HYACINTHOS 29057

[Antreas P. Hatzipolakis]:

Let ABC be a triangle and P a point.

Denote:

(Oa), (Ob), (Oc) = the circles with diameters AP, BP, CP, resp.

(O1) = the circle tangent the sides AB, AC of the angle A and internally the circle (Oa) at Ta

(O2) = the circle tangent the sides BC, BA of the angle B and internally the circle (Ob) at Tb  

(O3) = the circle tangent the sides CA, CB of the angle C and internally the circle (Oc) at Tc

Which is the locus of P such that:

1. The triangles ABC, TaTbTc are perspective?
2. The points P, Ta, Tb, Tc are concyclic?

H lies on the loci: Ercole Suppa, Romantics of Geometry 3128    

[César Lozada]:

 

Particular case: For P=H

 

Case 1: (Oa) is internally tangent to (O1)

Note: Barycentrics coordinates

Circle (O1):

·         Center = (-a^3+(b+c)*a^2+(b+c)^2*a-(b^2-c^2)*(b-c))/(2*(-a^2+b^2+c^2)) : b : c 

·         Squared-radius = (a+b-c)*(a-b+c)*(-a^2+b^2+c^2)^2/((a+b+c)^3*(-a+b+c))

 

(O1) touchpoints:

·         Ta = (-a^3-(b-c)^2*a-2*(b^2-c^2)*(b-c))/(a^2-b^2-c^2) : a-b+c : a-c+b 

·         Ab = 1/(-a^2+b^2+c^2) : 0 : 1/((a+b-c)*(a+c))  = (O1) ∩ AC

·         Ac = 1/(-a^2+b^2+c^2) : 1/((a-b+c)*(a+b)) : 0  = (O1) ∩ AB

 

Perspector ABC, TaTbTc = X(8)

 

Circle (Ta,Tb,Tc,H) = Fuhrmann circle:

·         Center =  X(355) 

·         Squared-radius = R*(R-2*r)

 

Ab, Ac and their cyclic points lie on the conic K₁ with center X(15476) and perspector P1. No ETC centers X(n) lie on K₁ for n<=32609.

 

Case 2: (Oa) is externally tangent to (O1)

Circle (O1):

·         Center =- (-a^3-(b+c)*a^2+(b+c)^2*a+(b^2-c^2)*(b-c))/(2*(-a^2+b^2+c^2)) : b : c 

·         Squared-radius = (a+b-c)*(a-b+c)*(-a^2+b^2+c^2)^2/((a+b+c)*(-a+b+c)^3)

 

(O1) touchpoints:  

·         Ta = (-a^3-(b-c)^2*a+2*(b^2-c^2)*(b-c))/(a^2-b^2-c^2) : a+b-c : a-b+c 

·         Ab = 1/(-a^2+b^2+c^2) : 0 : 1/((a-b+c)*(a-c))  = (O1) ∩ AC

·         Ac = 1/(-a^2+b^2+c^2) : 1/((a+b-c)*(a-b)) : 0 = (O1) ∩ AB

 

Perspector ABC, TaTbTc = X(7)

 

Circle (Ta, Tb, Tc, H) through ETCs 4, 7, 18343:

·         Center =  X(5805) 

·         Squared-radius = R*(-(2*R-r)*S^2+2*(4*R+r)^2*R*r^2)/(2*(r*(4*R+r))^2)

 

Ab, Ac and their cyclic points are not co-conic.

 

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There are two other triads of circles tangent to one of the circles (Oa), (Ob), and (Oc) and the other two correspondent sides of ABC. But analogous perspectivities or concyclicities are not present..

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Related centers:

P0 = X(35)X(79) ∩ X(5010)X(20277)

= a^2*(2*a^7-2*(2*b^2+b*c+2*c^2)*a^5-2*(b^3+c^3)*a^4+2*(b^2+c^2)*(b^2+b*c+c^2)*a^3+(b+c)*(4*b^4+4*c^4-(2*b^2-b*c+2*c^2)*b*c)*a^2+(b^2-c^2)*(b-c)*(-2*b^4-2*c^4-(4*b^2+5*b*c+4*c^2)*b*c)) : : (barys)

= lies on these lines: {35, 79}, {5010, 20277}

= isogonal conjugate of P1.

= {X(35), X(8606)}-harmonic conjugate of X(226)

= [ -5.9633323273683780, -5.4351629239329240, 10.1557768111079600 ]

 

P1 = ISOGONAL CONJUGATE OF P0

= (2*a^7+2*b*a^6-(b^2+4*c^2)*a^5-(3*b^3+2*b*c^2+2*c^3)*a^4-(3*b^4-2*c^4-(b-2*c)*b*c^2)*a^3-(b^2-c^2)*(b^3+4*c^3)*a^2+2*(b^3-c^3)*(b^2-c^2)*b*a+2*(b^3-c^3)*(b^2-c^2)^2)*(2*a^7+2*c*a^6-(4*b^2+c^2)*a^5-(2*b^3+2*b^2*c+3*c^3)*a^4+(2*b^4-3*c^4-(2*b-c)*b^2*c)*a^3+(b^2-c^2)*(4*b^3+c^3)*a^2+2*(b^3-c^3)*(b^2-c^2)*c*a-2*(b^3-c^3)*(b^2-c^2)^2) : : (barys)

= lies on the line {500, 9642}

= isogonal conjugate of P0.

= [ 5.7429539125479540, 6.3010333269830380, -3.3721834733337020 ]

 

César Lozada