Τετάρτη 29 Μαΐου 2013

A NPC CENTER ON THE EULER LINE

Let ABC be a triangle and N1,N2,N3 the NPC centers of OBC, OCA, OAB, resp.

The NPC center of N1N2N3 lies on the Euler line of ABC.

Antreas P. Hatzipolakis, 30 May 2013

***************************************************** The barycentric coordinates of the NPC center of N1N2N3 are:

( 2a^10 - 5a^8(b^2+c^2) + 2a^6(b^4+5b^2c^2+c^4) + a^4(4b^6-5b^4c^2-5b^2c^4+4c^6) - a^2(b^2-c^2)^2(4b^4+5b^2c^2+4c^4) + (b^2-c^2)^4(b^2+c^2) : ... : ...)

with (6-9-13)-search number: 4.7800096839999025703058

Angel Montesdeoca, Anopolis #331

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ETC1

X(73160) = X(2)X(6120)∩X(15859)X(66554) Barycentrics    Sin[B/3]*Sin[(2*B)/3]*Sin[B] + (Sin[A]*Sin[B] + Sin[C/3]*Sin[(2*C)/3])*Sin[C] : : ...