- 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 - --- In Hyacinthos@yahoogroups.com, "shokoshu2" <fc3a501@...> wrote:
>

Ceva, since Brianchon. OK, the question was lame :-)

> 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?

>

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 - "shokoshu2" wrote:

> > Suppose a triangle inconic touches AB in X, BC in Y and CA in Z.

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

> > 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 :-)

(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

This conic is a parabola iff it is tangent to the line at infinity

> 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

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 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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