## Re: [EMHL] A generalization...

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• ... It s a cubic, 0=cyclic sum of (u(u+v)bb+u(u+w)cc-(u+v)(u+w)aa)aa(yw-zv)yz ... It s a quartic, 0=cyclic sum of a^4 SA (bbzz-ccyy)yz ... -- Barry Wolk
Message 1 of 13 , Mar 1, 2011
> Let ABC be a triangle P = (x:y:z) a point, A'B'C'
> the cevian triangle of a point Q = (u:v:w).
>
> Which is the locus of P such that the circles
> (APA'),(BPB'),(CPC') are coaxial?

It's a cubic, 0=cyclic sum of
(u(u+v)bb+u(u+w)cc-(u+v)(u+w)aa)aa(yw-zv)yz
>
> Specialcase: P,Q are isogonal conjugate points.

It's a quartic, 0=cyclic sum of a^4 SA (bbzz-ccyy)yz

>
> Antreas

--
Barry Wolk
• Dear Antreas and Barry, ... this is the circular pivotal cubic with pivot Q, isopivot igQ (inverse of the isogonal conjugate of Q) Q must be different of H.
Message 2 of 13 , Mar 1, 2011
Dear Antreas and Barry,

> > Let ABC be a triangle P = (x:y:z) a point, A'B'C'
> > the cevian triangle of a point Q = (u:v:w).
> >
> > Which is the locus of P such that the circles
> > (APA'),(BPB'),(CPC') are coaxial?
>
> It's a cubic, 0=cyclic sum of
> (u(u+v)bb+u(u+w)cc-(u+v)(u+w)aa)aa(yw-zv)yz

this is the circular pivotal cubic with pivot Q, isopivot igQ (inverse of the isogonal conjugate of Q)
Q must be different of H.

http://bernard.gibert.pagesperso-orange.fr/Tables/table12.html

> >
> > Specialcase: P,Q are isogonal conjugate points.
>
> It's a quartic, 0=cyclic sum of a^4 SA (bbzz-ccyy)yz

This is Q002, one of the most extraordinary curves in the plane of ABC.

http://bernard.gibert.pagesperso-orange.fr/curves/q002.html

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Barry and Bernard A further generalization of the problem is possible if we replace the point P with three points Pa,Pb,Pc related to it. For example:
Message 3 of 13 , Mar 2, 2011
Dear Barry and Bernard

A further generalization of the problem is possible if we replace the
point P with three points Pa,Pb,Pc related to it.

For example:

Let ABC be a triangle, P a point [not in Darboux cubic], P1P2P3 the
pedal triangle of P, Q a point and A'B'C' the cevian triangle of Q.

Denote

Pa = BP2 /\ CP3

Pb = CP3 /\ AP1

Pc = AP1 /\ BP2

Which is the locus of P such that the circumcircles of
(APaA'),(BPbB'),(CPcC') are concurrent / have collinear
circumcenters?

APH

On Wed, Mar 2, 2011 at 8:21 AM, Bernard Gibert <bg42@...> wrote:

>
>
> Dear Antreas and Barry,
>
>
> > > Let ABC be a triangle P = (x:y:z) a point, A'B'C'
> > > the cevian triangle of a point Q = (u:v:w).
> > >
> > > Which is the locus of P such that the circles
> > > (APA'),(BPB'),(CPC') are coaxial?
> >
> > It's a cubic, 0=cyclic sum of
> > (u(u+v)bb+u(u+w)cc-(u+v)(u+w)aa)aa(yw-zv)yz
>
> this is the circular pivotal cubic with pivot Q, isopivot igQ (inverse of
> the isogonal conjugate of Q)
> Q must be different of H.
>
> http://bernard.gibert.pagesperso-orange.fr/Tables/table12.html
>
>
> > >
> > > Specialcase: P,Q are isogonal conjugate points.
> >
> > It's a quartic, 0=cyclic sum of a^4 SA (bbzz-ccyy)yz
>
> This is Q002, one of the most extraordinary curves in the plane of ABC.
>
> http://bernard.gibert.pagesperso-orange.fr/curves/q002.html
>
> Best regards
>
> Bernard
>

[Non-text portions of this message have been removed]
• Dear friends, does anybody can give a simple proof to the following inequality? If p, q, r are arbitrary positive real numbers a,b,c the sides, F the area of
Message 4 of 13 , Mar 5, 2011
Dear friends,
does anybody can give a simple proof to the
following inequality?
If p, q, r are arbitrary positive real numbers
a,b,c the sides, F the area of triangle ABC
and U = p/(q+r), V = q/(r+p), W = r/(p+q)
then prove that
U(bc)^2 + V(ca)^2 + W(ab)^2 >= 8F^2.

Best regards
• Dear Nikos, By Nesbitt inequality, we get that U+V+W = 3/2, thus by Chebyshev inequality we have U(bc)^2 + V(ca)^2 + W(ab)^2 ... hence is enough to prove that
Message 5 of 13 , Mar 5, 2011
Dear Nikos,

By Nesbitt inequality, we get that U+V+W >= 3/2, thus by Chebyshev inequality we have

U(bc)^2 + V(ca)^2 + W(ab)^2
>= (1/3)(U+V+W)((bc)^2 + (ca)^2 + (ab)^2)
>= (1/2)((bc)^2 + (ca)^2 + (ab)^2),

hence is enough to prove that

(bc)^2 + (ca)^2 + (ab)^2 >= 16 F^2,

but (bc)^2 + (ca)^2 + (ab)^2 - 16 F^2 equals to

(1/2) ((b^2-c^2)^2 + (c^2-a^2)^2 + (a^2-b^2)^2),

which is always positive.

>
> Dear friends,
> does anybody can give a simple proof to the
> following inequality?
> If p, q, r are arbitrary positive real numbers
> a,b,c the sides, F the area of triangle ABC
> and U = p/(q+r), V = q/(r+p), W = r/(p+q)
> then prove that
> U(bc)^2 + V(ca)^2 + W(ab)^2 >= 8F^2.
>
> Best regards
>
• Dear Franscisco, How can you say that by Chebyshev Inequality, U(bc)^2 + V(ca)^2 + W(ab)^2 = (1/3)(U+V+W)((bc)^2 + (ca)^2 + (ab)^2) ?? U,V,W and bc,ca,ab are
Message 6 of 13 , Mar 5, 2011
Dear Franscisco,

How can you say that by Chebyshev Inequality, U(bc)^2 + V(ca)^2 + W(ab)^2 >= (1/3)(U+V+W)((bc)^2 + (ca)^2 + (ab)^2) ??
U,V,W and bc,ca,ab are not necessarily similarly sorted, aren't they?

--- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...> wrote:
>
> Dear Nikos,
>
> By Nesbitt inequality, we get that U+V+W >= 3/2, thus by Chebyshev inequality we have
>
> U(bc)^2 + V(ca)^2 + W(ab)^2
> >= (1/3)(U+V+W)((bc)^2 + (ca)^2 + (ab)^2)
> >= (1/2)((bc)^2 + (ca)^2 + (ab)^2),
>
> hence is enough to prove that
>
> (bc)^2 + (ca)^2 + (ab)^2 >= 16 F^2,
>
> but (bc)^2 + (ca)^2 + (ab)^2 - 16 F^2 equals to
>
> (1/2) ((b^2-c^2)^2 + (c^2-a^2)^2 + (a^2-b^2)^2),
>
> which is always positive.
>
>
>
> >
> > Dear friends,
> > does anybody can give a simple proof to the
> > following inequality?
> > If p, q, r are arbitrary positive real numbers
> > a,b,c the sides, F the area of triangle ABC
> > and U = p/(q+r), V = q/(r+p), W = r/(p+q)
> > then prove that
> > U(bc)^2 + V(ca)^2 + W(ab)^2 >= 8F^2.
> >
> > Best regards
> >
>
• Rijul, you are right, I didn t take this into account. My apologies.
Message 7 of 13 , Mar 5, 2011
Rijul, you are right, I didn't take this into account.

My apologies.

--- In Hyacinthos@yahoogroups.com, "RIJUL" <rijul.champ@...> wrote:
>
> Dear Franscisco,
>
> How can you say that by Chebyshev Inequality, U(bc)^2 + V(ca)^2 + W(ab)^2 >= (1/3)(U+V+W)((bc)^2 + (ca)^2 + (ab)^2) ??
> U,V,W and bc,ca,ab are not necessarily similarly sorted, aren't they?
>
>
> --- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@> wrote:
> >
> > Dear Nikos,
> >
> > By Nesbitt inequality, we get that U+V+W >= 3/2, thus by Chebyshev inequality we have
> >
> > U(bc)^2 + V(ca)^2 + W(ab)^2
> > >= (1/3)(U+V+W)((bc)^2 + (ca)^2 + (ab)^2)
> > >= (1/2)((bc)^2 + (ca)^2 + (ab)^2),
> >
> > hence is enough to prove that
> >
> > (bc)^2 + (ca)^2 + (ab)^2 >= 16 F^2,
> >
> > but (bc)^2 + (ca)^2 + (ab)^2 - 16 F^2 equals to
> >
> > (1/2) ((b^2-c^2)^2 + (c^2-a^2)^2 + (a^2-b^2)^2),
> >
> > which is always positive.
> >
> >
> >
> > >
> > > Dear friends,
> > > does anybody can give a simple proof to the
> > > following inequality?
> > > If p, q, r are arbitrary positive real numbers
> > > a,b,c the sides, F the area of triangle ABC
> > > and U = p/(q+r), V = q/(r+p), W = r/(p+q)
> > > then prove that
> > > U(bc)^2 + V(ca)^2 + W(ab)^2 >= 8F^2.
> > >
> > > Best regards
> > >
> >
>
• Dear Franscisco, I knew this approach but it is not always correct because Chebyshev s Inequality 3(a1b1 + a2b2 + a3b3) = (a1 + a2 + a3)(b1 + b2 + b3) holds
Message 8 of 13 , Mar 5, 2011
Dear Franscisco,
I knew this approach but it is not
always correct because
Chebyshev's Inequality
3(a1b1 + a2b2 + a3b3) >= (a1 + a2 + a3)(b1 + b2 + b3)
holds as said RIJUL only when
a1 > a2 > a3 we have b1 > b2 > b3.

Best regards

> Dear Franscisco,
>
> How can you say that by Chebyshev Inequality, U(bc)^2 +
> V(ca)^2 + W(ab)^2  >= (1/3)(U+V+W)((bc)^2 + (ca)^2 +
> (ab)^2) ??
> U,V,W and bc,ca,ab are not necessarily similarly sorted,
> aren't they?
>
>
> --- In Hyacinthos@yahoogroups.com,
> "Francisco Javier" <garciacapitan@...> wrote:
> >
> > Dear Nikos,
> >
> > By Nesbitt inequality, we get that U+V+W >= 3/2,
> thus by Chebyshev inequality we have
> >
> >    U(bc)^2 + V(ca)^2 + W(ab)^2
> > >= (1/3)(U+V+W)((bc)^2 + (ca)^2 + (ab)^2)
> > >= (1/2)((bc)^2 + (ca)^2 + (ab)^2),
> >
> > hence is enough to prove that
> >
> > (bc)^2 + (ca)^2 + (ab)^2 >= 16 F^2,
> >
> > but (bc)^2 + (ca)^2 + (ab)^2 - 16 F^2 equals to
> >
> > (1/2) ((b^2-c^2)^2 + (c^2-a^2)^2 + (a^2-b^2)^2),
> >
> > which is always positive.
> >
> >
> >
> > --- In Hyacinthos@yahoogroups.com,
> > >
> > > Dear friends,
> > > does anybody can give a simple proof to the
> > > following inequality?
> > > If p, q, r are arbitrary positive real numbers
> > > a,b,c the sides, F the area of triangle ABC
> > > and U = p/(q+r), V = q/(r+p), W = r/(p+q)
> > > then prove that
> > > U(bc)^2 + V(ca)^2 + W(ab)^2 >= 8F^2.
> > >
> > > Best regards
> > >
> >
>
>
>
>
> ------------------------------------
>