Browse Groups

• More specifically: Let ABC be a triangle, P a point and k a real number. Call A , B , C ,P the points on lines AP, BP, CP such that
Message 1 of 3 , Mar 28
View Source
More specifically:

Let ABC be a triangle, P a point and k a real number.

Call A', B', C',P' the points on lines AP, BP, CP such that AA':A'P=BB':B'P=CC':C'P=OO':O'P=k:1.

Then the radical center R of circles (A, AA'), (B, BB'), (C, CC') is the inverse of P with respect the circle (O, OO').

--- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...> wrote:
>
> For a general P instead of H, your locus is the line OP, thus for P=H we have the Euler line.
>
> Francisco Javier.
>
> --- In Hyacinthos@yahoogroups.com, Antreas Hatzipolakis <anopolis72@> wrote:
> >
> > Let ABC be a triangle and A*, B*, C* three points on AH,BH,CH, resp,
> > such that:
> >
> > AA* / AH = BB* / BH = CC* / CH = t
> >
> > Which is the locus of the radical center of the circles
> > (A, AA*), (B, BB*), (C,CC*) ?
> >
> > APH
> >
> > On Thu, Mar 28, 2013 at 2:04 AM, Antreas <anopolis72@> wrote:
> >
> > > **
> > >
> > >
> > > Let ABC be a triangle and A',B',C' the midpoints
> > > of AH,BH,CH, resp. (A'B'C' = Euler triangle)
> > >
> > > The radical center of the circles (A,AA'), (B,BB'), (C,CC')
> > > is the midpoint of NO
> > >
> > > aph
> > >
> > >
> > >
> > >
> >
> >
> > [Non-text portions of this message have been removed]
> >
>
Your message has been successfully submitted and would be delivered to recipients shortly.
• Changes have not been saved
Press OK to abandon changes or Cancel to continue editing
• Your browser is not supported
Kindly note that Groups does not support 7.0 or earlier versions of Internet Explorer. We recommend upgrading to the latest Internet Explorer, Google Chrome, or Firefox. If you are using IE 9 or later, make sure you turn off Compatibility View.