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7 Gs

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  • Antreas P. Hatzipolakis
    Let ABC be a triangle, P = (x:y:z) a point, and A B C the cevian triangle of P. Gab := the centroid of A BP, Gac := the centroid of A CP Gbc := the centroid
    Message 1 of 18 , Sep 6, 2004
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      Let ABC be a triangle, P = (x:y:z) a point, and A'B'C'
      the cevian triangle of P.

      Gab := the centroid of A'BP, Gac := the centroid of A'CP
      Gbc := the centroid of B'CP, Gba := the centroid of B'AP
      Gca := the centroid of C'AP, Gcb := the centroid of C'BP

      G1 := the centroid of GabGbcGca
      G2 := the centroid of GbaGcbGac

      We have G1 = G2 := P*.

      Which are the h. coordinates of P* ?


      APH
      --
    • peter_mows
      Dear Antreas, ... I think barys are P* = (y + z) (6 x^3 + 8 x^2 y + 2 x y^2 + 8 x^2 z + 8 x y z + y^2 z + 2 x z^2 + y z^2) : : Also, let Ga = midpoint[Gba,
      Message 2 of 18 , Sep 6, 2004
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        Dear Antreas,

        > Let ABC be a triangle, P = (x:y:z) a point, and A'B'C'
        > the cevian triangle of P.
        >
        > Gab := the centroid of A'BP, Gac := the centroid of A'CP
        > Gbc := the centroid of B'CP, Gba := the centroid of B'AP
        > Gca := the centroid of C'AP, Gcb := the centroid of C'BP
        >
        > G1 := the centroid of GabGbcGca
        > G2 := the centroid of GbaGcbGac
        >
        > We have G1 = G2 := P*.
        >
        > Which are the h. coordinates of P* ?
        >

        I think barys are

        P* = (y + z) (6 x^3 + 8 x^2 y + 2 x y^2 + 8 x^2 z + 8 x y z + y^2 z
        + 2 x z^2 + y z^2) : :

        Also, let

        Ga = midpoint[Gba, Gca]
        Gb = midpoint[Gcb, Gab]
        Ga = midpoint[Gac, Gbc]

        Ga' = midpoint[Gab, Gac]
        Gb' = midpoint[Gbc, Gba]
        Ga' = midpoint[Gca, Gcb]

        P* is the centroid of GaGbGc & Ga'Gb'Gc'

        Also Ga,Gb,Gc is perspective to ABC at
        x (y + z) / (x + 3 y + 3 z) : :

        Ga'' = Gba, Gbc /\ Gca, Gcb
        Gb'' = Gcb, Gca /\ Gab, Gac
        Gc'' = Gac, Gab /\ Gbc, Gba

        Ga''Gb''Gc'' perspective to ABC at P.

        Best regards

        Peter J. C .Moses.
      • Antreas P. Hatzipolakis
        Dear Peter, ... [PM] ... Thanks !! Now, Fab := the centroid of A BA, Fac := the centroid of A CA Fbc := the centroid of B CB, Fba := the centroid of B AB Fca
        Message 3 of 18 , Sep 7, 2004
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          Dear Peter,

          [APH]:
          >> Let ABC be a triangle, P = (x:y:z) a point, and A'B'C'
          >> the cevian triangle of P.
          >>
          >> Gab := the centroid of A'BP, Gac := the centroid of A'CP
          >> Gbc := the centroid of B'CP, Gba := the centroid of B'AP
          >> Gca := the centroid of C'AP, Gcb := the centroid of C'BP
          >>
          >> G1 := the centroid of GabGbcGca
          >> G2 := the centroid of GbaGcbGac
          >>
          >> We have G1 = G2 := P*.
          >>
          >> Which are the h. coordinates of P* ?
          >>

          [PM]
          (with little editing):
          >I think barys are
          >
          >P* = (y + z) (6 x^3 + 8 x^2 y + 2 x y^2 + 8 x^2 z + 8 x y z + y^2 z
          >+ 2 x z^2 + y z^2) : :
          >
          >Also, let
          >
          >Ga = midpoint[Gba, Gca]
          >Gb = midpoint[Gcb, Gab]
          >Gc = midpoint[Gac, Gbc]
          >
          >Ga' = midpoint[Gab, Gac]
          >Gb' = midpoint[Gbc, Gba]
          >Gc' = midpoint[Gca, Gcb]
          >
          >P* is the centroid of GaGbGc & Ga'Gb'Gc'
          >
          >Also GaGbGc is perspective to ABC at
          >x (y + z) / (x + 3 y + 3 z) : :
          >
          >Ga'' = GbaGbc /\ GcaGcb
          >Gb'' = GcbGca /\ GabGac
          >Gc'' = GacGab /\ GbcGba
          >
          >Ga''Gb''Gc'' is perspective to ABC at P.


          Thanks !!

          Now,

          Fab := the centroid of A'BA, Fac := the centroid of A'CA
          Fbc := the centroid of B'CB, Fba := the centroid of B'AB
          Fca := the centroid of C'AC, Fcb := the centroid of C'BC

          F1 := the centroid of FabFbcFca
          F2 := the centroid of FbaFcbFac

          We have F1 = F2 := F#

          Coordinates?

          I guess that we have analogous properties for Fa, Fa'
          (defined analogously to Ga, Ga')

          Also, for which Ps the points Gs or Fs are lying on conics ?


          Greetings

          Antreas

          --
        • peter_mows
          Dear Antreas, [APH] ... F# = (y + z) (4 x^2 + 3 x y + 3 x z + 2 y z) : : some ETC points, {2,2},{4,373},{30,1651},{99,1649} Fa = midpoint[Fba, Fca] Fb =
          Message 4 of 18 , Sep 7, 2004
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            Dear Antreas,

            [APH]
            > Now,
            >
            > Fab := the centroid of A'BA, Fac := the centroid of A'CA
            > Fbc := the centroid of B'CB, Fba := the centroid of B'AB
            > Fca := the centroid of C'AC, Fcb := the centroid of C'BC
            >
            > F1 := the centroid of FabFbcFca
            > F2 := the centroid of FbaFcbFac
            >
            > We have F1 = F2 := F#
            >
            > Coordinates?
            >
            > I guess that we have analogous properties for Fa, Fa'
            > (defined analogously to Ga, Ga')
            >
            > Also, for which Ps the points Gs or Fs are lying on conics ?

            F# = (y + z) (4 x^2 + 3 x y + 3 x z + 2 y z) : :

            some ETC points, {2,2},{4,373},{30,1651},{99,1649}

            Fa = midpoint[Fba, Fca]
            Fb = midpoint[Fcb, Fab]
            Fa = midpoint[Fac, Fbc]

            Fa' = midpoint[Fab, Fac]
            Fb' = midpoint[Fbc, Fba]
            Fa' = midpoint[Fca, Fcb]

            Fa'' = midpoint[Fcb, Fbc]
            Fb'' = midpoint[Fca, Fac]
            Fc'' = midpoint[Fab, Fba]

            I reckon that
            F# is the centroid of FaFbFc, Fa'Fb'Fc' & Fa''Fb''Fc''

            and if

            Fa* = Fba, Fbc /\ Fca, Fcb
            Fb* = Fcb, Fca /\ Fab, Fac
            Fc* = Fac, Fab /\ Fbc, Fba

            then
            Fa*= Fb*= Fc* = G.

            Does not seem that the F's have exactly analogous properties to the
            G's.

            Best regards
            Peter
          • Antreas P. Hatzipolakis
            Dear Peter We have seen Gs,Fs; now how about Ds? Defined as follows: A B C := the cevian triangle of P = (x:y:z) Dab := the centroid of B C B Dac := the
            Message 5 of 18 , Sep 7, 2004
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              Dear Peter

              We have seen Gs,Fs; now how about Ds?
              Defined as follows:

              A'B'C' := the cevian triangle of P = (x:y:z)

              Dab := the centroid of B'C'B
              Dac := the centroid of C'B'C

              Dbc := the centroid of C'A'C
              Dba := the centroid of A'C'A

              Dca := the centroid of A'B'A
              Dcb := the centroid of B'A'B


              D1 := the centroid of DabDbcDca
              D2 := the centroid of DbaDcbDac

              I think that D1 = D2
              If so, then let D1 = D2 := P'

              Coordinates of P'?

              We can also define Da, etc, analogously to Ga, etc.


              Greetings

              Antreas

              --
            • Bernard Gibert
              Dear friends, let Q be the centroid of the cevian triangle of P. to construct P with given point Q. Best regards Bernard
              Message 6 of 18 , Sep 7, 2004
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                Dear friends,

                let Q be the centroid of the cevian triangle of P.

                to construct P with given point Q.

                Best regards

                Bernard
              • Teacher
                Dear Bernard! ... Let P has homogenius barycentric coordinates P(x:y:z) (in base of ABC), then Q has coordinates x(s^2-x^2):y(s^2-y^2):z(s^2-z^2), where
                Message 7 of 18 , Sep 7, 2004
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                  Dear Bernard!
                  >
                  > let Q be the centroid of the cevian triangle of P.
                  >
                  > to construct P with given point Q.
                  >
                  >
                  Let P has homogenius barycentric coordinates P(x:y:z) (in base of ABC),
                  then Q has coordinates x(s^2-x^2):y(s^2-y^2):z(s^2-z^2), where s=x+y+z.
                  So, we have equation of 3-d degree to find x,y,z and I think that in
                  general we can't construct P by ruler and compass.
                  Best regards,
                  Alex
                • peter_mows
                  Dear Antreas, [APH] ... I also think D1 = D2. P = (y + z) (5 x^2 + 3 x y + 3 x z + y z) : : Da = midpoint[Dba, Dca] Db = midpoint[Dcb, Dab] Da = midpoint[Dac,
                  Message 8 of 18 , Sep 7, 2004
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                    Dear Antreas,

                    [APH]
                    >
                    > A'B'C' := the cevian triangle of P = (x:y:z)
                    >
                    > Dab := the centroid of B'C'B
                    > Dac := the centroid of C'B'C
                    >
                    > Dbc := the centroid of C'A'C
                    > Dba := the centroid of A'C'A
                    >
                    > Dca := the centroid of A'B'A
                    > Dcb := the centroid of B'A'B
                    >
                    >
                    > D1 := the centroid of DabDbcDca
                    > D2 := the centroid of DbaDcbDac
                    >
                    > I think that D1 = D2
                    > If so, then let D1 = D2 := P'
                    >
                    > Coordinates of P'?
                    >
                    > We can also define Da, etc, analogously to Ga, etc.
                    >

                    I also think D1 = D2.

                    P' = (y + z) (5 x^2 + 3 x y + 3 x z + y z) : :

                    Da = midpoint[Dba, Dca]
                    Db = midpoint[Dcb, Dab]
                    Da = midpoint[Dac, Dbc]

                    Da' = midpoint[Dab, Dac]
                    Db' = midpoint[Dbc, Dba]
                    Da' = midpoint[Dca, Dcb]

                    Da'' = midpoint[Dcb, Dbc]
                    Db'' = midpoint[Dca, Dac]
                    Dc'' = midpoint[Dab, Dba]

                    P'is the centroid of DaDbDc, Da'Db'Dc' & Da''Db''Dc''

                    Da* = Dba, Dbc /\ Dca, Dcb
                    Db* = Dcb, Dca /\ Dab, Dac
                    Dc* = Dac, Dab /\ Dbc, Dba

                    Da*= Db*= Dc* = The centroid of the cevian of P.
                    = x (y + z) (2 x + y + z) : :

                    Some ETC points,
                    {1,1962},{2,2},{4,51},{7,354},{8,210},{20,154},
                    {144,165},{253,1853}

                    looks like
                    Dac, Dab, Da', Da* are colinear and parallel to BC.

                    Similiarily
                    Dbc, Dba, Db', Db* // to CA
                    Dca, Dcb, Dc', Dc* // to AB

                    Best regards
                    Peter.
                  • Antreas P. Hatzipolakis
                    ... [alex_geom@mtu-net.ru] ... Which is the locus of P such that G, P, Q are collinear? G = (1:1:1) P = (x:y:z) Q = (x(s^2-x^2):y(s^2-y^2):z(s^2-z^2)), if the
                    Message 9 of 18 , Sep 7, 2004
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                      [BG]:
                      >> let Q be the centroid of the cevian triangle of P.

                      [alex_geom@...]
                      >Let P has homogenius barycentric coordinates P(x:y:z) (in base of ABC),
                      >then Q has coordinates x(s^2-x^2):y(s^2-y^2):z(s^2-z^2), where s=x+y+z.

                      Which is the locus of P such that G, P, Q are collinear?

                      G = (1:1:1)
                      P = (x:y:z)
                      Q = (x(s^2-x^2):y(s^2-y^2):z(s^2-z^2)), if the above is correct

                      G,P,Q are collinear ==> Det. of the matrix = 0

                      ==> etc

                      Antreas
                      --
                    • Alexei Myakishev
                      Dear Antreas! ... So, the equation is x^3(y-z)+y^3(z-x)+z^3(x-y)=0 or, another form: (y^2-z^2)/x+(z^2-x^2)/y+(x^2-y^2)/z=0 I don,t know what it is, even I
                      Message 10 of 18 , Sep 7, 2004
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                        Dear Antreas!
                        > [BG]:
                        >>> let Q be the centroid of the cevian triangle of P.
                        >
                        > [alex_geom@...]
                        >> Let P has homogenius barycentric coordinates P(x:y:z) (in base of
                        >> ABC),
                        >> then Q has coordinates x(s^2-x^2):y(s^2-y^2):z(s^2-z^2), where
                        >> s=x+y+z.
                        >
                        > Which is the locus of P such that G, P, Q are collinear?
                        >
                        > G = (1:1:1)
                        > P = (x:y:z)
                        > Q = (x(s^2-x^2):y(s^2-y^2):z(s^2-z^2)), if the above is correct
                        >
                        > G,P,Q are collinear ==> Det. of the matrix = 0
                        >
                        > ==> etc
                        >
                        >
                        So, the equation is
                        x^3(y-z)+y^3(z-x)+z^3(x-y)=0
                        or, another form:
                        (y^2-z^2)/x+(z^2-x^2)/y+(x^2-y^2)/z=0
                        I don,t know what it is, even I don't know what interesting points lay
                        on it (exsept of course centroid).
                        Yours,
                        Alex (Alexei Myakishev)
                        And sorry for so silly nikname "Teacher" - that is simply because I am
                        using now the comp from the school where I'm working.
                      • Antreas P. Hatzipolakis
                        Dear Alexei ... This quartic is not listed in Bernard s curves: (unless I didn t see it!) http://perso.wanadoo.fr/bernard.gibert/relatedcurves.html We can get
                        Message 11 of 18 , Sep 7, 2004
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                          Dear Alexei

                          >> [BG]:
                          >>>> let Q be the centroid of the cevian triangle of P.
                          >>
                          >> [Alexei Myakishev]
                          >>> Let P has homogenius barycentric coordinates P(x:y:z) (in base of
                          >>> ABC),
                          >>> then Q has coordinates x(s^2-x^2):y(s^2-y^2):z(s^2-z^2), where
                          >>> s=x+y+z.
                          >>
                          >> Which is the locus of P such that G, P, Q are collinear?
                          >>
                          >> G = (1:1:1)
                          >> P = (x:y:z)
                          >> Q = (x(s^2-x^2):y(s^2-y^2):z(s^2-z^2)), if the above is correct
                          >>
                          >> G,P,Q are collinear ==> Det. of the matrix = 0
                          >>
                          >> ==> etc

                          [Alexei Myakishev]:
                          >So, the equation is
                          >x^3(y-z)+y^3(z-x)+z^3(x-y)=0
                          >or, another form:
                          >(y^2-z^2)/x+(z^2-x^2)/y+(x^2-y^2)/z=0
                          >I don,t know what it is, even I don't know what interesting points lay
                          >on it (exsept of course centroid).

                          This quartic is not listed in Bernard's curves:
                          (unless I didn't see it!)

                          http://perso.wanadoo.fr/bernard.gibert/relatedcurves.html

                          We can get more curves if we consider the isotomic points tP, tQ
                          of the above ones and make C(5,3) = 10 combinations, namely:

                          G = (1:1:1)
                          P = (x:y:z)
                          Q = (x(s^2-x^2):y(s^2-y^2):z(s^2-z^2))

                          tP = (1/x:1/y:1/z)
                          tQ = (1/x(s^2-x^2):1/y(s^2-y^2):1/z(s^2-z^2))

                          Loci of P such that be collinear the points:

                          1. G, P, Q (the above one)

                          2. G, P, tP (isotomic cubic with pivot G)

                          3. G, P, tQ

                          4. G, Q, tP

                          5. G, Q, tQ

                          6. G, tP, tQ

                          7. P, Q, tP

                          8. P, Q, tQ

                          9. P, tP, tQ

                          10. Q, tP, tQ

                          [Alexei Myakishev]:
                          >And sorry for so silly nikname "Teacher" - that is simply because I am
                          >using now the comp from the school where I'm working.

                          Well... you may give the above to your students, as exercises in
                          the Linear Algebra :-)


                          Greetings

                          Antreas


                          --
                        • Alexei Myakishev
                          Dear Antreas! ... We can also get some lines, for example: 1. P^3=(x^3:y^3:z^3);P;Q (Ofcourse, P^3=P^2*P in sence of multiplication of points according Paul
                          Message 12 of 18 , Sep 7, 2004
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                            Dear Antreas!
                            >
                            >>> [BG]:
                            >>>>> let Q be the centroid of the cevian triangle of P.
                            >>>
                            >>> [Alexei Myakishev]
                            >>>> Let P has homogenius barycentric coordinates P(x:y:z) (in base of
                            >>>> ABC),
                            >>>> then Q has coordinates x(s^2-x^2):y(s^2-y^2):z(s^2-z^2), where
                            >>>> s=x+y+z.
                            >>>
                            >>> Which is the locus of P such that G, P, Q are collinear?
                            >>>
                            >>> G = (1:1:1)
                            >>> P = (x:y:z)
                            >>> Q = (x(s^2-x^2):y(s^2-y^2):z(s^2-z^2)), if the above is correct
                            >>>
                            >>> G,P,Q are collinear ==> Det. of the matrix = 0
                            >>>
                            >>> ==> etc
                            >
                            > We can get more curves if we consider the isotomic points tP, tQ
                            > of the above ones and make C(5,3) = 10 combinations, namely:
                            >
                            > G = (1:1:1)
                            > P = (x:y:z)
                            > Q = (x(s^2-x^2):y(s^2-y^2):z(s^2-z^2))
                            >
                            > tP = (1/x:1/y:1/z)
                            > tQ = (1/x(s^2-x^2):1/y(s^2-y^2):1/z(s^2-z^2))

                            We can also get some lines, for example:
                            1. P^3=(x^3:y^3:z^3);P;Q
                            (Ofcourse, P^3=P^2*P in sence of multiplication of points according
                            Paul Yiu - but may be somebody can invent "direct" way of construction
                            of cube of a point?)
                            2. Let R=(y+z,x+z,y+x) - homotety image of P with center G and ratio
                            k=2.
                            So we have collineary points P*tR;P;Q
                            3. And, moreover, P^2*tR;P;Q
                            -->P*tR;P^2*tR;P;Q are collinear
                            (* means barycentric multiplication)

                            Best regards,
                            sincerely,
                            Alexei
                          • Bernard Gibert
                            Dear Alexei, ... here is one : tX is the isotomic conjugate of X, cX is the complement of X G is the centroid of ABC P is a point [P,Q] is the P-Ceva conjugate
                            Message 13 of 18 , Sep 8, 2004
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                              Dear Alexei,

                              > but may be somebody can invent "direct" way of construction
                              > of cube of a point?

                              here is one :

                              tX is the isotomic conjugate of X,
                              cX is the complement of X
                              G is the centroid of ABC
                              P is a point
                              [P,Q] is the P-Ceva conjugate of Q.

                              the lines P, ctP and G, tcP intersect at the (barycentric) square P^2
                              of P.

                              P^3 is the harmonic conjugate of [P,ctP] with respect to P^2 and
                              [P,P^2].

                              I'm sure there's a simpler construction.

                              Best regards

                              Bernard

                              [Non-text portions of this message have been removed]
                            • Vladimir Dubrovsky
                              Dear Ricardo! ... G, a point N on AC, a point L on AB and the straight r in which it is BC ========== The desired construction can be performed in 2 steps:
                              Message 14 of 18 , Dec 27, 2004
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                                Dear Ricardo!

                                > To construct ABC knowing:
                                G,
                                a point N on AC,
                                a point L on AB
                                and the straight r in which it is BC
                                ==========

                                The desired construction can be performed in 2 steps:

                                (1) we draw an arbitrary triangle A'B'C' with centroid G' and construct the
                                points N' and M' on sides A'C' and A'B' that are images of N and M under the
                                affine map f that takes ABC onto A'B'C';

                                (2) we find the vertex A of the unknown triangle as the image of A' under
                                the map inverse to f, which is uniquely defined by the given images G, N, M
                                of G', N', M'.

                                Namely, the equation |N', B'C'| : |G', B'C'| = |N, r| : |G, r|, where |X, l|
                                denotes the distance from X to line l, yields the distance |N', B'C'|, which
                                enables us to construct N' on A'C'. M' is constructed similarly.
                                Then, let A'G' meet N'M' at P'. We construct P on NM such that NP:PM =
                                N'P':P'M' and then A (on the line GP) such that GA : GP = G'A' : G'P'.
                                AN and AM meet r at C and B.


                                With best New Year wishes,

                                Vladimir Dubrovsky
                              • Vladimir Dubrovsky
                                Dear friends, coming back again to Ricardo s problem -- ... -- let me notice that since we can easily find the ratios in which points L and N divide their
                                Message 15 of 18 , Dec 29, 2004
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                                  Dear friends,

                                  coming back again to Ricardo's problem --

                                  > To construct ABC knowing:
                                  > G,
                                  > a point N on AC,
                                  > a point L on AB
                                  > and the straight r in which it is BC

                                  -- let me notice that since we can easily find the ratios in which points L
                                  and N divide their respective sides (they are the u and v used in Francois's
                                  solution), we know the barycentrics of these points. So the problem
                                  naturally generalizes to the following one:

                                  construct triangle ABC given three points X,Y,Z in its plane and their
                                  barycentric coordinates wrt to ABC.

                                  This boils down to solving a system of linear equations and consructing the
                                  solutions.
                                  A somewhat more geometric construction can be given in the spirit of my
                                  solution of the original problem: we draw an arbitrary triangle A'B'C', then
                                  find points X', Y', Z' with the given barycentrics wrt A'B'C', and then
                                  affinely map these points on XYZ; ABC is the image of A'B'C' under this
                                  mapping.
                                  In practice, we find the ratios in which, say, line A'B' divides segments
                                  X'Y' and X'Z', and construct points P and Q on lines XY and XZ that divide
                                  XY and XZ in these ratios, respectively. The line PQ will be just the side
                                  line AB. Two other side lines are constructed similarly.

                                  Best wishes for 2005,

                                  Vladimir Dubrovsky
                                • Ricardo Barroso
                                  Dear François, Vladimir, Jean. Pierre, Hyacinthos: In http://www.personal.us.es/rbarroso/trianguloGNLr2.htm is the solution of François with applet de
                                  Message 16 of 18 , Dec 30, 2004
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                                    Dear François, Vladimir, Jean. Pierre, Hyacinthos:

                                    In

                                    http://www.personal.us.es/rbarroso/trianguloGNLr2.htm

                                    is the solution of François with applet de Cabri-Java.

                                    Thanks for yours interest.

                                    Ricardo





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                                  • Steve Sigur
                                    Bernard, Do you have a similar construction for PQ. the barycentric product of P and Q. I have searched for this with no success. From the (right now warm)
                                    Message 17 of 18 , Jan 5, 2005
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                                      Bernard,

                                      Do you have a similar construction for PQ. the barycentric product of P
                                      and Q. I have searched for this with no success.

                                      From the (right now warm) south of USA,
                                      Steve


                                      On Sep 8, 2004, at 7:36 AM, Bernard Gibert wrote:

                                      >
                                      >> but may be somebody can invent "direct" way of construction
                                      >> of cube of a point?
                                      >
                                      > here is one :
                                      >
                                      > tX is the isotomic conjugate of X,
                                      > cX is the complement of X
                                      > G is the centroid of ABC
                                      > P is a point
                                      > [P,Q] is the P-Ceva conjugate of Q.
                                      >
                                      > the lines P, ctP and G, tcP intersect at the (barycentric) square P^2
                                      > of P.
                                      >
                                      > P^3 is the harmonic conjugate of [P,ctP] with respect to P^2 and
                                      > [P,P^2].
                                      >
                                      > I'm sure there's a simpler construction.
                                      >
                                      > Best regards
                                      >
                                      > Bernard
                                    • Bernard Gibert
                                      Dear Steve, nice to hear from you. ... I doubt there is a similar construction because of the disymmetry introduced by P and G, but maybe I m wrong... best
                                      Message 18 of 18 , Jan 6, 2005
                                      • 0 Attachment
                                        Dear Steve,

                                        nice to hear from you.

                                        > Do you have a similar construction for PQ. the barycentric product of
                                        > P
                                        > and Q. I have searched for this with no success.

                                        I doubt there is a similar construction because of the disymmetry
                                        introduced by P and G, but maybe I'm wrong...

                                        best regards

                                        Bernard


                                        >
                                        > On Sep 8, 2004, at 7:36 AM, Bernard Gibert wrote:
                                        >
                                        > >
                                        > >> but may be somebody can invent "direct" way of construction
                                        > >> of cube of a point?
                                        > >
                                        > > here is one :
                                        > >
                                        > > tX is the isotomic conjugate of X,
                                        > > cX is the complement of X
                                        > > G is the centroid of ABC
                                        > > P is a point
                                        > > [P,Q] is the P-Ceva conjugate of Q.
                                        > >
                                        > > the lines P, ctP and G, tcP intersect at the (barycentric) square
                                        > P^2
                                        > > of P.
                                        > >
                                        > > P^3 is the harmonic conjugate of [P,ctP] with respect to P^2 and
                                        > > [P,P^2].
                                        > >
                                        > > I'm sure there's a simpler construction.
                                        > >
                                        > > Best regards
                                        > >
                                        > > Bernard
                                        >
                                        >
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