- Sorry I repeat my correction

Sorry,

I wrote> By the way this T you used is the Sqrt of the Det

the correct is -Det.

I forgot to say that if the point (x : y : z)

is on the conic then the second intersection of the line

with

the conic is

( yz(g(xv-yu)^2 + h(zu-xw)^2) : zx(h(yw-zv)^2 + f(xv-yu)^2)

: . . .)

where u = Q - R, v = R - P, w = P - Q.

Best regards

Nikos Dergiades

> > Dear Francisco,

> > thank you very much.

> > I will study your solution.

> > I was expecting to have some cyclicity.

> > I mean if in (f,g,h), (p,q,r), (P,Q,R)

> > there is a cyclicity relative to a,b,c

> > then I wanted to have as point of

> > intersection a point ( F(a,b,c):F(b,c,a):F(c,a,b) ).

> > Is this expectation false?

> > For example I found the following solution:

> > Let

> > -(h P Q + h Q^2 - p P R - 2 p Q R - q Q R + g R^2 + r

> R^2)/R

> > = F(a,b,c)

> > If

> > (h P^2 + h P Q - p P R - 2 P q R - q Q R + f R^2 + r

> R^2)/R

> > = F(b,c,a)

> > then the intersection points are

> > ( F(a,b,c) + T : F(b,c,a) + T : F(c,a,b) + T )

> > ( F(a,b,c) - T : F(b,c,a) - T : F(c,a,b) - T )

> > But this fine result is under condition and does not

> hold

> > in all cases.

> > By the way this T you used is the Sqrt of the Det

> > 0 P Q R

> > P f r q

> > Q r g p

> > R q p h

> >

> > Best regards

> > Nikos Dergiades

> >

> >

> > > They are the points:

> > >

> > > {h Q^2 - 2 p Q R + g R^2,

> > > -h P Q + p P R + q Q R - r R^2 - R T,

> > > p P Q - q Q^2 - g P R + Q r R + Q T},

> > >

> > > {h Q^2 - 2 p Q R + g R^2,

> > > -h P Q + p P R + q Q R - r R^2 + R T,

> > > p P Q - q Q^2 - g P R + Q r R - Q T}}

> > >

> > > where T is the square root of

> > >

> > > (p^2 - g h) P^2 + (q^2 - f h) Q^2 + (r^2 - f g)

> R^2 +

> > (2 f p

> > > -

> > > 2 q r) Q R + (2 h r - 2 p q) P Q + (2 g q -

> 2

> > > p r) P R

> > >

> > > Mathematica says.

> > >

> > > Best regards,

> > >

> > > Francisco Javier.

> > >

> > >

> > >

> > > --- In Hyacinthos@yahoogroups.com,

> > > Nikolaos Dergiades <ndergiades@...> wrote:

> > > >

> > > > Dear friends

> > > > if we know the line

> > > > Px + Qy + Rz = 0

> > > > and the conic

> > > > fxx + gyy + hzz + 2pyz + 2qzx + 2rxy = 0

> > > > is there a formula giving the barycentric

> > > > coordinates of their intersection points?

> > > > Best regards

> > > > Nikos Dergiades

> > > >

> > >

> > >

> > >

> > >

> > > ------------------------------------

> > >

> > > Yahoo! Groups Links

> > >

> > >

> > > Hyacinthos-fullfeatured@yahoogroups.com

> > >

> > >

> >

> >

> > ------------------------------------

> >

> > Yahoo! Groups Links

> >

> >

> > Hyacinthos-fullfeatured@yahoogroups.com

> >

> >

>

>

> ------------------------------------

>

> Yahoo! Groups Links

>

>

> Hyacinthos-fullfeatured@yahoogroups.com

>

> - Dear Nikolaos, my solution is non symmetric because I use Mathematica's

Solve in this way:

Simplify[Solve[{line=0,conic=0},{y,z}], x>0]

As a result we get the points

ptSa= {h Q^2 - 2 p Q R + g R^2, -h P Q + p P R + q Q R - r R^2 - R T,

p P Q - q Q^2 - g P R + Q r R + Q T}

ptTa= {h Q^2 - 2 p Q R + g R^2, -h P Q + p P R + q Q R - r R^2 + R T,

p P Q - q Q^2 - g P R + Q r R - Q T}

where T is the square root of

-g h P^2 + p^2 P^2 - 2 p P q Q - f h Q^2 + q^2 Q^2 + 2 h P Q r +

2 (g P q + f p Q - (p P + q Q) r) R + (-f g + r^2) R^2 .

This gives a result easier to simplify.

In order to symetrize these expressions we can use the sum of the

permutations of this solution

You can use this function included in Baricentricas.nb:

Simetrizar[ptP_]:=Simplificar[Total[Table[Nest[PermutarTerna,ptP,n],{n,0,2}]]]

In order to that this works properly you have to change ReglaDePermutacion

definition because the default value doesn't include p, q, r, P, Q, R , f,

g, h:

ReglaDePermutacion = {p -> q, q -> r, r -> p, P -> Q, Q -> R, R -> P,

f -> g, g -> h, h -> f};

For example,

ptS = Factor[Simetrizar[ptSa]]

gives

{-h P Q + p P Q + h Q^2 - q Q^2 - g P R + p P R - 2 p Q R + q Q R +

Q r R + g R^2 - r R^2 - Q T + R T,

h P^2 - p P^2 - h P Q + P q Q + p P R - 2 P q R - f Q R + q Q R +

P r R + f R^2 - r R^2 + P T - R T,

g P^2 - p P^2 + p P Q + P q Q + f Q^2 - q Q^2 - 2 P Q r - g P R -

f Q R + P r R + Q r R - P T + Q T}

2013/2/18 Nikolaos Dergiades <ndergiades@...>

> Sorry I repeat my correction

--

>

>

> Sorry,

> I wrote

> > By the way this T you used is the Sqrt of the Det

> the correct is -Det.

> I forgot to say that if the point (x : y : z)

> is on the conic then the second intersection of the line

> with

> the conic is

> ( yz(g(xv-yu)^2 + h(zu-xw)^2) : zx(h(yw-zv)^2 + f(xv-yu)^2)

> : . . .)

>

> where u = Q - R, v = R - P, w = P - Q.

>

> Best regards

> Nikos Dergiades

>

>

>

> > > Dear Francisco,

> > > thank you very much.

> > > I will study your solution.

> > > I was expecting to have some cyclicity.

> > > I mean if in (f,g,h), (p,q,r), (P,Q,R)

> > > there is a cyclicity relative to a,b,c

> > > then I wanted to have as point of

> > > intersection a point ( F(a,b,c):F(b,c,a):F(c,a,b) ).

> > > Is this expectation false?

> > > For example I found the following solution:

> > > Let

> > > -(h P Q + h Q^2 - p P R - 2 p Q R - q Q R + g R^2 + r

> > R^2)/R

> > > = F(a,b,c)

> > > If

> > > (h P^2 + h P Q - p P R - 2 P q R - q Q R + f R^2 + r

> > R^2)/R

> > > = F(b,c,a)

> > > then the intersection points are

> > > ( F(a,b,c) + T : F(b,c,a) + T : F(c,a,b) + T )

> > > ( F(a,b,c) - T : F(b,c,a) - T : F(c,a,b) - T )

> > > But this fine result is under condition and does not

> > hold

> > > in all cases.

> > > By the way this T you used is the Sqrt of the Det

> > > 0 P Q R

> > > P f r q

> > > Q r g p

> > > R q p h

> > >

> > > Best regards

> > > Nikos Dergiades

> > >

> > >

> > > > They are the points:

> > > >

> > > > {h Q^2 - 2 p Q R + g R^2,

> > > > -h P Q + p P R + q Q R - r R^2 - R T,

> > > > p P Q - q Q^2 - g P R + Q r R + Q T},

> > > >

> > > > {h Q^2 - 2 p Q R + g R^2,

> > > > -h P Q + p P R + q Q R - r R^2 + R T,

> > > > p P Q - q Q^2 - g P R + Q r R - Q T}}

> > > >

> > > > where T is the square root of

> > > >

> > > > (p^2 - g h) P^2 + (q^2 - f h) Q^2 + (r^2 - f g)

> > R^2 +

> > > (2 f p

> > > > -

> > > > 2 q r) Q R + (2 h r - 2 p q) P Q + (2 g q -

> > 2

> > > > p r) P R

> > > >

> > > > Mathematica says.

> > > >

> > > > Best regards,

> > > >

> > > > Francisco Javier.

> > > >

> > > >

> > > >

> > > > --- In Hyacinthos@yahoogroups.com,

> > > > Nikolaos Dergiades <ndergiades@...> wrote:

> > > > >

> > > > > Dear friends

> > > > > if we know the line

> > > > > Px + Qy + Rz = 0

> > > > > and the conic

> > > > > fxx + gyy + hzz + 2pyz + 2qzx + 2rxy = 0

> > > > > is there a formula giving the barycentric

> > > > > coordinates of their intersection points?

> > > > > Best regards

> > > > > Nikos Dergiades

> > > > >

> > > >

> > > >

> > > >

> > > >

> > > > ------------------------------------

> > > >

> > > > Yahoo! Groups Links

> > > >

> > > >

> > > > Hyacinthos-fullfeatured@yahoogroups.com

> > > >

> > > >

> > >

> > >

> > > ------------------------------------

> > >

> > > Yahoo! Groups Links

> > >

> > >

> > > Hyacinthos-fullfeatured@yahoogroups.com

> > >

> > >

> >

> >

> > ------------------------------------

> >

> > Yahoo! Groups Links

> >

> >

> > Hyacinthos-fullfeatured@yahoogroups.com

> >

> >

>

>

> ------------------------------------

>

> Yahoo! Groups Links

>

>

>

>

---

Francisco Javier García Capitán

http://garciacapitan.99on.com

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

There were several replies, including asking for cyclically symmetric coordinates for those intersection points.

> if we know the line

> Px + Qy + Rz = 0

> and the conic

> fxx + gyy + hzz + 2pyz + 2qzx + 2rxy = 0

> is there a formula giving the barycentric

> coordinates of their intersection points?

> Best regards

> Nikos Dergiades

The points on the line Px+Qy+Rz=0 can be parametrized as

(x, y, z)=((Q-R)(QR+t), (R-P)(RP+t), (P-Q)(PQ+t))

Substitute these formulas for x,y,z into the equation of the conic, to get a cyclically symmetric quadratic equation for the parameter t. Its two roots will give the intersection points in the requested form.

--

Barry Wolk- Dear friends,

Thanks Barry!

This brings me to a new problem.

I never got into this.

Given an equation f(x,y,z).

How to convert it to a point in time (t): (x(t) : y(t) : z(t))?

To make it visual and more appealing. When we have a conical orbit of a planet or comet.

How can we describe its location as a function of time in barycentric coordinates ?

Any ideas?

Chris van Tienhoven

--- In Hyacinthos@yahoogroups.com, Barry Wolk <wolkbarry@...> wrote:

>

> > Dear friends

> > if we know the line

> > Px + Qy + Rz = 0

> > and the conic

> > fxx + gyy + hzz + 2pyz + 2qzx + 2rxy = 0

> > is there a formula giving the barycentric

> > coordinates of their intersection points?

> > Best regards

> > Nikos Dergiades

>

> There were several replies, including asking for cyclically symmetric coordinates for those intersection points.

>

> The points on the line Px+Qy+Rz=0 can be parametrized as

> (x, y, z)=((Q-R)(QR+t), (R-P)(RP+t), (P-Q)(PQ+t))

>

> Substitute these formulas for x,y,z into the equation of the conic, to get a cyclically symmetric quadratic equation for the parameter t. Its two roots will give the intersection points in the requested form.

> --

> Barry Wolk

>