triangle centres A simple approach to the nine-point conic

We present a proof in barycentrics of the existence and fundamental properties of the nine-point conic
associated with a triangle and two points not on any sideline. The proof involves only hand-calculation.

Theorem
Given ΔABC and points U,P not on any sideline, there exists a
conic containing the vertices of the cevian triangles of U and P.

We shall obtain this theorem by a devious method - we consider a conic associated with U and P which
then turns out to contain the required points. In the case U = G, the centroid of ΔABC, this approach
goes back to Mineur's Cubiques Anallagmatiques.

Note We will identify not nine but eighteen points on each such conic.

We work in barycentric coordinates throughout. If a point has coordinates which are symmetric in the
variables we give the first enclosed in square brackets. Thus, the complement of x:y:z is written [y+z].

In a similar vein, we write Σf(x,y,z) for the sum over all cyclic permutations of variables and coefficients.
Thus the circumcircle has equation Σ a2yz = 0.

Recall that, if P = p:q:r, X = x:y:z, the P-Ceva conjugate of X is [x(-x/p+y/q+z/r)].
We write T(P) for the tripolar of the point P. This is the line x/p+y/q+z/r = 0.

Now, following Mineur, define the locus

C(U,P) = { X : P-Ceva conjugate of X is on T(U) }.

Let us define two points on each cevian of P and U as follows :
PA = 0:q:r, the foot of the cevian AP,
PA = u(2p/u+q/v+r/w):q:r, a point on AP,
with the obvious analogues.

We also define two points which will play important roles :
U' = [u(2p/u+q/v+r/w)].
P' = [p(2u/p+v/q+w/r)].

Theorem 1
C(U,P) is a conic.
C(U,P) = C(P,U).
C(U,P) contains PA,PB,PC,UA,UB,UC, PA,PB,PC,UA,UB,UC.

Proof
Using the above formulae for the P-Ceva conjugate, this has equation Σ (x2/pu-(1/qw+1/rv)yz) = 0.
This is clearly a conic. By the symmetry in U, P, C(U,P) = C(P,U).
It meets the line BC at 0:q:r, and at 0:v:w, i.e. PA,UA, and similarly for CA, AB.
From its definition, C(U,P) contains the P-Ceva conjugates of all points on T(U).
The intersection of AP and T(U) is -u(q/v+r/w):q:r.
The P-Ceva conjugate of this is then (a multiple of) , PA, so PA is on C(U,P).
Since C(U,P) = C(P,U) it also contains UA, the U-Ceva conjugate of the intersection of AU and T(P).
It therefore contains the twelve named points.

Theorem 2
The lines UAPA, UBPB, UCPC concur at U'.
The lines PAUA, PBUB, PCUC concur at P'.
U,P,U',P' are collinear.

Proof
Simply observe that, with the coordinates above U' = (p/u+q/v+w/r)UA + PA, so U' is on UAPA.
The first two results now follow by symmetry.
Now observe that U' = P +(p/u+q/v+w/r)U, so U' is on UP. Similarly P' is on UP.

We use some non-standard notation. This may be regarded as convenient shorthand for combinations
which occur frequently in our work. In fact, I think of them as the basic operations.

Suppose that we have points P = p:q:r, X = x:y:z, U = u:v:w.
The P-complement of X is pc(X) = [p(y/q+z/r)].
The P-anticomplement of X is pa(X) = [p(-x/p+y/q+z/r)].
The P-isoconjugate of X is pi(X) = [p2yz].

These are related to the more usual functions as follows. These are easily verified by hand.
The P-Ceva conjugate of X is pc(pi(pa(X))).
The crosspoint of P and X is pc(pi(U)), and also uc(ui(P)). This is [pu(qw+rv)].
Note that pc(X) is then the P-Ceva conjugate of the crosspoint of P and X.

Some more non-standard definitions.
The P-centre of a conic is the pole of T(P) in the conic.
For points X, Y = x':y':z', not both on T(P), the P-harmonic of X and Y is the point ph(X,Y)
where (X,Y,ph(X,Y),Z) is harmonic, with Z the intersection of XY and T(P).

Observe that the G-centre is the usual centre, and the G-harmonic of X,Y is the mid-point.

It is again easy to verify that
The P-centre of C(U) is the P-Ceva conjugate of U. It is best to check that the P-Ceva conjugate has
the required polar!
The ph(X,Y) is [x(x'/p+y'/q+z'/r)+x'(x/p+y/q+z/r)]. To see this, simply find the intersection
of the line XY with T(P), Observe that the coordinates are the differences of the terms in the ph(X,Y),
so the given point is the harmonic.

Now, pc fixes P and the points of T(P). It is a projective transformation. Thus,

Lemma
If X is the P-centre of a conic D, then pc(X) is the P-centre of pc(D).

We write C(P) for the circumconic with perspector P. This has equation pyz+qzx+rxy = 0.

Theorem 3
Let Q be the crosspoint of U and P.
C(U,P) = uc(C(Q))= pc(C(Q)).

Proof
By definition, C(U,P) is the P-Ceva conjugate of T(P), i.e. uc(ui(ua(T(P)))).
Thus C(U,P) is the uc(D), where D=ui(ua(T(P))). By an easy calculation, D = C(Q). To see this,
first replace X = x:y:z by uc(X) in T(P) to get ua(T(P)). Then replace X by ui(X) in this to get D.

Theorem 4
The U-centre of C(Q) is uc(P).
The U-centre of C(U,P) is uc(uc(P), this is the point U'.
The P-centre of C(Q) is pc(U).
The P-centre of C(U,P) is pc(pc(U)), this is the point P'.

Proof
The U-centre of C(Q) is the U-Ceva conjugate of Q, i.e. uc(ui(ua(Q))).
But, as the crosspoint of U and P, Q is uc(ui(P))
Thus the U-centre of C(Q) is uc(P).
By the Lemma, and Theorem 3, the U-centre of C(U,P) is uc(uc(P))).
By easy calculation, this is the point U'.

Theorem 5
The points PA,PB,PC are uh(A,P), uh(B,P), uh(C,P).
The points UA,UB,UB are ph(A,U), ph(B,U), ph(C,U).

Proof
The calculation of uh(A,P) is easy. It yields the point PA.

Note
Theorems 5 and 2 show that that points PA,PB,PC, UA,UB,UB can be constructed in two ways :
(1) as harmonics, for example PA is uh(A,P), or
(2) as intersections, for example PA is the intersection of UAU' and AP.

Theorem 6
U' = uh(PA,UA) = uh(PB, UB) = uh(PC,UC) = uh(U,uh(U,P)) = uc(uc(P)).
P' = ph(UA,PA) = ph(UB, PB) = ph(UC,PC) = ph(P,ph(P,U)) = pc(pc(U)).

Proof
The results are proved by straightforward calculation of the harmonics.
The last version was given in Theorem 4.

Theorem 7
C(U,P) contains the intersections of C(Q) with T(P) and T(Q).

Proof
Just observe that the equation of C(U,P) can be rewritten as
(x/p+y/q+z/r)(x/u+y/v+z/w) - 2Σ(1/qw+1/rv)yz = 0.

Note
This gives a further four points on C(U,P).

Now for a result where the calculation requires care.

Theorem 8
For any point X on C(Q), ph(X,U) and uh(X,P) lie on C(U,P).

Proof
To verify this, we need to check when uh(X,P) is on C(U,P).
We take the equation of C(U,P) as X'X"-2Σx2/pu = 0,
where X' = x/u+y/v+z/w, and X" = x/p+y/q+z/r.
If we write P = p/u+q/v+r/w, then uh(X,P) = xP+pX':yP+qX':zP+rX'.
This is on C(U,P) if and only if
2PX'(PX"+3X') -2Σ x2P2/pu -4PXΣx/u - 2X2Σp/u = 0.
The last two terms each simplify to PX'2, so the condition becomes
2P2(XX' - Σx2/pu) = 0.
From the second factor, this holds for x:y:z on C(Q).

Finally, two other points on C(U,P) with fairly nice coordinates :

Theorem 9
Let R be the intersection of T(U) and T(P).
Then C(U,P) contains the P-Ceva and the U-Ceva conjugates of R.
These are V = [p(q/v-r/w)2] and W = [u(v/q-w/r)2].
Each lies on T(R) and on one of the inconics with perspectors U and P.

Proof
As R is on T(U), its P-Ceva conjugate is on C(U,P) by the definition of C(U,P).
As C(U,P) = C(P,U), the conic also contains the U-Ceva conjugate.
Direct calculation gives the coordinates of these points quite easily.
The fact that each lies on one of the inconics relies on the useful observation
that x:y:z is on the inconic perspector P if and only if x/p:y/q:z/r = f2:g2:h2,
with f+g+h = 0. It is easy to check that they lie on T(R) from its equation.

Note
C(U,P) is a circumconic of the Cevian triangle of P, as is the inconic with perspector P.
Thus these two conics have exactly one further intersection. Theorem 9 identifies it.
It may also be checked that T(R) is the tangent to the inconics at these points.

We now have our stated eighteen points on C(U,P), at least fourteen are real.

The sketch below shows a case with just fourteen real points. The conics C(U,P), C(Q)
have no real intersections.

Footnote

In Hyacinthos message #11490, Ayres gives a result on the diagram showing ΔABC and
the cevians of points P and Q. In a simple construction, he obtains two perspectors. These are
the P-complement and the P-isoconjugate of Q. The P-anticomplement operation is simply the
inverse of the P-complement operation.

A new result
This is due to Bernard Gibert. We give here a barycentric proof.

Gibert's Theorem
Let T be the cevapoint of U,P, so T = [1/(qw+rv)].
Let R be the intersection of T(U) and T(P), as above.
C(U,P) and I(T), the inconic with perspector T, are bitangent, with contacts on PU.
PU is the polar of R in both C(U,P) and I(T) so the common tangents meet at R.

Proof
Write C(U,P) as F(x,y,z) = Σ(qrvwx2-pu(qw+rv)yz) = 0,
and I(R) as G(x,y,z) = Σ((qw+rv)2x2-2(pv+qu)(pw+ru)yz) = 0.
Then G(x,y,z)-4F(x,y,z) = H(x,y,z)2, where H(x,y,z) = 0 is the line PQ.
Thus the conics are bitangent at their intersections with PU.
The second part is simply a tedious calculation of the polars of R.