**Notation**

For a given point P,

T(P) is the tripolar of P,

C(P) is the circumconic with perspector P,

I(P) is the incoic with perspector P.

For a fixed point W, P* is the image of P under the isoconjugation with pole W.

With hindsight, it appears that the most important result on the subject of non-pivotal

isocubics invariant under a harmonic homology can be stated as follows.

**Main Theorem**

Suppose that we have line L = T(X) and a point U not on L. Let *h* be the harmonic

homology with centre U and axis L = T(X).

Let V be any point on L. Let W be the pole of the isoconjugation swapping U and V.

Let R be the crosspoint of X and the tripole of UV. Then the cubic nK(W,R,U) is

(a) { M : VM, V*h*(M), VU, L is a harmonic pencil },

(b) { M : VM, VM*, VU, L is a harmonic pencil },

(c) { M : V, M*, *h*(M) are collinear },

(d) { M : *h*(M*) = *h*(M)* }.

Part (a) is equivalent to saying that the cubic is invariant under *h*.

The cubic meets L in V and the intersections with C(X*).

**Theorem 1**

Suppose that we have line L = T(X) and a point U not on L. Let *h* be the harmonic

homology with centre U and axis L = T(X).

The non-degenerate non-pivotal cubics nK(W,R,U) fixed by *h* have

(a) root R on the line RL, the polar of U in I(X),

(b) pole W on the line WL, the tripolar of the pole of the isoconjugation swapping

the points X and U.

(c) root R on the polar of U* in I(X) (and in C(X)).

(d) the U*-ceva conjugate of U lies on T(X).

Note that (b) is equivalent to saying that U* is on L = T(X). We then have the following

**Corollary 1.1**

If a non-degenerate non-pivotal isocubic nK(W,R,U) has a harmonic homology with

centre U, then the axis is the line joining U* and the U*-ceva conjugate of U.

This is clear from (d) and the remark following the Theorem.

Note that the two points of the corollary coincide only when U is fixed under the

isoconjugation. The cubic is then of type cK.

We will assume that U is not on a sideline, or L through a vertex of ΔABC.

An outline proof appears below. First we find some points which must lie on such a cubic.

**Three points on nK(W,R,U)**

These are the images of A,B,C with respect to the harmonic homology of U and L.

They are the second intersections of AU, BU, CU with the circumconic whose perspector

P is the pole of L in C(U), also known as the X-Ceva conjugate of U. Then U is the pole

of L in C(P). These clearly lie on all the nK(W,R,U) in the family (and on C(P)).

If X = x:y:z, and U is u:v:w, then these intersections are

A1 = x(-u/x+v/y+w/z):2v:2w , B1 = 2u:y(u/x-v/y+w/z:2w , C1 = 2u:2v:z(u/x+v/y-w/z).

**Proof notes for Theorem 1**

Suppose we take the general cubic nK(W,R,U). The condition that it contains A1 can be

expressed as an equation f1(W,R) = 0. This is linear in the coordinates of W and of R,

quadratic in those of X, and cubic in those of U. By algebraic symmetry, we get equations

f2(W,R) = 0, f3(W,R) = 0, expressing the conditions that B1 and C1 lie on the cubic.

If we treat these as equations in the coordinates of U, then they are consistent precisely

when one of the following conditions holds :

(a) U is on T(X) - but we have excluded this case since we do not get a homology,

(b) U is on T(R) - but T(R) already meets the cubic three times, so the cubic degenerates,

(c) U, R are conjugate with respect to I(X).

The last gives part (a) of the Theorem.

Now regard the equations as equations in the coordinates of R. We get consistency when

(d) U is on T(X) - see (a) above,

(e) U* is on the C(P) - but this meets the cubic in A,B,C,A1,B1,C1, so we have degeneracy,

(f) U* is on T(X).

The last was noted as equivalent to (b) of the Theorem.

If X = x:y:z, then the equation xf1(W,R)+yf2(W,R)+zf3(W,R) = 0 is satisfied when

(g) U is on T(X) - see (a) above,

(h) U*, R are conjugate in C(X).

But, from (f), U* is on T(X), so the polar of U* in C(X) is also the polar in IX). Thus (h) gives

(i) U*, R are conjugate in I(X).

This gives (c) in the Theorem.

Finally, we may rewite our equations in terms of the coordinates of U* instead of those of W.

They are then actually simpler, linear in the coordinates of U, U*, R and the isotomic of X.

The combination f1+f2+f3 = 0 shows that X lies on the circumconic whose perspector is

the stated ceva conjugate. Thus, the ceva conjugate lies on T(X). This is (d).

We can use the results of Theorem 1, and some of the above argument to determine the cubics

with given homology from just one piece of information.

**Theorem 2**

Each nK(W,R,U) in the family is determined by any of the triple {U*,R,W}

Choose U* on L

(1) R = pole of UU* in I(X) = crosspoint of X and the tripolar of UU*,

(2) W = pole of isoconjugation swapping U and U*.

Choose R on RL

(3) U* = intersection of L and the polar of R in I(X),

(4) W defined as in (2).

Choose W on WL

(5) U* = (W)-isoconjugate of U,

(6) R defined as in (1).

**Harmonic definition of nK(W,R,U)**

This appears in harmonic definitions. The main features are the following

**Theorem 3**

If L1 = T(X) and L2 = T(Y) meet in Z, and W is a point, then the locus

K(W,L1,L2) = { M : ZM, ZM*, L1, L2 is a harmonic pencil } is nK(W,R,Z),

where R is the crosspoint of X and Y.

Further,

(a) the cubic meets each line in Z, and in and a W-isoconjugate pair,

(b) the isoconjugate pair {P,P*} on L1 has P* on L2 if and only if P is a flex.

Now compare this with part (1) of Theorem 2. There, R is the crosspoint of X and

the tripole of UU*. Also, U is a flex, so (b) above proves yet again that U* is on

the line T(X). Theorem 3 then shows that K(W,UU*,T(X)) = nK(W,R,U*) = nK(W,R,U).

We can sum up these observations as

**Theorem 4**

If nK(W,R,U) is invariant under the harmonic homology, centre U, axis L = T(X), then

(a) nK(W,R,U) = { M : U*M, U*M*, U*U, T(X) is a harmonic pencil },

(b) nK(W,R,U) meets T(X) in an isoconjugate pair - the meets with C(X*).

**Cubics of type nK0(W,R,U)**

When U ≠ X, Maple calculation shows that these occur precisely when U* is an

intersection of L with the circumconic through U and X, so there are two in general.

When U = X, *all* are of type nK0.

**Remarks on Polar Conics**

We can check by Maple that the Polar Conic at U is the union of L and a line L'.

The line L' is the tangent at U to the circumconic through U and U*.

Thus we *do* have a flex at U, and L' is the inflexional tangent.

The Polar Conic of U* is complicated. We know by general theory that it meets the

cubic twice at U*, and at four other points P1,P2,P3,P4. The tangents at these pass

through U. We shall return to these four shortly.

**Theorem 5**

Suppose that P is on the cubic nK(U&U*,R,U) with harmonic homology h as above.

Suppose also that U*P meets the cubic again, at point Q other than U*,P.

(1) the lines U*P, U*P*, UU*, L form a harmonic pencil,

(2) U*, h(P), P* (the UU*-isoconjugate) are collinear,

(3) h(Q) = P*, Q* = h(P), so h(P*) = h(P)*,

(4) h(P) is the intersection of UP and U*P*,

(5) Q can be constructed as the intersection of U*P with UP*.

Proof :

By Theorem 1, we know that U*P, U*P*, UU*, L form a harmonic pencil.

From the definition of harmonic homology, U*P, U*h(P), UU*, L also do so.

At once, we have (2).

The line U*P* contains no further points on the cubic.

The line U*Q is the line U*P, so Q* and h(Q) are on U*P*.

They must be the points P*, h(P) in some order.

But P ≠ Q, so Q* ≠ P*. It must be h(P).

Likewise h(Q) ≠ h(P), so h(Q) = P*.

For (4), note that h(P) is on UP by definition.

**Theorem 6**

The points P1,P2,P3,P4 are

(1) precisely the points for which h(P) = P*.

(2) the intersections other than A,B,C,U,U* of nK(U&U*,R,U) with pK(U&U*,U).

(3) the intersections other than U*,U" of polar conic of U* with pK(U&U*,U),

where U" is the U-Ceva conjugate of U*.

(4) the intersections other than U*(twice) of polar conic of U* with nK(U&U*,R,U).

Proof :

In Theorem 5, we excluded the cases where U*P was the tangent at P, when Q = P.

The argument of the theorem shows that then h(P) = P*, h(P*) = P.

The condition that h(P) = P* amounts to P lying on two conics, so there are four

solutions. By their definition, the Pi have tangent U*Pi, so have h(P) = P*.

For (2), observe that U is on Pih(Pi), but here this is PiPi*.

Thus, Pi is on pK(U&U*,U). We now have all nine intersections.

**Degenerate cases**

The cubic nK(W,R,U) degenerates when it contains T(R).

As it contains U, we have two possibilities :

(a) U on T(R) - then R is on C(U). Hence we have degeneracy precisely when

R is an intersection of RL with C(U), there are two of these in general.

(b) U* on T(R). Now Maple calculation shows that this occurs precisely when

U* is an intersection of L with a sideline. There are three of these, but a little

thought shows that an intersection with a sideline can only be an isoconjugate

if it is a triangle vertex, and then U would have to be on a sideline.