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A CPCC point is a point M having Coincident Pedal and Cevian Circles. H is obviously a CPCC point since the cevian and pedal triangles of H coincide. A CPCC point is equivalently the orthocenter of its cevian triangle hence they are also called H-cevian points. Compare them to the Ix-anticevian points in Table 23. This is generalized in : Table 28 : cevian and anticevian points.

Recall that the pedal circle of M is also the pedal circle of its isogonal conjugate M* and that the cevian circle of M is also the cevian circle of its cyclocevian conjugate M#. From this, we see that the pedal triangle of M* must be a cevian triangle hence M* and therefore M must lie on the Darboux cubic K004. Similarly, the cevian triangle of M# must be a pedal triangle hence M# and therefore M must lie on the Lucas cubic K007. Thus there are exactly four CPCC points (apart H) since these two cubics intersect in 9 points among which A, B, C, H, L = X(20). See figure below.

For more information on the Darboux-Lucas pencil, see table 15.

table11cercle

 

Orion cubics

If we consider that the cevian triangle of M must be the pedal triangle of M*, we find that M must also lie on three cubics Ka, Kb, Kc which are all nodal cubics.

More precisely, Ka is a nodal circum-cubic with node A and nodal tangents the bisectors at A. It contains Ga (anticomplement of A), the foot Ha of the altitude through A, the antipode Oa of A on the circumcircle, the reflection A' of A in the perpendicular bisector of BC and the four CPCC points (red points on the figure). The equation of Ka is :

x(c^2y^2 - b^2z^2) + (SB y - SC z)yz = 0,

and those of Kb, Kc similarly. Note that the "sum" of these three nodal cubics is the Lucas cubic.

Ka, Kb, Kc generate a net of cubics all passing through A, B, C and the four CPCC points. This net contains the Darboux and Lucas cubics and many other cubics we shall meet in the sequel.

If we denote by P the point with barycentric coordinates (p:q:r), any cubic of the net is written under the form :

K(P) = p Ka + q Kb + r Kc

For instance, K(X2) is the Lucas cubic and K(X3) is the Darboux cubic.

Any such cubic is called an Orion cubic since it is the locus of point M such that P, M and the Orion transform OT(M) of M are collinear. See ETC (preamble before X2055) or Hyacinthos #7999 & sq.

• K(P) meets the line at infinity at the same points as the cubic pK(X6, aaP) where aaP denotes the anticomplement of the anticomplement of P.

It follows that there is one and only one equilateral K(P) obtained with aaP = X3 hence P = X(140) and there is a pencil of circular K(P) obtained when aaP hence P lie on the line at infinity. See below.

K(P) and pK(X6, aaP) meet at six other (finite) points that lie on the circumconic passing through X2 and the cevian product (cevapoint) of X6 and P.

• K(P) meets the circumcircle of ABC at the same (six) points as the isogonal pivotal cubic with pivot Q we shall call the Orion pivot of P here denoted by OP(P).

The first coordinate of Q is : a^2 (-b^2 c^2 p + a^2 c^2 q + a^2 b^2 r - c^4 q - b^4 r).

A construction and a study of this Orion pivot transformation is given below.

K(P) and pK(X6, Q) meet at three other points that lie on the line passing through X6 and the isogonal conjugate of the cevian product of X6 and P.

• The two pivotal cubics above coincide if and only if P is one of the points X3, X2574, X2575.

When P = X3, we have aaP = Q = X20 and K(P) is the Darboux cubic. The two other points give two circular cubics.

When P = (a^2:SC:SB) – the midpoint of the A-altitude – K(P) decomposes into the line BC and a hyperbola Ca. Cb and Cc are defined likewise. The "sum" of these three decomposed cubics is the cubic K168 = pK(X3, X2) which therefore also contains the four CPCC points.

Ca contains A, the midpoint Ma. Its two asymptotes are parallel to the lines BC and AO.

These three hyperbolas belong to the same pencil of conics which contains a rectangular hyperbola (H) passing through X(5), X(6), X(20), X(69), X(1498) whose asymptotes are parallel to those of the Jerabek hyperbola.

 

Special Orion cubics

This net of cubics K(P) (which are all K0 circum-cubics i.e. without term in xyz) contains a large number of remarkable cubics.

Pivotal K(P)

K(P) is a pK if and only if P lies on the Thomson cubic. The pole also lies on the Thomson cubic and is the G-Ceva conjugate of P. The pivot lies on the Lucas cubic and is the anticomplement of the isogonal conjugate of P. The following table shows a selection of these cubics. See also K(Ma), K(Mb), K(Mc) below.

P

pole

pivot

cubic

points on the cubic

X(1)

X(9)

X(8)

K199 Soddy-Nagel cubic

see the page

X(2)

X(2)

X(69)

K007 Lucas cubic

see the page

X(3)

X(6)

X(20)

K004 Darboux cubic

see the page

X(4)

X(1249)

X(4)

K329

see the page

X(6)

X(3)

X(2)

K168

see the page

X(9)

X(1)

X(329)

K332

see the page

X(57)

X(223)

X(7)

K333

see the page

X(223)

X(57)

E(623)

 

X(174), X(189), X(223), E(623)

X(282)

X(3341)

X(189)

 

X(84), X(189), X(282), X(1490)

X(1073)

X(3343)

X(253)

 

X(64), X(253), X(1073), X(1498)

X(1249)

X(4)

E(624)

 

X(253), X(1249), E(624)

X(3343)

X(1073)

E(625)

 

X(1032), X(2130)

 

 

 

 

 

 

 

 

 

 

Non-pivotal K(P)

K(P) is a nK0 if and only if P lies on an isogonal nK with root X(20) without any known center on it.

Circular K(P)

K(P) is a circular cubic if and only if P lies at infinity. Since all circular K(P) have nine common points, they form a pencil and have their singular focus on the circle with center X(140), radius R/4, passing through X(620) i.e. the complement of the nine-point circle. Here is a selection of such circular K(P) :

P

points on the cubic

singular focus

X(30)

X(4), X(13), X(14), X(20), X(30), X(1294) see K313

complement of X(125)

X(511)

X(3), X(511), X(1297)

X(620) = complement of X(115)

X(516)

X(175), X(176), X(516)

complement of X(116)

X(517)

X(1), X(40), X(517), X(1295)

complement of X(11)

X(519)

X(8), X(519), X(2370)

 

X(524)

X(2), X(69), X(524), X(2373)

 

X(527)

X(7), X(329), X(527)

 

X(539)

X(5), X(539), X(1141)

 

X(971)

X(84), X(971), X(1490)

 

Equilateral K(P)

K(P) is an equilateral cubic if and only if P = X(140), the nine-point center of the medial triangle. The corresponding cubic is K268.

K(P)+

K(P) is a K+ (i.e. has three concurring asymptotes) if and only if P lies on the cubic K071' which is the complement of the cubic Kconc = K071. The following table gives a selection of such cubics.

P

K(P)

points on the cubic / remark

X(3)

K004 Darboux cubic

see the page / central pK++

X(4)

K329 = pK(X1249, X4)

see above / pK+

X(39)

 

X(3), X(39)

X(140)

K268

see the page / K60+

X(389)

 

X(3), X(389)

Since the cubic K071' contains the midpoints Ma, Mb, Mc of ABC, there are three corresponding cubics which are all central K++ whose "sum" is the Lucas cubic.

K(Ma) is a central pK++ with center Ma and inflexional tangent at this point through K = X(6). Its pole is the midpoint of the altitude AH and its pivot is G.

The three asymptotes are the parallels at Ma to the lines AH, AB, AC i.e. two sidelines of the medial triangle and the perpendicular bisector of BC. The tangents at A, B, C, Ga are all perpendicular to the sideline BC.

 

Other Orion cubics

The following table gives several other interesting Orion cubics

P

K(P)

points on the cubic / remark

X(230)

K518

see the page

X(1147)

K519

see the page

X(378)

K520

see the page / nodal cubic

?

K521

see the page / nodal cubic

X(631)

K522

see the page

 

Other curves

These CPCC points lie on the circular quartic Q056, on the Euler-Morley quintic Q003, both also containing the Ix-anticevian points (see Table 23) and on Q058.

The four points also lie on Q034, the X(370)-quintic which also contains X(2), X(3), X(13), X(14), X(370) and its five mates (blue points).

In order to find Q034, it is enough to express that the vertices of the pedal triangle of M lie on the cevian circle of M. This gives three quintics which must contain the four CPCC points. The "sum" of these quintics is Q034.

 

Orion conjugation

Wilson Stothers (I quote him) observes that there is a conjugation implicit in the material above. Indeed, any K(P) passing through a given point Z belongs to a pencil of cubics already containing 8 points and then must pass through a nineth point Z' which only depends of Z. This point Z' is called the Orion conjugate of Z.

In an obvious notation, the conjugate of a point is given by Ca/Ka : Cb/Kb : Cc/Kc.

Some conjugate pairs are : {X1,X40}, {X2,X69}, {X3,X3}, {X4,X20}, {X6, X193*}, {X7,X329}, {X8,X8}.

Note that X3 and X8 (and hence its extraversions) are fixed. X193* is the isogonal conjugate of X193.

The set of fixed points is actually the sextic S(H).

The sextic S(H)

(1) has double points at A, B, C, the tangents being the angle bisectors

(2) contains :

the vertices of the medial triangle,

the points X3, X8 and its extraversions,

the four CPCC points,

(3) has six real asymptotes parallel to the sidelines and the altitudes.

Wilson notes a similarity in Table 11 and Table 23 which are reversing the roles of H and Ix. The analogous sextic is S(Ix).

 

Isogonal conjugates of the CPCC points

The isogonal conjugates of the four CPCC points lie on :

  • the Darboux cubic, hence each CPCC point, its isogonal conjugate and X(20) are collinear,
  • K172 = pK(X32, X3), the isogonal transform of the Lucas cubic,
  • K233 = pK(X25, X4), the isogonal transform of K168,
  • K632 = pK(X604, X1), the isogonal transform of K199,
  • K523, a circular cubic, the isogonal transform of K313,
  • the rectangular hyperbola (H') passing through X(3), X(6), X(20), X(193), X(2574), X(2575), X(2293), X(3057),
  • the isogonal transforms of the nodal cubics Ka, Kb, Kc which are three conics,
  • obviously and more generally, the isogonal transforms of all the curves seen above.
table11K004K523

The CPCC points are represented in red, their isogonal conjugates in green.

 

Orion pivot transformation

Recall that the Orion cubic K(P) meets the circumcircle at the same points as pK(X6, Q) where Q is the Orion pivot of P given by Q = a^2 (-b^2 c^2 p + a^2 c^2 q + a^2 b^2 r - c^4 q - b^4 r) : : .

The corresponding transformation is denoted OP : P –> Q.

The reciprocal transformation is given by : Q(p:q:r) –> P = a^2 SA [a^2 (p + q + r) - c^2 q - b^2 r] : : .

Construction of Q

table11Orionpivot

Let (Ha) be the conic passing through A, the reflection A' of A in the perpendicular bisector of BC, the reflection Ao of A in O, the traces V, W of X99 on the sidelines AC, AB.

(Ha) is a rectangular hyperbola that also contains X110 and whose asymptotes are the parallels to those of the Jerabek hyperbola passing through the midpoint of X110-A'.

Two other rectangular hyperbolas (Hb), (Hc) are defined similarly.

Now, given the point P, let us draw the line PA meeting (Ha) again at Sa and the line (La) passing through Sa and A'. See figure.

The lines (Lb), (Lc) are defined similarly. These three lines concur at Q.

A reversed construction with given point Q easily gives P.

Obviously, if P is one of the points X110, X2574, X2575 (infinite points of the Jerabek hyperbola) the points Q and P coincide since these three points lie on the three rectangular hyperbolas.

Note that the six points Sa, Sb, Sc, X110, X2574, X2575 lie on a same rectangular hyperbola (Hp) that passes through the two points P and Q. Q is the antipode of X110 on (Hp).

table11Orionpivot2

Properties

The mapping OP : P –> Q has three fixed points as already said. Since it is a mapping of degree 1, it preserves the degree of any curve i.e. the image of a line is another line, the image of a conic is another conic, etc. In particular, the line at infinity is globally transformed since it contains X2574 and X2575, similarly the axes of the Stammler hyperbola since they are the lines X110-X2574 and X110-X2575.

There is one and only one circumconic whose image is another circumconic, actually the circumcircle : this is that of perspector X577 (the barycentic square of O) and center X1147 (whose image is O). It contains A', B', C' (the images of A, B, C), X110.

OP has the remarkable property to transform two points symmetric about X110 into two other points also symmetric about X110. This is the case of the pairs X3,X399 and X20,X140.

***

The following list gives a selection of such pairings P,Q = OP(P) namely X(i),X(j) for i,j =

2,2979 – 3,20 – 6,69 – 32,1975 – 58,1043 – 101,664 – 155,4 – 182,1350 – 184,22 – 195,2888 – 371,490 – 372,489 – 394,1370 – 399,146 – 511,1503 – 519,2390 – 542,2781 – 1147,3 – 3167,2 – 3292,858.

Several very common ETC centers have unlisted images :

OP(X1) lies on 1,21 – 2,65 – 4,8 – 9,1405 – 10,908 – etc

OP(X4) lies on 2,185 – 3,74 – 4,52 – 8,2807 – etc

OP(X5) lies on 2,389 – 3,49 – 4,69 – 5,51 – etc