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22007TCS: P21

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  • Antreas Hatzipolakis
    Sep 25, 2013
    • 0 Attachment
      [Seiichi Kirikami]:

      Given a triangle ABC, a point P and its isoconjugate P*,

      let Oa, Ob, Oc be the intersections of the circumcircles of PBC, PCA, PAB and PA, PB, PC respectively other than P,

      let O1, O2, O3 be the intersections of the circumcircles of P*BC, P*CA, P*AB and P*A, P*B, P*C respectively other than P*,

      1. OaO2O3, ObO3O1, OcO1O2,

      2. O1ObOc, O2OcOa, O3OaOb,

      are concurrent.

      Reference: Anopolis #1046


      [Cesar Lozada]:


      Let

      Y(P) = OaO2O3 /\ ObO3O1 /\ OcO1O2

      Z(P) = O1ObOc /\ O2OcOa /\ O3OaOb

       

      then:

       

      For P=X(2)

      Y(P) = (5*a^2*(-b^2-c^2+a^2)-(b^2+c^2)^2+9*b^2*c^2)/a :: (trilinears)
             = (6*(SA^2+2*SB*SC)-(3*S^2+SW^2))/a : : (trilinears)

             = Anticomplementary conjugate of X(316)
             = Anticomplement of X(671)

             = Reflection of (2 / 99), (148 / 2), (671 / 2482)  
             = On lines (1,2796), (2,99), (20,542), (23,2936), (30,147), (114,3839), (194,1992), (376,2782), (384,597), (530,617), (531,616), (599,1975), (627,5463), (628,5464), (3314,5077)

             =  ( 4.248090114807508, -4.78657861274721, 4.993792699282826 )

      Z(P) = X(2930)

       

      For P=X(3)

      Y(P) = X(399)
      Z(P) = (S^6-SB*SC*(2*S^4+3*SA^2*(2*S^2-3*SB*SC)))/(a*SA) : : (trilinears)
             = Reflection of (4 / 107), (1294 / 3184)

             = on Neuberg cubic and lines (1,2816), (3,3462), (4,74), (19,2822), (20,1075), (112,376), (122,631), (146,4240), (399,2133), (1138,1157), (1148,4302), (1263,3481), (2790,3186), (3087,3269),  (3183,3529), (3324,4293), (3440,3479), (3441,3480)   

             = ( 6.167335386797941, 5.88646950050122, -3.281046120038592 )


      Reference: Anopolis #1049