HYACINTHOS 25690

 

[Antreas P Hatzipolakis]:

Let ABC be a triangle and A'B'C' the cevian triangle of H.

Denote:

Ma, Mb, Mc = the midpoints of BC, CA, AB, resp.
Aa, Bb, Cc = the midpoints of AH, BH, CH, resp.

Na, Nb, Nc = the NPC centers of MaBbCc, MbCcAa, McAaBb, resp.

ΑΝ ( = Euler line of MaBbCc) intersects AaNb, AaNc at B1, C1, resp.
ΒΝ ( = Euler line of MbCcAa) intersects BbNc, BbNa at C2, A2, resp.

CN ( = Euler line of McAaBb) intersects CcNa, CcNb at A3, B3, resp.

L1, L2, L3 = the Euler lines of AaB1C1, BbC2A2, CcA3B3, resp.

P, Pa, Pb, Pc = same points on the Euler lines L1,L2,L3 of AaB1C1, BbC2A2, CcA3B3, resp.

 

1. L1, L2, L3 are concurrent (parallels). Point of intersection (on the line at infinity)?

2.The centroid of PaPbPc is a fixed point on the Euler line of ABC. Which point is it?

 

 

[Peter Moses]:

Hi Antreas,

 

1) (b^2-c^2) (-3 a^6+7 a^4 b^2-5 a^2 b^4+b^6+7 a^4 c^2+3 a^2 b^2 c^2-b^4 c^2-5 a^2 c^4-b^2 c^4+c^6)::
at infinity,
the isogonal of {{74,12307},{550,1141},{2383, 3520},{5966,6636}}.

2) X(5066).

 

Best regards,

Peter Moses.