HYACINTHOS 24898

 [Antreas P. Hatzipolakis]:

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

Denote:

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

Oa, Ob, Oc = the circumcenters of A'MbMc, B'McMa, C'MaMb, resp.

O1, O2, O3 = the circumcenters of AObOc, BOcOa, COaOb, resp.

1. ABC, OaObOc are perspective (homothetic and share the same Euler line). Perspector?

2. ABC, O1O2O3 are orthologic

The orthologic center (O1O2O3, ABC) is the midpoint of HN. The other one?

3. OaObOc, O1O2O3 are orthologic.

The orthologic center (O1O2O3, OaObOc) is the midpoint of HN. The other one?

[Peter Moses]:

Hi Antreas,

 

>1. ABC, OaObOc are perspective (homothetic and share the same Euler line). Perspector?
X(3091).

 

>2. ABC, O1O2O3 are orthologic

>The

orthologic center (O1O2O3, ABC) is the midpoint of HN = X(546). The other one?
(a^2+b^2-c^2) (a^2-b^2+c^2) (a^4-4 a^2 b^2+3 b^4-2 a^2 c^2-4 b^2 c^2+c^4) (5 a^4-6 a^2 b^2+b^4-6 a^2 c^2-2 b^2 c^2+c^4) (a^4-2 a^2 b^2+b^4-4 a^2 c^2-4 b^2 c^2+3 c^4)::
on lines {{4,3527},{3089,4994},{3091, 8797}}.
{X(4), X(3527)}-harmonic conjugate of X(8796).

X(3523) X(8796).

 

>3. OaObOc, O1O2O3 are orthologic.

>The orthologic center (O1O2O3, OaObOc) is the midpoint of HN. The other one?
a^16-5 a^14 b^2+17 a^12 b^4-45 a^10 b^6+75 a^8 b^8-71 a^6 b^10+35 a^4 b^12-7 a^2 b^14-5 a^14 c^2+22 a^12 b^2 c^2-83 a^10 b^4 c^2+108 a^8 b^6 c^2+37 a^6 b^8 c^2-154 a^4 b^10 c^2+83 a^2 b^12 c^2-8 b^14 c^2+17 a^12 c^4-83 a^10 b^2 c^4+18 a^8 b^4 c^4+34 a^6 b^6 c^4+173 a^4 b^8 c^4-207 a^2 b^10 c^4+48 b^12 c^4-45 a^10 c^6+108 a^8 b^2 c^6+34 a^6 b^4 c^6-108 a^4 b^6 c^6+131 a^2 b^8 c^6-120 b^10 c^6+75 a^8 c^8+37 a^6 b^2 c^8+173 a^4 b^4 c^8+131 a^2 b^6 c^8+160 b^8 c^8-71 a^6 c^10-154 a^4 b^2 c^10-207 a^2 b^4 c^10-120 b^6 c^10+35 a^4 c^12+83 a^2 b^2 c^12+48 b^4 c^12-7 a^2 c^14-8 b^2 c^14::
on lines {{5,1498},{3091,8797}}.

 

Best regards,

Peter Moses.