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Incenter sweep

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  • Bernard Gibert
    Dear friends let P be a point in the plane of triangle ABC let I be the incenter of the cevian triangle of P what are the regions of the plane which cannot
    Message 1 of 4 , Jan 1, 2006
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      Dear friends

      let P be a point in the plane of triangle ABC

      let I be the incenter of the cevian triangle of P

      what are the regions of the plane which cannot contain I ?

      all the best for 2006 to all of you

      Best regards

      Bernard



      [Non-text portions of this message have been removed]
    • 黃嘉麟
      Dear all, Sorry, my mistake, there is no typo in the description of X(1276) and X(1277) in ETC. For the concyclic properties of X(1), X(484), X(1276), X(1277),
      Message 2 of 4 , Jan 1, 2006
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        Dear all,

        Sorry, my mistake, there is no typo in the description of X(1276) and X(1277) in ETC.

        For the concyclic properties of X(1), X(484), X(1276), X(1277), here is another proof.
        X(1277) = Xd( 15| 1), X(1276) = Xd( 16| 1),
        X( 1) = Xd( 4| 1), X( 484) = Xd(186| 1). d = "anticevian triangle"
        Since X(15)=inverse-in-circumcircle of X(16) and X(4)=inverse-in-circumcircle of X(186), it is easy to show X(4),X(186),X(15),X(16) are concyclic. Then it follows that X(1), X(484), X(1276), X(1277) are also concyclic.

        Note: For a circle with center O, radius R, if (P1,P2) and (Q1,Q2) are inverse pairs in circle O, then OP1*OP2=OQ1*OQ2=R^2. So triangle OP1Q2 is similar to triangle OP2Q1 and P1,P2,Q1,Q2 are concyclic.

        Happy new year
        Best regards
        Chia-Lin Hwang

        space <hwangd2000@...> 說:
        Dear all,

        In ETC, X(1276) and X(1277)(2nd and 3rd EVANS PERSPECTOR)
        are described as perspectors of excentral triangle and special
        triangles T, T'(some typoes in the description I believe).
        Moses(8/9/2004),further gave a relationship of these two points,
        X(1276) = inverse-in-Bevan-circle of X(1277).

        Here give another way to look at X(1276) and X(1277).
        Actually they are ISODYNAMIC POINTs X(15), X(16) wrt excentral
        triangle.

        Since X(15)=inverse-in-circumcircle of X(16) and the Bevan circle is
        the circumcircle of the excentral triangle,then the result of Moses
        follows obviously.

        I would like to suggest "double-index notation," Xd( p| q),
        to describe those notable points wrt special triangle.
        =====================================================
        Xd( p| q) = X(p) wrt X(q)-special triangle,
        d denotes some kind of special triangle.
        =====================================================
        I believe it is helpful to build new concept and extend Kimberling's
        points. For example:
        =====================================================
        X(1277) = Xd(15| 1) = X(15)-of-excentral triangle
        X(1276) = Xd(16| 1) = X(16)-of-excentral triangle
        =====================================================
        where d = "anticevian triangle" and
        excentral triangle= anticevian triangle of X(1).

        The properties of X(1276) and X(1277) in ETC is not many,
        now we know how to add more(with the help of X(15),X(16)). For
        example the pedal triangle of X(1276),X(1277) wrt excentral triangle
        is equilateral.

        Best regards
        Chia-Lin Hwang
        Taipei, Taiwan







        Yahoo! Groups Links









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        [Non-text portions of this message have been removed]
      • Steve Sigur
        Bernard, What a wonderful problem. Let the seven regions of the Plane of ABC be numbered as 0 for the central region 1,2,3 for the regions found by crossing
        Message 3 of 4 , Jan 1, 2006
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          Bernard,

          What a wonderful problem.

          Let the seven regions of the Plane of ABC be numbered as

          0 for the central region
          1,2,3 for the regions found by crossing (from the central region) the
          a,b, c segments of ABC.
          4,5,6 for the regions found by crossing the vertices.

          The incenter can be anywhere in the central region. There is always
          an excenter in the central region as well (from observation).


          For 1,2,3
          Consider the triangle plane with the ex-median lines added.

          If the point you mean is the incenter of the triangle formed by the
          three lines that form the Cevian triangle of P, then this center
          changes discontinuously as P crosses an exmedian line and a Cevian
          vertex travels through the line at infinity. So the incenter (defined
          thusly) will never be in a region between an exmedian and an edge.

          For 4, 5, 6

          There seems to be a curve they cannot cross so the incenter cannot be
          everywhere in this region, but I cannot figure out what the crurve is.

          What a wonderful problem,

          Back from Princeton,

          Steve





          On Jan 1, 2006, at 3:34 AM, Bernard Gibert wrote:

          > Dear friends
          >
          > let P be a point in the plane of triangle ABC
          >
          > let I be the incenter of the cevian triangle of P
          >
          > what are the regions of the plane which cannot contain I ?
          >
          > all the best for 2006 to all of you
          >
          > Best regards
          >
          > Bernard
          >
          >
          >
          > [Non-text portions of this message have been removed]
          >
          >
          >
          >
          > Yahoo! Groups Links
          >
          >
          >
          >
          >
          >
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