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

Re: [EMHL] LATO problems

Expand Messages
  • Darij Grinberg
    ... It is an irreducible sextic. The equation is a simple horror. Now I will continue with some other loci from ... For (7.1), the locus is the whole plane,
    Message 1 of 22 , Sep 1, 2003
    • 0 Attachment
      In Hyacinthos message #7720, I wrote:

      >> >> Let Q'a,Q'b,Q'c be the reflections
      >> >> of Qa,Qb,Qc in BC,CA,AB.
      >> >>
      >> >> Which are is locus of P such that
      >> >>
      >> >> (5.2) ABC, Q'aQ'bQ'c are perspective?
      >>
      >> From the first 6 triangle centers, I have found
      >> only X(1) on this locus. The locus seems to
      >> meet the Euler line 4 times, so it cannot be a
      >> cubic!

      It is an irreducible sextic. The equation is a
      simple horror.

      Now I will continue with some other loci from
      Antreas' message #7714:

      >> Let Ha, Ga, (to ask only for two simple points)
      >> be the Orthocenter, Centroid of QaBC
      >> and similarly Hb, Gb; Hc,Gc
      >>
      >> Which are the loci of P such that:
      >>
      >> (7.1) ABC, HaHbHc
      >> (7.2) ABC, GaGbGc
      >>
      >> are perspective?

      For (7.1), the locus is the whole plane, since the
      quadrilateral BPCHa is a parallelogram, and thus
      Ha is the reflection of P in the midpoint of BC,
      and similarly for Hb and Hc. It is well-known that
      now the perspector of triangles ABC and HaHbHc
      will be the complement of P.

      The locus in (7.2)... I am seeing that X(4) lies
      on the locus, while X(1) and X(3) don't lie, but I
      have not tested more points.

      Here is a related locus with anticevian instead of
      antipedal triangle: If Pa, Pb, Pc are the vertices
      of the anticevian triangle of a point P, and Ga,
      Gb, Gc are the centroids of triangles BPaC, CPbA,
      APcB, then the triangles ABC and GaGbGc are
      perspective if and only if the point P lies on a
      degenerate cubic consisting of the three medians
      of triangle ABC.

      Sincerely,
      Darij Grinberg
    • Darij Grinberg
      Dear Antreas Hatzipolakis, ... After thinking about this point some time ago, I had submitted it for Kimberling s ETC, and maybe it will appear there in some
      Message 2 of 22 , Sep 1, 2003
      • 0 Attachment
        Dear Antreas Hatzipolakis,

        In Hyacinthos message #7716, you wrote:

        >> Let ABC be a triangle and O its Circumcenter,
        >> and Oa,Ob,Oc the circumcenters of OBC, OCA, OAB.
        >>
        >> What point is the circumcenter of OaObOc ?

        After thinking about this point some time ago, I
        had submitted it for Kimberling's ETC, and maybe
        it will appear there in some time.

        --------------------------------------------------------------

        X(????) = CIRCUMCENTER OF KOSNITA TRIANGLE
        Trilinears f(A,B,C) : f(B,C,A) : f(C,A,B),
        where f(A,B,C) = cos(B-C)
        (tan A + tan B + tan C - 4 sin A sin B sin C)
        + 4 sin A sin B sin C
        (sec A - 2 cos A) (4 cos A cos B cos C - 1)
        Barycentrics (sin A)f(A,B,C) : (sin B)f(B,C,A) : (sin C)f(C,A,B)

        The Kosnita triangle is the triangle formed by the
        circumcenters of triangles BOC, COA, AOB, where
        O = X(3).

        X(????) lies on the following lines: 2,3 ...
        X(????) = midpoint of X(3) and X(26)

        --------------------------------------------------------------

        BTW, does anybody have simpler trilinears??

        Darij Grinberg
      • xpolakis
        [APH] ... Dear Darij I am not quite sure why you name the triangle as KOSNITA triangle. If this triangle deserves a name, then, in my opinion, it should be
        Message 3 of 22 , Sep 1, 2003
        • 0 Attachment
          [APH]

          >> Let ABC be a triangle and O its Circumcenter,
          >> and Oa,Ob,Oc the circumcenters of OBC, OCA, OAB.
          >>
          >> What point is the circumcenter of OaObOc ?

          [DG]:

          >After thinking about this point some time ago, I
          >had submitted it for Kimberling's ETC, and maybe
          >it will appear there in some time.
          >
          >--------------------------------------------------------------
          >
          >X(????) = CIRCUMCENTER OF KOSNITA TRIANGLE
          >Trilinears       f(A,B,C) : f(B,C,A) : f(C,A,B),
          >   where f(A,B,C) = cos(B-C)
          >               (tan A + tan B + tan C - 4 sin A sin B sin C)
          >           + 4 sin A sin B sin C
          >                (sec A - 2 cos A) (4 cos A cos B cos C - 1)
          >Barycentrics  (sin A)f(A,B,C) : (sin B)f(B,C,A) : (sin C)f(C,A,B)
          >
          >The Kosnita triangle is the triangle formed by the
          >circumcenters of triangles BOC, COA, AOB, where
          >O = X(3).
          >
          >X(????) lies on the following lines:      2,3      ...
          >X(????) = midpoint of X(3) and X(26)
          >
          >--------------------------------------------------------------
          >
          >BTW, does anybody have simpler trilinears??

          Dear Darij

          I am not quite sure why you name the triangle as
          KOSNITA triangle.

          If this triangle deserves a name, then, in my opinion,
          it should be NEUBERG triangle.

          It was Neuberg who first discovered the locus of
          P such that the circumcenter triangle
          of PBC, PCA, PAB (ie the triangle with vertices the
          Circumcenters of PBC, PCA, PAB) and the triangle ABC
          are perspective (Neuberg Cubic + +).
          That particular triangle ("KOSNITA" tr.)
          is for P = O.

          As for the trilinears:

          Since the point lies on the Euler line,
          its trilinears are (cosA - TcosBcosC ::)
          for some fixed real number T, but I don't know
          which number is it !

          Antreas
           
        • Darij Grinberg
          Dear Antreas Hatzipolakis, ... Do you remember the Kosnita theorem? It states that the triangle formed by the circumcenters of triangles BOC, COA, AOB is
          Message 4 of 22 , Sep 1, 2003
          • 0 Attachment
            Dear Antreas Hatzipolakis,

            In Hyacinthos message #7725, you wrote:

            >> I am not quite sure why you name the triangle as
            >> KOSNITA triangle.

            Do you remember the Kosnita theorem? It states that
            the triangle formed by the circumcenters of triangles
            BOC, COA, AOB is perspective to triangle ABC.

            This is, of course, a special case of the Neuberg
            cubic property, but in our particular case, the
            perspector is the isogonal conjugate of the
            nine-point center. This point and the triangle have
            some remarkable properties and deserve a name,
            hence "Kosnita triangle".

            Sincerely,
            Darij Grinberg
          • jpehrmfr
            Dear Darij and Antreas ... PBC) ... I think that there is a typo : PaPbPc is homothetic to the pedal triangle of the isogonal conjugate of P. Hence PaPbPc are
            Message 5 of 22 , Sep 2, 2003
            • 0 Attachment
              Dear Darij and Antreas
              > >> Finally, for which points P
              > >> the triangle PaPbPc is Similar to ABC? (Pa = circumcenter of
              PBC)

              > This is easy: the pedal triangle of P must be
              > similar to triangle ABC. The point P is the
              > circumcenter of triangle ABC. (This is if
              > A and Pa, B and Pb, C and Pc are corresponding
              > vertices of the similarity.)

              I think that there is a typo : PaPbPc is homothetic to the pedal
              triangle of the isogonal conjugate of P. Hence PaPbPc are similar
              only when P = H.
              Friendly. Jean-Pierre
            • Antreas P. Hatzipolakis
              I think this is a theorem we have discussed here: Let ABC be a triangle H its orthocenter, AHa, BHb, CHc its altitudes, and P a point. The circumcircles of
              Message 6 of 22 , Sep 2, 2003
              • 0 Attachment
                I think this is a theorem we have discussed here:

                Let ABC be a triangle H its orthocenter, AHa, BHb, CHc
                its altitudes, and P a point.

                The circumcircles of PAHa, PBHb, PCHc are concurrent
                and their circumcenters lie on a line perpendicular to HP.

                Now,

                1. If P moves on the circumcircle of ABC,
                then which is the locus of the point of
                concurrence of the circumcircles of PAHa, PBHb, PCHc,
                and which is the the envelope of the line of
                their circumcenters ?

                2. If P = (x:y:z) then which are the coordinates
                of the interesection point of the above perpendicular lines?
                (ie HP and line of circumcenters of PAHa, PBHb, PCHc).
                Particular case: P = O


                APH
                --
              • jpehrmfr
                Dear Antreas ... Q is the second common point of the line PH and the NP-circle(the other one is the midpoint of PH) The central line envelopes the conic
                Message 7 of 22 , Sep 2, 2003
                • 0 Attachment
                  Dear Antreas

                  > I think this is a theorem we have discussed here:
                  >
                  > Let ABC be a triangle H its orthocenter, AHa, BHb, CHc
                  > its altitudes, and P a point.
                  >
                  > The circumcircles of PAHa, PBHb, PCHc are concurrent
                  > and their circumcenters lie on a line perpendicular to HP.
                  >
                  > Now,
                  >
                  > 1. If P moves on the circumcircle of ABC,
                  > then which is the locus of the point of
                  > concurrence of the circumcircles of PAHa, PBHb, PCHc,
                  > and which is the the envelope of the line of
                  > their circumcenters ?

                  Q is the second common point of the line PH and the NP-circle(the
                  other one is the midpoint of PH)
                  The central line envelopes the conic inscribed in the medial
                  triangle with focii O and the NP-center.

                  > 2. If P = (x:y:z) then which are the coordinates
                  > of the interesection point of the above perpendicular lines?
                  > (ie HP and line of circumcenters of PAHa, PBHb, PCHc).
                  They are too tedious for me.
                  > Particular case: P = O

                  Friendly. Jean-Pierre
                • Antreas P. Hatzipolakis
                  Let ABC be a triangle H its orthocenter, HaHbHc its orthic triangle, EaEbEc its Euler triangle (ie Ea,Eb,Ec = midpoints of AH, BH, CH, resp.) and P a point
                  Message 8 of 22 , Sep 2, 2003
                  • 0 Attachment
                    Let ABC be a triangle H its orthocenter, HaHbHc
                    its orthic triangle, EaEbEc its Euler triangle (ie Ea,Eb,Ec =
                    midpoints of AH, BH, CH, resp.)
                    and P a point

                    [aph]
                    >The circumcircles of PAHa, PBHb, PCHc are concurrent
                    >and their circumcenters lie on a line perpendicular to HP.

                    The circumcircles of PEaHa, PEbHb, PEcHc are concurrent
                    and their circumcenters lie on a line perpendicular to HP.

                    [aph]:
                    >Now,
                    >
                    >1. If P moves on the circumcircle of ABC,
                    >then which is the locus of the point of
                    >concurrence of the circumcircles of PAHa, PBHb, PCHc,
                    >and which is the the envelope of the line of
                    >their circumcenters ?

                    The same questions, ie
                    1. If P moves on the circumcircle of ABC,
                    then which is the locus of the point of
                    concurrence of the circumcircles of PEaHa, PEbHb, PEcHc,
                    and which is the the envelope of the line of
                    their circumcenters ?

                    [aph]:
                    >2. If P = (x:y:z) then which are the coordinates
                    >of the intersection point of the above perpendicular lines?
                    >(ie HP and line of circumcenters of PAHa, PBHb, PCHc).
                    >Particular case: P = O

                    The same questions ie
                    2. If P = (x:y:z) then which are the coordinates
                    of the intersection point of the above perpendicular lines?
                    (ie HP and line of circumcenters of PEaHa, PEbHb, PEcHc).
                    Particular case: P = O


                    Antreas


                    --
                  • Nikolaos Dergiades
                    ... C) ... Since tanA+tanB+tanC = tanA*tanB*tanC the trilinears become f(A,B,C) = cos(B-C)+4cos(2A)*cosB*cosC Best regards Nikolaos Dergiades
                    Message 9 of 22 , Sep 2, 2003
                    • 0 Attachment
                      Dear Hyacinthists, Darij Grinberg wrote:

                      >X(????) = CIRCUMCENTER OF KOSNITA TRIANGLE
                      >Trilinears f(A,B,C) : f(B,C,A) : f(C,A,B),
                      > where f(A,B,C) = cos(B-C)
                      > (tan A + tan B + tan C - 4 sin A sin B sin
                      C)
                      > + 4 sin A sin B sin C
                      > (sec A - 2 cos A) (4 cos A cos B cos C - 1)

                      >
                      >X(????) lies on the following lines: 2,3 ...
                      >X(????) = midpoint of X(3) and X(26)
                      > does anybody have simpler trilinears??


                      Since tanA+tanB+tanC = tanA*tanB*tanC
                      the trilinears become
                      f(A,B,C) = cos(B-C)+4cos(2A)*cosB*cosC

                      Best regards
                      Nikolaos Dergiades
                    • Darij Grinberg
                      Dear Jean-Pierre and Antreas, ... [...] ... Yes, my apologies for this. Additionally, for P = H, the triangle PaPbPc is the reflection of triangle ABC in the
                      Message 10 of 22 , Sep 2, 2003
                      • 0 Attachment
                        Dear Jean-Pierre and Antreas,

                        In Hyacinthos message #7738, Jean-Pierre Ehrmann wrote:

                        >> Dear Darij and Antreas
                        >> > >> Finally, for which points P
                        >> > >> the triangle PaPbPc is Similar to ABC? (Pa =
                        >> > >> circumcenter of PBC)
                        >>
                        >> > This is easy: the pedal triangle of P must be
                        >> > similar to triangle ABC. The point P is the
                        >> > circumcenter of triangle ABC.
                        [...]
                        >> I think that there is a typo : PaPbPc is homothetic
                        >> to the pedal triangle of the isogonal conjugate of
                        >> P. Hence PaPbPc are similar only when P = H.

                        Yes, my apologies for this. Additionally, for P = H,
                        the triangle PaPbPc is the reflection of triangle ABC
                        in the nine-point center. Some people call this
                        triangle the "Carnot triangle" of triangle ABC.

                        Sincerely,
                        Darij Grinberg
                      • Antreas P. Hatzipolakis
                        Let ABC be a triangle P a point and PaPbPc the pedal triangle of P. Let Oa, Ob, Oc be the circumcenters of the triangles APbPc, BPcPa, CPaPb, resp. Which is
                        Message 11 of 22 , Feb 3 12:41 AM
                        • 0 Attachment
                          Let ABC be a triangle P a point and PaPbPc the pedal
                          triangle of P.

                          Let Oa, Ob, Oc be the circumcenters of the triangles
                          APbPc, BPcPa, CPaPb, resp.

                          Which is the locus of P such that PaPbPc, OaObOc
                          are perspective?

                          Greetings from Athens

                          Antreas

                          --
                        • Antreas P. Hatzipolakis
                          Let ABC be a triangle P,P* two isogonal conjugate points and PaPbPc, P*aP*bP*c their pedal triangles. Let Oa, Ob, Oc be the circumcenters of the triangles
                          Message 12 of 22 , Feb 3 3:33 AM
                          • 0 Attachment
                            Let ABC be a triangle P,P* two isogonal conjugate points and
                            PaPbPc, P*aP*bP*c their pedal triangles.

                            Let Oa, Ob, Oc be the circumcenters of the triangles
                            APbPc, BPcPa, CPaPb, resp. and Qa,Qb,Qc the circumcenters
                            of the triangles AP*bP*c, BP*cP*a, CP*aP*b, resp.

                            Which is the locus of P such that:

                            1. The lines OaQa, ObQb, OcQc are concurrent?

                            2. The Triangles ABC, (OaQa, ObQb, OcQc) are orthologic?


                            Enough problems for today!

                            Good Night from Athens

                            Antreas

                            --
                          • Paul Yiu
                            Dear Antreas, [APH]: Let ABC be a triangle P a point and PaPbPc the pedal triangle of P. Let Oa, Ob, Oc be the circumcenters of the triangles APbPc, BPcPa,
                            Message 13 of 22 , Feb 3 1:26 PM
                            • 0 Attachment
                              Dear Antreas,

                              [APH]: Let ABC be a triangle P a point and PaPbPc the pedal
                              triangle of P.

                              Let Oa, Ob, Oc be the circumcenters of the triangles
                              APbPc, BPcPa, CPaPb, resp.

                              Which is the locus of P such that PaPbPc, OaObOc
                              are perspective?

                              *** This is the Neuberg cubic.

                              Best regards
                              Sincerely
                              Paul
                            • Paul Yiu
                              Dear Antreas, [APH]: Let ABC be a triangle P a point and PaPbPc the pedal triangle of P. Let Oa, Ob, Oc be the circumcenters of the triangles APbPc, BPcPa,
                              Message 14 of 22 , Feb 3 2:41 PM
                              • 0 Attachment
                                Dear Antreas,

                                [APH]: Let ABC be a triangle P a point and PaPbPc the pedal
                                triangle of P.

                                Let Oa, Ob, Oc be the circumcenters of the triangles
                                APbPc, BPcPa, CPaPb, resp.

                                Which is the locus of P such that PaPbPc, OaObOc
                                are perspective?

                                [PY]: This is the Neuberg cubic.

                                *** This is already in Bernard's CTC, K001, attributed to Jean-Pierre.

                                Best regards
                                Sincerely
                                Paul
                              • Antreas P. Hatzipolakis
                                Dear Paul ... Probably this variation is interesting: Let PaPbPc, P*aP*bP*c be the pedal triangles of the isogonal points P,P*. And let Oa,Ob,Oc / Qa, Qb, Qc
                                Message 15 of 22 , Feb 3 4:57 PM
                                • 0 Attachment
                                  Dear Paul

                                  >[APH]: Let ABC be a triangle P a point and PaPbPc the pedal
                                  > triangle of P.
                                  >
                                  > Let Oa, Ob, Oc be the circumcenters of the triangles
                                  > APbPc, BPcPa, CPaPb, resp.
                                  >
                                  > Which is the locus of P such that PaPbPc, OaObOc
                                  > are perspective?
                                  >
                                  > [PY]: This is the Neuberg cubic.
                                  >
                                  > [PY] This is already in Bernard's CTC, K001, attributed to Jean-Pierre.

                                  Probably this variation is interesting:

                                  Let PaPbPc, P*aP*bP*c be the pedal triangles of the
                                  isogonal points P,P*.


                                  And let Oa,Ob,Oc / Qa, Qb, Qc be the circumcenters of the
                                  triangles APbPc, BPcPa, CPaPb / AP*bP*c, BP*cP*a, CP*aP*b

                                  Which is the locus of P such that the triangles

                                  1. PaPbPc, QaQbQc

                                  2. P*aP*bP*c, OaObOc

                                  are perspective?


                                  Greetings from Athens

                                  Antreas
                                  --
                                Your message has been successfully submitted and would be delivered to recipients shortly.