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

Three Perpendiculars To Sidelines From Incenter

Expand Messages
  • Quang Tuan Bui
    Dear All My Friends, From incenter I we construct one line perpendicular to sidelines BC. This line cuts AB at Ab, cuts AC at Ac and cuts incircle (other than
    Message 1 of 9 , Nov 7, 2006
      Dear All My Friends,
      From incenter I we construct one line perpendicular to sidelines BC. This line cuts AB at Ab, cuts AC at Ac and cuts incircle (other than touch point with BC) at Ai. Similarly we construct Bc, Ba, Bi, and Ca, Cb, Ci.
      Sometimes ago we have proved that three segmens AbAc, BcBa, CaCb have a feature: one segment is sum of other two. Now we can see some more interesting.
      There are three pairs of congruent segments. In fact:
      AiAc = BiBc; BiBa = CiCa; CiCb = AiAb
      Moreover, it is very easy to show that there are three isosceles trapezia here:
      BiCiCaBa; CiAiAbCb; AiBiBcAc
      Also very easy to show that AAi, BBi, CCi are concurrent at Nagel point Na.
      But there are three pairs of congruent segments which I still can not proof: let A'B'C' is Cevian triangle of Nagel point Na. Prove that:
      AAi = A'Na; BBi = B'Na; CCi = C'Na
      Please give me a simple proof!
      Thank you and best regards,
      Bui Quang Tuan


      ---------------------------------
      Sponsored Link

      Try Netflix today! With plans starting at only $5.99 a month what are you waiting for?

      [Non-text portions of this message have been removed]
    • Quang Tuan Bui
      Dear All My Friends, By calculations I have found: - Triangle AiBiCi has barycentrics: Ai = (a^2 : sa*sb : sa*sc) Bi = (sb*sa : b^2 : sb*sc) Ci = (sc*sa :
      Message 2 of 9 , Nov 8, 2006
        Dear All My Friends,
        By calculations I have found:
        - Triangle AiBiCi has barycentrics:
        Ai = (a^2 : sa*sb : sa*sc)
        Bi = (sb*sa : b^2 : sb*sc)
        Ci = (sc*sa : sc*sb : c^2)
        - Segments AAi = A'Na = sa*sqr((a*sa + (b - c)^2)/(s*a)). (Similarly with B, C)
        There are some things interesting with triangle AiBiCi:
        - Medial triangle of AiBiCi and intouch triangle of ABC are perspective at X(3057)
        - Three circumcircles of IAAi, IBBi, ICCi are concurrent at one point (other than I). This point is X(1319).
        - Let three lines AcBc, BaCa, CbAb bound one triangle A1B1C1. Let B1C1 cuts BC at A2. Similarly define B2, C2 then three lines AA2, BB2, CC2 are concurrent at X(651).
        - There is one circumconic (Co) passing through A, B, C, A1, B1, C1, X(651), X(109) with perspector = X(56) and center = X(478) (center of Yiu conic).
        This circumconic has barycentric equation:
        a^2/(b+c-a)*y*z + b^2/(c+a-b)*z*x + c^2/(a+b-c)*x*y = 0
        - Two triangles A1B1C1, AiBiCi are homothetic at one point with barycentrics:
        a^2*(b^3 + c^3 - a^3 - (a^2 + 3*b*c)*(b + c) + a*(b^2 + c^2) + 4*a*b*c) / (b+c-a) : :
        Search value: +0.165893366771994
        This point is on the line X(56)X(109)
        - Centroid of AiBiCi has barycentrics:
        a*((b + c)*(a^2 - (b - c)^2) - 8*a*b*c) : :
        Search value: +2.07716924144103
        This point is on the line X(1)X(3) and is reflection of X(210) in X(392).
        Please give me some advices and references:
        - There is special name for circumconic (Co)?
        - There are some another well known centers on this conic?
        Thank you and best regards,
        Bui Quang Tuan


        Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
        From incenter I we construct one line perpendicular to sidelines BC. This line cuts AB at Ab, cuts AC at Ac and cuts incircle (other than touch point with BC) at Ai. Similarly we construct Bc, Ba, Bi, and Ca, Cb, Ci.
        Sometimes ago we have proved that three segmens AbAc, BcBa, CaCb have a feature: one segment is sum of other two. Now we can see some more interesting.
        There are three pairs of congruent segments. In fact:
        AiAc = BiBc; BiBa = CiCa; CiCb = AiAb
        Moreover, it is very easy to show that there are three isosceles trapezia here:
        BiCiCaBa; CiAiAbCb; AiBiBcAc
        Also very easy to show that AAi, BBi, CCi are concurrent at Nagel point Na.
        But there are three pairs of congruent segments which I still can not proof: let A'B'C' is Cevian triangle of Nagel point Na. Prove that:
        AAi = A'Na; BBi = B'Na; CCi = C'Na
        Please give me a simple proof!
        Thank you and best regards,
        Bui Quang Tuan

        ---------------------------------
        Sponsored Link

        Try Netflix today! With plans starting at only $5.99 a month what are you waiting for?

        [Non-text portions of this message have been removed]






        ---------------------------------
        Sponsored Link

        For just $24.99/mo., Vonage offers unlimited local and long- distance calling. Sign up now.

        [Non-text portions of this message have been removed]
      • Quang Tuan Bui
        Dear All My Friends We can generalize the circumconic (Co) as follow: Suppose I is any point P with barycentrics (p : q : r). Construct A1B1C1 as before. Six
        Message 3 of 9 , Nov 9, 2006
          Dear All My Friends
          We can generalize the circumconic (Co) as follow:
          Suppose I is any point P with barycentrics (p : q : r). Construct A1B1C1 as before. Six points A, B, C, A1, B1, C1 are on one circumconic. The barycentric equation of this circumconic:
          P*y*z + Q*z*x + R*x*y = 0
          here:
          P = (p*SC + a^2*q)*(p*SB + a^2*r)
          Q = (q*SA + b^2*r)*(q*SC + b^2*p)
          R = (r*SB + c^2*p)*(r*SA + c^2*q)
          One example:
          If P = centroid then this circumconic has barycentrics of:
          - Perspector:
          (b^2 - c^2)^2 - 9*a^4 : :
          Search value: + 0.102966916856477
          - Fourth intersection point with circumcircle
          1/((b^2 - c^2)*(SA^2 + 2*b^2*c^2)) : :
          Search value:+ 0.713049056090128
          - Center:
          (SC + a^2)*(SB + a^2)*(2*SA^2 - 5*(b^4 + c^4 - a^4)) : :
          Search value: - 0.208642148311811
          Thank you and best regards,
          Bui Quang Tuan

          Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
          By calculations I have found:
          - Triangle AiBiCi has barycentrics:
          Ai = (a^2 : sa*sb : sa*sc)
          Bi = (sb*sa : b^2 : sb*sc)
          Ci = (sc*sa : sc*sb : c^2)
          - Segments AAi = A'Na = sa*sqr((a*sa + (b - c)^2)/(s*a)). (Similarly with B, C)
          There are some things interesting with triangle AiBiCi:
          - Medial triangle of AiBiCi and intouch triangle of ABC are perspective at X(3057)
          - Three circumcircles of IAAi, IBBi, ICCi are concurrent at one point (other than I). This point is X(1319).
          - Let three lines AcBc, BaCa, CbAb bound one triangle A1B1C1. Let B1C1 cuts BC at A2. Similarly define B2, C2 then three lines AA2, BB2, CC2 are concurrent at X(651).
          - There is one circumconic (Co) passing through A, B, C, A1, B1, C1, X(651), X(109) with perspector = X(56) and center = X(478) (center of Yiu conic).
          This circumconic has barycentric equation:
          a^2/(b+c-a)*y*z + b^2/(c+a-b)*z*x + c^2/(a+b-c)*x*y = 0
          - Two triangles A1B1C1, AiBiCi are homothetic at one point with barycentrics:
          a^2*(b^3 + c^3 - a^3 - (a^2 + 3*b*c)*(b + c) + a*(b^2 + c^2) + 4*a*b*c) / (b+c-a) : :
          Search value: +0.165893366771994
          This point is on the line X(56)X(109)
          - Centroid of AiBiCi has barycentrics:
          a*((b + c)*(a^2 - (b - c)^2) - 8*a*b*c) : :
          Search value: +2.07716924144103
          This point is on the line X(1)X(3) and is reflection of X(210) in X(392).
          Please give me some advices and references:
          - There is special name for circumconic (Co)?
          - There are some another well known centers on this conic?
          Thank you and best regards,
          Bui Quang Tuan

          Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
          From incenter I we construct one line perpendicular to sidelines BC. This line cuts AB at Ab, cuts AC at Ac and cuts incircle (other than touch point with BC) at Ai. Similarly we construct Bc, Ba, Bi, and Ca, Cb, Ci.
          Sometimes ago we have proved that three segmens AbAc, BcBa, CaCb have a feature: one segment is sum of other two. Now we can see some more interesting.
          There are three pairs of congruent segments. In fact:
          AiAc = BiBc; BiBa = CiCa; CiCb = AiAb
          Moreover, it is very easy to show that there are three isosceles trapezia here:
          BiCiCaBa; CiAiAbCb; AiBiBcAc
          Also very easy to show that AAi, BBi, CCi are concurrent at Nagel point Na.
          But there are three pairs of congruent segments which I still can not proof: let A'B'C' is Cevian triangle of Nagel point Na. Prove that:
          AAi = A'Na; BBi = B'Na; CCi = C'Na
          Please give me a simple proof!
          Thank you and best regards,
          Bui Quang Tuan

          ---------------------------------
          Sponsored Link

          Try Netflix today! With plans starting at only $5.99 a month what are you waiting for?

          [Non-text portions of this message have been removed]

          ---------------------------------
          Sponsored Link

          For just $24.99/mo., Vonage offers unlimited local and long- distance calling. Sign up now.

          [Non-text portions of this message have been removed]






          ---------------------------------
          Cheap Talk? Check out Yahoo! Messenger's low PC-to-Phone call rates.

          [Non-text portions of this message have been removed]
        • Quang Tuan Bui
          Dear All My Friends, If we choose P as circumcenter then we have one very simple circumconic: - The barycentrics equation: a^4*y*z + b^4*z*x + c^4*x*y = 0 -
          Message 4 of 9 , Nov 9, 2006
            Dear All My Friends,
            If we choose P as circumcenter then we have one very simple circumconic:
            - The barycentrics equation:
            a^4*y*z + b^4*z*x + c^4*x*y = 0
            - Perspector = X(32)
            - Fourth intersection point = X(110)
            - Center = X(206)
            Please give me some advices and references about this circumconic!
            Are there any other well known centers on this conic?
            Thank you and best regards,
            Bui Quang Tuan

            Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends
            We can generalize the circumconic (Co) as follow:
            Suppose I is any point P with barycentrics (p : q : r). Construct A1B1C1 as before. Six points A, B, C, A1, B1, C1 are on one circumconic. The barycentric equation of this circumconic:
            P*y*z + Q*z*x + R*x*y = 0
            here:
            P = (p*SC + a^2*q)*(p*SB + a^2*r)
            Q = (q*SA + b^2*r)*(q*SC + b^2*p)
            R = (r*SB + c^2*p)*(r*SA + c^2*q)
            One example:
            If P = centroid then this circumconic has barycentrics of:
            - Perspector:
            (b^2 - c^2)^2 - 9*a^4 : :
            Search value: + 0.102966916856477
            - Fourth intersection point with circumcircle
            1/((b^2 - c^2)*(SA^2 + 2*b^2*c^2)) : :
            Search value:+ 0.713049056090128
            - Center:
            (SC + a^2)*(SB + a^2)*(2*SA^2 - 5*(b^4 + c^4 - a^4)) : :
            Search value: - 0.208642148311811
            Thank you and best regards,
            Bui Quang Tuan

            Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
            By calculations I have found:
            - Triangle AiBiCi has barycentrics:
            Ai = (a^2 : sa*sb : sa*sc)
            Bi = (sb*sa : b^2 : sb*sc)
            Ci = (sc*sa : sc*sb : c^2)
            - Segments AAi = A'Na = sa*sqr((a*sa + (b - c)^2)/(s*a)). (Similarly with B, C)
            There are some things interesting with triangle AiBiCi:
            - Medial triangle of AiBiCi and intouch triangle of ABC are perspective at X(3057)
            - Three circumcircles of IAAi, IBBi, ICCi are concurrent at one point (other than I). This point is X(1319).
            - Let three lines AcBc, BaCa, CbAb bound one triangle A1B1C1. Let B1C1 cuts BC at A2. Similarly define B2, C2 then three lines AA2, BB2, CC2 are concurrent at X(651).
            - There is one circumconic (Co) passing through A, B, C, A1, B1, C1, X(651), X(109) with perspector = X(56) and center = X(478) (center of Yiu conic).
            This circumconic has barycentric equation:
            a^2/(b+c-a)*y*z + b^2/(c+a-b)*z*x + c^2/(a+b-c)*x*y = 0
            - Two triangles A1B1C1, AiBiCi are homothetic at one point with barycentrics:
            a^2*(b^3 + c^3 - a^3 - (a^2 + 3*b*c)*(b + c) + a*(b^2 + c^2) + 4*a*b*c) / (b+c-a) : :
            Search value: +0.165893366771994
            This point is on the line X(56)X(109)
            - Centroid of AiBiCi has barycentrics:
            a*((b + c)*(a^2 - (b - c)^2) - 8*a*b*c) : :
            Search value: +2.07716924144103
            This point is on the line X(1)X(3) and is reflection of X(210) in X(392).
            Please give me some advices and references:
            - There is special name for circumconic (Co)?
            - There are some another well known centers on this conic?
            Thank you and best regards,
            Bui Quang Tuan

            Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
            From incenter I we construct one line perpendicular to sidelines BC. This line cuts AB at Ab, cuts AC at Ac and cuts incircle (other than touch point with BC) at Ai. Similarly we construct Bc, Ba, Bi, and Ca, Cb, Ci.
            Sometimes ago we have proved that three segmens AbAc, BcBa, CaCb have a feature: one segment is sum of other two. Now we can see some more interesting.
            There are three pairs of congruent segments. In fact:
            AiAc = BiBc; BiBa = CiCa; CiCb = AiAb
            Moreover, it is very easy to show that there are three isosceles trapezia here:
            BiCiCaBa; CiAiAbCb; AiBiBcAc
            Also very easy to show that AAi, BBi, CCi are concurrent at Nagel point Na.
            But there are three pairs of congruent segments which I still can not proof: let A'B'C' is Cevian triangle of Nagel point Na. Prove that:
            AAi = A'Na; BBi = B'Na; CCi = C'Na
            Please give me a simple proof!
            Thank you and best regards,
            Bui Quang Tuan

            ---------------------------------
            Sponsored Link

            Try Netflix today! With plans starting at only $5.99 a month what are you waiting for?

            [Non-text portions of this message have been removed]

            ---------------------------------
            Sponsored Link

            For just $24.99/mo., Vonage offers unlimited local and long- distance calling. Sign up now.

            [Non-text portions of this message have been removed]

            ---------------------------------
            Cheap Talk? Check out Yahoo! Messenger's low PC-to-Phone call rates.

            [Non-text portions of this message have been removed]






            ---------------------------------
            Everyone is raving about the all-new Yahoo! Mail beta.

            [Non-text portions of this message have been removed]
          • Quang Tuan Bui
            Dear All My Friends, The general circumconic (Co) (with any point P) has one interesting feature: Let Q is perspector of (Co), gQ is isogonal conjugate of Q,
            Message 5 of 9 , Nov 9, 2006
              Dear All My Friends,
              The general circumconic (Co) (with any point P) has one interesting feature:
              Let Q is perspector of (Co), gQ is isogonal conjugate of Q, tgQ is isotomic conjugate of gQ, F is fourth intersection point of (Co) with circumcircle.
              Let X is any point on circumcircle, X' is barycentric product of X and tgQ then:
              1. X' is on the conic (Co).
              2. X, X' and F are collinear.

              We also can show that:
              If X is any point on circumcircle, R is one fixed point, X' is barycentric product of X and R then line XX' is always passing through one fixed point F on circumcircle (other than X). Moreover, when X moving, the locus of X' is one fixed circumconic taken F as fourth intersection point. The perspector of this conic is g(tR).

              Our circumconic (Co) belongs to this type of circumconics.

              Thank you and best regards,
              Bui Quang Tuan


              Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
              If we choose P as circumcenter then we have one very simple circumconic:
              - The barycentrics equation:
              a^4*y*z + b^4*z*x + c^4*x*y = 0
              - Perspector = X(32)
              - Fourth intersection point = X(110)
              - Center = X(206)
              Please give me some advices and references about this circumconic!
              Are there any other well known centers on this conic?
              Thank you and best regards,
              Bui Quang Tuan

              Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends
              We can generalize the circumconic (Co) as follow:
              Suppose I is any point P with barycentrics (p : q : r). Construct A1B1C1 as before. Six points A, B, C, A1, B1, C1 are on one circumconic. The barycentric equation of this circumconic:
              P*y*z + Q*z*x + R*x*y = 0
              here:
              P = (p*SC + a^2*q)*(p*SB + a^2*r)
              Q = (q*SA + b^2*r)*(q*SC + b^2*p)
              R = (r*SB + c^2*p)*(r*SA + c^2*q)
              One example:
              If P = centroid then this circumconic has barycentrics of:
              - Perspector:
              (b^2 - c^2)^2 - 9*a^4 : :
              Search value: + 0.102966916856477
              - Fourth intersection point with circumcircle
              1/((b^2 - c^2)*(SA^2 + 2*b^2*c^2)) : :
              Search value:+ 0.713049056090128
              - Center:
              (SC + a^2)*(SB + a^2)*(2*SA^2 - 5*(b^4 + c^4 - a^4)) : :
              Search value: - 0.208642148311811
              Thank you and best regards,
              Bui Quang Tuan

              Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
              By calculations I have found:
              - Triangle AiBiCi has barycentrics:
              Ai = (a^2 : sa*sb : sa*sc)
              Bi = (sb*sa : b^2 : sb*sc)
              Ci = (sc*sa : sc*sb : c^2)
              - Segments AAi = A'Na = sa*sqr((a*sa + (b - c)^2)/(s*a)). (Similarly with B, C)
              There are some things interesting with triangle AiBiCi:
              - Medial triangle of AiBiCi and intouch triangle of ABC are perspective at X(3057)
              - Three circumcircles of IAAi, IBBi, ICCi are concurrent at one point (other than I). This point is X(1319).
              - Let three lines AcBc, BaCa, CbAb bound one triangle A1B1C1. Let B1C1 cuts BC at A2. Similarly define B2, C2 then three lines AA2, BB2, CC2 are concurrent at X(651).
              - There is one circumconic (Co) passing through A, B, C, A1, B1, C1, X(651), X(109) with perspector = X(56) and center = X(478) (center of Yiu conic).
              This circumconic has barycentric equation:
              a^2/(b+c-a)*y*z + b^2/(c+a-b)*z*x + c^2/(a+b-c)*x*y = 0
              - Two triangles A1B1C1, AiBiCi are homothetic at one point with barycentrics:
              a^2*(b^3 + c^3 - a^3 - (a^2 + 3*b*c)*(b + c) + a*(b^2 + c^2) + 4*a*b*c) / (b+c-a) : :
              Search value: +0.165893366771994
              This point is on the line X(56)X(109)
              - Centroid of AiBiCi has barycentrics:
              a*((b + c)*(a^2 - (b - c)^2) - 8*a*b*c) : :
              Search value: +2.07716924144103
              This point is on the line X(1)X(3) and is reflection of X(210) in X(392).
              Please give me some advices and references:
              - There is special name for circumconic (Co)?
              - There are some another well known centers on this conic?
              Thank you and best regards,
              Bui Quang Tuan

              Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
              From incenter I we construct one line perpendicular to sidelines BC. This line cuts AB at Ab, cuts AC at Ac and cuts incircle (other than touch point with BC) at Ai. Similarly we construct Bc, Ba, Bi, and Ca, Cb, Ci.
              Sometimes ago we have proved that three segmens AbAc, BcBa, CaCb have a feature: one segment is sum of other two. Now we can see some more interesting.
              There are three pairs of congruent segments. In fact:
              AiAc = BiBc; BiBa = CiCa; CiCb = AiAb
              Moreover, it is very easy to show that there are three isosceles trapezia here:
              BiCiCaBa; CiAiAbCb; AiBiBcAc
              Also very easy to show that AAi, BBi, CCi are concurrent at Nagel point Na.
              But there are three pairs of congruent segments which I still can not proof: let A'B'C' is Cevian triangle of Nagel point Na. Prove that:
              AAi = A'Na; BBi = B'Na; CCi = C'Na
              Please give me a simple proof!
              Thank you and best regards,
              Bui Quang Tuan

              ---------------------------------
              Sponsored Link

              Try Netflix today! With plans starting at only $5.99 a month what are you waiting for?

              [Non-text portions of this message have been removed]

              ---------------------------------
              Sponsored Link

              For just $24.99/mo., Vonage offers unlimited local and long- distance calling. Sign up now.

              [Non-text portions of this message have been removed]

              ---------------------------------
              Cheap Talk? Check out Yahoo! Messenger's low PC-to-Phone call rates.

              [Non-text portions of this message have been removed]


              ---------------------------------
              Everyone is raving about the all-new Yahoo! Mail beta.

              [Non-text portions of this message have been removed]






              ---------------------------------
              Want to start your own business? Learn how on Yahoo! Small Business.

              [Non-text portions of this message have been removed]
            • Quang Tuan Bui
              Dear All My Friends, I am very sorry. The features related with perspector Q, gQ, tgQ, X, X of one conic and operation barycentric product are true for any
              Message 6 of 9 , Nov 10, 2006
                Dear All My Friends,
                I am very sorry. The features related with perspector Q, gQ, tgQ, X, X' of one conic and operation barycentric product are true for any circumconic.
                This fact is interesting for me because it explains why our circumconic (Co) is related with Nagel point (when P = incenter). In fact, its perspector X(56) = gX(8) and tgX(8) = Gergonne point. If we take barycentric product of any point on circumcircle with Gergonne point then result point is on the (Co) conic centered at X(478).
                It also explains why starting point of my messages is Nagel point.
                Thank you and best regards,
                Bui Quang Tuan

                Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                The general circumconic (Co) (with any point P) has one interesting feature:
                Let Q is perspector of (Co), gQ is isogonal conjugate of Q, tgQ is isotomic conjugate of gQ, F is fourth intersection point of (Co) with circumcircle.
                Let X is any point on circumcircle, X' is barycentric product of X and tgQ then:
                1. X' is on the conic (Co).
                2. X, X' and F are collinear.

                We also can show that:
                If X is any point on circumcircle, R is one fixed point, X' is barycentric product of X and R then line XX' is always passing through one fixed point F on circumcircle (other than X). Moreover, when X moving, the locus of X' is one fixed circumconic taken F as fourth intersection point. The perspector of this conic is g(tR).

                Our circumconic (Co) belongs to this type of circumconics.

                Thank you and best regards,
                Bui Quang Tuan

                Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                If we choose P as circumcenter then we have one very simple circumconic:
                - The barycentrics equation:
                a^4*y*z + b^4*z*x + c^4*x*y = 0
                - Perspector = X(32)
                - Fourth intersection point = X(110)
                - Center = X(206)
                Please give me some advices and references about this circumconic!
                Are there any other well known centers on this conic?
                Thank you and best regards,
                Bui Quang Tuan

                Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends
                We can generalize the circumconic (Co) as follow:
                Suppose I is any point P with barycentrics (p : q : r). Construct A1B1C1 as before. Six points A, B, C, A1, B1, C1 are on one circumconic. The barycentric equation of this circumconic:
                P*y*z + Q*z*x + R*x*y = 0
                here:
                P = (p*SC + a^2*q)*(p*SB + a^2*r)
                Q = (q*SA + b^2*r)*(q*SC + b^2*p)
                R = (r*SB + c^2*p)*(r*SA + c^2*q)
                One example:
                If P = centroid then this circumconic has barycentrics of:
                - Perspector:
                (b^2 - c^2)^2 - 9*a^4 : :
                Search value: + 0.102966916856477
                - Fourth intersection point with circumcircle
                1/((b^2 - c^2)*(SA^2 + 2*b^2*c^2)) : :
                Search value:+ 0.713049056090128
                - Center:
                (SC + a^2)*(SB + a^2)*(2*SA^2 - 5*(b^4 + c^4 - a^4)) : :
                Search value: - 0.208642148311811
                Thank you and best regards,
                Bui Quang Tuan

                Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                By calculations I have found:
                - Triangle AiBiCi has barycentrics:
                Ai = (a^2 : sa*sb : sa*sc)
                Bi = (sb*sa : b^2 : sb*sc)
                Ci = (sc*sa : sc*sb : c^2)
                - Segments AAi = A'Na = sa*sqr((a*sa + (b - c)^2)/(s*a)). (Similarly with B, C)
                There are some things interesting with triangle AiBiCi:
                - Medial triangle of AiBiCi and intouch triangle of ABC are perspective at X(3057)
                - Three circumcircles of IAAi, IBBi, ICCi are concurrent at one point (other than I). This point is X(1319).
                - Let three lines AcBc, BaCa, CbAb bound one triangle A1B1C1. Let B1C1 cuts BC at A2. Similarly define B2, C2 then three lines AA2, BB2, CC2 are concurrent at X(651).
                - There is one circumconic (Co) passing through A, B, C, A1, B1, C1, X(651), X(109) with perspector = X(56) and center = X(478) (center of Yiu conic).
                This circumconic has barycentric equation:
                a^2/(b+c-a)*y*z + b^2/(c+a-b)*z*x + c^2/(a+b-c)*x*y = 0
                - Two triangles A1B1C1, AiBiCi are homothetic at one point with barycentrics:
                a^2*(b^3 + c^3 - a^3 - (a^2 + 3*b*c)*(b + c) + a*(b^2 + c^2) + 4*a*b*c) / (b+c-a) : :
                Search value: +0.165893366771994
                This point is on the line X(56)X(109)
                - Centroid of AiBiCi has barycentrics:
                a*((b + c)*(a^2 - (b - c)^2) - 8*a*b*c) : :
                Search value: +2.07716924144103
                This point is on the line X(1)X(3) and is reflection of X(210) in X(392).
                Please give me some advices and references:
                - There is special name for circumconic (Co)?
                - There are some another well known centers on this conic?
                Thank you and best regards,
                Bui Quang Tuan

                Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                From incenter I we construct one line perpendicular to sidelines BC. This line cuts AB at Ab, cuts AC at Ac and cuts incircle (other than touch point with BC) at Ai. Similarly we construct Bc, Ba, Bi, and Ca, Cb, Ci.
                Sometimes ago we have proved that three segmens AbAc, BcBa, CaCb have a feature: one segment is sum of other two. Now we can see some more interesting.
                There are three pairs of congruent segments. In fact:
                AiAc = BiBc; BiBa = CiCa; CiCb = AiAb
                Moreover, it is very easy to show that there are three isosceles trapezia here:
                BiCiCaBa; CiAiAbCb; AiBiBcAc
                Also very easy to show that AAi, BBi, CCi are concurrent at Nagel point Na.
                But there are three pairs of congruent segments which I still can not proof: let A'B'C' is Cevian triangle of Nagel point Na. Prove that:
                AAi = A'Na; BBi = B'Na; CCi = C'Na
                Please give me a simple proof!
                Thank you and best regards,
                Bui Quang Tuan

                ---------------------------------
                Sponsored Link

                Try Netflix today! With plans starting at only $5.99 a month what are you waiting for?

                [Non-text portions of this message have been removed]

                ---------------------------------
                Sponsored Link

                For just $24.99/mo., Vonage offers unlimited local and long- distance calling. Sign up now.

                [Non-text portions of this message have been removed]

                ---------------------------------
                Cheap Talk? Check out Yahoo! Messenger's low PC-to-Phone call rates.

                [Non-text portions of this message have been removed]

                ---------------------------------
                Everyone is raving about the all-new Yahoo! Mail beta.

                [Non-text portions of this message have been removed]

                ---------------------------------
                Want to start your own business? Learn how on Yahoo! Small Business.

                [Non-text portions of this message have been removed]






                ---------------------------------
                Check out the all-new Yahoo! Mail beta - Fire up a more powerful email and get things done faster.

                [Non-text portions of this message have been removed]
              • Quang Tuan Bui
                Dear All My Friends, Some properties of this circumconic: 1. P and reflection of P in circumcenter share one common (Co) conic 2. If P on circumcirle then
                Message 7 of 9 , Nov 11, 2006
                  Dear All My Friends,
                  Some properties of this circumconic:
                  1. P and reflection of P in circumcenter share one common (Co) conic
                  2. If P on circumcirle then perspector of (Co) conic = barycentric product of P and antipode of P wrt circumcircle.
                  3. Locus of all perspector of (Co) conic when P is on circumcircle is one circumconic with perspector X(184) passing through:
                  X(2966) = (Co) perspector of X(98), X(99)
                  X(112) = (Co) perspector of X(1113), X(1114)
                  X(1576) = (Co) perspector of X(1379), X(1380)
                  X(1415) = (Co) perspector of X(1381), X(1382)
                  This locus is also locus of all barycentric product of X and antipode of X wrt circumcircle when X is on circumcircle.
                  Thank you and best regards,
                  Bui Quang Tuan

                  Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                  I am very sorry. The features related with perspector Q, gQ, tgQ, X, X' of one conic and operation barycentric product are true for any circumconic.
                  This fact is interesting for me because it explains why our circumconic (Co) is related with Nagel point (when P = incenter). In fact, its perspector X(56) = gX(8) and tgX(8) = Gergonne point. If we take barycentric product of any point on circumcircle with Gergonne point then result point is on the (Co) conic centered at X(478).
                  It also explains why starting point of my messages is Nagel point.
                  Thank you and best regards,
                  Bui Quang Tuan

                  Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                  The general circumconic (Co) (with any point P) has one interesting feature:
                  Let Q is perspector of (Co), gQ is isogonal conjugate of Q, tgQ is isotomic conjugate of gQ, F is fourth intersection point of (Co) with circumcircle.
                  Let X is any point on circumcircle, X' is barycentric product of X and tgQ then:
                  1. X' is on the conic (Co).
                  2. X, X' and F are collinear.

                  We also can show that:
                  If X is any point on circumcircle, R is one fixed point, X' is barycentric product of X and R then line XX' is always passing through one fixed point F on circumcircle (other than X). Moreover, when X moving, the locus of X' is one fixed circumconic taken F as fourth intersection point. The perspector of this conic is g(tR).

                  Our circumconic (Co) belongs to this type of circumconics.

                  Thank you and best regards,
                  Bui Quang Tuan

                  Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                  If we choose P as circumcenter then we have one very simple circumconic:
                  - The barycentrics equation:
                  a^4*y*z + b^4*z*x + c^4*x*y = 0
                  - Perspector = X(32)
                  - Fourth intersection point = X(110)
                  - Center = X(206)
                  Please give me some advices and references about this circumconic!
                  Are there any other well known centers on this conic?
                  Thank you and best regards,
                  Bui Quang Tuan

                  Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends
                  We can generalize the circumconic (Co) as follow:
                  Suppose I is any point P with barycentrics (p : q : r). Construct A1B1C1 as before. Six points A, B, C, A1, B1, C1 are on one circumconic. The barycentric equation of this circumconic:
                  P*y*z + Q*z*x + R*x*y = 0
                  here:
                  P = (p*SC + a^2*q)*(p*SB + a^2*r)
                  Q = (q*SA + b^2*r)*(q*SC + b^2*p)
                  R = (r*SB + c^2*p)*(r*SA + c^2*q)
                  One example:
                  If P = centroid then this circumconic has barycentrics of:
                  - Perspector:
                  (b^2 - c^2)^2 - 9*a^4 : :
                  Search value: + 0.102966916856477
                  - Fourth intersection point with circumcircle
                  1/((b^2 - c^2)*(SA^2 + 2*b^2*c^2)) : :
                  Search value:+ 0.713049056090128
                  - Center:
                  (SC + a^2)*(SB + a^2)*(2*SA^2 - 5*(b^4 + c^4 - a^4)) : :
                  Search value: - 0.208642148311811
                  Thank you and best regards,
                  Bui Quang Tuan

                  Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                  By calculations I have found:
                  - Triangle AiBiCi has barycentrics:
                  Ai = (a^2 : sa*sb : sa*sc)
                  Bi = (sb*sa : b^2 : sb*sc)
                  Ci = (sc*sa : sc*sb : c^2)
                  - Segments AAi = A'Na = sa*sqr((a*sa + (b - c)^2)/(s*a)). (Similarly with B, C)
                  There are some things interesting with triangle AiBiCi:
                  - Medial triangle of AiBiCi and intouch triangle of ABC are perspective at X(3057)
                  - Three circumcircles of IAAi, IBBi, ICCi are concurrent at one point (other than I). This point is X(1319).
                  - Let three lines AcBc, BaCa, CbAb bound one triangle A1B1C1. Let B1C1 cuts BC at A2. Similarly define B2, C2 then three lines AA2, BB2, CC2 are concurrent at X(651).
                  - There is one circumconic (Co) passing through A, B, C, A1, B1, C1, X(651), X(109) with perspector = X(56) and center = X(478) (center of Yiu conic).
                  This circumconic has barycentric equation:
                  a^2/(b+c-a)*y*z + b^2/(c+a-b)*z*x + c^2/(a+b-c)*x*y = 0
                  - Two triangles A1B1C1, AiBiCi are homothetic at one point with barycentrics:
                  a^2*(b^3 + c^3 - a^3 - (a^2 + 3*b*c)*(b + c) + a*(b^2 + c^2) + 4*a*b*c) / (b+c-a) : :
                  Search value: +0.165893366771994
                  This point is on the line X(56)X(109)
                  - Centroid of AiBiCi has barycentrics:
                  a*((b + c)*(a^2 - (b - c)^2) - 8*a*b*c) : :
                  Search value: +2.07716924144103
                  This point is on the line X(1)X(3) and is reflection of X(210) in X(392).
                  Please give me some advices and references:
                  - There is special name for circumconic (Co)?
                  - There are some another well known centers on this conic?
                  Thank you and best regards,
                  Bui Quang Tuan

                  Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                  From incenter I we construct one line perpendicular to sidelines BC. This line cuts AB at Ab, cuts AC at Ac and cuts incircle (other than touch point with BC) at Ai. Similarly we construct Bc, Ba, Bi, and Ca, Cb, Ci.
                  Sometimes ago we have proved that three segmens AbAc, BcBa, CaCb have a feature: one segment is sum of other two. Now we can see some more interesting.
                  There are three pairs of congruent segments. In fact:
                  AiAc = BiBc; BiBa = CiCa; CiCb = AiAb
                  Moreover, it is very easy to show that there are three isosceles trapezia here:
                  BiCiCaBa; CiAiAbCb; AiBiBcAc
                  Also very easy to show that AAi, BBi, CCi are concurrent at Nagel point Na.
                  But there are three pairs of congruent segments which I still can not proof: let A'B'C' is Cevian triangle of Nagel point Na. Prove that:
                  AAi = A'Na; BBi = B'Na; CCi = C'Na
                  Please give me a simple proof!
                  Thank you and best regards,
                  Bui Quang Tuan

                  ---------------------------------
                  Sponsored Link

                  Try Netflix today! With plans starting at only $5.99 a month what are you waiting for?

                  [Non-text portions of this message have been removed]

                  ---------------------------------
                  Sponsored Link

                  For just $24.99/mo., Vonage offers unlimited local and long- distance calling. Sign up now.

                  [Non-text portions of this message have been removed]

                  ---------------------------------
                  Cheap Talk? Check out Yahoo! Messenger's low PC-to-Phone call rates.

                  [Non-text portions of this message have been removed]

                  ---------------------------------
                  Everyone is raving about the all-new Yahoo! Mail beta.

                  [Non-text portions of this message have been removed]

                  ---------------------------------
                  Want to start your own business? Learn how on Yahoo! Small Business.

                  [Non-text portions of this message have been removed]

                  ---------------------------------
                  Check out the all-new Yahoo! Mail beta - Fire up a more powerful email and get things done faster.

                  [Non-text portions of this message have been removed]






                  ---------------------------------
                  Everyone is raving about the all-new Yahoo! Mail beta.

                  [Non-text portions of this message have been removed]
                • Quang Tuan Bui
                  Dear All My Friends, Center of this locus circumconic passing through X(2966), X(112), X(1576), X(1415) with perspector X(184) has barycentrics: a^4*(a^8 + b^8
                  Message 8 of 9 , Nov 11, 2006
                    Dear All My Friends,
                    Center of this locus circumconic passing through X(2966), X(112), X(1576), X(1415) with perspector X(184) has barycentrics:
                    a^4*(a^8 + b^8 + c^8 - 2*b^4*c^4 - 2*a^6*(b^2 + c^2) + 2*a^4*(b^4 + b^2*c^2 + c^4) - 2*a^2*(b^6 + c^6)) : :
                    Search value: +0.362080697920292
                    Thank you and best regards,
                    Bui Quang Tuan

                    Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                    Some properties of this circumconic:
                    1. P and reflection of P in circumcenter share one common (Co) conic
                    2. If P on circumcirle then perspector of (Co) conic = barycentric product of P and antipode of P wrt circumcircle.
                    3. Locus of all perspector of (Co) conic when P is on circumcircle is one circumconic with perspector X(184) passing through:
                    X(2966) = (Co) perspector of X(98), X(99)
                    X(112) = (Co) perspector of X(1113), X(1114)
                    X(1576) = (Co) perspector of X(1379), X(1380)
                    X(1415) = (Co) perspector of X(1381), X(1382)
                    This locus is also locus of all barycentric product of X and antipode of X wrt circumcircle when X is on circumcircle.
                    Thank you and best regards,
                    Bui Quang Tuan

                    Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                    I am very sorry. The features related with perspector Q, gQ, tgQ, X, X' of one conic and operation barycentric product are true for any circumconic.
                    This fact is interesting for me because it explains why our circumconic (Co) is related with Nagel point (when P = incenter). In fact, its perspector X(56) = gX(8) and tgX(8) = Gergonne point. If we take barycentric product of any point on circumcircle with Gergonne point then result point is on the (Co) conic centered at X(478).
                    It also explains why starting point of my messages is Nagel point.
                    Thank you and best regards,
                    Bui Quang Tuan

                    Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                    The general circumconic (Co) (with any point P) has one interesting feature:
                    Let Q is perspector of (Co), gQ is isogonal conjugate of Q, tgQ is isotomic conjugate of gQ, F is fourth intersection point of (Co) with circumcircle.
                    Let X is any point on circumcircle, X' is barycentric product of X and tgQ then:
                    1. X' is on the conic (Co).
                    2. X, X' and F are collinear.

                    We also can show that:
                    If X is any point on circumcircle, R is one fixed point, X' is barycentric product of X and R then line XX' is always passing through one fixed point F on circumcircle (other than X). Moreover, when X moving, the locus of X' is one fixed circumconic taken F as fourth intersection point. The perspector of this conic is g(tR).

                    Our circumconic (Co) belongs to this type of circumconics.

                    Thank you and best regards,
                    Bui Quang Tuan

                    Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                    If we choose P as circumcenter then we have one very simple circumconic:
                    - The barycentrics equation:
                    a^4*y*z + b^4*z*x + c^4*x*y = 0
                    - Perspector = X(32)
                    - Fourth intersection point = X(110)
                    - Center = X(206)
                    Please give me some advices and references about this circumconic!
                    Are there any other well known centers on this conic?
                    Thank you and best regards,
                    Bui Quang Tuan

                    Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends
                    We can generalize the circumconic (Co) as follow:
                    Suppose I is any point P with barycentrics (p : q : r). Construct A1B1C1 as before. Six points A, B, C, A1, B1, C1 are on one circumconic. The barycentric equation of this circumconic:
                    P*y*z + Q*z*x + R*x*y = 0
                    here:
                    P = (p*SC + a^2*q)*(p*SB + a^2*r)
                    Q = (q*SA + b^2*r)*(q*SC + b^2*p)
                    R = (r*SB + c^2*p)*(r*SA + c^2*q)
                    One example:
                    If P = centroid then this circumconic has barycentrics of:
                    - Perspector:
                    (b^2 - c^2)^2 - 9*a^4 : :
                    Search value: + 0.102966916856477
                    - Fourth intersection point with circumcircle
                    1/((b^2 - c^2)*(SA^2 + 2*b^2*c^2)) : :
                    Search value:+ 0.713049056090128
                    - Center:
                    (SC + a^2)*(SB + a^2)*(2*SA^2 - 5*(b^4 + c^4 - a^4)) : :
                    Search value: - 0.208642148311811
                    Thank you and best regards,
                    Bui Quang Tuan

                    Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                    By calculations I have found:
                    - Triangle AiBiCi has barycentrics:
                    Ai = (a^2 : sa*sb : sa*sc)
                    Bi = (sb*sa : b^2 : sb*sc)
                    Ci = (sc*sa : sc*sb : c^2)
                    - Segments AAi = A'Na = sa*sqr((a*sa + (b - c)^2)/(s*a)). (Similarly with B, C)
                    There are some things interesting with triangle AiBiCi:
                    - Medial triangle of AiBiCi and intouch triangle of ABC are perspective at X(3057)
                    - Three circumcircles of IAAi, IBBi, ICCi are concurrent at one point (other than I). This point is X(1319).
                    - Let three lines AcBc, BaCa, CbAb bound one triangle A1B1C1. Let B1C1 cuts BC at A2. Similarly define B2, C2 then three lines AA2, BB2, CC2 are concurrent at X(651).
                    - There is one circumconic (Co) passing through A, B, C, A1, B1, C1, X(651), X(109) with perspector = X(56) and center = X(478) (center of Yiu conic).
                    This circumconic has barycentric equation:
                    a^2/(b+c-a)*y*z + b^2/(c+a-b)*z*x + c^2/(a+b-c)*x*y = 0
                    - Two triangles A1B1C1, AiBiCi are homothetic at one point with barycentrics:
                    a^2*(b^3 + c^3 - a^3 - (a^2 + 3*b*c)*(b + c) + a*(b^2 + c^2) + 4*a*b*c) / (b+c-a) : :
                    Search value: +0.165893366771994
                    This point is on the line X(56)X(109)
                    - Centroid of AiBiCi has barycentrics:
                    a*((b + c)*(a^2 - (b - c)^2) - 8*a*b*c) : :
                    Search value: +2.07716924144103
                    This point is on the line X(1)X(3) and is reflection of X(210) in X(392).
                    Please give me some advices and references:
                    - There is special name for circumconic (Co)?
                    - There are some another well known centers on this conic?
                    Thank you and best regards,
                    Bui Quang Tuan

                    Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                    From incenter I we construct one line perpendicular to sidelines BC. This line cuts AB at Ab, cuts AC at Ac and cuts incircle (other than touch point with BC) at Ai. Similarly we construct Bc, Ba, Bi, and Ca, Cb, Ci.
                    Sometimes ago we have proved that three segmens AbAc, BcBa, CaCb have a feature: one segment is sum of other two. Now we can see some more interesting.
                    There are three pairs of congruent segments. In fact:
                    AiAc = BiBc; BiBa = CiCa; CiCb = AiAb
                    Moreover, it is very easy to show that there are three isosceles trapezia here:
                    BiCiCaBa; CiAiAbCb; AiBiBcAc
                    Also very easy to show that AAi, BBi, CCi are concurrent at Nagel point Na.
                    But there are three pairs of congruent segments which I still can not proof: let A'B'C' is Cevian triangle of Nagel point Na. Prove that:
                    AAi = A'Na; BBi = B'Na; CCi = C'Na
                    Please give me a simple proof!
                    Thank you and best regards,
                    Bui Quang Tuan

                    ---------------------------------
                    Sponsored Link

                    Try Netflix today! With plans starting at only $5.99 a month what are you waiting for?

                    [Non-text portions of this message have been removed]

                    ---------------------------------
                    Sponsored Link

                    For just $24.99/mo., Vonage offers unlimited local and long- distance calling. Sign up now.

                    [Non-text portions of this message have been removed]

                    ---------------------------------
                    Cheap Talk? Check out Yahoo! Messenger's low PC-to-Phone call rates.

                    [Non-text portions of this message have been removed]

                    ---------------------------------
                    Everyone is raving about the all-new Yahoo! Mail beta.

                    [Non-text portions of this message have been removed]

                    ---------------------------------
                    Want to start your own business? Learn how on Yahoo! Small Business.

                    [Non-text portions of this message have been removed]

                    ---------------------------------
                    Check out the all-new Yahoo! Mail beta - Fire up a more powerful email and get things done faster.

                    [Non-text portions of this message have been removed]

                    ---------------------------------
                    Everyone is raving about the all-new Yahoo! Mail beta.

                    [Non-text portions of this message have been removed]






                    ---------------------------------
                    Access over 1 million songs - Yahoo! Music Unlimited.

                    [Non-text portions of this message have been removed]
                  • Quang Tuan Bui
                    Dear All My Friends, In the case P = incenter, three lines AA2, BB2, CC2 are concurrent at X(651). In general these three lines are not concurrent. By some
                    Message 9 of 9 , Nov 12, 2006
                      Dear All My Friends,
                      In the case P = incenter, three lines AA2, BB2, CC2 are concurrent at X(651). In general these three lines are not concurrent. By some complicate calculations we can show that AA2, BB2, CC2 are concurrent if P is on the Darboux cubic. In the concurrent case we denote concurrent point of AA2, BB2, CC2 as R then R is on the (Co) conic of this case.
                      With each well known center on Darboux cubic we can calculate barycentrics of concurrent point R:
                      1. P = incenter
                      R = X(651)
                      2. P = circumcenter
                      R = X(110)
                      3. P = de Longchamp point
                      R = X(648)
                      4. P = Bevan point
                      R = X(651) (as P = incenter)
                      5. P = X(64)
                      R = a^2 / ((b^2 - c^2)*(-3*a^4 + (b^2 - c^2)^2 + 2*a^2*(b^2 + c^2))) : :
                      Search value: -1.71822904800113
                      6. P = X(84)
                      R = a / ((b - c)*(a^3 + a^2*(b + c) - (b - c)^2*(b + c) - a*(b + c)^2)) : :
                      Search value: -2.0268140785765
                      7. P = X(1490) ( as P = X(84) )
                      8. P = X(2130) or P = X(2131) then the coordinates of R are very complicated.

                      It is interesting that all these concurrent points are on MacBeath circumconic. (Center = symmedian point, perspector = circumcenter).
                      Thank you and best regards,
                      Bui Quang Tuan


                      Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                      Center of this locus circumconic passing through X(2966), X(112), X(1576), X(1415) with perspector X(184) has barycentrics:
                      a^4*(a^8 + b^8 + c^8 - 2*b^4*c^4 - 2*a^6*(b^2 + c^2) + 2*a^4*(b^4 + b^2*c^2 + c^4) - 2*a^2*(b^6 + c^6)) : :
                      Search value: +0.362080697920292
                      Thank you and best regards,
                      Bui Quang Tuan

                      Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                      Some properties of this circumconic:
                      1. P and reflection of P in circumcenter share one common (Co) conic
                      2. If P on circumcirle then perspector of (Co) conic = barycentric product of P and antipode of P wrt circumcircle.
                      3. Locus of all perspector of (Co) conic when P is on circumcircle is one circumconic with perspector X(184) passing through:
                      X(2966) = (Co) perspector of X(98), X(99)
                      X(112) = (Co) perspector of X(1113), X(1114)
                      X(1576) = (Co) perspector of X(1379), X(1380)
                      X(1415) = (Co) perspector of X(1381), X(1382)
                      This locus is also locus of all barycentric product of X and antipode of X wrt circumcircle when X is on circumcircle.
                      Thank you and best regards,
                      Bui Quang Tuan

                      Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                      I am very sorry. The features related with perspector Q, gQ, tgQ, X, X' of one conic and operation barycentric product are true for any circumconic.
                      This fact is interesting for me because it explains why our circumconic (Co) is related with Nagel point (when P = incenter). In fact, its perspector X(56) = gX(8) and tgX(8) = Gergonne point. If we take barycentric product of any point on circumcircle with Gergonne point then result point is on the (Co) conic centered at X(478).
                      It also explains why starting point of my messages is Nagel point.
                      Thank you and best regards,
                      Bui Quang Tuan

                      Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                      The general circumconic (Co) (with any point P) has one interesting feature:
                      Let Q is perspector of (Co), gQ is isogonal conjugate of Q, tgQ is isotomic conjugate of gQ, F is fourth intersection point of (Co) with circumcircle.
                      Let X is any point on circumcircle, X' is barycentric product of X and tgQ then:
                      1. X' is on the conic (Co).
                      2. X, X' and F are collinear.

                      We also can show that:
                      If X is any point on circumcircle, R is one fixed point, X' is barycentric product of X and R then line XX' is always passing through one fixed point F on circumcircle (other than X). Moreover, when X moving, the locus of X' is one fixed circumconic taken F as fourth intersection point. The perspector of this conic is g(tR).

                      Our circumconic (Co) belongs to this type of circumconics.

                      Thank you and best regards,
                      Bui Quang Tuan

                      Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                      If we choose P as circumcenter then we have one very simple circumconic:
                      - The barycentrics equation:
                      a^4*y*z + b^4*z*x + c^4*x*y = 0
                      - Perspector = X(32)
                      - Fourth intersection point = X(110)
                      - Center = X(206)
                      Please give me some advices and references about this circumconic!
                      Are there any other well known centers on this conic?
                      Thank you and best regards,
                      Bui Quang Tuan

                      Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends
                      We can generalize the circumconic (Co) as follow:
                      Suppose I is any point P with barycentrics (p : q : r). Construct A1B1C1 as before. Six points A, B, C, A1, B1, C1 are on one circumconic. The barycentric equation of this circumconic:
                      P*y*z + Q*z*x + R*x*y = 0
                      here:
                      P = (p*SC + a^2*q)*(p*SB + a^2*r)
                      Q = (q*SA + b^2*r)*(q*SC + b^2*p)
                      R = (r*SB + c^2*p)*(r*SA + c^2*q)
                      One example:
                      If P = centroid then this circumconic has barycentrics of:
                      - Perspector:
                      (b^2 - c^2)^2 - 9*a^4 : :
                      Search value: + 0.102966916856477
                      - Fourth intersection point with circumcircle
                      1/((b^2 - c^2)*(SA^2 + 2*b^2*c^2)) : :
                      Search value:+ 0.713049056090128
                      - Center:
                      (SC + a^2)*(SB + a^2)*(2*SA^2 - 5*(b^4 + c^4 - a^4)) : :
                      Search value: - 0.208642148311811
                      Thank you and best regards,
                      Bui Quang Tuan

                      Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                      By calculations I have found:
                      - Triangle AiBiCi has barycentrics:
                      Ai = (a^2 : sa*sb : sa*sc)
                      Bi = (sb*sa : b^2 : sb*sc)
                      Ci = (sc*sa : sc*sb : c^2)
                      - Segments AAi = A'Na = sa*sqr((a*sa + (b - c)^2)/(s*a)). (Similarly with B, C)
                      There are some things interesting with triangle AiBiCi:
                      - Medial triangle of AiBiCi and intouch triangle of ABC are perspective at X(3057)
                      - Three circumcircles of IAAi, IBBi, ICCi are concurrent at one point (other than I). This point is X(1319).
                      - Let three lines AcBc, BaCa, CbAb bound one triangle A1B1C1. Let B1C1 cuts BC at A2. Similarly define B2, C2 then three lines AA2, BB2, CC2 are concurrent at X(651).
                      - There is one circumconic (Co) passing through A, B, C, A1, B1, C1, X(651), X(109) with perspector = X(56) and center = X(478) (center of Yiu conic).
                      This circumconic has barycentric equation:
                      a^2/(b+c-a)*y*z + b^2/(c+a-b)*z*x + c^2/(a+b-c)*x*y = 0
                      - Two triangles A1B1C1, AiBiCi are homothetic at one point with barycentrics:
                      a^2*(b^3 + c^3 - a^3 - (a^2 + 3*b*c)*(b + c) + a*(b^2 + c^2) + 4*a*b*c) / (b+c-a) : :
                      Search value: +0.165893366771994
                      This point is on the line X(56)X(109)
                      - Centroid of AiBiCi has barycentrics:
                      a*((b + c)*(a^2 - (b - c)^2) - 8*a*b*c) : :
                      Search value: +2.07716924144103
                      This point is on the line X(1)X(3) and is reflection of X(210) in X(392).
                      Please give me some advices and references:
                      - There is special name for circumconic (Co)?
                      - There are some another well known centers on this conic?
                      Thank you and best regards,
                      Bui Quang Tuan

                      Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
                      From incenter I we construct one line perpendicular to sidelines BC. This line cuts AB at Ab, cuts AC at Ac and cuts incircle (other than touch point with BC) at Ai. Similarly we construct Bc, Ba, Bi, and Ca, Cb, Ci.
                      Sometimes ago we have proved that three segmens AbAc, BcBa, CaCb have a feature: one segment is sum of other two. Now we can see some more interesting.
                      There are three pairs of congruent segments. In fact:
                      AiAc = BiBc; BiBa = CiCa; CiCb = AiAb
                      Moreover, it is very easy to show that there are three isosceles trapezia here:
                      BiCiCaBa; CiAiAbCb; AiBiBcAc
                      Also very easy to show that AAi, BBi, CCi are concurrent at Nagel point Na.
                      But there are three pairs of congruent segments which I still can not proof: let A'B'C' is Cevian triangle of Nagel point Na. Prove that:
                      AAi = A'Na; BBi = B'Na; CCi = C'Na
                      Please give me a simple proof!
                      Thank you and best regards,
                      Bui Quang Tuan

                      ---------------------------------
                      Sponsored Link

                      Try Netflix today! With plans starting at only $5.99 a month what are you waiting for?

                      [Non-text portions of this message have been removed]

                      ---------------------------------
                      Sponsored Link

                      For just $24.99/mo., Vonage offers unlimited local and long- distance calling. Sign up now.

                      [Non-text portions of this message have been removed]

                      ---------------------------------
                      Cheap Talk? Check out Yahoo! Messenger's low PC-to-Phone call rates.

                      [Non-text portions of this message have been removed]

                      ---------------------------------
                      Everyone is raving about the all-new Yahoo! Mail beta.

                      [Non-text portions of this message have been removed]

                      ---------------------------------
                      Want to start your own business? Learn how on Yahoo! Small Business.

                      [Non-text portions of this message have been removed]

                      ---------------------------------
                      Check out the all-new Yahoo! Mail beta - Fire up a more powerful email and get things done faster.

                      [Non-text portions of this message have been removed]

                      ---------------------------------
                      Everyone is raving about the all-new Yahoo! Mail beta.

                      [Non-text portions of this message have been removed]

                      ---------------------------------
                      Access over 1 million songs - Yahoo! Music Unlimited.

                      [Non-text portions of this message have been removed]






                      ---------------------------------
                      Everyone is raving about the all-new Yahoo! Mail beta.

                      [Non-text portions of this message have been removed]
                    Your message has been successfully submitted and would be delivered to recipients shortly.