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X(1), X(3), X(4), X(20), X(40), X(64), X(84), X(1490), X(1498), X(2130), X(2131), X(3182), X(3183), X(3345), X(3346), X(3347), X(3348), X(3353), X(3354), X(3355), X(3472), X(3473), X(3637)

antipodes of A, B, C on the circumcircle

excenters, infinite points of the altitudes

CPCC or H-cevian points, see Table 11, their reflections about X(3), their isogonal conjugates

foci of the inconic with center X(3)

other points below

The Darboux cubic is the isogonal pK with pivot L = X(20). See Table 27. It is the only isogonal central pK. Its center is O. Hence, it is a pK++. See also table 15. It is anharmonically equivalent to the Thomson cubic. See Table 21. K004 is a member of the class CL043 : it meets the circumcircle at A, B, C and at their antipodes where the tangents are concurrent at the point X(1498). See also Q063. The isotomic transform of K004 is K183.

Locus properties (see also Table 6)

  1. Locus of point P whose pedal triangle is a cevian triangle, the perspector being on the Lucas cubic. For this reason, the Darboux cubic is called 1-pedal cubic in Pinkernell's paper.
  2. Locus of point P whose pedal and anticevian triangles are perspective, the perspector being on the Thomson cubic.
  3. Denote by PaPbPc the pedal triangle of point P and by T1, T2, T3, T4, T5, T6 the areas of triangles PBPa, PPaC, PCPb, PPbA, PAPc, PPcB respectively. The locus of point P such that T1*T3*T5 = T2*T4*T6 is the Darboux cubic. See also K003 (property 3) and Q082.
  4. The internal bisector at P of the triangle PPbPc meets PbPc at A1. B1 and C1 are defined similarly. The triangles ABC and A1B1C1 are perspective if and only if P lies on the Darboux cubic (Antreas Hatzipolakis, Hyacinthos #1945).
  5. The perpendicular at P to the line PA meets the lines AC, AB at Bc, Cb respectively. Define Ca, Ac, Ab, Ba similarly. The triangles AbBcCa and AcBaCb have equal areas if and only if P lies on the Darboux cubic (Paul Yiu, Hyacinthos #1959).
  6. Let P be a point with traces A_P, B_P, C_P. Denote by L1 the line joining the perpendicular feet of A_P on the cevians BB_P and CC_P. Let L2, L3 be the two lines analogously defined. The triangle bounded by L1, L2, L3 is perspective with ABC if and only if P lies on the Darboux cubic (together with the line at infinity, the three circles with diameters BC, CA, AB, and the Yiu quintic Q006) (Paul Yiu, Hyacinthos #1967). Jean-Pierre Ehrmann observes that these three lines are in fact concurrent if and only if P lies on Q006.
  7. Denote by PaPbPc the pedal triangle of point P and by Qa, Qb, Qc the intersections of the lines AP and PbPc, BP and PcPa, CP and PaPb respectively. The triangles PaPbPc and QaQbQc are perspective if and only if P lies on the Darboux cubic (together with the line at infinity and the circumcircle). This question was raised by Nikolaos Dergiades and answered by Antreas Hatzipolakis (Hyacinthos #8336 & sq.) who asks two other questions answered at Q007 and Q009.
  8. Locus of point P such that the pedal and antipedal triangles of P are perspective (together with the line at infinity and the circumcircle).
  9. Locus of point P such that the six vertices of the pedal and cevian (or antipedal and anticevian) triangles of P lie on a same conic. See also Table 11.
  10. The locus of P whose anticevian or antipedal triangle is perspective to the hexyl triangle is K343. In the anticevian case, the locus of the perspector is the Darboux cubic.
  11. The locus of P whose cevian triangle is perspective to the hexyl triangle is K344. The locus of the perspector is the Darboux cubic.
  12. Locus of point P such that the pedal triangle of P and the anticevian triangle of the isogonal conjugate of P are perspective.
  13. Locus of point P such that the antipedal triangle of P and the cevian triangle of the isogonal conjugate of P are perspective.
  14. Denote by PaPbPc the pedal triangle of point P. The circles PAPa, PBPb, PCPc form a pencil if and if P lies on the Darboux cubic (together with the line at infinity) (Hyacinthos #6329 and related messages). In this case, they have two common (real) points P and P' which lies on Q071, the Darboux quintic.
  15. The lines joining the vertices of ABC with the midpoints of the sides of the pedal triangle of P concur at Q if and only if P is on the Darboux cubic. Q lies on the Thomson cubic. (Francisco Garcia Capitan, private message, 04/01/2009)
  16. Locus of P such that ABC and the pedal of the pedal of P are perspective (Nikolaos Dergiades, ADGEOM #132).
  17. See K645.

Other properties

(Ha) is the hyperbola centered at A, passing through the antipode Ao of A in the circumcircle, through its reflection A1 in A and whose asymptotes are the perpendiculars at A to the sidelines AB and AC (these are the lines ABo and ACo). (Ha) meets the circle C(O,3R) at four points A1 and U, V, W. Naturally, U, V, W also lie on two similar hyperbolas (Hb) and (Hc) and the isogonal conjugates U*, V*, W* of U, V, W lie on the Darboux cubic.

U, V, W and their reflections U', V', W' in O are six points on the Darboux cubic. Furthermore, the triangle UVW has orthocenter L, centroid X(376) (the reflection of G in O) and the midpoints U", V", W" also lie on the Darboux cubic. They are the isogonal conjugates of U', V', W' and the midpoints of LU', LV', LW'.

U', V', W', A1, B1, C1 are the six points where C(O,3R) meets (T), the homothetic of the Thomson cubic under h(O,3), thus they are the images of the common points of the Thomson cubic and the circumcircle under the same homothety. U', V', W' also lie on three rectangular hyperbolas (H'a), (H'b), (H'c) passing through H. (H'a) contains A, the reflection of L in A and its asymptotes are parallel to the bisectors at A. Its center is the homothetic of L under h(A,-1/2). Its tangent at A is the line AX(64) i.e. it is tangent at A to the Darboux cubic.

 

A remarkable pencil of cubics

The Darboux cubic K004 and the decomposed cubic which is the union of the circumcircle and the Euler line generate a pencil of central cubics with center O passing through H, X(20) and the reflections A', B', C' of A, B, C about O. Each cubic is spK(P, Q) in CL055 where P is a point on the Euler line and Q is the midpoint of PX(20). spK(P,Q) contains the infinite points of pK(X6, P) and the isogonal conjugate of P.

This pencil also contains (apart K004 which is the only pK) the cubics K047, K080, K426, K443, K566 corresponding to K002, K003, pK(X6, X3146), K006, K005 respectively. Note that K426 and K443 are psK cubics. See Pseudo-Pivotal Cubics and Poristic Triangles.

 

Group structure and organization of points on the Darboux cubic

The usual group structure is easily adapted on K004 : the sum M+N of two points is the isogonal conjugate of the third intersection S of K004 with the line MN and, obviously, X(20), S, M+N are collinear. The pivot X(20) is the neutral element 0 of the group. Following Fred Lang (Forum Geometricorum, vol.2, 2002, pp.135--146), we construct the following table (splitted into a negative part and a positive part) taking into account that K004 is stable under four transformations (detailed underneath) namely isogonal conjugation, reflection about X(3), X(30)-Ceva conjugation, X(64)-crossconjugation.

n

-8

-7

-6

-5

-4

-3

-2

-1

0

P

X(3355)

X(3472)

X(2130)

X(3353)

X(3183)

X(3182)

X(1498)

X(1490)

X(20)

Q

 

 

 

 

X(3637)

X(2131)

X(3348)

X(3346)

X(64)

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

X(20)

X(40)

X(3)

X(1)

X(4)

X(84)

X(64)

X(3345)

X(3346)

X(3347)

X(3348)

X(3354)

X(2131)

X(3473)

X(3637)

X(64)

X(4)

X(3)

X(20)

X(1498)

X(3183)

X(2130)

X(3355)

 

 

 

 

 

 

 

Each point P on K004 corresponds to an integer n and Q = tg(P) denotes the tangential of P.

Three points on K004 are collinear if and only if their sum is 6. This gives the folllowing assertions generalized below.

• the sum of two isogonal conjugates P, P* is 6 since they are collinear with the neutral element X(20).

• the sum of two points P, P' symmetric about X(3) is 4 since X(3) is 2. Equivalently, the arithmetic mean is 2.

• the sum of P and its X(30)-Ceva conjugate X30/P is 0 since they are collinear with X(64), the tangential of X(20) with value 6.

• the sum of P and its X(64)-cross conjugate X64©P is 12 since they are collinear with X(2130), the tangential of X(64) with value -6.

The left-hand table shows a selection of points deduced from P with their corresponding integer. The right-hand table shows pairs of points with sum s hence collinear with a third point S of K004.

P

n

 

P'*

2 + n

P*

6 - n

P*'

-2 + n

P'

4 - n

tg(P*)

-6 + 2n

Q = tg(P)

6 - 2n

Q*

2n

X20/P

- n

tg(P')

-2 + 2n

X20/P*

-6 + n

Q'

-2 + 2n

X20/P'

-4 + n

X64©P

12 - n

s

S

example / remark

0

X(64)

X(30)-Ceva conjugates

1

X(84)

X(1) and X(1498)

2

X(4)

X(1) and X(1490)

3

X(1)

X(3) and X(40)

4

X(3)

symmetric about X(3)

5

X(40)

X(1) and X(3)

6

X(20)

isogonal conjugates

7

X(1490)

X(1) and X(4)

8

X(1498)

X(1) and X(84)

9

X(3182)

X(1) and X(64)

12

X(2130)

X(64)-cross conjugates

Remark : tg(P') = tg(P)' with same number -2 + 2n which is obvious geometrically.

In other words, this gives a quick and easy method to find the third point of K004 on a line passing through two given points. For example, the third point P on X(3183)-X(3473) is given by -4+13+ n = 6 hence n = -3 so P = X(3182).

Similarly, it is easy to identify the tangential of any point on K004.