Dear Angel

I think that the general problem is this:

Let P be a point and A'B'C' the pedal triangle of P.

Let Ab,Ac be the orthogonal projections of

A' on AB, AC, resp., Bc,Ba of B' on BC,BA, resp.

and Ca,Cb of C' on CA,CB, resp.

Which is the locus of P such that the circles (A'AbAc),(B'BcBa)

and (C'CaCb) are concurrent ?

Part of the locus is the circumcircle of ABC.

APH

On Thu, Dec 2, 2010 at 3:05 PM, Angel <amontes1949@...> wrote:

>

>

> Dear Antreas

>

> If L: px + qy + rz = 0, the circle (A'AbAc), whose equation

> a^2yz+b^2zx+c^2xy - (x+y+z) ((q*SB)/(q-r) y - (r*SC)/(q-r) z)=0, passing

> through A. The circles (A'AbAc), (B'BcBa) and (C'CaCb) are coaxial, radical

> axis d: p(q-r)SAx + q(r-p)SBy + r(p-q)zSC= 0, which passes through the

> orthocenter, for any line L

>

> Angel

>

> --- In Hyacinthos@yahoogroups.com <Hyacinthos%40yahoogroups.com>, Antreas

> Hatzipolakis <anopolis72@...> wrote:

>

> >

> > On Wed, Dec 1, 2010 at 12:47 AM, Antreas <anopolis72@...> wrote:

> >

> > >

> > >

> > > Let ABC be a trianle, L a line intersecting

> > > BC,CA,AB at A',B',C' resp.

> > > Let Ab,Ac be the orthogonal projections of

> > > A' on AB, AC, resp., Bc,Ba of B' on BC,BA, resp.

> > > and Ca,Cb of C' on CA,CB, resp.

> > > The circles (A'AbAc),(B'BcBa) and (C'CaCb)

> > > are concurrent.

> > >

> > > For which lines L the point of concurrence

> > > is on the circumcircle of ABC ?

> > >

> > > APH

> > >

> > >

> > >

> > Let P = (x:y:z) be a point. As L moves around P, which is the locus

> > of the point of concurrence? (Special case P = O)

> >

> > APH

> >

> >

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

> >

>

>

>

--

http://anopolis72000.blogspot.com/

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