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Θέμα: [EMHL] Re: two excircles

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  • Nikolaos Dergiades
    Dear Alexei and Jean-Pierre another approach that needs an elementary construction is the following: but I can not give a synthetic proof. If X, Y are points
    Message 1 of 17 , Jul 1, 2007
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      Dear Alexei and Jean-Pierre
      another approach that needs an elementary
      construction is the following:
      but I can not give a synthetic proof.
      If X, Y are points on the sides AB, AC
      and the perpendicular bisector of XY meets
      the radical axis of the excircles (Ib), (Ic)
      or the radical axis of (I), (Ia) at the point Z
      then the circumcircle of XYZ is tangent to
      (Ib), (Ic) or to (I), (Ia) if and only if
      XY is parallel of antiparallel to BC.

      Jean-Pierre what is the general statement
      of the property of hyperbola you used?

      Best regards
      Nikos Dergiades


      > [Alexei]
      > > Let the points X, Y be on the sidelines AC, BC of
      > the triangle ABC
      > > and XY is parallel to AB. Then there exists a
      > circle passing through
      > > X, Y and touching two excircles of the triangle.
      >
      >[JPE] More precisely :
      > - there exists a circle through X, Y touching the
      > A-excircle and the B-
      > excircle
      > - there exists a circle through X, Y touching the
      > incircle and the C-
      > excircle
      > We have the same result if XY is antiparallel to AB
      >
      > Here are some remarks (I've changed the notations
      > and I didn't find a
      > synthetic proof)
      > Let I,Ia,Ib,Ic be the incenter and excenters; U,U'
      > the feet on BC of
      > the internal and external A-bisectors
      > Then there exists
      > - an hyperbola h with center U going through the 4
      > projections of I and
      > Ia upon AB and AC (the asymptots are BC and the
      > other common tangent
      > through U to the incircle and the A-excircle)
      > - an hyperbola h' with center U' going through the 4
      > projections of Ib
      > and Ic upon AB and AC (the asymptots are BC and the
      > other common
      > tangent through U' to the B- and C-excircles)
      >
      > Consider a point M lying on h; the parallel and
      > antiparallel to BC
      > through M intersect AB and AC at four points lying
      > on a circle, and
      > this circle touches the incircle and the A-excircle
      > (and the line
      > through the contact points goes through U)
      > Consider a point M' lying on h'; the parallel and
      > antiparallel to BC
      > through M' intersect AB and AC at four points lying
      > on a circle, and
      > this circle touches the B-excircle and the
      > C-excircle (and the line
      > through the contact points goes through U')







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    • Nikolaos Dergiades
      I repeat to corredt my previous message Dear Alexei and Jean-Pierre another approach that needs an elementary construction is the following: but I can not give
      Message 2 of 17 , Jul 1, 2007
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        I repeat to corredt my previous message

        Dear Alexei and Jean-Pierre
        another approach that needs an elementary
        construction is the following:
        but I can not give a synthetic proof.
        If X, Y are points on the sides AB, AC
        XY is parallel of antiparallel to BC
        and the perpendicular bisector of XY meets
        the radical axis of the excircles (Ib), (Ic)
        or the radical axis of (I), (Ia) at the point Z
        then the circumcircle of XYZ is tangent to
        (Ib), (Ic) or to (I), (Ia).

        Jean-Pierre what is the general statement
        of the property of hyperbola you used?

        Best regards
        Nikos Dergiades


        > [Alexei]
        > > Let the points X, Y be on the sidelines AC, BC of
        > the triangle ABC
        > > and XY is parallel to AB. Then there exists a
        > circle passing through
        > > X, Y and touching two excircles of the triangle.
        >
        >[JPE] More precisely :
        > - there exists a circle through X, Y touching the
        > A-excircle and the B-
        > excircle
        > - there exists a circle through X, Y touching the
        > incircle and the C-
        > excircle
        > We have the same result if XY is antiparallel to AB
        >
        > Here are some remarks (I've changed the notations
        > and I didn't find a
        > synthetic proof)
        > Let I,Ia,Ib,Ic be the incenter and excenters; U,U'
        > the feet on BC of
        > the internal and external A-bisectors
        > Then there exists
        > - an hyperbola h with center U going through the 4
        > projections of I and
        > Ia upon AB and AC (the asymptots are BC and the
        > other common tangent
        > through U to the incircle and the A-excircle)
        > - an hyperbola h' with center U' going through the 4
        > projections of Ib
        > and Ic upon AB and AC (the asymptots are BC and the
        > other common
        > tangent through U' to the B- and C-excircles)
        >
        > Consider a point M lying on h; the parallel and
        > antiparallel to BC
        > through M intersect AB and AC at four points lying
        > on a circle, and
        > this circle touches the incircle and the A-excircle
        > (and the line
        > through the contact points goes through U)
        > Consider a point M' lying on h'; the parallel and
        > antiparallel to BC
        > through M' intersect AB and AC at four points lying
        > on a circle, and
        > this circle touches the B-excircle and the
        > C-excircle (and the line
        > through the contact points goes through U')








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      • Luís Lopes
        Dear Hyacinthists, I don t know what happens/has hapenned but many messages related to this thread, including all mine, didn t show up on my inbox I will reply
        Message 3 of 17 , Jul 2, 2007
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          Dear Hyacinthists,

          I don't know what happens/has hapenned but many
          messages related to this thread, including all mine,
          didn't show up on my inbox

          I will reply to François directly from the archive.

          Dear Martin Acosta,

          That is exactly what my problem is and your solution
          is what I am looking for. I still have to understand it
          so I could construct I'.

          Best regards,
          Luis


          >From: "Martin Acosta" <maedu@...>
          >Reply-To: Hyacinthos@yahoogroups.com
          >To: Hyacinthos@yahoogroups.com
          >Subject: Re: [EMHL] line/conic intersection
          >Date: Sat, 30 Jun 2007 10:07:44 +0000
          >
          >I think your problem is to draw a line by the focus and the intersection
          >between d and r, when this intersection is out of reach. Let name I this
          >intersection point.
          >
          >You could draw a point D on d and a point R on r.
          >So You have the triangle DRI.
          >You cand find a homothetic of this triangle around F, so that I' is on
          >screen.
          >the line through F and I' passes through I.
          >
          >Martin
          >
          >
          >
          > >From: "Francois Rideau" <francois.rideau@...>
          > >Reply-To: Hyacinthos@yahoogroups.com
          > >To: Hyacinthos@yahoogroups.com
          > >Subject: Re: [EMHL] line/conic intersection
          > >Date: Fri, 29 Jun 2007 18:17:17 +0200
          > >
          > >Dear Luis

          [clip]

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        • qedtexte
          Construct (with R&C) the intersection of a line with a conic (parabola) given by its focus and directrix . Dear Hyacinthists, Dear François, I will
          Message 4 of 17 , Jul 2, 2007
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            Construct (with R&C) the intersection of a line <r>
            with a conic (parabola) given by its focus <F> and
            directrix <d>.

            Dear Hyacinthists,

            Dear François,

            I will reply directly from the archives.
            I have never received this message.

            ===
            >The center of C is one of the intersections sought.
            >Otherwise perform a homothecy of C of pole P.
            > Please, gimme exactly your construction with your homothecy!

            I will suppose the "ideal" position of the data.

            Let P=d/\r. With P_0 on r as center draw a circle C
            tangent to d. Let P_1 be the contact point.
            Draw s=(P,F) and let F_1 and F_2 be the intersections
            between <s> and <C>. Draw t=(P_0,F_1) and u=(P_0,F_2).
            Draw from <F> t'||t and u'||u. The intersections O=r/\t'
            and O'=r/\u' are the intersections of <r> and the parabola.

            Your solution seems nice but I still have to
            understand it.

            Friendly,
            Luis



            --- In Hyacinthos@yahoogroups.com, "Francois Rideau"
            <francois.rideau@...> wrote:
            >
            > Dear Luis
            > ===
            > I think we have a big problem when <d> and <r> are
            > near orthogonal
            > ===
            > I don't think so. In this case one has P=d/\r and with P
            > as pole one makes a homothecy. Draw a circle C with center
            > in r and tangent do d. If C goes through F fine.
            >
            > >
            > > That's just the problem of this construction: to find circles
            tangent to
            > > the directrix <d> and through the focus <F> of which the centers
            are on line
            > > <r>. These centers are the sought points of intersection. These
            circles are
            > > also on point <F'> symmetric of <F> wrt line <r>. So these
            circles are in a
            > > pencil and their intersections with <d> are in involution,
            (Desargues
            > > theorem). For example, if <Gamma> is a circle in this pencil
            cutting <d> in
            > > <P> and <Q>, then JP.JQ = JF.JF' = power of J wrt <Gamma>,
            (product of
            > > oriented segments), where J = FF' /\ <d>. In case of a touching
            circle in T,
            > > we have JT² = JF.JF'. The circle of center J and radius JT is
            orthogonal
            > > to all circles of the pencil. So draw any circle of the pencil
            and draw
            > > segments tangents to this circle through J and so on...
          • Luís Lopes
            Construct (with R&C) the intersection of a line with a conic (parabola) given by its focus and directrix . Dear Hyacinthists, Dear François, I will
            Message 5 of 17 , Jul 2, 2007
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              Construct (with R&C) the intersection of a line <r>
              with a conic (parabola) given by its focus <F> and
              directrix <d>.

              Dear Hyacinthists,

              Dear François,

              I will reply directly from the archives.
              I have never received the message below.

              To help me (you) in the future: look for msgs
              15338-342, 15344, 15352.
              I am still experiecing problems sending/receiving msgs.
              Sorry for double postings.

              ===
              The center of C is one of the intersections sought.
              Otherwise perform a homothecy of C of pole P.
              Please, gimme exactly your construction with your homothecy!


              I will suppose the "ideal" position of the data.

              Let P=d/\r. With P_0 on r as center draw a circle C
              tangent to d. Let P_1 be the contact point.
              Draw s=(P,F) and let F_1 and F_2 be the intersections
              between <s> and <C>. Draw t=(P_0,F_1) and u=(P_0,F_2).
              Draw from <F> t'||t and u'||u. The intersections O=r/\t'
              and O'=r/\u' are the intersections of <r> and the parabola.

              Your solution seems nice but I still have to
              understand it.

              Friendly,
              Luis



              --- In Hyacinthos@yahoogroups.com, "Francois Rideau" <francois.rideau@...>
              wrote:

              Dear Luis
              ===
              I think we have a big problem when <d> and <r> are
              near orthogonal
              ===
              I don't think so. In this case one has P=d/\r and with P
              as pole one makes a homothecy. Draw a circle C with center
              in r and tangent do d. If C goes through F fine.

              >
              >That's just the problem of this construction: to find circles tangent to
              >the directrix <d> and through the focus <F> of which the centers are on
              >line <r>. These centers are the sought points of intersection. These
              >circles are also on point <F'> symmetric of <F> wrt line <r>. So these
              >circles are in a pencil and their intersections with <d> are in involution,
              >(Desargues theorem). For example, if <Gamma> is a circle in this pencil
              >cutting <d> in <P> and <Q>, then JP.JQ = JF.JF' = power of J wrt <Gamma>,
              >(product of oriented segments), where J = FF' /\ <d>. In case of a touching
              >circle in T, we have JT² = JF.JF'. The circle of center J and radius JT is
              >orthogonal to all circles of the pencil. So draw any circle of the pencil
              >and draw segments tangents to this circle through J and so on...

              _________________________________________________________________
              MSN Busca: fácil, rápido, direto ao ponto. http://search.msn.com.br
            • Francois Rideau
              Yes, nice solution and now I remember that I learned it more than fifty years ago. Of course the most important is the discussion which must follows on the
              Message 6 of 17 , Jul 2, 2007
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                Yes, nice solution and now I remember that I learned it more than fifty
                years ago.
                Of course the most important is the discussion which must follows on the
                number of solutions.
                About the teaching of the theory of conics, I think the best book was
                written by the great Henri Lebesgue himself: Les coniques, published in 1942
                at Gauthier-Villars, just after his death, during the nazi occupation of
                France.
                Poor Henri, nowadays, the only conic still in use in France is the circle on
                roadsigns just before runabouts!
                Friendly
                Francois


                On 7/2/07, Luís Lopes <qed_texte@...> wrote:
                >
                > Construct (with R&C) the intersection of a line <r>
                > with a conic (parabola) given by its focus <F> and
                > directrix <d>.
                >
                > Dear Hyacinthists,
                >
                > Dear François,
                >
                > I will reply directly from the archives.
                > I have never received the message below.
                >
                > To help me (you) in the future: look for msgs
                > 15338-342, 15344, 15352.
                > I am still experiecing problems sending/receiving msgs.
                > Sorry for double postings.
                >
                > ===
                > The center of C is one of the intersections sought.
                > Otherwise perform a homothecy of C of pole P.
                > Please, gimme exactly your construction with your homothecy!
                >
                > I will suppose the "ideal" position of the data.
                >
                > Let P=d/\r. With P_0 on r as center draw a circle C
                > tangent to d. Let P_1 be the contact point.
                > Draw s=(P,F) and let F_1 and F_2 be the intersections
                > between <s> and <C>. Draw t=(P_0,F_1) and u=(P_0,F_2).
                > Draw from <F> t'||t and u'||u. The intersections O=r/\t'
                > and O'=r/\u' are the intersections of <r> and the parabola.
                >
                > Your solution seems nice but I still have to
                > understand it.
                >
                > Friendly,
                > Luis
                >
                > --- In Hyacinthos@yahoogroups.com <Hyacinthos%40yahoogroups.com>,
                > "Francois Rideau" <francois.rideau@...>
                > wrote:
                >
                > Dear Luis
                > ===
                > I think we have a big problem when <d> and <r> are
                > near orthogonal
                > ===
                > I don't think so. In this case one has P=d/\r and with P
                > as pole one makes a homothecy. Draw a circle C with center
                > in r and tangent do d. If C goes through F fine.
                >
                > >
                > >That's just the problem of this construction: to find circles tangent to
                > >the directrix <d> and through the focus <F> of which the centers are on
                > >line <r>. These centers are the sought points of intersection. These
                > >circles are also on point <F'> symmetric of <F> wrt line <r>. So these
                > >circles are in a pencil and their intersections with <d> are in
                > involution,
                > >(Desargues theorem). For example, if <Gamma> is a circle in this pencil
                > >cutting <d> in <P> and <Q>, then JP.JQ = JF.JF' = power of J wrt <Gamma>,
                >
                > >(product of oriented segments), where J = FF' /\ <d>. In case of a
                > touching
                > >circle in T, we have JT² = JF.JF'. The circle of center J and radius JT
                > is
                > >orthogonal to all circles of the pencil. So draw any circle of the pencil
                >
                > >and draw segments tangents to this circle through J and so on...
                >
                > __________________________________________________________
                > MSN Busca: fácil, rápido, direto ao ponto. http://search.msn.com.br
                >
                >
                >


                [Non-text portions of this message have been removed]
              • Luís Lopes
                Dear Hyacinthists, Dear François, It seems the messages are leaving/arriving normally again. I see now your solution is to draw a circle through two given
                Message 7 of 17 , Jul 2, 2007
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                  Dear Hyacinthists,

                  Dear François,

                  It seems the messages are leaving/arriving normally again.

                  I see now your solution is to draw a circle through two
                  given points and tangent to a line. This is classical.

                  One can have two solutions. These are the circumcenters
                  of the triangles ABC given <h_a,d_a,m_b>.

                  Barukh Ziv some 5 years ago sent me a message showing
                  that O belongs to a parabola <P=(F,d)> and to h'_a,
                  reflection of h_a on d_a.

                  The directrix <d> is parallel to (B,C) and distant m_b^2/4h_a
                  from it. The focus <F> is distant \ell/4 from (A,H_a) and
                  9h_a/8 from <d>. Here \ell=\sqrt{m_b^2-(h_a/2)^2}.

                  Data to have 2 solutions (from Die Konstruktion von
                  Dreiecken, Kurt Herterich).

                  <h_a=6 cm>, <m_b=5 cm>, and <d_a=6.5 cm>;

                  <h_a=3 cm>, <m_b=3.9 cm>, and <d_a=5 cm>;

                  Best regards,
                  Luis

                  >From: "Francois Rideau" <francois.rideau@...>
                  >Reply-To: Hyacinthos@yahoogroups.com
                  >To: Hyacinthos@yahoogroups.com
                  >Subject: Re: [EMHL] line/conic intersection
                  >Date: Mon, 2 Jul 2007 20:21:59 +0200
                  >
                  >Yes, nice solution and now I remember that I learned it more than fifty
                  >years ago.
                  >Of course the most important is the discussion which must follows on the
                  >number of solutions.
                  >About the teaching of the theory of conics, I think the best book was
                  >written by the great Henri Lebesgue himself: Les coniques, published in
                  >1942
                  >at Gauthier-Villars, just after his death, during the nazi occupation of
                  >France.
                  >Poor Henri, nowadays, the only conic still in use in France is the circle
                  >on
                  >roadsigns just before runabouts!
                  >Friendly
                  >Francois
                  >
                  >
                  >On 7/2/07, Luís Lopes <qed_texte@...> wrote:
                  [clip]

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