## NPC tangent to (O)

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• For which P s the NPC of PBC is tangent to the circumcircle of ABC? APH
Message 1 of 8 , Mar 8, 2013
For which P's the NPC of PBC is tangent to the circumcircle of ABC?

APH
• It is a quartic trhough B and C intersecting line CA at two points Y1,Y2 and line AB on two points Z1,Z2. Equation: a^8 x^4 - 2 a^4 b^4 x^4 + b^8 x^4 - 2 a^4
Message 2 of 8 , Mar 8, 2013
It is a quartic trhough B and C intersecting line CA at two points Y1,Y2 and line AB on two points Z1,Z2.

Equation:

a^8 x^4 - 2 a^4 b^4 x^4 + b^8 x^4 - 2 a^4 c^4 x^4 - 2 b^4 c^4 x^4 +
c^8 x^4 + 4 a^8 x^3 y - 4 a^6 b^2 x^3 y - 4 a^4 b^4 x^3 y +
4 a^2 b^6 x^3 y - 4 a^4 c^4 x^3 y - 4 a^2 b^2 c^4 x^3 y +
4 a^8 x^2 y^2 - 8 a^6 b^2 x^2 y^2 + 4 a^4 b^4 x^2 y^2 +
4 a^8 x^3 z - 4 a^4 b^4 x^3 z - 4 a^6 c^2 x^3 z -
4 a^2 b^4 c^2 x^3 z - 4 a^4 c^4 x^3 z + 4 a^2 c^6 x^3 z +
12 a^8 x^2 y z - 8 a^6 b^2 x^2 y z - 4 a^4 b^4 x^2 y z -
8 a^6 c^2 x^2 y z - 4 a^4 c^4 x^2 y z + 8 a^8 x y^2 z -
8 a^6 b^2 x y^2 z + 4 a^8 x^2 z^2 - 8 a^6 c^2 x^2 z^2 +
4 a^4 c^4 x^2 z^2 + 8 a^8 x y z^2 - 8 a^6 c^2 x y z^2 + 4 a^8 y^2 z^2 =0.

--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...> wrote:
>
> For which P's the NPC of PBC is tangent to the circumcircle of ABC?
>
> APH
>
• So we have three quartics Q(a),Q(b),Q(c), respective to sidelines and would be interesting if they have one or more points in common. APH 2013/3/9 Francisco
Message 3 of 8 , Mar 9, 2013
So we have three quartics Q(a),Q(b),Q(c), respective to sidelines and would
be interesting if
they have one or more points in common.

APH

2013/3/9 Francisco Javier <garciacapitan@...>

> **
>
>
> It is a quartic trhough B and C intersecting line CA at two points Y1,Y2
> and line AB on two points Z1,Z2.
>
> Equation:
>
> a^8 x^4 - 2 a^4 b^4 x^4 + b^8 x^4 - 2 a^4 c^4 x^4 - 2 b^4 c^4 x^4 +
> c^8 x^4 + 4 a^8 x^3 y - 4 a^6 b^2 x^3 y - 4 a^4 b^4 x^3 y +
> 4 a^2 b^6 x^3 y - 4 a^4 c^4 x^3 y - 4 a^2 b^2 c^4 x^3 y +
> 4 a^8 x^2 y^2 - 8 a^6 b^2 x^2 y^2 + 4 a^4 b^4 x^2 y^2 +
> 4 a^8 x^3 z - 4 a^4 b^4 x^3 z - 4 a^6 c^2 x^3 z -
> 4 a^2 b^4 c^2 x^3 z - 4 a^4 c^4 x^3 z + 4 a^2 c^6 x^3 z +
> 12 a^8 x^2 y z - 8 a^6 b^2 x^2 y z - 4 a^4 b^4 x^2 y z -
> 8 a^6 c^2 x^2 y z - 4 a^4 c^4 x^2 y z + 8 a^8 x y^2 z -
> 8 a^6 b^2 x y^2 z + 4 a^8 x^2 z^2 - 8 a^6 c^2 x^2 z^2 +
> 4 a^4 c^4 x^2 z^2 + 8 a^8 x y z^2 - 8 a^6 c^2 x y z^2 + 4 a^8 y^2 z^2 =0.
>
>
> --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...>
> wrote:
> >
> > For which P's the NPC of PBC is tangent to the circumcircle of ABC?
> >
> > APH
> >
>

[Non-text portions of this message have been removed]
• Well, there is no common points for the three loci. The B-locus and C-locus intersect at the antipode of A on the circumcircle, and they have another common
Message 4 of 8 , Mar 9, 2013
Well, there is no common points for the three loci.

The B-locus and C-locus intersect at the antipode of A on the circumcircle, and they have another common point, interior to the triangle whose coordinates are more complicated.

Best regards,

Francisco Javier.

--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...> wrote:
>
> For which P's the NPC of PBC is tangent to the circumcircle of ABC?
>
> APH
>
• In the cases of loci respective to sides, with no common point , we can ask for possible perspectivities. So here, if qA is the common point of Q(b), Q(c)
Message 5 of 8 , Mar 9, 2013
In the cases of loci respective to sides, with no common point , we can ask
for
possible perspectivities.
So here, if qA is the common point of Q(b), Q(c) inside the triangle
and similarly qB and qC, we can ask whether ABC (or some other triangle of
the reference triangle)
and qAqBqC are perspective (or orthologic etc).

aph

On Sat, Mar 9, 2013 at 4:15 PM, Francisco Javier <garciacapitan@...>wrote:

> **
>
>
> Well, there is no common points for the three loci.
>
> The B-locus and C-locus intersect at the antipode of A on the
> circumcircle, and they have another common point, interior to the triangle
> whose coordinates are more complicated.
>
> Best regards,
>
> Francisco Javier.
>
>
> --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...>
> wrote:
> >
> > For which P's the NPC of PBC is tangent to the circumcircle of ABC?
> >
> > APH
> >
>
>
>

--
http://anopolis72000.blogspot.com/

[Non-text portions of this message have been removed]
• In the way I arrived to it, the calculation of the coordinates of these points of intersection requires solving a quintic. I don t know if it could be done in
Message 6 of 8 , Mar 9, 2013
In the way I arrived to it, the calculation of the coordinates of these
points of intersection requires solving a quintic. I don't know if it could
be done in another way.

2013/3/9 Antreas Hatzipolakis <anopolis72@...>

> In the cases of loci respective to sides, with no common point , we can ask
> for
> possible perspectivities.
> So here, if qA is the common point of Q(b), Q(c) inside the triangle
> and similarly qB and qC, we can ask whether ABC (or some other triangle of
> the reference triangle)
> and qAqBqC are perspective (or orthologic etc).
>
> aph
>
> On Sat, Mar 9, 2013 at 4:15 PM, Francisco Javier <garciacapitan@...
> >wrote:
>
> > **
> >
> >
> > Well, there is no common points for the three loci.
> >
> > The B-locus and C-locus intersect at the antipode of A on the
> > circumcircle, and they have another common point, interior to the
> triangle
> > whose coordinates are more complicated.
> >
> > Best regards,
> >
> > Francisco Javier.
> >
> >
> > --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...>
> > wrote:
> > >
> > > For which P's the NPC of PBC is tangent to the circumcircle of ABC?
> > >
> > > APH
> > >
> >
> >
> >
>
>
>
> --
> http://anopolis72000.blogspot.com/
>
>
> [Non-text portions of this message have been removed]
>
>
>
> ------------------------------------
>
>
>
>
>

--
---
Francisco Javier García Capitán
http://garciacapitan.99on.com

[Non-text portions of this message have been removed]
• ... Dear Antreas It is a quartic (cA) through B and C (cusps). Equation of (cA) (García Capitán,
Message 7 of 8 , Mar 11, 2013
--- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...> wrote:
>
> For which P's the NPC of PBC is tangent to the circumcircle of ABC?
>
> APH
>

Dear Antreas

It is a quartic (cA) through B and C (cusps).

Equation of (cA) (García Capitán, http://tech.groups.yahoo.com/group/Hyacinthos/message/21696):

a^8 x^4 - 2 a^4 b^4 x^4 + b^8 x^4 - 2 a^4 c^4 x^4 - 2 b^4 c^4 x^4 +
c^8 x^4 + 4 a^8 x^3 y - 4 a^6 b^2 x^3 y - 4 a^4 b^4 x^3 y +
4 a^2 b^6 x^3 y - 4 a^4 c^4 x^3 y - 4 a^2 b^2 c^4 x^3 y +
4 a^8 x^2 y^2 - 8 a^6 b^2 x^2 y^2 + 4 a^4 b^4 x^2 y^2 +
4 a^8 x^3 z - 4 a^4 b^4 x^3 z - 4 a^6 c^2 x^3 z -
4 a^2 b^4 c^2 x^3 z - 4 a^4 c^4 x^3 z + 4 a^2 c^6 x^3 z +
12 a^8 x^2 y z - 8 a^6 b^2 x^2 y z - 4 a^4 b^4 x^2 y z -
8 a^6 c^2 x^2 y z - 4 a^4 c^4 x^2 y z + 8 a^8 x y^2 z -
8 a^6 b^2 x y^2 z + 4 a^8 x^2 z^2 - 8 a^6 c^2 x^2 z^2 +
4 a^4 c^4 x^2 z^2 + 8 a^8 x y z^2 - 8 a^6 c^2 x y z^2 + 4 a^8 y^2 z^2 =0.

The tangent ((a^2 - b^2)x + a^2z=0) in the cusp B meet again (cA) in Ab, and the tangent ((a^2 - c^2)x + a^2y=0) in the cusp C meet again (cA) in Ac.

Define Ba, Bc, Ca, and Cb simillary, then the triangle bounded by (AbAc, BcBa, CaCb) is perspective with ABC; the perspector is:

(SA/((a^4 - 2a^2b^2 + b^4 + 2a^2b*c - 2b^3c -
2a^2c^2 + 4b^2c^2 - 2b*c^3 + c^4)*
(a^4 - 2a^2b^2 + b^4 - 2a^2b*c + 2b^3*c -
2a^2c^2 + 4b^2c^2 + 2b*c^3 + c^4)): ... : ....).

Or:

( SA/( 4a^6(b^2+c^2) - 6a^4(b^2+c^2)^2+ 4a^2(b^2+c^2)^3
-(a^8 + b^8+ c^8 +2b^2c^2(2b^2+c^2)(b^2+2c^2)) ): ... : ...).

With (6-9-13)-search number -0.688951377609844758609146059...

Best regards
Angel Montesdeoca
• Dear Angel, Francisco In general we have two problems with tangent circles: Let A,B be two fixed points and (Q) a given circle. Which is the locus of P such
Message 8 of 8 , Mar 11, 2013
Dear Angel, Francisco

In general we have two problems with tangent circles:

Let A,B be two fixed points and (Q) a given circle.

Which is the locus of P such that:

1. The circumcircle of PAB is tangent to (Q)
2. The NPC of PAB is tangent to (Q).

I guess that, in general (when there is a solution, since restrictions
apply for the circle),
both the loci are again quartics.

In triangle geometry:

Let ABC be a triangle, D a fixed point and (Q) the pedal or cevian circle
of D.

Which is the locus of P such that the (1) circumcircle or (2) the NPC of PBC
is tangent to (Q).

Antreas

On Mon, Mar 11, 2013 at 3:43 PM, Angel <amontes1949@...> wrote:

> **
>
>
>
>
> --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@...>
> wrote:
> >
> > For which P's the NPC of PBC is tangent to the circumcircle of ABC?
> >
> > APH
> >
>
> Dear Antreas
>
> It is a quartic (cA) through B and C (cusps).
>
> Equation of (cA) (García Capitán,
> http://tech.groups.yahoo.com/group/Hyacinthos/message/21696):
>
>
> a^8 x^4 - 2 a^4 b^4 x^4 + b^8 x^4 - 2 a^4 c^4 x^4 - 2 b^4 c^4 x^4 +
> c^8 x^4 + 4 a^8 x^3 y - 4 a^6 b^2 x^3 y - 4 a^4 b^4 x^3 y +
> 4 a^2 b^6 x^3 y - 4 a^4 c^4 x^3 y - 4 a^2 b^2 c^4 x^3 y +
> 4 a^8 x^2 y^2 - 8 a^6 b^2 x^2 y^2 + 4 a^4 b^4 x^2 y^2 +
> 4 a^8 x^3 z - 4 a^4 b^4 x^3 z - 4 a^6 c^2 x^3 z -
> 4 a^2 b^4 c^2 x^3 z - 4 a^4 c^4 x^3 z + 4 a^2 c^6 x^3 z +
> 12 a^8 x^2 y z - 8 a^6 b^2 x^2 y z - 4 a^4 b^4 x^2 y z -
> 8 a^6 c^2 x^2 y z - 4 a^4 c^4 x^2 y z + 8 a^8 x y^2 z -
> 8 a^6 b^2 x y^2 z + 4 a^8 x^2 z^2 - 8 a^6 c^2 x^2 z^2 +
> 4 a^4 c^4 x^2 z^2 + 8 a^8 x y z^2 - 8 a^6 c^2 x y z^2 + 4 a^8 y^2 z^2 =0.
>
> The tangent ((a^2 - b^2)x + a^2z=0) in the cusp B meet again (cA) in Ab,
> and the tangent ((a^2 - c^2)x + a^2y=0) in the cusp C meet again (cA) in
> Ac.
>
> Define Ba, Bc, Ca, and Cb simillary, then the triangle bounded by (AbAc,
> BcBa, CaCb) is perspective with ABC; the perspector is:
>
> (SA/((a^4 - 2a^2b^2 + b^4 + 2a^2b*c - 2b^3c -
> 2a^2c^2 + 4b^2c^2 - 2b*c^3 + c^4)*
> (a^4 - 2a^2b^2 + b^4 - 2a^2b*c + 2b^3*c -
> 2a^2c^2 + 4b^2c^2 + 2b*c^3 + c^4)): ... : ....).
>
> Or:
>
> ( SA/( 4a^6(b^2+c^2) - 6a^4(b^2+c^2)^2+ 4a^2(b^2+c^2)^3
> -(a^8 + b^8+ c^8 +2b^2c^2(2b^2+c^2)(b^2+2c^2)) ): ... : ...).
>
> With (6-9-13)-search number -0.688951377609844758609146059...
>
> Best regards
> Angel Montesdeoca
>
> /
>

[Non-text portions of this message have been removed]
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