HYACINTHOS 28911

[Tran Quang Hung]:

Let ABC be a triangle.with NPC center N.

A'B'C' is the circumcevian triangle of N.

N1, N2, N3, N4, N5, N6 are the NPC centers of the triangles AC'B', B'AC, CB'A', A'CB, BA'C', C'BA, respectively.

Let H1, H2, H3 be the orthocenters of the triangles N6N1N2, N2N3N4, N4N5N6, respectively.

Then the NPC center of the triangle H1H2H3 lies on the Euler line of ABC.  

[Peter Moses]:

 

Hi Antreas,

X(10285) on lines {2,3}, {54,1263}, {1154,20327}, {11671,16766}, {12026,14051}, {14140,24573}, {19552,21230}, {24385,25044}

Best regards,
Peter Moses.

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