HYACINTHOS 25601

[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