**Notation**

P = u:v:w, Q = p:q:r.

T(M), the tripolar of a point M,

C(M), the circumconic with perspector M,

I(M), the inconic with perspector M.

R, the cevapoint of P and Q,

T, the crosspoint of P and Q,

Y, the intersection of T(P) and T(Q), also the perspector of the circumconic through P and Q.

Z, the tripole of the line PQ.

C(P,Q), the common circumconic of the Cevian Triangles of P and Q,

I(P,Q), the common inconic of the Cevian Triangles of P and Q, a diagonal conic.

C'(P,Q), the common circumconic of the Anticevian Triangles of P and Q, the diagonal conic through P and Q.

I'(P,Q), the common inconic of the Anticevian Triangles of P and Q.

D(P,Q), the circumconic through P and Q, = C(Y),

D'(P,Q), the inconic touching T(P) and T(Q), = I(Z).

G(P,Q), the diagonal conic x^{2}/pu + y^{2}/qv + z^{2}/rw = 0.

**Duality**

For any conic **C**, duality in **C** maps a point to its polar in **C**, a line to its pole in **C**.

Composing dualities in two conics gives a projective mapping from points to points.

By La Hire's Theorem, applying duality in the same conic twice gives the identity mapping.

Duality in G(P,Q) is particularly interesting, it swaps P and T(Q), and also Q and T(P).

After the above remarks, we instantly have :

**Lemma 1**

The following pairs are dual in G(P,Q).

(1) C(P,Q) and I'(P,Q),

(2) I(P,Q) and C'(P,Q),

(3) D(P,Q) and D'(P,Q), *i.e.* C(Y) and I(Z).

(4) C(T) and I(R),

(5) Y and PQ.

**Note**

The conic G(P,Q) may have no real points. However, we can implement the duality in the

following way :

Let h denote the isoconjugation swapping P and Q.

The dual of a point M is the tripolar of h(M).

The dual of a line L is h(N), where N is the tripole of L.

Though it is less easy to prove, we also have :

**Lemma 2**

(1) C(P,Q) and I(P,Q) are dual in D(P,Q),

(2) C'(P,Q) and I'(P,Q) are dual in D'(P,Q).

The two parts of Lemma 2 are easily seen to be equivalent by applying G(P,Q)-duality.

The first proof of this used the fact that, for M on C(P,Q), the line joining the P-Ceva and Q-Ceva conjugates

of M, is the polar of M in D(P,Q). Once known, the result may be verified by a routine computation since we

can write down matrices for each of the conics.

The vertices of the Anticevian triangle of x:y:z are -x:y,z, x:-y:z and x:y:-z, so that, trivially,

**Lemma 3**

If a point M lies on a *diagonal* conic **C**, then

the vertices of the Anticevian triangle of M also lie on **C**.

**Some projective mappings**

From Lemmata 1 and 2, we have some useful projective mappings :

dg = composite of D(P,Q) and G(P,Q)-dualites maps C'(P,Q) to C(P,Q) (and I'(P,Q) to I(P,Q)).

gd = composite of G(P,Q) and D(P,Q)-dualites maps C(P,Q) to C'(P,Q) (and I(P,Q) to I'(P,Q)).

In an obvious notation, gd' = dg and d'g = gd, so there are no new mappings here.

Recall that C'(P,Q) and I(P,Q) are diagonal conics.

Thus each point on one of these gives a further three.

Our projective mappings give similar structure on C(P,Q) and I'(P,Q).

In more detail :

**Theorem 1**

(1) If C(P,Q) contains a point M, then it also contains the vertices of the Anticevian triangle

of M, taken with respect to the Anticevian triangle of the point Y.

(2) If I'(P,Q) contains a point N, then it also contains the vertices of the Anticevian triangle

of N, taken with respect to the Cevian triangle of the point Z.

Proof

(1) The point M is the image of a point M' on C'(P,Q) under the map dg.

Then C'(P,Q) also contains the vertices of the Anticevian triangle of M' with respect to ΔABC.

Since the mapping is projective, C(P,Q) also contains the vertices of the Anticevian triangle

of M with respect to the image of ΔABC under dg. This is the Anticevian triangle of Y, by an

elementary calculation.

(2) This is proved similarly. The calculation gives the Cevain triangle in the statement.

**Note. **
In the notation of Bernard Gibert's paper

P1 on C(P,Q) gives the vertices of the Cevian triangle of P,

P2 on C(P,Q) gives the vertices of the Cevian triangle of Q.

If we have a triangle Δ and a conic **C** which contains the vertices of the Anticevian triangles of

each of its points, then **C** is a diagonal conic. Hence Δ is self-polar with respect to **C**.

**Corollary 1.1**

(1) The Anticevian triangle of Y is self-polar with respect to C(P,Q).

(2) The Cevian triangle of Z is self-polar with respect to I'(P,Q).

In fact, each instance of a triangle self-polar with respect to a conic arises in this way for

any of its cevian triangles.

**Theorem 2**

Suppose that the triangle Δ is self-polar with respect to the conic **C**.

Let U be a point not on a sideline of Δ.

We work with respect to the triangle Δ.

Let the tripolar of U meet **C** at V and W.

Let P be the U-Ceva conjugate of V,

Let Q be the U-Ceva conjugate of W.

(1) **C** = C(P,Q) relative to the Cevian triangle of U in Δ.

(2) P and Q lie on

(a) the polar of U in **C**,

(b) I(U) (relative to Δ),

(c) C(U) relative to the Cevian triangle of U.

(3) VW is also the tripolar of U relative to the Cevian triangle of U.

Proof

Let U = u:v:w, and let Δ' be the Cevian triangle of U.

As Δ is self-polar in **C**, **C** has equation Kx^{2} + Ly^{2} + Mz^{2} = 0.

Let V = p:q:r.

As V is on T(U), p/u+q/v+r/w = 0.

As V is on **C**, Kp^{2} + Lq^{2} + Mr^{2} = 0. (*)

(1) Each vertex of Δ", the Anticevian triangle of V has the form p:±q:±r, so

lies on **C**, and on a line of the form p/u±q/v±r/w = 0.

The latter are the sidelines of Δ', so Δ" is inscribed in Δ'.

Now Δ' is a Cevian triangle, so, by the Cevian Nest Theorem,

Δ' and Δ" are perspective. Further, the perspector is the point P,

the U-Ceva conjugate of V. So Δ" is the Cevian triangle of P in Δ'.

Using W, we get the point Q as perspector.

Now **C** is seen to be C(P,Q) relative to Δ'.

(2) By calculation, as V is on T(U), P = p^{2}/u:q^{2}/v:r^{2}/w.

From this it is clear that it lies on I(U). This is just C(U) relative to Δ'.

The polar of U in **C** is Kux + Lvy +Mwz = 0.

Now, V is also on **C**, so (*) holds, and so P is on the polar of U.

Likewise, Q lies on the given conic and line.

(3) This is just the observation that the tripolar is the perspectrix of Δ and Δ'.

Note that we can recover U, and hence P,Q, if we know the line L on which P,Q lie.

Then U is the pole of L in **C**. On the other hand, if we know just P, then there may

be as many as sixteen possible U, and hence sixteen Q. We know that P is on I(U),

so that tU, the isotomic conjugate lies on I(tP). Also, P is the U-Ceva conjugate of

a point on **C**, so U is the cevapoint of P and a point on **C**. Each condition puts the

point
U on a quadric.

In elementary proofs, we met a theorem of Bernard Gibert :

**Theorem**

C(P,Q) and I(R) are bitangent. The line of contacts is PQ, the common pole is Y.

With Lemma 1, we now have the

**Corollary**

I'(P,Q) and C(T) are bitangent. The line of contacts is PQ, the common pole is Y.