HYACINTHOS 26878

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle and A'B'C', A"B"C" the cevian, antipedal triangle of I, resp .

Denote:

Ma, Mb, Mc = the midpoints of AA', BB', CC', resp. 

Mab, Mac = the orthogonal projections of Ma on BB', CC', resp.

Mbc, Mba = the orthogonal projections of Mb on CC', AA', resp.

Mca, Mcb = the orthogonal projections of Mc on AA', BB', resp.

La, Lb, Lc = the Euler lines of MaMabMac, MbMbcMba, McMcaMcb, resp.

L1, L2, L3 = the reflections of La, Lb, Lc in BC, CA, AB, resp.

 

A*B*C* = the triangle bounded by L1,L2, L3

 

 

A"B"C", A*B*C* are parallelogic.

 

Parallelogic centers?

 

 

 

[Peter Moses]:


Hi Antreas,

(A"B"C", A*B*C*) = X(484).

(A*B*C*, A"B"C") = 

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

on lines {{3,5718},{11,30},{46,500},{100,524},{141,11322},{404,5241},{442,4278},{511,1155},{851,3286},{1211,13588},{1503,5078},{2352,3782},{4188,5233},{5124,7465},{5204,9840},{5347,7411},{5453,5903},{6097,13408}}.

Best regards,
Peter Moses.