HYACINTHOS 24906

 [Antreas P. Hatzipolakis]:

Let ABC be a triangle and IaIbIc the antipedal triangle of I.

A'B'C' = the pedal triangle of O of ABC  [= NPC center of IaIbIc] wrt triangle IaIbIc.

N* = the NPC center of A'B'C'.

N* is lying on the Euler line of ABC.

[Peter Moses]:

2 a^7-2 a^6 b-5 a^5 b^2+5 a^4 b^3+4 a^3 b^4-4 a^2 b^5-a b^6+b^7-2 a^6 c-6 a^5 b c+a^4 b^2 c+5 a^3 b^3 c+2 a^2 b^4 c+a b^5 c-b^6 c-5 a^5 c^2+a^4 b c^2+4 a^3 b^2 c^2+2 a^2 b^3 c^2+a b^4 c^2-3 b^5 c^2+5 a^4 c^3+5 a^3 b c^3+2 a^2 b^2 c^3-2 a b^3 c^3+3 b^4 c^3+4 a^3 c^4+2 a^2 b c^4+a b^2 c^4+3 b^3 c^4-4 a^2 c^5+a b c^5-3 b^2 c^5-a c^6-b c^6+c^7::

is on lines {{2,3},{79,5432},{484,3649},{2 771,6684},{3035,3647},{5433,54 41},{6690,6701}}.

Midpoint of X(i) and X(j) for these {i,j}: {{3, 5499}, {5, 3651}, {6175, 8703}}.

Reflection of X(i) in X(j) for these {i,j}: {{5428, 3530}, {6841, 3628}, {10021, 140}}.

X[21] - 3 X[549], 3 X[3] + X[2475], X[2475] - 3 X[5499], 3 X[140] - 2 X[6675], 4 X[6675] - 3 X[10021].

{X(i),X(j)}-harmonic conjugate of X(k) for these (i,j,k): (3,6951,550),(5,550,6895).

 

Best regards,

Peter Moses.