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A concyclic problem

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  • lovemathforever
    Dear friends of Hyacinthos, I found an interesting concyclic problem. Can anyone solve it synthetically? Given triangle ABC with its circumcircle (O) and its
    Message 1 of 21 , Apr 10, 2010
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      Dear friends of Hyacinthos,
      I found an interesting concyclic problem. Can anyone solve it synthetically?
      Given triangle ABC with its circumcircle (O) and its orthocenter H. Let H_aH_bH_c be the orthic triangle of triangle ABC. P is an arbitrary point on Euler line. AP, BP, CP intersect (O) again at A_1, B_1, C_1. Let A_2, B_2, C_2 be the reflections of A_1, B_1, C_1 wrt H_a, H_b, H_c, respectively. Prove that H, A_2,B_2,C_2 are concyclic.
      Best regard,
      Linh Nguyen Van
    • Francisco Javier
      Dear Linh Nguyen Van, We have that the result is also true when P is on the circumcircle. In this case A1=B1=C1=P and the circumcircle of A2B2C2 is image of
      Message 2 of 21 , Apr 10, 2010
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        Dear Linh Nguyen Van,

        We have that the result is also true when P is on the circumcircle.
        In this case A1=B1=C1=P and the circumcircle of A2B2C2 is image of the nine point circle by a homothety with P as center and ratio 2. Is easy that H is on this circumcircle.

        Finally, we have a degenerate case when P is on an altitude, say AH. In this case we have A2=H, thus H,A_2,B_2,C_2 are trivially concyclic.

        --- In Hyacinthos@yahoogroups.com, "lovemathforever" <lovemathforever@...> wrote:
        >
        > Dear friends of Hyacinthos,
        > I found an interesting concyclic problem. Can anyone solve it synthetically?
        > Given triangle ABC with its circumcircle (O) and its orthocenter H. Let H_aH_bH_c be the orthic triangle of triangle ABC. P is an arbitrary point on Euler line. AP, BP, CP intersect (O) again at A_1, B_1, C_1. Let A_2, B_2, C_2 be the reflections of A_1, B_1, C_1 wrt H_a, H_b, H_c, respectively. Prove that H, A_2,B_2,C_2 are concyclic.
        > Best regard,
        > Linh Nguyen Van
        >
      • armpist
        Dear Linh Nguyen Van See Hagge circle construction in FG 2007 pp. 231-247 M.T.
        Message 3 of 21 , Apr 11, 2010
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          Dear Linh Nguyen Van

          See Hagge circle construction in FG 2007 pp. 231-247

          M.T.

          --- In Hyacinthos@yahoogroups.com, "lovemathforever" <lovemathforever@...> wrote:
          >
          > Dear friends of Hyacinthos,
          > I found an interesting concyclic problem. Can anyone solve it synthetically?
          > Given triangle ABC with its circumcircle (O) and its orthocenter H. Let H_aH_bH_c be the orthic triangle of triangle ABC. P is an arbitrary point on Euler line. AP, BP, CP intersect (O) again at A_1, B_1, C_1. Let A_2, B_2, C_2 be the reflections of A_1, B_1, C_1 wrt H_a, H_b, H_c, respectively. Prove that H, A_2,B_2,C_2 are concyclic.
          > Best regard,
          > Linh Nguyen Van
          >
        • lovemathforever
          Dear Francisco, Yes, two cases you mentioned are trivival. But my problem is stated that P lies on Euler line. Dear Armpist, Please read the message above
          Message 4 of 21 , Apr 11, 2010
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            Dear Francisco,
            Yes, two cases you mentioned are trivival. But my problem is stated that P lies on Euler line.

            Dear Armpist,
            Please read the message above carefully! My problem is only similar to Hagge circle, but it is not Hagge.

            Best regard,
            Linh Nguyen Van
          • Francisco Javier
            Dear Linh Nguyen Van, I haven t a synthetic solution at the moment, but perhaps is useful to know that there are another (although trivial) solutions of the
            Message 5 of 21 , Apr 11, 2010
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              Dear Linh Nguyen Van,

              I haven't a synthetic solution at the moment, but perhaps is useful to know that there are another (although trivial) solutions of the problem.

              --- In Hyacinthos@yahoogroups.com, "lovemathforever" <lovemathforever@...> wrote:
              >
              > Dear friends of Hyacinthos,
              > I found an interesting concyclic problem. Can anyone solve it synthetically?
              > Given triangle ABC with its circumcircle (O) and its orthocenter H. Let H_aH_bH_c be the orthic triangle of triangle ABC. P is an arbitrary point on Euler line. AP, BP, CP intersect (O) again at A_1, B_1, C_1. Let A_2, B_2, C_2 be the reflections of A_1, B_1, C_1 wrt H_a, H_b, H_c, respectively. Prove that H, A_2,B_2,C_2 are concyclic.
              > Best regard,
              > Linh Nguyen Van
              >
            • Antreas
              [Linh Nguyen Van] ... As locus problem: Let ABC be a triangle, HaHbHc the pedal triangle of H (orthic triangle), P a point, A1B1C1 the circumcevian triangle of
              Message 6 of 21 , Apr 12, 2010
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                [Linh Nguyen Van]
                > Given triangle ABC with its circumcircle (O) and its orthocenter
                > H. Let H_aH_bH_c be the orthic triangle of triangle ABC. P is
                > an arbitrary point on Euler line. AP, BP, CP intersect (O) again
                > at A_1, B_1, C_1. Let A_2, B_2, C_2 be the reflections of A_1,
                > B_1, C_1 wrt H_a, H_b, H_c, respectively. Prove that H, A_2,B_2,C_2
                > are concyclic.

                As locus problem:

                Let ABC be a triangle, HaHbHc the pedal triangle of H (orthic
                triangle), P a point, A1B1C1 the circumcevian triangle of P
                and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc, resp.

                Which is the locus of P such that the circumcircle of
                A2B2C2 passes through H?

                Questions:

                1. Is the answer: "Circumcircle of ABC + Three Altitudes + Euler
                Line of ABC" complete?

                2. As P moves on the Euler line, where is moving the circumcenter
                of A2B2C2?

                Variation:

                Let MaMbMc be the pedal triangle of O (medial triangle) and A2,
                B2,C2 the reflections of A1,B1,C1 in Ma,Mb,Mc, resp.
                Which is the locus of P such that the circumcircle of
                A2B2C2 passes through H?

                Antreas
              • Nikolaos Dergiades
                Dear Antreas ... It seems to be part of the Euler line. Best regards Nikos __________________________________________________ Χρησιμοποιείτε
                Message 7 of 21 , Apr 12, 2010
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                  Dear Antreas

                  > As locus problem:
                  >
                  > Let ABC be a triangle, HaHbHc the pedal triangle of H
                  > (orthic
                  > triangle), P a point, A1B1C1 the circumcevian triangle of P
                  >
                  > and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
                  > resp.
                  >
                  > Which is the locus of P such that the circumcircle of
                  > A2B2C2 passes through H?
                  >
                  > Questions:
                  >
                  >
                  > 2. As P moves on the Euler line, where is moving the
                  > circumcenter
                  > of A2B2C2?

                  It seems to be part of the Euler line.

                  Best regards
                  Nikos

                  __________________________________________________
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                • Francisco Javier
                  We may want to know what is the locus of points P such that the orthocenter lies on one of the sides of A2B2C2. Answer: For example, B2, C2 and H are collinear
                  Message 8 of 21 , Apr 12, 2010
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                    We may want to know what is the locus of points P such that the orthocenter lies on one of the sides of A2B2C2.

                    Answer: For example, B2, C2 and H are collinear if and only if P is on the inverse with respect to the circumcircle of ABC of the circumcircle of BCH.

                    --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
                    >
                    > [Linh Nguyen Van]
                    > > Given triangle ABC with its circumcircle (O) and its orthocenter
                    > > H. Let H_aH_bH_c be the orthic triangle of triangle ABC. P is
                    > > an arbitrary point on Euler line. AP, BP, CP intersect (O) again
                    > > at A_1, B_1, C_1. Let A_2, B_2, C_2 be the reflections of A_1,
                    > > B_1, C_1 wrt H_a, H_b, H_c, respectively. Prove that H, A_2,B_2,C_2
                    > > are concyclic.
                    >
                    > As locus problem:
                    >
                    > Let ABC be a triangle, HaHbHc the pedal triangle of H (orthic
                    > triangle), P a point, A1B1C1 the circumcevian triangle of P
                    > and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc, resp.
                    >
                    > Which is the locus of P such that the circumcircle of
                    > A2B2C2 passes through H?
                    >
                    > Questions:
                    >
                    > 1. Is the answer: "Circumcircle of ABC + Three Altitudes + Euler
                    > Line of ABC" complete?
                    >
                    > 2. As P moves on the Euler line, where is moving the circumcenter
                    > of A2B2C2?
                    >
                    > Variation:
                    >
                    > Let MaMbMc be the pedal triangle of O (medial triangle) and A2,
                    > B2,C2 the reflections of A1,B1,C1 in Ma,Mb,Mc, resp.
                    > Which is the locus of P such that the circumcircle of
                    > A2B2C2 passes through H?
                    >
                    > Antreas
                    >
                  • Nikolaos Dergiades
                    Dear Linh Nguyen Van, you wrote ... I tried to give a synthetic proof but did nothing. I tried to give a proof with barycentrics but also did nothing because
                    Message 9 of 21 , Apr 12, 2010
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                      Dear Linh Nguyen Van,

                      you wrote
                      > Given triangle ABC with its circumcircle (O) and
                      > its orthocenter H. Let HaHbHc be the orthic triangle of
                      > triangle ABC. P is an arbitrary point on Euler line.
                      > AP, BP, CP intersect (O) again at A1, B1, C1.
                      > Let A2, B2, C2 be the reflections of A1, B1, C1 wrt Ha, Hb, Hc,
                      > respectively. Prove that H, A2, B2, C2 are concyclic.

                      I tried to give a synthetic proof but did nothing.
                      I tried to give a proof with barycentrics but also did nothing
                      because the computations were very complicated.

                      Here is a proof with complex numbers.
                      Let the circumcircle of ABC be the unit circle and at the
                      points A, B, C, O be the complex numbers a, b, c, 0.
                      Since the conjugate(a) = 1/a ... we can do all we need.
                      Then it is known and easy to prove that at H is the number h=a+b+c.
                      At the point P on the Euler line is the number t*(a+b+c).
                      At the point A1 is the number a1=bc[a-t(a+b+c)]/[t(ab+bc+ca)-bc].
                      At the point Ha is the number ha=[a(a+b+c)-bc]/(2a).
                      The reflection of O in Ha is the point O1 at the number o1=2ha.
                      The point A2 is at the number a2 = 2ha - a1.
                      The triangle O1A2H is isosceles because
                      O1A2 = OA1 = R = OH1 = O1H where H1 (on the circumcircle)
                      is the reflection of H in Ha.
                      The mid point M1 of A2H is at the point m1 = (a2+h)/2.
                      Finally the line O1M1 meets the Euler line at the point Q
                      that is equidistant from A2 and H.
                      This point Q is at the number q = m*(a+b+c)
                      where m = [(s+2abc)t-2abc]/[(s+abc)t-abc]
                      where s = (a+b)(b+c)(c+a).
                      Since this point Q is symmetric relative to a,b,c
                      is equidistant from B2, H and from C2, H.
                      Hence the points A2, B2, C2, H are concyclic
                      with center Q on the Euler line.

                      If P = H then t = 1 and Q = H
                      If P = O then t = 0 and Q = X382 the reflection of O in H.

                      Best regards
                      Nikos Dergiades




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                    • Nikolaos Dergiades
                      Dear friends, Francisco mentioned me that the points A2, B2, C2, H are collinear if P is the inverse of H in (O). Yes this is true if Q lies at infinity or the
                      Message 10 of 21 , Apr 12, 2010
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                        Dear friends,
                        Francisco mentioned me that the points
                        A2, B2, C2, H are collinear if P is the inverse
                        of H in (O).
                        Yes this is true if Q lies at infinity
                        or the denominator of q is zero
                        that is (s+abc)t-abc = 0 or
                        t = abc / [(a+b)(b+c)(c+a)+abc]
                        and for this t the point P is the inverse
                        of H in (O).
                        Thank you Francisco.
                        Best regards
                        Nikos Dergiades


                        > Dear Linh Nguyen Van,
                        >
                        > you wrote
                        > > Given triangle ABC with its circumcircle (O) and
                        > > its orthocenter H. Let HaHbHc be the orthic triangle
                        > of
                        > > triangle ABC. P is an arbitrary point on Euler line.
                        > > AP, BP, CP intersect (O) again at A1, B1, C1.
                        > > Let A2, B2, C2 be the reflections of A1, B1, C1 wrt
                        > Ha, Hb, Hc,
                        > > respectively. Prove that H, A2, B2, C2 are concyclic.
                        >
                        > I tried to give a synthetic proof but did nothing.
                        > I tried to give a proof with barycentrics but also did
                        > nothing
                        > because the computations were very complicated.
                        >
                        > Here is a proof with complex numbers.
                        > Let the circumcircle of ABC be the unit circle and at the
                        > points A, B, C, O be the complex numbers a, b, c, 0.
                        > Since the conjugate(a) = 1/a ... we can do all we need.
                        > Then it is known and easy to prove that at H is the number
                        > h=a+b+c.
                        > At the point P on the Euler line is the number t*(a+b+c).
                        > At the point A1 is the number
                        > a1=bc[a-t(a+b+c)]/[t(ab+bc+ca)-bc].
                        > At the point Ha is the number ha=[a(a+b+c)-bc]/(2a).
                        > The reflection of O in Ha is the point O1 at the number
                        > o1=2ha.
                        > The point A2 is at the number a2 = 2ha - a1.
                        > The triangle O1A2H is isosceles because
                        > O1A2 = OA1 = R = OH1 = O1H  where H1 (on the
                        > circumcircle)
                        > is the reflection of H in Ha.
                        > The mid point M1 of A2H is at the point m1 = (a2+h)/2.
                        > Finally the line O1M1 meets the Euler line at the point Q
                        > that is equidistant from A2 and H.
                        > This point Q is at the number q = m*(a+b+c)
                        > where m = [(s+2abc)t-2abc]/[(s+abc)t-abc]
                        > where s = (a+b)(b+c)(c+a).
                        > Since this point Q is symmetric relative to a,b,c
                        > is equidistant from B2, H  and from C2, H.
                        > Hence the points A2, B2, C2, H are concyclic
                        > with center Q on the Euler line.
                        >
                        > If P = H then t = 1 and Q = H
                        > If P = O then t = 0 and Q = X382 the reflection of O in H.
                        >
                        > Best regards
                        > Nikos Dergiades
                        >
                        >
                        >  
                        >
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                      • lovemathforever
                        Thanks you dear Nikolaos for your solution. I am also finding the synthetic solution but false. ... Dear Francisco, For this variation you can see Hagge
                        Message 11 of 21 , Apr 12, 2010
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                          Thanks you dear Nikolaos for your solution. I am also finding the synthetic solution but false.

                          >
                          >[Francisco Javier]
                          > Variation:
                          >
                          > Let MaMbMc be the pedal triangle of O (medial triangle) and A2,
                          > B2,C2 the reflections of A1,B1,C1 in Ma,Mb,Mc, resp.
                          > Which is the locus of P such that the circumcircle of
                          > A2B2C2 passes through H?
                          >
                          Dear Francisco,
                          For this variation you can see 'Hagge circle'. Every points P on the plane satisfy this condition.
                        • Alexey Zaslavsky
                          Dear Linh Nguyen Van! I haven t a synthetic proof, but using complex numbers it is easy to prove your assertion and also obtain that the circumcenter of
                          Message 12 of 21 , Apr 12, 2010
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                            Dear Linh Nguyen Van!
                            I haven't a synthetic proof, but using complex numbers it is easy to prove your assertion and also obtain that the circumcenter of A_2B_2C_2 lie on the Euelr line.

                            Sincerely Alexey



                            Dear friends of Hyacinthos,
                            I found an interesting concyclic problem. Can anyone solve it synthetically?
                            Given triangle ABC with its circumcircle (O) and its orthocenter H. Let H_aH_bH_c be the orthic triangle of triangle ABC. P is an arbitrary point on Euler line. AP, BP, CP intersect (O) again at A_1, B_1, C_1. Let A_2, B_2, C_2 be the reflections of A_1, B_1, C_1 wrt H_a, H_b, H_c, respectively. Prove that H, A_2,B_2,C_2 are concyclic.
                            Best regard,
                            Linh Nguyen Van



                            .



                            [Non-text portions of this message have been removed]
                          • Antreas
                            [APH] ... [ND] ... Dear Nikos So if P lies on the Euler line, then the circumcenter of A2B2C2 lies also on the Euler line. Reversely: If the circumcenter of
                            Message 13 of 21 , Apr 13, 2010
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                              [APH]
                              > > As locus problem:
                              > >
                              > > Let ABC be a triangle, HaHbHc the pedal triangle of H
                              > > (orthic
                              > > triangle), P a point, A1B1C1 the circumcevian triangle of P
                              > >
                              > > and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
                              > > resp.
                              > >
                              > > Which is the locus of P such that the circumcircle of
                              > > A2B2C2 passes through H?
                              > >
                              > > Questions:
                              > >
                              > >
                              > > 2. As P moves on the Euler line, where is moving the
                              > > circumcenter
                              > > of A2B2C2?

                              [ND]
                              > It seems to be part of the Euler line.

                              Dear Nikos

                              So if P lies on the Euler line, then the circumcenter
                              of A2B2C2 lies also on the Euler line.

                              Reversely: If the circumcenter of A2B2C2 is on the
                              Euler line, then where is lying P? That is:

                              Let ABC be a triangle, HaHbHc the pedal triangle of H
                              (orthic triangle), P a point, A1B1C1 the circumcevian triangle
                              of P and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
                              resp.
                              Which is the locus of P such that the circumcenter of
                              A2B2C2 is lying on the Euler Line of ABC?

                              It is the Euler Line + ?????

                              APH
                            • Antreas
                              [APH] ... Variation (easier?): Let ABC be a triangle, HaHbHc the pedal triangle of H (orthic triangle), P a point, and A ,B , and C the reflections of P in
                              Message 14 of 21 , Apr 13, 2010
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                                [APH]
                                > Let ABC be a triangle, HaHbHc the pedal triangle of H
                                > (orthic triangle), P a point, A1B1C1 the circumcevian triangle
                                > of P and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
                                > resp.
                                > Which is the locus of P such that the circumcenter of
                                > A2B2C2 is lying on the Euler Line of ABC?
                                >
                                > It is the Euler Line + ?????

                                Variation (easier?):

                                Let ABC be a triangle, HaHbHc the pedal triangle of H
                                (orthic triangle), P a point, and A',B', and C' the reflections
                                of P in Ha, Hb, Hc, resp.
                                Which is the locus of P such that the circumcenter of
                                A'B'C' is lying on the Euler Line of ABC?

                                Is it Euler Line +???

                                APH
                              • Francisco Javier
                                Thear Antreas, ... This is the Euler line + A quintic though A, B, C, H, Ha, Hb, Hc Equation: -a^4 c^2 x^3 y^2 + 2 a^2 b^2 c^2 x^3 y^2 - b^4 c^2 x^3 y^2 + a^2
                                Message 15 of 21 , Apr 13, 2010
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                                  Thear Antreas,

                                  > [APH]
                                  > > Let ABC be a triangle, HaHbHc the pedal triangle of H
                                  > > (orthic triangle), P a point, A1B1C1 the circumcevian triangle
                                  > > of P and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
                                  > > resp.
                                  > > Which is the locus of P such that the circumcenter of
                                  > > A2B2C2 is lying on the Euler Line of ABC?
                                  > >
                                  > > It is the Euler Line + ?????

                                  This is the Euler line + A quintic though A, B, C, H, Ha, Hb, Hc

                                  Equation:

                                  -a^4 c^2 x^3 y^2 + 2 a^2 b^2 c^2 x^3 y^2 - b^4 c^2 x^3 y^2 +
                                  a^2 c^4 x^3 y^2 - b^2 c^4 x^3 y^2 - a^4 c^2 x^2 y^3 +
                                  2 a^2 b^2 c^2 x^2 y^3 - b^4 c^2 x^2 y^3 - a^2 c^4 x^2 y^3 +
                                  b^2 c^4 x^2 y^3 + a^6 x^3 y z - 2 a^4 b^2 x^3 y z +
                                  a^2 b^4 x^3 y z - 2 a^4 c^2 x^3 y z + 4 a^2 b^2 c^2 x^3 y z -
                                  2 b^4 c^2 x^3 y z + a^2 c^4 x^3 y z - 2 b^2 c^4 x^3 y z +
                                  a^6 x^2 y^2 z - a^4 b^2 x^2 y^2 z - a^2 b^4 x^2 y^2 z +
                                  b^6 x^2 y^2 z - 2 a^4 c^2 x^2 y^2 z + 4 a^2 b^2 c^2 x^2 y^2 z -
                                  2 b^4 c^2 x^2 y^2 z + c^6 x^2 y^2 z + a^4 b^2 x y^3 z -
                                  2 a^2 b^4 x y^3 z + b^6 x y^3 z - 2 a^4 c^2 x y^3 z +
                                  4 a^2 b^2 c^2 x y^3 z - 2 b^4 c^2 x y^3 z - 2 a^2 c^4 x y^3 z +
                                  b^2 c^4 x y^3 z - a^4 b^2 x^3 z^2 + a^2 b^4 x^3 z^2 +
                                  2 a^2 b^2 c^2 x^3 z^2 - b^4 c^2 x^3 z^2 - b^2 c^4 x^3 z^2 +
                                  a^6 x^2 y z^2 - 2 a^4 b^2 x^2 y z^2 + b^6 x^2 y z^2 -
                                  a^4 c^2 x^2 y z^2 + 4 a^2 b^2 c^2 x^2 y z^2 - a^2 c^4 x^2 y z^2 -
                                  2 b^2 c^4 x^2 y z^2 + c^6 x^2 y z^2 + a^6 x y^2 z^2 -
                                  2 a^2 b^4 x y^2 z^2 + b^6 x y^2 z^2 + 4 a^2 b^2 c^2 x y^2 z^2 -
                                  b^4 c^2 x y^2 z^2 - 2 a^2 c^4 x y^2 z^2 - b^2 c^4 x y^2 z^2 +
                                  c^6 x y^2 z^2 + a^4 b^2 y^3 z^2 - a^2 b^4 y^3 z^2 -
                                  a^4 c^2 y^3 z^2 + 2 a^2 b^2 c^2 y^3 z^2 - a^2 c^4 y^3 z^2 -
                                  a^4 b^2 x^2 z^3 - a^2 b^4 x^2 z^3 + 2 a^2 b^2 c^2 x^2 z^3 +
                                  b^4 c^2 x^2 z^3 - b^2 c^4 x^2 z^3 - 2 a^4 b^2 x y z^3 -
                                  2 a^2 b^4 x y z^3 + a^4 c^2 x y z^3 + 4 a^2 b^2 c^2 x y z^3 +
                                  b^4 c^2 x y z^3 - 2 a^2 c^4 x y z^3 - 2 b^2 c^4 x y z^3 +
                                  c^6 x y z^3 - a^4 b^2 y^2 z^3 - a^2 b^4 y^2 z^3 + a^4 c^2 y^2 z^3 +
                                  2 a^2 b^2 c^2 y^2 z^3 - a^2 c^4 y^2 z^3



                                  >
                                  > Variation (easier?):
                                  >
                                  > Let ABC be a triangle, HaHbHc the pedal triangle of H
                                  > (orthic triangle), P a point, and A',B', and C' the reflections
                                  > of P in Ha, Hb, Hc, resp.
                                  > Which is the locus of P such that the circumcenter of
                                  > A'B'C' is lying on the Euler Line of ABC?
                                  >
                                  > Is it Euler Line +???
                                  >

                                  This is Euler line + nothing



                                  > APH
                                  >
                                • Alexey Zaslavsky
                                  Dear Linh Nguyen Van and Nikos! If triangle ABC is regular we obtain next interesting result. Given regular triangle ABC and point P. A_1B_1C_1 is the
                                  Message 16 of 21 , Apr 13, 2010
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                                    Dear Linh Nguyen Van and Nikos!

                                    If triangle ABC is regular we obtain next interesting result.
                                    Given regular triangle ABC and point P. A_1B_1C_1 is the circucmevian
                                    triangle of P, points A_2, B_2, C_2 are the reflections of A_1, B_1, C_1 in
                                    the midpoints of BC, CA, AB
                                    Then the circumcircle of A_2B_2C_2 pass through the center O of ABC and its
                                    center is the reflection of P in O.

                                    Proof. Points O and A_2 lie on the circle with center in p[oint A_0 opposite
                                    to A. Thus the bisector of segment OA_2 is also the bisectrix of angle
                                    OA_0A_2 which is parallel to AP.
                                    It is clear that this line passes through point Q symmetric to P wrt O.
                                    Similarly the bisectors of OB_2 and OC_2 pass through Q.

                                    Sincerely
                                    Alexey
                                  • Antreas
                                    Another problem with circumcenters: Let ABC be a triangle and X,Y,Z three collinear points and XaXbXc, YaYbYc the pedal triangles of X,Y resp. Let A1,B1,C1 be
                                    Message 17 of 21 , Apr 13, 2010
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                                      Another problem with circumcenters:

                                      Let ABC be a triangle and X,Y,Z three collinear points and
                                      XaXbXc, YaYbYc the pedal triangles of X,Y resp.

                                      Let A1,B1,C1 be the reflections of Z in Xa,Xb,Xc, resp.
                                      and A2,B2,C2 the reflections of A1,B1,C1 in Ya,Yb,Yc, resp.

                                      The circumcircle of A2B2C2 passes through Z and has center
                                      on the line XYZ.

                                      True??

                                      Antreas
                                    • Francisco Javier
                                      Yes, it is true, If XZ:ZY = t and W is the circumcenter of A2B2C2 then W is on line XY and XW:WY = -2 - 1/t. Francisco Javier.
                                      Message 18 of 21 , Apr 13, 2010
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                                        Yes, it is true,

                                        If XZ:ZY = t and W is the circumcenter of A2B2C2 then W is on line XY and XW:WY = -2 - 1/t.

                                        Francisco Javier.





                                        --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
                                        >
                                        >
                                        > Another problem with circumcenters:
                                        >
                                        > Let ABC be a triangle and X,Y,Z three collinear points and
                                        > XaXbXc, YaYbYc the pedal triangles of X,Y resp.
                                        >
                                        > Let A1,B1,C1 be the reflections of Z in Xa,Xb,Xc, resp.
                                        > and A2,B2,C2 the reflections of A1,B1,C1 in Ya,Yb,Yc, resp.
                                        >
                                        > The circumcircle of A2B2C2 passes through Z and has center
                                        > on the line XYZ.
                                        >
                                        > True??
                                        >
                                        > Antreas
                                        >
                                      • Antreas Hatzipolakis
                                        ... Is it listed in Bernard s list of 5ics? APH [Non-text portions of this message have been removed]
                                        Message 19 of 21 , Apr 14, 2010
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                                          >
                                          >
                                          > > [APH]
                                          > > > Let ABC be a triangle, HaHbHc the pedal triangle of H
                                          > > > (orthic triangle), P a point, A1B1C1 the circumcevian triangle
                                          > > > of P and A2,B2,C2 the reflections of A1,B1,C1 in Ha,Hb,Hc,
                                          > > > resp.
                                          > > > Which is the locus of P such that the circumcenter of
                                          > > > A2B2C2 is lying on the Euler Line of ABC?
                                          > > >
                                          > > > It is the Euler Line + ?????
                                          >
                                          > [Francisco]


                                          > This is the Euler line + A quintic though A, B, C, H, Ha, Hb, Hc
                                          >
                                          > Equation:
                                          >

                                          Is it listed in Bernard's list of 5ics?

                                          APH


                                          [Non-text portions of this message have been removed]
                                        • Antreas
                                          [APH] ... Another variation: Let ABC be a triangle, HaHbHc the pedal triangle of H (orthic triangle), Q a fixed point on the Euler line, and P a point. Let
                                          Message 20 of 21 , Apr 14, 2010
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                                            [APH]
                                            > > Let ABC be a triangle, HaHbHc the pedal triangle of H
                                            > > (orthic triangle), P a point, and A',B', and C' the reflections
                                            > > of P in Ha, Hb, Hc, resp.
                                            > > Which is the locus of P such that the circumcenter of
                                            > > A'B'C' is lying on the Euler Line of ABC?
                                            > >
                                            > > Is it Euler Line +???

                                            [Francisco]:
                                            > This is Euler line + nothing

                                            Another variation:

                                            Let ABC be a triangle, HaHbHc the pedal triangle of H
                                            (orthic triangle), Q a fixed point on the Euler line, and
                                            P a point.

                                            Let A1,B1,C1 be the reflections of Q in Ha,Hb,Hc, resp.
                                            and A2,B2,C2 the reflections of A1,B1,C1 in P, resp.

                                            Which is the locus of P such that the circumcenter
                                            of A2B2C2 is on the Euler line of ABC?

                                            Is it the Euler Line ??

                                            Perspectivity:

                                            Are the triangles HaHbHc, A2B2C2 perspective?

                                            In general:

                                            Let X,Y,Z be three collinear points, XaXbXc the
                                            pedal triangle of X, and A1,B1,C1 the reflections
                                            of Y in Xa,Xb,Xc resp. and A2,B2,C2 the reflections
                                            of A1,B1,C1 in Z, resp.

                                            Are the triangles A2B2C2, XaXbXc perspective
                                            (with perspector on the line XYZ)?

                                            APH
                                          • lovemathforever
                                            Dear friends of Hyacinthos: The synthetics solutions of this problem can be found at here:
                                            Message 21 of 21 , May 2, 2010
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                                              Dear friends of Hyacinthos:
                                              The synthetics solutions of this problem can be found at here:
                                              http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&p=1863882&sid=ce1cf3b9df63a1477cc4699f020559dc#p1863882

                                              Best regard,
                                              Linh Nguyen Van
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