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Foci of steiner inelipse

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  • Milorad Stevanovic
    Dear friends, by using of barycentrics I have proved the following Theorem. The unique points that are the Lemoine points of their pedal triangles are the foci
    Message 1 of 15 , Nov 28, 2007
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      Dear friends,

      by using of barycentrics I have proved the following

      Theorem.
      The unique points that are the Lemoine points
      of their pedal triangles are the foci of Steiner inellipse of triangle.

      Remark 1.
      In case of a>b>c we have the above points Q1 and Q2
      Q1(1+(b^2-c^2)/E1:1+(a^2-c^2)/E2:1-(a^2-b^2)/E3),
      Q2(1-(b^2-c^2)/E1:1-(a^2-c^2)/E2:1+(a^2-b^2)/E3),
      where
      E1=sqrt[(a^2+m)^2-(b*c)^2],
      E2=sqrt[(b^2+m)^2-(c*a)^2],
      E3=sqrt[(c^2+m)^2-(a*b)^2]
      and
      m^2=a^4+b^4+c^4-[(a*b)^2+(b*c)^2+(c*a)^2].

      Remark 2.
      As we know the Lemoine point is centroid of its pedal triangle,
      and centroid of triangle is the center of homothety for
      triangle ABC and its medial triangle, and also for excribed
      and for inscribed Steiner ellipses of triangle ABC, and between
      focii of these ellipses( i.e. between Bickhart points and these
      pointsQ1 and Q2) we can see that centroid, Lemoine point,
      pedal triangles and these two ellipses are interconnected.
      I feel that there exist more undiscoverd relations here.

      Best regards
      Milorad R.Stevanovic

      [Non-text portions of this message have been removed]
    • jpehrmfr
      Dear Milorad ... triangle. ... This is an immediate consequence of the three following remarks : 1) If f is the affine map A,B,C - vertices of the pedal
      Message 2 of 15 , Nov 28, 2007
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        Dear Milorad
        > by using of barycentrics I have proved the following
        >
        > Theorem.
        > The unique points that are the Lemoine points
        > of their pedal triangles are the foci of Steiner inellipse of
        triangle.
        >
        > Remark 1.
        > In case of a>b>c we have the above points Q1 and Q2
        > Q1(1+(b^2-c^2)/E1:1+(a^2-c^2)/E2:1-(a^2-b^2)/E3),
        > Q2(1-(b^2-c^2)/E1:1-(a^2-c^2)/E2:1+(a^2-b^2)/E3),
        > where
        > E1=sqrt[(a^2+m)^2-(b*c)^2],
        > E2=sqrt[(b^2+m)^2-(c*a)^2],
        > E3=sqrt[(c^2+m)^2-(a*b)^2]
        > and
        > m^2=a^4+b^4+c^4-[(a*b)^2+(b*c)^2+(c*a)^2].
        >
        > Remark 2.
        > As we know the Lemoine point is centroid of its pedal triangle,
        > and centroid of triangle is the center of homothety for
        > triangle ABC and its medial triangle, and also for excribed
        > and for inscribed Steiner ellipses of triangle ABC, and between
        > focii of these ellipses( i.e. between Bickhart points and these
        > pointsQ1 and Q2) we can see that centroid, Lemoine point,
        > pedal triangles and these two ellipses are interconnected.
        > I feel that there exist more undiscoverd relations here.

        This is an immediate consequence of the three following remarks :
        1) If f is the affine map A,B,C -> vertices of the pedal triangle of
        P, then f(P*) = P where P* = isogonal conjugate of P
        2) The sidelines of the pedal triangle of P are in ratio
        a.AP : b.BP : c.CP
        3) P is a focus of the Steiner inellipse <=> P is barycentric
        1/AP^2 : 1/BP^2 : 1/CP^2

        Now,
        P is the Lemoine point of his pedal triangle <=>
        P = f[(a.AP)^2 : (b.BP)^2 : (c.CP)^2]) <=>
        P* = (a.AP)^2 : (b.BP)^2 : (c.CP)^2 <=> P = 1/AP^2 : 1/BP^2 : 1/CP^2
        Friendly.Jean-Pierre
      • Alexey.A.Zaslavsky
        Dear Milorad and Jean-Pierre! Arsenij Akopjan proved another property of these foci. Let a, b, c be the complex numbers corresponding to the vertex of given
        Message 3 of 15 , Nov 28, 2007
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          Dear Milorad and Jean-Pierre!
          Arsenij Akopjan proved another property of these foci. Let a, b, c be the complex numbers corresponding to the vertex of given triangle. Consider the polynom P(z)=(z-a)(z-b)(z-c). Then P'(z)=3(z-f_1)(z-f_2), where f_1, f_2 correspond to the foci of Steiner inellipse. The proof is in our book published (in Russian) by Moscow center of mathematical education. Also it will be published in English by AMS.

          Sincerely Alexey

          Antivirus scanning: Symantec Mail Security for SMTP.

          [Non-text portions of this message have been removed]
        • jpehrmfr
          Dear Alexei ... the complex numbers corresponding to the vertex of given triangle. Consider the polynom P(z)=(z-a)(z-b)(z-c). Then P (z)=3(z-f_1)(z-f_2), where
          Message 4 of 15 , Nov 29, 2007
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            Dear Alexei

            > Arsenij Akopjan proved another property of these foci. Let a, b, c be
            the complex numbers corresponding to the vertex of given triangle.
            Consider the polynom P(z)=(z-a)(z-b)(z-c). Then P'(z)=3(z-f_1)(z-f_2),
            where f_1, f_2 correspond to the foci of Steiner inellipse. The proof
            is in our book published (in Russian) by Moscow center of mathematical
            education. Also it will be published in English by AMS.

            If M as affix z
            1) P'(z)=0 <=> 1/(z-a)+1/(z-b)+1/(z-c) = 0
            <=> (z-a)/|z-a|^2 + (z-b)/|z-b|^2 +(z-c)/|z-c|^2 = 0
            <=> M has barycentrics 1/AM^2 : 1/BM^2 : 1/CM^2
            2) As the roots of P'(z) = 0 give a pair of isogonal conjugate points
            with midpoint G, they are the affixes of the focii of the Steiner
            inellipse
            I think that 1) and 2) are old and classical results.
            Have you any reference?
            Friendly. Jean-Pierre
          • Bernard Gibert
            Dear Jean-Pierre, ... Is there something similar for the two other imaginary foci ? Best regards Bernard [Non-text portions of this message have been removed]
            Message 5 of 15 , Nov 29, 2007
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              Dear Jean-Pierre,

              > If M as affix z
              > 1) P'(z)=0 <=> 1/(z-a)+1/(z-b)+1/(z-c) = 0
              > <=> (z-a)/|z-a|^2 + (z-b)/|z-b|^2 +(z-c)/|z-c|^2 = 0
              > <=> M has barycentrics 1/AM^2 : 1/BM^2 : 1/CM^2
              > 2) As the roots of P'(z) = 0 give a pair of isogonal conjugate points
              > with midpoint G, they are the affixes of the focii of the Steiner
              > inellipse


              Is there something similar for the two other imaginary foci ?


              Best regards

              Bernard



              [Non-text portions of this message have been removed]
            • jpehrmfr
              Dear Bernard ... points ... The problem is that we cannot represent an imaginary point of the plane with a complex number; we must use 2 complex numbers or a
              Message 6 of 15 , Nov 29, 2007
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                Dear Bernard

                > > If M as affix z
                > > 1) P'(z)=0 <=> 1/(z-a)+1/(z-b)+1/(z-c) = 0
                > > <=> (z-a)/|z-a|^2 + (z-b)/|z-b|^2 +(z-c)/|z-c|^2 = 0
                > > <=> M has barycentrics 1/AM^2 : 1/BM^2 : 1/CM^2
                > > 2) As the roots of P'(z) = 0 give a pair of isogonal conjugate
                points
                > > with midpoint G, they are the affixes of the focii of the Steiner
                > > inellipse
                >
                >
                > Is there something similar for the two other imaginary foci ?

                The problem is that we cannot represent an imaginary point of the plane
                with a complex number; we must use 2 complex numbers or a quaternion or
                something like that.
                As any one an idea?
                Jean-Pierre
              • Ignacio Larrosa Cañestro
                ... Best regards, Ignacio Larrosa Cañestro A Coruña (España) ilarrosa@mundo-r.com
                Message 7 of 15 , Nov 29, 2007
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                  Alexey.A.Zaslavsky wrote:
                  > Dear Milorad and Jean-Pierre!
                  > Arsenij Akopjan proved another property of these foci. Let a, b, c be
                  > the complex numbers corresponding to the vertex of given triangle.
                  > Consider the polynom P(z)=(z-a)(z-b)(z-c). Then
                  > P'(z)=3(z-f_1)(z-f_2), where f_1, f_2 correspond to the foci of
                  > Steiner inellipse. The proof is in our book published (in Russian) by
                  > Moscow center of mathematical education. Also it will be published in
                  > English by AMS.
                  >
                  > Sincerely Alexey

                  On 8/11/2000 Jim Parish wrote:
                  > Ignacio Larrosa Cañestro asked:
                  >> A theorem of complex variable (Lucas' theorem?) say that if {a_1,
                  >> ... ,a_n}are the roots of a polinomal P(z) of degree n, then the n-1
                  >> roots {r_1, ...,r_(n-1)} of P'(z) are inside of the convex capsule
                  >> of {a_1, ...,a_n}.
                  >>
                  >> The configuration {a_1, ..., a_n, r_1, ..., r_(n-1)} is invariant
                  >> under isometries or dilatations (similitudes?).
                  >>
                  >> If n=3, the points r_1, r_2 are
                  >>
                  >> {r_1, r_2} = {(a+b+c)/3 +/- sqrt((a^2+b^2+c^2)/6 -
                  >> ((a+b+c)/3)^2/2)}
                  >>
                  >> where {a, b, c} are the complex numbers correspond to the vertices.
                  >>
                  >> Are there {r_1, r_2} known points of the triangle?
                  >
                  > Yes; they are the foci of the Steiner inellipse.
                  > Measure everything from the centroid; then a+b+c=0. Let z, w be
                  > such that a=z+w, b=oz+oow, c=ooz+ow, where o is a primitive
                  > cube root of unity. Then r_1 and r_2 work out to be the square
                  > roots of zw.
                  > The midpoint G1 of the side BC is -(b+c)/2 = 1/2(z+w). Therefore
                  > G1-r_1=1/2(Sqrt[z]-Sqrt[w])^2 and G1-r_2=1/2(Sqrt[z]+Sqrt[w])^2.
                  > An easy computation then shows that the sum dist(r_1,G1)+dist(r_2,G1)
                  > is |z|+|w|, and the same result obtains for the other midpoints G2,
                  > G3. Thus r_1, r_2 are the foci of the Steiner inellipse.
                  >
                  > Jim Parish
                  >

                  Best regards,

                  Ignacio Larrosa Cañestro
                  A Coruña (España)
                  ilarrosa@...
                • Jim Parish
                  I ve proven a generalization of this result to any polygon which is affine- equivalent to a regular polygon:
                  Message 8 of 15 , Nov 29, 2007
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                    I've proven a generalization of this result to any polygon which is affine-
                    equivalent to a regular polygon:
                    http://forumgeom.fau.edu/FG2006volume6/FG200633index.html

                    Jim Parish

                    Ignacio Larrosa Cañestro wrote:
                    > Alexey.A.Zaslavsky wrote:
                    > > Dear Milorad and Jean-Pierre!
                    > > Arsenij Akopjan proved another property of these foci. Let a, b, c be
                    > > the complex numbers corresponding to the vertex of given triangle.
                    > > Consider the polynom P(z)=(z-a)(z-b)(z-c). Then
                    > > P'(z)=3(z-f_1)(z-f_2), where f_1, f_2 correspond to the foci of
                    > > Steiner inellipse. The proof is in our book published (in Russian) by
                    > > Moscow center of mathematical education. Also it will be published in
                    > > English by AMS.
                    > >
                    > > Sincerely Alexey
                    >
                    > On 8/11/2000 Jim Parish wrote:
                    > > Ignacio Larrosa Cañestro asked:
                    > >> A theorem of complex variable (Lucas' theorem?) say that if {a_1,
                    > >> ... ,a_n}are the roots of a polinomal P(z) of degree n, then the n-1
                    > >> roots {r_1, ...,r_(n-1)} of P'(z) are inside of the convex capsule
                    > >> of {a_1, ...,a_n}.
                    > >>
                    > >> The configuration {a_1, ..., a_n, r_1, ..., r_(n-1)} is invariant
                    > >> under isometries or dilatations (similitudes?).
                    > >>
                    > >> If n=3, the points r_1, r_2 are
                    > >>
                    > >> {r_1, r_2} = {(a+b+c)/3 +/- sqrt((a^2+b^2+c^2)/6 -
                    > >> ((a+b+c)/3)^2/2)}
                    > >>
                    > >> where {a, b, c} are the complex numbers correspond to the vertices.
                    > >>
                    > >> Are there {r_1, r_2} known points of the triangle?
                    > >
                    > > Yes; they are the foci of the Steiner inellipse.
                    > > Measure everything from the centroid; then a+b+c=0. Let z, w be
                    > > such that a=z+w, b=oz+oow, c=ooz+ow, where o is a primitive
                    > > cube root of unity. Then r_1 and r_2 work out to be the square
                    > > roots of zw.
                    > > The midpoint G1 of the side BC is -(b+c)/2 = 1/2(z+w). Therefore
                    > > G1-r_1=1/2(Sqrt[z]-Sqrt[w])^2 and G1-r_2=1/2(Sqrt[z]+Sqrt[w])^2.
                    > > An easy computation then shows that the sum dist(r_1,G1)+dist(r_2,G1)
                    > > is |z|+|w|, and the same result obtains for the other midpoints G2,
                    > > G3. Thus r_1, r_2 are the foci of the Steiner inellipse.
                    > >
                    > > Jim Parish
                    > >
                    >
                    > Best regards,
                    >
                    > Ignacio Larrosa Cañestro
                    > A Coruña (España)
                    > ilarrosa@...
                    >



                    -------------------------------------------------
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                  • Nikolaos Dergiades
                    Dear Friends, thanks to Jean-Pierre s nice proof, I ve learned why the barycentrics of the focii of Steiner inellipse are 1/AM^2 : 1/BM^2 : 1/CM^2. For a
                    Message 9 of 15 , Nov 29, 2007
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                      Dear Friends,
                      thanks to Jean-Pierre's nice proof,
                      I've learned why the barycentrics of the focii
                      of Steiner inellipse are 1/AM^2 : 1/BM^2 : 1/CM^2.
                      For a generalization to any inconic in ABC
                      (the conic that is tangent to the sides of ABC
                      at the vertices of the cevian triangle A'B'C'
                      of P=(x : y : z) in barycentrics)
                      let p,a',b',c' be the affixes of the perspector P
                      and the tangency points A',B',C' then
                      b'=(xa + zc)/(x + z)
                      c'=(xa + yb)/(x + y)
                      Let f be the affix of a focus F of the inconic.

                      From Poncelet's theorem that
                      a tangent to a conic segment XY
                      between two other tangents, is seen from F with a
                      constant angle XFY, we get that
                      angle(AFC') - angle(B'FA) = 0 (mod pi) or
                      arg((c'-f)/(a-f)) - arg((a-f)/(b'-f)) = 0 (mod pi) or
                      the number Q = ((c'-f)/(a-f))/((a-f)/(b'-f)) is real.
                      Substituting b', c' we get that the number
                      [xy(f-a)(f-b)+yz(f-b)(f-c)+zx(f-c)(f-a)]/(a-f)^2
                      must be real. Similarly the numbers
                      [xy(f-a)(f-b)+yz(f-b)(f-c)+zx(f-c)(f-a)]/(b-f)^2
                      [xy(f-a)(f-b)+yz(f-b)(f-c)+zx(f-c)(f-a)]/(c-f)^2
                      must be real.
                      Hence
                      xy(f-a)(f-b)+yz(f-b)(f-c)+zx(f-c)(f-a) = 0 (1)
                      because otherwise two of the numbers
                      (a-f)^2, (b-f)^2, (c-f)^2 would have the same
                      argument or two of the numbers
                      a-f, b-f, c-f would have arguments that differ by pi
                      or that F lies on one side of ABC that is impossible.
                      From (1) we get that f is a root of the equation
                      Kff - Lf + N = 0 where
                      K = xy + yz + zx
                      L = x(y+z)a + y(z+x)b + z(x+y)c
                      N = xyab + yzbc + zxca
                      If P is on the Steiner circumellipse K = xy+yz+zx = 0
                      then we have only one focus and the inconic is
                      a parabola. If K <> 0 the equation gives two roots
                      f, f' hence two focii F, F' the projections of which
                      on the tangent sides of ABC are lying on the same
                      circle and hence F, F' are isogonal conjugate.
                      The center O of the inconic
                      that is the mid point of the focii has affix
                      L/2K = [(b+c)/x+(c+a)/y+(a+b)/z]/2(1/x+1/y+1/z)
                      that is the affix of the complement of the
                      isotomic conjugate of P.
                      From (1) we get also that
                      (f-a)/x(f-a)^2)+(f-b)/y(f-b)^2+(f-c)/z(f-c)^2 = 0
                      or
                      (f-a)/x.AF^2+(f-b)/y.BF^2+(f-c)/z.CF^2 = 0
                      or that F has barycentrics
                      1/x.AF^2 : 1/y.BF^2 : 1/z.CF^2.
                      I don't like my proof. Is there a simpler?

                      Best regards
                      Nikos Dergiades


                      [JPE]
                      > Let a, b, c be
                      > the complex numbers corresponding to the vertex of
                      > given triangle.
                      > Consider the polynom P(z)=(z-a)(z-b)(z-c). Then
                      > P'(z)=3(z-f_1)(z-f_2),
                      > where f_1, f_2 correspond to the foci of Steiner
                      > inellipse. The proof
                      > is in our book published (in Russian) by Moscow
                      > center of mathematical
                      > education. Also it will be published in English by
                      > AMS.
                      >
                      > If M as affix z
                      > 1) P'(z)=0 <=> 1/(z-a)+1/(z-b)+1/(z-c) = 0
                      > <=> (z-a)/|z-a|^2 + (z-b)/|z-b|^2 +(z-c)/|z-c|^2 = 0
                      > <=> M has barycentrics 1/AM^2 : 1/BM^2 : 1/CM^2
                      > 2) As the roots of P'(z) = 0 give a pair of isogonal
                      > conjugate points
                      > with midpoint G, they are the affixes of the focii
                      > of the Steiner
                      > inellipse
                      > I think that 1) and 2) are old and classical
                      > results.
                      > Have you any reference?
                      > Friendly. Jean-Pierre




                      ___________________________________________________________
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                    • Nikolaos Dergiades
                      Oops. I had to go to bed. I misunderstood why ? P (z)=0 1/(z-a)+1/(z-b)+1/(z-c) = 0 (z-a)/|z-a|^2 + (z-b)/|z-b|^2 +(z-c)/|z-c|^2 = 0 So I repeat my
                      Message 10 of 15 , Nov 29, 2007
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                        Oops. I had to go to bed.
                        I misunderstood why ?

                        P'(z)=0 <=> 1/(z-a)+1/(z-b)+1/(z-c) = 0
                        <=> (z-a)/|z-a|^2 + (z-b)/|z-b|^2 +(z-c)/|z-c|^2 = 0

                        So I repeat my previous message
                        in the part I think I understand it.
                        Best regards
                        Nikos Dergiades



                        > Dear Alexei
                        >
                        > > Arsenij Akopjan proved another property of these
                        > foci. Let a, b, c be
                        > the complex numbers corresponding to the vertex of
                        > given triangle.
                        > Consider the polynom P(z)=(z-a)(z-b)(z-c). Then
                        > P'(z)=3(z-f_1)(z-f_2),
                        > where f_1, f_2 correspond to the foci of Steiner
                        > inellipse. The proof
                        > is in our book published (in Russian) by Moscow
                        > center of mathematical
                        > education. Also it will be published in English by
                        > AMS.
                        >
                        > If M as affix z
                        > 1) P'(z)=0 <=> 1/(z-a)+1/(z-b)+1/(z-c) = 0
                        > <=> (z-a)/|z-a|^2 + (z-b)/|z-b|^2 +(z-c)/|z-c|^2 = 0
                        > <=> M has barycentrics 1/AM^2 : 1/BM^2 : 1/CM^2
                        > 2) As the roots of P'(z) = 0 give a pair of isogonal
                        > conjugate points
                        > with midpoint G, they are the affixes of the focii
                        > of the Steiner
                        > inellipse
                        > I think that 1) and 2) are old and classical
                        > results.
                        > Have you any reference?
                        > Friendly. Jean-Pierre
                        >
                        >
                        >
                        >
                        > Yahoo! Groups Links
                        >
                        >
                        >
                        >
                        >




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                      • Nikolaos Dergiades
                        Dear Friends, For a generalization to any inconic in ABC (the conic that is tangent to the sides of ABC at the vertices of the cevian triangle A B C of P=(x :
                        Message 11 of 15 , Nov 29, 2007
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                          Dear Friends,

                          For a generalization to any inconic in ABC
                          (the conic that is tangent to the sides of ABC
                          at the vertices of the cevian triangle A'B'C'
                          of P=(x : y : z) in barycentrics)
                          let p,a',b',c' be the affixes of the perspector P
                          and the tangency points A',B',C' then
                          b'=(xa + zc)/(x + z)
                          c'=(xa + yb)/(x + y)
                          Let f be the affix of a focus F of the inconic.

                          From Poncelet's theorem that
                          a tangent to a conic segment XY
                          between two other tangents, is seen from F with a
                          constant angle XFY, we get that
                          angle(AFC') - angle(B'FA) = 0 (mod pi) or
                          arg((c'-f)/(a-f)) - arg((a-f)/(b'-f)) = 0 (mod pi) or
                          the number Q = ((c'-f)/(a-f))/((a-f)/(b'-f)) is real.
                          Substituting b', c' we get that the number
                          [xy(f-a)(f-b)+yz(f-b)(f-c)+zx(f-c)(f-a)]/(a-f)^2
                          must be real. Similarly the numbers
                          [xy(f-a)(f-b)+yz(f-b)(f-c)+zx(f-c)(f-a)]/(b-f)^2
                          [xy(f-a)(f-b)+yz(f-b)(f-c)+zx(f-c)(f-a)]/(c-f)^2
                          must be real.
                          Hence
                          xy(f-a)(f-b)+yz(f-b)(f-c)+zx(f-c)(f-a) = 0 (1)
                          because otherwise two of the numbers
                          (a-f)^2, (b-f)^2, (c-f)^2 would have the same
                          argument or two of the numbers
                          a-f, b-f, c-f would have arguments that differ by pi
                          or that F lies on one side of ABC that is impossible.
                          From (1) we get that f is a root of the equation
                          Kff - Lf + N = 0 where
                          K = xy + yz + zx
                          L = x(y+z)a + y(z+x)b + z(x+y)c
                          N = xyab + yzbc + zxca
                          If P is on the Steiner circumellipse K = xy+yz+zx = 0
                          then we have only one focus and the inconic is
                          a parabola. If K <> 0 the equation gives two roots
                          f, f' hence two focii F, F' the projections of which
                          on the tangent sides of ABC are lying on the same
                          circle and hence F, F' are isogonal conjugate.
                          The center O of the inconic
                          that is the mid point of the focii has affix
                          L/2K = [(b+c)/x+(c+a)/y+(a+b)/z]/2(1/x+1/y+1/z)
                          that is the affix of the complement of the
                          isotomic conjugate of P.

                          I don't like my proof. Is there a simpler?

                          Best regards
                          Nikos Dergiades


                          [JPE]
                          > Let a, b, c be
                          > the complex numbers corresponding to the vertex of
                          > given triangle.
                          > Consider the polynom P(z)=(z-a)(z-b)(z-c). Then
                          > P'(z)=3(z-f_1)(z-f_2),
                          > where f_1, f_2 correspond to the foci of Steiner
                          > inellipse. The proof
                          > is in our book published (in Russian) by Moscow
                          > center of mathematical
                          > education. Also it will be published in English by
                          > AMS.
                          >
                          > If M as affix z
                          > 1) P'(z)=0 <=> 1/(z-a)+1/(z-b)+1/(z-c) = 0
                          > <=> (z-a)/|z-a|^2 + (z-b)/|z-b|^2 +(z-c)/|z-c|^2 = 0
                          > <=> M has barycentrics 1/AM^2 : 1/BM^2 : 1/CM^2
                          > 2) As the roots of P'(z) = 0 give a pair of isogonal
                          > conjugate points
                          > with midpoint G, they are the affixes of the focii
                          > of the Steiner
                          > inellipse
                          > I think that 1) and 2) are old and classical
                          > results.
                          > Have you any reference?
                          > Friendly. Jean-Pierre


















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                        • Alexey.A.Zaslavsky
                          Dear Jean-Pierre! I don t know first reference, may be Arsenij can help you. His mail: akopjan@gmail.com. But it is interesting that the result of Milorad
                          Message 12 of 15 , Nov 29, 2007
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                            Dear Jean-Pierre!
                            I don't know first reference, may be Arsenij can help you. His mail: akopjan@.... But it is interesting that the result of Milorad immediately follows from your assertion 1)

                            Sincerely Alexey

                            Dear Alexei

                            > Arsenij Akopjan proved another property of these foci. Let a, b, c be
                            the complex numbers corresponding to the vertex of given triangle.
                            Consider the polynom P(z)=(z-a)(z-b)(z-c). Then P'(z)=3(z-f_1)(z-f_2),
                            where f_1, f_2 correspond to the foci of Steiner inellipse. The proof
                            is in our book published (in Russian) by Moscow center of mathematical
                            education. Also it will be published in English by AMS.

                            If M as affix z
                            1) P'(z)=0 <=> 1/(z-a)+1/(z-b)+1/(z-c) = 0
                            <=> (z-a)/|z-a|^2 + (z-b)/|z-b|^2 +(z-c)/|z-c|^2 = 0
                            <=> M has barycentrics 1/AM^2 : 1/BM^2 : 1/CM^2
                            2) As the roots of P'(z) = 0 give a pair of isogonal conjugate points
                            with midpoint G, they are the affixes of the focii of the Steiner
                            inellipse
                            I think that 1) and 2) are old and classical results.
                            Have you any reference?
                            Friendly. Jean-Pierre




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                          • Milorad Stevanovic
                            Dear Jean-Pierre, and all other friends 1. I have considered the problem to find the points P(x:y:z) with property x(AP^2)=y(BP^2)=z(CP^2). The coordinate
                            Message 13 of 15 , Nov 30, 2007
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                              Dear Jean-Pierre,
                              and all other friends

                              1.
                              I have considered the problem
                              to find the points P(x:y:z) with property
                              x(AP^2)=y(BP^2)=z(CP^2).

                              The coordinate solution of this problem I have
                              found about 15 years ago.
                              2.
                              I have considered the problem
                              to find the points that are Lemoine points
                              of their pedal triangles.

                              I have found the coordinates of solutions of
                              this problem not long ago..
                              3.
                              I have worked on a known problem of equal cevian
                              points, ie of the points P with propertiy
                              AA1=BB1=CC1
                              where A1,B1,C1 are the points of intersections
                              of AP,BP and CP with sides of triangle ABC.

                              It has been discovered that the solutions
                              of the problem 3 were the focii of Steiner exellipse.

                              When I connected the sollutions of problems
                              2 and 3 I have found that the solutions of problem
                              2 are the focii of Steiner inellipse.

                              When I connected the solutions of problems 1.
                              and new solution of problem 2 I have found that
                              the solutions of problem 1 are the focii of Steiner
                              inellipse.
                              That I had for some other message.

                              It is interesting to see that by various roads,
                              in various times,guided by the same or different idea,
                              we came to the same place.

                              Friendly
                              All the best

                              Milorad R.Stevanovic



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                            • Jeff
                              Dear Jean-Pierre, ... I ve wondered about using complicated numbers like those defined in [1] below. The author writes in the introduction At first sight it
                              Message 14 of 15 , Dec 1, 2007
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                                Dear Jean-Pierre,

                                > The problem is that we cannot represent an imaginary point of the
                                > plane with a complex number; we must use 2 complex numbers or a
                                > quaternion or something like that.
                                > As any one an idea?

                                I've wondered about using 'complicated numbers' like those defined in
                                [1] below.
                                The author writes in the introduction "At first sight it may therefore
                                seem paradoxical that in the present paper a Cauchy formula will be
                                established for an algebra that has almost the inevitability of
                                quaternions and Clifford matrices, but which is commutative and has zero
                                divisors."

                                I'm not even close to presenting an argument for the use of the
                                complicated numbers. But, please consider this as an idea nevertheless.

                                [1] I.J. Good, "A simple generalization of analytic function theory",
                                Expositiones Mathematicae, Expo. Math 6 (1988), 289-311, Published by
                                Bibliographisches Institut & F.A. Brockhaus AG 1988.
                              • Milorad Stevanovic
                                Dear Jean-Pierre and all friends As I said in earlier message, I have considered the problem to find the points P(x:y:z)(given in barycentrics), with property
                                Message 15 of 15 , Dec 4, 2007
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                                  Dear Jean-Pierre and all friends

                                  As I said in earlier message,
                                  I have considered the problem
                                  to find the points P(x:y:z)(given
                                  in barycentrics), with property

                                  x(AP^2)=y(BP^2)=z(CP^2).

                                  I have found in my papers, the following solutions
                                  (under the condition a>b>c)

                                  1.Q1(x1:y1:z1)
                                  x1=c^2+m-sqrt(D3),
                                  y1=b^2,
                                  z1=a^2+m-sqrt(D1).

                                  2.Q2(x2:y2:z2)
                                  z2=c^2+m+sqrt(D3),
                                  y2=b^2,
                                  z2=a^2+m+sqrt(D1)

                                  where

                                  m^2=a^4+b^4+c^4-(ab)^2-(bc)^2-(ca)^2,(m>0),
                                  D1=(a^2+m)^2-(bc)^2,
                                  D3=(c^2+m)^2-(ab)^2.

                                  These two points are real and are focii of Steiner inellipse.

                                  There are also two complex points

                                  3.Q3(x3:y3:z3)
                                  x3=c^2-m+i sqrt(E3),
                                  y3=b^2,
                                  z3=a^2-m+i sqrt(E1).

                                  4.Q4(x4:y4:z4)
                                  x4=c^2-m-i sqrt(E3),
                                  y4=b^2,
                                  z4=a^2-m-i sqrt(E1)

                                  where

                                  m^2=a^4+b^4+c^4-(ab)^2-(bc)^2-(ca)^2,(m>0),
                                  E1=(bc)^2-(a^2-m)^2,
                                  E3=(ab)^2-(c^2-m)^2..

                                  The points Q3 and Q4 are collinear
                                  with centroid G of triangle ABC.

                                  How these points Q3 and Q4 are related
                                  to the focii of Steiner inellipse?

                                  What is the geometric meaning of these points Q3 and Q4?

                                  Friendly
                                  Yours

                                  Milorad R.Stevanovic



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