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

EL GRECO CIRCLE ?? [EMHL] Re: X3542 : A construction

Expand Messages
  • Antreas
    .... and the most natural question in the world is what properties the A2,B2,C2 have?? Do we have a Picasso-like circle, which, If true (as I think), I will
    Message 1 of 13 , Feb 10, 2012
    • 0 Attachment
      .... and the most natural question in the world is what properties the
      A2,B2,C2 have??

      Do we have a Picasso-like circle, which, If true (as I think), I will
      name it EL GRECO Circle ?

      See

      http://anthrakitis.blogspot.com/2012/02/picasso-like-circles.html
      <http://anthrakitis.blogspot.com/2012/02/picasso-like-circles.html>


      APH

      --- In Hyacinthos@yahoogroups.com, Chandan Banerjee <chandana.snbv@...>
      wrote:
      >
      > Yes, it is true even if you replace O by P,
      > It follows from the following problem:-
      > It follows from the following problem which is a generalisation of a
      > problem you posted earlier:- Given a triangle ABC, A1B1C1 be the
      > circumcevian triangle of a point R and A2B2C2 be the circumcevian
      triangle
      > of a point Q. If A3B3C3 is the circumcevian triangle of R wrt A2B2C2,
      then
      > A1A3,B1B3,C1C3 concurr on RQ.
      >
      > Now take R= reflection of H on P and Q=H and the result follows.
      >
      > On Fri, Feb 10, 2012 at 3:19 AM, Antreas Hatzipolakis
      > anopolis72@...wrote:
      >
      > > **
      > >
      > >
      > > Generalization: Is it true if we replace O with a point P?
      > > (The perspector on the HP line)
      > >
      > > APH
      > >
      > > On Thu, Feb 9, 2012 at 9:08 PM, Francisco Javier garciacapitan@...
      > > >wrote:
      > >
      > > > **
      > >
      > > >
      > > >
      > > > Yes, so, here is more precise information: The line A"H intersect
      the
      > > > circumcircle at point A1, A2 with coordinates
      > > >
      > > > A1:
      > > >
      > > > {a^2 (a^2+b^2-c^2) (a^2-b^2+c^2) (a^6-a^4 b^2-a^2 b^4+b^6-3 a^4
      c^2+2 a^2
      > > > b^2 c^2-3 b^4 c^2+3 a^2 c^4+3 b^2 c^4-c^6) (a^6-3 a^4 b^2+3 a^2
      > > b^4-b^6-a^4
      > > > c^2+2 a^2 b^2 c^2+3 b^4 c^2-a^2 c^4-3 b^2 c^4+c^6),-(a^2-b^2-c^2)
      > > > (a^2+b^2-c^2) (a^6-3 a^4 b^2+3 a^2 b^4-b^6-a^4 c^2+2 a^2 b^2 c^2+3
      b^4
      > > > c^2-a^2 c^4-3 b^2 c^4+c^6) (a^6 b^2-a^4 b^4-a^2 b^6+b^8+a^6 c^2-2
      a^4 b^2
      > > > c^2+a^2 b^4 c^2-4 b^6 c^2-a^4 c^4+a^2 b^2 c^4+6 b^4 c^4-a^2 c^6-4
      b^2
      > > > c^6+c^8),-(a^2-b^2-c^2) (a^2-b^2+c^2) (a^6-a^4 b^2-a^2 b^4+b^6-3
      a^4
      > > c^2+2
      > > > a^2 b^2 c^2-3 b^4 c^2+3 a^2 c^4+3 b^2 c^4-c^6) (a^6 b^2-a^4
      b^4-a^2
      > > > b^6+b^8+a^6 c^2-2 a^4 b^2 c^2+a^2 b^4 c^2-4 b^6 c^2-a^4 c^4+a^2
      b^2 c^4+6
      > > > b^4 c^4-a^2 c^6-4 b^2 c^6+c^8)}
      > > >
      > > > A2:
      > > >
      > > > {(a^4-2 a^2 b^2-b^4+2 b^2 c^2-c^4) (a^4-b^4-2 a^2 c^2+2 b^2
      c^2-c^4),-2
      > > > b^2 (b-c) (b+c) (-a^4+b^4+2 a^2 c^2-2 b^2 c^2+c^4),-2 c^2 (-b+c)
      (b+c)
      > > > (-a^4+2 a^2 b^2+b^4-2 b^2 c^2+c^4)}
      > > >
      > > > and cyclicaly for B1, B2, and C1, C2.
      > > >
      > > > With these coordinates, A1B1C1 and ABC are always perspective at
      X3542.
      > > >
      > > >
      > > > --- In Hyacinthos@yahoogroups.com, "rhutson2" rhutson2@ wrote:
      > > > >
      > > > > The choice of A1 and A2, etc. appears to depend on the shape of
      ABC.
      > > For
      > > > example, for the ETC ref. triangle, it is A1B1C2 that is
      perspective to
      > > ABC
      > > > at X(3542). And it does not seem to depend on acute/obtuse, as
      some acute
      > > > triangles I have tried need the roles of A1/A2, B1/B2, and/or
      C1/C2
      > > > reversed.
      > > > >
      > > > > Randy
      > > > >
      > > > > --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@>
      wrote:
      > > > > >
      > > > > >
      > > > > > http://anthrakitis.blogspot.com/2012/02/perspective.html
      > > > > > <http://anthrakitis.blogspot.com/2012/02/perspective.html>
      > > > > >
      > > > > > APH
      > > >
      > >
      > > [Non-text portions of this message have been removed]
      > >
      > >
      > >
      >
      >
      >
      > --
      > CHANDAN
      >
      >
      > [Non-text portions of this message have been removed]
      >



      [Non-text portions of this message have been removed]
    • Chandan Banerjee
      Suppose, A3B3C3 is the circumcevian triangle of H wrt ABC and A4B4C4 be the antipodal triangle of ABC. A2B2C2 is the cevian triangle of the De-Longchamps point
      Message 2 of 13 , Feb 10, 2012
      • 0 Attachment
        Suppose, A3B3C3 is the circumcevian triangle of H wrt ABC and A4B4C4 be the
        antipodal triangle of ABC. A2B2C2 is the cevian triangle of the
        De-Longchamps point of ABC wrt A3B3C3, then if we can prove that A4B4C4 and
        A5B5C5 are perspective with the perspector lying on OH, then the result
        follows from picasso circles problem.

        To prove it, note that, A5B5C5 be the cirumcevian triangle of the
        De-longchamps point of ABC wrt A4B4C4. Clearly, A5B5C5 is the antipodal
        triangle of A3B3C3. Now if we use the generalised problem I posted in the
        proof of picasso circles problem earlier on triangle A3B3C3(and take
        R=De-Longchamps Point of ABC, Q= O), then we get the required result.

        --
        CHANDAN


        [Non-text portions of this message have been removed]
      • Francisco Javier
        With respect to A1B1C1 and A2B2C2, I find that if we change O by another point P, A1B1C1 is then A2B2C2 are perspective if and only P is P lies on NPC or in
        Message 3 of 13 , Feb 10, 2012
        • 0 Attachment
          With respect to A1B1C1 and A2B2C2, I find that if we change O by another point P, A1B1C1 is then A2B2C2 are perspective if and only P is P lies on NPC or in the trilinear polar of X2052.

          A1:

          {SB SC (SB + SC) (-SA u + SB v - SC w) (SA u + SB v - SC w),
          SA SC (SA u - SB v + SC w) (SA SB u + SA SC u + 2 SB SC u + SB^2 v -
          SB SC v - SB SC w + SC^2 w),
          SA SB (SA u + SB v - SC w) (SA SB u + SA SC u + 2 SB SC u + SB^2 v -
          SB SC v - SB SC w + SC^2 w)}

          A2: {(SA SB u + SA SC u + SB SC u + SB^2 v - SB SC w) (SA SB u + SA SC u +
          SB SC u - SB SC v + SC^2 w), -(SA + SC) (-SB v + SC w) (SA SB u +
          SA SC u + SB SC u + SB^2 v - SB SC w), -(SA + SB) (SB v -
          SC w) (SA SB u + SA SC u + SB SC u - SB SC v + SC^2 w)}




          --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...> wrote:
          >
          > Yes, so, here is more precise information: The line A"H intersect the circumcircle at point A1, A2 with coordinates
          >
          > A1:
          >
          > {a^2 (a^2+b^2-c^2) (a^2-b^2+c^2) (a^6-a^4 b^2-a^2 b^4+b^6-3 a^4 c^2+2 a^2 b^2 c^2-3 b^4 c^2+3 a^2 c^4+3 b^2 c^4-c^6) (a^6-3 a^4 b^2+3 a^2 b^4-b^6-a^4 c^2+2 a^2 b^2 c^2+3 b^4 c^2-a^2 c^4-3 b^2 c^4+c^6),-(a^2-b^2-c^2) (a^2+b^2-c^2) (a^6-3 a^4 b^2+3 a^2 b^4-b^6-a^4 c^2+2 a^2 b^2 c^2+3 b^4 c^2-a^2 c^4-3 b^2 c^4+c^6) (a^6 b^2-a^4 b^4-a^2 b^6+b^8+a^6 c^2-2 a^4 b^2 c^2+a^2 b^4 c^2-4 b^6 c^2-a^4 c^4+a^2 b^2 c^4+6 b^4 c^4-a^2 c^6-4 b^2 c^6+c^8),-(a^2-b^2-c^2) (a^2-b^2+c^2) (a^6-a^4 b^2-a^2 b^4+b^6-3 a^4 c^2+2 a^2 b^2 c^2-3 b^4 c^2+3 a^2 c^4+3 b^2 c^4-c^6) (a^6 b^2-a^4 b^4-a^2 b^6+b^8+a^6 c^2-2 a^4 b^2 c^2+a^2 b^4 c^2-4 b^6 c^2-a^4 c^4+a^2 b^2 c^4+6 b^4 c^4-a^2 c^6-4 b^2 c^6+c^8)}
          >
          > A2:
          >
          > {(a^4-2 a^2 b^2-b^4+2 b^2 c^2-c^4) (a^4-b^4-2 a^2 c^2+2 b^2 c^2-c^4),-2 b^2 (b-c) (b+c) (-a^4+b^4+2 a^2 c^2-2 b^2 c^2+c^4),-2 c^2 (-b+c) (b+c) (-a^4+2 a^2 b^2+b^4-2 b^2 c^2+c^4)}
          >
          > and cyclicaly for B1, B2, and C1, C2.
          >
          > With these coordinates, A1B1C1 and ABC are always perspective at X3542.
          >
          >
          >
          > --- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@> wrote:
          > >
          > > The choice of A1 and A2, etc. appears to depend on the shape of ABC. For example, for the ETC ref. triangle, it is A1B1C2 that is perspective to ABC at X(3542). And it does not seem to depend on acute/obtuse, as some acute triangles I have tried need the roles of A1/A2, B1/B2, and/or C1/C2 reversed.
          > >
          > > Randy
          > >
          > > --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
          > > >
          > > >
          > > > http://anthrakitis.blogspot.com/2012/02/perspective.html
          > > > <http://anthrakitis.blogspot.com/2012/02/perspective.html>
          > > >
          > > > APH
          > > >
          > > >
          > > >
          > > > [Non-text portions of this message have been removed]
          > > >
          > >
          >
        • Chandan Banerjee
          There are some typo in my previous post. In the first paragraph, it will be A2B2C2 instead of A5B5C5. I have defined A5B5C5 in the first line of second
          Message 4 of 13 , Feb 10, 2012
          • 0 Attachment
            There are some typo in my previous post. In the first paragraph, it will be
            A2B2C2 instead of A5B5C5. I have defined A5B5C5 in the first line of second
            paragraph.

            --
            CHANDAN


            [Non-text portions of this message have been removed]
          • Francisco Javier
            I m sorry. I meant A1B1C1 and ABC always perspective, A2B2C2 and ABC perspective if and only P is P lies on NPC or in the trilinear polar of X2052.
            Message 5 of 13 , Feb 10, 2012
            • 0 Attachment
              I'm sorry. I meant

              A1B1C1 and ABC always perspective,

              A2B2C2 and ABC perspective if and only P is P lies on NPC or in the trilinear polar of X2052.


              --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...> wrote:
              >
              > With respect to A1B1C1 and A2B2C2, I find that if we change O by another point P, A1B1C1 is then A2B2C2 are perspective if and only P is P lies on NPC or in the trilinear polar of X2052.
              >
              > A1:
              >
              > {SB SC (SB + SC) (-SA u + SB v - SC w) (SA u + SB v - SC w),
              > SA SC (SA u - SB v + SC w) (SA SB u + SA SC u + 2 SB SC u + SB^2 v -
              > SB SC v - SB SC w + SC^2 w),
              > SA SB (SA u + SB v - SC w) (SA SB u + SA SC u + 2 SB SC u + SB^2 v -
              > SB SC v - SB SC w + SC^2 w)}
              >
              > A2: {(SA SB u + SA SC u + SB SC u + SB^2 v - SB SC w) (SA SB u + SA SC u +
              > SB SC u - SB SC v + SC^2 w), -(SA + SC) (-SB v + SC w) (SA SB u +
              > SA SC u + SB SC u + SB^2 v - SB SC w), -(SA + SB) (SB v -
              > SC w) (SA SB u + SA SC u + SB SC u - SB SC v + SC^2 w)}
              >
              >
              >
              >
              > --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@> wrote:
              > >
              > > Yes, so, here is more precise information: The line A"H intersect the circumcircle at point A1, A2 with coordinates
              > >
              > > A1:
              > >
              > > {a^2 (a^2+b^2-c^2) (a^2-b^2+c^2) (a^6-a^4 b^2-a^2 b^4+b^6-3 a^4 c^2+2 a^2 b^2 c^2-3 b^4 c^2+3 a^2 c^4+3 b^2 c^4-c^6) (a^6-3 a^4 b^2+3 a^2 b^4-b^6-a^4 c^2+2 a^2 b^2 c^2+3 b^4 c^2-a^2 c^4-3 b^2 c^4+c^6),-(a^2-b^2-c^2) (a^2+b^2-c^2) (a^6-3 a^4 b^2+3 a^2 b^4-b^6-a^4 c^2+2 a^2 b^2 c^2+3 b^4 c^2-a^2 c^4-3 b^2 c^4+c^6) (a^6 b^2-a^4 b^4-a^2 b^6+b^8+a^6 c^2-2 a^4 b^2 c^2+a^2 b^4 c^2-4 b^6 c^2-a^4 c^4+a^2 b^2 c^4+6 b^4 c^4-a^2 c^6-4 b^2 c^6+c^8),-(a^2-b^2-c^2) (a^2-b^2+c^2) (a^6-a^4 b^2-a^2 b^4+b^6-3 a^4 c^2+2 a^2 b^2 c^2-3 b^4 c^2+3 a^2 c^4+3 b^2 c^4-c^6) (a^6 b^2-a^4 b^4-a^2 b^6+b^8+a^6 c^2-2 a^4 b^2 c^2+a^2 b^4 c^2-4 b^6 c^2-a^4 c^4+a^2 b^2 c^4+6 b^4 c^4-a^2 c^6-4 b^2 c^6+c^8)}
              > >
              > > A2:
              > >
              > > {(a^4-2 a^2 b^2-b^4+2 b^2 c^2-c^4) (a^4-b^4-2 a^2 c^2+2 b^2 c^2-c^4),-2 b^2 (b-c) (b+c) (-a^4+b^4+2 a^2 c^2-2 b^2 c^2+c^4),-2 c^2 (-b+c) (b+c) (-a^4+2 a^2 b^2+b^4-2 b^2 c^2+c^4)}
              > >
              > > and cyclicaly for B1, B2, and C1, C2.
              > >
              > > With these coordinates, A1B1C1 and ABC are always perspective at X3542.
              > >
              > >
              > >
              > > --- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@> wrote:
              > > >
              > > > The choice of A1 and A2, etc. appears to depend on the shape of ABC. For example, for the ETC ref. triangle, it is A1B1C2 that is perspective to ABC at X(3542). And it does not seem to depend on acute/obtuse, as some acute triangles I have tried need the roles of A1/A2, B1/B2, and/or C1/C2 reversed.
              > > >
              > > > Randy
              > > >
              > > > --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
              > > > >
              > > > >
              > > > > http://anthrakitis.blogspot.com/2012/02/perspective.html
              > > > > <http://anthrakitis.blogspot.com/2012/02/perspective.html>
              > > > >
              > > > > APH
              > > > >
              > > > >
              > > > >
              > > > > [Non-text portions of this message have been removed]
              > > > >
              > > >
              > >
              >
            • Chandan Banerjee
              I think there is a mistake in the defintion of A1,A2,B1,B2 and so. Because if we take HA1
              Message 6 of 13 , Feb 10, 2012
              • 0 Attachment
                I think there is a mistake in the defintion of A1,A2,B1,B2 and so. Because
                if we take HA1<HA2 and so then it will not be always true. But if we take
                B" on the same side of B2 and B1 on the other side, then it will be always
                true.

                --
                CHANDAN


                [Non-text portions of this message have been removed]
              • Chandan Banerjee
                I have got that if ABC and A2B2C2 are concurrent, then P lies on NPC or on a line which maps to the polar of H wrt the circumcircle of ABC under a homothety of
                Message 7 of 13 , Feb 10, 2012
                • 0 Attachment
                  I have got that if ABC and A2B2C2 are concurrent, then P lies on NPC or on a line which maps to the polar of H wrt the circumcircle of ABC under a homothety of ratio 2. And Also the perspector lies on the polar of H.
                  I think its equivalent to your statement, but I don't know the properties of X2052, so I can't prove that they are equivalent now.

                  Proof:-

                  Suppose, A3B3C3 is the circumcevian triangle of H wrt ABC. P' be the reflection of H on P. Suppose, perspector of ABC and A2B2C2 is Q. Clearly, A2B2C2 is the circumcevian triangle of P' wrt A3B3C3. Applying Pascal's theorem on AA2A3BB2B3 we get that the intersection point of AB3 and BA3 lies on P'Q and similar for others. But we also have that AB3 and BA3 intersects on the polar of H wrt the circumcircle of ABC. So P'Q is the polar of H. So P' lies on the polar of H. And its easy to see that if it lies on H, then it can be proved using the same argument.

                  --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...> wrote:
                  >
                  > With respect to A1B1C1 and A2B2C2, I find that if we change O by another point P, A1B1C1 is then A2B2C2 are perspective if and only P is P lies on NPC or in the trilinear polar of X2052.
                  >
                  > A1:
                  >
                  > {SB SC (SB + SC) (-SA u + SB v - SC w) (SA u + SB v - SC w),
                  > SA SC (SA u - SB v + SC w) (SA SB u + SA SC u + 2 SB SC u + SB^2 v -
                  > SB SC v - SB SC w + SC^2 w),
                  > SA SB (SA u + SB v - SC w) (SA SB u + SA SC u + 2 SB SC u + SB^2 v -
                  > SB SC v - SB SC w + SC^2 w)}
                  >
                  > A2: {(SA SB u + SA SC u + SB SC u + SB^2 v - SB SC w) (SA SB u + SA SC u +
                  > SB SC u - SB SC v + SC^2 w), -(SA + SC) (-SB v + SC w) (SA SB u +
                  > SA SC u + SB SC u + SB^2 v - SB SC w), -(SA + SB) (SB v -
                  > SC w) (SA SB u + SA SC u + SB SC u - SB SC v + SC^2 w)}
                  >
                  >
                  >
                  >
                  > --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@> wrote:
                  > >
                  > > Yes, so, here is more precise information: The line A"H intersect the circumcircle at point A1, A2 with coordinates
                  > >
                  > > A1:
                  > >
                  > > {a^2 (a^2+b^2-c^2) (a^2-b^2+c^2) (a^6-a^4 b^2-a^2 b^4+b^6-3 a^4 c^2+2 a^2 b^2 c^2-3 b^4 c^2+3 a^2 c^4+3 b^2 c^4-c^6) (a^6-3 a^4 b^2+3 a^2 b^4-b^6-a^4 c^2+2 a^2 b^2 c^2+3 b^4 c^2-a^2 c^4-3 b^2 c^4+c^6),-(a^2-b^2-c^2) (a^2+b^2-c^2) (a^6-3 a^4 b^2+3 a^2 b^4-b^6-a^4 c^2+2 a^2 b^2 c^2+3 b^4 c^2-a^2 c^4-3 b^2 c^4+c^6) (a^6 b^2-a^4 b^4-a^2 b^6+b^8+a^6 c^2-2 a^4 b^2 c^2+a^2 b^4 c^2-4 b^6 c^2-a^4 c^4+a^2 b^2 c^4+6 b^4 c^4-a^2 c^6-4 b^2 c^6+c^8),-(a^2-b^2-c^2) (a^2-b^2+c^2) (a^6-a^4 b^2-a^2 b^4+b^6-3 a^4 c^2+2 a^2 b^2 c^2-3 b^4 c^2+3 a^2 c^4+3 b^2 c^4-c^6) (a^6 b^2-a^4 b^4-a^2 b^6+b^8+a^6 c^2-2 a^4 b^2 c^2+a^2 b^4 c^2-4 b^6 c^2-a^4 c^4+a^2 b^2 c^4+6 b^4 c^4-a^2 c^6-4 b^2 c^6+c^8)}
                  > >
                  > > A2:
                  > >
                  > > {(a^4-2 a^2 b^2-b^4+2 b^2 c^2-c^4) (a^4-b^4-2 a^2 c^2+2 b^2 c^2-c^4),-2 b^2 (b-c) (b+c) (-a^4+b^4+2 a^2 c^2-2 b^2 c^2+c^4),-2 c^2 (-b+c) (b+c) (-a^4+2 a^2 b^2+b^4-2 b^2 c^2+c^4)}
                  > >
                  > > and cyclicaly for B1, B2, and C1, C2.
                  > >
                  > > With these coordinates, A1B1C1 and ABC are always perspective at X3542.
                  > >
                  > >
                  > >
                  > > --- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@> wrote:
                  > > >
                  > > > The choice of A1 and A2, etc. appears to depend on the shape of ABC. For example, for the ETC ref. triangle, it is A1B1C2 that is perspective to ABC at X(3542). And it does not seem to depend on acute/obtuse, as some acute triangles I have tried need the roles of A1/A2, B1/B2, and/or C1/C2 reversed.
                  > > >
                  > > > Randy
                  > > >
                  > > > --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
                  > > > >
                  > > > >
                  > > > > http://anthrakitis.blogspot.com/2012/02/perspective.html
                  > > > > <http://anthrakitis.blogspot.com/2012/02/perspective.html>
                  > > > >
                  > > > > APH
                  > > > >
                  > > > >
                  > > > >
                  > > > > [Non-text portions of this message have been removed]
                  > > > >
                  > > >
                  > >
                  >
                • Chandan Banerjee
                  Sorry, I made a typo again. In the first paragraph it will be under a homothety of ratio 2 wrt H . On Sat, Feb 11, 2012 at 3:52 AM, Chandan Banerjee ... --
                  Message 8 of 13 , Feb 10, 2012
                  • 0 Attachment
                    Sorry, I made a typo again. In the first paragraph it will be "under a
                    homothety of ratio 2 wrt H".

                    On Sat, Feb 11, 2012 at 3:52 AM, Chandan Banerjee
                    <chandana.snbv@...>wrote:

                    > **
                    >
                    >
                    > I have got that if ABC and A2B2C2 are concurrent, then P lies on NPC or on
                    > a line which maps to the polar of H wrt the circumcircle of ABC under a
                    > homothety of ratio 2. And Also the perspector lies on the polar of H.
                    > I think its equivalent to your statement, but I don't know the properties
                    > of X2052, so I can't prove that they are equivalent now.
                    >
                    > Proof:-
                    >
                    > Suppose, A3B3C3 is the circumcevian triangle of H wrt ABC. P' be the
                    > reflection of H on P. Suppose, perspector of ABC and A2B2C2 is Q. Clearly,
                    > A2B2C2 is the circumcevian triangle of P' wrt A3B3C3. Applying Pascal's
                    > theorem on AA2A3BB2B3 we get that the intersection point of AB3 and BA3
                    > lies on P'Q and similar for others. But we also have that AB3 and BA3
                    > intersects on the polar of H wrt the circumcircle of ABC. So P'Q is the
                    > polar of H. So P' lies on the polar of H. And its easy to see that if it
                    > lies on H, then it can be proved using the same argument.
                    >
                    >
                    > --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...>
                    > wrote:
                    > >
                    > > With respect to A1B1C1 and A2B2C2, I find that if we change O by another
                    > point P, A1B1C1 is then A2B2C2 are perspective if and only P is P lies on
                    > NPC or in the trilinear polar of X2052.
                    > >
                    > > A1:
                    > >
                    > > {SB SC (SB + SC) (-SA u + SB v - SC w) (SA u + SB v - SC w),
                    > > SA SC (SA u - SB v + SC w) (SA SB u + SA SC u + 2 SB SC u + SB^2 v -
                    > > SB SC v - SB SC w + SC^2 w),
                    > > SA SB (SA u + SB v - SC w) (SA SB u + SA SC u + 2 SB SC u + SB^2 v -
                    > > SB SC v - SB SC w + SC^2 w)}
                    > >
                    > > A2: {(SA SB u + SA SC u + SB SC u + SB^2 v - SB SC w) (SA SB u + SA SC u
                    > +
                    > > SB SC u - SB SC v + SC^2 w), -(SA + SC) (-SB v + SC w) (SA SB u +
                    > > SA SC u + SB SC u + SB^2 v - SB SC w), -(SA + SB) (SB v -
                    > > SC w) (SA SB u + SA SC u + SB SC u - SB SC v + SC^2 w)}
                    > >
                    > >
                    > >
                    > >
                    > > --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@>
                    > wrote:
                    > > >
                    > > > Yes, so, here is more precise information: The line A"H intersect the
                    > circumcircle at point A1, A2 with coordinates
                    > > >
                    > > > A1:
                    > > >
                    > > > {a^2 (a^2+b^2-c^2) (a^2-b^2+c^2) (a^6-a^4 b^2-a^2 b^4+b^6-3 a^4 c^2+2
                    > a^2 b^2 c^2-3 b^4 c^2+3 a^2 c^4+3 b^2 c^4-c^6) (a^6-3 a^4 b^2+3 a^2
                    > b^4-b^6-a^4 c^2+2 a^2 b^2 c^2+3 b^4 c^2-a^2 c^4-3 b^2
                    > c^4+c^6),-(a^2-b^2-c^2) (a^2+b^2-c^2) (a^6-3 a^4 b^2+3 a^2 b^4-b^6-a^4
                    > c^2+2 a^2 b^2 c^2+3 b^4 c^2-a^2 c^4-3 b^2 c^4+c^6) (a^6 b^2-a^4 b^4-a^2
                    > b^6+b^8+a^6 c^2-2 a^4 b^2 c^2+a^2 b^4 c^2-4 b^6 c^2-a^4 c^4+a^2 b^2 c^4+6
                    > b^4 c^4-a^2 c^6-4 b^2 c^6+c^8),-(a^2-b^2-c^2) (a^2-b^2+c^2) (a^6-a^4
                    > b^2-a^2 b^4+b^6-3 a^4 c^2+2 a^2 b^2 c^2-3 b^4 c^2+3 a^2 c^4+3 b^2 c^4-c^6)
                    > (a^6 b^2-a^4 b^4-a^2 b^6+b^8+a^6 c^2-2 a^4 b^2 c^2+a^2 b^4 c^2-4 b^6
                    > c^2-a^4 c^4+a^2 b^2 c^4+6 b^4 c^4-a^2 c^6-4 b^2 c^6+c^8)}
                    > > >
                    > > > A2:
                    > > >
                    > > > {(a^4-2 a^2 b^2-b^4+2 b^2 c^2-c^4) (a^4-b^4-2 a^2 c^2+2 b^2
                    > c^2-c^4),-2 b^2 (b-c) (b+c) (-a^4+b^4+2 a^2 c^2-2 b^2 c^2+c^4),-2 c^2
                    > (-b+c) (b+c) (-a^4+2 a^2 b^2+b^4-2 b^2 c^2+c^4)}
                    > > >
                    > > > and cyclicaly for B1, B2, and C1, C2.
                    > > >
                    > > > With these coordinates, A1B1C1 and ABC are always perspective at X3542.
                    > > >
                    > > >
                    > > >
                    > > > --- In Hyacinthos@yahoogroups.com, "rhutson2" <rhutson2@> wrote:
                    > > > >
                    > > > > The choice of A1 and A2, etc. appears to depend on the shape of ABC.
                    > For example, for the ETC ref. triangle, it is A1B1C2 that is perspective to
                    > ABC at X(3542). And it does not seem to depend on acute/obtuse, as some
                    > acute triangles I have tried need the roles of A1/A2, B1/B2, and/or C1/C2
                    > reversed.
                    > > > >
                    > > > > Randy
                    > > > >
                    > > > > --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
                    > > > > >
                    > > > > >
                    > > > > > http://anthrakitis.blogspot.com/2012/02/perspective.html
                    > > > > > <http://anthrakitis.blogspot.com/2012/02/perspective.html>
                    > > > > >
                    > > > > > APH
                    > > > > >
                    > > > > >
                    > > > > >
                    > > > > > [Non-text portions of this message have been removed]
                    > > > > >
                    > > > >
                    > > >
                    > >
                    >
                    >
                    >



                    --
                    CHANDAN


                    [Non-text portions of this message have been removed]
                  Your message has been successfully submitted and would be delivered to recipients shortly.