HYACINTHOS 27117

[Antreas P. Hatzipolakis]:

Let ABC be a triangle an P a point.

Denote:

(Oa), (Ob), (Oc) = the circles with diameters BC, CA, AB, resp.

(O1), (O2), (O3) = the reflections of (Oa), (Ob), (Oc) in AP, BP, CP, resp.

Which is the locus of P such that the radical center of (O1), (O2), (O3) lies on the OP line ?.

 

H, I lie on the locus.

[César Lozada ]:

 

Excentral circum-septic q7 through ETC’s 1, 2, 3, 4, 1113, 1114.

 

Q7 = ∑ [ y*z*(-(b^2-c^2)*(-a^2+b^2+c^2) ^2*b^2*c^2*x^5+(-(a^6-(b^2+2* c^2)*a^4-(b^2-6*c^2)*b^2*a^2+( b^2-c^2)*(b^4-3*b^2*c^2-c^4))* b^2*c^2*y+(a^6-(2*b^2+c^2)*a^ 4+(6*b^2-c^2)*c^2*a^2+(b^2-c^ 2)*(b^4+3*b^2*c^2-c^4))*b^2*c^ 2*z)*x^4-8*a^2*b^2*c^2*(a^2-b^ 2-c^2)*(b^2-c^2)*z*y*x^3-8*(b^ 2-c^2)*a^4*b^2*c^2*x*y^2*z^2+ a^4*c^2*(a^2-b^2+c^2)*(a^2-c^ 2)*z*y^4-(2*a^4+(4*b^2-3*c^2)* a^2-(b^2-c^2)*(2*b^2+c^2))*a^ 4*c^2*z^2*y^3+(2*a^4-(3*b^2-4* c^2)*a^2+(b^2-c^2)*(b^2+2*c^2) )*a^4*b^2*z^3*y^2-(a^2-b^2)*a^ 4*b^2*(a^2+b^2-c^2)*z^4*y) ] = 0  (barys)

 

ETC pairs (P,Q=radical center) : (1,1482), (2,4), (4,4)

 

Limit point for P->X(3):

Q(X(3)) = X(3)X(12278) ∩ X(4)X(13289)

= (SB+SC)*((5*R^2-2*SW)*SA^2-2*( 35*R^2*(2*R^2-SW)+4*SW^2)*SA+( 7*R^2-2*SW)*S^2) : : (barys)

= on lines: {3, 12278}, {4, 13289}, {74, 11250}, {184, 3357}, {186, 11704}, {378, 5895}, {1147, 12281}, {1614, 15062}, {2071, 7689}, {3448, 12118}, {4550, 10539}, {5622, 8537}, {11457, 12254}, {14059, 14385}

= [ 9.8914185916117340, 8.1150763448580180, -6.5427354145073550 ]

 

César Lozada