HYACINTHOS 25939

[Antreas P. Hatzipolakis]:

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

Denote:

Ba, Ca = the orthogonal projections of B', C' on BA, CA, resp.

Cb, Ab = the orthogonal projections of C', A' on CB, AB, resp.

Ac, Bc = the orthogonal projections of A', B' on AC, BC, resp.

 

 

A" = AA' Intersection BaCa
B" = BB' Intersection CbAb
C" = CC' Intersection AcBc


A* = BC Intersection B"C"
B* = CA Intersection C"A"
C* = AB Intersection A"B"

A*, B*, C* are coillinear.

A** = B'C' Intersection B"C"
B** = C'A' Intersection C"A"
C** = A'B' Intersection A"B"

A**, B**, C** are collinear.

A*** = B'C' Intersection BC
B*** = C'A' Intersection CA
C*** = A'B' Intersection AB

A***, B***, C*** are collinear (orthic axis)

The lines A*B*C*, A**B**C**, A***B***C*** are concurrent.
ie the lines A*B*C*, A**B**C** intersect on the orthic axis.

Point of intersection ?

 

[Peter Moses]:

Hi Antreas,

 

(b^2-c^2) (-a^2+b^2-c^2) (a^2+b^2-c^2) (-a^6+3 a^4 b^2-3 a^2 b^4+b^6+3 a^4 c^2+6 a^2 b^2 c^2-b^4 c^2-3 a^2 c^4-b^2 c^4+c^6):: 

on line {230,231}.

{X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (2501,6753,6587).

barycentric product X(i)X(j) for these {i,j}: {{523, 3089}, {2501, 11433}}.

barycentric quotient X(i)/X(j) for these {i,j}: {{3089, 99}, {8573, 4558}, {11433, 4563}}.

 

Best regards,

Peter Moses.