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

Re: [EMHL] Reflecting circles on the altitudes

Expand Messages
  • Antreas
    Dear Peter Thanks! I am leaving now, so can t check if the following are true and interesting: Let (Oa), (Ob), (Oc) be the reflections of the circles (N,NA),
    Message 1 of 5 , Feb 22, 2013
    • 0 Attachment
      Dear Peter

      Thanks!

      I am leaving now, so can't check if the
      following are true and interesting:

      Let (Oa), (Ob), (Oc) be the reflections
      of the circles (N,NA), (N,NB), (N,NC) in
      BC,CA,AB, resp.

      I guess that:

      a). (Oa),(Ob),(Oc) are concurrent.

      b). ABC, OaObOc are perspective

      c). HaHbHc, OaObOc are perspective

      d). NaNbNc, OaObOc are perspective.

      Greetings

      APH



      --- In Hyacinthos@yahoogroups.com, "Moses, Peter J. C." <mows@...> wrote:
      >
      > Hi Antreas,
      >
      > 1) P = X(265)
      > 2) Q = X(52)
      >
      > Best regards,
      > Peter.
      >
      > ----- Original Message -----
      > From: Antreas
      > To: Hyacinthos@yahoogroups.com
      > Sent: Friday, February 22, 2013 11:50 AM
      > Subject: [EMHL] Reflecting circles on the altitudes
      >
      >
      >
      > Let ABC be a triangle and HaHbHc the orthic triangle.
      >
      > Denote:
      >
      > (Na) = the reflection of the circle (N,NA) in AHa
      >
      > (Nb) = the reflection of the circle (N,NB) in BHb
      >
      > (Nc) = the reflection of the circle (N,NC) in CHc
      >
      > 1. The circles (Na),(Nb),(Nc) are concurrent at a point P
      >
      > 2. The triangles HaHbHc, NaNbNc are perspective at a point Q.
      >
      > Coordinates of P, Q ?
      >
      > APH
      >
      >
      >
      >
      >
      > [Non-text portions of this message have been removed]
      >
    • Moses, Peter J. C.
      Hi Antreas, a) X(3) b) No. c) X(52) d) X(52) also OaObOc is perspective to the circumorthic and Carnot triangles at X(110) and to the anticomplementary
      Message 2 of 5 , Feb 22, 2013
      • 0 Attachment
        Hi Antreas,

        a) X(3)
        b) No.
        c) X(52)
        d) X(52)

        also
        OaObOc is perspective to the circumorthic and Carnot triangles at X(110)
        and to the anticomplementary triangle at 2 X[5] - 3 X[195]

        Best regards,
        Peter.

        ----- Original Message -----
        From: Antreas
        To: Hyacinthos@yahoogroups.com
        Sent: Friday, February 22, 2013 1:02 PM
        Subject: Re: [EMHL] Reflecting circles on the altitudes



        Dear Peter

        Thanks!

        I am leaving now, so can't check if the
        following are true and interesting:

        Let (Oa), (Ob), (Oc) be the reflections
        of the circles (N,NA), (N,NB), (N,NC) in
        BC,CA,AB, resp.

        I guess that:

        a). (Oa),(Ob),(Oc) are concurrent.

        b). ABC, OaObOc are perspective

        c). HaHbHc, OaObOc are perspective

        d). NaNbNc, OaObOc are perspective.

        Greetings

        APH

        --- In Hyacinthos@yahoogroups.com, "Moses, Peter J. C." wrote:
        >
        > Hi Antreas,
        >
        > 1) P = X(265)
        > 2) Q = X(52)
        >
        > Best regards,
        > Peter.
        >
        > ----- Original Message -----
        > From: Antreas
        > To: Hyacinthos@yahoogroups.com
        > Sent: Friday, February 22, 2013 11:50 AM
        > Subject: [EMHL] Reflecting circles on the altitudes
        >
        >
        >
        > Let ABC be a triangle and HaHbHc the orthic triangle.
        >
        > Denote:
        >
        > (Na) = the reflection of the circle (N,NA) in AHa
        >
        > (Nb) = the reflection of the circle (N,NB) in BHb
        >
        > (Nc) = the reflection of the circle (N,NC) in CHc
        >
        > 1. The circles (Na),(Nb),(Nc) are concurrent at a point P
        >
        > 2. The triangles HaHbHc, NaNbNc are perspective at a point Q.
        >
        > Coordinates of P, Q ?
        >
        > APH
        >
        >
        >
        >
        >
        > [Non-text portions of this message have been removed]
        >





        [Non-text portions of this message have been removed]
      • Antreas
        Let ABC be a triangle and A B C the circumcevian triangle of I. The circle (I,IA ) intersects again the circumcircle at A , the (I,IB ) at B and the (I,IC )
        Message 3 of 5 , Feb 22, 2013
        • 0 Attachment
          Let ABC be a triangle and A'B'C' the circumcevian triangle of I.

          The circle (I,IA') intersects again the circumcircle at A",
          the (I,IB') at B" and the (I,IC') at C".

          1. ABC, A"B"C" are perspective on OI line

          2. Antipedal triangle of I (excentral triangle), A"B"C" are perspective.

          Orthic triangle Version:

          Let ABC be a triangle, HaHbHc the orthic triangle and
          A'B'C' the circumcevian triangle of H wrt NPC.

          The circle (H,HA') intersects again the NPC at A",
          the (H,HB') at B" and the (H, HC') at C".

          a. HaHbHc, A"B"C" are perspective (on OI line of orthic = Euler line of ABC)

          b. ABC, A"B"C" are perspective.

          Perspectors?

          APH
        • Angel
          Dear Antreas, 1. ABC, A B C are perspective on OI line. Perspector: X(36) 2. Antipedal triangle of I (excentral triangle),A B C are perspective. The first
          Message 4 of 5 , Feb 22, 2013
          • 0 Attachment
            Dear Antreas,

            1. ABC, A"B"C" are perspective on OI line. Perspector: X(36)


            2. Antipedal triangle of I (excentral triangle),A"B"C" are perspective. The first barycentric coordinate of the perspector is:

            a(3a^6 - 4(b + c)a^5 + (8b*c - 3(b^2 + c^2))a^4 +
            (10(b^3 + c^3) - 4b*c(b + c))a^3 +
            (9b^2c^2 - 10b*c(b^2 + c^2) - b^4 - c^4)a^2 +
            (8b*c(b^3 + c^3) - 6(b^5 + c^5))a +
            b^6 + 2b^5c - b^4c^2 - 4b^3c^3 - b^2c^4 + 2b*c^5 + c^6)


            a. HaHbHc, A"B"C" are perspective (on OI line of orthic=Euler line of ABC). Perspector: Infinity Point of the Euler line



            b. ABC,A"B"C" are perspective.

            The first barycentric coordinate of the perspector is (S, SA, SB, SC usual Conway notation):

            SA/( (S^2-3SA^2) (SA(c^2-b^2)^2 + a^2(SB*SC-SA^2)) )

            This point is the barycentric product X(265)X(2986) or X(69)X(1989))X(2986)

            Best regards,
            Angel Montesdeoca



            --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
            >
            > Let ABC be a triangle and A'B'C' the circumcevian triangle of I.
            >
            > The circle (I,IA') intersects again the circumcircle at A",
            > the (I,IB') at B" and the (I,IC') at C".
            >
            > 1. ABC, A"B"C" are perspective on OI line
            >
            > 2. Antipedal triangle of I (excentral triangle), A"B"C" are perspective.
            >
            > Orthic triangle Version:
            >
            > Let ABC be a triangle, HaHbHc the orthic triangle and
            > A'B'C' the circumcevian triangle of H wrt NPC.
            >
            > The circle (H,HA') intersects again the NPC at A",
            > the (H,HB') at B" and the (H, HC') at C".
            >
            > a. HaHbHc, A"B"C" are perspective (on OI line of orthic = Euler line of ABC)
            >
            > b. ABC, A"B"C" are perspective.
            >
            > Perspectors?
            >
            > APH
            >
          Your message has been successfully submitted and would be delivered to recipients shortly.