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.