The orthopole of a line is defined using perpendiculars. Perpendicular
directions can be

thought of as conjugate directions in a circle. If we
replace the circle by another conic,

then we will get a notion with similar
properties. Since the idea of conjugacy depends

only on the homothety class
of a conic, we may as well choose a circumconic.

Suppose we choose C(K), the circumconic with perspector K. Two directions defined by

the infinte points P and Q are K-*conjugate* if P lies on the polar of Q in C(K). Of course

this is equivalent to the condition that Q lies on the polar of P in C(K).

We will choose notation to remind us of the classical orthopole.

Let D be
any point.

Let H = tD (the isotomic conjugate of D),

Let L = aH (the
anticomplement of H) = atD,

Let K = cD (the complement of D) = taK : K is the
centre of I(H),

Let O = ctD =ctaK : O is the centre of I(D) and of C(K).

*Note* We can begin with any of the points D, H, L, K, O and derive the
others.

**The generalized orthopole.**

**Theorem 1**

Let C(K) be a fixed circumconic of triangle ABC. Let H =
taK.

Let m be a line.

Let m' be a line conjugate with m in C(K).

Let Am
be the intersection of m with the line through A parallel to m', and Bm, Cm
similarly.

Let mA be the line through Am parallel to the Cevian AH, and mB,
mC similarly.

Then mA, mB and mC concur.

**Definition**

The point of concurrence guaranteed by Theorem 1 is the
K-orthopole of m.

*Conway's notation*

Suppose that we write K = p:q:r. Then sA = (-p+q+r)/2, sB = (p-q+r)/2, sC = (p+q-r)/2.

Note that H is 1/sA:1/sB:1/sC = sBsC:sCsA:sAsB, and O = psA:qsB:rsC.

If m is the line ux+vy+wz = 0, then the K-orthopole of m has first barycentric

(usBsC - vqsB - wrsC)(up-vsC-wsB) = (usBsC - vqsB - wrsC)(upsA-vsCsA-wsAsB).

Obviously, if K = X(6), we get the usual orthopole.

*Notes in proof*

The usual geometric proofs that I know of break down.
But we can prove the theorem purely by

algebra. Since the coordinates
a^{2}:b^{2}:c^{2} satisfy no algebraic relations, we can
replace them

by any three variables, such as the coordinates of the given K.

*Geometry*

If m is a line not through O, the centre of C(K), then a
suitable m' is the line joining O with the

pole of m in C(K). To the
triangle geometer, this pole is the X-Ceva conjugate of K, where X is

the
tripole of m.

If m is through O, then a suitable m' is the polar of any point
on m. To the triangle geometer,

the polar of a point Y is the tripolar of
the cevapoint Y and K.

In any case, we could take m' as the polar of the
infinite point of m.

We use the cevians AH, BH, CH as these are conjugate to
BC, CA, AB in C(K).

We have natural analogues of other notions associated with perpendicularity.

The K-*pedal triangle* of a point Z has vertices AHnBC, BHnCA, CHnAB.

The K-*circumcevian triangle* has vertices oA, the second meet of AO and C(K), oB, oC similarly.

Simply using the algebra from the case K = X(6), we have the following result.

**Theorem 2**

If Z is a fixed point, then the locus of K-orthopoles of lines through Z is a conic **C**.

The locus contains the vertices of the K-pedal triangle of Z.

The locus is degenerate if and only if Z is on C(K) or the Line at Infinity.

The locus is an inconic if and only if Z = L. Then the locus is I(D).

Otherwise, the centre of **C** is the mid-point of Z and H = taK.

*Proof notes*

Since we have formulae for the coordinates of the K-orthopole, the equation of the locus can

be established by a straight-forward calculation. From the formulae, we find the lines which

give K-orthopoles on the sidelines, and hence find the intersection of the locus with these.

The lines through Z parallel to AH, BH, CH give the vertices of the K-pedal triangle of Z.

This is immediate from the geometry - the K-conjugate of AH is BC, so the construction

of Bm and Cm from B and C each give rise to the A-vertex of the K-pedal triangle of Z.

The points on each sideline coincide when Z is L, or is on C(K) or the Line at Infinity.

Obviously each such case must be an inconic or a conic degenerate as a repeated line.

The case Z = L gives I(D). The latter types give degenerate conics.

*Observation*

We noted that the lines through Z parallel to AH, BH, CH give the vertices of the K-pedal

triangle of Z. The other intersections of the locus with the sidelines come from the lines

through Z and the antipodes oA, oB, oC mentioned above. This can be seen as follows

We know that, if X and Y the Circumcircle, then the orthopole of XY is the intersection

of the Simson lines of X and Y. For the antipode of A, the perpendiculars to AB and AC

meet the sides at B and C, so the Simson Line of the antipode is BC. Hence the orthopole

of any line through this point must lie on BC. The general case follows by the algebra.

*A desmic structure*

The points oA, oB, oC and the infinite points of AH, BH, CH which we have used are part

of a desmic structure including A, B, C. We note that the points are K-isoconjugate pairs.

The desmic structure gives the Grassmann cubics pK(K,L) and the degenerate nK(K,G)

which is the union of C(K) and the Line at Infinity.

The following facts are easily verified by algebra. They are easliy observed in Cabri.

We write C(P,Q) for the nine-point conic through the vertices of the Cevian triangles

of of the points P and Q.

(1) For Z = L, the locus is I(D) = C(D,D), centre O.

(2) For Z = O, the locus is C(G,H), centre cO.

(3) For Z = H, the locus is C(H,?), centre H.

The second perspector in (3) is the barycentric product of H and tL.

This raises the general question of identifying Z for which the locus is a nine-point conic.

The answer involves an unexpected relation to the theory of central pivotal isocubics.

To begin with, we look at results of Ceva and Carnot, written in triangle geometry notation.

These are very easy to prove by algebra.

**Ceva's Theorem**

The points 0:u2:u3, v1:0:v3, w1:w2:0 are the vertices of a cevian triangle if and only if

u2v3w1 - u3v1w2 = 0.

**Carnot's Theorem**

The points 0:u2:u3, v1:0:v3, w1:w2:0, 0:U2:U3, V1:0:V3, W1:W2:0 lie on a conic

if and only if u2v3w1U2V3W1 - u3v1w2U3V1W2 = 0.

Then we have trivial, but nice, consequences.

**Corollary A**

The vertices of two cevian triangles lie on a conic.

**Corollary B**

If a conic cuts the sidelines at the vertices of a cevian triangle, then the other three

intersections with the sidelines are also the vertices of a cevian triangle.

In our context, we have six intersections of a conic with the sidelines of ΔABC. Three are

the vertices of the K-pedal triangle of Z. These can be grouped into two sets of three, each

set having one on each sideline, in four ways. We could take the vertices of the K-pedal

triangle as one set. The other possibilities have one set with two vertices of the K-pedal

triangle and the appropriate one of the others. There are clearly three such cases.

The first possibility is the more interesting. For K = X(6), the theory is well-known.

The general case was observed by Yiu, and appears in Gibert's Isocubic Book.

**Theorem 3**

The K-pedal triangle of Z is a cevian triangle if and only if Z is on pK(K,L).

For such Z, the perspector Z' lies on pK(G,D), and on the line ZL.

Further, if W is the reflection of Z in O, W' = tZ'.

The cubic pK(K,L) is central, with centre O. It contains oA, oB, oC and their K-isoconjugates,

which are the infinite points (the infinite points on AH, BH, CH). It contains O, H, L.

Each central cubic is of this form.

Now we can observe that, when Z is on pK(K,L), the locus for Z has the vertices of the K-pedal

triangle those of a cevian triangle. Then Corollary B shows that the locus is a nine-point conic.

Our examples above arise since L, O and H are on pK(K,L). By some algebra, we can establish

all Z which give rise to nine-point conics.

This needs some new notation. Recall that oA, oB, oC are the antipodes of A, B, C on C(K).

Let **CA** be the conic through oA, oB, oC, L and A. Define **CB**, **CC** similarly.

**Theorem 4**

The locus for Z is a nine-point conic if and only if

Z lies on pK(K,L), or

Z lies on one of **CA**, **CB**, **CC**.

*Proof notes*

We have coordinates for the six intersections with the sidelines.

Applying our version of Ceva's Theorem first to the vertices of the K-pedal triangle gives the

first examples. Taking just two of these vertices gives the other cases. It is easy to check

that the conics which arise as the condition in the later cases are through the given points.

*Locating the centres*

We know that the centre of the locus for Z is the mid-point of Z and H. Thus, if Z lies on a

curve **C**, then the centre is on the curve homothetic with **C** by h(H,1/2). This can also be

described as the complement of the reflection of **C** in O. Thus :

If Z is on pK(K,L), then the centre is on the complement of pK(K,L) (as this is central).

If Z is on **CA**, the the centre is on the complement of the circumconic through A and H.

In the latter case, the circumconic, and hence its complement, are "K-rectangular" in

the sense that the asymptotes are K-conjugate.

**Examples**

Some examples of nine-point conics with K = X(6).

For given Z, the locus is C(Z1,Z2), with centre Z3.

Each Z is on the Darboux Cubic.

Z1 is on the Lucas Cubic, and on ZX(20).

Z3 is the mid-point of Z and X(4).

Z | Z1 | Z2 | Z3 |

X(1) | X(7) | X(1440) | X(946) |

X(3) | X(2) | X(4) | X(5) |

X(4) | X(4) | Z4 | X(4) |

X(20) | X(69) | X(69) | X(3) |

X(40) | X(8) | X(2) | X(10) |

X(64) | X(253) | ? | cX(1498) |

X(1498) | X(20) | ? | cX(64) |

X(84) | X(189) | ? | cX(1490) |

X(1490) | X(329) | ? | cX(84) |

Z4 is the barycentric product of X(4) and the isotomic conjugate of X(20).

Note that, if Q is the reflection of P in O, then the centre for P is cQ, the

complement of Q. and the Z1 for P and Q are isotomic conjugates.