HYACINTHOS 25222

[Antreas P. Hatzipolakis]:

 

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

Denote:

Na, Nb, Nc = the NPC centers of IBC, ICA, IAB, resp.

Oa, Ob, Oc = the circumcentεrs of IBC, ICA, IAB, resp. = circumcenters of IaBC, IbCA, IcAB, resp.

 

The circumcircles of IOaNa, IObNb, IOcNc are coaxial.

Second point of intersection?

[Peter Moses]:

Hi Antreas,

 

a (2 a^9-a^8 b-6 a^7 b^2+2 a^6 b^3+6 a^5 b^4-2 a^3 b^6-2 a^2 b^7+b^9-a^8 c+2 a^7 b c-2 a^5 b^3 c+2 a^4 b^4 c-2 a^3 b^5 c+2 a b^7 c-b^8 c-6 a^7 c^2+6 a^5 b^2 c^2-3 a^4 b^3 c^2-2 a^3 b^4 c^2+5 a^2 b^5 c^2+2 a b^6 c^2-2 b^7 c^2+2 a^6 c^3-2 a^5 b c^3-3 a^4 b^2 c^3+6 a^3 b^3 c^3-3 a^2 b^4 c^3-2 a b^5 c^3+2 b^6 c^3+6 a^5 c^4+2 a^4 b c^4-2 a^3 b^2 c^4-3 a^2 b^3 c^4-4 a b^4 c^4-2 a^3 b c^5+5 a^2 b^2 c^5-2 a b^3 c^5-2 a^3 c^6+2 a b^2 c^6+2 b^3 c^6-2 a^2 c^7+2 a b c^7-2 b^2 c^7-b c^8+c^9)::
on lines {{1, 399}, {3, 9904}, {10, 10272}, {36, 11670}, {110, 517}, {125, 11230}, {265, 9955}, {542, 1386}, {942, 10091}, {1125, 10264}, {1385, 5663}, {1482, 2948}, {1511, 3579}, {1986, 11363}, {2646, 7727}, {3448, 5886}, {3576, 10620}, {3656, 9143}, {5126, 10081}, {5609, 10222}, {5972, 11231}, {9957, 10088}}.

Search -1.638464218267965509294330138 26.

Midpoint of X(i) and X(j) for these {i,j}: {{1, 399}, {1482, 2948}, {3656, 9143}}.

Reflection of X(i) in X(j) for these {i,j}: {{10, 10272}, {265, 9955}, {3579, 1511}, {10264, 1125}}.

X[3448] - 3 X[5886], 3 X[110] + X[7978], 3 X[3] - X[9904], 2 X[5609] + X[10222], 3 X[3576] - X[10620], 2 X[125] - 3 X[11230], 4 X[5972] - 3 X[11231].

 

 

Best regards,

Peter Moses.