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RE: [EMHL] IMO training 2004 homework problem sets I, II, III

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  • Nikolaos Dergiades
    Dear Darij and Vladimir For the case that the quadrilateral is not convex from a sketch I concluded (perhaps this is not correct) that the diagonals BD, AC of
    Message 1 of 12 , Apr 11, 2004
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      Dear Darij and Vladimir

      For the case that the quadrilateral is not convex from a sketch
      I concluded (perhaps this is not correct) that the diagonals
      BD, AC of the quadrilateral must be parallel and the point O
      must lie on the line joining the mid points of the diagonals.

      For example if the points A,B,D,C are concyclic and the
      quadrilateral ABDC is isosceles trapezium AB=DC, AC || BD
      then the quadrilateral ABCD is not convex and if
      the cartesian coordinates are
      B(-a , 0) , D(a , 0) , O( -d, 0)
      where O is the point satisfying the "halfside" condition
      A(-p , q) , C( p , q), a, d, p, q > 0
      then p , q can be found from p/d = q/a
      (aa+dd)pp + 2ad(a-2d)p + ddaa = 0

      Special Case :
      d = 2a and p = 2a q = a or
      d = 2a and p = 2a/5 q = a/5

      Best regards
      Nikolaos Dergiades
    • Alexey.A.Zaslavsky
      Dear Darij! III.8. Let ABC be a triangle with circumradius R and inradius r. It is well-known that there exist infinitely many triangles having the same
      Message 2 of 12 , Apr 11, 2004
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        Dear Darij!

        III.8. Let ABC be a triangle with circumradius R and
        inradius r. It is well-known that there exist
        infinitely many triangles having the same circumcircle
        and the same incircle as triangle ABC.

        (a) Show that the orthocenters of all of these
        triangles lie on a common circle.

        (b) What can be said about the locus of the nine-point
        centers of all of these triangles in the case R = 4r ?

        My pupils D.Kosov and M.Muzafarof examined the trajectories of some centers
        of Ponsele's triangle. Their results are typed in "Kvant",
        N 2, 2003 (www.kvant.mccme.ru).

        Sincerely Alexey
      • Darij Grinberg
        Dear Alexey, ... By a curious coincidence, I have found this paper yesterday before receiving your mail! It was indeed nice to see my conjecture verified that
        Message 3 of 12 , Apr 12, 2004
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          Dear Alexey,

          In Hyacinthos message #9659, you wrote:

          >> My pupils D.Kosov and M.Muzafarof examined the
          >> trajectories of some centers of Ponsele's
          >> triangle. Their results are typed in "Kvant",
          >> N 2, 2003 (www.kvant.mccme.ru).

          By a curious coincidence, I have found this
          paper yesterday before receiving your mail! It
          was indeed nice to see my conjecture verified
          that the poristic locus (or "trajectory") of
          the symmedian point is an ellipse. However,
          concerning the proofs of the rather simple
          poristic loci (i. e., those of the Nagel point,
          nine-point center, orthocenter, centroid), I
          prefer using the Feuerbach theorem rather than
          calculating the distances using barycentric
          coordinates.

          Sincerely,
          Darij Grinberg
        • Alexey.A.Zaslavsky
          Dear Darij! ... You are right, for this points there are a synthetical proof. But without coordinates we could not prove the most interesting our result: the
          Message 4 of 12 , Apr 12, 2004
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            Dear Darij!
            >
            >By a curious coincidence, I have found this
            >paper yesterday before receiving your mail! It
            >was indeed nice to see my conjecture verified
            >that the poristic locus (or "trajectory") of
            >the symmedian point is an ellipse. However,
            >concerning the proofs of the rather simple
            >poristic loci (i. e., those of the Nagel point,
            >nine-point center, orthocenter, centroid), I
            >prefer using the Feuerbach theorem rather than
            >calculating the distances using barycentric
            >coordinates.
            >
            You are right, for this points there are a synthetical proof. But without
            coordinates we could not prove the most interesting our result: the
            trajectory of Gergonne point is the circle coaxial with the circumcircle and
            the incircle.

            Sincerely Alexey
          • ndergiades
            Dear Darij and Vladimir sorry a correction. In my 9658 message ... ******** The point O is (0, -d) and not (-d , 0) Best regards Nikolaos Dergiades
            Message 5 of 12 , Apr 12, 2004
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              Dear Darij and Vladimir
              sorry a correction. In my 9658 message
              I wrote:

              > For the case that the quadrilateral is not convex from a sketch
              > I concluded (perhaps this is not correct) that the diagonals
              > BD, AC of the quadrilateral must be parallel and the point O
              > must lie on the line joining the mid points of the diagonals.
              >
              > For example if the points A,B,D,C are concyclic and the
              > quadrilateral ABDC is isosceles trapezium AB=DC, AC || BD
              > then the quadrilateral ABCD is not convex and if
              > the cartesian coordinates are
              > B(-a , 0) , D(a , 0) , O( -d, 0)
              > where O is the point satisfying the "halfside" condition
              > A(-p , q) , C( p , q), a, d, p, q > 0
              > then p , q can be found from p/d = q/a
              > (aa+dd)pp + 2ad(a-2d)p + ddaa = 0
              >
              > Special Case :
              > d = 2a and p = 2a q = a or
              > d = 2a and p = 2a/5 q = a/5

              ********

              The point O is (0, -d) and not (-d , 0)

              Best regards
              Nikolaos Dergiades
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