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25601Re: Fermat point lies on NPC

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  • Antreas Hatzipolakis
    Mar 10, 2017

      [Tran Quang Hung]:

       

      Let ABC be a triangle

       

      A',B',C' are midpoints of BC,CA,AB.

       

      Fa,Fb,Fc are the first Fermat points of the triangles AB'C',BC'A',CA'B'.

       

      Then the second Fermat point of triangle FaFbFc lies on NPC of triangle ABC.


       

      [César Lozada]:


      If F1=X(13) and F2=X(14) (Fermat points) then the point Q is Q=X(115) = midpoint(F1,F2)

      Inverting F1 and F2 in the construction, we also get that it is X(115).

       

      For F1=X(15) and F2=X(16) (isodynamic points), Q=X(187)= midpoint(F1,F2)

       

      For F1=X(17) and F2=X(18) (Napoleon points), Q= midpoint(F1,F2) =

       

      Q = midpoint of X(17) and X(18)

      = 4*a^4-6*(b^2+c^2)*a^2+5*(b^2- c^2)^2 : : (barycentrics)

      = On lines: {2,7765}, {4,5206}, {6,17}, {32,5056}, {115,140}, {187,3850}, {532,8260}, {533,8259}, {547,5007}, {550,3054}, {574,3533}, {629,6674}, {630,6673}, {1504,10195}, {1505,10194}, {3090,7753}, {3523,7756}, {3525,11648}, {3628,9698}, {3851,7747}, {5059,8588}, {5067,7772}, {5070,5309}, {5461,7824}, {6292,6722}

      = midpoint of X(17) and X(18)

      = reflection of X(i) in X(j) for these (i,j): (629,6674), (630,6673)

      = [ -7.447009723854339, 7.40521126726905, 1.951061169423193 ]

       

      For F1=X(590) and F2=X(615) (isodynamic points), Q=X(3054)= midpoint(F1,F2)

      For F1=X(3071) and F2=X(3072) (isodynamic points), Q=X(5254)= midpoint(F1,F2)

       

       

      Generalization???

      Let A’B’C’ be the medial triangle of ABC.

      A line through X(6) cuts Evans conic at F1 and F2.

       

      Denote Fa1=F1-of-A’BC and define Fb1, Fc1 cyclically; let F12 = F2-of-Fa1Fb1Fc1.

      Denote Fa2=F2-of-A’BC and define Fb2, Fc2 cyclically; let F21 = F1-of-Fa2Fb2Fc2.

       

      Conjecture:

      F12 = F21 = midpoint(F1,F2)

       

      César Lozada