## 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
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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 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
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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
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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
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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
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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
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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
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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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