HYACINTHOS 26705

 [Antreas P. Hatzipolakis]:

 
Let ABC be a triangle.

Denote:

A', B', C' = the reflections of N in BC, CA, AB, resp.

 

Ab, Ac = the orthogonal projections of A' on AB, AC, resp.

Abb, Acc = the orthogonal projections of Ab, Ac on NB', NC', resp.

Bc, Ba = the orthogonal projections of B' on BC, BA, resp.

Bcc, Baa = the orthogonal projections of Bc, Ba on NC', NA', resp.

 

Ca, Cb = the orthogonal projections of C' on CA, CB, resp.

Caa, Cbb = the orthogonal projections of Ca, Cb on NA', NB', resp.

We have: 

Baa = Caa = : A*

Cbb = Abb = : B*

Acc = Bcc = : C*

ABC, A*B*C* are orthologic.
The orthologic center (A*B*C*, ABC) is the N.
The other one (ABC, A*B*C*) ?

Other properties of A*B*C*?

 

 

 

[César Lozada]:

 

 

> The other one (ABC, A*B*C*) ?

Q(A->A*) = isogonal conjugate of X(1199)

= (SB^2-4*R^2*SB-3*S^2)*(SC^2-4* R^2*SC-3*S^2) : : (barys)

= on lines: {5, 97}, {233, 14627}, {394, 1656}, {1073, 5070}, {3926, 14786}

= isogonal conjugate of X(1199)

= [ 0.131017493206534, 0.09948246456820, 3.511322394187991 ]

 

A*B*C* is persective at X(5) to these triangles: 3rd Euler, 4th Euler, submedial.

A*B*C* is parallelogic to 1st and 2nd Parry triangles. Parallelogic center A* to these triangles is X(5) and reciprocal centers are of few interest.

 

César Lozada