RE: [EMHL] Cevian triangles and the circumcircle
- Dear Eric,
> I don't know if the following theorem is knownThis is true with any circle, or even conic. The theorem follows from
> Let A'B'C' be the cevian triangle of a point P wrt a triangle ABC.
> B'C' intersects the circumcircle of ABC at A1 and A2.
> It is not important on which side of BC you choose A1 or A2.
> Define B1, B2, C1 and C2 similarly.
> The lines A1B1, A2C1 and B2C2 form a triangle A*B*C* perspective
> with A'B'C'
> In fact there are 8 possible triangles A*B*C* by changing A1 with
> A2, B1 with B2 and C1 with C2
> However they only lead to 4 different perspectors
Pascal's hexagon theorem, combined with Desargues' two triangle theorem.
See for an explanation (also why there are only 4 perspectors) my paper
Equilateral Chordal Triangles in FG 2, 33-37.
> QuestionWith P(x:y:z) we find the quartic
> What is the area of the plane so that for a point inside this area
> all three sides of its cevian triangle intersect (or at least touch)
> the circumcircle ?
a^4y^2z^2 + b^4x^2z^2 + c^4x^2y^2
- 2a^2b^2xyz^2 + 2a^2c^2xy^2z + 2b^2c^2x^2yz = 0
to let the line A'B' be tangent to the circumcircle.
Similar quartics for the other Cevians. The point P has to be inside all
three of them.