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  • xpolakis@otenet.gr
    Let ABC be a triangle. Draw a cevian AD so that: the intouch (antitangential) triangles of the triangles ADB, ADC be in perspective. That is: A / / /
    Message 1 of 18 , Jun 29, 2000
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      Let ABC be a triangle. Draw a cevian AD so that:
      the intouch (antitangential) triangles of the triangles ADB, ADC be
      in perspective.

      That is:


      A
      /\
      / \
      / \
      / \
      / \
      / X' \
      / \
      Z X Z'

      / \
      / \

      B----Y----D----Y'---C




      The incircle of ADB touches AD, DB, BA in X,Y,Z, respectively.
      ADC AD, DC, CA in X',Y',Z', respectively.

      XX',YZ', ZY' : concurrent.


      Antreas
    • xpolakis@otenet.gr
      Let ABC be a triangle and P a point. Draw the perpendicular bisectors of PA,PB,PC, forming triangle A ,B ,C [ie A = (perp. bis. of PB) / (perp. bis. of PB)]
      Message 2 of 18 , Sep 3, 2000
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        Let ABC be a triangle and P a point. Draw the perpendicular bisectors
        of PA,PB,PC, forming triangle A',B',C'
        [ie A' = (perp. bis. of PB) /\ (perp. bis. of PB)]
        Obviously A', B', C' are the circumcenters of triangles PBC,PCA,PAB, resp.
        Obviously again, if P = H, then A'B'C' is homothetic to ABC (the homothetic
        center is the 9PC center of ABC).

        A question (which perhaps we have already ansewred before,
        but I am not sure) is:

        Which is the locus of P such that ABC is in perspective with A'B'C'?

        Antreas
      • Bernard Gibert
        Dear Antreas, ... If my computation is OK, it is the reunion of : ­ the line at infinity ­ the circumcircle ­ the Neuberg cubic Best regards Bernard
        Message 3 of 18 , Sep 3, 2000
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          Dear Antreas,
          >
          > Let ABC be a triangle and P a point. Draw the perpendicular bisectors
          > of PA,PB,PC, forming triangle A',B',C'
          > [ie A' = (perp. bis. of PB) /\ (perp. bis. of PB)]
          > Obviously A', B', C' are the circumcenters of triangles PBC,PCA,PAB, resp.
          > Obviously again, if P = H, then A'B'C' is homothetic to ABC (the homothetic
          > center is the 9PC center of ABC).
          >
          > A question (which perhaps we have already ansewred before,
          > but I am not sure) is:
          >
          > Which is the locus of P such that ABC is in perspective with A'B'C'?
          >
          If my computation is OK, it is the reunion of :
          ­ the line at infinity
          ­ the circumcircle
          ­ the Neuberg cubic

          Best regards

          Bernard
        • yiu@fau.edu
          Dear Antreas, [APH]: Let ABC be a triangle and P a point. Draw the perpendicular bisectors of PA,PB,PC, forming triangle A ,B ,C [ie A = (perp. bis. of PB)
          Message 4 of 18 , Sep 3, 2000
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            Dear Antreas,

            [APH]: Let ABC be a triangle and P a point. Draw the perpendicular
            bisectors of PA,PB,PC, forming triangle A',B',C'
            [ie A' = (perp. bis. of PB) /\ (perp. bis. of PC)]
            ...

            Which is the locus of P such that ABC is in perspective with A'B'C'?

            Never, in the following sense:

            the line AA', BB', CC' are concurrent if and only if P
            lies on the circumcircle of ABC, in which case, A', B', C'
            all coincide with the circumcenter O.


            Best regards.
            Sincerely,
            Paul
          • yiu@fau.edu
            Dear Antreas and Bernard, [APH]: Let ABC be a triangle and P a point. Draw the perpendicular bisectors of PA,PB,PC, forming triangle A ,B ,C [ie A = (perp.
            Message 5 of 18 , Sep 3, 2000
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              Dear Antreas and Bernard,

              [APH]: Let ABC be a triangle and P a point. Draw the perpendicular
              bisectors of PA,PB,PC, forming triangle A',B',C'
              [ie A' = (perp. bis. of PB) /\ (perp. bis. of PC)]
              Which is the locus of P such that ABC is in perspective with A'B'C'?

              [BG]: If my computation is OK, it is the reunion of :
              ­ the line at infinity
              ­ the circumcircle
              ­ the Neuberg cubic

              You are right! Bernard, my claim in message 1336 was wrong.

              Best regards.
              Sincerely,
              Paul
            • xpolakis@otenet.gr
              Let AA ,BB ,CC be the altitudes of a triangle ABC. Let Ab,Ac be the orth. projections of A on AB, AC resp., and A b, A c the orth. projections of Ab,Ac on
              Message 6 of 18 , Dec 6, 2000
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                Let AA',BB',CC' be the altitudes of a triangle ABC.
                Let Ab,Ac be the orth. projections of A' on AB, AC resp.,
                and A'b, A'c the orth. projections of Ab,Ac on AA' resp.
                Let A1 = A'Ab /\ BA'c and A2 = A'Ac /\ CA'b.
                Similarly we define the points B1,B2; C1,C2.
                Then the triangle bounded by the lines A1A2,B1B2,C1C2
                and the triangle ABC are perspective (homothetic).
                (PDEME, Feb. 1961, p. 136)

                A
                /|\
                / | \
                / | \
                / | \
                / | \
                Ab A'b \
                / | \
                / | \
                / A'c Ac
                / A1 | A2 \
                / | \
                / | \
                B------------A'-----------C


                Antreas
              • yiu@fau.edu
                Dear Antreas, [APH]: Let AA ,BB ,CC be the altitudes of a triangle ABC. Let Ab,Ac be the orth. projections of A on AB, AC resp., and A b, A c the orth.
                Message 7 of 18 , Dec 7, 2000
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                  Dear Antreas,

                  [APH]:
                  Let AA',BB',CC' be the altitudes of a triangle ABC.
                  Let Ab,Ac be the orth. projections of A' on AB, AC resp.,
                  and A'b, A'c the orth. projections of Ab,Ac on AA' resp.
                  Let A1 = A'Ab /\ BA'c and A2 = A'Ac /\ CA'b.
                  Similarly we define the points B1,B2; C1,C2.
                  Then the triangle bounded by the lines A1A2,B1B2,C1C2
                  and the triangle ABC are perspective (homothetic).
                  (PDEME, Feb. 1961, p. 136)

                  A
                  /|\
                  / | \
                  / | \
                  / | \
                  / | \
                  Ab A'b \
                  / | \
                  / | \
                  / A'c Ac
                  / A1 | A2 \
                  / | \
                  / | \
                  B------------A'-----------C

                  **********

                  Let X1 and X2 be the traces of A1 and A2 on the side BC.

                  In triangle AA'B, we have

                  AA1/AX1 = AAb/AbB + AA'c/A'cA' = AAb/AbB + AAc/AcC.

                  For the same reason,

                  AA2/AX2 = same expression.

                  This means that A1A2 is paralllel to BC.

                  For the same reason, the lines B1B2 and C1C2 are parallel
                  to CA and AB respectively. It follows that the three lines
                  bound a triangle homothetic to ABC.

                  Where is the homothetic center?

                  My sketch suggests that the line A1A2 passes through the
                  orthocenter H. But this cannot be true, for otherwise there
                  would be no triangle!

                  Best regards.
                  Sincerely,
                  Paul
                • yiu@fau.edu
                  Dear Antreas, A variant of the problem you posed: Let AA ,BB ,CC be the altitudes of a triangle ABC. Let Ab,Ac be the orth. projections of A on AB, AC resp.,
                  Message 8 of 18 , Dec 7, 2000
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                    Dear Antreas,

                    A variant of the problem you posed:

                    Let AA',BB',CC' be the altitudes of a triangle ABC.
                    Let Ab,Ac be the orth. projections of A' on AB, AC resp.,
                    Define Bc, Ba, Ca, Cb analogously.

                    A
                    /|\
                    / | \
                    / | \
                    / | \
                    / | \
                    Ab | \
                    / | \
                    / | \
                    / | Ac
                    / | \
                    / | \
                    / | \
                    B------------A'-----------C



                    The triangle bounded by AbAc, BcBa, CaCb is perspector,
                    and the perspector is one of the most basic triangle centers!


                    Best regards.
                    Sincerely,
                    Paul
                  • yiu@fau.edu
                    Dear Antreas, Jean-Pierre, Fred and friends, Let P be a point with traces X,Y,Z on the sides of triangle ABC. B_(X), C_(X) the perpendicular feet of X on CA
                    Message 9 of 18 , Dec 7, 2000
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                      Dear Antreas, Jean-Pierre, Fred and friends,

                      Let P be a point with traces X,Y,Z on the sides of triangle ABC.
                      B_(X), C_(X) the perpendicular feet of X on CA and AB respectively.
                      Similarly define C_(Y), A_(Y), A_(Z), and B_(Z). The locus of
                      P for which the triangle bounded by the three lines B_(X)C_(X),
                      C_(Y)A_(Y), and A_(Z)B_(Z) is perspective is the union of the
                      following three curves.

                      (1) The circumcircle, In this case, the three lines
                      are parallel to the Simson line of P.

                      (2) The Darboux cubic.

                      (3) The quartic with equation

                      sum aaSBC v^2w^2 = (sum aaSBC)(u+v+w)uvw.

                      In this case, the three lines are concurrent.

                      The quartic in (3) can be constructed as follows. It is the
                      ISOTOMIC conjugate of a conic C, which is, with respect to the
                      anticomplementary triangle, the circumconic with perspector
                      the pro-orthocenter, i.e., the point with homogeneous barycentric
                      coordinates

                      (a^2/SA : b^2/SB : c^2/SC).

                      Best regards.
                      Sincerely,
                      Paul
                    • yiu@fau.edu
                      Dear friends, [PY]: Let P be a point with traces X,Y,Z on the sides of triangle ABC. B_(X), C_(X) the perpendicular feet of X on CA and AB respectively.
                      Message 10 of 18 , Dec 7, 2000
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                        Dear friends,

                        [PY]:
                        Let P be a point with traces X,Y,Z on the sides of triangle ABC.
                        B_(X), C_(X) the perpendicular feet of X on CA and AB respectively.
                        Similarly define C_(Y), A_(Y), A_(Z), and B_(Z). The locus of
                        P for which the triangle bounded by the three lines B_(X)C_(X),
                        C_(Y)A_(Y), and A_(Z)B_(Z) is perspective is the union of the
                        following three curves.

                        (1) The circumcircle, In this case, the three lines
                        are parallel to the Simson line of P.

                        (2) The Darboux cubic.

                        Here are some examples

                        Point on Darboux cubic Perspector
                        ---------------------------------------
                        Orthocenter symmedian point
                        Incenter X(79) = [1/(bb+bc+cc-aa):...:...]
                        de Longchamps point Orthocenter
                        Circumcenter X(275) homothetic.

                        Best regards.
                        Sincerely,
                        Paul

                        (3) The quartic with equation

                        sum aaSBC v^2w^2 = (sum aaSBC)(u+v+w)uvw.

                        In this case, the three lines are concurrent.

                        The quartic in (3) can be constructed as follows. It is the
                        ISOTOMIC conjugate of a conic C, which is, with respect to the
                        anticomplementary triangle, the circumconic with perspector
                        the pro-orthocenter, i.e., the point with homogeneous barycentric
                        coordinates

                        (a^2/SA : b^2/SB : c^2/SC).

                        Best regards.
                        Sincerely,
                        Paul
                      • alpercay2000
                        Let ABC be an acute triangle,T the midpoint of arc BC of the circle circumscribing ABC.Let G and K be the projections of A and T respectively on BC,let H and L
                        Message 11 of 18 , Aug 12, 2003
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                          Let ABC be an acute triangle,T the midpoint of arc BC of the circle
                          circumscribing ABC.Let G and K be the projections of A and T
                          respectively on BC,let H and L be the projections of B and C on
                          AT,and let E be the midpoint of AB.Prove that:

                          (a)KH//AC,GL//BT,GH//TC,and LK//AB
                          (b)G,H,K and L are con-cyclic.
                          (c)The center of the circle through G,H and K lies on the Euler
                          circle of ABC.
                          Bogdan Suceava,California State University
                        • Darij Grinberg
                          Dear Alper Cay, ... See the geometry-college threads http://mathforum.org/epigone/geometry-college/whixhaxskeld
                          Message 12 of 18 , Aug 12, 2003
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                            Dear Alper Cay,

                            In Hyacinthos message #7487, you wrote:

                            >> Let ABC be an acute triangle,T the midpoint of arc BC of the
                            >> circle circumscribing ABC.Let G and K be the projections of
                            >> A and T respectively on BC,let H and L be the projections of
                            >> B and C on AT,and let E be the midpoint of AB.Prove that:
                            >>
                            >> (a)KH//AC,GL//BT,GH//TC,and LK//AB
                            >> (b)G,H,K and L are con-cyclic.
                            >> (c)The center of the circle through G,H and K lies on the Euler
                            >> circle of ABC.
                            >> Bogdan Suceava,California State University

                            See the geometry-college threads

                            http://mathforum.org/epigone/geometry-college/whixhaxskeld
                            http://mathforum.org/epigone/geometry-college/nurshoustray

                            especially the last three messages of the first thread and the
                            first two messages of the second thread. (However, notions like
                            "anticenter" are explained in the previous messages.)

                            Darij Grinberg
                          • Алексей Мякишев
                            Dear friends! One construction: Given tiangle ABC. On BC line we mark two points: A1 and A2, so that AA1=AB and AA2=AC. And perpendicylar bissector to that
                            Message 13 of 18 , May 14, 2007
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                              Dear friends!
                              One construction:
                              Given tiangle ABC. On BC line we mark two points: A1 and A2, so that AA1=AB and AA2=AC. And perpendicylar bissector to that segment.
                              So we do with two others sides.
                              Point of intersection is symmetry with center O of H.
                              Is it obvious?
                              Best regards,
                              Alexei


                              [Non-text portions of this message have been removed]
                            • Francois Rideau
                              Dear Alexei I think it is only a matter about product of central symmetries along each sides of ABC. For example, look at the side BC. Let A be the middle of
                              Message 14 of 18 , May 15, 2007
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                                Dear Alexei
                                I think it is only a matter about product of central symmetries along each
                                sides of ABC.
                                For example, look at the side BC. Let A' be the middle of segment BC and L
                                be the orthogonal projection on side BC of the orthocenter H.
                                Let s(A') and s(L) be the central symmetry respectively wrt A' and L .
                                Then A2 = s(L).s(A').s(L) (A1) = s(U) (A1)
                                where S(U) is the central symmetry wrt U = s(L)(A') and we are done.
                                Friendly
                                François


                                On 5/14/07, Алексей Мякишев <alex_geom@...> wrote:
                                >
                                > Dear friends!
                                > One construction:
                                > Given tiangle ABC. On BC line we mark two points: A1 and A2, so that
                                > AA1=AB and AA2=AC. And perpendicylar bissector to that segment.
                                > So we do with two others sides.
                                > Point of intersection is symmetry with center O of H.
                                > Is it obvious?
                                > Best regards,
                                > Alexei
                                >
                                > [Non-text portions of this message have been removed]
                                >
                                >
                                >


                                [Non-text portions of this message have been removed]
                              • Nikolaos Dergiades
                                Dear Alexei ... Perhaps you have a typo. A1 is the symmetric of B wrt the altitude AH A2 is the symmetric of C wrt the altitude AH The segment A1A2 is the
                                Message 15 of 18 , May 15, 2007
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                                  Dear Alexei

                                  > One construction:
                                  > Given tiangle ABC. On BC line we mark two points:
                                  > A1 and A2, so that AA1=AB and AA2=AC. And
                                  > perpendicylar bissector to that segment.
                                  > So we do with two others sides.
                                  > Point of intersection is symmetry with center O of
                                  > H.
                                  > Is it obvious?

                                  Perhaps you have a typo.
                                  A1 is the symmetric of B wrt the altitude AH
                                  A2 is the symmetric of C wrt the altitude AH
                                  The segment A1A2 is the symmetric of BC
                                  wrt the altitude AH and hence the perpendicular
                                  bisector of BC that passes through O
                                  is symmetric to the perpendicular bisector of A1A2
                                  that passes through the symmetric of O with center H.

                                  Best regards
                                  Nikos Dergiades






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                                • Huub van Kempen
                                  Dear Hyacinthos, Given a circle (M,r) and two lines l and m outside the circle. The points A and B lie on the circumference of the circle. Question: To
                                  Message 16 of 18 , May 2, 2008
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                                    Dear Hyacinthos,

                                    Given a circle (M,r) and two lines l and m outside the circle.
                                    The points A and B lie on the circumference of the circle.
                                    Question: To construct the points C and D on the circumference so that
                                    AC and BD intersect on line l and CD is parallel to line m.

                                    Regards,

                                    Huub van Kempen
                                    _________________________________________________________________
                                    Probeer Live Search: de zoekmachine van de makers van MSN!
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                                  • jpehrmfr
                                    Dear Huub van Kempen ... [UV] is the diameter of the circle perpendicular to m. If CD is a variable chord parallel to m, the locus of the common point of AC
                                    Message 17 of 18 , May 2, 2008
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                                      Dear Huub van Kempen

                                      > Given a circle (M,r) and two lines l and m outside the circle.
                                      > The points A and B lie on the circumference of the circle.
                                      > Question: To construct the points C and D on the circumference so that
                                      > AC and BD intersect on line l and CD is parallel to line m.

                                      [UV] is the diameter of the circle perpendicular to m.
                                      If CD is a variable chord parallel to m, the locus of the common point
                                      of AC and BD is the rectangular hyperbola through A,B,U,V (the center
                                      of the hyperbola is the midpoint of AB)
                                      So you have just to find the common points of l with the hyperbola
                                      above.
                                      Friendly. Jean-Pierre
                                    • maedu@hotmail.com
                                      If you put C on circle, and a parallel to m by C, then the locus of the intersection of AC and BD is a hyperbole passing by A, B, and the points of contact of
                                      Message 18 of 18 , May 2, 2008
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                                        If you put C on circle, and a parallel to m by C, then the locus of the intersection of AC and BD is a hyperbole passing by A, B, and the points of contact of two parallels to m tangents to the circle.
                                        The points of intersection of this hyperbola with line l are the solutions.

                                        Martin


                                        From: Huub van Kempen
                                        Sent: Friday, May 02, 2008 11:04 AM
                                        To: hyacinthos@yahoogroups.com
                                        Subject: [EMHL] Problem


                                        Dear Hyacinthos,

                                        Given a circle (M,r) and two lines l and m outside the circle.
                                        The points A and B lie on the circumference of the circle.
                                        Question: To construct the points C and D on the circumference so that
                                        AC and BD intersect on line l and CD is parallel to line m.

                                        Regards,

                                        Huub van Kempen
                                        __________________________________________________________
                                        Probeer Live Search: de zoekmachine van de makers van MSN!
                                        http://www.live.com/?searchOnly=true

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