## Circle in Barycentrics

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• What is the condition for a conic to be circle, preferably as a condition imposed on the matrix of conic? Sung Hyun [Non-text portions of this message have
Message 1 of 25 , Jan 10, 2011
What is the condition for a conic to be circle, preferably as a condition
imposed on the matrix of conic?

Sung Hyun

[Non-text portions of this message have been removed]
• Dear Sung Hyun, in normal coordinates, if the symmetric matrix of the conic is {{alpha,p,q},{p,beta,r},{q,r,gamma}} I have the following conditions: beta*a^6 -
Message 2 of 25 , Jan 10, 2011
Dear Sung Hyun,
in normal coordinates, if the symmetric matrix of the conic is

{{alpha,p,q},{p,beta,r},{q,r,gamma}}

I have the following conditions:

beta*a^6 - 2*b*p*a^5 +
2*b*c*r*a^4 + b^2*alpha*a^4 -
2*b^2*beta*a^4 - 2*c^2*beta*a^4 +
4*b^3*p*a^3 + 2*b*c^2*p*a^3 -
2*b^2*c*q*a^3 - 2*b*c^3*r*a^2 -
2*b^3*c*r*a^2 - 2*b^4*alpha*a^2 +
b^4*beta*a^2 + c^4*beta*a^2 +
2*b^2*c^2*gamma*a^2 - 2*b^5*p*a +
2*b^3*c^2*p*a - 2*b^2*c^3*q*a +
2*b^4*c*q*a + b^6*alpha +
b^2*c^4*alpha - 2*b^4*c^2*alpha = 0

AND

-beta*a^4 + 2*b*p*a^3 -
2*b*c*r*a^2 - b^2*alpha*a^2 +
b^2*beta*a^2 + c^2*beta*a^2 -
2*b^3*p*a + 2*b^2*c*q*a +
b^4*alpha - b^2*c^2*alpha = 0

certainly, it is possible to find more elegant and symmetric conditions...

Regards

Fred

--- In Hyacinthos@yahoogroups.com, Lim Sung Hyun <bbbbbblow@...> wrote:
>
> What is the condition for a conic to be circle, preferably as a condition
> imposed on the matrix of conic?
>
> Sung Hyun
>
>
> [Non-text portions of this message have been removed]
>
• Dear Sung Hyun and Fred, I get that a conic f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0 is a circle when (p - q + r - h) SB - (p + q - r - g) SC =
Message 3 of 25 , Jan 10, 2011
Dear Sung Hyun and Fred,

I get that a conic

f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0

is a circle when

(p - q + r - h) SB - (p + q - r - g) SC = 0

and

(f + h + 2 q) SB^2 + (f + g + 2 r) SC^2 + (2 p - g -
h) a^2 SA + (2 f - g - h + 4 p + 2 q + 2 r) SB SC = 0.

Best regards, at 11-1-11.

--- In Hyacinthos@yahoogroups.com, Lim Sung Hyun <bbbbbblow@...> wrote:
>
> What is the condition for a conic to be circle, preferably as a condition
> imposed on the matrix of conic?
>
> Sung Hyun
>
>
> [Non-text portions of this message have been removed]
>
• Dear Lim Sung Hyun ... Suppose that the radical axis of the circle C0 and the circumcircle (O) is ux+vy+wz=0; as the union of the line at infinity and the
Message 4 of 25 , Jan 11, 2011
Dear Lim Sung Hyun
>
> What is the condition for a conic to be circle, preferably as a condition
> imposed on the matrix of conic?

Suppose that the radical axis of the circle C0 and the circumcircle (O) is ux+vy+wz=0; as the union of the line at infinity and the radical axis is a degenerated member of the pencil (O),C0, the equation of C0 is (1) :
k(a^2*yz+b^2zx+c^2xy)+(x+y+z)(ux+vy+wz)=0 for some k
It follows that,if M is the matrix of a conic and V the column matrix
M[2,2]-M[3,3]
M[3,3]-M[1,1]
M[1,1]-M[2,2]
The conic is a circle when MV and SV are colinear, where S is the matrix of the circumcircle
(or equivalently V is an eigenvector of S^(-1)M)
Jean-Pierre
• Dear Sung Hyun The conic f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0 with matrix f r q r g p q p h is a circle when (g + h -2p)/aa = (h + f -2q)/bb
Message 5 of 25 , Jan 11, 2011
Dear Sung Hyun

The conic
f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0
with matrix
f r q
r g p
q p h

is a circle when
(g + h -2p)/aa = (h + f -2q)/bb = (f + q -2r)/cc.

Best regards

> Lim Sung Hyun <bbbbbblow@...> wrote:
> >
> > What is the condition for a conic to be circle,
> preferably as a condition
> > imposed on the matrix of conic?
> >
> > Sung Hyun
• I correct the formula from (g + h -2p)/aa = (h + f -2q)/bb = (f + q -2r)/cc. to (g + h -2p)/aa = (h + f -2q)/bb = (f + g -2r)/cc. Sorry. ND
Message 6 of 25 , Jan 11, 2011
I correct the formula from
(g + h -2p)/aa = (h + f -2q)/bb = (f + q -2r)/cc.
to
(g + h -2p)/aa = (h + f -2q)/bb = (f + g -2r)/cc.

Sorry. ND

> Dear Sung Hyun
>
> The conic
> f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0
> with matrix
> f r q
> r g p
> q p h
>
> is a circle when
> (g + h -2p)/aa = (h + f -2q)/bb = (f + q -2r)/cc.
>
> Best regards
>
>
> > Lim Sung Hyun <bbbbbblow@...> wrote:
> > >
> > > What is the condition for a conic to be circle,
> > preferably as a condition
> > > imposed on the matrix of conic?
> > >
> > > Sung Hyun
>
>
>
>
>
> ------------------------------------
>
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
>
• Dear Nikos, I think that your formulas are really nice. Thank you. 2011/1/11 Nikolaos Dergiades ... -- ... Francisco Javier García
Message 7 of 25 , Jan 11, 2011
Dear Nikos, I think that your formulas are really nice. Thank you.

> I correct the formula from
> (g + h -2p)/aa = (h + f -2q)/bb = (f + q -2r)/cc.
> to
> (g + h -2p)/aa = (h + f -2q)/bb = (f + g -2r)/cc.
>
> Sorry. ND
>
> > Dear Sung Hyun
> >
> > The conic
> > f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0
> > with matrix
> > f r q
> > r g p
> > q p h
> >
> > is a circle when
> > (g + h -2p)/aa = (h + f -2q)/bb = (f + q -2r)/cc.
> >
> > Best regards
> >
> >
> > > Lim Sung Hyun <bbbbbblow@...> wrote:
> > > >
> > > > What is the condition for a conic to be circle,
> > > preferably as a condition
> > > > imposed on the matrix of conic?
> > > >
> > > > Sung Hyun
> >
> >
> >
> >
> >
> > ------------------------------------
> >
> >
> >
> > Hyacinthos-fullfeatured@yahoogroups.com
> >
> >
> >
>
>
>
>
> ------------------------------------
>
>
>
>
>

--
---
Francisco Javier García Capitán
http://garciacapitan.auna.com

[Non-text portions of this message have been removed]
• Dear Nikolaos [ND] ... Thank you very much for this so nice form. My previous message gave a very bad form of the condition but it gives an immediate proof of
Message 8 of 25 , Jan 11, 2011
Dear Nikolaos
[ND]
> The conic
> f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0
> with matrix
> f r q
> r g p
> q p h
> is a circle when
> (g + h -2p)/aa = (h + f -2q)/bb = (f + g -2r)/cc.

Thank you very much for this so nice form.
My previous message gave a very bad form of the condition but it gives an immediate proof of yours (I didn't notice that) :
As the equation of a circle is
k(a^2yz+b^2zx+c^2xy)+(x+y+z)(ux+vy+wz) = 0,
we have (g + h -2p)/aa = (h + f -2q)/bb = (f + g -2r)/cc = -k
Many thanks again
JP
• Nikos, great formula indeed. Your condition is equivalent to: (g+h-2p : h+f-2q : f+g-2r) = K I think this can be simplified further. Any idea? Sung Hyun
Message 9 of 25 , Jan 11, 2011
Nikos, great formula indeed.

(g+h-2p : h+f-2q : f+g-2r) = K

I think this can be simplified further. Any idea?

Sung Hyun

[Non-text portions of this message have been removed]
• Dear everyone, There seems to be a very intimate relation between K and circle. 1. Perspector of circumcircle = K 2. For bicevian conic through cevian
Message 10 of 25 , Jan 11, 2011
Dear everyone,

There seems to be a very intimate relation between K and circle.

1. Perspector of circumcircle = K
2. For bicevian conic through cevian triangles of P and Q, (ctP)(ctQ)=K
3. Polar circle, the only diagonal circle has equation S_Ax^2 + S_By^2 +
S_Cz^2 = 0 ( Note that (S_A:S_B:S_C) = aK)
4. Conic is circle iff (g+h-2p : h+f-2q : f+g-2r) = K

Can anyone give a general interpretation for this?

Sung Hyun

[Non-text portions of this message have been removed]
• Dear friends, thank you very much for your messages. This formula is not mine. I found it in the first volume of George Kapetis Triangle Geometry (in Greek)
Message 11 of 25 , Jan 11, 2011
Dear friends,
thank you very much for your messages.
This formula is not mine.
I found it in the first volume of George Kapetis'
Triangle Geometry (in Greek) dating from 1996.
The writer gives a not simple proof in a more
general context.
He ignores the general circle formula
aayz+bbzx+ccxy-(x+y+z)(Px+Qy+Rz) = 0
and I had early noted that from the above equality we
can have the simple proof that Jean-Pierre mentioned.

From this book also I found that the kind of the
conic is given by the sign of the det of the mattrix
0 1 1 1
1 f r q
1 r g p
1 q p h

Best regards

[SH]
> Nikos, great formula indeed.
> Your condition is equivalent to:
>
> (g+h-2p : h+f-2q : f+g-2r) = K
>
> I think this can be simplified further. Any idea?

[JP]
> Thank you very much for this so nice form.
> My previous message gave a very bad form of the condition
> but it gives an immediate proof of yours (I didn't notice
> that) :
> As the equation of a circle is
> k(a^2yz+b^2zx+c^2xy)+(x+y+z)(ux+vy+wz) = 0,
> we have (g + h -2p)/aa = (h + f -2q)/bb = (f + g -2r)/cc =
> -k
• Dear Nikolaos ... If M is the matrix of the conic C0, M* the adjugate matrix of M, ie gh-pp pq-hr pr-gq pq-hr hf-qq rq-fp pr-gq rq-fp fg-rr and U the
Message 12 of 25 , Jan 11, 2011
Dear Nikolaos
> From this book also I found that the kind of the
> conic is given by the sign of the det of the mattrix
> 0 1 1 1
> 1 f r q
> 1 r g p
> 1 q p h

If M is the matrix of the conic C0, M* the adjugate matrix of M, ie
gh-pp pq-hr pr-gq
pq-hr hf-qq rq-fp
pr-gq rq-fp fg-rr
and U the column matrix
u
v
w
then U* = M*U is the pole wrt C0 of the line L of equation ux+vy+wz=0
Put T(u,v,w) = t(U)U* (t for transposition) T = quadratic form with matrix M*
Then L and C0 have 0,1,2 real common points according to the fact that T>0, T=0, T<0
More over, the determinant of the matrix
0 u v w
u f r q
v r g p
w q p h
is -T.
The matrix of the singular conic union of the tangents to C0 from the pole of L is T.M-det(M).U.t(U)
With u=v=w=1, we get the center of C0, the number of real infinite points ie the kind of the conic and the equation of the pair of asymptots.
Friendly. Jean-Pierre
• Dear Jean-Pierre thank you for this proof and especialy for equation of the pair of asymptots. M* means the adjoint matrix of M U* what is? Best regards Nikos
Message 13 of 25 , Jan 12, 2011
Dear Jean-Pierre
thank you for this proof
and especialy for equation of
the pair of asymptots.

M* means the adjoint matrix of M
U* what is?

Best regards

> Dear Nikolaos
> > From this book also I found that the kind of the
> > conic is given by the sign of the det of the mattrix
> > 0 1 1 1
> > 1 f r q
> > 1 r g p
> > 1 q p h
>
> If M is the matrix of the conic C0, M* the adjugate matrix
> of M, ie
> gh-pp  pq-hr  pr-gq
> pq-hr  hf-qq  rq-fp
> pr-gq  rq-fp  fg-rr
> and U the column matrix
> u
> v
> w
> then U* = M*U is the pole wrt C0 of the line L of equation
> ux+vy+wz=0
> Put T(u,v,w) = t(U)U*  (t for transposition) T =
> quadratic form with matrix M*
> Then L and C0 have 0,1,2 real common points according to
> the fact that T>0, T=0, T<0
> More over, the determinant of the matrix
> 0 u v w
> u f r q
> v r g p
> w q p h
> is -T.
> The matrix of the singular conic union of the tangents to
> C0 from the pole of L is T.M-det(M).U.t(U)
> With u=v=w=1, we get the center of C0, the number of real
> infinite points ie the kind of the conic and the equation of
> the pair of asymptots.
> Friendly. Jean-Pierre
>
>
>
>
> ------------------------------------
>
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
>
• Dear Nikoalaos ... U* = M*.U If U is the column matrix u v w, U* is the column of the coordinates of the pole of the line ux+vy+wz=0 JP
Message 14 of 25 , Jan 12, 2011
Dear Nikoalaos
> M* means the adjoint matrix of M
> U* what is?
U* = M*.U
If U is the column matrix u v w, U* is the column of the coordinates of the pole of the line ux+vy+wz=0
JP
• Dear Friends, What is the condition for a conic to be an orthogonal hyperbola, related to the conic f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0
Message 15 of 25 , Jan 15, 2011
Dear Friends,

What is the condition for a conic to be an orthogonal hyperbola,
related to the conic
f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0
with matrix
| f r q |
| r g p |
| q p h |

best regards,

Chris van Tienhoven

>[ND]
> The conic
> f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0
> with matrix
> f r q
> r g p
> q p h
> is a circle when
> (g + h -2p)/aa = (h + f -2q)/bb = (f + q -2r)/cc.

> > Lim Sung Hyun <bbbbbblow@> wrote:
> > > What is the condition for a conic to be circle,
> > preferably as a condition
> > > imposed on the matrix of conic?
• Dear Chris, ... a^2 f + b^2 g + c^2 h - 2 (SA p + SB q + SC r)=0 best regards Bernard [Non-text portions of this message have been removed]
Message 16 of 25 , Jan 15, 2011
Dear Chris,

> What is the condition for a conic to be an orthogonal hyperbola,
> related to the conic
> f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y = 0
> with matrix
> | f r q |
> | r g p |
> | q p h |

a^2 f + b^2 g + c^2 h - 2 (SA p + SB q + SC r)=0

best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Cris, A little explanation to Bernard s message. If (x1 : y1 : z1) and (x2 : y2 : z2) in bar.. are the infinite points of the hyperbola then these points
Message 17 of 25 , Jan 15, 2011
Dear Cris,
A little explanation
to Bernard's message.
If (x1 : y1 : z1) and (x2 : y2 : z2) in bar..
are the infinite points of the hyperbola
then these points are orthogonal
if x1x2S_A + y1y2S_B + z1z2S_C = 0. (1)
If we solve the system of equations
x + y + z = 0 and
fx^2 + gy^2 + hz^2 + 2pyz + 2qzx + 2rxy = 0
by putting z = - x - y
the discriminant of the resulting equation F(x/y) = 0
must be negative. Hence if d = det of
0 1 1 1
1 f r q
1 r g p
1 q p h
we must have d < 0 and since from F(x/y) = 0
x1x2 / y1y2 = (g + h - 2p) / (h + p - 2q)
we conclude that (1) becomes
(g + h - 2p)S_A + (h + p - 2q)S_B + (f + g - 2r)S_C = 0
or
(f+p-q-r)aa + (g+q-r-p)bb + (h+r-p-q)cc = 0
Best regards

[CVT]
> > What is the condition for a conic to be an orthogonal
> hyperbola,
> > related to the conic
> > f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x y =
> 0
> > with matrix
> > | f r q |
> > | r g p |
> > | q p h |
>
[BG]
> a^2 f + b^2 g + c^2 h - 2 (SA p + SB q + SC r)=0
>
>
• Sorry. the discriminant of the resulting equation F(x/y) = 0 must be positive. Hence if d = det of 0 1 1 1 1 f r q 1 r g p 1 q p h we must have d 0 ND
Message 18 of 25 , Jan 15, 2011
Sorry.
the discriminant of the resulting equation F(x/y) = 0
must be positive. Hence if d = det of
0 1 1 1
1 f r q
1 r g p
1 q p h
we must have d > 0

ND

> Θέμα: Re: [EMHL] Re: Circle in Barycentrics
> Προς: Hyacinthos@yahoogroups.com
> Ημερομηνία: Σάββατο, 15 Ιανουάριος 2011, 12:16
> Dear Cris,
> A little explanation
> to Bernard's message.
> If (x1 : y1 : z1) and (x2 : y2 : z2) in bar..
> are the infinite points of the hyperbola
> then these points are orthogonal
> if x1x2S_A + y1y2S_B + z1z2S_C = 0. (1)
> If we solve the system of equations
> x + y + z = 0 and
> fx^2 + gy^2 + hz^2 + 2pyz + 2qzx + 2rxy = 0
> by putting z = - x - y
> the discriminant of the resulting equation F(x/y) = 0
> must be negative. Hence if d = det of
> 0 1 1 1
> 1 f r q
> 1 r g p
> 1 q p h
> we must have d < 0 and since from F(x/y) = 0
> x1x2 / y1y2 = (g + h - 2p) / (h + p - 2q)
> we conclude that (1) becomes
> (g + h - 2p)S_A + (h + p - 2q)S_B + (f + g - 2r)S_C = 0
> or
> (f+p-q-r)aa + (g+q-r-p)bb + (h+r-p-q)cc = 0
> Best regards
>
> [CVT]
> > > What is the condition for a conic to be an
> orthogonal
> > hyperbola,
> > > related to the conic
> > > f x^2 + g y^2 + h z^2 + 2 p y z + 2 q z x + 2 r x
> y =
> > 0
> > > with matrix
> > > | f r q |
> > > | r g p |
> > > | q p h |
> >
> [BG]
> > a^2 f + b^2 g + c^2 h - 2 (SA p + SB q + SC r)=0
> >
> >
>
>
>
>
>
> ------------------------------------
>
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
>
• Dear all, What is the condition for conic to have eccentricity e? We might infer from the following: Circle - infty - (g+h-2p::)=K Rect.Hyp - sqrt(2) -
Message 19 of 25 , Jan 15, 2011
Dear all,

What is the condition for conic to have eccentricity e?

We might infer from the following:

Circle -> \infty -> (g+h-2p::)=K
Rect.Hyp -> sqrt(2) -> (f+p-q-r::)K lies on line at infinity

Sung Hyun

[Non-text portions of this message have been removed]
• Dear Lim Sung Hyun ... Put U=a^2(f+p-q-r)+b^2(g+q-r-p)+c^2(h+r-p-q) u=(g+h-2p)/a^2; v=(h+f-2q)/b^2; w=(f+g-2r)/c^2; V=SA(v-w)^2+SB(w-u)^2+SC(u-v)^2 Then (if I
Message 20 of 25 , Jan 15, 2011
Dear Lim Sung Hyun

> What is the condition for conic to have eccentricity e?
>
> We might infer from the following:
>
> Circle -> \infty -> (g+h-2p::)=K
> Rect.Hyp -> sqrt(2) -> (f+p-q-r::)K lies on line at infinity

Put U=a^2(f+p-q-r)+b^2(g+q-r-p)+c^2(h+r-p-q)
u=(g+h-2p)/a^2; v=(h+f-2q)/b^2; w=(f+g-2r)/c^2;
V=SA(v-w)^2+SB(w-u)^2+SC(u-v)^2
Then (if I didn't mistake)
a^2.b^2.c^2.(e^2-2)^2.V = e^2.U^2
Friendly. Jean-Pierre
• Dear Jean-Pierre, Put U=a^2(f+p-q-r)+b^2(g+q-r-p)+c^2(h+r-p-q) ... a^2.b^2.c^2.(e^2-2)^2.V = e^2.U^2 shouldn t e^2 be e^4 instead? Also, may I know how you got
Message 21 of 25 , Jan 15, 2011
Dear Jean-Pierre,

Put U=a^2(f+p-q-r)+b^2(g+q-r-p)+c^2(h+r-p-q)
> u=(g+h-2p)/a^2; v=(h+f-2q)/b^2; w=(f+g-2r)/c^2;
> V=SA(v-w)^2+SB(w-u)^2+SC(u-v)^2
> Then (if I didn't mistake)
> a^2.b^2.c^2.(e^2-2)^2.V = e^2.U^2
> Friendly. Jean-Pierre
>

a^2.b^2.c^2.(e^2-2)^2.V = e^2.U^2

Also, may I know how you got the expressions?

Regards,
Sung Hyun

[Non-text portions of this message have been removed]
• Dear Lim Sung Hyun [JPE] ... [LDH] ... Yes, you re right; thank you for the correction and sorry for the typo ; the right member should be (e^2.U)^2 ... The
Message 22 of 25 , Jan 15, 2011
Dear Lim Sung Hyun
[JPE]
> Put U=a^2(f+p-q-r)+b^2(g+q-r-p)+c^2(h+r-p-q)
> > u=(g+h-2p)/a^2; v=(h+f-2q)/b^2; w=(f+g-2r)/c^2;
> > V=SA(v-w)^2+SB(w-u)^2+SC(u-v)^2
> > Then (if I didn't mistake)
> > a^2.b^2.c^2.(e^2-2)^2.V = e^2.U^2
> > Friendly. Jean-Pierre
[LDH]
> a^2.b^2.c^2.(e^2-2)^2.V = e^2.U^2
> shouldn't e^2 be e^4 instead?

Yes, you're right; thank you for the correction and sorry for the typo ; the right member should be (e^2.U)^2

> Also, may I know how you got the expressions?

The conic has the same points at infinity than the circumconic with perspector g+h-2p:h+f-2q:f+g-2r because
fxx+gyy+hzz+2pyz+2qzx+2rxy+(g+h-2p)yz+(h+f-2q)zx+(f+g-2r)xy=
(x+y+z)(fx+gy+hz)
Now, we get a problem that we have discussed some weeks ago.
The formula a^2.b^2.c^2.(e^2-2)^2.V = e^4.U^2 is not the best one but I have tried, as you wanted, to have in the formula the conditions for the conic to be a circle and to be a rectangular hyperbola
We have a shorter formula if we use the condition for the conic to be a parabola :
Put Q = cyclicsum(hg-pp+2qr-2pf) (3 terms)
Q is the sum of the 9 elements of the adjugate matrix M* of the matrix M of the conic, so Q is the image of (1,1,1) under the quadratic form with matrix M* and Q =0 means that the conic touches the line at infinity (Q>0 for an ellipse, Q=0 for a parabola, Q<0 for an hyperbola)
Then we have
(1-e^2).U^2 = 4.S^2.Q.(2-e^2)^2 with S =area(ABC)
Friendly. Jean-Pierre
• Dear Lim Sung Hyun and Nikolaos Just a little remark I wrote ... (x+y+z)(fx+gy+hz) This proves immediately the Nikos condition for a conic to be a circle (the
Message 23 of 25 , Jan 15, 2011
Dear Lim Sung Hyun and Nikolaos
Just a little remark
I wrote
> The conic has the same points at infinity than the circumconic with perspector g+h-2p:h+f-2q:f+g-2r because
> fxx+gyy+hzz+2pyz+2qzx+2rxy+(g+h-2p)yz+(h+f-2q)zx+(f+g-2r)xy=
(x+y+z)(fx+gy+hz)

This proves immediately the Nikos condition for a conic to be a circle (the perspector of the circumconic must be aa:bb:cc)

Jean-Pierre
• Hello everyone Hyacinthos In http://personal.us.es/rbarroso/trianguloscabri/ has published 600 journal issue. It contains four research articles on the
Message 24 of 25 , Jan 16, 2011
Hello everyone
Hyacinthos

In

http://personal.us.es/rbarroso/trianguloscabri/

has published 600 journal issue.

It contains four research articles on the geometry of the triangle and an interview with a Spanish problemist.

Best regards

Ricardo

[Non-text portions of this message have been removed]
• Two possible alternatives (I did not do the actual work): * Let Gamma be the equation of the hyperbola. Then, the homogeneous pairs h1=[p1:q1] and h2=[p2:q2]
Message 25 of 25 , Jan 16, 2011
Two possible alternatives (I did not do the actual work):

* Let Gamma be the equation of the hyperbola. Then, the homogeneous
pairs h1=[p1:q1] and h2=[p2:q2] corresponding to the points at infinity
H1=[p1:q1:-p1-q1] and H2=[p2:q2:-p2-q2] of the hyperbola are found as
the solutions of the equation Gamma([p:q:-p-q])=0. For the hyperbola to
be a right hyperbola, H1 and H2 now have to be conjugate with respect to
the circumcircle of the base triangle. As the condition for conjugacy
can be expressed in terms of the coefficients of the equation for h1 and
h2 (or so it seems), we get our condition for a right hyperbola. I am
only wondering how this can lead to a set of equations...

* For any right hyperbola all conjugate diameters are perpendicular to
each other. As we know the center C of the hyperbola, we can take any
line l through it, find its pole L and determine the line CL. As the
points at infinity of l and CL now have to be conjugate with respect to
the circumcircle of the base triangle, we have a condition that has to
be satisfied for all possible l. This should imply a vanishing matrix
and hence a set of conditions on the coefficients of the equation of the
hyperbola.

In general, if we want to know the angle between the asymptotes of a
hyperbola, or the eccentricity of an ellipse for that matter, we can use
the Laguerre/Cayley/Klein formula, i.e the equality between twice the
angle between two lines and the absolute value of logarithm of the cross
ratio of the points at infinity of the lines and the two circular points
of the plane. The computations, however, obviously are not going to be
pretty. In the case of an ellipse this leads to an expression for
(a-b)/(a+b) (or its reciprocal) (where a, b are the lengths of the
semi-axes) which gives us 4 candidates for the eccentricity.

Eisso

On 1/16/11 2:39 AM, jpehrmfr wrote:
>
>
>
> Dear Lim Sung Hyun and Nikolaos
> Just a little remark
> I wrote
> > The conic has the same points at infinity than the circumconic with
> perspector g+h-2p:h+f-2q:f+g-2r because
> > fxx+gyy+hzz+2pyz+2qzx+2rxy+(g+h-2p)yz+(h+f-2q)zx+(f+g-2r)xy=
> (x+y+z)(fx+gy+hz)
>
> This proves immediately the Nikos condition for a conic to be a circle
> (the perspector of the circumconic must be aa:bb:cc)
>
> Jean-Pierre
>
>

--

========================================
Eisso J. Atzema, Ph.D.
Department of Mathematics& Statistics
University of Maine
Orono, ME 04469
Tel.: (207) 581-3928 (office)
(207) 866-3871 (home)
Fax.: (207) 581-3902
E-mail: atzema@...
========================================

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