HYACINTHOS 26403

[Le Viet An]:

Let ABC be a triangle and HaHbHc, DaDbDc the cevian, circumcevian triangles of H, resp.
Let P be a point lying on the circumcircle.

Denote:

Ba, Ca = the orthogonal projections of B, C on PDa, resp.
La = the Euler line of HaBaCa. Similarly Lb, Lc.

Then La,Lb,Lc are concurrent.

Which is the point of concurrence for P = X(74), X(110),... ?


[Peter Moses]:

Hi Antreas,

X(74):
4 a^16-16 a^14 b^2+21 a^12 b^4-4 a^10 b^6-15 a^8 b^8+16 a^6 b^10-9 a^4 b^12+4 a^2 b^14-b^16-16 a^14 c^2+46 a^12 b^2 c^2-44 a^10 b^4 c^2+17 a^8 b^6 c^2-11 a^6 b^8 c^2+17 a^4 b^10 c^2-13 a^2 b^12 c^2+4 b^14 c^2+21 a^12 c^4-44 a^10 b^2 c^4+32 a^8 b^4 c^4-5 a^6 b^6 c^4-15 a^4 b^8 c^4+15 a^2 b^10 c^4-4 b^12 c^4-4 a^10 c^6+17 a^8 b^2 c^6-5 a^6 b^4 c^6+14 a^4 b^6 c^6-6 a^2 b^8 c^6-4 b^10 c^6-15 a^8 c^8-11 a^6 b^2 c^8-15 a^4 b^4 c^8-6 a^2 b^6 c^8+10 b^8 c^8+16 a^6 c^10+17 a^4 b^2 c^10+15 a^2 b^4 c^10-4 b^6 c^10-9 a^4 c^12-13 a^2 b^2 c^12-4 b^4 c^12+4 a^2 c^14+4 b^2 c^14-c^16:: 

on lines {{54,125},{110,10112},{113,137},{539,5642},{2777,12254},{2888,5972},{2914,12112},{10212,10610},{10619,10628},{12121,12316}}.
Midpoint of X(12121) and X(12316).
Reflection of X(i) in X(j) for these {i,j}: {{113, 11702}, {125, 54}, {2888, 5972}}.
3 X[5642] - 4 X[11597], 4 X[2914] - X[13202].
{X(3043),X(10114)}-harmonic conjugate of X(125).

X(110): X(125).

Best regards,
Peter Moses.