HYACINTHOS 25048

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle, A'B'C' the pedal triagle of I and IaIbIc the
antipedal triangle of I (excentral triangle)

Denote:

A", B", C" = the reflections of I in BC, CA, AB, resp.

Aa, Ab, Ac = the orthogonal projections of A" on AI, BI, CI, resp.
Ba, Bb, Bc = the orthogonal projections of B" on AI, BI, CI, resp.
Ca, Cb, Cc = the orthogonal projections of C" on AI, BI, CI, resp.

La, Lb, Lc = the Euler lines of AaAbAc, BaBbBc, CaCbCc, resp.

1. La,Lb, Lc are concurrent.

2. The parallels to La, Lb, Lc through A, B, C, resp. are concurrent.

3. The parallels to La, Lb, Lc through A', B', C', resp. are concurrent.

4. The parallels to La, Lb, Lc through A", B", C", resp. are concurrent.

5. The parallels to La, Lb, Lc through Ia, Ib, Ic, resp. are concurrent.

[Peter Moses]:

Hi Antreas,

1) a (a^2-b^2+b c-c^2) (a^3 b-a^2 b^2-a b^3+b^4+a^3 c+a b^2 c-a^2 c^2+a b c^2-2 b^2 c^2-a c^3+c^4)::
on lines {{1,104},{3,10093},{4,10052},{ 7,80},{8,10940},
{11,113},{12,5885},...}.
midpoint {{100,3868},{5903,7972},{9803, 9964}}.
reflection of X(i) in X(j) for these: {{1,5083},{11,942},{72,3035},{ 6326,9946}}.

2) X(104).

3) Same as 1).

4) a (a^5 b-a^4 b^2-2 a^3 b^3+2 a^2 b^4+a b^5-b^6+a^5 c-a^4 b c+2 a^3 b^2 c-3 a b^4 c+b^5 c-a^4 c^2+2 a^3 b c^2-3 a^2 b^2 c^2+2 a b^3 c^2+b^4 c^2-2 a^3
c^3+2 a b^2 c^3-2 b^3 c^3+2 a^2 c^4-3 a b c^4+b^2 c^4+a c^5+b c^5-c^6)::
on lines {{1,104},{11,5902},...}.
midpoint {3901,5541}.
reflection of X(i) in X(j) for these: {{80,65},{104,5884},{1320, 3874},{3869,214},{5693,119},{ 5697,1317},{5904,1145}}.

5) X(6326).

Best regards,
Peter Moses.