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Re: [EMHL] Cevian Trace Equal Area Points

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  • Francois Rideau
    Dear Tuan Now, I tell you why I am so concerned by your results in choo-choo theory. One of the main result of this theory is the following: Given any 2
    Message 1 of 18 , Jul 31, 2007
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      Dear Tuan
      Now, I tell you why I am so concerned by your results in choo-choo theory.
      One of the main result of this theory is the following:
      Given any 2 inscribed triangles PaPbPc and UaUbUc of the reference triangle
      ABC, (cevian triangles are not needed!), there exists (in general), a unique
      point Q in the plane, such that:

      Area(QPaUa) = Area(QPbUb) = Area(QPcUc)

      In this formula , area means signed area of course!

      Now I give you a way (without proof for lack of room) to construct this
      point.
      1° Let E be the fixed point of the affine map f sending PaPbPc on UaUbUc.
      E is the equicenter, following Neuberg notation. Why equicenter? Simply
      because E has same barycentrics in PaPbPc and UaUbUc.
      There is a very known construction of E, due to Sollertinski. I have already
      talked about it in some previous posts.

      2° Let g be the affine map sending PaPbPc on ABC, then Q = g(E).
      It is also very easy to have a construction of Q.

      This area equality is the reason why I have called Q the areal center in
      choo-choo theory.



      Now in case triangles PaPbPc and QaQbQc are cevian triangles of P and U,
      your formula gives barycentrics of Q from the barycentrics of P and U in a
      very compact form though I don't like your using the cross difference which
      is an euclidian and not an affine notion.

      In the other way, the knowledge of the equicenter E and the areal center Q
      determines entirely affine one to one maps between sides of ABC such that if
      abc and a'b'c' are any 2 inscribed homologous triangles, then we have:
      Area (Qaa') = Area(Qbb') = Area(Qcc')

      But this is another story!

      Friendly
      Francois







      On 7/27/07, Quang Tuan Bui <bqtuan1962@...> wrote:
      >
      > Dear All My Friends,
      > Given triangle ABC and two points P, U with barycentrics P = (p : q : r),
      > U = (u : v : w). PaPbPc and UaUbUc are Cevian triangles of P, U
      > respectively.
      > We construct point Q = X(6)*c(P)*c(U)*t(cd(P, U))
      > here:
      > c(P) = complement of P
      > c(U) = complement of U
      > t(cd(P, U)) = isotomic conjugate of cross difference of P, U
      > * = barycentric product
      > The results:
      > a). Barycentrics of Q = (q + r)*(v + w)/(q*w - r*v) : :
      > b).
      > c). If P is on Lucas cubic then the locus of U such that Q is on
      > circumcircle is also Lucas cubic.
      > d). If P = t(U) then the locus of U such that Q is on circumcircle is
      > Lucas cubic (and three sidelines of antimedian triangle).
      > Best regards,
      > Bui Quang Tuan
      >
      >
      > ---------------------------------
      > Park yourself in front of a world of choices in alternative vehicles.
      > Visit the Yahoo! Auto Green Center.
      >
      > [Non-text portions of this message have been removed]
      >
      >
      >


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    • Nikolaos Dergiades
      Dear Francois thank you very much for explaining what is equicenter E and areal point Q. Another simple construction for Q comes from the following property:
      Message 2 of 18 , Jul 31, 2007
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        Dear Francois
        thank you very much for explaining
        what is equicenter E and areal point Q.
        Another simple construction for Q comes from the
        following property:

        If A'B'C', A"B"C" are two inscribed triangles
        in ABC we construct the line La that passes
        through A and is parallel to the line that connects
        the mid points of B'C", B"C' and similarly
        construct the lines Lb, Lc. These three lines
        are concurrent at Q the areal point such that
        the signed area (QA'A") = (QB'B") = (QC'C").

        If in barycentrics the points are
        A' = (0 : p : 1-p)
        B' = (1-q : 0 : q)
        C' = (r : 1-r : 0)
        A" = (0 : p' : 1-p')
        B" = (1-q' : 0 : q')
        C" = (r' : 1-r' : 0)
        then the point Q is the isotomic conjugate
        of (p-p' : q-q' : r-r') and the signed area is
        (ABC).(p-p').(q-q').(r-r')/K where
        K = pq+qr+rp+p'q'+q'r'+r'p'+pp'+qq'+rr'-
        -(p+q+r)(p'+q'+r').

        Best regards
        Nikos Dergiades

        > Given any 2 inscribed triangles PaPbPc and UaUbUc of
        > the reference triangle
        > ABC, (cevian triangles are not needed!), there
        > exists (in general), a unique
        > point Q in the plane, such that:
        >
        > Area(QPaUa) = Area(QPbUb) = Area(QPcUc)
        >
        > In this formula , area means signed area of course!
        >
        > Now I give you a way (without proof for lack of
        > room) to construct this
        > point.
        > 1° Let E be the fixed point of the affine map f
        > sending PaPbPc on UaUbUc.
        > E is the equicenter, following Neuberg notation. Why
        > equicenter? Simply
        > because E has same barycentrics in PaPbPc and
        > UaUbUc.
        > There is a very known construction of E, due to
        > Sollertinski. I have already
        > talked about it in some previous posts.
        >
        > 2° Let g be the affine map sending PaPbPc on ABC,
        > then Q = g(E).
        > It is also very easy to have a construction of Q.
        >
        > This area equality is the reason why I have called Q
        > the areal center in
        > choo-choo theory.




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      • Francois Rideau
        Anh Tuan oi Now, I criticize your formula: We construct point Q = X(6)*c(P)*c(U)*t(cd(P, U)) I find it too much complex for X(6) is not needed. You deduct it
        Message 3 of 18 , Aug 1, 2007
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          Anh Tuan oi
          Now, I criticize your formula:

          We construct point Q = X(6)*c(P)*c(U)*t(cd(P, U))

          I find it too much complex for X(6) is not needed.

          You deduct it from the formula:

          a). Barycentrics of Q = (q + r)*(v + w)/(q*w - r*v) : :



          I look at ETC glossary the meaning of the point:
          M = q*w - r*v::
          where P = (p:q:r) and Q = (u:v:w)
          M is the cross difference of P and Q only in case of using trilinears.
          But here, you are using barycentrics so X(6) appareance is artificial though
          needed to get a correct formula.

          I don't find in ETC glossary the point M = q*w -r*v:: where P =(p:q:r) and Q
          = (u:v:w) in using barycentrics;
          M is simply the dual point of the line through P and Q.
          Is there a special name for this point with such a projective definition?
          Calling it AT(P,Q) for the moment, then we will have:
          Q = c(P)*c(U)*t(AT(P,U))
          Friendly
          Francois

          >
          >
          >


          [Non-text portions of this message have been removed]
        • Quang Tuan Bui
          Dear Francois and Nikos, Thank you very much for your interesting and general analyses from general view point. I learn a lot from your messages. In fact I
          Message 4 of 18 , Aug 1, 2007
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            Dear Francois and Nikos,
            Thank you very much for your interesting and general analyses from general view point. I learn a lot from your messages. In fact I construct point Q as following (some more complicated than Nikos):
            PaPbPc, UaUbUc are two inscribed triangles in ABC
            Rb = reflection of Ub in midpoint of APb
            Rc = reflection of Uc in midpoint of APc
            La = line connected A and midpoint of RbRc
            Similarly construct Lb, Lc. Three lines La, Lb, Lc bound one triangle A'B'C'.
            Two triangles ABC and A'B'C' are perspective at a point Q.
            We can very easy to show that
            Area(QPaUa) = Area(QPbUb) = Area(QPcUc)
            Moreover:
            Area(A'PaUa) = -Area(A'PbUb) = -Area(A'PcUc)
            I have found this point and this construction by combine following small well known elementary locus problem into triangle:
            Two segments XY, X'Y' are on given two lines. To find the locus of Z such that Area(ZXY) = Area(ZX'Y').
            If areas are not signed then the locus is two lines passing intersection of two given lines. I apply this result for each vertex with two sidelines then I have got six lines as in my construction before.
            From this construction we can easy to get barycentrics of Q. In order to get triangle center I choose two Cevian triangles of P, U and get compact expression:
            Q = (q + r)*(v + w)/(q*w - r*v) : :
            From this expression I see that Q = X(6)*c(P)*c(U)*t(cd(P, U)).
            All another results I have got from this formula and by barycentric calculations.

            Thank you again and best regards,
            Bui Quang Tuan


            ---------------------------------
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          • Nikolaos Dergiades
            Dear friends, the point Q is on the line at infinity if the point (p-p : q-q : r-r ) is on the Steiner circumellipse. Hence the denominator K can be written
            Message 5 of 18 , Aug 1, 2007
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              Dear friends,
              the point Q is on the line at infinity if the
              point (p-p' : q-q' : r-r') is on the
              Steiner circumellipse.
              Hence the denominator K can be written another way
              K = xy+yz+zx where x=p-p', y=q-q', z=r-r'.
              or
              area = (ABC)/[1/(p-p')+1/(q-q')+1/(r-r')]
              = (ABC)/(BC/A"A' + CA/B"B' + AB/C"C')

              Best regards
              Nikos Dergiades

              > If A'B'C', A"B"C" are two inscribed triangles
              > in ABC we construct the line La that passes
              > through A and is parallel to the line that connects
              > the mid points of B'C", B"C' and similarly
              > construct the lines Lb, Lc. These three lines
              > are concurrent at Q the areal point such that
              > the signed area (QA'A") = (QB'B") = (QC'C").
              >
              > If in barycentrics the points are
              > A' = (0 : p : 1-p)
              > B' = (1-q : 0 : q)
              > C' = (r : 1-r : 0)
              > A" = (0 : p' : 1-p')
              > B" = (1-q' : 0 : q')
              > C" = (r' : 1-r' : 0)
              > then the point Q is the isotomic conjugate
              > of (p-p' : q-q' : r-r') and the signed area is
              > (ABC).(p-p').(q-q').(r-r')/K where
              > K = pq+qr+rp+p'q'+q'r'+r'p'+pp'+qq'+rr'-
              > -(p+q+r)(p'+q'+r').
              >
              > Best regards
              > Nikos Dergiades




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            • Nikolaos Dergiades
              Dear Tuan, ... Yes the same method I also used. If A is the intersection of XY, X Y and M, M are the mid points of XY , X Y then the locus such that the
              Message 6 of 18 , Aug 1, 2007
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                Dear Tuan,

                > I have found this point and this construction by
                > combine following small well known elementary locus
                > problem into triangle:
                > Two segments XY, X'Y' are on given two lines. To
                > find the locus of Z such that Area(ZXY) =
                > Area(ZX'Y').
                > If areas are not signed then the locus is two
                > lines passing intersection of two given lines.

                Yes the same method I also used.
                If A is the intersection of XY, X'Y' and
                M, M' are the mid points of XY', X'Y
                then the locus such that the signed areas
                are equal is the line parallel to MM' that
                passes through A.

                Best regards
                Nikos Dergiades




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              • Francois Rideau
                Dear Nikos and Tuan Thanks for your interesting remarks. Of course in choo-choo theory, these lines La, Lb, Lc play a central role and I have called them equal
                Message 7 of 18 , Aug 1, 2007
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                  Dear Nikos and Tuan
                  Thanks for your interesting remarks.
                  Of course in choo-choo theory, these lines La, Lb, Lc play a central role
                  and I have called them equal area axis.
                  Notice that their construction is affine! I had also my own construction
                  slightly different from yours.
                  As for Tuan formula, I would be happy to have a name if any for the dual
                  point M(q*w - r*v::)of the line through P(p:q:r) and Q(u:v:w) given by their
                  barycentrics.
                  Friendly
                  Francois
                  PS As Nikos notice, point isotomic of the areal center also plays an
                  important role in choo-choo theory.
                  As I go away from Paris in Britanny for several weeks even months far away
                  from the web, I give you a choo-choo configuration, so you can think about
                  me in this summer time:
                  Instead of cevian tiangles, I will look at pedal triangles PaPbPc and QaQbQc
                  of points P and Q wrt triangle ABC and I call L the line through P and Q.
                  Let E be the equicenter and S the areal center of the pedal triangles.
                  Then:
                  1° E is the orthopole of line L wrt ABC.
                  2° S is on the ABC-circumcircle and its Simson line is parallel to line L.

                  The first point was knew by Neuberg for a very very long time and maybe
                  that's why he found the orthopole. The first proof I saw was due to
                  V.Thebault
                  As for the second point, I would be happy to have some reference if any.
                  Of course these properties of points E and S are shared by any pair of
                  triangles of points on line L but this is another (choo-choo) story.


                  [Non-text portions of this message have been removed]
                • Francois Rideau
                  Dear Nikos and Tuan ... So if abc and a b c are 2 inscribed triangles in ABC and if we use ordinary area ( i.e not signed), the Tuan-Nikos six lines cut
                  Message 8 of 18 , Aug 2, 2007
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                    Dear Nikos and Tuan



                    > > Two segments XY, X'Y' are on given two lines. To
                    > > find the locus of Z such that Area(ZXY) =
                    > > Area(ZX'Y').
                    > > If areas are not signed then the locus is two
                    > > lines passing intersection of two given lines.
                    >
                    > I note this fact:
                    >
                    > These 2 lines of this locus are harmonic conjugate wrt lines XY and X'Y'.


                    So if abc and a'b'c' are 2 inscribed triangles in ABC and if we use ordinary
                    area ( i.e not signed), the Tuan-Nikos six lines cut themselves in 4
                    points forming an harmonic quad wrt ABC, each of these four points beeing an
                    equal area center wrt abc and a'b'c'. I will have to think about it in
                    choo-choo theory!
                    friendly
                    Francois

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                  • Francois Rideau
                    Dear Nikos I give you my own construction of the equal area axis based on vectors. So you could compare it with yours. Given 4 points A, B, A , B such that
                    Message 9 of 18 , Aug 2, 2007
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                      Dear Nikos
                      I give you my own construction of the equal area axis based on vectors.
                      So you could compare it with yours.
                      Given 4 points A, B, A', B' such that line AB and A'B' are through point O,
                      let L+ be the line locus of points M such that:
                      Area(MAB) = Area(MA'B')
                      and L- be the line locus of points M such that:
                      Area(MAB) = -Area(MA'B')
                      where Area means signed area.
                      Then L+ is the line through O directed by the vector V(AB) - V(A'B')
                      and L- is the line through O directed by the vector V(AB) + V(A'B').
                      That's why the four lines AB, A'B', L+, L- are in an harmonic range.
                      Friendly
                      Francois


                      On 8/2/07, Francois Rideau <francois.rideau@...> wrote:
                      >
                      > Dear Nikos and Tuan
                      >
                      >
                      >
                      > > > Two segments XY, X'Y' are on given two lines. To
                      > > > find the locus of Z such that Area(ZXY) =
                      > > > Area(ZX'Y').
                      > > > If areas are not signed then the locus is two
                      > > > lines passing intersection of two given lines.
                      > >
                      > > I note this fact:
                      > >
                      > > These 2 lines of this locus are harmonic conjugate wrt lines XY and
                      > > X'Y'.
                      >
                      >
                      > So if abc and a'b'c' are 2 inscribed triangles in ABC and if we use
                      > ordinary area ( i.e not signed), the Tuan-Nikos six lines cut themselves
                      > in 4 points forming an harmonic quad wrt ABC, each of these four points
                      > beeing an equal area center wrt abc and a'b'c'. I will have to think about
                      > it in choo-choo theory!
                      > friendly
                      > Francois
                      >
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                    • Nikolaos Dergiades
                      Dear Francois, [Tuan] ... [Francois] ... Yes. The line La+ is parallel to the segment of mid points of bc , b c and gives equal areas The line La- is parallel
                      Message 10 of 18 , Aug 2, 2007
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                        Dear Francois,

                        [Tuan]
                        > > > Two segments XY, X'Y' are on given two lines.
                        > To
                        > > > find the locus of Z such that Area(ZXY) =
                        > > > Area(ZX'Y').
                        > > > If areas are not signed then the locus is two
                        > > > lines passing intersection of two given lines.
                        > >
                        > > I note this fact:
                        > >
                        > > These 2 lines of this locus are harmonic conjugate
                        > wrt lines XY and X'Y'.
                        >
                        [Francois]
                        > So if abc and a'b'c' are 2 inscribed triangles in
                        > ABC and if we use ordinary
                        > area ( i.e not signed), the Tuan-Nikos six lines
                        > cut themselves in 4
                        > points forming an harmonic quad wrt ABC, each of
                        > these four points beeing an
                        > equal area center wrt abc and a'b'c'. I will have to
                        > think about it in
                        > choo-choo theory!

                        Yes.
                        The line La+ is parallel to the segment of
                        mid points of bc', b'c and gives equal areas
                        The line La- is parallel to the segment of
                        mid points of bb', cc' and gives opposite areas
                        La+ and La- pass through A. Similarly consider
                        the lines Lb+, Lb-, Lc+, Lc-.
                        The lines La+, Lb+, LC+ are concurrent at Q
                        The lines La+, Lb-, LC- are concurrent at Qa
                        The lines La-, Lb+, LC- are concurrent at Qb
                        The lines La-, Lb-, LC+ are concurrent at Qc
                        If the areas we discuss (Paa'), (Pbb'), (Pcc')
                        are S1(P), S2(P), S3(P) then
                        S1(Q) = S2(Q) = S3(Q)
                        -S1(Qa) = S2(Qa) = S3(Qa)
                        S1(Qb) = -S2(Qb) = S3(Qb)
                        S1(Qc) = S2(Qc) = -S3(Qc)

                        Best regards
                        Nikos Dergiades






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                      • Nikolaos Dergiades
                        Dear Francois, I ve sent you a message and now I saw your message that you sent it before mine. As you see we use almost the same symbolism. You conclude the
                        Message 11 of 18 , Aug 2, 2007
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                          Dear Francois,
                          I've sent you a message
                          and now I saw your message that you sent it
                          before mine.
                          As you see we use almost the same symbolism.
                          You conclude the harmonic relation using vectors
                          and I conclude it because my lines La+, La-, AB, AC
                          are parallel to the sides and diagonals of a
                          parallelogram.
                          Now I have an interesting result.
                          I investigated pedal triangles of
                          isogonal conjugate points.
                          If abc is the pedal triangle of P and
                          a'b'c' is the pedal triangle of P' isogonal
                          conjugate of P then the areal point Q of these
                          triangles lies on the circumcircle of ABC.
                          This is from a proof.
                          The following is from observation:
                          The point Q is the fourth intersection
                          of circumcircle with the rectangular circumhyperbola
                          of the circumcevian triangle of P, that passes
                          through P.
                          Similarly Q lies on the rectangular circumhyperbola
                          of the circumcevian triangle of P', that passes
                          through P'.

                          Best regards
                          Nikos Dergiades

                          > Dear Nikos
                          > I give you my own construction of the equal area
                          > axis based on vectors.
                          > So you could compare it with yours.
                          > Given 4 points A, B, A', B' such that line AB and
                          > A'B' are through point O,
                          > let L+ be the line locus of points M such that:
                          > Area(MAB) = Area(MA'B')
                          > and L- be the line locus of points M such that:
                          > Area(MAB) = -Area(MA'B')
                          > where Area means signed area.
                          > Then L+ is the line through O directed by the
                          > vector V(AB) - V(A'B')
                          > and L- is the line through O directed by the vector
                          > V(AB) + V(A'B').
                          > That's why the four lines AB, A'B', L+, L- are in an
                          > harmonic range.
                          > Friendly
                          > Francois




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                        • Francois Rideau
                          Dear Nikos Cheer up, I will look at your beautiful proof but notice that the areal point of any 2 pedal triangles is always on the circumcircle. I have several
                          Message 12 of 18 , Aug 2, 2007
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                            Dear Nikos
                            Cheer up, I will look at your beautiful proof but notice that the areal
                            point of any 2 pedal triangles is always on the circumcircle.
                            I have several proofs of this fact and if you like calculus, one of them is
                            based on the Morley trick, i.e to choose the circumcircle as the unit circle
                            in the complex plane.
                            For other Hyacinthists, I give a sketch of my construction of the equal area
                            axis.
                            It is based in the following result of linear algebra:
                            S(L,M,N) = f(LM,LN)
                            where S is the signed area function , f is some determinant function that is
                            to say a bilinear skew symmetric function of 2 vectors and LM and LN beeing
                            vectors.
                            So if lines AB and A'B' are on O, we have:
                            S(M,A,B) = f(MA,MB) = f(MO + OA, MO + OB) = f(MO,OB) + f(OA,MO) = f(MO,OB -
                            OA) = f(MO,AB)
                            In the same way: S(M,A',B') = f(MO, A'B')
                            Now if k is any real, we have:
                            S(M,A,B) -k.S(M,A',B') = f(MO, AB -k.A'B') and we are done.
                            Friendly
                            Francois

                            On 8/2/07, Nikolaos Dergiades <ndergiades@...> wrote:
                            >
                            > Dear Francois,
                            > I've sent you a message
                            > and now I saw your message that you sent it
                            > before mine.
                            > As you see we use almost the same symbolism.
                            > You conclude the harmonic relation using vectors
                            > and I conclude it because my lines La+, La-, AB, AC
                            > are parallel to the sides and diagonals of a
                            > parallelogram.
                            > Now I have an interesting result.
                            > I investigated pedal triangles of
                            > isogonal conjugate points.
                            > If abc is the pedal triangle of P and
                            > a'b'c' is the pedal triangle of P' isogonal
                            > conjugate of P then the areal point Q of these
                            > triangles lies on the circumcircle of ABC.
                            > This is from a proof.
                            > The following is from observation:
                            > The point Q is the fourth intersection
                            > of circumcircle with the rectangular circumhyperbola
                            > of the circumcevian triangle of P, that passes
                            > through P.
                            > Similarly Q lies on the rectangular circumhyperbola
                            > of the circumcevian triangle of P', that passes
                            > through P'.
                            >
                            > Best regards
                            > Nikos Dergiades
                            >
                            > > Dear Nikos
                            > > I give you my own construction of the equal area
                            > > axis based on vectors.
                            > > So you could compare it with yours.
                            > > Given 4 points A, B, A', B' such that line AB and
                            > > A'B' are through point O,
                            > > let L+ be the line locus of points M such that:
                            > > Area(MAB) = Area(MA'B')
                            > > and L- be the line locus of points M such that:
                            > > Area(MAB) = -Area(MA'B')
                            > > where Area means signed area.
                            > > Then L+ is the line through O directed by the
                            > > vector V(AB) - V(A'B')
                            > > and L- is the line through O directed by the vector
                            > > V(AB) + V(A'B').
                            > > That's why the four lines AB, A'B', L+, L- are in an
                            > > harmonic range.
                            > > Friendly
                            > > Francois
                            >
                            >
                            >
                            >
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                            >
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                          • Quang Tuan Bui
                            Dear Francois and Nikos, If we choose two inscribed triangles as Cevian triangles of two Brocard points then Q can be one triangle center with barycentrics: Q
                            Message 13 of 18 , Aug 3, 2007
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                              Dear Francois and Nikos,

                              If we choose two inscribed triangles as Cevian triangles of two Brocard points then Q can be one triangle center with barycentrics:
                              Q = 1/((a^4 - b^2*c^2)*(b^2 + c^2)) : :
                              Search value: -0.248703594444072
                              Q is on Steiner circumellipse, collinear with Brocard midpoint X(39) and Steiner point X(99).
                              Q is isotomic conjugate of X(732), one infinite point.

                              Have you all very nice weekend!
                              Best regards,
                              Bui Quang Tuan


                              Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                              Given triangle ABC and two points P, U with barycentrics P = (p : q : r), U = (u : v : w). PaPbPc and UaUbUc are Cevian triangles of P, U respectively.
                              We construct point Q = X(6)*c(P)*c(U)*t(cd(P, U))
                              here:
                              c(P) = complement of P
                              c(U) = complement of U
                              t(cd(P, U)) = isotomic conjugate of cross difference of P, U
                              * = barycentric product
                              The results:
                              a). Barycentrics of Q = (q + r)*(v + w)/(q*w - r*v) : :
                              b). Area(QPaUa) = Area(QPbUb) = Area(QPcUc)
                              c). If P is on Lucas cubic then the locus of U such that Q is on circumcircle is also Lucas cubic.
                              d). If P = t(U) then the locus of U such that Q is on circumcircle is Lucas cubic (and three sidelines of antimedian triangle).
                              Best regards,
                              Bui Quang Tuan


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                            • Andreas P. Hatzipolakis
                              Dear Nikos It was not an answer to me, but a comment to me as list-owner. [ND] ... [FR] ... ... and was [ the post] incomplete APH -- [Non-text portions of
                              Message 14 of 18 , Aug 4, 2007
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                                Dear Nikos

                                It was not an answer to me, but a comment to me as list-owner.

                                [ND]
                                >
                                >Dear Francois,
                                >You answered to Antreas, I think by mistake.
                                >I am not Antreas.


                                [FR]

                                > > Dear Antreas
                                > > Sorry, I send my post too soon!


                                ... and was [ the post] incomplete

                                APH





                                --

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                              • Quang Tuan Bui
                                Dear Francois and Nikos, We denote S(T1, T2) = equal area point of two inscribed triangles T1, T2. Of course S(T1, T2) = S(T2, T1). Suppose T1, T2, T3 are
                                Message 15 of 18 , Aug 5, 2007
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                                  Dear Francois and Nikos,

                                  We denote S(T1, T2) = equal area point of two inscribed triangles T1, T2. Of course S(T1, T2) = S(T2, T1).
                                  Suppose T1, T2, T3 are three inscribed triangles then three equal area points S(T1, T2), S(T2, T3), S(T3, T1) are always on one circumconic. Denote this circumconic as C(T1, T2, T3).

                                  Suppose now T1, T2, T3 are three Cevian triangles of P=X(2), U=X(4) and X respectively.
                                  There is one special locus:
                                  The locus of X such that circumconic C(T1, T2, T3) is passing also through X is one quintic:
                                  CyclicSum[ (b^2 - c^2)*y*z*(a^2*SA*(x^3 - y*z*(y + z - x)) + b^2*c^2*(x^3 - y*z*x)) ] = 0
                                  This circumquintic passes:
                                  - Vetices of antimedial triangle.
                                  - X(2), X(4), X(69), X(110), X(2574), X(2575)

                                  Best regards,
                                  Bui Quang Tuan


                                  Quang Tuan Bui <bqtuan1962@...> wrote: Given triangle ABC and two points P, U with barycentrics P = (p : q : r), U = (u : v : w). PaPbPc and UaUbUc are Cevian triangles of P, U respectively.
                                  We construct point Q = X(6)*c(P)*c(U)*t(cd(P, U))
                                  here:
                                  c(P) = complement of P
                                  c(U) = complement of U
                                  t(cd(P, U)) = isotomic conjugate of cross difference of P, U
                                  * = barycentric product
                                  The results:
                                  a). Barycentrics of Q = (q + r)*(v + w)/(q*w - r*v) : :
                                  b). Area(QPaUa) = Area(QPbUb) = Area(QPcUc)


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