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Circle in Barycentrics

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  • Lim Sung Hyun
    What is the condition for a conic to be circle, preferably as a condition imposed on the matrix of conic? Sung Hyun [Non-text portions of this message have
    Message 1 of 25 , Jan 10, 2011
      What is the condition for a conic to be circle, preferably as a condition
      imposed on the matrix of conic?

      Sung Hyun


      [Non-text portions of this message have been removed]
    • fredlangch
      Dear Sung Hyun, in normal coordinates, if the symmetric matrix of the conic is {{alpha,p,q},{p,beta,r},{q,r,gamma}} I have the following conditions: beta*a^6 -
      Message 2 of 25 , Jan 10, 2011
        Dear Sung Hyun,
        in normal coordinates, if the symmetric matrix of the conic is

        {{alpha,p,q},{p,beta,r},{q,r,gamma}}

        I have the following conditions:

        beta*a^6 - 2*b*p*a^5 +
        2*b*c*r*a^4 + b^2*alpha*a^4 -
        2*b^2*beta*a^4 - 2*c^2*beta*a^4 +
        4*b^3*p*a^3 + 2*b*c^2*p*a^3 -
        2*b^2*c*q*a^3 - 2*b*c^3*r*a^2 -
        2*b^3*c*r*a^2 - 2*b^4*alpha*a^2 +
        b^4*beta*a^2 + c^4*beta*a^2 +
        2*b^2*c^2*gamma*a^2 - 2*b^5*p*a +
        2*b^3*c^2*p*a - 2*b^2*c^3*q*a +
        2*b^4*c*q*a + b^6*alpha +
        b^2*c^4*alpha - 2*b^4*c^2*alpha = 0

        AND

        -beta*a^4 + 2*b*p*a^3 -
        2*b*c*r*a^2 - b^2*alpha*a^2 +
        b^2*beta*a^2 + c^2*beta*a^2 -
        2*b^3*p*a + 2*b^2*c*q*a +
        b^4*alpha - b^2*c^2*alpha = 0

        certainly, it is possible to find more elegant and symmetric conditions...

        Regards

        Fred

        --- In Hyacinthos@yahoogroups.com, Lim Sung Hyun <bbbbbblow@...> wrote:
        >
        > What is the condition for a conic to be circle, preferably as a condition
        > imposed on the matrix of conic?
        >
        > Sung Hyun
        >
        >
        > [Non-text portions of this message have been removed]
        >
      • Francisco Javier
        Dear Sung Hyun and Fred, I get that a conic f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0 is a circle when (p - q + r - h) SB - (p + q - r - g) SC =
        Message 3 of 25 , Jan 10, 2011
          Dear Sung Hyun and Fred,

          I get that a conic

          f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0

          is a circle when

          (p - q + r - h) SB - (p + q - r - g) SC = 0

          and

          (f + h + 2 q) SB^2 + (f + g + 2 r) SC^2 + (2 p - g -
          h) a^2 SA + (2 f - g - h + 4 p + 2 q + 2 r) SB SC = 0.

          Best regards, at 11-1-11.

          --- In Hyacinthos@yahoogroups.com, Lim Sung Hyun <bbbbbblow@...> wrote:
          >
          > What is the condition for a conic to be circle, preferably as a condition
          > imposed on the matrix of conic?
          >
          > Sung Hyun
          >
          >
          > [Non-text portions of this message have been removed]
          >
        • jpehrmfr
          Dear Lim Sung Hyun ... Suppose that the radical axis of the circle C0 and the circumcircle (O) is ux+vy+wz=0; as the union of the line at infinity and the
          Message 4 of 25 , Jan 11, 2011
            Dear Lim Sung Hyun
            >
            > What is the condition for a conic to be circle, preferably as a condition
            > imposed on the matrix of conic?

            Suppose that the radical axis of the circle C0 and the circumcircle (O) is ux+vy+wz=0; as the union of the line at infinity and the radical axis is a degenerated member of the pencil (O),C0, the equation of C0 is (1) :
            k(a^2*yz+b^2zx+c^2xy)+(x+y+z)(ux+vy+wz)=0 for some k
            It follows that,if M is the matrix of a conic and V the column matrix
            M[2,2]-M[3,3]
            M[3,3]-M[1,1]
            M[1,1]-M[2,2]
            The conic is a circle when MV and SV are colinear, where S is the matrix of the circumcircle
            (or equivalently V is an eigenvector of S^(-1)M)
            Jean-Pierre
          • Nikolaos Dergiades
            Dear Sung Hyun The conic f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0 with matrix f r q r g p q p h is a circle when (g + h -2p)/aa = (h + f -2q)/bb
            Message 5 of 25 , Jan 11, 2011
              Dear Sung Hyun

              The conic
              f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0
              with matrix
              f r q
              r g p
              q p h

              is a circle when
              (g + h -2p)/aa = (h + f -2q)/bb = (f + q -2r)/cc.

              Best regards
              Nikos Dergiades


              > Lim Sung Hyun <bbbbbblow@...> wrote:
              > >
              > > What is the condition for a conic to be circle,
              > preferably as a condition
              > > imposed on the matrix of conic?
              > >
              > > Sung Hyun
            • Nikolaos Dergiades
              I correct the formula from (g + h -2p)/aa = (h + f -2q)/bb = (f + q -2r)/cc. to (g + h -2p)/aa = (h + f -2q)/bb = (f + g -2r)/cc. Sorry. ND
              Message 6 of 25 , Jan 11, 2011
                I correct the formula from
                (g + h -2p)/aa = (h + f -2q)/bb = (f + q -2r)/cc.
                to
                (g + h -2p)/aa = (h + f -2q)/bb = (f + g -2r)/cc.

                Sorry. ND

                > Dear Sung Hyun
                >
                > The conic
                > f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0
                > with matrix
                > f r q
                > r g p
                > q p h
                >
                > is a circle when
                > (g + h -2p)/aa = (h + f -2q)/bb = (f + q -2r)/cc.
                >
                > Best regards
                > Nikos Dergiades
                >
                >
                > > Lim Sung Hyun <bbbbbblow@...> wrote:
                > > >
                > > > What is the condition for a conic to be circle,
                > > preferably as a condition
                > > > imposed on the matrix of conic?
                > > >
                > > > Sung Hyun
                >
                >
                >
                >
                >
                > ------------------------------------
                >
                > Yahoo! Groups Links
                >
                >
                >     Hyacinthos-fullfeatured@yahoogroups.com
                >
                >
                >
              • Francisco Javier García Capitán
                Dear Nikos, I think that your formulas are really nice. Thank you. 2011/1/11 Nikolaos Dergiades ... -- ... Francisco Javier García
                Message 7 of 25 , Jan 11, 2011
                  Dear Nikos, I think that your formulas are really nice. Thank you.

                  2011/1/11 Nikolaos Dergiades <ndergiades@...>

                  > I correct the formula from
                  > (g + h -2p)/aa = (h + f -2q)/bb = (f + q -2r)/cc.
                  > to
                  > (g + h -2p)/aa = (h + f -2q)/bb = (f + g -2r)/cc.
                  >
                  > Sorry. ND
                  >
                  > > Dear Sung Hyun
                  > >
                  > > The conic
                  > > f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0
                  > > with matrix
                  > > f r q
                  > > r g p
                  > > q p h
                  > >
                  > > is a circle when
                  > > (g + h -2p)/aa = (h + f -2q)/bb = (f + q -2r)/cc.
                  > >
                  > > Best regards
                  > > Nikos Dergiades
                  > >
                  > >
                  > > > Lim Sung Hyun <bbbbbblow@...> wrote:
                  > > > >
                  > > > > What is the condition for a conic to be circle,
                  > > > preferably as a condition
                  > > > > imposed on the matrix of conic?
                  > > > >
                  > > > > Sung Hyun
                  > >
                  > >
                  > >
                  > >
                  > >
                  > > ------------------------------------
                  > >
                  > > Yahoo! Groups Links
                  > >
                  > >
                  > > Hyacinthos-fullfeatured@yahoogroups.com
                  > >
                  > >
                  > >
                  >
                  >
                  >
                  >
                  > ------------------------------------
                  >
                  > Yahoo! Groups Links
                  >
                  >
                  >
                  >


                  --
                  ---
                  Francisco Javier García Capitán
                  http://garciacapitan.auna.com


                  [Non-text portions of this message have been removed]
                • jpehrmfr
                  Dear Nikolaos [ND] ... Thank you very much for this so nice form. My previous message gave a very bad form of the condition but it gives an immediate proof of
                  Message 8 of 25 , Jan 11, 2011
                    Dear Nikolaos
                    [ND]
                    > The conic
                    > f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0
                    > with matrix
                    > f r q
                    > r g p
                    > q p h
                    > is a circle when
                    > (g + h -2p)/aa = (h + f -2q)/bb = (f + g -2r)/cc.

                    Thank you very much for this so nice form.
                    My previous message gave a very bad form of the condition but it gives an immediate proof of yours (I didn't notice that) :
                    As the equation of a circle is
                    k(a^2yz+b^2zx+c^2xy)+(x+y+z)(ux+vy+wz) = 0,
                    we have (g + h -2p)/aa = (h + f -2q)/bb = (f + g -2r)/cc = -k
                    Many thanks again
                    JP
                  • Lim Sung Hyun
                    Nikos, great formula indeed. Your condition is equivalent to: (g+h-2p : h+f-2q : f+g-2r) = K I think this can be simplified further. Any idea? Sung Hyun
                    Message 9 of 25 , Jan 11, 2011
                      Nikos, great formula indeed.
                      Your condition is equivalent to:

                      (g+h-2p : h+f-2q : f+g-2r) = K

                      I think this can be simplified further. Any idea?

                      Sung Hyun


                      [Non-text portions of this message have been removed]
                    • Lim Sung Hyun
                      Dear everyone, There seems to be a very intimate relation between K and circle. 1. Perspector of circumcircle = K 2. For bicevian conic through cevian
                      Message 10 of 25 , Jan 11, 2011
                        Dear everyone,

                        There seems to be a very intimate relation between K and circle.

                        1. Perspector of circumcircle = K
                        2. For bicevian conic through cevian triangles of P and Q, (ctP)(ctQ)=K
                        3. Polar circle, the only diagonal circle has equation S_Ax^2 + S_By^2 +
                        S_Cz^2 = 0 ( Note that (S_A:S_B:S_C) = aK)
                        4. Conic is circle iff (g+h-2p : h+f-2q : f+g-2r) = K

                        Can anyone give a general interpretation for this?

                        Sung Hyun


                        [Non-text portions of this message have been removed]
                      • Nikolaos Dergiades
                        Dear friends, thank you very much for your messages. This formula is not mine. I found it in the first volume of George Kapetis Triangle Geometry (in Greek)
                        Message 11 of 25 , Jan 11, 2011
                          Dear friends,
                          thank you very much for your messages.
                          This formula is not mine.
                          I found it in the first volume of George Kapetis'
                          Triangle Geometry (in Greek) dating from 1996.
                          The writer gives a not simple proof in a more
                          general context.
                          He ignores the general circle formula
                          aayz+bbzx+ccxy-(x+y+z)(Px+Qy+Rz) = 0
                          and I had early noted that from the above equality we
                          can have the simple proof that Jean-Pierre mentioned.

                          From this book also I found that the kind of the
                          conic is given by the sign of the det of the mattrix
                          0 1 1 1
                          1 f r q
                          1 r g p
                          1 q p h

                          Best regards
                          Nikos Dergiades

                          [SH]
                          > Nikos, great formula indeed.
                          > Your condition is equivalent to:
                          >
                          > (g+h-2p : h+f-2q : f+g-2r) = K
                          >
                          > I think this can be simplified further. Any idea?

                          [JP]
                          > Thank you very much for this so nice form.
                          > My previous message gave a very bad form of the condition
                          > but it gives an immediate proof of yours (I didn't notice
                          > that) :
                          > As the equation of a circle is
                          > k(a^2yz+b^2zx+c^2xy)+(x+y+z)(ux+vy+wz) = 0,
                          > we have (g + h -2p)/aa = (h + f -2q)/bb = (f + g -2r)/cc =
                          > -k
                        • jpehrmfr
                          Dear Nikolaos ... If M is the matrix of the conic C0, M* the adjugate matrix of M, ie gh-pp pq-hr pr-gq pq-hr hf-qq rq-fp pr-gq rq-fp fg-rr and U the
                          Message 12 of 25 , Jan 11, 2011
                            Dear Nikolaos
                            > From this book also I found that the kind of the
                            > conic is given by the sign of the det of the mattrix
                            > 0 1 1 1
                            > 1 f r q
                            > 1 r g p
                            > 1 q p h

                            If M is the matrix of the conic C0, M* the adjugate matrix of M, ie
                            gh-pp pq-hr pr-gq
                            pq-hr hf-qq rq-fp
                            pr-gq rq-fp fg-rr
                            and U the column matrix
                            u
                            v
                            w
                            then U* = M*U is the pole wrt C0 of the line L of equation ux+vy+wz=0
                            Put T(u,v,w) = t(U)U* (t for transposition) T = quadratic form with matrix M*
                            Then L and C0 have 0,1,2 real common points according to the fact that T>0, T=0, T<0
                            More over, the determinant of the matrix
                            0 u v w
                            u f r q
                            v r g p
                            w q p h
                            is -T.
                            The matrix of the singular conic union of the tangents to C0 from the pole of L is T.M-det(M).U.t(U)
                            With u=v=w=1, we get the center of C0, the number of real infinite points ie the kind of the conic and the equation of the pair of asymptots.
                            Friendly. Jean-Pierre
                          • Nikolaos Dergiades
                            Dear Jean-Pierre thank you for this proof and especialy for equation of the pair of asymptots. M* means the adjoint matrix of M U* what is? Best regards Nikos
                            Message 13 of 25 , Jan 12, 2011
                              Dear Jean-Pierre
                              thank you for this proof
                              and especialy for equation of
                              the pair of asymptots.

                              M* means the adjoint matrix of M
                              U* what is?

                              Best regards
                              Nikos Dergiades


                              > Dear Nikolaos
                              > > From this book also I found that the kind of the
                              > > conic is given by the sign of the det of the mattrix
                              > > 0 1 1 1
                              > > 1 f r q
                              > > 1 r g p
                              > > 1 q p h
                              >
                              > If M is the matrix of the conic C0, M* the adjugate matrix
                              > of M, ie
                              > gh-pp  pq-hr  pr-gq
                              > pq-hr  hf-qq  rq-fp
                              > pr-gq  rq-fp  fg-rr
                              > and U the column matrix
                              > u
                              > v
                              > w
                              > then U* = M*U is the pole wrt C0 of the line L of equation
                              > ux+vy+wz=0
                              > Put T(u,v,w) = t(U)U*  (t for transposition) T =
                              > quadratic form with matrix M*
                              > Then L and C0 have 0,1,2 real common points according to
                              > the fact that T>0, T=0, T<0
                              > More over, the determinant of the matrix
                              > 0 u v w
                              > u f r q
                              > v r g p
                              > w q p h
                              > is -T.
                              > The matrix of the singular conic union of the tangents to
                              > C0 from the pole of L is T.M-det(M).U.t(U)
                              > With u=v=w=1, we get the center of C0, the number of real
                              > infinite points ie the kind of the conic and the equation of
                              > the pair of asymptots.
                              > Friendly. Jean-Pierre
                              >
                              >
                              >
                              >
                              > ------------------------------------
                              >
                              > Yahoo! Groups Links
                              >
                              >
                              >     Hyacinthos-fullfeatured@yahoogroups.com
                              >
                              >
                              >
                            • jpehrmfr
                              Dear Nikoalaos ... U* = M*.U If U is the column matrix u v w, U* is the column of the coordinates of the pole of the line ux+vy+wz=0 JP
                              Message 14 of 25 , Jan 12, 2011
                                Dear Nikoalaos
                                > M* means the adjoint matrix of M
                                > U* what is?
                                U* = M*.U
                                If U is the column matrix u v w, U* is the column of the coordinates of the pole of the line ux+vy+wz=0
                                JP
                              • Chris Van Tienhoven
                                Dear Friends, What is the condition for a conic to be an orthogonal hyperbola, related to the conic f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0
                                Message 15 of 25 , Jan 15, 2011
                                  Dear Friends,

                                  What is the condition for a conic to be an orthogonal hyperbola,
                                  related to the conic
                                  f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0
                                  with matrix
                                  | f r q |
                                  | r g p |
                                  | q p h |

                                  best regards,

                                  Chris van Tienhoven

                                  >[ND]
                                  > The conic
                                  > f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0
                                  > with matrix
                                  > f r q
                                  > r g p
                                  > q p h
                                  > is a circle when
                                  > (g + h -2p)/aa = (h + f -2q)/bb = (f + q -2r)/cc.

                                  > > Lim Sung Hyun <bbbbbblow@> wrote:
                                  > > > What is the condition for a conic to be circle,
                                  > > preferably as a condition
                                  > > > imposed on the matrix of conic?
                                • Bernard Gibert
                                  Dear Chris, ... a^2 f + b^2 g + c^2 h - 2 (SA p + SB q + SC r)=0 best regards Bernard [Non-text portions of this message have been removed]
                                  Message 16 of 25 , Jan 15, 2011
                                    Dear Chris,

                                    > What is the condition for a conic to be an orthogonal hyperbola,
                                    > related to the conic
                                    > f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0
                                    > with matrix
                                    > | f r q |
                                    > | r g p |
                                    > | q p h |


                                    a^2 f + b^2 g + c^2 h - 2 (SA p + SB q + SC r)=0

                                    best regards

                                    Bernard

                                    [Non-text portions of this message have been removed]
                                  • Nikolaos Dergiades
                                    Dear Cris, A little explanation to Bernard s message. If (x1 : y1 : z1) and (x2 : y2 : z2) in bar.. are the infinite points of the hyperbola then these points
                                    Message 17 of 25 , Jan 15, 2011
                                      Dear Cris,
                                      A little explanation
                                      to Bernard's message.
                                      If (x1 : y1 : z1) and (x2 : y2 : z2) in bar..
                                      are the infinite points of the hyperbola
                                      then these points are orthogonal
                                      if x1x2S_A + y1y2S_B + z1z2S_C = 0. (1)
                                      If we solve the system of equations
                                      x + y + z = 0 and
                                      fx^2 + gy^2 + hz^2 + 2pyz + 2qzx + 2rxy = 0
                                      by putting z = - x - y
                                      the discriminant of the resulting equation F(x/y) = 0
                                      must be negative. Hence if d = det of
                                      0 1 1 1
                                      1 f r q
                                      1 r g p
                                      1 q p h
                                      we must have d < 0 and since from F(x/y) = 0
                                      x1x2 / y1y2 = (g + h - 2p) / (h + p - 2q)
                                      we conclude that (1) becomes
                                      (g + h - 2p)S_A + (h + p - 2q)S_B + (f + g - 2r)S_C = 0
                                      or
                                      (f+p-q-r)aa + (g+q-r-p)bb + (h+r-p-q)cc = 0
                                      Best regards
                                      Nikos Dergiades

                                      [CVT]
                                      > > What is the condition for a conic to be an orthogonal
                                      > hyperbola,
                                      > > related to the conic
                                      > > f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y =
                                      > 0
                                      > > with matrix
                                      > > | f r q |
                                      > > | r g p |
                                      > > | q p h |
                                      >
                                      [BG]
                                      > a^2 f + b^2 g + c^2 h - 2 (SA p + SB q + SC r)=0
                                      >
                                      >
                                    • Nikolaos Dergiades
                                      Sorry. the discriminant of the resulting equation F(x/y) = 0 must be positive. Hence if d = det of 0 1 1 1 1 f r q 1 r g p 1 q p h we must have d 0 ND
                                      Message 18 of 25 , Jan 15, 2011
                                        Sorry.
                                        the discriminant of the resulting equation F(x/y) = 0
                                        must be positive. Hence if d = det of
                                        0 1 1 1
                                        1 f r q
                                        1 r g p
                                        1 q p h
                                        we must have d > 0

                                        ND

                                        --- Στις Σάβ., 15/01/11, ο/η Nikolaos Dergiades <ndergiades@...> έγραψε:

                                        > Από: Nikolaos Dergiades <ndergiades@...>
                                        > Θέμα: Re: [EMHL] Re: Circle in Barycentrics
                                        > Προς: Hyacinthos@yahoogroups.com
                                        > Ημερομηνία: Σάββατο, 15 Ιανουάριος 2011, 12:16
                                        > Dear Cris,
                                        > A little explanation
                                        > to Bernard's message.
                                        > If (x1 : y1 : z1) and (x2 : y2 : z2) in bar..
                                        > are the infinite points of the hyperbola
                                        > then these points are orthogonal
                                        > if x1x2S_A + y1y2S_B + z1z2S_C = 0. (1)
                                        > If we solve the system of equations
                                        > x + y + z = 0 and
                                        > fx^2 + gy^2 + hz^2 + 2pyz + 2qzx + 2rxy = 0
                                        > by putting z = - x - y
                                        > the discriminant of the resulting equation F(x/y) = 0
                                        > must be negative. Hence if d = det of
                                        > 0 1 1 1
                                        > 1 f r q
                                        > 1 r g p
                                        > 1 q p h
                                        > we must have d < 0 and since from F(x/y) = 0
                                        > x1x2 / y1y2 = (g + h - 2p) / (h + p - 2q)
                                        > we conclude that (1) becomes
                                        > (g + h - 2p)S_A + (h + p - 2q)S_B + (f + g - 2r)S_C = 0
                                        > or
                                        > (f+p-q-r)aa + (g+q-r-p)bb + (h+r-p-q)cc = 0
                                        > Best regards
                                        > Nikos Dergiades
                                        >
                                        > [CVT]
                                        > > > What is the condition for a conic to be an
                                        > orthogonal
                                        > > hyperbola,
                                        > > > related to the conic
                                        > > > f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x
                                        > y =
                                        > > 0
                                        > > > with matrix
                                        > > > | f r q |
                                        > > > | r g p |
                                        > > > | q p h |
                                        > >
                                        > [BG]
                                        > > a^2 f + b^2 g + c^2 h - 2 (SA p + SB q + SC r)=0
                                        > >
                                        > >
                                        >
                                        >
                                        >
                                        >
                                        >
                                        > ------------------------------------
                                        >
                                        > Yahoo! Groups Links
                                        >
                                        >
                                        >     Hyacinthos-fullfeatured@yahoogroups.com
                                        >
                                        >
                                        >
                                      • Lim Sung Hyun
                                        Dear all, What is the condition for conic to have eccentricity e? We might infer from the following: Circle - infty - (g+h-2p::)=K Rect.Hyp - sqrt(2) -
                                        Message 19 of 25 , Jan 15, 2011
                                          Dear all,

                                          What is the condition for conic to have eccentricity e?

                                          We might infer from the following:

                                          Circle -> \infty -> (g+h-2p::)=K
                                          Rect.Hyp -> sqrt(2) -> (f+p-q-r::)K lies on line at infinity

                                          Sung Hyun


                                          [Non-text portions of this message have been removed]
                                        • jpehrmfr
                                          Dear Lim Sung Hyun ... Put U=a^2(f+p-q-r)+b^2(g+q-r-p)+c^2(h+r-p-q) u=(g+h-2p)/a^2; v=(h+f-2q)/b^2; w=(f+g-2r)/c^2; V=SA(v-w)^2+SB(w-u)^2+SC(u-v)^2 Then (if I
                                          Message 20 of 25 , Jan 15, 2011
                                            Dear Lim Sung Hyun

                                            > What is the condition for conic to have eccentricity e?
                                            >
                                            > We might infer from the following:
                                            >
                                            > Circle -> \infty -> (g+h-2p::)=K
                                            > Rect.Hyp -> sqrt(2) -> (f+p-q-r::)K lies on line at infinity

                                            Put U=a^2(f+p-q-r)+b^2(g+q-r-p)+c^2(h+r-p-q)
                                            u=(g+h-2p)/a^2; v=(h+f-2q)/b^2; w=(f+g-2r)/c^2;
                                            V=SA(v-w)^2+SB(w-u)^2+SC(u-v)^2
                                            Then (if I didn't mistake)
                                            a^2.b^2.c^2.(e^2-2)^2.V = e^2.U^2
                                            Friendly. Jean-Pierre
                                          • Lim Sung Hyun
                                            Dear Jean-Pierre, Put U=a^2(f+p-q-r)+b^2(g+q-r-p)+c^2(h+r-p-q) ... a^2.b^2.c^2.(e^2-2)^2.V = e^2.U^2 shouldn t e^2 be e^4 instead? Also, may I know how you got
                                            Message 21 of 25 , Jan 15, 2011
                                              Dear Jean-Pierre,

                                              Put U=a^2(f+p-q-r)+b^2(g+q-r-p)+c^2(h+r-p-q)
                                              > u=(g+h-2p)/a^2; v=(h+f-2q)/b^2; w=(f+g-2r)/c^2;
                                              > V=SA(v-w)^2+SB(w-u)^2+SC(u-v)^2
                                              > Then (if I didn't mistake)
                                              > a^2.b^2.c^2.(e^2-2)^2.V = e^2.U^2
                                              > Friendly. Jean-Pierre
                                              >

                                              a^2.b^2.c^2.(e^2-2)^2.V = e^2.U^2
                                              shouldn't e^2 be e^4 instead?

                                              Also, may I know how you got the expressions?

                                              Regards,
                                              Sung Hyun


                                              [Non-text portions of this message have been removed]
                                            • jpehrmfr
                                              Dear Lim Sung Hyun [JPE] ... [LDH] ... Yes, you re right; thank you for the correction and sorry for the typo ; the right member should be (e^2.U)^2 ... The
                                              Message 22 of 25 , Jan 15, 2011
                                                Dear Lim Sung Hyun
                                                [JPE]
                                                > Put U=a^2(f+p-q-r)+b^2(g+q-r-p)+c^2(h+r-p-q)
                                                > > u=(g+h-2p)/a^2; v=(h+f-2q)/b^2; w=(f+g-2r)/c^2;
                                                > > V=SA(v-w)^2+SB(w-u)^2+SC(u-v)^2
                                                > > Then (if I didn't mistake)
                                                > > a^2.b^2.c^2.(e^2-2)^2.V = e^2.U^2
                                                > > Friendly. Jean-Pierre
                                                [LDH]
                                                > a^2.b^2.c^2.(e^2-2)^2.V = e^2.U^2
                                                > shouldn't e^2 be e^4 instead?

                                                Yes, you're right; thank you for the correction and sorry for the typo ; the right member should be (e^2.U)^2

                                                > Also, may I know how you got the expressions?

                                                The conic has the same points at infinity than the circumconic with perspector g+h-2p:h+f-2q:f+g-2r because
                                                fxx+gyy+hzz+2pyz+2qzx+2rxy+(g+h-2p)yz+(h+f-2q)zx+(f+g-2r)xy=
                                                (x+y+z)(fx+gy+hz)
                                                Now, we get a problem that we have discussed some weeks ago.
                                                The formula a^2.b^2.c^2.(e^2-2)^2.V = e^4.U^2 is not the best one but I have tried, as you wanted, to have in the formula the conditions for the conic to be a circle and to be a rectangular hyperbola
                                                We have a shorter formula if we use the condition for the conic to be a parabola :
                                                Put Q = cyclicsum(hg-pp+2qr-2pf) (3 terms)
                                                Q is the sum of the 9 elements of the adjugate matrix M* of the matrix M of the conic, so Q is the image of (1,1,1) under the quadratic form with matrix M* and Q =0 means that the conic touches the line at infinity (Q>0 for an ellipse, Q=0 for a parabola, Q<0 for an hyperbola)
                                                Then we have
                                                (1-e^2).U^2 = 4.S^2.Q.(2-e^2)^2 with S =area(ABC)
                                                Friendly. Jean-Pierre
                                              • jpehrmfr
                                                Dear Lim Sung Hyun and Nikolaos Just a little remark I wrote ... (x+y+z)(fx+gy+hz) This proves immediately the Nikos condition for a conic to be a circle (the
                                                Message 23 of 25 , Jan 15, 2011
                                                  Dear Lim Sung Hyun and Nikolaos
                                                  Just a little remark
                                                  I wrote
                                                  > The conic has the same points at infinity than the circumconic with perspector g+h-2p:h+f-2q:f+g-2r because
                                                  > fxx+gyy+hzz+2pyz+2qzx+2rxy+(g+h-2p)yz+(h+f-2q)zx+(f+g-2r)xy=
                                                  (x+y+z)(fx+gy+hz)

                                                  This proves immediately the Nikos condition for a conic to be a circle (the perspector of the circumconic must be aa:bb:cc)

                                                  Jean-Pierre
                                                • Ricardo Barroso
                                                  Hello everyone Hyacinthos In http://personal.us.es/rbarroso/trianguloscabri/ has published 600 journal issue. It contains four research articles on the
                                                  Message 24 of 25 , Jan 16, 2011
                                                    Hello everyone
                                                    Hyacinthos


                                                    In

                                                    http://personal.us.es/rbarroso/trianguloscabri/



                                                    has published 600 journal issue.



                                                    It contains four research articles on the geometry of the triangle and an interview with a Spanish problemist.



                                                    Best regards



                                                    Ricardo















                                                    [Non-text portions of this message have been removed]
                                                  • Eisso Atzema
                                                    Two possible alternatives (I did not do the actual work): * Let Gamma be the equation of the hyperbola. Then, the homogeneous pairs h1=[p1:q1] and h2=[p2:q2]
                                                    Message 25 of 25 , Jan 16, 2011
                                                      Two possible alternatives (I did not do the actual work):

                                                      * Let Gamma be the equation of the hyperbola. Then, the homogeneous
                                                      pairs h1=[p1:q1] and h2=[p2:q2] corresponding to the points at infinity
                                                      H1=[p1:q1:-p1-q1] and H2=[p2:q2:-p2-q2] of the hyperbola are found as
                                                      the solutions of the equation Gamma([p:q:-p-q])=0. For the hyperbola to
                                                      be a right hyperbola, H1 and H2 now have to be conjugate with respect to
                                                      the circumcircle of the base triangle. As the condition for conjugacy
                                                      can be expressed in terms of the coefficients of the equation for h1 and
                                                      h2 (or so it seems), we get our condition for a right hyperbola. I am
                                                      only wondering how this can lead to a set of equations...

                                                      * For any right hyperbola all conjugate diameters are perpendicular to
                                                      each other. As we know the center C of the hyperbola, we can take any
                                                      line l through it, find its pole L and determine the line CL. As the
                                                      points at infinity of l and CL now have to be conjugate with respect to
                                                      the circumcircle of the base triangle, we have a condition that has to
                                                      be satisfied for all possible l. This should imply a vanishing matrix
                                                      and hence a set of conditions on the coefficients of the equation of the
                                                      hyperbola.

                                                      In general, if we want to know the angle between the asymptotes of a
                                                      hyperbola, or the eccentricity of an ellipse for that matter, we can use
                                                      the Laguerre/Cayley/Klein formula, i.e the equality between twice the
                                                      angle between two lines and the absolute value of logarithm of the cross
                                                      ratio of the points at infinity of the lines and the two circular points
                                                      of the plane. The computations, however, obviously are not going to be
                                                      pretty. In the case of an ellipse this leads to an expression for
                                                      (a-b)/(a+b) (or its reciprocal) (where a, b are the lengths of the
                                                      semi-axes) which gives us 4 candidates for the eccentricity.

                                                      Eisso

                                                      On 1/16/11 2:39 AM, jpehrmfr wrote:
                                                      >
                                                      >
                                                      >
                                                      > Dear Lim Sung Hyun and Nikolaos
                                                      > Just a little remark
                                                      > I wrote
                                                      > > The conic has the same points at infinity than the circumconic with
                                                      > perspector g+h-2p:h+f-2q:f+g-2r because
                                                      > > fxx+gyy+hzz+2pyz+2qzx+2rxy+(g+h-2p)yz+(h+f-2q)zx+(f+g-2r)xy=
                                                      > (x+y+z)(fx+gy+hz)
                                                      >
                                                      > This proves immediately the Nikos condition for a conic to be a circle
                                                      > (the perspector of the circumconic must be aa:bb:cc)
                                                      >
                                                      > Jean-Pierre
                                                      >
                                                      >


                                                      --

                                                      ========================================
                                                      Eisso J. Atzema, Ph.D.
                                                      Department of Mathematics& Statistics
                                                      University of Maine
                                                      Orono, ME 04469
                                                      Tel.: (207) 581-3928 (office)
                                                      (207) 866-3871 (home)
                                                      Fax.: (207) 581-3902
                                                      E-mail: atzema@...
                                                      ========================================



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