HYACINTHOS 25279

[Antreas P. Hatzipolakis]:

 

Let ABC  be a triangle.

Denote:

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

 

Ab, Ac = the orthogonal projections of A" on NB', NC', resp.
Bc, Ba = the orthogonal projections of B" on NC', NA', resp.

Ca, Cb = the orthogonal projections of C" on NA', NB', resp.

 

La, Lb, Lc = the Euler lines of A"AbAc, B"BcBa, C"CaCb, resp.

A*B*C* = the triangle bounded by La, Lb, Lc

ABC, A*B*C* are parallelogic.

The parallelogic center (ABC, A*B*C*) is de Longchamps point X20 (reflection of H in O)

The other one?

[César Lozada]:

 

Z(A*->A) = complement of X(389)

 

[....]

GENERALIZATION:

Let ABC  be a triangle.

Denote:

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

 

Ab, Ac = the orthogonal projections of A" on NB', NC', resp.
Bc, Ba = the orthogonal projections of B" on NC', NA', resp.

Ca, Cb = the orthogonal projections of C" on NA', NB', resp.

A'b, A'c = points on A"Ab, A"Ac, resp.  such that: A'bAb/A'bA" = A'cAc/A'cA" = t

B'c, B'a = points on B"Bc, B"Ba, resp. such that: B'cBc/B'cB" = B'aBa/B'aB" = t

C'a, C'b = points on C"Ca, C"Cb, resp. such that: C'aCa/C'aC" = C'bCb/C'bC" = t

La, Lb, Lc = the Euler lines of A"A'bA'c, B"B'cB'a, C"C'aC'b, resp.

A*B*C* = the triangle bounded by La, Lb, Lc

ABC, A*B*C* are parallelogic.

The parallelogic center (ABC, A*B*C*) is de Longchamps point X20 (reflection of H in O)

Locus of the other one (A*B*C*, ABC) as t varies?

 

[César Lozada]:

 

 

> GENERALIZATION:

> The parallelogic center (ABC, A*B*C*) is de Longchamps point X20 (reflection of H in O)

Confirmed.

 

Locus of the other one (A*B*C*, ABC) as t varies?

Line {5,141} through ETC’s { 5, 141, 211, 511, 623, 624, 625, 626, 635, 636, 639, 640, 1216, 2039, 2040, 3454, 3613, 3934, 5031, 5103, 5403, 5404, 5446, 5480, 6101, 7683, 7684, 7685, 7849, 8262, 9822, 9823, 9824, 9969, 10110, 10170, 10263, 11675} = trilinear polar of:

 

Q = isogonal conjugate of X(3050)

= 1/((b^2-c^2)*(a^4-(b^2+c^2)*a^ 2-b^2*c^2)) : : (barycentrics)

= On lines: {110,9514}, {194,11002}, {2421,4576}, {3448,9513}, {3613,5169}

= isogonal conjugate of X(3050)

= Trilinear pole of the line {5,141}

= [ -1.126522332473350, 0.65912950741095, 3.704277437918342 ]

 

César Lozada