HYACINTHOS 26330

[Antreas P. Hatzipolakis]:

General Theorem (*)

Let ABC be a triangle, L a line and P a point on L.

 

Denote:

A1, B1, C1 = the other than P intersections of the circumcircles of PBC, PCA, PAB and L, resp. 

 

The tangents to circumcircles of PBC, PCA, PAB at A1, B1, C1, resp. bound a triangle A'B'C'.

The circumcircles of ABC and A'B'C' are tangent. 

(*)  AoPS

Question: 
If L = the Euler line and P = O, H, N, ...., then which are the touchpoints of the circumcircles of ABC and A'B'C' ?

 

[Peter Moses]:

Hi Antreas,

 

P = O

X(1304).

 

P = H

X(10420).

 

P = N

a^2 (a-b) (a+b) (a-c) (a+c) (a^2+b^2-c^2) (a^2-b^2+c^2) (a^10-4 a^8 b^2+4 a^6 b^4+2 a^4 b^6-5 a^2 b^8+2 b^10-3 a^8 c^2+9 a^6 b^2 c^2-3 a^4 b^4 c^2+2 a^2 b^6 c^2-5 b^8 c^2+2 a^6 c^4-10 a^4 b^2 c^4-3 a^2 b^4 c^4+2 b^6 c^4+2 a^4 c^6+9 a^2 b^2 c^6+4 b^4 c^6-3 a^2 c^8-4 b^2 c^8+c^10) (a^10-3 a^8 b^2+2 a^6 b^4+2 a^4 b^6-3 a^2 b^8+b^10-4 a^8 c^2+9 a^6 b^2 c^2-10 a^4 b^4 c^2+9 a^2 b^6 c^2-4 b^8 c^2+4 a^6 c^4-3 a^4 b^2 c^4-3 a^2 b^4 c^4+4 b^6 c^4+2 a^4 c^6+2 a^2 b^2 c^6+2 b^4 c^6-5 a^2 c^8-5 b^2 c^8+2 c^10):: 

on line {{1141,13619},...}.

isoconjugate of X(656) and X(10096).

barycentric quotient X(112)/X(10096).

 

P = G

X(10098).

 

Best regards,

Peter Moses.