Yes, they are concyclic.

The center X of the circle has trilinears:

2*cos(A)+4*sin(3*A/2)*cos(B/2-C/2)+ cos(B-C)+2 : :

ETC search: 2.387773069046934.., 2.38593313995937.., 0.886815506990847..

X=Midpoint of X(I),X(J) for these (I,J):

(1,500)

X lies on line X(I),X(J) for these (I,J):

(1,30), (3,81), (5,581), (21,323), (30,79), (58,5428), (79,500),

(140,3216),

(186,2906), (237,1896), (285,1082), (358,1882), (386,549), (500,554),

(511,1385),

(550,991), (554,1081), (1036,4022), (1081,1464), (1154,2646), (1464,1717),

(1675,4280), (1717,1836), (1836,3058), (2071,4309), (2072,3746),

(2771,3743),

(2943,4341), (3058,3649), (3108,3784), (3649,3655), (3655,3656),

(3656,3782),

(3704,3990), (3782,4654), (4654,4854), (4854,5160), (5160,5434),

(5434,5441)

Regards

César Lozada

_____

De:

Hyacinthos@yahoogroups.com [mailto:

Hyacinthos@yahoogroups.com] En nombre

de Antreas Hatzipolakis

Enviado el: Miércoles, 17 de Abril de 2013 05:34 a.m.

Para: Hyacinthos

Asunto: Re: [EMHL] NPC. locus.

A CIRCLE:

Let ABC be a triangle, A'B'C' the cevian triangle of I and N1, N2, N3

the NPC centers of IB'C', IC'A', IA'B', resp.

The points I, N1,N2,N3 are concyclic.

Center of the circle?

LOCUS:

Let ABC be a triangle, P a point, A'B'C' the cevian triangle of P and N1,

N2, N3

the NPC centers of PB'C', PC'A', PA'B', resp.

Which is the locus of P such that the points P, N1,N2,N3 are concyclic ?

Antreas

On Wed, Apr 17, 2013 at 1:43 AM, Antreas <

anopolis72@...
<mailto:anopolis72%40gmail.com> > wrote:

> **

>

>

> Let ABC be a triangle and P a point.

>

> Which is the locus of P such that the NPC center of PBC

> lies on the line AP ?

>

> Ceva triangle variation:

>

> Let ABC be a triangle, P a point and A'B'C'

> the cevian triangle of P.

> Which is the locus of P such that the NPC center

> of PB'C' lies on the line APA' ?

> (The I is on the locus)

>

> APH

>

>

>

>

>

http://anopolis72000.blogspot.com/
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