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

This shows that the intersection point H = B+rC on BC is independent of the particular member c

CubicFromTwoConics.html

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