## Inconic question

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• Suppose a triangle inconic touches AB in X, BC in Y and CA in Z. Counting parameters implies there is exact one equation in the values of AX,BX,BY,CY,CZ,AZ.
Message 1 of 4 , Jul 17 2:24 AM
Suppose a triangle inconic touches AB in X, BC in Y and CA in Z.
Counting parameters implies there is exact one equation
in the values of AX,BX,BY,CY,CZ,AZ. Which is?

Hauke
• ... Ceva, since Brianchon. OK, the question was lame :-) OK, let s up it a bit. Can any point inside ABC be the Brianchon point of an inconic? (I suspect yes.)
Message 2 of 4 , Jul 25 4:48 AM
--- In Hyacinthos@yahoogroups.com, "shokoshu2" <fc3a501@...> wrote:
>
> Suppose a triangle inconic touches AB in X, BC in Y and CA in Z.
> Counting parameters implies there is exact one equation
> in the values of AX,BX,BY,CY,CZ,AZ. Which is?
>
Ceva, since Brianchon. OK, the question was lame :-)
OK, let's up it a bit. Can any point inside ABC be the
Brianchon point of an inconic? (I suspect yes.)
Which is the locus of all inparabola Brianchon points?
(This surely is listed on Mathworld...somewhere.)

Hauke
• ... So we can take X=(0,y ,z ), Y=(x ,0,z ), Z=(x ,y ,0), abd the conic is (x/x )^2 + (y/y )^2 + (z/z )^2 - 2yz/y z - 2zx/z x - 2xy/x y = 0. ... This conic
Message 3 of 4 , Jul 27 10:49 AM
"shokoshu2" wrote:

> > Suppose a triangle inconic touches AB in X, BC in Y and CA in Z.
> > Counting parameters implies there is exact one equation
> > in the values of AX,BX,BY,CY,CZ,AZ. Which is?
>
> Ceva, since Brianchon. OK, the question was lame :-)

So we can take X=(0,y',z'), Y=(x',0,z'), Z=(x',y',0), abd the conic is

(x/x')^2 + (y/y')^2 + (z/z')^2 - 2yz/y'z' - 2zx/z'x' - 2xy/x'y' = 0.

> OK, let's up it a bit. Can any point inside ABC be the
> Brianchon point of an inconic? (I suspect yes.)
> Which is the locus of all inparabola Brianchon points?
> (This surely is listed on Mathworld...somewhere.)
>
> Hauke

This conic is a parabola iff it is tangent to the line at infinity
iff y'z'+z'x'+x'y'=0, so (x',y',z') is on the Steiner circumellipse.
--
Barry Wolk
• The locus of all inparabola Brianchon points is the Steiner circumellipse. For any inparabola, the join of the focus and the Brianchon point is on a very well
Message 4 of 4 , Jul 28 6:52 AM
The locus of all inparabola Brianchon points is the Steiner circumellipse.
For any inparabola, the join of the focus and the Brianchon point is on a
very well known point in ETC.
Friendly Francois

On Tue, Jul 27, 2010 at 7:49 PM, Barry Wolk <wolkbarry@...> wrote:

>
>
> "shokoshu2" wrote:
>
> > > Suppose a triangle inconic touches AB in X, BC in Y and CA in Z.
> > > Counting parameters implies there is exact one equation
> > > in the values of AX,BX,BY,CY,CZ,AZ. Which is?
> >
> > Ceva, since Brianchon. OK, the question was lame :-)
>
> So we can take X=(0,y',z'), Y=(x',0,z'), Z=(x',y',0), abd the conic is
>
> (x/x')^2 + (y/y')^2 + (z/z')^2 - 2yz/y'z' - 2zx/z'x' - 2xy/x'y' = 0.
>
> > OK, let's up it a bit. Can any point inside ABC be the
> > Brianchon point of an inconic? (I suspect yes.)
> > Which is the locus of all inparabola Brianchon points?
> > (This surely is listed on Mathworld...somewhere.)
> >
> > Hauke
>
> This conic is a parabola iff it is tangent to the line at infinity
> iff y'z'+z'x'+x'y'=0, so (x',y',z') is on the Steiner circumellipse.
> --
> Barry Wolk
>
>
>

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