- Dear Antreas!

It is an equivalence of loci. Properties of the cubic?

This cubic contains four vertices of quadrilateral, two common points of its opposite sidelines, Mickel point, an infinite point of Gauss line and two circular points.

Sincerely Alexey

[Non-text portions of this message have been removed] - Here is a variation that is symmetric in ABC.

Given D, for what P do the circumcenters of the six triangles PBC, PCA, PAB, PDA, PDB, and PDC all lie on a conic?

This locus will contain the circumcircles of triangles BCD, ACD, ABD, and ABC. The rest of the locus should be of small degree.

Francisco Javier wrote:

> Dear Antreas,

--

>

> Let D=(u:v:w) be the coordinates of $D$ with respect $ABC$

>

> If P is on some of the circumcircles of DAB, ABC, BCD, CDA

> then we have only three circumcenters and they will be

> concyclic. On the remaining cases I get that P=(x:y:z)

> should be on the cubic

>

> c^2 v x^2 y - c^2 u x y^2 - a^2 v x^2 z + b^2 v x^2 z + c^2

> v x^2 z +

> b^2 w x^2 z + a^2 u x y z - b^2 u x y z - c^2 u x y z +

> a^2 w x y z +

> b^2 w x y z - c^2 w x y z + a^2 w y^2 z - b^2 u x

> z^2 -

> a^2 v x z^2 - b^2 v x z^2 + c^2 v x z^2 - a^2 v y z^2 =

> 0.

>

> Perhaps Bernard Gibert can explain us the meaning of this cubic.

>

> I am able to check that this cubic passes through A, B, C,

> D, but I have not more information on it.

>

> Best regards,

>

> Francisco Javier.

>

> "Antreas" <anopolis72@...> wrote:

> >

> > Let ABCD be a quadrilateral.

> >

> > For which points P, the circumcenters of PAB,

> > PBC, PCD, PDA are concyclic?

> >

> > APH

Barry Wolk - Dear Barry,

> Given D, for what P do the circumcenters of the six triangles PBC, PCA, PAB, PDA, PDB, and PDC all lie on a conic?

I far prefer this more symmetrical approach but I'm just wondering how can you predict this "small" degree ?

>

> This locus will contain the circumcircles of triangles BCD, ACD, ABD, and ABC. The rest of the locus should be of small degree.

Anything to do with Lemoyne's theories ?

Best regards

Bernard

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

Because I was able to keep reducing the degree by canceling some common factors while attempting (unsuccessfully) to do the calculation. And symmetry meant there would be other common factors which hadn't occurred yet.

>

> > Given D, for what P do the circumcenters of the six

> triangles PBC, PCA, PAB, PDA, PDB, and PDC all lie on a conic?

> >

> > This locus will contain the circumcircles of triangles

> BCD, ACD, ABD, and ABC. The rest of the locus should be of

> small degree.

>

> I far prefer this more symmetrical approach but I'm just

> wondering how can you predict this "small" degree ?

This turned out to be trivial. The perpendicular bisectors of the 4 segments PA, PB, PC, PD each contain 3 of those 6 circumcenters. And the 6 intersection points of 4 lines can lie on a conic only when they partially collapse, where 2 of its points or lines coincide.

So the expected "small degree" turned out to be degree 0.

>

I don't know what that theory is.

> Anything to do with Lemoyne's theories ?

>

--

> Best regards

>

> Bernard

Barry Wolk- --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>

Variations:

> Let ABCD be a quadrilateral.

>

> For which points P, the circumcenters of PAB,

> PBC, PCD, PDA are concyclic?

We can replace circumcenters with orthocenters, or

even concyclicity with orthocentricity.

That is:

For which points P, the orthocenters of

of PAB, PBC, PCD, PDA form an orthocentric

tetrad?

APH