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RE: [EMHL] Re: AIG construction

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  • Nikolaos Dergiades
    Dear Luis, A construction for the first problem is the following: Construct an angle xAy as the given and on the bisector of this angle the point I such that
    Message 1 of 8 , Mar 24, 2005
      Dear Luis,
      A construction for the first problem is the following:
      Construct an angle xAy as the given and on the bisector of this
      angle the point I such that the distance from Ay is IK = r.
      The circle (I, IK) meets the segment AI at L and the parallel from
      L to Ay meets IK at N.
      On Ay we take the point D such that ID = m_a
      and on the extention of AI we take the segment IE = KD.
      If F is the orthogonal projection of E on the line LN
      the line IF meets the circle (I, IN) at two points X, X'
      Take on AE the point H such that AH = FX (FX > FX')
      The perpendicular to AH at H meets the circle (A, m_a)
      at M the mid point of BC and hence BC can be constructed
      as tangent to the incircle (I, IK).

      I think an analogus construction we can have for the other problem
      but I have no time to think about it.

      Best regards
      Nikolaos Dergiades

      > Dear Hyacinthists,
      >
      > Dear Eric,
      >
      > I have been looking for a "true" geometric construction of
      > both problems A(\alpha) , m_a, r and A(\alpha), m_a, r_a,
      > where m_a is the median from A.
      >
      > One can then construct A, I and draw the incircle. And
      > similarly, A, I_a and draw the excircle. Now would it be
      > possible to construct the point G ?
      >
      > Best regards,
      > Luis
      >
    • Francois Rideau
      There another riddles, affine ones just to forget Euclid during some time! Let A_{0}, A_[1}, A_{2}, 3 points in an affine ( possibly euclidian for Euclid
      Message 2 of 8 , Mar 25, 2005
        There another riddles, affine ones just to forget Euclid
        during some time!
        Let A_{0}, A_[1}, A_{2}, 3 points in an affine ( possibly euclidian
        for Euclid lovers) real plane not in the same line.

        Let A_{3} another point acting as a parameter.

        Let f the affine map defined by:
        f(A_{0}) = A_{1}; f(A_{1}) = A_{2}; f(A_{2}) = A_{3}

        We also define recursively the sequence { n --> A_{n} } by:
        A_{n+1} = f(A_{n}).
        Answer the few following questions and you can imagine
        scores of other ones:
        1° What are the sets of A_{3} points so that
        f has just one fixed point O and examine the map
        A_{3} --> O.
        2° What is the locus of A_{3} so that f has no fixed points?
        3° What is the set of A_{3} so that f has a line of fixed points?
        4° What is the locus of A_{3} so that f is parallelogic?
        5° If the plane is euclidian, (cheer Euclid!), what is the locus of
        A_3 so that f is orthologic and in that case what is the locus of the
        fixed point of f?
        6°What is the set of A_{3} so that
        the sequence { n --> A{n} } is bounded?
        7° What is the set of A_{3} so that
        the sequence { n --> A_{n} } converges (to the fixed point of f) ?
        8° What is the set of A_{3} so that
        the sequence { n --> A_{n} } is periodic.
        9° What is the set of A_{3} so that
        the sequence { n --> A_{n} } has accumulation points?
        10° What is the locus of A_{3} so that the range
        of the sequence { n --> A_{n} } is on a conic?
        Separate on that locus the elliptic, hyperbolic and
        parabolic cases.
        And so one....
        I hope you will find these riddles interesting.
        I have all the answers if you want!
        Friendly yours
        François
      • Luís Lopes
        Dear Hyacinthists, Dear Nikolaos Dergiades, Thank you very much for this solution. It is exactly this kind of solution I was looking for. Unfortunately I
        Message 3 of 8 , Mar 28, 2005
          Dear Hyacinthists,

          Dear Nikolaos Dergiades,

          Thank you very much for this solution. It is exactly this
          kind of solution I was looking for.

          Unfortunately I wasn't able to justify it. Could you please
          provide the general ideas leading to this elegant solution?

          Maybe afterwards I would be able to understand it and mimic
          it to solve the second (with r_a) problem.

          Best regards,
          Luis


          >From: "Nikolaos Dergiades" <ndergiades@...>
          >Reply-To: Hyacinthos@yahoogroups.com
          >To: <Hyacinthos@yahoogroups.com>
          >Subject: RE: [EMHL] Re: AIG construction
          >Date: Thu, 24 Mar 2005 21:38:57 +0200
          >
          >Dear Luis,
          >A construction for the first problem is the following:
          >Construct an angle xAy as the given and on the bisector of this
          >angle the point I such that the distance from Ay is IK = r.
          >The circle (I, IK) meets the segment AI at L and the parallel from
          >L to Ay meets IK at N.
          >On Ay we take the point D such that ID = m_a
          >and on the extention of AI we take the segment IE = KD.
          >If F is the orthogonal projection of E on the line LN
          >the line IF meets the circle (I, IN) at two points X, X'
          >Take on AE the point H such that AH = FX (FX > FX')
          >The perpendicular to AH at H meets the circle (A, m_a)
          >at M the mid point of BC and hence BC can be constructed
          >as tangent to the incircle (I, IK).
          >
          >I think an analogus construction we can have for the other problem
          >but I have no time to think about it.
          >
          >Best regards
          >Nikolaos Dergiades
          >
        • Nikolaos Dergiades
          Dear Luis, [LL] ... [ND] ... [LL] ... I worked as follows: In Cartesian coordinates we consider the points A(0,0), I(d,0), M(g,h). Let S(2g,2h) be the
          Message 4 of 8 , Mar 30, 2005
            Dear Luis,

            [LL]
            >I have been looking for a "true" geometric construction of
            >both problems A(\alpha) , m_a, r and A(\alpha), m_a, r_a,
            >where m_a is the median from A.

            [ND]
            > >A construction for the first problem is the following:
            > >Construct an angle xAy as the given and on the bisector of this
            > >angle the point I such that the distance from Ay is IK = r.
            > >The circle (I, IK) meets the segment AI at L and the parallel from
            > >L to Ay meets IK at N.
            > >On Ay we take the point D such that ID = m_a
            > >and on the extention of AI we take the segment IE = KD.
            > >If F is the orthogonal projection of E on the line LN
            > >the line IF meets the circle (I, IN) at two points X, X'
            > >Take on AE the point H such that AH = FX (FX > FX')
            > >The perpendicular to AH at H meets the circle (A, m_a)
            > >at M the mid point of BC and hence BC can be constructed
            > >as tangent to the incircle (I, IK).

            [LL]
            > Thank you very much for this solution. It is exactly this
            > kind of solution I was looking for.
            >
            > Unfortunately I wasn't able to justify it. Could you please
            > provide the general ideas leading to this elegant solution?
            >
            > Maybe afterwards I would be able to understand it and mimic
            > it to solve the second (with r_a) problem.


            I worked as follows:
            In Cartesian coordinates we consider the points
            A(0,0), I(d,0), M(g,h).
            Let S(2g,2h) be the reflection of A in M.

            The tangent of the angle IAB is t = r/sqrt(d^2-r^2).
            the equation of the line AB is y= tx
            the equation of the line AC is y= -tx
            The line SC: y-2h = t(x-2g) is parallel to AB
            Hence solving the system of SC, AC we find
            the point C(g-h/t,h-tg). Similarly we find the point
            B(g+h/t,h+tg) and then the equation of the line
            BC: t^2gx-hy+h^2-t^2g^2=0
            From the formula of the distance r of I from BC
            substituting t and h from h^2=m^2-g^2 where m = AM
            we get the equation
            (dg)^2=m^2(d^2-r^2) which is not true
            or the equation
            dg(dg-2r^2)=(d^2-r^2)(m^2-r^2) or
            g(g-2r^2/d)=(1-(r/d)^2)(m^2-r^2) or
            g(g-2IN)=cos^2(A/2)(m^2-r^2) or
            g(g-XX')=(cos(A/2)IE)^2 or
            g(g-XX')=NF^2 or g = FX = AH
            Since g is the first coordinate of M the
            perpendicular to AH at H meets the circle (A, m)
            at M the mid point of BC and hence BC can be constructed
            as tangent to the circle (I, IK).
            This proof and construction is the same
            if r is the inradius but we must have d<m or
            if r is the a_exradius but we must have d>m.

            Best regards
            Nikolaos Dergiades
          • Luís Lopes
            Dear Hyacinthists, Dear Nikolaos Dergiades, Thank you for your reply. I will work through your justification and the solution that followed. Now an easier and
            Message 5 of 8 , Mar 30, 2005
              Dear Hyacinthists,

              Dear Nikolaos Dergiades,

              Thank you for your reply. I will work through your
              justification and the solution that followed.

              Now an easier and nice construction problem.
              Or rather two.

              Construct ABC given (a, h_a, m_b\pm m_c) where
              \pm = + -

              Taken from Court.

              Best regards,
              Luis


              >From: "Nikolaos Dergiades" <ndergiades@...>
              >Reply-To: Hyacinthos@yahoogroups.com
              >To: <Hyacinthos@yahoogroups.com>
              >Subject: RE: [EMHL] alpha, m_a, r (r_a) construction [was AIG construction]
              >Date: Wed, 30 Mar 2005 17:07:28 +0300
              >
              >
              >Dear Luis,
              >
              > > Unfortunately I wasn't able to justify it. Could you please
              > > provide the general ideas leading to this elegant solution?
              > >
              [....]
              >
              >I worked as follows:
              >In Cartesian coordinates we consider the points
              >A(0,0), I(d,0), M(g,h).
              >Let S(2g,2h) be the reflection of A in M.

              [....]

              >Best regards
              >Nikolaos Dergiades
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