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Re: Concurrent Radical Axes

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  • rhutson2
    It also lies on line X155-X195, and is X(1147) of the Euler triangle.
    Message 1 of 8 , Apr 9, 2013
      It also lies on line X155-X195, and is X(1147) of the Euler triangle.

      --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...> wrote:
      >
      > It also lies at least lines X4-X110 and X5-X389.
      >
      > --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@> wrote:
      > >
      > > This point P, not in ETC, lies on line OQ with Q=X(1568) and satisfies the ratio OP:PQ=-OH^2/R^2.
      > >
      > > Francisco Javier.
      > >
      > > --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
      > > >
      > > > Let ABC be a triangle, A'B'C' the orthic triangle and
      > > > A"B"C" the Euler triangle (ie A",B",C" are the second
      > > > intersections of NPC with AA',BB',CC', resp. = midpoints
      > > > of AH,BH,CH, resp.)
      > > >
      > > > Denote:
      > > >
      > > > Ra = the radical axis of ((B', B'C"),(C',C'B"))
      > > >
      > > > Rb = the radical axis of ((C', C'A"),(A',A'C"))
      > > >
      > > > Rc = the radical axis of ((A', A'B"),(B',B'A"))
      > > >
      > > > The Ra,Rb,Rc are concurrent.
      > > >
      > > > Point?
      > > >
      > > > Antreas
      > > >
      > >
      >
    • Antreas
      Variations and Points http://anthrakitis.blogspot.gr/2013/04/concurrent-radical-axes.html APH
      Message 2 of 8 , Apr 10, 2013
        Variations and Points

        http://anthrakitis.blogspot.gr/2013/04/concurrent-radical-axes.html

        APH

        --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
        >
        > Let ABC be a triangle, A'B'C' the orthic triangle and
        > A"B"C" the Euler triangle (ie A",B",C" are the second
        > intersections of NPC with AA',BB',CC', resp. = midpoints
        > of AH,BH,CH, resp.)
        >
        > Denote:
        >
        > Ra = the radical axis of ((B', B'C"),(C',C'B"))
        >
        > Rb = the radical axis of ((C', C'A"),(A',A'C"))
        >
        > Rc = the radical axis of ((A', A'B"),(B',B'A"))
        >
        > The Ra,Rb,Rc are concurrent.
        >
        > Point?
        >
        > Antreas
        >
      • rhutson2
        Antreas, Your point P1.2 (Search = 2.3145425702586469385) is also the complement of X(1147) and the centroid of ABCX(68). Randy
        Message 3 of 8 , Apr 10, 2013
          Antreas,

          Your point P1.2 (Search = 2.3145425702586469385) is also the complement of X(1147) and the centroid of ABCX(68).

          Randy

          --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
          >
          > Variations and Points
          >
          > http://anthrakitis.blogspot.gr/2013/04/concurrent-radical-axes.html
          >
          > APH
          >
          > --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
          > >
          > > Let ABC be a triangle, A'B'C' the orthic triangle and
          > > A"B"C" the Euler triangle (ie A",B",C" are the second
          > > intersections of NPC with AA',BB',CC', resp. = midpoints
          > > of AH,BH,CH, resp.)
          > >
          > > Denote:
          > >
          > > Ra = the radical axis of ((B', B'C"),(C',C'B"))
          > >
          > > Rb = the radical axis of ((C', C'A"),(A',A'C"))
          > >
          > > Rc = the radical axis of ((A', A'B"),(B',B'A"))
          > >
          > > The Ra,Rb,Rc are concurrent.
          > >
          > > Point?
          > >
          > > Antreas
          > >
          >
        • Antreas Hatzipolakis
          Let ABC be a triangle and P a point. Denote: Ba, Ca = the orthogonal projections of B, C on AP, resp. R1 = the radical axis of (ABaC), (ACaB) Similarly R2, R3
          Message 4 of 8 , Sep 4, 2016
            Let ABC be a triangle and P a point.

            Denote:

            Ba, Ca = the orthogonal projections of B, C on AP, resp.

            R1 = the radical axis of (ABaC), (ACaB)

            Similarly R2, R3

            Which is the locus of P such that R1, R2, R3 are concurrent?
            The entire plane?

            APH
          • Antreas Hatzipolakis
            [APH]: Let ABC be a triangle and P a point. Denote: Ba, Ca = the orthogonal projections of B, C on AP, resp. R1 = the radical axis of (ABaC), (ACaB) Similarly
            Message 5 of 8 , Sep 5, 2016

               

               

               [APH]:

              Let ABC be a triangle and P a point.

              Denote:

              Ba, Ca = the orthogonal projections of B, C on AP, resp.

              R1 = the radical axis of (ABaC), (ACaB)

              Similarly R2, R3

              Which is the locus of P such that R1, R2, R3 are concurrent?

              The entire plane?


              [César Lozada]:

              > Which is the locus of P such that R1, R2, R3 are concurrent? The entire plane?


              Yes. For P=u:v:w (trilinears) they concur at Z(P)=u^2*cos(A) : :

              ETC pairs(P,Z(P)):

              (1,1), (2,75), (3,255), (4,158), (5,1087), (6,31), (7,1088), (8,341), (9,200), (10,1089), (11,1090), (12,1091), (15,1094), (16,1095), (19,1096), (20,1097), (21,1098), (30,1099), (31,560), (32,1917), (37,756), (40,1103), (42,872), (44,678), (46,1079), (55,1253), (56,1106), (57,269), (58,849), (63,326), (65,1254), (73,7138), (75,561), (76,1928), (81,757), (84,1256), (86,873), (88,679), (90,7042), (100,765), (101,1110), (110,1101), (174,7), (188,8), (190,7035), (192,8026), (238,8300), (259,55), (266,56), (365,6), (366,2), (483,179), (507,174), (508,85), (509,57), (513,244), (514,1111), (518,4712), (519,4738), (523,1109), (556,3596), (649,3248), (650,2310), (651,7045), (652,2638), (656,2632), (661,2643), (758,4736), (798,4117), (1049,1085), (1077,1028), (1125,6533), (1488,7002), (2089,7022), (2238,4094), (2292,6042), (3082,400), (4146,6063), (4166,220), (4179,594), (4182,346), (4367,7207), (6724,12), (6725,6057), (6726,480), (6727,60), (6728,11), (6729,3271), (6730,4081), (6731,5423), (6733,59), (7025,188), (7039,7044), (7041,7036), (7370,7023), (7371,479), (7591,7066), (9326,2226)

               

              For Fermat points:

              Z(X(13))= cos(A)*csc(A+Pi/3)^2 : : (trilinears)

              = on cubics K420b, K638 and these lines: (5,8919), (13,15), (470,8838)

              = [ -0.003162528585600, -0.00276260060789, 3.644036680137045 ]

               

              Z(X(14))= cos(A)*csc(A-Pi/3)^2 : : (trilinears)

              = on cubics K420a, K638 and these lines: (5,8918), (14,16), (471,8836), (5619,6774)

              = [ 2.242952306040436, 4.57457502811627, -0.561557755730165 ]

               

              César Lozada




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