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