Tuesday, May 4, 2021

1693. A Circle Centered at Incenter

1 comment:

  1. The construction can be generalized. Let P=(u:v:w) be a point in the plane of triangle ABC, and let DEF be its cevian triangle. Circle (PEF) intersects line AB at Ac and AC at Ab. Define Ba, Bc, Cb, Ca cyclically. The six points lie on a conic. The equation of the conic is not nice, but the center of the conic has a simple form

    ( a^2 (u+v)(u+w) v w : b^2 (v+w)(v+u) u w : : c^2 (u+w)(v+w) u v ).

    It is the cevapoint of X(6) and the isogonal conjugate of P.

    Elias M. Hagos

    ReplyDelete

1988. Circumcenters On Euler Line

  Let ABC be a triangle with P, Q are two points on Euler line of ABC. A',B',C' are inverse images of A,B,C through circle diame...