Thursday, May 6, 2021

1694.

1 comment:

  1. Let P=(u:v:w) be a point in the plane of triangle ABC. Let A1, A2 be the circumcircle intercepts of the line through P parallel to sideline BC. Let A3 be the point of intersection of the Simson lines of A1 and A2. Define B3 and C3 cyclically. Let H be the orthocenter of ABC and A'B'C' the orthic triangle.

    1) A3 lies on the A-altitude (line AA'), B3 on the B-altitude and C3 on the C-altitude. Specifically,

    A3 = (u+v+w) H + u A - u A',
    B3 = (u+v+w) H + v B - v B',
    C3 = (u+v+w) H + w C - w C'.

    Thus, A3B3C3 is perspective to ABC at the orthocenter.

    2) The circumcircle intercepts may not be real but A3, B3, C3 are always real;

    3) For P in { X(15), X(16) }, A3B3C3 is equilateral; In both cases the center lies on the Euler line;

    4) Inspired by your result, I found the following:

    a) The locus of P such that the symmedian point of A3B3C3 is on the Euler line is the K018 cubic with barycentric equation

    Cyclic Sum ( a^2 v w ( (a^2-b^2) v - (a^2-c^2) w ) = 0;

    b) The locus of P such that the centroid of A3B3C3 is on the Euler line is the Brocard axis (line X(3)X(6) ).


    Elias M. Hagos

    ReplyDelete

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