HYACINTHOS 27094

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and A'B'C' the pedal triangle of a point P.

Denote:

Na, Nb, Nc = the NPC centers of PBC, PCA, PAB, resp.

 

A1, Ab, Ac = the orthogonal projections of Na on PA', PB, PC, resp.
Ba, B2, Bc = the orthogonal projections of Nb on PA, PB', PC, resp.
Ca, Cb, C3 = the orthogonal projections of Nc on PA, PB, PC', resp.

Oa, Ob, Oc = the circumcenters of A1AbAc, BaB2Bc, CaCbC3, resp.

 

1. P = O

In this case A1, B2, C3 = Na, Nb, Nc, resp. (since Na, Nb, Nc lie on OA', OB', OC', resp.)
1.1. The NPC center of OaObOc lies on the Euler line of ABC.

1.2. ABC, OaObOc are orthologic.
The orthologic center (OaObOc, ABC) is the O.
The other one (ABC, OaObOc)?

2. P = N
2.1. The circumcenter of OaObOc lies on trhe Euler line of ABC.
2.2. ABC, OaObOc are orthologic.
The orthologic center (OaObOc, ABC)  lies on the Euler line of ABC
Orthologic centers?


[Peter Moses]:


Hi Antreas,

1.1) X(10212).
1.2) X(265).

2.1) 

2 a^16-9 a^14 b^2+9 a^12 b^4+21 a^10 b^6-65 a^8 b^8+73 a^6 b^10-41 a^4 b^12+11 a^2 b^14-b^16-9 a^14 c^2+10 a^12 b^2 c^2+29 a^10 b^4 c^2-30 a^8 b^6 c^2-63 a^6 b^8 c^2+118 a^4 b^10 c^2-69 a^2 b^12 c^2+14 b^14 c^2+9 a^12 c^4+29 a^10 b^2 c^4-32 a^8 b^4 c^4-19 a^6 b^6 c^4-64 a^4 b^8 c^4+141 a^2 b^10 c^4-64 b^12 c^4+21 a^10 c^6-30 a^8 b^2 c^6-19 a^6 b^4 c^6-26 a^4 b^6 c^6-83 a^2 b^8 c^6+146 b^10 c^6-65 a^8 c^8-63 a^6 b^2 c^8-64 a^4 b^4 c^8-83 a^2 b^6 c^8-190 b^8 c^8+73 a^6 c^10+118 a^4 b^2 c^10+141 a^2 b^4 c^10+146 b^6 c^10-41 a^4 c^12-69 a^2 b^2 c^12-64 b^4 c^12+11 a^2 c^14+14 b^2 c^14-c^16:: 

on lines {{2,3},{1263,7604},...}.
midpoint of X(5) and X(5501).
reflection of X(i) in X(j) for these {i,j}: {{12056,140},{12811,15335},{13469,3628}}.

2.2a) (ABC, OaObOc): X(1263).
2.2b) (OaObOc, ABC): X(3628).

Best regards,
Peter Moses.