## Circumcenter of N1N2N3 [EMHL] Re: Radical axes, NPCs, Os

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• For P = O, Op = complement of X(157). In general, Op = complement of the nine-point-center of the antipedal triangle of P. Randy
Message 1 of 18 , Apr 15, 2013
For P = O, Op = complement of X(157).

In general, Op = complement of the nine-point-center of the antipedal triangle of P.

Randy

--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
> Let ABC be a triangle, P a point and N1,N2,N3 the NPC centers
> of PBC, PCA, PAB, resp.
>
> Which is the circumcenter Op of N1N2N3 ?
>
> P = H. Op = N [N1 = N2 = N3 = N (N1N2N3 is degenerated)]
>
> P = I. Op = N
>
> P = O. Op lies on the line NU, where U is the Poncelet point of O
> wrt ABC (ie the point where concur the NPCs of OBC,OCA,OAB and ABC)
> Coordinates???
>
> P = N. Op lies on the Euler line of ABC.
> Coordinates???
>
> If P describes a line (Euler line, etc) which is the locus of Op?
>
> If P describes the circumcircle, Op is the Poncelet point of P
> wrt ABC. If P describes other curves?
>
> Figures:
> http://anthrakitis.blogspot.gr/2013/04/circumcenter-of-n1n2n3.html
>
> APH
>
> [APH]
> > Let ABC be a triangle.
> >
> > Denote:
> >
> > N1,N2,N3 = The NPC centers of IBC, ICA, IAB, resp.
>
> >
> > 2. The Circumcenter of N1N2N3 is N, the NPC center of ABC.
>
• ... Euler line (where A1,B1,C1 are the NPC centers of PBC, PCA, PAB)? ... I think it is the same with Bernard s Q038:
Message 2 of 18 , Apr 16, 2013
> [APH]
>
> Which is the locus of P such that the circumcenter of A1B1C1 is on the
Euler line (where A1,B1,C1 are the NPC centers of PBC, PCA, PAB)?

[Francisco]:

> It is a circular quintic through I, H and N.
> The other three intersection points with Euler line are X30
> (the infinite point) and the instersections with the
> circumcircle X1113 and X1114.
> Equation:
>
> a^6 c^2 x^3 y^2 - 3 a^4 b^2 c^2 x^3 y^2 + 3 a^2 b^4 c^2 x^3 y^2 -
> b^6 c^2 x^3 y^2 - a^4 c^4 x^3 y^2 + a^2 b^2 c^4 x^3 y^2 -
> a^2 c^6 x^3 y^2 + c^8 x^3 y^2 + a^6 c^2 x^2 y^3 -
> 3 a^4 b^2 c^2 x^2 y^3 + 3 a^2 b^4 c^2 x^2 y^3 - b^6 c^2 x^2 y^3 -
> a^2 b^2 c^4 x^2 y^3 + b^4 c^4 x^2 y^3 + b^2 c^6 x^2 y^3 -
> c^8 x^2 y^3 + 3 a^2 b^4 c^2 x^3 y z - 3 b^6 c^2 x^3 y z -
> 3 a^2 b^2 c^4 x^3 y z + 3 b^2 c^6 x^3 y z + 2 a^6 c^2 x^2 y^2 z -
> 2 b^6 c^2 x^2 y^2 z - 4 a^4 c^4 x^2 y^2 z + 4 b^4 c^4 x^2 y^2 z +
> 2 a^2 c^6 x^2 y^2 z - 2 b^2 c^6 x^2 y^2 z + 3 a^6 c^2 x y^3 z -
> 3 a^4 b^2 c^2 x y^3 z + 3 a^2 b^2 c^4 x y^3 z - 3 a^2 c^6 x y^3 z -
> a^6 b^2 x^3 z^2 + a^4 b^4 x^3 z^2 + a^2 b^6 x^3 z^2 - b^8 x^3 z^2 +
> 3 a^4 b^2 c^2 x^3 z^2 - a^2 b^4 c^2 x^3 z^2 - 3 a^2 b^2 c^4 x^3 z^2 +
> b^2 c^6 x^3 z^2 - 2 a^6 b^2 x^2 y z^2 + 4 a^4 b^4 x^2 y z^2 -
> 2 a^2 b^6 x^2 y z^2 + 2 b^6 c^2 x^2 y z^2 - 4 b^4 c^4 x^2 y z^2 +
> 2 b^2 c^6 x^2 y z^2 + 2 a^6 b^2 x y^2 z^2 - 4 a^4 b^4 x y^2 z^2 +
> 2 a^2 b^6 x y^2 z^2 - 2 a^6 c^2 x y^2 z^2 + 4 a^4 c^4 x y^2 z^2 -
> 2 a^2 c^6 x y^2 z^2 + a^8 y^3 z^2 - a^6 b^2 y^3 z^2 -
> a^4 b^4 y^3 z^2 + a^2 b^6 y^3 z^2 + a^4 b^2 c^2 y^3 z^2 -
> 3 a^2 b^4 c^2 y^3 z^2 + 3 a^2 b^2 c^4 y^3 z^2 - a^2 c^6 y^3 z^2 -
> a^6 b^2 x^2 z^3 + b^8 x^2 z^3 + 3 a^4 b^2 c^2 x^2 z^3 +
> a^2 b^4 c^2 x^2 z^3 - b^6 c^2 x^2 z^3 - 3 a^2 b^2 c^4 x^2 z^3 -
> b^4 c^4 x^2 z^3 + b^2 c^6 x^2 z^3 - 3 a^6 b^2 x y z^3 +
> 3 a^2 b^6 x y z^3 + 3 a^4 b^2 c^2 x y z^3 - 3 a^2 b^4 c^2 x y z^3 -
> a^8 y^2 z^3 + a^2 b^6 y^2 z^3 + a^6 c^2 y^2 z^3 -
> a^4 b^2 c^2 y^2 z^3 - 3 a^2 b^4 c^2 y^2 z^3 + a^4 c^4 y^2 z^3 +
> 3 a^2 b^2 c^4 y^2 z^3 - a^2 c^6 y^2 z^3 = 0

I think it is the same with Bernard's Q038:
http://bernard.gibert.pagesperso-orange.fr/curves/q038.html

If yes, Bernard may add that property of the quintic.

APH

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