## Re: [EMHL] Mickel point

Expand Messages
• Dear Alexey Now I am looking at your proof using angles and I don t understand it! As the triangles AMX and XMC are similar we receive that angle
Message 1 of 29 , Jan 30, 2008
Dear Alexey
Now I am looking at your proof using angles and I don't understand it!

As the triangles AMX and XMC are similar we receive that angle
AXC=pi-AMC=AI1C.
May be you should mean: AXC = pi - (AMC/2) ?

And the last angle equality : ...= AI1C is also not obvious.

Friendly
Francois

[Non-text portions of this message have been removed]
• Dear Friends, I encountered a trivial property of the parabola, not found in my text books. It gives a very nice description of the parabolas circumscribing a
Message 2 of 29 , Jan 31, 2008
Dear Friends, I encountered a trivial property of the parabola, not
found in my text books. It gives a very nice description of the
parabolas circumscribing a given triangle.
Besides, using an appropriate projectivity, generalizes to a nice
structural description of all conics circumscribing a given triangle and
tangent to a given line. Here are the remarks:

Let the vertices of our basic triangle, points {a,b,c}, be on a
parabola. Consider also points {a'=b+c-a, b'=c+a-b, c'=a+b-c} which are
the vertices of the anticomplementary triangle.
The parallels from {a',b',c'} to the axis of the parabola meet the
opposite sides of the anticomplementary triangle at points lying on the
parabola.

This is a special case of the following more general setting that should
be treated in text-books on conics (may be I have the wrong ones).

Consider a triangle ABC and a line L . The set of all circumconics of
ABC tangent to L has the following structure.

[1] Given the triangle ABC and the arbitrary line L construct its
trilinear pole P w.r. to ABC and the corresponding precevian triangle
A0B0C0.
[2] Consider an arbitrary point Q on L and the lines {QA0,QB0,QC0}
joining Q to the vertices of A0B0C0. The intersection points
{A'',B'',C''} of these lines with the opposite sides of triangle A0B0C0
and points {A,B,C} are six points on a conic c(Q).
[3] c(Q) passes through Q and is also tangent to L at Q. Thus, it is the
unique conic passing through the vertices of ABC and tangent to L at its
point Q.
[4] The tripole of L with respect to triangle A''B''C'' is a point Q'
on line QP.
[5] The tangential triangle A1B1C1 of ABC relative to the conic c(Q) is
line-perspective to A''B''C'', the perspectrix being line L.
[6] The locus of perspectors of conics c(Q) is the inscribed conic of
triangle ABC with perspector P.

Friendly
Paris Pamfilos

Internal Virus Database is out-of-date.
Checked by AVG Free Edition.
Version: 7.5.516 / Virus Database: 269.19.5/1228 - Release Date:
16/1/2008 9:01 ?i

[Non-text portions of this message have been removed]
• Dear Paris This projective configuration of a conic through 3 points A, B, C and tangent at Q to a line (L) is very beautiful! I am sure there is a nice proof
Message 3 of 29 , Feb 1, 2008
Dear Paris

This projective configuration of a conic through 3 points A, B, C and
tangent at Q to a line (L) is very beautiful!

I am sure there is a nice proof in the general case using Desargues and
Plücker
involution theorems known for a long time that we can find as an exercise in
some old book.

But the idea to send line (L) at infinity and to prove the theorem in this
special case is very educational.

Another (educational?) idea should be to send both points B and C at
infinity that is to say looking at the hyperbola through point A with given
asymptotic directions and tangent to a line (L) at Q. I am sure that
elementary affine properties of hyperbola can also do the job.

Friendly
Francois

[Non-text portions of this message have been removed]
• Dear Eisso, François and Alexei ... such ... in ... angle ... imaginary ... the ... for ... Of course, there is a little problem because there are 4 common
Message 4 of 29 , Feb 3, 2008
Dear Eisso, François and Alexei
> What I think you are saying, François, is that the circular map f
such
> that f(A) = A', f(B) = B', f(C) = C' can be represented by complex
> conjugation w.r.t. to the complex plane if we place the quadrilateral
in
> such a way that the Miquel point is at the origin and the (common)
angle
> bisectors of AMA', BMB' and CMC' coincide with the real and
imaginary
> axes. I think that does not quite answer the concern that I had. In
> fact, I think you would run into some difficulties too for non-convex
> cases as it is not immediately evident (at least not to me) which of
the
> common angle bisectors should be taken for the real axis and which
for
> the complex axis.

Of course, there is a little problem because there are 4 common points
of a A-bisector and a A'-bisector, lying on 2 circles going through A
and A'. We have to choose the right circles through A,A' and B,B' to
get the fixed points of f.
So, it is better to use the circles FAA' and FBB' (F = Miquel point)
because they are the 9P-circles of the orthocentric systems with
vertices the 4 common points of the A- and A'- bisectors and the 4
common points of the B- and B'-bisectors; so, there is no more
ambiguity.
T = midpoint of the circular arc AFA', T' = midpoint of the circular
arc BFB'; the fixed points are the common points of the circle with
center T going through A and the circle with center T' going through B.
Friendly
Jean-Pierre
• Dear Paris and Francois! It seems that these problems were considered by Lev and Tatyana Emelyanoff. They presented the problems devoted to trypolars at Summer
Message 5 of 29 , Feb 4, 2008
Dear Paris and Francois!
It seems that these problems were considered by Lev and Tatyana Emelyanoff. They presented the problems devoted to trypolars at Summer conference of Towns tournament-2005.
Also in our with Arsenij Akopyan book next theorem is proved. Given the triangle ABC and the point C0. An arbitrary line passing through C0 intersect AC and BC in B' and A' resp. P is the common point of AA' and BB'. Then
1. The locus of P is the circumconic touching CC0.
2. If CP intersect AB in C' then all lines A'C' have the common point B0 and BB0 touches the circumconic.

Sincerely Alexey

Dear Paris

This projective configuration of a conic through 3 points A, B, C and
tangent at Q to a line (L) is very beautiful!

I am sure there is a nice proof in the general case using Desargues and
Plücker
involution theorems known for a long time that we can find as an exercise in
some old book.

But the idea to send line (L) at infinity and to prove the theorem in this
special case is very educational.

Another (educational?) idea should be to send both points B and C at
infinity that is to say looking at the hyperbola through point A with given
asymptotic directions and tangent to a line (L) at Q. I am sure that
elementary affine properties of hyperbola can also do the job.

Friendly
Francois

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

Antivirus scanning: Symantec Mail Security for SMTP.

[Non-text portions of this message have been removed]
• Dear Alexey Very interesting problem I had had to tackle when I was a young boy more than fifty years ago. I change your notation a little bit to avoid
Message 6 of 29 , Feb 5, 2008
Dear Alexey
Very interesting problem I had had to tackle when I was a young boy more
than fifty years ago.

I change your notation a little bit to avoid subscript and I name D your
point C0.
I think the locus of P is the ABC-circumconic tangent at A to line AD and
tangent at B to line BD.
Let A"B"C" be the ABC-cevian triangle of D.
Then line B'C' is on the harmonic conjugate of D wrt BB"and line A'C' is on
the harmonic conjugate of D wrt AA".

Friendly
Francois
On Feb 5, 2008 8:29 AM, Alexey.A.Zaslavsky <zasl@...> wrote:

> Dear Paris and Francois!
> It seems that these problems were considered by Lev and Tatyana
> Emelyanoff. They presented the problems devoted to trypolars at Summer
> conference of Towns tournament-2005.
> Also in our with Arsenij Akopyan book next theorem is proved. Given the
> triangle ABC and the point C0. An arbitrary line passing through C0
> intersect AC and BC in B' and A' resp. P is the common point of AA' and BB'.
> Then
> 1. The locus of P is the circumconic touching CC0.
> 2. If CP intersect AB in C' then all lines A'C' have the common point B0
> and BB0 touches the circumconic.
>
> Sincerely Alexey
>
>
> Dear Paris
>
> This projective configuration of a conic through 3 points A, B, C and
> tangent at Q to a line (L) is very beautiful!
>
> I am sure there is a nice proof in the general case using Desargues and
> Plücker
> involution theorems known for a long time that we can find as an exercise
> in
> some old book.
>
> But the idea to send line (L) at infinity and to prove the theorem in this
>
> special case is very educational.
>
> Another (educational?) idea should be to send both points B and C at
> infinity that is to say looking at the hyperbola through point A with
> given
> asymptotic directions and tangent to a line (L) at Q. I am sure that
> elementary affine properties of hyperbola can also do the job.
>
> Friendly
> Francois
>
> [Non-text portions of this message have been removed]
>
>
>
>
> Antivirus scanning: Symantec Mail Security for SMTP.
>
> [Non-text portions of this message have been removed]
>
>
>

[Non-text portions of this message have been removed]
• Dear Francois! Thank you for your correstion. Of course the conic is inscribed on the triangle A0B0C0. Sincerely
Message 7 of 29 , Feb 5, 2008
Dear Francois!
Thank you for your correstion. Of course the conic is inscribed on the triangle A0B0C0.

Sincerely Alexey

Dear Alexey
Very interesting problem I had had to tackle when I was a young boy more
than fifty years ago.

I change your notation a little bit to avoid subscript and I name D your
point C0.
I think the locus of P is the ABC-circumconic tangent at A to line AD and
tangent at B to line BD.
Let A"B"C" be the ABC-cevian triangle of D.
Then line B'C' is on the harmonic conjugate of D wrt BB"and line A'C' is on
the harmonic conjugate of D wrt AA".

Friendly
Francois
On Feb 5, 2008 8:29 AM, Alexey.A.Zaslavsky <zasl@...> wrote:

> Dear Paris and Francois!
> It seems that these problems were considered by Lev and Tatyana
> Emelyanoff. They presented the problems devoted to trypolars at Summer
> conference of Towns tournament-2005.
> Also in our with Arsenij Akopyan book next theorem is proved. Given the
> triangle ABC and the point C0. An arbitrary line passing through C0
> intersect AC and BC in B' and A' resp. P is the common point of AA' and BB'.
> Then
> 1. The locus of P is the circumconic touching CC0.
> 2. If CP intersect AB in C' then all lines A'C' have the common point B0
> and BB0 touches the circumconic.
>
> Sincerely Alexey
>
>
> Dear Paris
>
> This projective configuration of a conic through 3 points A, B, C and
> tangent at Q to a line (L) is very beautiful!
>
> I am sure there is a nice proof in the general case using Desargues and
> Plücker
> involution theorems known for a long time that we can find as an exercise
> in
> some old book.
>
> But the idea to send line (L) at infinity and to prove the theorem in this
>
> special case is very educational.
>
> Another (educational?) idea should be to send both points B and C at
> infinity that is to say looking at the hyperbola through point A with
> given
> asymptotic directions and tangent to a line (L) at Q. I am sure that
> elementary affine properties of hyperbola can also do the job.
>
> Friendly
> Francois
>
> [Non-text portions of this message have been removed]
>
>
>
>
> Antivirus scanning: Symantec Mail Security for SMTP.
>
> [Non-text portions of this message have been removed]
>
>
>

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

Antivirus scanning: Symantec Mail Security for SMTP.

[Non-text portions of this message have been removed]
• Dear Alexey If I name Db the harmonic conjugate of D wrt BB and Da the harmonic conjugate of D wrt AA , then the tangent at C to the circumconic locus of P is
Message 8 of 29 , Feb 5, 2008
Dear Alexey
If I name Db the harmonic conjugate of D wrt BB" and Da the harmonic
conjugate of D wrt AA", then the tangent at C to the circumconic locus of P
is just the line Cab.

1° What is the locus of D in order to the locus of P be a rectangular
hyperbola? : ( a line)
2° What is the locus of D in order to the locus of P be a parabola?: ( a
hyperbola)

Friendly
Francois

On Feb 5, 2008 9:59 AM, Francois Rideau <francois.rideau@...> wrote:

> Dear Alexey
> Very interesting problem I had had to tackle when I was a young boy more
> than fifty years ago.
>
> I change your notation a little bit to avoid subscript and I name D your
> point C0.
> I think the locus of P is the ABC-circumconic tangent at A to line AD and
> tangent at B to line BD.
> Let A"B"C" be the ABC-cevian triangle of D.
> Then line B'C' is on the harmonic conjugate of D wrt BB"and line A'C' is
> on the harmonic conjugate of D wrt AA".
>
> Friendly
> Francois
>
> On Feb 5, 2008 8:29 AM, Alexey.A.Zaslavsky <zasl@...> wrote:
>
> > Dear Paris and Francois!
> > It seems that these problems were considered by Lev and Tatyana
> > Emelyanoff. They presented the problems devoted to trypolars at Summer
> > conference of Towns tournament-2005.
> > Also in our with Arsenij Akopyan book next theorem is proved. Given the
> > triangle ABC and the point C0. An arbitrary line passing through C0
> > intersect AC and BC in B' and A' resp. P is the common point of AA' and BB'.
> > Then
> > 1. The locus of P is the circumconic touching CC0.
> > 2. If CP intersect AB in C' then all lines A'C' have the common point B0
> > and BB0 touches the circumconic.
> >
> > Sincerely Alexey
> >
> >
> > Dear Paris
> >
> > This projective configuration of a conic through 3 points A, B, C and
> > tangent at Q to a line (L) is very beautiful!
> >
> > I am sure there is a nice proof in the general case using Desargues and
> > Plücker
> > involution theorems known for a long time that we can find as an
> > exercise in
> > some old book.
> >
> > But the idea to send line (L) at infinity and to prove the theorem in
> > this
> > special case is very educational.
> >
> > Another (educational?) idea should be to send both points B and C at
> > infinity that is to say looking at the hyperbola through point A with
> > given
> > asymptotic directions and tangent to a line (L) at Q. I am sure that
> > elementary affine properties of hyperbola can also do the job.
> >
> > Friendly
> > Francois
> >
> > [Non-text portions of this message have been removed]
> >
> >
> >
> >
> > Antivirus scanning: Symantec Mail Security for SMTP.
> >
> > [Non-text portions of this message have been removed]
> >
> >
> >
>
>

[Non-text portions of this message have been removed]
• Dear Francois! 1 is the partial case of next general fact: if the pencil of conics determined by the points A, B, C, D is considered then the locus of poles of
Message 9 of 29 , Feb 6, 2008
Dear Francois!
1 is the partial case of next general fact: if the pencil of conics determined by the points A, B, C, D is considered then the locus of poles of line AB is the line. This line is the polar of common point of AB and CD wrt any conic of the pencil. In our case it is the common point of AB and the altitude CH.
The locus 2 is the hyperbola with asymptotes CA and CB touching AB in its midpoint.

Sincerely Alexey

Dear Alexey
If I name Db the harmonic conjugate of D wrt BB" and Da the harmonic
conjugate of D wrt AA", then the tangent at C to the circumconic locus of P
is just the line Cab.

1° What is the locus of D in order to the locus of P be a rectangular
hyperbola? : ( a line)
2° What is the locus of D in order to the locus of P be a parabola?: ( a
hyperbola)

Friendly
Francois

On Feb 5, 2008 9:59 AM, Francois Rideau <francois.rideau@...> wrote:

> Dear Alexey
> Very interesting problem I had had to tackle when I was a young boy more
> than fifty years ago.
>
> I change your notation a little bit to avoid subscript and I name D your
> point C0.
> I think the locus of P is the ABC-circumconic tangent at A to line AD and
> tangent at B to line BD.
> Let A"B"C" be the ABC-cevian triangle of D.
> Then line B'C' is on the harmonic conjugate of D wrt BB"and line A'C' is
> on the harmonic conjugate of D wrt AA".
>
> Friendly
> Francois
>
> On Feb 5, 2008 8:29 AM, Alexey.A.Zaslavsky <zasl@...> wrote:
>
> > Dear Paris and Francois!
> > It seems that these problems were considered by Lev and Tatyana
> > Emelyanoff. They presented the problems devoted to trypolars at Summer
> > conference of Towns tournament-2005.
> > Also in our with Arsenij Akopyan book next theorem is proved. Given the
> > triangle ABC and the point C0. An arbitrary line passing through C0
> > intersect AC and BC in B' and A' resp. P is the common point of AA' and BB'.
> > Then
> > 1. The locus of P is the circumconic touching CC0.
> > 2. If CP intersect AB in C' then all lines A'C' have the common point B0
> > and BB0 touches the circumconic.
> >
> > Sincerely Alexey
> >
> >
> > Dear Paris
> >
> > This projective configuration of a conic through 3 points A, B, C and
> > tangent at Q to a line (L) is very beautiful!
> >
> > I am sure there is a nice proof in the general case using Desargues and
> > Plücker
> > involution theorems known for a long time that we can find as an
> > exercise in
> > some old book.
> >
> > But the idea to send line (L) at infinity and to prove the theorem in
> > this
> > special case is very educational.
> >
> > Another (educational?) idea should be to send both points B and C at
> > infinity that is to say looking at the hyperbola through point A with
> > given
> > asymptotic directions and tangent to a line (L) at Q. I am sure that
> > elementary affine properties of hyperbola can also do the job.
> >
> > Friendly
> > Francois
> >
> > [Non-text portions of this message have been removed]
> >
> >
> >
> >
> > Antivirus scanning: Symantec Mail Security for SMTP.
> >
> > [Non-text portions of this message have been removed]
> >
> >
> >
>
>

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

Antivirus scanning: Symantec Mail Security for SMTP.

[Non-text portions of this message have been removed]
• Dear Alexey Cheer up! Thanks to your problem, it was true nostalgia for me! Friendly Francois ... [Non-text portions of this message have been removed]
Message 10 of 29 , Feb 6, 2008
Dear Alexey
Cheer up!
Thanks to your problem, it was true nostalgia for me!
Friendly
Francois

On Feb 6, 2008 11:07 AM, Alexey.A.Zaslavsky <zasl@...> wrote:

> Dear Francois!
> 1 is the partial case of next general fact: if the pencil of conics
> determined by the points A, B, C, D is considered then the locus of poles of
> line AB is the line. This line is the polar of common point of AB and CD wrt
> any conic of the pencil. In our case it is the common point of AB and the
> altitude CH.
> The locus 2 is the hyperbola with asymptotes CA and CB touching AB in its
> midpoint.
>
> Sincerely Alexey
>
>
> Dear Alexey
> If I name Db the harmonic conjugate of D wrt BB" and Da the harmonic
> conjugate of D wrt AA", then the tangent at C to the circumconic locus of
> P
> is just the line Cab.
>
> 1° What is the locus of D in order to the locus of P be a rectangular
> hyperbola? : ( a line)
> 2° What is the locus of D in order to the locus of P be a parabola?: ( a
> hyperbola)
>
> Friendly
> Francois
>
> On Feb 5, 2008 9:59 AM, Francois Rideau <francois.rideau@...<francois.rideau%40gmail.com>>
> wrote:
>
> > Dear Alexey
> > Very interesting problem I had had to tackle when I was a young boy more
>
> > than fifty years ago.
> >
> > I change your notation a little bit to avoid subscript and I name D your
>
> > point C0.
> > I think the locus of P is the ABC-circumconic tangent at A to line AD
> and
> > tangent at B to line BD.
> > Let A"B"C" be the ABC-cevian triangle of D.
> > Then line B'C' is on the harmonic conjugate of D wrt BB"and line A'C' is
>
> > on the harmonic conjugate of D wrt AA".
> >
> > Friendly
> > Francois
> >
> > On Feb 5, 2008 8:29 AM, Alexey.A.Zaslavsky <zasl@...<zasl%40cemi.rssi.ru>>
> wrote:
> >
> > > Dear Paris and Francois!
> > > It seems that these problems were considered by Lev and Tatyana
> > > Emelyanoff. They presented the problems devoted to trypolars at Summer
>
> > > conference of Towns tournament-2005.
> > > Also in our with Arsenij Akopyan book next theorem is proved. Given
> the
> > > triangle ABC and the point C0. An arbitrary line passing through C0
> > > intersect AC and BC in B' and A' resp. P is the common point of AA'
> and BB'.
> > > Then
> > > 1. The locus of P is the circumconic touching CC0.
> > > 2. If CP intersect AB in C' then all lines A'C' have the common point
> B0
> > > and BB0 touches the circumconic.
> > >
> > > Sincerely Alexey
> > >
> > >
> > > Dear Paris
> > >
> > > This projective configuration of a conic through 3 points A, B, C and
> > > tangent at Q to a line (L) is very beautiful!
> > >
> > > I am sure there is a nice proof in the general case using Desargues
> and
> > > Plücker
> > > involution theorems known for a long time that we can find as an
> > > exercise in
> > > some old book.
> > >
> > > But the idea to send line (L) at infinity and to prove the theorem in
> > > this
> > > special case is very educational.
> > >
> > > Another (educational?) idea should be to send both points B and C at
> > > infinity that is to say looking at the hyperbola through point A with
> > > given
> > > asymptotic directions and tangent to a line (L) at Q. I am sure that
> > > elementary affine properties of hyperbola can also do the job.
> > >
> > > Friendly
> > > Francois
> > >
> > > [Non-text portions of this message have been removed]
> > >
> > >
> > >
> > >
> > > Antivirus scanning: Symantec Mail Security for SMTP.
> > >
> > > [Non-text portions of this message have been removed]
> > >
> > >
> > >
> >
> >
>
> [Non-text portions of this message have been removed]
>
>
>
>
> Antivirus scanning: Symantec Mail Security for SMTP.
>
> [Non-text portions of this message have been removed]
>
>
>

[Non-text portions of this message have been removed]
• Dear Alexey and Francois I was in leave last week and could not participate. I ll read your letters soon and think about. Thank you for your interest. Paris
Message 11 of 29 , Feb 9, 2008
Dear Alexey and Francois
I was in leave last week and could not participate.
Paris

-----Original Message-----
From: Hyacinthos@yahoogroups.com [mailto:Hyacinthos@yahoogroups.com] On
Behalf Of Francois Rideau
Sent: Wednesday, February 06, 2008 2:02 PM
To: Hyacinthos@yahoogroups.com
Subject: [SPAM] Re: [EMHL] Circumconics

Dear Alexey
Cheer up!
Thanks to your problem, it was true nostalgia for me!
Friendly
Francois

On Feb 6, 2008 11:07 AM, Alexey.A.Zaslavsky <HYPERLINK

> Dear Francois!
> 1 is the partial case of next general fact: if the pencil of conics
> determined by the points A, B, C, D is considered then the locus of
poles of
> line AB is the line. This line is the polar of common point of AB and
CD wrt
> any conic of the pencil. In our case it is the common point of AB and
the
> altitude CH.
> The locus 2 is the hyperbola with asymptotes CA and CB touching AB in
its
> midpoint.
>
> Sincerely Alexey
>
>
> Dear Alexey
> If I name Db the harmonic conjugate of D wrt BB" and Da the harmonic
> conjugate of D wrt AA", then the tangent at C to the circumconic locus
of
> P
> is just the line Cab.
>
> 1° What is the locus of D in order to the locus of P be a rectangular
> hyperbola? : ( a line)
> 2° What is the locus of D in order to the locus of P be a parabola?: (
a
> hyperbola)
>
> Friendly
> Francois
>
> On Feb 5, 2008 9:59 AM, Francois Rideau <HYPERLINK
"mailto:francois.rideau%40gmail.com"francois.rideau@...<francois.
rideau%-40gmail.com>->
> wrote:
>
> > Dear Alexey
> > Very interesting problem I had had to tackle when I was a young boy
more
>
> > than fifty years ago.
> >
> > I change your notation a little bit to avoid subscript and I name D
your
>
> > point C0.
> > I think the locus of P is the ABC-circumconic tangent at A to line
> and
> > tangent at B to line BD.
> > Let A"B"C" be the ABC-cevian triangle of D.
> > Then line B'C' is on the harmonic conjugate of D wrt BB"and line
A'C' is
>
> > on the harmonic conjugate of D wrt AA".
> >
> > Friendly
> > Francois
> >
> > On Feb 5, 2008 8:29 AM, Alexey.A.Zaslavsky <HYPERLINK
> wrote:
> >
> > > Dear Paris and Francois!
> > > It seems that these problems were considered by Lev and Tatyana
> > > Emelyanoff. They presented the problems devoted to trypolars at
Summer
>
> > > conference of Towns tournament-2005.
> > > Also in our with Arsenij Akopyan book next theorem is proved.
Given
> the
> > > triangle ABC and the point C0. An arbitrary line passing through
C0
> > > intersect AC and BC in B' and A' resp. P is the common point of
AA'
> and BB'.
> > > Then
> > > 1. The locus of P is the circumconic touching CC0.
> > > 2. If CP intersect AB in C' then all lines A'C' have the common
point
> B0
> > > and BB0 touches the circumconic.
> > >
> > > Sincerely Alexey
> > >
> > >
> > > Dear Paris
> > >
> > > This projective configuration of a conic through 3 points A, B, C
and
> > > tangent at Q to a line (L) is very beautiful!
> > >
> > > I am sure there is a nice proof in the general case using
Desargues
> and
> > > Plücker
> > > involution theorems known for a long time that we can find as an
> > > exercise in
> > > some old book.
> > >
> > > But the idea to send line (L) at infinity and to prove the theorem
in
> > > this
> > > special case is very educational.
> > >
> > > Another (educational?-) idea should be to send both points B and C
at
> > > infinity that is to say looking at the hyperbola through point A
with
> > > given
> > > asymptotic directions and tangent to a line (L) at Q. I am sure
that
> > > elementary affine properties of hyperbola can also do the job.
> > >
> > > Friendly
> > > Francois
> > >
> > > [Non-text portions of this message have been removed]
> > >
> > >
> > >
> > >
> > > Antivirus scanning: Symantec Mail Security for SMTP.
> > >
> > > [Non-text portions of this message have been removed]
> > >
> > >
> > >
> >
> >
>
> [Non-text portions of this message have been removed]
>
>
>
>
> Antivirus scanning: Symantec Mail Security for SMTP.
>
> [Non-text portions of this message have been removed]
>
>
>

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

Internal Virus Database is out-of-date.
Checked by AVG Free Edition.
Version: 7.5.516 / Virus Database: 269.19.5/1228 - Release Date:
16/1/2008 9:01 ðì

Internal Virus Database is out-of-date.
Checked by AVG Free Edition.
Version: 7.5.516 / Virus Database: 269.19.5/1228 - Release Date:
16/1/2008 9:01 ðì

[Non-text portions of this message have been removed]
• Dear Alexey, 1) Could I have some concrete reference on Lev and Tatyana s presentation? 2) The theorem you proved with Arsenij is the old and classical way of
Message 12 of 29 , Feb 14, 2008
Dear Alexey,
1) Could I have some concrete reference on Lev and Tatyana's
presentation?
2) The theorem you proved with Arsenij is the old and classical way of
"Maclaurin's generation of conics"
"http://www.math.uoc.gr/~pamfilos/eGallery/problems/Maclaurin.html"http:
//www.math.uoc.gr/~pamfilos/eGallery/problems/Maclaurin.html)
3) This gives of course a direct proof of the proposition I posted.
4) I prefer though the reduction to parabola and the stress of this
property of anticomplementary
and more general the precevian triangle. They clarify greatly I think
the structure of the
circumconics tangent to a given line, and especially the circumparabolas
of a triangle.
Sincerely Paris

-----Original Message-----
From: Hyacinthos@yahoogroups.com [mailto:Hyacinthos@yahoogroups.com] On
Behalf Of Alexey.A.Zaslavsky
Sent: Tuesday, February 05, 2008 7:29 AM
To: Hyacinthos@yahoogroups.com
Subject: [SPAM] Re: [EMHL] Circumconics

Dear Paris and Francois!
It seems that these problems were considered by Lev and Tatyana
Emelyanoff. They presented the problems devoted to trypolars at Summer
conference of Towns tournament-2005.
Also in our with Arsenij Akopyan book next theorem is proved. Given the
triangle ABC and the point C0. An arbitrary line passing through C0
intersect AC and BC in B' and A' resp. P is the common point of AA' and
BB'. Then
1. The locus of P is the circumconic touching CC0.
2. If CP intersect AB in C' then all lines A'C' have the common point B0
and BB0 touches the circumconic.

Sincerely Alexey

Dear Paris

This projective configuration of a conic through 3 points A, B, C and
tangent at Q to a line (L) is very beautiful!

I am sure there is a nice proof in the general case using Desargues and
Plücker
involution theorems known for a long time that we can find as an
exercise in
some old book.

But the idea to send line (L) at infinity and to prove the theorem in
this
special case is very educational.

Another (educational?-) idea should be to send both points B and C at
infinity that is to say looking at the hyperbola through point A with
given
asymptotic directions and tangent to a line (L) at Q. I am sure that
elementary affine properties of hyperbola can also do the job.

Friendly
Francois

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

Antivirus scanning: Symantec Mail Security for SMTP.

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

Internal Virus Database is out-of-date.
Checked by AVG Free Edition.
Version: 7.5.516 / Virus Database: 269.19.5/1228 - Release Date:
16/1/2008 9:01 ðì

No virus found in this outgoing message.
Checked by AVG Free Edition.
Version: 7.5.516 / Virus Database: 269.20.4/1276 - Release Date:
13/2/2008 9:41 ðì

[Non-text portions of this message have been removed]
• Dear Alexey and Paris As for the circumparabolas of a triangle ABC, it is also useful to transform them by isotomy. By isotomy wrt ABC, such a ABC-
Message 13 of 29 , Feb 15, 2008
Dear Alexey and Paris
As for the circumparabolas of a triangle ABC, it is also useful to
transform them by isotomy.
By isotomy wrt ABC, such a ABC- circumparabola (P) is transformed in a line
(L) tangent to the ABC-Steiner circumellipse (S).
For example to get the parabolas through 4 points A, B, C, D, you look at
the ABC-isotomic conjugate D' of D.
We construct lines (L1) and (L2) through D' tangent to the ellipse (S) and
we get the 2 parabolas (P1) and (P2) on A, B, C, D transformed by
ABC-isotomy from the lines (L1) and (L2).
Of course there is some discussion to do!
It is interesting to compare this affine solution with the original
Newton's euclidian construction.
Friendly
Francois

[Non-text portions of this message have been removed]
• Dear Paris! The Emelyanov s paper was published (in Russsian) in Matematicheskoe prosveschenije , 2002, N 6. May be it also was published in FG, but I amn t
Message 14 of 29 , Feb 17, 2008
Dear Paris!
The Emelyanov's paper was published (in Russsian) in "Matematicheskoe prosveschenije", 2002, N 6. May be it also was published in FG, but I amn't sure.

Sincerely Alexey

Dear Alexey,
1) Could I have some concrete reference on Lev and Tatyana's
presentation?
2) The theorem you proved with Arsenij is the old and classical way of
"Maclaurin's generation of conics"
"http://www.math.uoc.gr/~pamfilos/eGallery/problems/Maclaurin.html"http:
//www.math.uoc.gr/~pamfilos/eGallery/problems/Maclaurin.html)
3) This gives of course a direct proof of the proposition I posted.
4) I prefer though the reduction to parabola and the stress of this
property of anticomplementary
and more general the precevian triangle. They clarify greatly I think
the structure of the
circumconics tangent to a given line, and especially the circumparabolas
of a triangle.
Sincerely Paris

-----Original Message-----
From: Hyacinthos@yahoogroups.com [mailto:Hyacinthos@yahoogroups.com] On
Behalf Of Alexey.A.Zaslavsky
Sent: Tuesday, February 05, 2008 7:29 AM
To: Hyacinthos@yahoogroups.com
Subject: [SPAM] Re: [EMHL] Circumconics

Dear Paris and Francois!
It seems that these problems were considered by Lev and Tatyana
Emelyanoff. They presented the problems devoted to trypolars at Summer
conference of Towns tournament-2005.
Also in our with Arsenij Akopyan book next theorem is proved. Given the
triangle ABC and the point C0. An arbitrary line passing through C0
intersect AC and BC in B' and A' resp. P is the common point of AA' and
BB'. Then
1. The locus of P is the circumconic touching CC0.
2. If CP intersect AB in C' then all lines A'C' have the common point B0
and BB0 touches the circumconic.

Sincerely Alexey

Dear Paris

This projective configuration of a conic through 3 points A, B, C and
tangent at Q to a line (L) is very beautiful!

I am sure there is a nice proof in the general case using Desargues and
Plücker
involution theorems known for a long time that we can find as an
exercise in
some old book.

But the idea to send line (L) at infinity and to prove the theorem in
this
special case is very educational.

Another (educational?-) idea should be to send both points B and C at
infinity that is to say looking at the hyperbola through point A with
given
asymptotic directions and tangent to a line (L) at Q. I am sure that
elementary affine properties of hyperbola can also do the job.

Friendly
Francois

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

Antivirus scanning: Symantec Mail Security for SMTP.

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

Internal Virus Database is out-of-date.
Checked by AVG Free Edition.
Version: 7.5.516 / Virus Database: 269.19.5/1228 - Release Date:
16/1/2008 9:01 ðì

No virus found in this outgoing message.
Checked by AVG Free Edition.
Version: 7.5.516 / Virus Database: 269.20.4/1276 - Release Date:
13/2/2008 9:41 ðì

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

Antivirus scanning: Symantec Mail Security for SMTP.

[Non-text portions of this message have been removed]
• Dear Francois thank you for the nice remarks on the alternative description of circumparabolas. Dear Alexei thank you for the reference. Is Matematicheskoe
Message 15 of 29 , Feb 18, 2008
Dear Francois
thank you for the nice remarks on the alternative description
of circumparabolas.
Dear Alexei
thank you for the reference.
Is "Matematicheskoe prosveschenije" electronically accessible?
Do you know?
Best regards to all
Paris

No virus found in this outgoing message.
Checked by AVG Free Edition.
Version: 7.5.516 / Virus Database: 269.20.7/1286 - Release Date:
18/2/2008 6:49 ii

[Non-text portions of this message have been removed]
• Dear Paris! Yes in www.mccme.ru you can find the numbers of MP but I think that there is only russian text. Sincerely
Message 16 of 29 , Feb 19, 2008
Dear Paris!
Yes in www.mccme.ru you can find the numbers of "MP" but I think that there is only russian text.

Sincerely Alexey

Dear Francois
thank you for the nice remarks on the alternative description
of circumparabolas.
Dear Alexei
thank you for the reference.
Is "Matematicheskoe prosveschenije" electronically accessible?
Do you know?
Best regards to all
Paris

No virus found in this outgoing message.
Checked by AVG Free Edition.
Version: 7.5.516 / Virus Database: 269.20.7/1286 - Release Date:
18/2/2008 6:49 ii

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

Antivirus scanning: Symantec Mail Security for SMTP.

[Non-text portions of this message have been removed]
Your message has been successfully submitted and would be delivered to recipients shortly.