Loading ...
Sorry, an error occurred while loading the content.

RE: [EMHL] Cevian triangles and the circumcircle

Expand Messages
  • Floor en Lyanne van Lamoen
    Dear Eric, ... This is true with any circle, or even conic. The theorem follows from Pascal s hexagon theorem, combined with Desargues two triangle theorem.
    Message 1 of 2 , Oct 31, 2004
      Dear Eric,

      > I don't know if the following theorem is known
      > 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

      This is true with any circle, or even conic. The theorem follows from
      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.

      > Question
      > 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 ?

      With P(x:y:z) we find the quartic

      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.

      Kind regards,
    Your message has been successfully submitted and would be delivered to recipients shortly.