Let {P

The discussion here is a sequel to the one in ConicsTangentIntersecting.html where the first part of the claim is proved. For the seek of convenience I repeat some facts from this discussion here:

Let D be the intersection point of the common tangent (m) with line BC. I use a projective base consisting of {A,B,C,K}, K being some point on the harmonic conjugate of line AD with respect to the pair of lines {AB,AC}. This simplifies some calculations done below.

In this system (see ProjectiveBase.html ) points {A,B,C,K} have correspondingly coordinates {(1,0,0),(0,1,0),(0,0,1),(1,1,1)}.

The intersection point D' of AK with BC has in this system the representation D' = B+C and D = B-C.

Then line m = AD is described by the equation : y + z = 0 and line m' = AD' by the equation y - z = 0.

Conics c

c : yz = 0 (lines AC, AB) and c' : x(y+z) = 0 (lines BC, m).

The general conic c

c

Let E be a point on line BC : E = B + kC. The second intersection point F of c

c(F,F) = 0 => c(A+pE,A+pE)=0 => 2pc(A,E) + p

This replacing E = B+kC and making the matrix multiplications implies

p = -2c(A,B+kC)/c(B+kC,B+kC) = -c(A,B+kC)/(kc(B,C)) = -t(1+k)/k.

On the other side the tangent of the conic at F is given by equation c(F,X) = 0.

Its intersection point with line BC is found by replacing H=B+rC into last equation:

c(F,B+rC) = 0 => c(A+pE,B+rC) = 0 => c(A+p(B+kC),B+rC)=0 => c(A,B)+rc(A,C)+pkc(C,B)+rpc(B,C)=0.

Introducing the values from the matrix we obtain: t + rt + pr + pk = 0 => r = -(pk+t)/(p+t) = -k

Taking point E = B+kC on BC as before, the above calculations show that

{p

Point A in terms of F's (modulo multiplicative constant) is

A = p

I = p

Using the calculation of the previous section for H = B-k

the intersection point J of line HI with line m : y+z = 0 results by applying y+z = 0 to the parametric expression for line

HI : (B -k

t -k

Thus J = Ht + I => H = (1/t)(J - I) =>

G = (1/t)(J + I) = (1/t)(Ht + 2I) = H+(2/t)I = H + ((k-1)/(p

G(k) = [k(1-k)(t

It is readily seen that for k one of {-1, 0, infinity, 1} the value of G is correspondingly {A,B,C,D}.

D is an inflexion point of the cubic.

The claim follows by computing the direction of the tangent at a point of the curve using the parameterization of the previous section. In fact, the tangent at a point (x

These coefficients can be found from the parametric form of the curve G(k)=(G

The first results from the Euler formula for homogeneous functions and the second is the chain rule. Thus, the coefficients of the tangent at G(k) are given by a vector product (G

G'(k) = [(1-2k)(t

we find that (for brevity taking T=(t

- Tangent at A (k=-1) : y+z = 0,

- Tangent at B (k=0) : 2x + Tz = 0,

- Tangent at C (k=infinity) : 2x + Ty = 0,

- Tangent at D (k=1) : 4x +Ty + Tz = 0.

First tangent is line m=AD. The other three intersect at the same point (-T, 2, 2) lying on AD' : y-z=0.

Applying the equation of the tangent at D : f(x,y,z)=4x+Ty+Tz at a point G(k) of the cubic we find f(G(k))=(k-1)

In fact, since the cubic is a rational one it is singular and has a node. To prove the claim apply the formula y-z=0 for line AD' to G(k) to obtain the equation (k+1)(k

The first solution corresponds to point A and the two other solutions define the same point (-T,3,3) on the curve. This is seen by a simple calculation.

ProjectiveBase.html

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