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Two Emile Hyacinthe Lemoine centers

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  • jpehrmfr
    Dear Hyacinthists looking at the beginning of the very interesting from EH Lemoyne : Suite de theoremes et de resultats concernant la geometrie du triangle
    Message 1 of 3 , Dec 1, 2003
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      Dear Hyacinthists
      looking at the beginning of the very interesting from EH Lemoyne :
      Suite de theoremes et de resultats concernant la geometrie du
      triangle
      avalaible at
      http://www.hti.umich.edu/u/umhistmath/
      I see that he discovered the following centers :
      1) Ma is the unique point on the segment BC such as
      MaB + d(Ma,AB) = MaC + d(Ma, AC) with d for distance; define Mb, Mc
      cyclically. Then MaMbMc and ABC are perspective at the point of the
      line IG with trilinear 1+2R/a:...
      2) Na is the unique point on the segment BC such as
      NaB + d(Na,AC) = NaC + d(Na, AB) with d for distance; define Nb, Nc
      cyclically. Then NaNbNc and ABC are perspective at the point of the
      line IG with trilinear (2R/a)-1:...
      I don't think - but, may be, I'm wrong - that these points are in
      the current ETC.
      Friendly. Jean-Pierre
    • Antreas P. Hatzipolakis
      Dear Jean-Pierre ... You mean Lemoine (as in the subject line) ... The perpendicular to BC at Ma intersects AB at Ab, and AC at Ac. Define A = BAc / CAb.
      Message 2 of 3 , Dec 2, 2003
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        Dear Jean-Pierre

        >looking at the beginning of the very interesting from EH Lemoyne:

        You mean Lemoine (as in the subject line)

        >Suite de theoremes et de resultats concernant la geometrie du
        >triangle avalaible at http://www.hti.umich.edu/u/umhistmath/
        >I see that he discovered the following centers :
        >1) Ma is the unique point on the segment BC such as
        >MaB + d(Ma,AB) = MaC + d(Ma, AC) with d for distance; define Mb, Mc
        >cyclically. Then MaMbMc and ABC are perspective at the point of the
        >line IG with trilinear 1+2R/a:...


        The perpendicular to BC at Ma intersects AB at Ab, and AC at Ac.

        Define

        A' = BAc /\ CAb. Similarly B', C'.

        If my midnight calculations were correct, then these three
        points are collinear.

        And probably the same is true for the other case below
        but didn't try it.


        Good night from Athens

        Antreas


        >2) Na is the unique point on the segment BC such as
        >NaB + d(Na,AC) = NaC + d(Na, AB) with d for distance; define Nb, Nc
        >cyclically. Then NaNbNc and ABC are perspective at the point of the
        >line IG with trilinear (2R/a)-1:...
        >I don't think - but, may be, I'm wrong - that these points are in
        >the current ETC.
        >Friendly. Jean-Pierre


        --
      • Antreas P. Hatzipolakis
        Dear Jean - Pierre ... In other words: Let Ma be *the* point on the line segment BC such that: MaB + MaMab = MaC + MaMac where Mab, Mac are the orthogonal
        Message 3 of 3 , Dec 3, 2003
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          Dear Jean - Pierre

          You wrote:
          >I see that he [=Lemoine]discovered the following centers :
          >1) Ma is the unique point on the segment BC such as
          >MaB + d(Ma,AB) = MaC + d(Ma, AC) with d for distance; define Mb, Mc
          >cyclically. Then MaMbMc and ABC are perspective at the point of the
          >line IG with trilinear 1+2R/a:...

          In other words:

          Let Ma be *the* point on the line segment BC such that:

          MaB + MaMab = MaC + MaMac

          where Mab, Mac are the orthogonal projections of Ma on
          AB, AC, resp. Similarly Mb, Mc.

          The triangles ABC, MaMbMc are perspective with trilinear perspector


          ((1 + sinA) / sinA ::)

          Now,

          1. Let Ka be *the* point on the line segment BC such that:

          KaB + BKab = KaC + CKac

          where Kab, Kac are the orthogonal projections of Ka on
          AB, AC, resp. Similarly Kb, Kc.

          The triangles ABC, KaKbKc are perspective with trilinear perspector:

          ((1 + cosA) / sinA ::) = (cot(A/2) ::) [mittenpunkt]

          2. Let La be *the* point on the line segment BC such that:

          LaLab + BLab = LaLaC + CLac

          where Lab, Lac are the orthogonal projections of La on
          AB, AC, resp. Similarly Lb, Lc.

          The triangles ABC, LaLbLc are perspective with trilinear perspector:

          ((sinA + cosA) / sinA ::) = (1 + cotA ::)

          3. Let Pa be *the* point on the line segment BC such that:

          perimeter(BPaPab) = perimeter(CPaPac)

          where Pab, Pac are the orthogonal projections of Pa on
          AB, AC resp. Similarly Pb, Pc.

          The triangles ABC, PaPbPc are perspective with trilinear perspector:

          ((1 + sinA + cosA) / sinA ::)


          4. Let Qa be *the* point on the line segment BC such that:

          area(BQaQab) = area(CQaQac)

          where Qab, Qac are the orthogonal projections of Qa on
          AB, AC, res. Similarly Qb, Qc.

          The triangles ABC, QaQbQc are perspective with trilinear perspector:

          (sqrt(cotA) ::)


          >2) Na is the unique point on the segment BC such as
          >NaB + d(Na,AC) = NaC + d(Na, AB) with d for distance; define Nb, Nc

          That is, NaB - d(Na, AB) = NaC - d(Na,AC)

          >cyclically. Then NaNbNc and ABC are perspective at the point of the
          >line IG with trilinear (2R/a)-1:...

          I think that we get similar results (for 1 and 2) just by changing
          the "+" to "-" in the above trilinears.


          Antreas
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