[Antreas P. Hatzipolakis]:
Let ABC be a triangle.
Denote:
Ab, Ac = the orthogonal projections of A on BI, CI, resp.
Bc, Ba = the orthogonal projections of B on CI, AI, resp.
Ca, Cb = the orthogonal projections of C on AI, BI, resp.
La, Lb, Lc = the Euler lines of AAbAc, BBcBa, CCaCb, resp.
A->A* = X(80)
Denote:
Ab, Ac = the orthogonal projections of A on BI, CI, resp.
Bc, Ba = the orthogonal projections of B on CI, AI, resp.
Ca, Cb = the orthogonal projections of C on AI, BI, resp.
La, Lb, Lc = the Euler lines of AAbAc, BBcBa, CCaCb, resp.
La1, Lb1, Lc1 = the reflections of La, Lb, Lc in BC, CA, AB, resp.
La2, Lb2, Lc2 = the parallels to La1, Lb1, Lc1 through A, B, C, resp.
La3, Lb3, Lc3 = the reflections of La2, Lb2, Lc2 in BC, CA, AB, resp.
La4, Lb4, Lc4 = the parallels to La3, Lb3, Lc3 through A, B, C, resp.
La2, Lb2, Lc2 = the parallels to La1, Lb1, Lc1 through A, B, C, resp.
La3, Lb3, Lc3 = the reflections of La2, Lb2, Lc2 in BC, CA, AB, resp.
La4, Lb4, Lc4 = the parallels to La3, Lb3, Lc3 through A, B, C, resp.
La4, Lb4, Lc4 are concurrent.
or if A*B*C* = the triangle bounded by La3, Lb3, Lc3, then ABC, A*B*C* are parallelogic
[César Lozada]:
Parallelogic centers:
A->A* = X(80)
A*->A = reflection of X(80) in X(15906)
= a*((b+c)*a^5-b*c*a^4-(b+c)*( b^2+c^2)*a^3+(b^4+b^2*c^2+c^4) *a^2-(b^2-c^2)*(b-c)*(b^3+c^3) ) :: (barys)
= on lines: {1, 1283}, {65, 1421}, {80, 15906}, {399, 16014}, {517, 3689}, {2262, 5540}, {2773, 4707}, {2835, 11570}, {2841, 11571}, {2842, 3868}, {3216, 4674}
= reflection of X(80) in X(15906)
= [ -2.7817312918942880, -9.8618600661913350, 11.7519820470680500 ]
César Lozada
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