HYACINTHOS 24611

[Antreas P. Hatzipolakis]:

 

Let ABC be a triangle.

Denote:

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

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

N1, N2, N3 = the reflections of Na, Nb, Nc in B'C', C'A', A'B', resp.

The centroid of N1N2N3 lies on the OI line of ABC.

 

[Peter Moses]:

Hi Antreas,

 

a (3 a^5 b-3 a^4 b^2-6 a^3 b^3+6 a^2 b^4+3 a b^5-3 b^6+3 a^5 c-6 a^4 b c+8 a^3 b^2 c-11 a b^4 c+6 b^5 c-3 a^4 c^2+8 a^3 b c^2-16 a^2 b^2 c^2+8 a b^3 c^2+3 b^4 c^2-6 a^3 c^3+8 a b^2 c^3-12 b^3 c^3+6 a^2 c^4-11 a b c^4+3 b^2 c^4+3 a c^5+6 b c^5-3 c^6)::
on lines {{1,3},{5,4004},{355,10107},{1 864,6797},{2800,3817},{3698,56 94},{3754,5887},{3922,9956},{ 4018,5690},{4323,6961},{4848,6 842},{5927,9952}}.

 

Midpoint of X(3576) and X(5903).

Reflection of X(i) in X(j) for these {i,j}: {{5887, 10175}, {10175, 3754}, {10247, 942}}.

 

(3 r + 2 R) X[1] - (3 r - R) X[3].

2 X[942]-X[10247],X[3057]-4 X[5885],X[3576]+X[5903],3 X[10202]-2 X[10246],2 X[5]-5 X[4004],X[355]-4 X[10107],X[3817]-3 X[3919],5 X[3698]-2 X[5694],4 X[3754]-X[5887],2 X[3754]-X[10175],X[5887]-2 X[10175],7 X[3922]-4 X[9956],X[4018]+2 X[5690].

 

Best regards,

Peter Moses.