## Re: [EMHL] Re: Mccay ?

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• Dear Francisco, ... Your S is probably the actual area of ABC and not Conway s S i.e. twice the area. Anyway, your cubic is a nK(X6, X6, ?) without known
Message 1 of 10 , Apr 5 11:41 PM
Dear Francisco,

> No, I see that the cubic with equation
>
> 4 S^2 xyz + CyclicSum(a^2 y z(c^2y+b^2z)) = 0 (*)
>
> also passes through the centers of Apollonian circles, but it is not
> K192.

Your "S" is probably the actual area of ABC and not Conway's S i.e.
twice the area.

Anyway, your cubic is a nK(X6, X6, ?) without known center on it and
without any remarkable other property as far as I can see.

Best regards

Bernard

[Non-text portions of this message have been removed]
• Yes, S was the area of ABC, like in your K192 page.
Message 2 of 10 , Apr 5 11:52 PM
Yes, S was the area of ABC, like in your K192 page.

--- In Hyacinthos@yahoogroups.com, Bernard Gibert <bg42@...> wrote:
>
> Dear Francisco,
>
> > No, I see that the cubic with equation
> >
> > 4 S^2 xyz + CyclicSum(a^2 y z(c^2y+b^2z)) = 0 (*)
> >
> > also passes through the centers of Apollonian circles, but it is not
> > K192.
>
> Your "S" is probably the actual area of ABC and not Conway's S i.e.
> twice the area.
>
> Anyway, your cubic is a nK(X6, X6, ?) without known center on it and
> without any remarkable other property as far as I can see.
>
> Best regards
>
> Bernard
>
>
>
>
> [Non-text portions of this message have been removed]
>
• Dear Francisco Thanks! Here is a similar problem: Let ABC be a triangle, P a point and A B C , A B C its pedal, circumcevian triangles, resp. Let O be the
Message 3 of 10 , Apr 6 1:29 AM
Dear Francisco

Thanks!

Here is a similar problem:

Let ABC be a triangle, P a point and A'B'C', A"B"C"
its pedal, circumcevian triangles, resp.

Let O' be the center of the pedal circle of P, and
A*B*C* the circumcevian of O' wrt A'B'C' in the pedal
circle of P. (ie A*,B*,C* are the antipodes of A',B',C'
in the pedal circle).

Which is the locus of P such that A"B"C", A*B*C* are
perspective?

Antreas

[APH]
> > Let ABC be a triangle, P,P* two isogonal conjugate points
> > and A*B*C* the pedal triangle of P*.
> >
> > Denote:
> >
> > A'B'C' := the circumcevian triangle of P wrt ABC
> > (in the circumcircle of ABC)
> >
> > A"B"C" := the circumcevian triangle of P* wrt A*B*C* in
> > the pedal circle of P*
> >
> > (ie A',B',C' are the second intersections of AP,BP,CP
> > with
> > the circumcircle of ABC, and A",B",C" are the second
> > intersections
> > of P*A*,PB*,P*C* with the circumcircle of A*B*C*).
> >
> > Which is the locus of P such that A'B'C', A"B"C" are perspective?
> >
> > (For P = O or H the perspector is G)

[Francisco Javier]
> Yes, I get the McCay cubic, also with the line at infinity,
> the circumcircle and the cubic with equation
>
> 4 S^2 xyz + CyclicSum(a^2 yz(c^2y+b^2z)) = 0
>
> Surely Bernard can give this cubic a name.
• In this case we get the sidelines, the line at infinity, the circumcircle and a septic trhough the orthocenter, the incenter, the excenters and the feet of the
Message 4 of 10 , Apr 6 2:10 AM
In this case we get the sidelines, the line at infinity, the circumcircle and a septic trhough the orthocenter, the incenter, the excenters and the feet of the altitudes

2 a^4 c^6 x^4 y^3 - 2 b^4 c^6 x^4 y^3 - 4 a^2 c^8 x^4 y^3 +
2 c^10 x^4 y^3 + 2 a^4 c^6 x^3 y^4 - 2 b^4 c^6 x^3 y^4 +
4 b^2 c^8 x^3 y^4 - 2 c^10 x^3 y^4 - a^6 b^2 c^2 x^4 y^2 z +
3 a^4 b^4 c^2 x^4 y^2 z - 3 a^2 b^6 c^2 x^4 y^2 z +
b^8 c^2 x^4 y^2 z + 3 a^4 b^2 c^4 x^4 y^2 z +
6 a^2 b^4 c^4 x^4 y^2 z - 9 b^6 c^4 x^4 y^2 z -
7 a^2 b^2 c^6 x^4 y^2 z + 3 b^4 c^6 x^4 y^2 z + 5 b^2 c^8 x^4 y^2 z -
a^8 c^2 x^3 y^3 z + 2 a^6 b^2 c^2 x^3 y^3 z -
2 a^2 b^6 c^2 x^3 y^3 z + b^8 c^2 x^3 y^3 z + 7 a^6 c^4 x^3 y^3 z +
3 a^4 b^2 c^4 x^3 y^3 z - 3 a^2 b^4 c^4 x^3 y^3 z -
7 b^6 c^4 x^3 y^3 z - 11 a^4 c^6 x^3 y^3 z + 11 b^4 c^6 x^3 y^3 z +
5 a^2 c^8 x^3 y^3 z - 5 b^2 c^8 x^3 y^3 z - a^8 c^2 x^2 y^4 z +
3 a^6 b^2 c^2 x^2 y^4 z - 3 a^4 b^4 c^2 x^2 y^4 z +
a^2 b^6 c^2 x^2 y^4 z + 9 a^6 c^4 x^2 y^4 z -
6 a^4 b^2 c^4 x^2 y^4 z - 3 a^2 b^4 c^4 x^2 y^4 z -
3 a^4 c^6 x^2 y^4 z + 7 a^2 b^2 c^6 x^2 y^4 z - 5 a^2 c^8 x^2 y^4 z +
a^6 b^2 c^2 x^4 y z^2 - 3 a^4 b^4 c^2 x^4 y z^2 +
7 a^2 b^6 c^2 x^4 y z^2 - 5 b^8 c^2 x^4 y z^2 -
3 a^4 b^2 c^4 x^4 y z^2 - 6 a^2 b^4 c^4 x^4 y z^2 -
3 b^6 c^4 x^4 y z^2 + 3 a^2 b^2 c^6 x^4 y z^2 + 9 b^4 c^6 x^4 y z^2 -
b^2 c^8 x^4 y z^2 + 12 a^4 b^4 c^2 x^3 y^2 z^2 -
12 a^2 b^6 c^2 x^3 y^2 z^2 - 12 a^4 b^2 c^4 x^3 y^2 z^2 +
12 a^2 b^2 c^6 x^3 y^2 z^2 + 12 a^6 b^2 c^2 x^2 y^3 z^2 -
12 a^4 b^4 c^2 x^2 y^3 z^2 + 12 a^2 b^4 c^4 x^2 y^3 z^2 -
12 a^2 b^2 c^6 x^2 y^3 z^2 + 5 a^8 c^2 x y^4 z^2 -
7 a^6 b^2 c^2 x y^4 z^2 + 3 a^4 b^4 c^2 x y^4 z^2 -
a^2 b^6 c^2 x y^4 z^2 + 3 a^6 c^4 x y^4 z^2 +
6 a^4 b^2 c^4 x y^4 z^2 + 3 a^2 b^4 c^4 x y^4 z^2 -
9 a^4 c^6 x y^4 z^2 - 3 a^2 b^2 c^6 x y^4 z^2 + a^2 c^8 x y^4 z^2 -
2 a^4 b^6 x^4 z^3 + 4 a^2 b^8 x^4 z^3 - 2 b^10 x^4 z^3 +
2 b^6 c^4 x^4 z^3 + a^8 b^2 x^3 y z^3 - 7 a^6 b^4 x^3 y z^3 +
11 a^4 b^6 x^3 y z^3 - 5 a^2 b^8 x^3 y z^3 -
2 a^6 b^2 c^2 x^3 y z^3 - 3 a^4 b^4 c^2 x^3 y z^3 +
5 b^8 c^2 x^3 y z^3 + 3 a^2 b^4 c^4 x^3 y z^3 -
11 b^6 c^4 x^3 y z^3 + 2 a^2 b^2 c^6 x^3 y z^3 +
7 b^4 c^6 x^3 y z^3 - b^2 c^8 x^3 y z^3 -
12 a^6 b^2 c^2 x^2 y^2 z^3 + 12 a^2 b^6 c^2 x^2 y^2 z^3 +
12 a^4 b^2 c^4 x^2 y^2 z^3 - 12 a^2 b^4 c^4 x^2 y^2 z^3 +
5 a^8 b^2 x y^3 z^3 - 11 a^6 b^4 x y^3 z^3 + 7 a^4 b^6 x y^3 z^3 -
a^2 b^8 x y^3 z^3 - 5 a^8 c^2 x y^3 z^3 + 3 a^4 b^4 c^2 x y^3 z^3 +
2 a^2 b^6 c^2 x y^3 z^3 + 11 a^6 c^4 x y^3 z^3 -
3 a^4 b^2 c^4 x y^3 z^3 - 7 a^4 c^6 x y^3 z^3 -
2 a^2 b^2 c^6 x y^3 z^3 + a^2 c^8 x y^3 z^3 + 2 a^10 y^4 z^3 -
4 a^8 b^2 y^4 z^3 + 2 a^6 b^4 y^4 z^3 - 2 a^6 c^4 y^4 z^3 -
2 a^4 b^6 x^3 z^4 + 2 b^10 x^3 z^4 - 4 b^8 c^2 x^3 z^4 +
2 b^6 c^4 x^3 z^4 + a^8 b^2 x^2 y z^4 - 9 a^6 b^4 x^2 y z^4 +
3 a^4 b^6 x^2 y z^4 + 5 a^2 b^8 x^2 y z^4 -
3 a^6 b^2 c^2 x^2 y z^4 + 6 a^4 b^4 c^2 x^2 y z^4 -
7 a^2 b^6 c^2 x^2 y z^4 + 3 a^4 b^2 c^4 x^2 y z^4 +
3 a^2 b^4 c^4 x^2 y z^4 - a^2 b^2 c^6 x^2 y z^4 -
5 a^8 b^2 x y^2 z^4 - 3 a^6 b^4 x y^2 z^4 + 9 a^4 b^6 x y^2 z^4 -
a^2 b^8 x y^2 z^4 + 7 a^6 b^2 c^2 x y^2 z^4 -
6 a^4 b^4 c^2 x y^2 z^4 + 3 a^2 b^6 c^2 x y^2 z^4 -
3 a^4 b^2 c^4 x y^2 z^4 - 3 a^2 b^4 c^4 x y^2 z^4 +
a^2 b^2 c^6 x y^2 z^4 - 2 a^10 y^3 z^4 + 2 a^6 b^4 y^3 z^4 +
4 a^8 c^2 y^3 z^4 - 2 a^6 c^4 y^3 z^4

Best regards,

Francisco Javier.

--- In Hyacinthos@yahoogroups.com, "xpolakis" <xpolakis@...> wrote:
>
> Dear Francisco
>
> Thanks!
>
> Here is a similar problem:
>
> Let ABC be a triangle, P a point and A'B'C', A"B"C"
> its pedal, circumcevian triangles, resp.
>
> Let O' be the center of the pedal circle of P, and
> A*B*C* the circumcevian of O' wrt A'B'C' in the pedal
> circle of P. (ie A*,B*,C* are the antipodes of A',B',C'
> in the pedal circle).
>
> Which is the locus of P such that A"B"C", A*B*C* are
> perspective?
>
> Antreas
>
> [APH]
> > > Let ABC be a triangle, P,P* two isogonal conjugate points
> > > and A*B*C* the pedal triangle of P*.
> > >
> > > Denote:
> > >
> > > A'B'C' := the circumcevian triangle of P wrt ABC
> > > (in the circumcircle of ABC)
> > >
> > > A"B"C" := the circumcevian triangle of P* wrt A*B*C* in
> > > the pedal circle of P*
> > >
> > > (ie A',B',C' are the second intersections of AP,BP,CP
> > > with
> > > the circumcircle of ABC, and A",B",C" are the second
> > > intersections
> > > of P*A*,PB*,P*C* with the circumcircle of A*B*C*).
> > >
> > > Which is the locus of P such that A'B'C', A"B"C" are perspective?
> > >
> > > (For P = O or H the perspector is G)
>
> [Francisco Javier]
> > Yes, I get the McCay cubic, also with the line at infinity,
> > the circumcircle and the cubic with equation
> >
> > 4 S^2 xyz + CyclicSum(a^2 yz(c^2y+b^2z)) = 0
> >
> > Surely Bernard can give this cubic a name.
>
• Dear Francisco [APH] ... Hmmm... I don t see O in the locus. If P = O, then A*B*C* is the Euler triangle (A*,B*,C* are the midpoints of AH,BH,CH) and A ,B ,C
Message 5 of 10 , Apr 6 4:16 AM
Dear Francisco

[APH]
> > Here is a similar problem:
> >
> > Let ABC be a triangle, P a point and A'B'C', A"B"C"
> > its pedal, circumcevian triangles, resp.
> >
> > Let O' be the center of the pedal circle of P, and
> > A*B*C* the circumcevian of O' wrt A'B'C' in the pedal
> > circle of P. (ie A*,B*,C* are the antipodes of A',B',C'
> > in the pedal circle).
> >
> > Which is the locus of P such that A"B"C", A*B*C* are
> > perspective?

[Francisco Javier]:
> In this case we get the sidelines, the line at infinity, the >circumcircle and a septic trhough the orthocenter, the incenter, the >excenters and the feet of the altitudes

Hmmm... I don't see O in the locus.

If P = O, then A*B*C* is the Euler triangle
(A*,B*,C* are the midpoints of AH,BH,CH)
and A",B",C" are the antipodes of A,B,C
in the circumcircle)
I think that A"B"C", A*B*C* are perspective
at G (as in the first case)

APH
• Dear Antreas, yes, you are right, O is on the locus, but is not true the general case that if P is on the locus then the same occur with its isogonal
Message 6 of 10 , Apr 6 4:24 AM
Dear Antreas, yes, you are right, O is on the locus, but is not true the general case that if P is on the locus then the same occur with its isogonal conjugate.

--- In Hyacinthos@yahoogroups.com, "xpolakis" <xpolakis@...> wrote:
>
> Dear Francisco
>
> [APH]
> > > Here is a similar problem:
> > >
> > > Let ABC be a triangle, P a point and A'B'C', A"B"C"
> > > its pedal, circumcevian triangles, resp.
> > >
> > > Let O' be the center of the pedal circle of P, and
> > > A*B*C* the circumcevian of O' wrt A'B'C' in the pedal
> > > circle of P. (ie A*,B*,C* are the antipodes of A',B',C'
> > > in the pedal circle).
> > >
> > > Which is the locus of P such that A"B"C", A*B*C* are
> > > perspective?
>
> [Francisco Javier]:
> > In this case we get the sidelines, the line at infinity, the >circumcircle and a septic trhough the orthocenter, the incenter, the >excenters and the feet of the altitudes
>
> Hmmm... I don't see O in the locus.
>
> If P = O, then A*B*C* is the Euler triangle
> (A*,B*,C* are the midpoints of AH,BH,CH)
> and A",B",C" are the antipodes of A,B,C
> in the circumcircle)
> I think that A"B"C", A*B*C* are perspective
> at G (as in the first case)
>
> APH
>
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