- Dear Randy and Angel

And in this case I think that A'B'C', N1N2N3 are perspective:

Denote:

HaHbHc = the orthic triangle.

(N1), (N2), (N3) = the reflections of (N) in HaO, HbO, HcO, resp.

A'B'C' = the triangle bounded by the radical axes of ((O),(N1)),

((O),(N2)), ((O),(N3)),

GENERALIZATION:

Denote:

PaPbPc = the pedal triangle of a point P on the Euler line

(N1), (N2), (N3) = the reflections of (N) in PaO, PbO, PcO, resp.

A'B'C' = the triangle bounded by the radical axes of

((O),(N1)), ((O),(N2)),((O),(N3)).

A'B'C', N1N2N3 are perspective.

True???

APH

On Wed, Feb 13, 2013 at 12:13 AM, Angel amontes1949@...> wrote:

> Dear Randy,

>

>

> A'B'C' is also perspective to N1N2N3 at non-ETC 3.427184657834269

> (coordinates?)

>

> Its coordinates are very complicated; the first coordinate of its center

> is:

>

> a^20(b^2+c^2)-

> 2a^18(3b^4+7b^2c^2+3c^4)+

> a^16(b^2+c^2)(13b^4+37b^2c^2+13c^4)-

> 2a^14(4b^8+39b^6c^2+68b^4c^4+39b^2c^6+4c^8)-

> a^12(b^2+c^2)(14b^8-71b^6c^2-106b^4c^4-71b^2c^6+14c^8)+

> 2a^10(14b^12-3b^10c^2-64b^8c^4-74b^6c^6- 64b^4c^8-3b^2c^10+14c^12)-

> a^8(b^2+c^2)(14b^12+27b^10c^2-110b^8c^4+ 66b^6c^6-

> 110b^4c^8+27b^2c^10+14c^12)-

> 2a^6(b^2-c^2)^2(4b^12-27b^10c^2- 8b^8c^4-2b^6c^6-8b^4c^8-27b^2c^10+4c^12)+

> a^4(b^2-c^2)^4(b^2+c^2)(13b^8-23b^6c^2+4b^4c^4- 23b^2c^6+13c^8) -

> 2a^2(b^2-c^2)^6(b^2+c^2)^2(3b^4-2b^2c^2+3c^4)+

> (b^2-c^2)^8(b^2+c^2)^3

>

> Best regards.

> Angel M.

>

>

> --- In Hyacinthos@yahoogroups.com, "rhutson2" wrote:

> >

> > Dear Antreas and Angel,

> >

> > A'B'C' is also perspective to N1N2N3 at non-ETC 3.427184657834269

> (coordinates?)

> >

> > N1N2N3 is inversely similar to ABC, with similitude center X(3).

> > N1N2N3 is perspective to ABC at X(3519).

> > X(3) of N1N2N3 = X(3)

> > X(186) of N1N2N3 = X(1511)

> > X(2070) of N1N2N3 = X(110)

> > The Euler line of N1N2N3 is line X(3)X(74).

> > Line IO of N1N2N3 is line X(3)X(191).

> > The Brocard axis of N1N2N3 is line X(3)X(67).

> >

> > Randy

> >

> > --- In Hyacinthos@yahoogroups.com, "Angel" wrote:

> > >

> > > Dear Antreas,

> > >

> > > The triangles ABC, A'B'C' are perspective.

> > >

> > > Perspector: X(1177) = 1st SARAGOSSA POINT OF X(67) (M.Iliev, 5/25/07)

> > >

> > > Best regards.

> > > Angel M.

> > >

> > > --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis wrote:

> > > >

> > > > 3. Let OaObOc be the medial triangle [pedal tr. of O], (N1),(N2),(N3)

> > > > the reflections of the NPC (N) in the perp. bisectors

> > > > OOa,OOb,OOc, resp. and and A'B'C' the triangle bounded by the radical

> > > > axes of ((O),(N1)), ((O),(N2)), ((O),(N3)), resp.

> > > > The triangles ABC, A'B'C' are perspective.

> > > >

> > > > Perspector?

> > > >

> > > > APH - Are the triangles ABC, A'B'C' perspective?Feuerbach point Fc.The NPC of CHIc intersects the excircle (Ic) at C' other than theFeuerbach point Fb,The NPC of BHIb intersects the excircle (Ib) at B' other than theFeuerbach point Fa.Let ABC be a triangle with excenters Ia,Ib,Ic.

The NPC of AHIa interscts the excircle (Ia) at A' other than theIn any case, has the triangle A'B'C' any interesting properties?APH