## Re: [EMHL] Transform Isogonal Circular pK Cubics Into Circumcircle

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• Dear All My Friends, In fact, the transform gT(U) is conjugate transform for whole plane and we can construct it by mentioned construction with P = X(1) and
Message 1 of 9 , Sep 13, 2009
Dear All My Friends,

In fact, the transform gT(U) is conjugate transform for whole plane and we can construct it by mentioned construction with P = X(1) and then make isogonal transform. Easy get barycentrics of gT(U) transform and it is:

a/(a*(-a*v*w + b*w*u + c*u*v) + b*c*u*(v + w)) : :

This conjugate transforms:
- infinite point to infinite point
- circumcircle to circumconic passing X(100), X(643), X(644), X(664), X(1120), X(1280), X(1320), X(1897)

Some conjugate pairs (i,j) in ETC such that gT(X(i)) = X(j):
(1,1) (2,1002) (4,8) (6,996) (7,1000) (10,65) (21,1389)

Is there any relation of this conjugate with well-known transform?

Thank you and best regards,
Bui Quang Tuan

--- On Sun, 9/13/09, Quang Tuan Bui <bqtuan1962@...> wrote:

> From: Quang Tuan Bui <bqtuan1962@...>
> Subject: Re: [EMHL] Transform Isogonal Circular pK Cubics Into Circumcircle
> To: Hyacinthos@yahoogroups.com
> Date: Sunday, September 13, 2009, 6:29 PM
> Dear All My Friends,
>
> When P=X(1) we denote the transform as Q=T(U) where U is
> one infinite point. Because Q is on circumcircle so gQ =
> isogonal conjugate of Q = one another infinite point.
> Naturally we can take gQ as U and make the transform T(gQ).
> It is interesting that we will get T(gQ) = gU = isogonal
> conjugate of U, of course on circumcircle.
> So, if P = X(1), U is one infinite point then:
> T(gT(U)) = gU or gT(gT(U)) = U
> It means that gT is conjugate transform between infinite
> points.
>
> Best regards,
> Bui Quang Tuan
>
> --- On Sun, 9/13/09, Quang Tuan Bui <bqtuan1962@...>
> wrote:
>
> > From: Quang Tuan Bui <bqtuan1962@...>
> > Subject: [EMHL] Transform Isogonal Circular pK Cubics
> Into Circumcircle
> > To: hyacinthos@yahoogroups.com
> > Date: Sunday, September 13, 2009, 2:41 AM
> > Dear All My Friends,
> >
> > Given triangle ABC, point P with barycentrics (p : q :
> r),
> > point U with barycentrics (u : v : w).
> > A line from P parallel with UA intersects BC at A1
> > A line from P parallel with BC intersects UA at A2
> > La is the line connected A1, A2. Similarly define Lb,
> Lc.
> > (This configuration can be constructed also with U is
> > infinite point)
> >
> > Results:
> >
> > 1. Three lines La, Lb, Lc are concurrent at one point
> Q
> > with barycentrics:
> > p^2*(-p*v*w + q*w*u + r*u*v) + p*q*r*u*(v + w)
> > :  :
> >
> > 2. If U is infinite point then the locus of P such
> > that  Q is on circumcircle is one isogonal circular
> pK
> > cubic. See Table 1 in:
> > http://pagesperso-orange.fr/bernard.gibert/Classes/cl035.html
> >
> > 3. Therefore, if P=X(1) and U is one infinite point
> then Q
> > is on circumcircle. From now, all results are with
> P=X(1).
> >
• Dear All My Friends, I have found that gT conjugate is one special case of following more general conjugate (May be well-known already?). Suppose P, U are any
Message 2 of 9 , Sep 13, 2009
Dear All My Friends,

I have found that gT conjugate is one special case of following more general conjugate (May be well-known already?).

Suppose P, U are any two distinct points; gP, gU are isogonal conjugate of P, U respectively. L is a line passing through U and parallel with gPgU. Q is second intersection of L with circumconic (A, B, C, P, U) (other than P).

The transform (depending on fixed point P) Q = TP(U) is one conjugate transform.

In the case gT conjugate: P = X(1).

Conjugate transform Q=TP(U) can be constructed by ruler and compass. If F is fourth intersection of circumconic (A, B, C, P, U) with circumcircle then Q=TP(U) is also on line FgU

First barycentrics of Q = TP(U):

p*(q*w - r*v) / (a^2*p*v*w*(c^2*q*v - b^2*r*w) + a^2*q*r*u*(b^2*w^2 - c^2*v^2) + b^2*c^2*p*u*(v + w)*(q*w - r*v))

Best regards,
Bui Quang Tuan

--- On Mon, 9/14/09, Quang Tuan Bui <bqtuan1962@...> wrote:

> From: Quang Tuan Bui <bqtuan1962@...>
> Subject: Re: [EMHL] Transform Isogonal Circular pK Cubics Into Circumcircle
> To: Hyacinthos@yahoogroups.com
> Date: Monday, September 14, 2009, 2:09 AM
> Dear All My Friends,
>
> In fact, the transform gT(U) is conjugate transform for
> whole plane and we can construct it by mentioned
> construction with P = X(1) and then make isogonal transform.
> Easy get barycentrics of gT(U) transform and it is:
>
> a/(a*(-a*v*w + b*w*u + c*u*v) + b*c*u*(v + w))
> :  :
>
> This conjugate transforms:
> - infinite point to infinite point
> - circumcircle to circumconic passing X(100), X(643),
> X(644), X(664), X(1120), X(1280), X(1320), X(1897)
>
> Some conjugate pairs (i,j) in ETC such that gT(X(i)) =
> X(j):
> (1,1)  (2,1002)  (4,8)  (6,996)
> (7,1000)  (10,65)  (21,1389)
>
> Is there any relation of this conjugate with well-known
> transform?
>
> Thank you and best regards,
> Bui Quang Tuan
>
• Dear All My Friends, Following are some loci related this TP(U) conjugate transform. 1. The locus of U such that TP(U) = P is one cubic C(P) with barycentric
Message 3 of 9 , Sep 15, 2009
Dear All My Friends,

Following are some loci related this TP(U) conjugate transform.

1. The locus of U such that TP(U) = P is one cubic C(P) with barycentric equation:
CyclicSum[ a^2*p*y*z*((b^2*r^2 + c^2*q*p + c^2*q*r)*y -(c^2*q^2 + b^2*r*p + b^2*r*q)*z) + b^2*c^2*p^2*(q - r)*x*y*z ] = 0

Some properties of these cubics
a). P=X(1), the locus is only X(1)
b). If gP = isogonal conjugate of P then C(gP) = g(C(P))
c). The locus of P such that C(P) is K0 cubic (without term x*y*z in equation) is K102 (Grebe cubic)
d). Some interesting cubics:
C(X(2)) = K280, C(X(6)) = K281, C(X(4)) = K028

2. The locus of U such that TP(U) = U is one quintic Q(P) with barycentric equation:
CyclicSum[ a^2*p^2*(c^2*q + b^2*r)*y^2*z^2*(r*y - q*z) + p*q*r*x*y*z*(a^2*y*z*(c^2*q - b^2*r) - b^2*c^2*x^2*(q - r)) ] = 0

a). If P = X(1), Q(P) is degenerated into three angle bisectors and one circumconic passing X(80), X(100), X(291), X(668), X(1018), X(1783)
b). Q(X(2)) passes X(1), X(2), X(39), X(291), X(2009), X(2010)
c). Q(X(4)) = Q038

3. The locus of U such that U, gU, TP(U) are collinear is union of circumcircle, quintic Q(P) mentioned in a) and one pK(X6, X6/P) cubic

Best regards,
Bui Quang Tuan

--- On Mon, 9/14/09, Quang Tuan Bui <bqtuan1962@...> wrote:

> From: Quang Tuan Bui <bqtuan1962@...>
> Subject: Re: [EMHL] Transform Isogonal Circular pK Cubics Into Circumcircle
> To: Hyacinthos@yahoogroups.com
> Date: Monday, September 14, 2009, 12:08 PM
> Dear All My Friends,
>
> I have found that gT conjugate is one special case of
> following more general conjugate (May be well-known
>
> Suppose P, U are any two distinct points; gP, gU are
> isogonal conjugate of P, U respectively. L is a line passing
> through U and parallel with gPgU. Q is second intersection
> of L with circumconic (A, B, C, P, U) (other than P).
>
> The transform (depending on fixed point P) Q = TP(U) is one
> conjugate transform.
>
> In the case gT conjugate: P = X(1).
>
> Conjugate transform Q=TP(U) can be constructed by ruler and
> compass. If F is fourth intersection of circumconic (A, B,
> C, P, U) with circumcircle then Q=TP(U) is also on line FgU
>
> First barycentrics of Q = TP(U):
>
> p*(q*w - r*v) / (a^2*p*v*w*(c^2*q*v - b^2*r*w) +
> a^2*q*r*u*(b^2*w^2 - c^2*v^2) + b^2*c^2*p*u*(v + w)*(q*w -
> r*v))
>
> Best regards,
> Bui Quang Tuan
>
• Dear Tuan, ... This is actually the locus of U such that the lines PU and gPgU are parallel. ... not really, C(X1) is the union of the cevian lines of X(1) ...
Message 4 of 9 , Sep 16, 2009
Dear Tuan,

> The locus of U such that TP(U) = P is one cubic C(P) with
> barycentric equation:
> CyclicSum[ a^2*p*y*z*((b^2*r^2 + c^2*q*p + c^2*q*r)*y -(c^2*q^2 +
> b^2*r*p + b^2*r*q)*z) + b^2*c^2*p^2*(q - r)*x*y*z ] = 0

This is actually the locus of U such that the lines PU and gPgU are
parallel.

>
> Some properties of these cubics
> a). P=X(1), the locus is only X(1)

not really, C(X1) is the union of the cevian lines of X(1)

> b). If gP = isogonal conjugate of P then C(gP) = g(C(P))
> c). The locus of P such that C(P) is K0 cubic (without term x*y*z in
> equation) is K102 (Grebe cubic)
> d). Some interesting cubics:
> C(X(2)) = K280, C(X(6)) = K281, C(X(4)) = K028

and K009 = C(X3).

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Bernard, Thank you for your remarks! Your three interesting remarks are true! By the way, please correct the equation of K281 in your site:
Message 5 of 9 , Sep 16, 2009
Dear Bernard,

By the way, please correct the equation of K281 in your site:

http://pagesperso-orange.fr/bernard.gibert/Exemples/k281.html

I see K281 equation is exact as K280 equation but in fact, K281 is isogonal conjugate of K280.

Thank you and best regards,
Bui Quang Tuan

--- On Wed, 9/16/09, Bernard Gibert <bg42@...> wrote:

> From: Bernard Gibert <bg42@...>
> Subject: Re: [EMHL] Transform Isogonal Circular pK Cubics Into Circumcircle
> To: Hyacinthos@yahoogroups.com
> Date: Wednesday, September 16, 2009, 6:28 PM
> Dear Tuan,
>
> >  The locus of U such that TP(U) = P is one cubic
> C(P) with
> > barycentric equation:
> > CyclicSum[ a^2*p*y*z*((b^2*r^2 + c^2*q*p + c^2*q*r)*y
> -(c^2*q^2 +
> > b^2*r*p + b^2*r*q)*z) + b^2*c^2*p^2*(q - r)*x*y*z ] =
> 0
>
> This is actually the locus of U such that the lines PU and
> gPgU are
> parallel.
>
> >
> > Some properties of these cubics
> > a). P=X(1), the locus is only X(1)
>
> not really, C(X1) is the union of the cevian lines of X(1)
>
>
> > b). If gP = isogonal conjugate of P then C(gP) =
> g(C(P))
> > c). The locus of P such that C(P) is K0 cubic (without
> term x*y*z in
> > equation) is K102 (Grebe cubic)
> > d). Some interesting cubics:
> > C(X(2)) = K280, C(X(6)) = K281, C(X(4)) = K028
>
> and K009 = C(X3).
>
>
> Best regards
>
> Bernard
>
>
• Dear Tuan, ... the equations are not identical ! Best regards Bernard [Non-text portions of this message have been removed]
Message 6 of 9 , Sep 16, 2009
Dear Tuan,

> By the way, please correct the equation of K281 in your site:
>
> http://pagesperso-orange.fr/bernard.gibert/Exemples/k281.html
>
> I see K281 equation is exact as K280 equation but in fact, K281 is
> isogonal conjugate of K280.

the equations are not identical !

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Bernard, Sorry, you are right! K280 and K281 equations are true. It is interesting that they are closely similar! Thank you and best regards, Bui Quang
Message 7 of 9 , Sep 16, 2009
Dear Bernard,

Sorry, you are right! K280 and K281 equations are true. It is interesting that they are closely similar!

Thank you and best regards,
Bui Quang Tuan

--- On Wed, 9/16/09, Bernard Gibert <bg42@...> wrote:

> From: Bernard Gibert <bg42@...>
> Subject: Re: [EMHL] Transform Isogonal Circular pK Cubics Into Circumcircle
> To: Hyacinthos@yahoogroups.com
> Date: Wednesday, September 16, 2009, 7:10 PM
> Dear Tuan,
>
> > By the way, please correct the equation of K281 in
> >
> > http://pagesperso-orange.fr/bernard.gibert/Exemples/k281.html
> >
> > I see K281 equation is exact as K280 equation but in
> fact, K281 is
> > isogonal conjugate of K280.
>
>
> the equations are not identical !
>
> Best regards
>
> Bernard
>
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