We show how pivotal isocubics occur in triples, related by perspectivity. The discussion

is motivated by the classical Lucas, Thomson and Darboux Cubics.

**Notation**

pK(W,P) is the isocubic with pole W and pivot P.

T(X) denotes the tripolar of the point X.

x(X,U) is the cross-point of X and U.

We also use the following non-standard notation.

Suppose that Δ(X) = ΔA'B'C' denotes the cevian triangle of X in ΔABC.

For a point U, let AU meet B'C' at A", BU meet C'A' at B", CU meet A'B' at C".

By general theory, ΔA"B"C" is also perspective with ΔA'B'C'.

The perspector is x(X,U).

The triangle Δ A"B"C" is denoted by Δ(X,U).

Also, suppose that U is not on T(X).

The lines AU, BU, CU meet T(X) at points A"', B"', C"'.

We write Δ'(X,U) for the degenerate triangle ΔA"'B"'C"'.

We will write "the (X,Y)-isoconjugate of Z" for the image of Z under the isoconjugation

which interchanges X and Y, and "W-isoconjugate" when the isoconjugation has pole W.

Suppose that X = x:y:z, and U = u:v:w. The following are useful in our descriptions :

the X-complement of U is x(v/y+w/z):y(w/z+u/x):z(u/x+v/y),

the X-anticomplement of U is x(v/y+w/z-u/x):y(w/z+u/x-v/y):z(u/x+v/y-w/z).

**Theorem 1**

Suppose that X = x:y:z, and U = u:v:w.

(1) The vertices of Δ(X,U) are x(u/y+w/z):v:w, u:y(w/z+u/x):w, u:v:z(u/x+w/z).

(2) The vertices of Δ'(X,U) are -x(u/y+w/z):v:w, u:-y(w/z+u/x):w, u:v:-z(u/x+w/z).

These are easy to *verify*.

**Grassmann Cubics**

We recall the definitions of pivotal isocubics as Grassmann loci.

Let Δ = ΔA*B*C* be a triangle perspective with ΔABC.

Let A^ = meet of MA* and BC, B^ = meet of MB* and CA, C^ = meet of MC* and AB.

GP(Δ) = { M : ΔA^B^C^ is perspective with ΔABC} is a pivotal isocubic.

Note that the condition is equivalent to saying that ΔA^B^C^ is a cevian triangle.

It is easy to prove that the perspectors that arise from points of GP(Δ) lie on another

pivotal isocubic GP^(Δ). This can be described as follows :

GP^(Δ) = { M : cevian triangle of M is perspective with Δ }.

Even if Δ comes from triangle centres, this cubic will not have triangle centres as pole

and pivot. We state the general result

**Theorem 2**

Suppose that Δ has vertices p:v:w, u:q:w, u:v:r. Then GP^(Δ) = pK(W,P),

with W = u/(qr-vw):v/(pr-uw):w/(pq-uv), and P = 1/(qr-vw):1/(pr-uw):1/(pq-uv).

If M = x:y:z is on GP(Δ), then the associated perspector lies on the line MN,

where N is the point u+p:v+q:w+r.

We now take a special case where triangle centres are obtained.

**Theorem 3**

Suppose that X = x:y:z, and U = u:v:w.

(1) **K0** = GP^(Δ(X,U)) =GP^(Δ'(X,U)) = pK(W,P), where

W = x^2:y^2:z^2, P = x^2/u:y^2/v:z^2/w.

(2) **K1** = GP(Δ(X,U)) = pK(W1,X),

**K2** = GP(Δ'(X,U)) = pK(W1,P2), where

W1 = xu(v/y+w/z):yv(w/z+u/x):zw(u/x+v/y) , and

P2 = x(-u/x+v/y+w/z):y(u/x-v/y+w/z):z(u/x+v/y-w/z).

(3) **K1** is the X-complement of **K0**.

(4) **K2** has a harmonic homology with axis T(X) and centre Z, where

Z = x(v/y+w/z):y(w/z+u/x):z(u/x+v/y).

(5) **K0, K1** meet at A, B, C, X(twice), U and on T(X).

The common tangent at X contains P and x(X,U).

(6) **K0, K2** meet at A,B,C, U, P2 and four others.

(7) **K1, K2** meet at A,B,C, U, Z and the fixed points of (U,Z)-isoconjugation

Note that

W is the barycentric square of X, so this isoconjugation fixes X,

W1 is the barycentric product of X and x(X,U), so this isoconjugation swaps X and x(X,U),

it also swaps U and Z.

P2 is the X-anticomplement of U,

Z is the X-complement of U.

From our descriptions, when we start with a point on one cubic, we get a configuration which

has a perspector on another of our cubics. In fact, we can identify the perspector precisely.

For example, if M is on **K0**, then the cevian triangle of M is perspective with Δ(X,U). The

perspector is then a point N on **K1**. Of course, we may equally obtain M from N.

**Theorem 4**

If M and N are associated points on **K0**, **K1**, then

(1) N is the W1-isoconjugate of the X-complement of M,

(2) M is the X-anticomplement of the W1-isoconjugate of N.

This is easy to verify by Maple. The situation for **K2** is less satisfactory.

**Theorem 5**

If M and N are associated points on **K0**, **K2**, then

(1) N is the intersection of the lines joining

(a) M and the X-anticomplement of U, the root of **K2**

(b) the W1-isoconjugate of the W-isoconjugate of M and Z, the centre of **K2**,

This also contains the W1-isoconjugate of M.

(c) U and x(X,M).

(2) M is the intersection of the lines joining

(a) N and the X-anticomplement of U,

(b) the W-isoconjugate of the W1-isoconjugate of N and the W-isoconjugate of U, the root of **K0**.

This also contains the W-isoconjugate of M.

Note.

The line in (1)(c) does not appear to have a simple equivalent in (2).

**Corollary 5.1**

(1) W-isoconjugate points on **K0** lead to homologous points on** K2**

(2) W1-isoconjugate points N1, N2 on **K2** lead to points M1, M2 on **K0** which lie the line through

N1, N2 (and therefore P2). Note that P2 is on **K0**, so this identifies M2 from M1.

Proof notes.

(1) From (1)(b) of the Theorem, M and its W-isoconjugate lead to points on **K0** which lie on a line

through the centre. These must be homologous as the only two on this line other than the centre

which lie on the cubic.

(2) This uses part (a) in each case. As the line contains the root of **K2**, N1, N2 give the same line.