Παρασκευή 25 Οκτωβρίου 2019

HYACINTHOS 26833

[Antreas P. Hatzipolakis]:
 
 
Let ABC be a triangle, P a point and A'B'C' the cevian triangle of Ο,

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.

R1, R2, R3 = the Euler lines of MaMabMac, MbMbaMbc, McMcaMcb, resp.
 
1. R1, R2, R3 are concurrent.
2. The parallels to R1, R2, R3 through Ma, Mb, Mc, resp. are concurrent.
3. The reflections of R1, R2, R3 in BC, CA, AB, resp. are concurrent (parallels)


[Peter Moses]:


Hi Antreas,

1). 2 a^12-7 a^10 b^2+11 a^8 b^4-12 a^6 b^6+10 a^4 b^8-5 a^2 b^10+b^12-7 a^10 c^2+12 a^8 b^2 c^2-5 a^6 b^4 c^2-6 a^4 b^6 c^2+10 a^2 b^8 c^2-4 b^10 c^2+11 a^8 c^4-5 a^6 b^2 c^4+4 a^4 b^4 c^4-5 a^2 b^6 c^4+7 b^8 c^4-12 a^6 c^6-6 a^4 b^2 c^6-5 a^2 b^4 c^6-8 b^6 c^6+10 a^4 c^8+10 a^2 b^2 c^8+7 b^4 c^8-5 a^2 c^10-4 b^2 c^10+c^12:: 
on lines {{2,14652},{111,930},{125,128},{137,6676},{6036,7499},{7568,13467}}.
complement X(14769).
midpoint of X(14652) and X(14769).
3 X[2] + X[14652].
{X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (2, 14652, 14769).
nine point circle of the medial triangle inverse of X(125).

2). a^14 b^2-5 a^12 b^4+11 a^10 b^6-15 a^8 b^8+15 a^6 b^10-11 a^4 b^12+5 a^2 b^14-b^16+a^14 c^2-4 a^12 b^2 c^2+6 a^10 b^4 c^2-2 a^8 b^6 c^2-11 a^6 b^8 c^2+24 a^4 b^10 c^2-20 a^2 b^12 c^2+6 b^14 c^2-5 a^12 c^4+6 a^10 b^2 c^4+2 a^6 b^6 c^4-17 a^4 b^8 c^4+30 a^2 b^10 c^4-16 b^12 c^4+11 a^10 c^6-2 a^8 b^2 c^6+2 a^6 b^4 c^6+8 a^4 b^6 c^6-15 a^2 b^8 c^6+26 b^10 c^6-15 a^8 c^8-11 a^6 b^2 c^8-17 a^4 b^4 c^8-15 a^2 b^6 c^8-30 b^8 c^8+15 a^6 c^10+24 a^4 b^2 c^10+30 a^2 b^4 c^10+26 b^6 c^10-11 a^4 c^12-20 a^2 b^2 c^12-16 b^4 c^12+5 a^2 c^14+6 b^2 c^14-c^16:: 
on lines {{4,14652},{5,49},{115,128},{131,137},{157,381},{2072,12095},{10276,14788},{11258,13512}}.
midpoint of X(4) and X(14652).
reflection of X(14769) in X(5).
nine point circle inverse of X(265).

3). X(1154).

Best regards,
Peter Moses.

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