HYACINTHOS 29369

[Antreas P. Hatzipolakis]:

 

GENERALIZATION  

Let ABC be a triangle, L a line passing through G, P a point and A'B'C', A"B"C" the pedal triangles of P, O, resp.

A*, B*, C* = the orthogonal projections of A', B', C' on L, resp.

A**, B**, C** = the orthogonal projections of A", B", C" on A'A*, B'B*, C'C*, resp.

The centroid of A**B**C** lies on the L line

If L = the Euler line, then which is the centroid in terms of P?


[Peter Moses]:

 

Hi Antreas,
 
b^2 c^2 (8 a^8-8 a^6 b^2-3 a^4 b^4+4 a^2 b^6-b^8-8 a^6 c^2+14 a^4 b^2 c^2-4 a^2 b^4 c^2+4 b^6 c^2-3 a^4 c^4-4 a^2 b^2 c^4-6 b^4 c^4+4 a^2 c^6+4 b^2 c^6-c^8) p+a^2 c^2 (2 a^8+a^6 b^2-5 a^2 b^6+2 b^8-5 a^6 c^2-a^4 b^2 c^2+11 a^2 b^4 c^2+b^6 c^2+3 a^4 c^4-7 a^2 b^2 c^4-9 b^4 c^4+a^2 c^6+7 b^2 c^6-c^8) q+a^2 b^2 (2 a^8-5 a^6 b^2+3 a^4 b^4+a^2 b^6-b^8+a^6 c^2-a^4 b^2 c^2-7 a^2 b^4 c^2+7 b^6 c^2+11 a^2 b^2 c^4-9 b^4 c^4-5 a^2 c^6+b^2 c^6+2 c^8) r : :

Example P = X(182) -> 

4*a^8 - 3*a^6*b^2 - 2*a^4*b^4 + b^8 - 3*a^6*c^2 + 4*a^4*b^2*c^2 + a^2*b^4*c^2 + 3*b^6*c^2 - 2*a^4*c^4 + a^2*b^2*c^4 - 8*b^4*c^4 + 3*b^2*c^6 + c^8 : : 

 

= lies on these lines: {2,3}, {476,7698}, {523,597}, {543,5972}, {1648,18907}, {2782,5642}, {3734,11053}, {4045,10160}, {7606,16324}, {7792,16092}, {7804,22104}, {10796,11657}, {11174,16316}, {14561,16279}, {15048,31945}

Best regards,
Peter Moses.