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Re: [EMHL] The pedal triangles of Ia, Ib, Ic

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  • ndergiades
    Thank you Bernard You are right. The three intersections are very close. For the triangle ABC with sin(A/2) = 3/5 sin(B/2) = 5/13 I found with MathCad for the
    Message 1 of 35 , Feb 28, 2002
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      Thank you Bernard

      You are right.
      The three intersections are very close.
      For the triangle ABC
      with sin(A/2) = 3/5 sin(B/2) = 5/13
      I found with MathCad for the three lines OI
      a rational determinant with
      nominator and denominator integer numbers
      with about 20 digits and the value of the
      fraction about 0.00005.

      Best regards
      Nik


      --- In Hyacinthos@y..., Bernard Gibert <b.gibert@f...> wrote:
      > Dear Antreas, Paul and Nik,
      >
      > >>> No surprise if the OI lines [after the OG and OK lines]
      > >>> of the pedal triangles of Ia,Ib,Ic
      > >>> are concurrent on some point on the Euler line of ABC as well!
      > >>>
      > >>> ***
      > >>> A quick sketch suggest that they are.
      >
      > Cabri says they are NOT !
      > although the three intersections are very close.
      >
      > >>>It would be interesting to know
      > >>> the
      > >>> point of concurrency.
      > >
      > [PY]
      > > Now I have no confidence that the three lines are concurrent.
      > > May be Edward can help check this.
      > >
      >
      > Best regards
      >
      > Bernard
    • Boutte Gilles
      Dear Antreas ... I apologize, you are right.. I am mistaked in my notations. Gilles
      Message 35 of 35 , Mar 3, 2002
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        Dear Antreas

        > Dear Gilles,
        >
        > [APH]
        > Let IaaIabIac be the pedal triangle of Ia.
        >
        > A' := IaIaa /\ IabIac
        >
        > Similarly B', C'
        >
        > The triangles ABC, A'B'C' are perspective.
        >
        > [...]
        >
        >
        > ==> the point of concurrence is the centroid G.
        >
        > [...]
        >
        > 1. A" := the orth. projection of Iaa on IabIac.
        >
        > Similarly B", C"
        >
        > Then the triangles ABC, A"B"C" are perspective.
        >
        > [GB]:
        > The triangle MaMbMc with sides IabIac, IbaIbc, IcaIcb is homothetic to
        > IaIbIc. A' is the feet on IabIac of the altitude, as we prove with
        > Nikolaos Dergiades.
        > A'B'C' is the orthic triangle of MaMbMc, and ABC is the orthic triangle
        > of IaIbIc. So ABC and A'B'C' are homothetic, therefore they are
        > perspective. Perspector is the Mittelpunkt X_9 of ABC.
        >
        > *********
        >
        > I guess your A'B'C' is mine A"B"C".


        I apologize, you are right.. I am mistaked in my notations.

        Gilles

        >
        >
        > Let's see what the perspector of ABC, A"B"C" is:
        >
        >
        > A
        > ..................
        > / \
        > / \
        > B----A1-Iaa--------C
        > / \
        > A3 A2
        > Iac \
        > / A" Iab
        > / \
        > Ia
        >
        > [...]
        >
        > ==> AA" is the a-cevian of Mittenpunkt (cot(A/2) ::)
        >
        >
        > Antreas
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