## Θέμα: Re: Θέμα: [EMHL] Areas Of Triangles Inscribed In Circumconic

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• Dear Francois, ... Yes you are right if the areas are signed. The circumcevian triangle A1B1C1 of P is similar to the pedal(P) and the circumcevian triangle
Message 1 of 16 , Feb 1, 2007
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Dear Francois,

> 1° In case of (Co) beeing the circumcircle, I think
> the Tuan locus is also
> the locus of the points P such that OP = OQ, where O
> is the circumcenter and
> Q the isogonal conjugate of P.

Yes you are right if the areas are signed.
The circumcevian triangle A1B1C1 of P is
similar to the pedal(P) and
the circumcevian triangle A2B2C2 of Q is
similar to the pedal(Q) and
Since P and Q are isogonal conjugates
pedal(P) and pedal(Q) have the same circumcircle.
the ratio of areas pedal(P)/A1B1C1 = pedal(Q)/A2B2C2
The triangles A1B1C1, A2B2C2 have equal areas
iff pedal(P), pedal(Q) have equal areas
but it is known that
(pedal(P))/(ABC) = |R^2-OP^2|/(4RR)
Hence |R^2-OP^2| = |R^2-OQ^2|.
If the areas are signed then then
we get OP = OQ as you mentioned
but if the areas are not signed then we can have
OP^2 + OQ^2 = 2R^2.

Best regards

___________________________________________________________
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• Dear Nikos Yes, I forgot this question of sign but may be you did the same and your locus must be: pqrxyz(x+y+z)^3 = (+)(-)(pyz+qzx+rxy)^3 So it has 2
Message 2 of 16 , Feb 1, 2007
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Dear Nikos
Yes, I forgot this question of sign but may be you did the same and your
locus must be:
pqrxyz(x+y+z)^3 = (+)(-)(pyz+qzx+rxy)^3
So it has 2 components of degree 6?
Friendly
Francois
Can you write explicitly the factorization when p = q = r?
Friendly
Francois
PS
In the circle case, complex numbers are not needed and metric inversion
formula are enough to do the job!
For example B'C' = |k|.BC/(PB.PC) and so on..
B'C'.C'A'.A'B' = |k|^3.BC.CA.AB/(PA.PB.PC)^2
Area(A'B'C') = |k|^3.Area(ABC)/(PA.PB.PC)^2
where k is the inversion modulus and P the pole and so on...
>
> Dear Francois,
>
> > 1� In case of (Co) beeing the circumcircle, I think
> > the Tuan locus is also
> > the locus of the points P such that OP = OQ, where O
> > is the circumcenter and
> > Q the isogonal conjugate of P.
>
> Yes you are right if the areas are signed.
> The circumcevian triangle A1B1C1 of P is
> similar to the pedal(P) and
> the circumcevian triangle A2B2C2 of Q is
> similar to the pedal(Q) and
> Since P and Q are isogonal conjugates
> pedal(P) and pedal(Q) have the same circumcircle.
> the ratio of areas pedal(P)/A1B1C1 = pedal(Q)/A2B2C2
> The triangles A1B1C1, A2B2C2 have equal areas
> iff pedal(P), pedal(Q) have equal areas
> but it is known that
> (pedal(P))/(ABC) = |R^2-OP^2|/(4RR)
> Hence |R^2-OP^2| = |R^2-OQ^2|.
> If the areas are signed then then
> we get OP = OQ as you mentioned
> but if the areas are not signed then we can have
> OP^2 + OQ^2 = 2R^2.
>
> Best regards
>
>
>
>
>
>
>
> ___________________________________________________________
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• Dear Francois, ... Yes. You are right. The locus must be: pqrxyz(x+y+z)^3 = (+)(-)(pyz+qzx+rxy)^3 What do you mean to write explicitly the factorization when p
Message 3 of 16 , Feb 1, 2007
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Dear Francois,

> Yes, I forgot this question of sign but may be you
> did the same and your
> locus must be:
> pqrxyz(x+y+z)^3 = (+)(-)(pyz+qzx+rxy)^3
> So it has 2 components of degree 6?
> Friendly
> Francois
> Can you write explicitly the factorization when p =
> q = r?
> Friendly
> Francois
> PS
> In the circle case, complex numbers are not needed
> and metric inversion
> formula are enough to do the job!
> For example B'C' = |k|.BC/(PB.PC) and so on..
> B'C'.C'A'.A'B' = |k|^3.BC.CA.AB/(PA.PB.PC)^2
> Area(A'B'C') = |k|^3.Area(ABC)/(PA.PB.PC)^2
> where k is the inversion modulus and P the pole and
> so on...

Yes. You are right.
The locus must be:
pqrxyz(x+y+z)^3 = (+)(-)(pyz+qzx+rxy)^3

What do you mean to write explicitly the
factorization when p = q = r?
My computer did the factorization.
if p = q = r the locus is
xyz(x+y+z)^3 = (+)(-)(yz+zx+xy)^3
or for (+)
(yz+zx+xy)^3 - xyz(x+y+z)^3 = 0 (1)
A factorization by hand is the following:
Since
(x+y+z)^3 = x^3+y^3+z^3+3(x+y)(y+z)(z+x)
(yz+zx+xy)^3=(yz)^3+(zx)^3+(xy)^3+3xyz(x+y)(y+z)(z+x)
(1) becomes
(yz)^3+(zx)^3+(xy)^3-xyz(x^3+y^3+z^3)=0 or
xy^3(xx-yz)+xz^3(xx-yz)-yz(x^4-(yz)^2)=0 or
(xx-yz)(xy^3+xz^3-yz(x^2+yz))=0 or
(xx-yz)(xy(yy-zx)-zz(yy-zx))=0 or
-(xx-yz)(yy-zx)(zz-xy)=0

Best regards

___________________________________________________________
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• ... I just want to know if the other component is also reducible or not? I have no Maple nor Mathematica on my computer and I am lazy! Friendly Francois
Message 4 of 16 , Feb 1, 2007
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>
> What do you mean to write explicitly the
> factorization when p = q = r?

I just want to know if the other component is also reducible or not?
I have no Maple nor Mathematica on my computer and I am lazy!

Friendly
Francois

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• The other component is not reducible. I use Mathcad 7. Friendly Nikos ... ___________________________________________________________
Message 5 of 16 , Feb 2, 2007
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The other component is not reducible.

Friendly
Nikos

> I just want to know if the other component is also
> reducible or not?
> I have no Maple nor Mathematica on my computer and I
> am lazy!

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• Dear Nikos and Francois, Thank you very much for nice results! I would like to inform you some my small efforts: With any point X we denote CT(X) as triangle
Message 6 of 16 , Feb 2, 2007
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Dear Nikos and Francois,
Thank you very much for nice results! I would like to inform you some my small efforts:
With any point X we denote CT(X) as triangle bounded by three circumcenters of three triangles XBC, XCA, XAB. By my calculations, in the case when circumconic is circumcircle, the sextic is also the locus of P such that
Area(CT(P)) = Area(CT(gP)) here gP = isogonal conjugate of P
May be we can find similar result with Steiner circumellipse and isotomic conjugate tP.
Another my idea is that to find the locus of homothetic center of ABC and triangle bounded by A1A2, B1B2, C1C2. The degree of this curve is may be less than six.
By the way, when working with this configuration I see one simple fact I think easy to prove. If you know about this fact so please give me some advices. I hope that this fact can be used somehow here. This fact is:
If through any point X on circumconic we draw three parallel lines with three sidelines of reference triangle then the triangle bounded by three second intesections (other than X) of three these lines with circumconic has area value the same as area value of reference triangle.
Thank you and best regards,
Bui Quang Tuan

> Given triangle ABC with its circumconic (Co). Let
> A1B1C1 and A2B2C2 are any two other triangles
> inscribed in (Co), both not perspective with ABC and
> A1A2//BC, B1B2//CA, C1C2//AB. T1 is a triangle
> bounded by three lines AA1, BB1, CC1. T2 is a
> triangle bounded by three lines AA2, BB2, CC2.
> Area(A1B1C1)/Area(T1) = Area(A2B2C2)/Area(T2)
> In the case when A1B1C1 is perspective with ABC
> with perspector P then A2B2C2 is also perpective
> with ABC with perspector, say Q. What is the locus
> of P such that:
> Area(A1B1C1) = Area(A2B2C2). In this case: what is
> the locus of Q?

If (p : q : r) is the perspector of the circumconic
and P = (x : y : z) then Q = (p/x : q/y : r/z)
the locus of P is the sextic
pqrxyz(x+y+z)^3 = (pyz+qzx+rxy)^3 and the locus of
Q is the same.
Note: In case of circumcircle
Q is the isogonal conjugate of P and
in case of Steiner circum ellipse
Q is the isotomic conjugate of P.

Best regards

---------------------------------

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• . ... Here some hint : We start with the triangle ABC, the circumconic {Gamma} and ... . ... [Non-text portions of this message have been removed]
Message 7 of 16 , Feb 2, 2007
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.
> This fact is:
> If through any point X on circumconic we draw three parallel lines with
> three sidelines of reference triangle then the triangle bounded by three
> second intesections (other than X) of three these lines with circumconic has
> area value the same as area value of reference triangle.
>

Here some hint : We start with the triangle ABC, the circumconic {Gamma} and
> any point M on {Gamma}. We get a new in-triangle A'B'C' such that MA' is
> parallel to BC, MB' is parallel to CA, MC' is parallel to AB. Then lines
> AA', BB', CC' are parallel and triangle A'B'C' is the image of ABC by some
> affine symmetry f, preserving {Gamma} of which the axis is some diameter of
> {Gamma} and the direction AA', BB' or CC'. As f change the signed area in
> its opposite, we are done! Friendly Francois
>

.
>
>
>

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• Dear Francois, Thank you very much for your hints. I think they are complete proof already and not only normal proof. It is special proof: proof by picture. By
Message 8 of 16 , Feb 2, 2007
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Dear Francois,
Thank you very much for your hints. I think they are complete proof already and not only normal proof. It is special proof: proof by picture.
By your instructions, we can draw four special lines: AA'//BB'//CC' and one line passing through midpoints of these segments, center of circumconic and is the collinear line of Pascal theorem for our six points. These four lines combine with sidelines of two triangles to make some equal area pieces so we can see the result right after drawing.
I love this simple picture!
Thank you again and best regards,
Bui Quang Tuan

Francois Rideau <francois.rideau@...> wrote: .
> This fact is:
> If through any point X on circumconic we draw three parallel lines with
> three sidelines of reference triangle then the triangle bounded by three
> second intesections (other than X) of three these lines with circumconic has
> area value the same as area value of reference triangle.
>

Here some hint : We start with the triangle ABC, the circumconic {Gamma} and
> any point M on {Gamma}. We get a new in-triangle A'B'C' such that MA' is
> parallel to BC, MB' is parallel to CA, MC' is parallel to AB. Then lines
> AA', BB', CC' are parallel and triangle A'B'C' is the image of ABC by some
> affine symmetry f, preserving {Gamma} of which the axis is some diameter of
> {Gamma} and the direction AA', BB' or CC'. As f change the signed area in
> its opposite, we are done! Friendly Francois
>

.
>
>
>

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