## 25601Re: Fermat point lies on NPC

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• 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.

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)