You can also take the point O (instead of K), draw the circle

(O,OA), intersecting Lb,Lc at B,C, and ask for

the locus of K as O moves on L = OK.

The Genelal problem is:

Let ABC be a triangle with the sidelines AB = Lb, AC = Lc,

two fixed lines.

Let P be a point lying on a fixed line L.

As B,C move on Lb, Lc, and P move on L so that P = (x:y:z) with

respect ABC, which is the locus of a point Q = (x':y':z') with

respect ABC?

So we can assume that P = O,H,G,N,I, ..., L = PX (with X

a fixed triangle center other than P : O,H,G,N,I, ... )

and ask for the locus of Q = O,H,G,N,I, ...

(but other than P)

APH

--- In Hyacinthos@yahoogroups.com, "Giovanni Artico" <zeroprof@...> wrote:

>

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

> >

> >

> > What do you mean by "same procedure"?

> >

>

> I take a variable point on the line L as K, construct the triangle and

> the corresponding O, and trace the locus of O as K moves on the line L.

>

> b.t.w. it seems that also IG, IH, IGe generate a conic: can you check

> this?

>

> Best regards

>

> Giovanni

>