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Perspector based on incenter and excircles

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  • Eric Danneels
    Dear Hyacinthians, here s a new point I ve discovered by chance while playing with extraversions. Let I be the incenter. Let A be the point where the
    Message 1 of 23 , Mar 21, 2004
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      Dear Hyacinthians,

      here's a new point I've discovered by chance while playing with
      extraversions.

      Let I be the incenter.
      Let A' be the point where the A-excircle touches BC and let A* be
      the second intersection of the line IA' with the A-excircle.
      Define B', C' and B*, C* cyclically.

      ==> Triangles ABC and A*B*C* are perspective

      The barycentric coordinates of the perspector P are

      P = ( a^5 - a^4.(b + c) - 2.a^3.(b + c)^2 + 2.a^2.(b + c)^3
      + a.(b + c)^4 - (b + c)^5 :
      b^5 - b^4.(c + a) - 2.b^3.(c + a)^2 + 2.b^2.(c + a)^3
      + b.(c + a)^4 - (c + a)^5 :
      c^5 - c^4.(a + b) - 2.c^3.(a + b)^2 + 2.c^2.(a + b)^3
      + c.(a + b)^4 - (a + b)^5 )

      It's amazing that:

      - the most basic points still hide new points
      - basic points and simple constructions are no guarantee for simple
      coordinates

      Greeting from Bruges

      Eric Danneels
    • Nikolaos Dergiades
      Dear Eric, I think that this property holds for every point P as follows: If P is a point and P1, P2, P3 are its extraversions If the line PX1 where X1 is the
      Message 2 of 23 , Mar 21, 2004
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        Dear Eric,
        I think that this property holds for every point P as follows:
        If P is a point and P1, P2, P3 are its extraversions
        If the line PX1 where X1 is the orthogonal projection of P1 on BC
        meets the pedal circle of P1 at A* and similarly define the points
        B*, C* then the triangles ABC, A*B*C* are perspective.

        See the analogous results about the incircle and the Prasolov point
        messages 6465 and before.
        Best regards
        Nikolaos Dergiades

        [ED]
        Let I be the incenter.
        Let A' be the point where the A-excircle touches BC and let A* be
        the second intersection of the line IA' with the A-excircle.
        Define B', C' and B*, C* cyclically.

        ==> Triangles ABC and A*B*C* are perspective

        The barycentric coordinates of the perspector P are

        P = ( a^5 - a^4.(b + c) - 2.a^3.(b + c)^2 + 2.a^2.(b + c)^3
        + a.(b + c)^4 - (b + c)^5 :
        b^5 - b^4.(c + a) - 2.b^3.(c + a)^2 + 2.b^2.(c + a)^3
        + b.(c + a)^4 - (c + a)^5 :
        c^5 - c^4.(a + b) - 2.c^3.(a + b)^2 + 2.c^2.(a + b)^3
        + c.(a + b)^4 - (a + b)^5 )

        It's amazing that:

        - the most basic points still hide new points
        - basic points and simple constructions are no guarantee for simple
        coordinates

        Greeting from Bruges

        Eric Danneels
      • Darij Grinberg
        Dear Eric, ... This is indeed fascinating. It reminds me on the Nagel-Schröder point X(1339) which is the point of intersection of the circles AIA , BIB ,
        Message 3 of 23 , Mar 22, 2004
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          Dear Eric,

          In Hyacinthos message #9595, you wrote:

          >> Let I be the incenter.
          >> Let A' be the point where the A-excircle touches BC
          >> and let A* be the second intersection of the line
          >> IA' with the A-excircle.
          >> Define B', C' and B*, C* cyclically.
          >>
          >> ==> Triangles ABC and A*B*C* are perspective

          This is indeed fascinating. It reminds me on the
          Nagel-Schröder point X(1339) which is the point of
          intersection of the circles AIA', BIB', CIC'
          different from I. By the way, I have no synthetic
          proof of the coaxality of these circles - can
          anybody help?

          Let me add some conjectures:

          The poles of the lines IA', IB', IC' with respect
          to the A-, B-, C-excircles are collinear. In other
          words, the tangents to the A-, B-, C-excircles at
          the points A*, B*, C* meet the lines BC, CA, AB at
          three collinear points.

          The polars of the incenter I with respect to the
          excircles of triangle ABC enclose a triangle in
          perspective to triangle ABC. The perspector is a
          point not in the ETC.

          Darij Grinberg
        • Eric Danneels
          Dear Nikolaos, ... What do you mean by the extraversions P1, P2 and P3 of P ? The recent messages about extraversions didn t make things clearer to me, on the
          Message 4 of 23 , Mar 22, 2004
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            Dear Nikolaos,

            you wrote:

            > I think that this property holds for every point P as follows:
            > If P is a point and P1, P2, P3 are its extraversions
            > If the line PX1 where X1 is the orthogonal projection of P1 on BC
            > meets the pedal circle of P1 at A* and similarly define the points
            > B*, C* then the triangles ABC, A*B*C* are perspective.
            >
            > See the analogous results about the incircle and the Prasolov point
            > messages 6465 and before.

            What do you mean by the extraversions P1, P2 and P3 of P ?
            The recent messages about extraversions didn't make things clearer
            to me, on the contrary...

            I also don't see the link with messages 6465 and before.

            However, about your generalization of the Prasolov point I would
            like to add another generalization. (I thought it was known but I
            couldn't find any reference).

            Let A1B1BC1 be the pedal triangle of P and let PA1 intersect the
            pedal circle again in A*. Define B* and C* cyclically.

            ==> AA*, BB* and CC* are also concurrent

            Greetings from Bruges

            Eric
          • Nikolaos Dergiades
            Dear Eric, [ED] ... Sorry. what I wrote about extraversions is nonsense. It doesn t hold. Take it as I didn t wrote it. As for your Prasolov generalization you
            Message 5 of 23 , Mar 22, 2004
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              Dear Eric,

              [ED]
              > What do you mean by the extraversions P1, P2 and P3 of P ?
              > The recent messages about extraversions didn't make things clearer
              > to me, on the contrary...
              >
              > I also don't see the link with messages 6465 and before.
              >
              > However, about your generalization of the Prasolov point I would
              > like to add another generalization. (I thought it was known but I
              > couldn't find any reference).
              >
              > Let A1B1BC1 be the pedal triangle of P and let PA1 intersect the
              > pedal circle again in A*. Define B* and C* cyclically.

              > ==> AA*, BB* and CC* are also concurrent

              Sorry. what I wrote about extraversions is nonsense.
              It doesn't hold. Take it as I didn't wrote it.

              As for your Prasolov generalization you are riqht
              because the point A* is antipode in the same pedal circle
              of the projection on BC of the point P* isogonal conjugate of P.

              Greetings from Thessaloniki
              Nikolaos Dergiades
            • jpehrmfr
              Dear Eric and Darij [ED] ... Eric gave hugly barycentric coordinates for the perspector. In fact, I think that they are (p-a)^3 : (p-b)^3 : (p-c)^3. ... They
              Message 6 of 23 , Mar 22, 2004
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                Dear Eric and Darij
                [ED]
                > >> Let I be the incenter.
                > >> Let A' be the point where the A-excircle touches BC
                > >> and let A* be the second intersection of the line
                > >> IA' with the A-excircle.
                > >> Define B', C' and B*, C* cyclically.
                > >>
                > >> ==> Triangles ABC and A*B*C* are perspective
                Eric gave hugly barycentric coordinates for the perspector. In fact,
                I think that they are (p-a)^3 : (p-b)^3 : (p-c)^3.

                > Let me add some conjectures:
                >
                > The poles of the lines IA', IB', IC' with respect
                > to the A-, B-, C-excircles are collinear. In other
                > words, the tangents to the A-, B-, C-excircles at
                > the points A*, B*, C* meet the lines BC, CA, AB at
                > three collinear points.
                They lie on the trilinear polar of X(346) = (p-a)^2:(p-b)^2:(p-c)^2
                >
                > The polars of the incenter I with respect to the
                > excircles of triangle ABC enclose a triangle in
                > perspective to triangle ABC. The perspector is a
                > point not in the ETC.

                This point is 4X(9)-3X(165), ie X(193) of the excentral triangle.
                He lies on the line X(9)X(165) (line GK of the excentral triangle)
                and on the parallel through the Nagel point to the Soddy line.
                He is barycentric a/((p-a)^2-(b-a)(c-a)):...
                Friendly. Jean-Pierre
              • jpehrmfr
                Dear Darij ... If we condider the polars of a point M wrt the three incircles : 1)they concur if and only if M lies on the radical circle of the three circles,
                Message 7 of 23 , Mar 23, 2004
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                  Dear Darij
                  > The polars of the incenter I with respect to the
                  > excircles of triangle ABC enclose a triangle in
                  > perspective to triangle ABC. The perspector is a
                  > point not in the ETC.

                  If we condider the polars of a point M wrt the three incircles :
                  1)they concur if and only if M lies on the radical circle of the
                  three circles, which is the circle with center the Spieker point,
                  radius 1/2*root(p^2+r^2)
                  2)they board a triangle perspective with ABC if and only if M lies
                  on the Spieker central cubic
                  http://perso.wanadoo.fr/bernard.gibert/Exemples/k033.html
                  Friendly. Jean-Pierre
                • dick tahta
                  About Nikolaos now withdrawn conjecture that ... Cabri suggests that for any point P there are a number of points P1 that have the above property. I have not
                  Message 8 of 23 , Mar 23, 2004
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                    About Nikolaos' now withdrawn conjecture that

                    > If P is a point and P1, P2, P3 are its extraversions
                    > If the line PX1 where X1 is the orthogonal projection of P1 on BC
                    > meets the pedal circle of P1 at A* and similarly define the points
                    > B*, C* then the triangles ABC, A*B*C* are perspective.

                    Cabri suggests that for any point P there are a number of points P1 that
                    have the above property. I have not been able to determine a locus.

                    Dick Tahta
                  • Paul Yiu
                    Dear Eric and friends, [ED]: Here s a new point I ve discovered by chance while playing with extraversions. Let I be the incenter. Let A be the point where
                    Message 9 of 23 , Mar 23, 2004
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                      Dear Eric and friends,

                      [ED]: Here's a new point I've discovered by chance while playing with
                      extraversions.

                      Let I be the incenter.
                      Let A' be the point where the A-excircle touches BC and let A* be
                      the second intersection of the line IA' with the A-excircle.
                      Define B', C' and B*, C* cyclically.

                      ==> Triangles ABC and A*B*C* are perspective

                      The barycentric coordinates of the perspector P are

                      P = ( a^5 - a^4.(b + c) - 2.a^3.(b + c)^2 + 2.a^2.(b + c)^3
                      + a.(b + c)^4 - (b + c)^5 :
                      b^5 - b^4.(c + a) - 2.b^3.(c + a)^2 + 2.b^2.(c + a)^3
                      + b.(c + a)^4 - (c + a)^5 :
                      c^5 - c^4.(a + b) - 2.c^3.(a + b)^2 + 2.c^2.(a + b)^3
                      + c.(a + b)^4 - (a + b)^5 )

                      *** These coordinates simplify, as Jean-Pierre has pointed out, into
                      ((b+c-a)^3 : (c+a-b)^3 : (a+b-c)^3)). P is indeed the ``barycentric
                      cube'' of the Nagel point.

                      I have also investigated the question with the incenter I replaced by
                      a general point Q with barycentric coordinates (u:v:w).
                      With A* = second intersection of QA' with the A-excircle etc., the
                      triangle A*B*C* is perspective with ABC if and only if Q lies on one
                      of the following curves:

                      (i) the line at infinity,
                      (ii) the inferior of the circumconic with center I, (which is the
                      conic through A', B', C' and the midpoints of the sides of ABC),
                      (iii) the Spieker central cubic, identified as K033 in Bernard's CTC).

                      The locus of the perspector is
                      (i) the isogonal conjugate of the incircle,
                      (ii) the inscribed conic with center X(1329), which is the
                      intersection of the lines joining the Nagel point to the Feuerbach
                      point, and the centroid to the outer Feuerbach point X(12),
                      (iii) to be determined, but here are the perspectors for various Q on
                      the Spieker central cubic:

                      Q perspector
                      ----------------------
                      I (b+c-a)^3:...:...
                      H (a^3+a^2(b+c)-a(b+c)^2-(b-c)^2(b+c))^2/(b+c-a):...:...
                      X(8) X(8)
                      X(10) (b+c-a)/a^2 : ... : ...
                      X(40) X(7)
                      X(65) X(1259)
                      X(72) X(55)

                      Best regards
                      Sincerely
                      Paul
                    • Eric Danneels
                      Dear Jean-Pierre, my barycentrics were really ugly I didn t notice they had a common divisor (a+b+c)^2 Greetings from Brussels Eric
                      Message 10 of 23 , Mar 23, 2004
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                        Dear Jean-Pierre,

                        my barycentrics were really ugly

                        I didn't notice they had a common divisor (a+b+c)^2

                        Greetings from Brussels

                        Eric
                      • Bernard Gibert
                        Dear Jean-Pierre ... If you replace the excircles with the three in-conics centered at the vertices of the anticevian of a point Q, then : 1) the polars of M
                        Message 11 of 23 , Mar 24, 2004
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                          Dear Jean-Pierre

                          > [JP] If we condider the polars of a point M wrt the three incircles :
                          > 1)they concur if and only if M lies on the radical circle of the
                          > three circles, which is the circle with center the Spieker point,
                          > radius 1/2*root(p^2+r^2)
                          > 2)they board a triangle perspective with ABC if and only if M lies
                          > on the Spieker central cubic

                          If you replace the excircles with the three in-conics centered at the
                          vertices of the anticevian of a point Q, then :

                          1) the polars of M wrt to these conics concur iff M lies on a conic (C)
                          centered at the complement of Q.
                          The three in-conics and (C) share the same directions of axis.

                          2) these polars bound a triangle perspective to ABC iff M lies on the
                          central pK
                          - with pole the center of the inconic with perspector Q,
                          - with pivot the anticomplement of Q,
                          - with center the complement of Q,

                          This is another way to approach our central pK in Special Isocubics §3.

                          Best regards

                          Bernard

                          [Non-text portions of this message have been removed]
                        • Paul Yiu
                          Dear Bernard and Jean-Pierre [JP] If we condider the polars of a point M wrt the three incircles : 1)they concur if and only if M lies on the radical circle of
                          Message 12 of 23 , Mar 24, 2004
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                            Dear Bernard and Jean-Pierre

                            [JP] If we condider the polars of a point M wrt the three incircles :
                            1)they concur if and only if M lies on the radical circle of the
                            three circles, which is the circle with center the Spieker point,
                            radius 1/2*root(p^2+r^2).
                            ...

                            [BG]: If you replace the excircles with the three in-conics centered
                            at the vertices of the anticevian of a point Q, then :

                            1) the polars of M wrt to these conics concur iff M lies on a conic (C)
                            centered at the complement of Q. The three in-conics and (C) share the
                            same directions of axis.
                            ...

                            *** Let Q' be the complement of Q. This conic (C) is the one through the
                            six intersections of the inconics with the polar of Q' with respect to
                            the three inconics. If M is a point on this conic, its polars with
                            respect to the inconic intersect at its antipode on the conic (C).

                            Best regards
                            Sincerely
                            Paul
                          • jpehrmfr
                            Dear Paul and Bernard ... centered ... conic (C) ... the ... through the ... respect to ... We know that : (1) The polar circle Cs of the three excircles is
                            Message 13 of 23 , Mar 24, 2004
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                              Dear Paul and Bernard
                              > [JP] If we condider the polars of a point M wrt the three
                              incircles :
                              > 1)they concur if and only if M lies on the radical circle of the
                              > three circles, which is the circle with center the Spieker point,
                              > radius 1/2*root(p^2+r^2).
                              > ...
                              >
                              > [BG]: If you replace the excircles with the three in-conics
                              centered
                              > at the vertices of the anticevian of a point Q, then :
                              >
                              > 1) the polars of M wrt to these conics concur iff M lies on a
                              conic (C)
                              > centered at the complement of Q. The three in-conics and (C) share
                              the
                              > same directions of axis.
                              > ...
                              >
                              > *** Let Q' be the complement of Q. This conic (C) is the one
                              through the
                              > six intersections of the inconics with the polar of Q' with
                              respect to
                              > the three inconics. If M is a point on this conic, its polars with
                              > respect to the inconic intersect at its antipode on the conic (C).
                              We know that :
                              (1) The polar circle Cs of the three excircles is centered at the
                              complement of I and the polar line of S wrt an excircle is the
                              radical axis of Cs and the excircle.
                              Your results are now obvious if you change in (1) a,b,c to the
                              barycentric coordinates of Q.
                              Friendly. Jean-Pierre
                            • Darij Grinberg
                              Dear Eric, Jean-Pierre and Paul, I have finally found time to return to the problem on the concurrent lines and the excircles. Thanks for the very interesting
                              Message 14 of 23 , Mar 26, 2004
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                                Dear Eric, Jean-Pierre and Paul,

                                I have finally found time to return to the problem
                                on the concurrent lines and the excircles. Thanks
                                for the very interesting data.

                                In Hyacinthos message #9607, Paul Yiu wrote:

                                >> These coordinates simplify, as Jean-Pierre has
                                >> pointed out, into
                                >> ((b+c-a)^3 : (c+a-b)^3 : (a+b-c)^3)). P is
                                >> indeed the ``barycentric cube'' of the Nagel
                                >> point.

                                In fact, now I see this coincides with another
                                configuration.

                                The point A* is just the point on the A-excircle
                                such that the circle BCA* is tangent to the
                                A-excircle!!

                                And now the fact that the lines AA*, BB*, CC*
                                concur turns out to be equivalent to what
                                Antreas wrote in Hyacinthos message #5636
                                (our A*, B*, C* are his A2, B2, C2).

                                Now it wouldn't be surprising if there is a
                                synthetic proof of this all!

                                Sincerely,
                                Darij Grinberg
                              • Antreas P. Hatzipolakis
                                Dear Darij ... That is, ... We can consider other [than the three excircles] well-defined triads of circles touching the sidelines, and ask similar questions.
                                Message 15 of 23 , Mar 26, 2004
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                                  Dear Darij

                                  [DG]:
                                  >And now the fact that the lines AA*, BB*, CC*
                                  >concur turns out to be equivalent to what
                                  >Antreas wrote in Hyacinthos message #5636
                                  >(our A*, B*, C* are his A2, B2, C2).

                                  That is,

                                  [APH]:
                                  >> Let A2, B2, C2 be the points of contact of the circles
                                  >> passing through (B,C), (C,A), (A,B) and touching
                                  >> the a-,b-c-excircles (resp.) [instead of the incircle].
                                  >> Are the triangles ABC, A2B2C2 perspective?

                                  We can consider other [than the three excircles] well-defined triads
                                  of circles touching the sidelines, and ask similar questions.

                                  Just an other one:

                                  Let (Ka), (Kb), (Kc) be the three circles with diameters the three
                                  altitudes AHa, BHb, CHc [==> they touch BC, CA, AB at Ha, Hb, Hc, resp.]

                                  Consider now the circle passing through (B, C) and touching (Ka) at A*.
                                  And similarly define the points B*, C*.

                                  Are the triangles ABC, A*B*C* perspective?

                                  If not :

                                  Denote :

                                  A# = BC /\ AA*, B# = CA /\ BB*, C# = AB /\ CC*

                                  Are the points A#, B#, C# collinear?

                                  Antreas
                                  --
                                • jpehrmfr
                                  Dear Antreas ... triads ... resp.] ... at A*. ... A# = BC inter HbHc = harmonic conjugate of Ha wrt B,C; hence A#, B#, C# lie on the trilinear polar of H
                                  Message 16 of 23 , Mar 26, 2004
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                                    Dear Antreas
                                    > We can consider other [than the three excircles] well-defined
                                    triads
                                    > of circles touching the sidelines, and ask similar questions.
                                    >
                                    > Just an other one:
                                    >
                                    > Let (Ka), (Kb), (Kc) be the three circles with diameters the three
                                    > altitudes AHa, BHb, CHc [==> they touch BC, CA, AB at Ha, Hb, Hc,
                                    resp.]
                                    >
                                    > Consider now the circle passing through (B, C) and touching (Ka)
                                    at A*.
                                    > And similarly define the points B*, C*.
                                    >
                                    > Are the triangles ABC, A*B*C* perspective?
                                    >
                                    > If not :
                                    >
                                    > Denote :
                                    >
                                    > A# = BC /\ AA*, B# = CA /\ BB*, C# = AB /\ CC*
                                    >
                                    > Are the points A#, B#, C# collinear?

                                    A# = BC inter HbHc = harmonic conjugate of Ha wrt B,C; hence A#, B#,
                                    C# lie on the trilinear polar of H (orthic axis)
                                    Of course, A* = projection of Ha upon AA#.
                                    Friendly. Jean-Pierre
                                  • jpehrmfr
                                    Dear Antreas ... triads ... If the circle Ca touches BC at U, let U = harmonic conjugate of U wrt (B,C) and U = antipode of U on Ca. Then A2 is the
                                    Message 17 of 23 , Mar 27, 2004
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                                      Dear Antreas
                                      > [APH]:
                                      > >> Let A2, B2, C2 be the points of contact of the circles
                                      > >> passing through (B,C), (C,A), (A,B) and touching
                                      > >> the a-,b-c-excircles (resp.) [instead of the incircle].
                                      > >> Are the triangles ABC, A2B2C2 perspective?
                                      >
                                      > We can consider other [than the three excircles] well-defined
                                      triads
                                      > of circles touching the sidelines, and ask similar questions.
                                      If the circle Ca touches BC at U, let U' = harmonic conjugate of U
                                      wrt (B,C) and U'' = antipode of U on Ca. Then A2 is the projection
                                      of U upon U'U''.
                                      I've tried two particular cases : the three circles Ca, Cb, Cc touch
                                      respectively BC, CA, AB at the vertices of the pedal - or cevian -
                                      triangle of P and go respectively through A, B, C. Unfortunately,
                                      I'm absolutely unable to find even one point P such as A2B2C2 and
                                      ABC are perspective (their locus is of degree 12 in the case of the
                                      pedal triangle and of degree 9 in the case of the cevian triangle).
                                      Friendly. Jean-Pierre
                                    • Antreas P. Hatzipolakis
                                      Dear Jean-Pierre ... Here is a variation: Each one of three circles Ka,Kb,Kc passes through TWO vertices of the pedal (cevian) triangle of P, and touches ONE
                                      Message 18 of 23 , Mar 27, 2004
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                                        Dear Jean-Pierre

                                        [APH]:

                                        >> We can consider other [than the three excircles] well-defined triads
                                        >> of circles touching the sidelines, and ask similar questions.

                                        [JPE]:
                                        >I've tried two particular cases : the three circles Ca, Cb, Cc touch
                                        >respectively BC, CA, AB at the vertices of the pedal - or cevian -
                                        >triangle of P and go respectively through A, B, C. Unfortunately,
                                        >I'm absolutely unable to find even one point P such as A2B2C2 and
                                        >ABC are perspective (their locus is of degree 12 in the case of the
                                        >pedal triangle and of degree 9 in the case of the cevian triangle).

                                        Here is a variation:

                                        Each one of three circles Ka,Kb,Kc passes through TWO vertices
                                        of the pedal (cevian) triangle of P, and touches ONE sideline of ABC.

                                        Let A2, B2, C2 be the points of contact of the circles
                                        passing through (B,C), (C,A), (A,B) and touching
                                        the Ka,Kb,Kc circles (resp.)

                                        Which is now the locus of P such that
                                        ABC, A2B2C2 are perspective?

                                        Greetings

                                        Antreas
                                        --
                                      • Nikolaos Dergiades
                                        Dear Antreas and Jean-Pierre ... Another variation: Let P be a point and A1B1C1 its pedal triangle. Define Ka the circle with center P and tangent to BC at
                                        Message 19 of 23 , Mar 27, 2004
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                                          Dear Antreas and Jean-Pierre

                                          [APH]:

                                          >> We can consider other [than the three excircles] well-defined triads
                                          >> of circles touching the sidelines, and ask similar questions.

                                          [JPE]:
                                          >I've tried two particular cases : the three circles Ca, Cb, Cc touch
                                          >respectively BC, CA, AB at the vertices of the pedal - or cevian -
                                          >triangle of P and go respectively through A, B, C. Unfortunately,
                                          >I'm absolutely unable to find even one point P such as A2B2C2 and
                                          >ABC are perspective (their locus is of degree 12 in the case of the
                                          >pedal triangle and of degree 9 in the case of the cevian triangle).

                                          Another variation:

                                          Let P be a point and A1B1C1 its pedal triangle. Define Ka the circle
                                          with center P and tangent to BC at A1. Similarly define the circles Kb,Kc.

                                          Let A2, B2, C2 be the points of contact of the circles
                                          passing through (B,C), (C,A), (A,B) and touching
                                          the Ka,Kb,Kc circles (resp.)

                                          Which is the locus of P such that
                                          ABC, A2B2C2 are perspective?

                                          An obvious point on this locus is the circumcenter O of ABC.
                                          It is easy to give a synthetic proof that the triangles
                                          ABC, A2B2C2 are homothetic and that the perspector is the
                                          De Longchamps point.

                                          Another point on this locus (I think) is the incenter I of ABC
                                          and that the perspector is not in ETC.

                                          Best regards
                                          Nikos
                                        • jpehrmfr
                                          Dear Nikolaos and Antreas ... circle ... circles Kb,Kc. ... I think that this point is barycentric 1/(p-a)^3:.., which is X(479) in ETC. It seems that your
                                          Message 20 of 23 , Mar 27, 2004
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                                            Dear Nikolaos and Antreas
                                            > Let P be a point and A1B1C1 its pedal triangle. Define Ka the
                                            circle
                                            > with center P and tangent to BC at A1. Similarly define the
                                            circles Kb,Kc.
                                            >
                                            > Let A2, B2, C2 be the points of contact of the circles
                                            > passing through (B,C), (C,A), (A,B) and touching
                                            > the Ka,Kb,Kc circles (resp.)
                                            >
                                            > Which is the locus of P such that
                                            > ABC, A2B2C2 are perspective?
                                            >
                                            > An obvious point on this locus is the circumcenter O of ABC.
                                            > It is easy to give a synthetic proof that the triangles
                                            > ABC, A2B2C2 are homothetic and that the perspector is the
                                            > De Longchamps point.
                                            >
                                            > Another point on this locus (I think) is the incenter I of ABC
                                            > and that the perspector is not in ETC.

                                            I think that this point is barycentric 1/(p-a)^3:.., which is X(479)
                                            in ETC.
                                            It seems that your locus is a ugly curve of degree 15.
                                            Friendly. Jean-Pierre
                                          • Nikolaos Dergiades
                                            Dear Jean-Pierre, [JPE] I think that this point is barycentric 1/(p-a)^3:.., which is X(479) in ETC. It seems that your locus is a ugly curve of degree 15.
                                            Message 21 of 23 , Mar 28, 2004
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                                              Dear Jean-Pierre,

                                              [JPE]
                                              I think that this point is barycentric 1/(p-a)^3:.., which is X(479)
                                              in ETC.
                                              It seems that your locus is a ugly curve of degree 15.

                                              *************

                                              Thanks for the locus.
                                              You are right for X(479) I was looking for a wrong search number.
                                              It seems to me that on the locus are also the three excenters of ABC.

                                              Best regards
                                              Nikolaos Dergiades
                                            • xpolakis
                                              Another triad of circles touching the sidelines of ABC. Let Abc be a triangle and A1B1C1 a triangle inscribed in ABC. Construct three CONGRUENT and CONCURRENT
                                              Message 22 of 23 , Mar 30, 2004
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                                                Another triad of circles touching the sidelines of ABC.

                                                Let Abc be a triangle and A1B1C1 a triangle inscribed in ABC.
                                                Construct three CONGRUENT and CONCURRENT circles Ka, Kb, Kc such that
                                                Ka touches BC at A1, Kb touches CA at B1, and Kc touches AB at C1.

                                                If A1B1C1 is the pedal (or cevian) triangle of P, and Ca, Cb, Cc
                                                are the three circles passing through (B,C), (C,A), (A,B)
                                                and touching Ka, Kb, Kc at A2,B2,C2 resp., then which is the locus
                                                of P such that ABC, A2B2C2 are perspective?

                                                APH
                                              • Antreas P. Hatzipolakis
                                                ... And which is the locus of P such that the triangles ABC, KaKbKc are perspective? (if A1B1C1 = pedal or cevian tr. of P) APH --
                                                Message 23 of 23 , Apr 1, 2004
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                                                  [APH]:

                                                  >Let ABC be a triangle and A1B1C1 a triangle inscribed in ABC.
                                                  >Construct three CONGRUENT and CONCURRENT circles (Ka), (Kb), (Kc) such that
                                                  >Ka touches BC at A1, Kb touches CA at B1, and Kc touches AB at C1.
                                                  >
                                                  >If A1B1C1 is the pedal (or cevian) triangle of P, and (Ca), (Cb), (Cc)
                                                  >are the three circles passing through (B,C), (C,A), (A,B)
                                                  >and touching (Ka), (Kb), (Kc) at A2,B2,C2 resp., then which is the locus
                                                  >of P such that ABC, A2B2C2 are perspective?

                                                  And which is the locus of P such that the triangles
                                                  ABC, KaKbKc are perspective? (if A1B1C1 = pedal or cevian tr. of P)

                                                  APH

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