HYACINTHOS 28432

[Antreas P. Hatzipolakis]:

 

 

Let ABC be a triangle, P a point and A'B'C' the cevian triangle of P.

Denote:

(Oab) = the circle near to B touching the cevianline AA', the sideline BC and the circumcircle externally

(Oac) = the circle near to C touching the cevianline AA', the sideline BC and the circumcircle externally  

Similarly (Obc), (Oba) and (Oca), (Ocb).

A1B1C1 = the triangle bounded by OabOac, ObcOba, OcaOcb.
A2B2C2 = the triangle bounded  by ObaOca, OcbOab, OacObc  
A3B3C3 = the triangle bounded by  ObcOcb, OcaOac, ObaOab

Which is the locus of P such that 

1.1. ABC, A1B1C1 are perspective?
1.2. ABC, A1B1C1 are orthologic?

 

2.1. ABC, A2B2C2 are perspective?
2.2. ABC, A2B2C2 are orthologic?

3.1. ABC, A3B3C3 are perspective?
3.2. ABC, A3B3C3 are orthologic?

For P = Gergonne point X(7) we have (Oab) = (Oac), (Obc) = (Oba), (Oca) = (Ocb) 

and ABC, A1B1C1 are homothetic (*)

(*) From the Russian journal "Mathematical Education" Series 3, 1(1997)
Republished in the Greek forum mathematica

PS There are more problems in that configuration:

For which P's the six circumcenters lie on a conic, and for which ones the conic is a circle?

Denote:
Ra, Rb, Rc = the radical axes of ((Oab), (Oac)), ((Obc), (Oba)), ((Oca), (Ocb)), resp

Sa, Sb, Sc = the radical axes of ((Oba), (Oca)), ((Ocb), (Oab)), ((Oac), (Obc)), resp 

Ta, Tb, Tc = the radical axes of ((Obc), (Ocb)), ((Oca), (Oac)), ((Oba), (Oab)), resp   

Which is the locus of P such that
-- Ra, Rb, Rc
-- Sa, Sb, Sc
-- Ta, Tb, Tc

are concurrent?

 

 

[César Lozada]: 

 

Very hard to solve for any P.

 

For P=X(7):

Radius(Oab)=Radius(Oac)=Radius(A-excircle), and similarly for the other circles.

 

Centers Oab, Oac have trilinears:

Oab = -2*a*b*c : ((b+c)*(a+b-c)+K(a,b,c))*c : ((b+c)*(a-b+c)- K(a,b,c))*b

Oac = -2*a*b*c : ((b+c)*(a+b-c)- K(a,b,c))*c : ((b+c)*(a-b+c)+ K(a,b,c)*b

 

where K(a,b,c) = sqrt(a*(-a+b+c)*(a*(a+b+c)-2*(b-c)^2))

 

The bisector perpendicular of (Oab, OAc) = the bisector perpendicular of BC

OabOac = | K(a,b,c)/(-a+b+c) |

 

Let Ab, Ac be the touchpoints of BC with (Oab) and (Oac), resp., and denote Bc, Ba, Ca, Cb cyclically. These six point lie on a conic with center:

O* = X(9)X(497) ∩ X(200)X(220)

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

= on lines: {2, 24181}, {9, 497}, {37, 3677}, {40, 15487}, {57, 16593}, {190, 1088}, {200, 220}, {440, 17742}, {518, 15490}, {644, 3870}, {1040, 2324}, {2325, 15286}, {3161, 10580}, {4370, 8557}, {4853, 15853}, {16572, 21096}, {17755, 20173}

= complement of the isogonal conjugate of X(21002)

= barycentric product X(i)*X(j) for these {i,j}: {8, 3174}, {220, 20946}, {341, 21002}, {346, 16572} , {728, 8732}

= trilinear product X(i)*X(j) for these {i,j}: {9, 3174}, {200, 16572}, {346, 21002}, {480, 8732}

= (medial triangle)-isotomic conjugate of-X(200)

= [ 45.9898914253600600, -21.1640125352149300, -2.9334305746483950 ]

 

The homothetic center ABC, A1B1C1 is X(57)

 

Ra, Rb, Rc concur at X(3)

Sa,Sb,Sc do not concur

Ta,Tb,Tc do not concur

 

César Lozada