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Re: Altitudes bisected etc

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  • Darij Grinberg
    Dear Antreas, ... ^ This should be Pa. Your condition is easily seen to be equivalent to the condition that the B-parallelian and the C-parallelian through Pa
    Message 1 of 3 , Dec 7, 2003
      Dear Antreas,

      In Hyacinthos message #8782, you wrote:

      >> Let ABC be a triangle and Pa a point.
      >>
      >> The parallel to AB through Pa intersects BC at Aab
      >> The parallel to AC through Pa intersects BC at Aac
      >>
      >> Let A'ab, A'ac be the orthogonal projections of
      >> Aab, Aac on AC, AB, resp.
      >>
      >> Which is the locus of P such that AabA'ab = AacA'ac ?
      ^
      This should be Pa.

      Your condition is easily seen to be equivalent to the
      condition that the B-parallelian and the C-parallelian
      through Pa have equal length. (The B-parallelian of a
      point P is the segment joining the points where the
      parallel to CA through P meets AB and BC.)

      Your locus is the line with trilinear equation

      b(cz+ax) = c(ax+by).

      >> The locus is a line La.
      >>
      >> Similarly define the lines Lb, Lc.
      >>
      >> The three lines La, Lb, Lc are concurrent.

      They concur at the congruent parallelians point
      X(192).

      Sincerely,
      Darij Grinberg
    • Darij Grinberg
      Dear Antreas, ... No, since your equation entails B C || BC, C A || CA and A B || AB, so that ABC is the antimedial triangle of A B C , and P is the
      Message 2 of 3 , Dec 7, 2003
        Dear Antreas,

        In Hyacinthos message #8781, you wrote:

        >> Let ABC be a triangle and AA', BB', CC' the cevian
        ^^^^^^^^^^^^^
        >> triangle of P.
        >>
        >> For which points P we have that:
        >>
        >> AP / PA' = BP / PB' = CP / PC' ?
        >>
        >> Are there other than the centroid points with that
        >> property?

        No, since your equation entails B'C' || BC, C'A' || CA
        and A'B' || AB, so that ABC is the antimedial triangle
        of A'B'C', and P is the centroid of triangle ABC.

        Sincerely,
        Darij Grinberg
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