Isoconjugation comes from the theory of projective geometry. In terms of points, it associates

with a point X = x:y:z a point pyz:qzx:rxy. The point P = p:q:r is the *pole* of the isoconjugation.

The literature on triangle centres is rather confusing.

In ETC glossary, Clark gives the definition of X-isoconjugate as the isoconjugation with pole the

barycentric product of X1 and the isogonal conjugate of X. I can see no geometry in this.

**Beware.** ETC is inconsistent. The Glossary claims that X(2)-isoconjugation is isotomic conjugation.

In the body of ETC several X(2)-isoconjugates are listed. These use the definition in the Glossary,

and make it clear that it is not isotomic conjugation. See, for example X2053 and X2054.

Several other authors define X-isoconjugation as the isoconjugation with pole X, while others as

the isoconjugation with X as a fixed point, so the pole is the barycentric square of X.

An isoconjugation h may be specified by giving a pair of points X,Y interchanged by h. The pole

is then the barycentric product of X and Y. The pole of h is obviously h(G), the image of G.

Here, we shall specify an isoconjugation by giving a pair X,Y, and talk of (X,Y)-isoconjugation.

Here is a construction which yields a number of useful points (and requires only a staight-edge).

**Construction 1**

Let X = x1 : x2 : x3, and write Xa, Xb, Xc for the traces of X on BC, CA, AB.

Given points P = p1:p2:p3, Q = q1:q2:q3 and R =r1:r2:r3.

Let B' be the meet of PaRc and AC

Let C" be the meet of QaRb and AB

Let A* be the meet of BB' and CC"

Define B*, C* by cyclic symmetry.

Then ABC and A*B*C* are perspective, with perspector S =s1:s2:s3, where si = pi.qi/ri, (i = 1,2,3).

Thus S is the image of R under (P,Q)-isoconjugation.

Some special cases where the construction yields useful output :

When P = Q = G S is the isotomic conjugate of R.

When P = Q = I, S is the isogonal conjugate of R.

(or P = G, Q = K)

When R = G, S is the barycentric product of P and Q.

When R = I, S is the trilinear product of P and Q.

When Q = G, S is the barycentric quotient of P and R.

When R = X(76), S is the trilinear quotient of P and R.

In the construction, we can replace Ra, Rb, Rc by the intersections
of BC, CA, AB with the tripolar of R.

This version is a generalisation of that of the barycentric square by Paul Yiu (barycentric notes).

Other constructions for the isoconjugation have been given by Keith Dean and Floor van Lamoen

(forum geometricorum)
and by Bernard Gibert (glossary of CTC). Their constructions each

require a point on PQ, so cannot be used when P = Q. Here, we have such no restriction.

We also have the dual notion of line isoconjugation.

Duality associates the point U = u:v:w with the line L(U) : ux+vy+wz = 0.

In triangle geometry, we more often meet the idea of the* tripolar* of U, the line T(U) = x/u+y/v+z/w = 0.

The point U is then the *tripole* of the line.

Given lines L = T(X) ,M = T(Y) ,N = T(Z) , we define the (L,M)-*isoconjugate* of N as the line T(W), where

W is the (X,Y)-isoconjugate of Z. We could replace each T(M) by L(M), with the same resulting line N.

We could now use the above construction to get line isoconjugates. But we have a simple *direct* method.

**Construction 2.**

For a line L, let aL, bL, cL denote the intersections of L with BC, CA, AB.

Given lines L, M, N.

Let A1 be the meet of the lines aLcN and CA.

Let A2 be the meet of the lines aMbN and AB.

Let A" be the meet of A1A2 and BC.

Define B", C" by cyclic symmetry.

Then A",B",C" are collinear, and define the (L,M)-isoconjugate of N.

**Application**

Suppose that F and R are fixed points. F, and that L is any line through F.

Then the locus of the intersection of L and the (L,L)-isoconjugate of T(R) is cK(#F,R).

This is quite different from the constructions in Special Isocubics.

**Notes.**

(1) This construction gives directly the third point on each line through F.

(2) The (L,L)-isoconjugates of T(R) envelop the circumconic with perspector the (F,F)-isoconjugate of R.

This circumconic is tangent to cK(#F,R) at A, B, C.

(3) The Maple calculation verifying the above locus consists of checking when the intersection of a line L

with the (L,L)-isoconjugate of T(R) lies on cK(#F,R). As above, this happens when F is on L. It also

happens when L is the dual of a point on the nine-point conic C(tF,tR), where tX denotes the isotomic

conjugate of X. This is equivalent to saying that L is tangent to the common inconic of the anticevian

triangles of the points F and R. See duality and projective maps.