HYACINTHOS 25236

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle.

The circle (B, BC) intersects the circumcircle again at Ab

The circle (C, CB) intersects the circumcircle again at Ac

The perpendicular from N to AAb intersects AB at A2

The perpendicular from N to AAc intersects AC at A3

B,C,A2,A3 are concyclic (*). Let (Oa) be the circle (B,C,A2,A3) and simiarly (Ob), (Oc).

1. Which is the radical center of (Oa), (Ob), (Oc)?

2. Are the triangles ABC, OaObOc perspective ?

3. ABC, OaObOc are orthologic.

The orthologic center (OaObOc, ABC) is the O.

Which is the orthologic center *ABC, OaObOc) ?

 

(*) Stathis Koutras' Problem 422 in "Romantics of Geometry"

[Peter Moses]:

Hi Antreas,

 

Ab = {a^2 (-a^2+b^2+c^2), (c^2-a^2) (-a^2+b^2+c^2), c^2 (a^2-c^2)}.

power (Oa) = {((-a^2+b^2+c^2)^2 - b ^2 c^2) / (2 (-a^2+b^2+c^2)), 0, 0}.

 

1) X(265).

 

2) No.

However, perspective to medial, tangential and many others at O.

Perspective to Euler triangle at a^8 b^2-a^6 b^4-3 a^4 b^6+5 a^2 b^8-2 b^10+a^8 c^2-a^6 b^2 c^2+3 a^4 b^4 c^2-9 a^2 b^6 c^2+6 b^8 c^2-a^6 c^4+3 a^4 b^2 c^4+8 a^2 b^4 c^4-4 b^6 c^4-3 a^4 c^6-9 a^2 b^2 c^6-4 b^4 c^6+5 a^2 c^8+6 b^2 c^8-2 c^10::
on lines {{2,9927},{3,10113},{4,11270}, {5,5890},{54,1656},{74,5895},{ 125,6241},...}.

Midpoint of X(4) and X(11270).

 

3) X(265).

 

Best regards,

Peter Moses.