HYACINTHOS 26484

 

[Antreas P. Hatzipolakis]:

Let ABC be a triangle and A'B'C' the cevian triangle of Η.
Let P, Pa, Pb, Pc be same points of the triangles ABC, AB'C', BC'A', CA'B', resp.

Denote:

A" = (Reflection of BB' in PaB') /\ (Reflection of CC' in PaC')

B" = (Reflection of CC' in PbC') /\ (Reflection of AA' in PbA')

C" = (Reflection of AA' in PcA') /\ (Reflection of BB' in PcB')

If Pa, Pb, Pc =  Orthocenters or Circumcenters of AB'C', BC'A', CA'B', resp., then the triangles A'B'C', A"B"C" are circumcyclologic.

Cyclologic centers?

 

[César Lozada]:

 

Case P=O

Q(A’->A”) = X(125)

Q(A”->A’) = X(403)

 

Case P=H

Q(A’->A”) = complement of X(12092)

= (3*SA^2-10*R^2*SA+3*S^2-4*SB* SC)*(SA^2+2*S^2+2*SB*SC-SW^2)* (SA+SW-5*R^2) : : (barycentrics)

= On the nine-points-circle and these lines: {2, 12092}, {113, 5876}, {296, 955}, {824, 12575}, {1131, 12913}, {2412, 6026}, {3819, 11764}, {4473, 4499}, {5031, 12838}, {5551, 11434}, {6126, 10395}, {6505, 6739}, {7200, 8338}, {7212, 10787}, {9329, 10080}, {9686, 12957}, {11375, 13925}

= complement of X(12092)

= [ 1.890888620412988, 3.35717029191522, 0.443751839621693 ]

 

Q(A”->A’) = Polar circle-inverse-of-X(5964)

= ((1+2*cos(2*A))^2-4*cos(A)^2)* ((2*cos(A)+cos(3*A))*cos(B-C)- (cos(2*A)+1)*cos(2*(B-C))+cos( A)*cos(3*(B-C))-cos(2*A)-1/2)* tan(A) : : (barycentrics)

= On lines: {4, 5964}, {65, 6881}, {425, 12560}, {1356, 13723}, {2122, 13562}, {2426, 8376}, {3559, 7114}, {3857, 11215}, {4180, 7869}, {6102, 6240}, {6651, 9657}, {7129, 8106}, {7720, 8675}, {8070, 11530}

= polar circle-inverse-of-X(5964)

= [ -2.777586987501765, 0.48016733386582, 4.590204168078090 ]

 

César Lozada