[Antreas P. Hatzipolakis]:
Let ABC be a triangle.
Denote:
A1, A2 = orthogonal projections of N on AH, AG, resp.
B1, B2 = orthogonal projections of N on BH, BG, resp.
A1, A2 = orthogonal projections of N on AH, AG, resp.
B1, B2 = orthogonal projections of N on BH, BG, resp.
C1, C2 = orthogonal projections of N on CH, CG, resp.
M1, M2, M3 = the midpoints of A1A2, B1B2, C1C2, resp.
N, M1, M2, M3 are concyclic.
The center of the circle lies on the Euler line.
M1, M2, M3 = the midpoints of A1A2, B1B2, C1C2, resp.
N, M1, M2, M3 are concyclic.
The center of the circle lies on the Euler line.
[Peter Moses]:
Hi Antreas,
>The center of the circle lies on the Euler line.
2 a^8 b^2-4 a^6 b^4+4 a^2 b^8-2 b^10+2 a^8 c^2-2 a^6 b^2 c^2-3 a^4 b^4 c^2-3 a^2 b^6 c^2+6 b^8 c^2-4 a^6 c^4-3 a^4 b^2 c^4-2 a^2 b^4 c^4-4 b^6 c^4-3 a^2 b^2 c^6-4 b^4 c^6+4 a^2 c^8+6 b^2 c^8-2 c^10:;
2 a^8 b^2-4 a^6 b^4+4 a^2 b^8-2 b^10+2 a^8 c^2-2 a^6 b^2 c^2-3 a^4 b^4 c^2-3 a^2 b^6 c^2+6 b^8 c^2-4 a^6 c^4-3 a^4 b^2 c^4-2 a^2 b^4 c^4-4 b^6 c^4-3 a^2 b^2 c^6-4 b^4 c^6+4 a^2 c^8+6 b^2 c^8-2 c^10:;
on lines {{2,3},{1216,11808},{5943,1155 7},{6146,8254},{10610,11572},{ 11264,12242}}.
3 X[5] + X[427], X[378] + 7 X[3851], X[22] - 9 X[5055], 3 X[547] - X[6676], 15 X[5071] + X[7391], 5 X[1656] - X[7502], 7 X[3090] - X[7555].
3 (2 J^2 - 7) X[2] - (2 J^2 - 9) X[3], J = OH / R.
{X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (5,1594,140),(5,2072,547),(5,3 845,10254),(5,5133,5066),(5,55 76,546),(5,10224,3628),(140,54 6,3575),(546,547,10096),(5169, 10254,3845),(7574,7579,1594).
Best regards,
Peter Moses.
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