HYACINTHOS 25252

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and A'B'C', A"B"C" the cevian triangles of H, G, resp.

Denote:

Ma, Mb, Mc = the midpoints of AA', BB', CC', resp.

M1, M2, M3 = the midpoints of AA", BB", CC", resp.

Na, Nb, Nc = the NPC centers of GMaM1, GMbM2, GMcM3, resp.

N1, N2, N3 = the NPC centers of HMaM1, HMbM2, HMcM3, resp.

[...]

Let Mi, Mii, Miii be the midpoints of NaN1, NbN2, NcN3, resp.

3. ABC, MiMiiMiii are orthologic

The orthologic center (MiMiiMiii, ABC) lies on the Euler line.

[Peter Moses]:

Hi Antreas,

 

>3. ABC, MiMiiMiii are orthologic

Orthologic center (ABC, MiMiiMiii):

a^2 (6 a^4-12 a^2 b^2+6 b^4+13 a^2 c^2+13 b^2 c^2-17 c^4) (6 a^4+13 a^2 b^2-17 b^4-12 a^2 c^2+13 b^2 c^2+6 c^4):: on lines {}.

Search -14.161795299591058133.

Its isogonal conjugate: 17 a^4-13 a^2 b^2-6 b^4-13 a^2 c^2+12 b^2 c^2-6 c^4::,
is on lines {{6,20},{187,3534},{376,5585}, {382,8589},{550,5210},{574,165 7},{1384,5346},{3054,3522},{ 3055,3146},{3763,7833},{5024,7 753}}.

 

>The orthologic center (MiMiiMiii, ABC) lies on the Euler line.
X(3860)

 

Best regards,

Peter Moses.