Loading ...
Sorry, an error occurred while loading the content.

Re: More 3D centers

Expand Messages
  • jpehrmfr
    Dear Jeff and Hauke [JB] ... locus ... [Hauke] ... line L ... straight ... is ... straight ... is ... is ... Except if t = 1, the locus of M (in the plane ABC)
    Message 1 of 11 , Nov 29, 2005
    • 0 Attachment
      Dear Jeff and Hauke

      [JB]
      > Anyway,...can you elaborate some more on your results below? This
      > looks interesting to me. Is there a simple way to see that the
      locus
      > exists only in a plane perpendicular to the plane of the triangle
      > (and why always through X3) using your examples?
      [Hauke]
      > > > P is a point in space.
      > > > Place P such that PA:PB:PC=a:b:c. The result is a straight
      line L
      > > which pierces ABC in #3, and the projection of L into ABC is the
      > > Euler line.
      > > >
      > > > Place P such that PA:PB:PC=1/a:1/b:1/c. The result is a
      straight
      > > line L which pierces ABC in #3, and the projection of L into ABC
      is
      > > the Brocard axis.
      > > >
      > > > Place P such that PA:PB:PC=s-a:s-b:s-c. The result is a
      straight
      > > line L which pierces ABC in #3, and the projection of L into ABC
      is
      > > the OI line (#1-#3).
      > > >
      > > > Place P such that PA:PB:PC=1/(s-a):1/(s-b):1/(s-c). The result
      is
      > a
      > > straight line L which pierces ABC in #3, and the projection of L
      > into
      > > ABC is the line through #3 and Mittenpunkt #9.

      Except if t = 1, the locus of M (in the plane ABC) such as MA = t.MB
      is a circle and the power of O wrt this circle is allways R^2.
      If u,v,w are positive and distinct, the points M (in the plane ABC)
      such as MA:MB:MC = u:v:w are the common points (if they are real) Q,
      Q' of three coaxal circles and the radical axis QQ' goes through O.
      Now the points P in the space such as PA:PB:PC = u:v:w will be the
      common points of three coaxal spheres, ie the circle with diameter
      QQ' in a plane perpendicular to the plane ABC.
      Obviously, the projection of this circle upon the plane ABC is the
      segment [QQ']; the line QQ' goes through O but O never lies in the
      segment [QQ'].

      To be more precise, if l = (ua+vb-wc)(-ua+vb+wc)(ua-vb+wc)
      If l>0 then Q,Q' are real and distinct; thus the locus of P is the
      circle above
      If l=0 then Q=Q' and the locus of P is the line perpendicular at Q
      to the plane ABC
      If l<0, a point P such as PA:PB:PC = u:v:w cannot exist

      Friendly. Jean-Pierre
    • Bernard Gibert
      Dear friends, let P be a point on the circumcircle of triangle ABC with incenter I. denote by Pa, Pb, Pc the incenters of triangles PBC, PCA, PAB. I, Pa, Pb,
      Message 2 of 11 , Dec 1, 2005
      • 0 Attachment
        Dear friends,

        let P be a point on the circumcircle of triangle ABC with incenter I.

        denote by Pa, Pb, Pc the incenters of triangles PBC, PCA, PAB.

        I, Pa, Pb, Pc are the vertices of a rectangle.

        Find P such that the rectangle becomes a square.

        Perspectivities ?...

        Best regards

        Bernard



        [Non-text portions of this message have been removed]
      • Paris Pamfilos
        Obviously the incenters Pa, Pb, etc. are on the bisectors of angles of the quadrangle ABCP and it seems that the condition is that the four bisectors pass
        Message 3 of 11 , Dec 2, 2005
        • 0 Attachment
          Obviously the incenters Pa, Pb, etc. are on the bisectors of angles
          of the quadrangle ABCP and it seems that the condition is that the
          four bisectors pass through a common point.
          Best regards
          Paris Pamfilos

          -----Original Message-----
          From: Hyacinthos@yahoogroups.com [mailto:Hyacinthos@yahoogroups.com]On
          Behalf Of Bernard Gibert
          Sent: Thursday, December 01, 2005 6:02 PM
          To: Hyacinthos@yahoogroups.com
          Subject: [EMHL] incenters & squares


          Dear friends,

          let P be a point on the circumcircle of triangle ABC with incenter I.

          denote by Pa, Pb, Pc the incenters of triangles PBC, PCA, PAB.

          I, Pa, Pb, Pc are the vertices of a rectangle.

          Find P such that the rectangle becomes a square.

          Perspectivities ?...

          Best regards

          Bernard



          [Non-text portions of this message have been removed]





          Yahoo! Groups Links








          --
          No virus found in this incoming message.
          Checked by AVG Free Edition.
          Version: 7.1.362 / Virus Database: 267.13.10/190 - Release Date: 1/12/2005
        • jpehrmfr
          Dear Bernard ... Suppose that P lies on the arc BC of the circumcircle opposite to A. The lines PaPb and PaPc go respectively through the excenters Ib and Ic.
          Message 4 of 11 , Dec 3, 2005
          • 0 Attachment
            Dear Bernard
            > let P be a point on the circumcircle of triangle ABC with incenter I.
            >
            > denote by Pa, Pb, Pc the incenters of triangles PBC, PCA, PAB.
            >
            > I, Pa, Pb, Pc are the vertices of a rectangle.
            >
            > Find P such that the rectangle becomes a square.

            Suppose that P lies on the arc BC of the circumcircle opposite to A.
            The lines PaPb and PaPc go respectively through the excenters Ib and
            Ic.
            The rectangle will be a square if PaI is bisector of IbPaIc.
            Hence the construction :
            Pa lies on the circle (Ca) of diameter IbIc (center Wa) such as <IbPaI
            = <IPaIc = 45°. P lies on the line WaPa.
            Do the same for P lying on the arcs CA or AB of the circumcircle.
            Friendly. Jean-Pierre
          • Hauke Reddmann
            ... I prefer to replace cannot exist by does not lie in real space ; I encountered many geometrical problems where e.g. a locus is imaginary except one
            Message 5 of 11 , Dec 3, 2005
            • 0 Attachment
              --- In Hyacinthos@yahoogroups.com, "jpehrmfr" <Jean-
              Pierre.Ehrmann.70@n...> wrote:
              >
              > To be more precise, if l = (ua+vb-wc)(-ua+vb+wc)(ua-vb+wc)
              > If l>0 then Q,Q' are real and distinct; thus the locus of P is the
              > circle above
              > If l=0 then Q=Q' and the locus of P is the line perpendicular at Q
              > to the plane ABC
              > If l<0, a point P such as PA:PB:PC = u:v:w cannot exist
              >
              I prefer to replace "cannot exist" by "does not lie in
              real space"; I encountered many geometrical problems
              where e.g. a locus is imaginary except one special real point.
              Also, I don't shriek back from using negative or imaginary
              distances inbetween if it gives a cute result :-)

              THX for verifying it's a circle - where (projected back on ABC)
              does its highest/lowest point lie?

              Hauke
            Your message has been successfully submitted and would be delivered to recipients shortly.