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Re: Homothetic triangles

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  • Antreas
    ... As Francisco pointed out: this follows inmediately from Desargues theorem for any ABC and A B C We can use it in the reference triangle ABC to get
    Message 1 of 7 , May 1, 2010
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      --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
      >
      > Let ABC, A'B'C' be two homothetic triangles
      > and X,Y,Z three collinear points.
      >
      > Are the triangles bounded by the lines
      > (AX, BY, CZ) and (A'X, B'Y, C'Z) perspective?

      As Francisco pointed out: "this follows inmediately from Desargues
      theorem for any ABC and A'B'C'"

      We can use it in the reference triangle ABC to get points,
      from two given triangles and three collinear points.

      Here is an example:

      Inside ABC exists a unique point P such that:
      Three Congruent (equal) circles (A'),(B'),(C') concur at P,
      and circle (A') touches the sides of the angle A,
      (B') of angle B and (C') of angle C.
      [A', B', C' are the centers of the circles]

      Now, let Lp a line passing through P and
      intersecting the circles (A'),(B'),(C') at
      X,Y,Z resp. (other than P).

      The triangles bounded by the lines
      (AX, BY, CZ) and (A'X, B'Y, C'Z) are perspective.

      Which is the locus of the perspectors
      as L moves around P?

      APH
    • Barry Wolk
      ... I get two such points P. If your 3 circles have radius r*R/(R+r) then P=X(55). And if their radius is r*R/(R-r) then P=X(56). -- Barry Wolk
      Message 2 of 7 , May 6, 2010
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        Antreas wrote:
        > Here is an example:
        >
        > Inside ABC exists a unique point P such that:
        > Three Congruent (equal) circles (A'),(B'),(C') concur at P,
        > and circle (A') touches the sides of the angle A,
        > (B') of angle B and (C') of angle C.
        > [A', B', C' are the centers of the circles]

        I get two such points P. If your 3 circles have radius r*R/(R+r)
        then P=X(55). And if their radius is r*R/(R-r) then P=X(56).
        --
        Barry Wolk
      • Antreas Hatzipolakis
        ... inside triangle (that s what I had in mind) APH [Non-text portions of this message have been removed]
        Message 3 of 7 , May 6, 2010
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          On Thu, May 6, 2010 at 11:18 PM, Barry Wolk <wolkbarry@...> wrote:

          >
          >
          > Antreas wrote:
          > > Here is an example:
          > >
          > > Inside ABC exists a unique point P such that:
          > > Three Congruent (equal) circles (A'),(B'),(C') concur at P,
          > > and circle (A') touches the sides of the angle A,
          > > (B') of angle B and (C') of angle C.
          > > [A', B', C' are the centers of the circles]
          >
          > I get two such points P. If your 3 circles have radius r*R/(R+r)
          > then P=X(55). And if their radius is r*R/(R-r) then P=X(56).
          > --
          > Barry Wolk
          >
          >
          > Indeed. Two points P. But I think only one point with circles centers
          inside triangle (that's what I had in mind)

          APH


          [Non-text portions of this message have been removed]
        • Antreas Hatzipolakis
          Let ABC be a triangle and A B C the cevian triangle of I. Denote: Bc, Cb = the reflections of B, C in CC , BB , resp. Oa = the circumcenter of ABcCb.
          Message 4 of 7 , Jul 25 5:04 PM
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            Let ABC be a triangle and A'B'C' the cevian triangle of I.

            Denote:

            Bc, Cb = the reflections of B, C in CC', BB', resp.

            Oa = the circumcenter of ABcCb. Similarly Ob, Oc.

            OaObOc and the excentral triangle IaIbIc are homothetic.

            Which point is the homothetic center wrt triangles:

            1. ABC (on its OI line)

            2. OaObOc (on its Euler line)

            3. IaIbIc (on its Euler line)  ??

            APH
          • Antreas Hatzipolakis
            [APH]: Let ABC be a triangle and A B C the cevian triangle of I. ... Locus: Let ABC be a triangle P a point and PaPbPc the antipedal triangle of P. Denote:
            Message 5 of 7 , Jul 25 6:05 PM
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              [APH]:

              Let ABC be a triangle and A'B'C' the cevian triangle of I.

              Denote:

              Bc, Cb = the reflections of B, C in CC', BB', resp.

              Oa = the circumcenter of ABcCb. Similarly Ob, Oc.

              OaObOc and the excentral triangle IaIbIc are homothetic.

              Which point is the homothetic center wrt triangles:

              1. ABC (on its OI line)

              2. OaObOc (on its Euler line)

              3. IaIbIc (on its Euler line)  ??

              APH


              Locus:

              Let ABC be a triangle P a point and PaPbPc the antipedal triangle of P.

              Denote:

              Bc, Cb = the reflections of B, C in CP, BP, resp.

              Oa = the circumcenter of ABcCb. Similarly Ob,Oc.

              Which is the locus of P such that OaObOc, PaPbPc are perspective?

              APH




            • Antreas Hatzipolakis
              [APH]: Let ABC be a triangle and A B C the cevian triangle of I. Denote: Bc, Cb = the reflections of B, C in CC , BB , resp. Oa = the circumcenter of ABcCb.
              Message 6 of 7 , Jul 26 4:39 AM
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                [APH]:

                 

                Let ABC be a triangle and A'B'C' the cevian triangle of I.

                Denote:

                Bc, Cb = the reflections of B, C in CC', BB', resp.

                Oa = the circumcenter of ABcCb. Similarly Ob, Oc.

                OaObOc and the excentral triangle IaIbIc are homothetic.

                Which point is the homothetic center wrt triangles:

                1. ABC (on its OI line)

                2. OaObOc (on its Euler line)

                3. IaIbIc (on its Euler line)  ??

                APH



                [César Lozada]


                1. ABC (on its OI line)
                X(3576)

                2. OaObOc (on its Euler line)
                X(381)

                3. IaIbIc (on its Euler line)   
                X(381)

                CL





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