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

Another locus

Expand Messages
  • xpolakis@otenet.gr
    let ABC be a triangle and A B C the pedal triangle of a point P. Let A be a point on PA (see figure) A / / B C / /
    Message 1 of 6 , Mar 1, 2001
    • 0 Attachment
      let ABC be a triangle and A'B'C' the pedal triangle of a point P.

      Let A" be a point on PA' (see figure)

      A
      /\
      / \ B"
      C" / \
      / \
      / \
      / \
      / \
      / \
      / P \
      / | \
      / | \
      B--------A'------------C
      |
      |
      |
      |
      |
      |
      A"

      such that PA" = a.

      Similarly we define the points B", C" on PB', PC', respectively
      (PB" = b, PC" = c)

      Which is the locus of P such that AA", BB", CC" are concurrent?


      APH
    • xpolakis
      Let ABC be a triangle, P a point and A B C the pedal triangle of P. The parallels through A,B,C to B C , C A , A B , resp. bound a triangle A B C . Which is
      Message 2 of 6 , Aug 16, 2009
      • 0 Attachment
        Let ABC be a triangle, P a point and A'B'C' the
        pedal triangle of P.

        The parallels through A,B,C to B'C', C'A', A'B', resp.
        bound a triangle A"B"C".

        Which is the locus of P such that the Euler Lines of
        A"BC, B"CA, C"AB are concurrent?

        Variation: A'B'C' : cevian triangle of P.


        APH
      • Francisco Javier
        Another unanswered post. In this case, the locus is a cubic with long equation (shown below) through X3 and X4. Best regards, Francisco Javier. ... S stands
        Message 3 of 6 , Oct 23, 2011
        • 0 Attachment
          Another unanswered post.

          In this case, the locus is a cubic with long equation (shown below) through X3 and X4.

          Best regards,

          Francisco Javier.

          ----

          S stands for twice the area of ABC.

          -8 a^4 b^4 c^2 x^3 + 8 a^2 b^6 c^2 x^3 + 8 a^4 b^2 c^4 x^3 -
          8 a^2 b^2 c^6 x^3 - 8 a^2 b^4 c^2 S x^3 + 8 b^6 c^2 S x^3 +
          8 a^2 b^2 c^4 S x^3 - 8 b^2 c^6 S x^3 + a^8 c^2 x^2 y -
          8 a^6 b^2 c^2 x^2 y - 14 a^4 b^4 c^2 x^2 y + 24 a^2 b^6 c^2 x^2 y -
          3 b^8 c^2 x^2 y - 4 a^6 c^4 x^2 y + 16 a^4 b^2 c^4 x^2 y -
          12 a^2 b^4 c^4 x^2 y + 8 b^6 c^4 x^2 y + 6 a^4 c^6 x^2 y -
          8 a^2 b^2 c^6 x^2 y - 6 b^4 c^6 x^2 y - 4 a^2 c^8 x^2 y +
          c^10 x^2 y - 4 a^6 c^2 S x^2 y - 20 a^4 b^2 c^2 S x^2 y +
          12 a^2 b^4 c^2 S x^2 y + 12 b^6 c^2 S x^2 y + 8 a^4 c^4 S x^2 y +
          8 a^2 b^2 c^4 S x^2 y - 8 b^4 c^4 S x^2 y - 4 a^2 c^6 S x^2 y -
          4 b^2 c^6 S x^2 y + 3 a^8 c^2 x y^2 - 24 a^6 b^2 c^2 x y^2 +
          14 a^4 b^4 c^2 x y^2 + 8 a^2 b^6 c^2 x y^2 - b^8 c^2 x y^2 -
          8 a^6 c^4 x y^2 + 12 a^4 b^2 c^4 x y^2 - 16 a^2 b^4 c^4 x y^2 +
          4 b^6 c^4 x y^2 + 6 a^4 c^6 x y^2 + 8 a^2 b^2 c^6 x y^2 -
          6 b^4 c^6 x y^2 + 4 b^2 c^8 x y^2 - c^10 x y^2 -
          12 a^6 c^2 S x y^2 - 12 a^4 b^2 c^2 S x y^2 +
          20 a^2 b^4 c^2 S x y^2 + 4 b^6 c^2 S x y^2 + 8 a^4 c^4 S x y^2 -
          8 a^2 b^2 c^4 S x y^2 - 8 b^4 c^4 S x y^2 + 4 a^2 c^6 S x y^2 +
          4 b^2 c^6 S x y^2 - 8 a^6 b^2 c^2 y^3 + 8 a^4 b^4 c^2 y^3 -
          8 a^2 b^4 c^4 y^3 + 8 a^2 b^2 c^6 y^3 - 8 a^6 c^2 S y^3 +
          8 a^4 b^2 c^2 S y^3 - 8 a^2 b^2 c^4 S y^3 + 8 a^2 c^6 S y^3 -
          a^8 b^2 x^2 z + 4 a^6 b^4 x^2 z - 6 a^4 b^6 x^2 z + 4 a^2 b^8 x^2 z -
          b^10 x^2 z + 8 a^6 b^2 c^2 x^2 z - 16 a^4 b^4 c^2 x^2 z +
          8 a^2 b^6 c^2 x^2 z + 14 a^4 b^2 c^4 x^2 z + 12 a^2 b^4 c^4 x^2 z +
          6 b^6 c^4 x^2 z - 24 a^2 b^2 c^6 x^2 z - 8 b^4 c^6 x^2 z +
          3 b^2 c^8 x^2 z + 4 a^6 b^2 S x^2 z - 8 a^4 b^4 S x^2 z +
          4 a^2 b^6 S x^2 z + 20 a^4 b^2 c^2 S x^2 z - 8 a^2 b^4 c^2 S x^2 z +
          4 b^6 c^2 S x^2 z - 12 a^2 b^2 c^4 S x^2 z + 8 b^4 c^4 S x^2 z -
          12 b^2 c^6 S x^2 z + a^10 y^2 z - 4 a^8 b^2 y^2 z +
          6 a^6 b^4 y^2 z - 4 a^4 b^6 y^2 z + a^2 b^8 y^2 z -
          8 a^6 b^2 c^2 y^2 z + 16 a^4 b^4 c^2 y^2 z - 8 a^2 b^6 c^2 y^2 z -
          6 a^6 c^4 y^2 z - 12 a^4 b^2 c^4 y^2 z - 14 a^2 b^4 c^4 y^2 z +
          8 a^4 c^6 y^2 z + 24 a^2 b^2 c^6 y^2 z - 3 a^2 c^8 y^2 z -
          4 a^6 b^2 S y^2 z + 8 a^4 b^4 S y^2 z - 4 a^2 b^6 S y^2 z -
          4 a^6 c^2 S y^2 z + 8 a^4 b^2 c^2 S y^2 z - 20 a^2 b^4 c^2 S y^2 z -
          8 a^4 c^4 S y^2 z + 12 a^2 b^2 c^4 S y^2 z + 12 a^2 c^6 S y^2 z -
          3 a^8 b^2 x z^2 + 8 a^6 b^4 x z^2 - 6 a^4 b^6 x z^2 + b^10 x z^2 +
          24 a^6 b^2 c^2 x z^2 - 12 a^4 b^4 c^2 x z^2 - 8 a^2 b^6 c^2 x z^2 -
          4 b^8 c^2 x z^2 - 14 a^4 b^2 c^4 x z^2 + 16 a^2 b^4 c^4 x z^2 +
          6 b^6 c^4 x z^2 - 8 a^2 b^2 c^6 x z^2 - 4 b^4 c^6 x z^2 +
          b^2 c^8 x z^2 + 12 a^6 b^2 S x z^2 - 8 a^4 b^4 S x z^2 -
          4 a^2 b^6 S x z^2 + 12 a^4 b^2 c^2 S x z^2 + 8 a^2 b^4 c^2 S x z^2 -
          4 b^6 c^2 S x z^2 - 20 a^2 b^2 c^4 S x z^2 + 8 b^4 c^4 S x z^2 -
          4 b^2 c^6 S x z^2 - a^10 y z^2 + 6 a^6 b^4 y z^2 - 8 a^4 b^6 y z^2 +
          3 a^2 b^8 y z^2 + 4 a^8 c^2 y z^2 + 8 a^6 b^2 c^2 y z^2 +
          12 a^4 b^4 c^2 y z^2 - 24 a^2 b^6 c^2 y z^2 - 6 a^6 c^4 y z^2 -
          16 a^4 b^2 c^4 y z^2 + 14 a^2 b^4 c^4 y z^2 + 4 a^4 c^6 y z^2 +
          8 a^2 b^2 c^6 y z^2 - a^2 c^8 y z^2 + 4 a^6 b^2 S y z^2 +
          8 a^4 b^4 S y z^2 - 12 a^2 b^6 S y z^2 + 4 a^6 c^2 S y z^2 -
          8 a^4 b^2 c^2 S y z^2 - 12 a^2 b^4 c^2 S y z^2 - 8 a^4 c^4 S y z^2 +
          20 a^2 b^2 c^4 S y z^2 + 4 a^2 c^6 S y z^2 + 8 a^6 b^2 c^2 z^3 -
          8 a^2 b^6 c^2 z^3 - 8 a^4 b^2 c^4 z^3 + 8 a^2 b^4 c^4 z^3 +
          8 a^6 b^2 S z^3 - 8 a^2 b^6 S z^3 - 8 a^4 b^2 c^2 S z^3 +
          8 a^2 b^4 c^2 S z^3 ==0.




          --- In Hyacinthos@yahoogroups.com, xpolakis@... wrote:
          >
          > let ABC be a triangle and A'B'C' the pedal triangle of a point P.
          >
          > Let A" be a point on PA' (see figure)
          >
          > A
          > /\
          > / \ B"
          > C" / \
          > / \
          > / \
          > / \
          > / \
          > / \
          > / P \
          > / | \
          > / | \
          > B--------A'------------C
          > |
          > |
          > |
          > |
          > |
          > |
          > A"
          >
          > such that PA" = a.
          >
          > Similarly we define the points B", C" on PB', PC', respectively
          > (PB" = b, PC" = c)
          >
          > Which is the locus of P such that AA", BB", CC" are concurrent?
          >
          >
          > APH
          >
        • Francisco Javier
          In the pedal case, the locus is (line at infinity) + (circumcircle) + (napoleon cubic). In the cevian case, the locus is a 15th degree curve through X2, X4 and
          Message 4 of 6 , Oct 23, 2011
          • 0 Attachment
            In the pedal case, the locus is (line at infinity) + (circumcircle) + (napoleon cubic).

            In the cevian case, the locus is a 15th degree curve through X2, X4 and X7.

            Best regards,

            Francisco Javier.

            --- In Hyacinthos@yahoogroups.com, "xpolakis" <anopolis72@...> wrote:
            >
            > Let ABC be a triangle, P a point and A'B'C' the
            > pedal triangle of P.
            >
            > The parallels through A,B,C to B'C', C'A', A'B', resp.
            > bound a triangle A"B"C".
            >
            > Which is the locus of P such that the Euler Lines of
            > A"BC, B"CA, C"AB are concurrent?
            >
            > Variation: A'B'C' : cevian triangle of P.
            >
            >
            > APH
            >
          • Barry Wolk
            ... [long equation snipped -- BW] ... Since points at infinity are extraneous roots, this becomes a conic, CyclicSum( (SB-SC) (SA x (x+y+z) + S y z) = 0. --
            Message 5 of 6 , Oct 27, 2011
            • 0 Attachment
              > Another unanswered post.
              >
              > In this case, the locus is a cubic with long equation (shown below) through X3
              > and X4.
              >
              > Best regards,
              >
              > Francisco Javier.
              >
              > ----
              >
              > S stands for twice the area of ABC.
              >

              [long equation snipped -- BW]



              > --- In Hyacinthos@yahoogroups.com, xpolakis@... wrote:
              >>
              >> let ABC be a triangle and A'B'C' the pedal triangle of a point
              > P.
              >>
              >> Let A" be a point on PA' (see figure)
              >>
              >>                               A
              >>                               /\
              >>                               /  \            B"
              >>               C"            /    \
              >>                             /      \
              >>                           /        \
              >>                           /          \
              >>                         /            \
              >>                         /              \
              >>                       /    P          \
              >>                       /      |          \
              >>                     /      |            \
              >>                     B--------A'------------C
              >>                             |
              >>                             |
              >>                             |
              >>                             |
              >>                             |
              >>                             |
              >>                             A"
              >>
              >> such that PA" = a.
              >>
              >> Similarly we define the points B", C" on PB', PC',
              > respectively
              >> (PB" = b, PC" = c)
              >>
              >> Which is the locus of P such that AA", BB", CC" are
              > concurrent?
              >>
              >>
              >> APH

              Since points at infinity are extraneous roots, this becomes a conic,
              CyclicSum( (SB-SC) (SA x (x+y+z) + S y z) = 0.
              --
              Barry Wolk
            • Francisco Javier
              Dear Barry Wolk: Yes, you are right. I misunderstood the statement, since I toke A satisfying A A = a, etc,instead of PA =a, etc, then A lying on the
              Message 6 of 6 , Oct 27, 2011
              • 0 Attachment
                Dear Barry Wolk:

                Yes, you are right. I misunderstood the statement, since I toke A'' satisfying A'A'' = a, etc,instead of PA''=a, etc, then A'' lying on the square constructed externally on BC.

                Best regards,

                Francisco Javier.

                --- In Hyacinthos@yahoogroups.com, Barry Wolk <wolkbarry@...> wrote:
                >
                > > Another unanswered post.
                > >
                > > In this case, the locus is a cubic with long equation (shown below) through X3
                > > and X4.
                > >
                > > Best regards,
                > >
                > > Francisco Javier.
                > >
                > > ----
                > >
                > > S stands for twice the area of ABC.
                > >
                >
                > [long equation snipped -- BW]
                >
                >
                > >�
                > > --- In Hyacinthos@yahoogroups.com, xpolakis@ wrote:
                > >>
                > >> let ABC be a triangle and A'B'C' the pedal triangle of a point
                > > P.
                > >>
                > >> Let A" be a point on PA' (see figure)
                > >>
                > >> � � � � � � � � � � � � � � � A
                > >> � � � � � � � � � � � � � � � /\
                > >> � � � � � � � � � � � � � � � /� \� � � � � � B"
                > >> � � � � � � � C"� � � � � � /� � \
                > >> � � � � � � � � � � � � � � /� � � \
                > >> � � � � � � � � � � � � � /� � � � \
                > >> � � � � � � � � � � � � � /� � � � � \
                > >> � � � � � � � � � � � � /� � � � � � \
                > >> � � � � � � � � � � � � /� � � � � � � \
                > >> � � � � � � � � � � � /� � P� � � � � \
                > >> � � � � � � � � � � � /� � � |� � � � � \
                > >> � � � � � � � � � � /� � � |� � � � � � \
                > >> � � � � � � � � � � B--------A'------------C
                > >> � � � � � � � � � � � � � � |
                > >> � � � � � � � � � � � � � � |
                > >> � � � � � � � � � � � � � � |
                > >> � � � � � � � � � � � � � � |
                > >> � � � � � � � � � � � � � � |
                > >> � � � � � � � � � � � � � � |
                > >> � � � � � � � � � � � � � � A"
                > >>
                > >> such that PA" = a.
                > >>
                > >> Similarly we define the points B", C" on PB', PC',
                > > respectively
                > >> (PB" = b, PC" = c)
                > >>
                > >> Which is the locus of P such that AA", BB", CC" are
                > > concurrent?
                > >>
                > >>
                > >> APH
                >
                > Since points at infinity are extraneous roots, this becomes a conic,
                > CyclicSum( (SB-SC) (SA x (x+y+z) + S y z) = 0.
                > --
                > Barry Wolk
                >
              Your message has been successfully submitted and would be delivered to recipients shortly.