I - CONCURRENT EULER LINES

Let ABC be a triangle and A'B'C' the cevian triangle of P = I (incenter).

Denote:

Oab, Oac = the circumcenters of ABA', ACA', resp.

Oa = the circumcenter of AOabOac. Similarly Ob, Oc

1. The Euler lines of ABC, AOabOac, BObcOba, COcaOcb are concurrent.

2. The Orthocenter of OaObOc is lying on the Euler Line of ABC.

Note:

For A'B'C' = cevian triangle of a point P, the circumcircles of AOabOac, BObcOba, COcaOcb are concurrent at O of ABC.

By Miquel theorem in the quadrilateral (AB,BC,CA,AA'), O is lying on the circumcircle of AOabOac. Similarly O is lying on the circumcircles of BObcOba, COcaOcb

Antreas P. Hatzipolakis, 3 October 2014

*** The Euler lines of ABC, AOabOac, BObcOba, COcaOcb are concurrent in X(186) = Inverse in circumcircle of orthocenter.

*** The Orthocenter of OaObOc is

X=(a^2 (a^2 - b^2 - b c - c^2) (a^5 (b + c) - 2 a^3 (b^3 + c^3) - a^2 b c (b^2 + c^2) + a (b^5 - b^4 c - b c^4 + c^5)+ b c (b^2 - c^2)^2) : ... :...)

lying on the Euler Line of ABC, and with (6-9-13)-search number 1.269044824987441168348286504.

X = (r^2 + 2 r R - R^2 + s^2)X(3) + R^2 X(4)

Angel Montesdeoca. Hyacinthos #22604

1) X(186)

2) a^2 (a^2-b^2-b c-c^2) (a^5 b-2 a^3 b^3+a b^5+a^5 c-a^2 b^3 c-a b^4 c+b^5 c-2 a^3 c^3-a^2 b c^3-2 b^3 c^3-a b c^4+a c^5+b c^5)::

on lines {{2,3},{35,500},{55,5453},{511,5495},{3724,5492}},

Search = 3.7222371386671506172

Agree with barys, however I think the search number is suspect. Reckon it should be 3.7222371386671506172.

Peter Moses

PARALLEL NN-LINES

10. Let ABC be a triangle and P a point.

Denote:

Ab, Ac = the orthogonal projections of A on PB,PC, resp.

Na1 = the NPC center of AAbAc

Na2 = the NPC center of Na1AbAc.

Similarly Nb1, Nb2 and Nc1, Nc2.

The lines Na1Na2, Nb1Nb2, Nc1Nc2 are parallel.

11. If P = I,

the lines Na1Na2, Nb1Nb2, Nc1Nc2 are parallel to Euler Line of ABC

21. Let ABC be a triangle.

Denote:

Na1 = the NPC center of IBC

Na2 = the NPC center of Na1BC.

Similarly Nb1,Nb2, Nc1,Nc2

The lines Na1Na2, Nb1Nb2, Nc1Nc2 are parallel to Euler Line of ABC

31. Let ABC be a triangle and IaIbIc the antipedal triangle of I (excentral triangle)

Denote:

Ab, Ac = the orthogonal projections of A on IaIc, IaIb, resp.

Na1 = the NPC center of AAbAc

Na2 = the NPC center of Na1AbAc

The lines Na1Na2, Nb1Nb2, Nc1Nc2 are parallel to Euler Line of ABC

41. Let ABC be a triangle and IaIbIc the antipedal triangle of I (excentral triangle)

Denote:

Na1 = the NPC center of IaBC

Oa = the circumcenter of IaBC

Nao1 = The NPC center of OaBC.

Similarly Nb1, Nbo1, Nc1, Nco1.

The lines Na1Nao1, Nb1Nbo1, Nc1Nco1 are parallel to OI line of ABC.

Antreas P. Hatzipolakis, 1 October 2014