HYACINTHOS 27230

[Tran Quang Hung]:

 

Let ABC be a triangle with the first Fermat point F1.

 

A1B1C1 is the circumcevian triangle of F1.

 

A'B'C' is the cevian triangle of F1 wrt ABC.

 

A1'B1'C1' is the cevian triangle of F1 wrt A1B1C1.

 

Let F2,F2',F21 and F21' be the second Fermat points of triangles ABC, A'B'C', A1B1C1 and A1'B1'C1'. resp. .

 

Then four points F2, F2', F21 and F21' are concyclic.

 

Which is this circle ?

 

And similarly, which is the first Fermat circle ?

[César Lozada]:

 

Quite complicated!

 

A numerical solution (graphically tested) suggests that the center of the circle could be:

 

O1 = reflection of X(5608) in X(11176)

= (2*S^2-sqrt(3)*(3*R^2-SB-SC)* S-3*(3*SA-2*SW)*R^2-3*SB*SC- SW^2)*(SB-SC): : (barys)

= on lines: {396, 11617}, {619, 1649}, {690, 6771}, {5608, 11176}

= reflection of X(5608) in X(11176)

= [ 12.0740131051576200, 0.8813578790584561, -2.5421277905903770 ]

 

Swapping the 1^st and 2^nd Fermat points in your construction:

 

O2 = reflection of X(5607) in X(11176)

= (2*S^2+sqrt(3)*(3*R^2-SB-SC)* S-3*(3*SA-2*SW)*R^2-3*SB*SC- SW^2)*(SB-SC) : : (barys)

= on lines: {395, 11618}, {618, 1649}, {690, 6774}, {5607, 11176}

= reflection of X(5607) in X(11176)

= [ 2.7822912540943070, 6.9776518578198250, -2.4741519984728700 ]

 

Have you noted that both circles degenerate to lines for some shapes of ABC, others than isosceles ABC?

 

The midpoint of O1 and O2 is:

= (SB-SC)*(2*S^2-3*(3*SA-2*SW)* R^2-3*SB*SC-SW^2)  (barys)

= 4*X(140)-X(11615), X(351)-3*X(5054), 5*X(631)-X(9147)

= on lines: {2, 2780}, {3, 9148}, {140, 11176}, {351, 5054}, {523, 7623}, {549, 804}, {620, 2793}, {631, 9147}, {690, 6036}, {3268, 16220}, {9188, 10168}

= midpoint of X(i) and X(j) for these {i,j}: {3, 9148}, {3268, 16220}

= reflection of X(9188) in X(10168)

= [ 7..4281521796259650, 3.9295048684391390, -2.5081398945316250 ]

 

César Lozada