) 4.
Here again the values of s, t in terms of u, v (section 2) have been used. It may also be checked
that 12 34 and 14 23 pass through P.
The equations of the tangents to the conic at the points T, U, V, W may now be determined.
L5 Tangent at T
(u2 – v2)2x – 4u2 (u2 – v2) y + 4u4z = 0.
(3.3)
4
2
2 2
L6 Tangent at U
(u + v) x – 4uv (u + v) y + 4u v z = 0.
(3.4)
2
2 2
2
2
2
4
L7 Tangent at V
(u – v ) x + 4v (u – v ) y + 4v z = 0.
(3.5)
4
2
2 2
L8 Tangent at W
(u – v) x + 4uv (u – v) y + 4u v z = 0.
(3.6)
In terms of u and v, the tangent at D, line L4 (see Equation (2.6)) also has equation
(u2 – v2)4x + 8u2v2 (u2 – v2)2y + 16u4v4z = 0.
4. The 28 points of intersection, four of which lie on the polar line
3
(3.7)
The eight lines L1 – L8 create 28 points on intersection, 4 of which lie on the polar line p. Their
co-ordinates may now be worked out and are:
12: x = 2, y = 1, z = 0.
13: x = 0, y = 1, z = 0. (On the polar line p.)
14: x = – 8u2v2, y = (u2 – v2)2, z = 0.
15: x = 4u2, y = u2 – v2, z = 0.
16: x = 4uv, y = (u + v) 2, z = 0.
17: x = 4v2, y = v2 – u2, z = 0.
18: x = – 4uv, y = (u – v) 2.
23: x = 0, y = 1, z = 2.
24: x = – 8u2v2, y = u4 – 6u2v2 + v4, z = 2(u2 – v2)2. (On the polar line p.)
25: x = 4u2, y = 3u2 – v2, z = 2(u2 – v2).
26: x = 4uv, y = u2 + 4uv + v2, z = 2(u + v) 2.
27: x = – 4v2, y = u2 – 3v2, z = 2(u2 – v2).
28: x = – 4uv, y = u2 – 4uv + v2, z = 2(u – v) 2.
34: x = 0, y = – 2u2v2, z = (u2 – v2)2
35: x = 0, y = u2, z = (u2 – v2).
36: x = 0, y = uv, z = (u + v) 2.
37: x = 0, y = v2, z = (v2 – u2).
38: x = 0, y = – uv, z = (u – v) 2.
45: x = – 8u4v2, y = u2 (u2 – v2) (u2 – 3v2), z = (u2 – v2)3.
46: x = – 8u3v3, y = uv(u + v)2(u2 – 4uv + v2), z = (u – v)2(u + v)4.
47: x = 8u2v4, y = v2 (v2 – u2)(3u2 – v2), z = (u2 – v2)3.
48: x = 8u3v3, y = – uv(u – v)2(u2 + 4uv + v2), z = (u + v)2(u – v)4.
56: x = 4u3v, y = u (u + v) (u2 + 2uv – v2), z = (u – v) (u + v) 3.
57: x = – 4u2v2, y = (u2 – v2)2, z = (u2 – v2)2. (On the polar line p.)
58: x = 4u3v, y = u (u2 – 2uv – v2)(v – u), z = (u + v)(v – u)3.
67: x = 4uv3, y = – v (u + v)(u2 – 2uv – v2), z = (v – u)(u + v)3.
68: x = – 4u2v2, y = – 4u2v2, z = (u2 – v2)2. (On the polar line p.)
78: x = 4uv3, y = v (v – u) (u2 + 2uv – v2), z = (u + v) (u – v) 3.
5. The three 8-point Conics
The 24 points of Section 4, not on the polar line, lie on three 8-point conics, with each point
lying on one of these conics. Their equations are of the form
lx2 + my2 + nz2 + 2fyz + 2gzx + 2hxy = 0.
(5.1)
We now provide the constants l, m, n, f, g, h for each of these conics.
4
Conic 12 58 14 78 34 67 23 56
l = (u2 – v2)4, m = – 16u2v2 (u2 – v2)2, n = 16u4v4, f = 4u2v2 (u4 – 6u2v2 + v4),
g = u8 + 14u4v4 + v8, h = – (u2 – v2)2(u4 – 6u2v2 + v4).
(5.2)
Conic 35 27 46 38 17 45 28 16
l = (u2 – v2)4, m = – 16uv3 (u + v) (u – v) 3, n = 16u4v4, f = 8u2v3 (u – v)(u2 – 2uv – v2),
g = 2v (u – v) (u6 – u4v2 + 6u3v3 + u2v4 + 2uv5 – v6),
h = 2v (u + v) (v – u) 3(u3 – u2v – 3uv2 – v3).
(5.3)
Conic 15 18 48 47 37 36 26 25
l = (u2 – v2)4, m = – 16u3v (u – v)(u + v)3, n = 16u4v4, f = 8u3v2(u3 – u2v – 3uv2 – v3),
g = 2u (u + v) (u6 + 2u5v – u4v2 + 6u3v3 + u2v4 – v6),
h = – 2u (u – v) 2(u + v) 3(u2 – 2uv – v2).
(5.4)
If one includes the vertices of the quadrangles there are other 8-point conics, for example
Conic A B U W 56 27 17 58 with
l = 0, m = 8v2 (u2 – v2), n = 8u2v2, f = – 12u2v2, g = – u4 + 6u2v2 + 3v4, h = (u2 – v2)4.
(5.5)
6. The above 12 pairs of points create 12 lines all passing through P
We remind the reader that P has co-ordinates P(s, 0, t), with s = (2uv) 2 and t = (u2 – v2)2. We
now give the equations of the 12 lines.
Line 23 14:
Line 12 34:
Line 15 37:
Line 35 17:
Line 25 47:
Line 27 45:
Line 26 48:
Line 28 46:
Line 58 67:
Line 56 78:
Line 38 16:
Line 18 36:
tx + 2sy – sz = 0.
(6.1)
tx – 2ty – sz = 0.
(6.2)
2
2
2
2 2
2
2 2
4u (u – v ) y + 4u v z – (u – v ) x = 0.
(6.3)
2
2 2
2
2
2
2 2
(u – v ) x + 4v (u – v ) y – 4u v z = 0.
(6.4)
2
2
2
2 2 2
2
2 2
2
4
2 2
4
(u + v )(u – v ) (v – 3u )x + 4u (u – v )(u – 2u v – 3v )y
+ 4u2v2 (u2 + v2) (3u2 – v2) z = 0.
(6.5)
2
2 2 2
2
2
2
2
2
2 2
2
2
(u – v ) (u – 3v ) x + (u – v ) (3u – v ) y + 4u v (3v – u )z = 0.
(6.6)
2
2 2 2
2
2 2
2
(u – v ) (u + 4uv + v ) x – 4uv (u + v) (u – 4uv + v ) y
– 4u2v2 (u2 + 4uv + v2) z = 0.
(6.7)
2
2 2 2
2
2
2
2 2
2
2
2
(u – v ) (u + v ) (u – 4uv + v ) x + 4uv (u – v) (u + v ) (u + 4uv + v ) y
– 4u2v2 (u2 + v2) (u2 – 4uv + v2) z = 0. (6.8)
(u2 – v2)2(u2 + v2)(u2 – 2uv – v2)x + 4uv(u2 – v2)(u4 + 2u3v + 2uv3 – v4)y
– 4u2v2 (u2 + v2) (u2 – 2uv – v2) z = 0. (6.9)
(u2 – v2)2(u2 + v2) (u2 + 2uv – v2) x – 4uv (u2 – v2) (u4 – 2u3 – 2uv3 – v4) y
– 4u2v2 (u2 + v2) (u2 + 2uv – v2) z = 0.(6.10)
(u2 – v2)2x – 4uv (u2 – v2) y – 4u2v2z = 0.
(6.11)
2
2 2
2
2 2
(u – v ) x + 4uv (u + v) y – 4u v z = 0.
(6.12)
5
It is straightforward to show that P lies on all of the above 12 lines.
7. The Quadrilaterals and Points on the line UW
The co-ordinates of the vertices of the quadrilaterals ABCD and TUVW have been obtained in
Sections 2 and 3, and these enable us to find four more points on the line UW. First the equation
of UW is
(u2 – v2)2x + 4u2v2 (2y – z) = 0.
(7.1)
We now give the equations of eight lines through the vertices.
AT:
(u2 – v2) y – 2u2z = 0.
AV:
(u2 – v2) y + 2v2z = 0.
BT:
(u2 – v2) x + (v2 – 3u2) y + 2u2z = 0.
BV:
(u2 – v2) x + (3v2 – u2) y – 2v2z = 0.
CT:
(u2 – v2) x – 2u2y = 0.
CV:
(u2 – v2) x – 2v2y = 0.
DT:
(u2 – v2)3x – 2u2 (u2 – v2) (u2 – 3v2) y – 8u4v2z = 0.
DV:
(u2 – v2)3x + 2v2 (u2 – v2) (3u2 – v2) y + 8u2v4z = 0.
(7.2)
(7.3)
(7.4)
(7.5)
(7.6)
(7.7)
(7.8)
(7.9)
These 8 lines provide 4 intersections that lie on the line UW. Their co-ordinates are:
atdv = AT^DV: x = 4u2v2(3u2 + v2), y = – 2u2(u2 – v2)2, z = – (u2 – v2)3.
avdt = AV^DT: x = 4u2v2(u2 + 3v2), y = – 2v2(u2 – v2)2, z = (u2 – v2)3.
btcv = BT^CV : x = 4u2v2, y = – 2u2(u2 – v2), z = – (u2 – v2)(3u2 + v2).
bvct = BV^CT: x = 4u2v2, y = 2v2 (u2 – v2), z = (u2 – v2) (u2 + 3v2).
It is left to the reader to determine in similar manner four further points lying on TV.
8. Ten more points on the polar line
The polar line p was first introduced in equation (2.7). It is also the polar line of P with respect to
each of the three conics in Section 5. We now give the co-ordinates of four more points on p
(other than the four 13 24 57 68 already introduced).
bvdt = BV^DT: x = 4u2v2(3v2 – u2), y = – 2v2(u2 – v2)(3u2 – v2), z = (u2 – v2)2(u2 – 3v2).
btdv = BT^DV: x = 4u2v2(3u2 – v2), y = – 2u2(u2 – v2)(u2 – 3v2), z = – (u2 – v2)2(3u2 – v2).
avct = AV^CT: x = 4u2v2, y = 2v2(u2 – v2), z = – (u2 – v2)2.
atcv = AT^CV: x = – 4u2v2, y = 2u2(u2 – v2), z = (u2 – v2)2.
We now give the equations of the four lines forming the quadrilateral ABCD.
AB:
y = z.
BC:
x = y.
6
(8.1)
(8.2)
CD:
DA:
tx + sy = 0.
ty + sz = 0.
(8.3)
(8.4)
We now give (again) the co-ordinates of 57 and 68. These are:
57 = AB^CD: x = s, y = – t, z = – t.
68 = AD^BC: x = s, y = s, z = – t.
We now give the equations of the four lines forming the quadrilateral TUVW.
TU: (u + v) 2(u2 – v2) x – 2u (u3 + 3u2v + uv2 – v3) y + 4u3vz = 0.
UV: (u + v) 2(u2 – v2) x – 2v (u3 – u2v – uv2 – v3) y – 4uv3z = 0.
VW: (u – v) 2 (u2 – v2) x + 2v (u3 + u2v – 3u2v + v3) y + 4uv3z = 0.
WT: (u – v) 2 (u2 – v2) x – 2u (u3 – 3u2v + u2v + v3) y – 4u3vz = 0.
(8.5)
(8.6)
(8.7)
(8.8)
We now give the co-ordinates of two more points on the polar line.
tuvw = TU^VW: x = – 4u2v2(u2 + 2uv – v2), y = 2uv(u2 – v2)(u2 – 2uv – v2),
z = (u2 – v2)2(u2 + 2uv – v2).
twuv = WT^UV; x = = – 4u2v2(u2 – 2uv – v2), y = – 2uv(u2 – v2)(u2 + 2uv – v2),
z = (u2 – v2)2(u2 – 2uv – v2).
We now give the equations of eight more lines relating vertices of the two quadrilaterals.
AU: (u + v) 2y – 2uvz = 0.
BU: (u + v) 2x – (u2 + 4uv + v2) y + 2uvz = 0.
CU: 2uv (u + v) 2x – 4u2v2y = 0.
DU: (u + v)(u2 – v2)2x – 2uv(u + v)2(u2 – 4uv + v2)y – 8u3v3z = 0.
AW: (u – v) 2y + 2uvz = 0.
BW: (u – v) 2x – (u2 – 4uv + v2) y – 2uvz = 0.
CW: (u – v) 2x + 2uvy = 0.
DW: (u – v)2(u2 – v2)2x + 2uv(u – v)2(u2 + 4uv + v2)y + 8u3v3z = 0.
(8.9)
(8.10)
(8.11)
(8.12)
(8.13)
(8.14)
(8.15)
(8.16)
Finally these lines provide four further intersections lying on the polar line. Their co-ordinates
are:
aucw = AU^CW: x = – 4u2v2, y = 2uv(u – v)2, z = (u2 – v2)2.
budw = BU^DW: x = – 4u2v2(u2 + 4uv + v2), y = 2uv(u + v)2(u2 – 4uv + v2),
z = (u2 – v2)2(u2 + 4uv + v2),
cuaw = CU^AW: x = – 4u2v2, y = – 2uv(u + v)2, z = (u2 – v2)2.
dubw = DU^BW: x = – 4u2v2(u2 – 4uv + v2), y = – 2uv(u – v)2(u2 + 4uv + v2),
z = (u2 – v2)2(u2 – 4uv + v2).
In conclusion it must be emphasized that although quadrilateral ABCD is arbitrary, TUVW is
related to it by the construction outlined in Section 3, which explains the large number of
7
concurrences. Also the analysis is limited to positive values of s and t, the reason for the choice
of u and v being to exclude square roots from the co-ordinates of many of the points.
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