## focus of circumscribing parabolas

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• Dear Garcia, Nikolaos, and François, here is an observation that might lead to a more simple construction of the focal points of the two parabolas
Message 1 of 11 , Sep 22, 2007
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Dear Garcia, Nikolaos, and François,

here is an observation that might lead to a more simple construction of
the focal points of the two parabolas circumscribing a (convex)

Note that for every triangle self-polar w.r.t. a parabola the three
lines connecting the midpoints of its sides are tangent lines to the
parabola. This means that the focus of a parabola lies on the nine-point
circle of any triangle which is self-polar with respect to that
parabola. Since the diagonal triangle of any quadrilateral is
self-polar w.r.t. any conic section circumscribing the quadrilateral, it
follows that the focal points of both circumscribing parabolas are on
the nine-point circle of the diagonal triangle. Moreover, in addition to
the four points on the parabola, we now have four lines tangent to the
parabola (the three lines connecting the midpoints of the diagonal
triangle and the line at infinity). Perhaps it is possible to now find
the focal points as the points of intersection of this nine-point circle
with a straight line.

best,
Eisso

--

========================================
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@...
========================================
• Dear friends and Eisso and Garcia To put it right, I try to find some relationship between the perspector S and the focus F of an ABC- circumparabola . I
Message 2 of 11 , Sep 22, 2007
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Dear friends and Eisso and Garcia

To put it right, I try to find some relationship between the perspector S
and the focus F of an ABC- circumparabola <P>.

I don't find it but I notice some interesting facts certainly known for a
long time.

As you know it, the locus of S is the inner Steiner ABC- ellipse.

The locus of F looks like some cubic with lines parallel to the sidelines
on the centroid as 3 real asymptotes.

The envelope of line SF is some high degree curve(?)

The envelope of the axis of <P> looks very nice as the medial triangle
Steiner deltoid.

In other words , axis of the ABC-circumparabola are just the Simson lines
of the medial triangle.

I would like to have a proof or any reference of this last fact. Any idea?

Friendly

Francois

On 9/22/07, Eisso J. Atzema <atzema@...> wrote:
>
> Dear Garcia, Nikolaos, and François,
>
> here is an observation that might lead to a more simple construction of
> the focal points of the two parabolas circumscribing a (convex)
>
> Note that for every triangle self-polar w.r.t. a parabola the three
> lines connecting the midpoints of its sides are tangent lines to the
> parabola. This means that the focus of a parabola lies on the nine-point
> circle of any triangle which is self-polar with respect to that
> parabola. Since the diagonal triangle of any quadrilateral is
> self-polar w.r.t. any conic section circumscribing the quadrilateral, it
> follows that the focal points of both circumscribing parabolas are on
> the nine-point circle of the diagonal triangle. Moreover, in addition to
> the four points on the parabola, we now have four lines tangent to the
> parabola (the three lines connecting the midpoints of the diagonal
> triangle and the line at infinity). Perhaps it is possible to now find
> the focal points as the points of intersection of this nine-point circle
> with a straight line.
>
> best,
> Eisso
>
> --
>
> ========================================
> 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@... <atzema%40math.umaine.edu>
> ========================================
>
>
>

[Non-text portions of this message have been removed]
• Dear Eisso Very nice, you locating the focus on the nine point circle of the diagonal triangle of ABCD. The end of the construction is now clear: the focus is
Message 3 of 11 , Sep 23, 2007
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Dear Eisso
Very nice, you locating the focus on the nine point circle of the diagonal
triangle of ABCD.
The end of the construction is now clear: the focus is the isogonal
conjugate of the point at infinity of the axis wrt the medial triangle of
<T>, following some Poncelet theorem and as you notice it, this point at
infinity is also a point at infinity of the nine point conic of ABCD.
Now if you summarize this construction, all the problem is to construct the
points at infinity of this nine point conic of ABCD. Newton solved it ,
(read Dorie!), and I am eager to know your construction. As for me, I have
an affine construction of these points at infinity for this conic have an
affine definition..
Friendly
Francois
On 9/22/07, Eisso J. Atzema <atzema@...> wrote:
>
> Dear Garcia, Nikolaos, and François,
>
> here is an observation that might lead to a more simple construction of
> the focal points of the two parabolas circumscribing a (convex)
>
> Note that for every triangle self-polar w.r.t. a parabola the three
> lines connecting the midpoints of its sides are tangent lines to the
> parabola. This means that the focus of a parabola lies on the nine-point
> circle of any triangle which is self-polar with respect to that
> parabola. Since the diagonal triangle of any quadrilateral is
> self-polar w.r.t. any conic section circumscribing the quadrilateral, it
> follows that the focal points of both circumscribing parabolas are on
> the nine-point circle of the diagonal triangle. Moreover, in addition to
> the four points on the parabola, we now have four lines tangent to the
> parabola (the three lines connecting the midpoints of the diagonal
> triangle and the line at infinity). Perhaps it is possible to now find
> the focal points as the points of intersection of this nine-point circle
> with a straight line.
>
> best,
> Eisso
>
> --
>
> ========================================
> 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@... <atzema%40math.umaine.edu>
> ========================================
>
>
>

[Non-text portions of this message have been removed]
• Dear Francois, ... I think you are right as I can conclude from sketch. Let H be the orthocenter of ABC and A1B1C1 be the medial triangle of ABC. There exists
Message 4 of 11 , Sep 23, 2007
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Dear Francois,

> In other words , axis of the ABC-circumparabola are
> just the Simson lines
> of the medial triangle.
>
> I would like to have a proof or any reference of
> this last fact. Any idea?

I think you are right as I can conclude from sketch.
Let H be the orthocenter of ABC and
A1B1C1 be the medial triangle of ABC.
There exists a unique point D on the circumcircle of
ABC such that the isogonal conjugate of the tangent
at D to the circumcircle is the ABC-circumparabola.
Let P be the mid point of HD.
The axis of the ABC-circumparabola is the
Simson line of P wrt the triangle A1B1C1.

Best regards

___________________________________________________________
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• Dear Nikos Nice remark! I go back to Paris in a few day. Better to quit the cider country where I uttered so much blunders! Here I think to find some
Message 5 of 11 , Sep 24, 2007
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Dear Nikos
Nice remark!
I go back to Paris in a few day. Better to quit the cider country where I
uttered so much blunders!
Here I think to find some references on this axis circumparabola
configuration as we know that nothing is new in Triangle Geometry!
Now I ask you and everyboby if you know and what you think about the book:
Divine proportions
Rational trigonometry to Universal Geometry
written by NJ Wildberger
Friendly
Francois
>
> Dear Francois,
>
> > In other words , axis of the ABC-circumparabola are
> > just the Simson lines
> > of the medial triangle.
> >
> > I would like to have a proof or any reference of
> > this last fact. Any idea?
>
> I think you are right as I can conclude from sketch.
> Let H be the orthocenter of ABC and
> A1B1C1 be the medial triangle of ABC.
> There exists a unique point D on the circumcircle of
> ABC such that the isogonal conjugate of the tangent
> at D to the circumcircle is the ABC-circumparabola.
> Let P be the mid point of HD.
> The axis of the ABC-circumparabola is the
> Simson line of P wrt the triangle A1B1C1.
>
> Best regards
>
>
>
> ___________________________________________________________
> �������������� Yahoo!;
> ���������� �� ���������� �������� (spam); �� Yahoo! Mail
> �������� ��� �������� ������ ��������� ���� ��� �����������
>
>
>
>
>
>
>
>
>

[Non-text portions of this message have been removed]
• Dear Eisso, Nikos, Francisco Eventually this problem of the construction of the parabolas on 4 points A, B, C, D is fascinating and you learn a lot of geometry
Message 6 of 11 , Sep 25, 2007
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Dear Eisso, Nikos, Francisco

Eventually this problem of the construction of the parabolas on 4 points A,
B, C, D is fascinating and you learn a lot of geometry solving it.

Cheer up for your using the diagonal triangle T of the 4 points A, B, C, D
after noticing it was self conjugate wrt the parabola.

I just want to make another remark.

As you told us, the medial triangle of T is tangent to the parabolas, so
both foci are on its circumcircle that is to say the nine point circle of T
but orthocenter of the medial triangle that is to say the circumcenter of T
is on both directrix. (Faure Theorem)

So both directrix are also easy to draw.

As I have already said it, the big problem in this construction is to find
the directions of the axis of the parabolas and it was Newton who solved it
first without knowing any Desargues theorem and any line at infinity.

I can elaborate here in this short post but I advise everybody to read Dorie

There is a modern way to look at this Newton theorem. In fact he proves
without noticing it of course that a parabola has a structure of an affine
line.

Francisco sends me a very interesting paper on this topic, that is to say
another « solution » of this construction , named Church's problem as if
nobody on the Island knew that Isaac had solved it a long time ago!

I nearly got a headache reading this paper and in fact I gave up in the
middle for I was not able to read the drawing it.

But the idea suggested by the writer, a Mr Hyde, is interesting.

The parabola problem takes place in some plane <P>, so Dr Jekill, sorry, Mr
Hyde, makes a conic projection from the plane <P> onto another plane <P'>.
So he solves a new problem that he finds easier in <P'>, ( and maybe he is
the only one to think it this way), and he projects back the solution in
<P>.

What a nightmare and in fact he also uses a making drawing ( epure in
French) to do the job!

Clearly Mr Hyde fears nothing nor nobody, even projective geometry!

In fact this parabola problem is only a special case of the following
problem in projective geometry that is to say construct the conics on 4
points A, B, C, D and tangent to a line L.

Every time you choose an affine chart of the projective plane, that is to
say that you choose some line of the projective plane as line at infinity,
then you get a new problem in the affine plane, image of the chart.

I think that Hyde finds easier to choose line CD as a new line at infinity.

So he get the new problem, projectively isomorphic to the old:

To construct hyperbolas on 2 points A and B, knowing the directions of the
asymptots and also tangent to some line L.

I think this problem is easier for you can take a frame in the affine plane
with both axis parallel to the known directions of the asymptots. In this
frame, hyperbolas are homographic graphs.

I let GSP users to find a metric solution and Cabri users to find an affine
solution but please, don't send me any making drawing!

I would like to have some discussion following the position of the tangent
line in the affine plane.

Friendly

Francois

[Non-text portions of this message have been removed]
• Dear François, Nikos and Francisco, I d be interested in reading Church s paper. Could anyone mail this paper to me as well? Section V of Book I of Newton s
Message 7 of 11 , Sep 27, 2007
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Dear François, Nikos and Francisco,

I'd be interested in reading Church's paper. Could anyone mail this
paper to me as well? Section V of Book I of Newton's Principia (the part
on fitting conic sections) has been lying on my desk for over a year
now, but I still have not had a change to look at all of it. I did look
at a few theorems in Section V and it is amazing work -- especially
since this Section V predates the rest of the Principia by almost a
decade, at which time Newton could not really be considered particularly
well-read in mathematics (in fact, he never really was). My own proof of
Newton's construction is a few lines, but it uses a lemma that requires
proof. The proof of Newton's construction relies heavily on the dual of
version of Steiner's Theorem (sometimes referred to as Chasles'
Theorem), which certainly was known to Halley (even though he did not
grasp the potential of this tool) and (implicitly) to Newton as well (he
seems to use something very similar in a few places in Section V, if I
remember correctly).

The lemma that I need is that for any quadrilateral ABCD with diagonal
points E=AC\capBD, F=AD\capBC, G=AB\capCD, the line connecting the
midpoints of EF and EG is a tangent line to the exparabola of ABCD.
Moreover, the lines connecting the points of intersection of this lines
with the sides of ABCD with E are each parallel to the side opposite the
side they are on. From here the proof of Newton's result is
straightforward. I uploaded a GSP drawing with a sketch of proof (or
sketchy proof if you will) of Newton's construction (the file is called
parabola.gsp). I leave the proof of the lemma to you. Judging from my
own proof, it seems to require the dual of Steiner's Theorem, as well as

best,
Eisso

Francois Rideau wrote:
>
> Dear Eisso, Nikos, Francisco
>
> Eventually this problem of the construction of the parabolas on 4
> points A,
> B, C, D is fascinating and you learn a lot of geometry solving it.
>
> Cheer up for your using the diagonal triangle T of the 4 points A, B, C, D
> after noticing it was self conjugate wrt the parabola.
>
> I just want to make another remark.
>
> As you told us, the medial triangle of T is tangent to the parabolas, so
> both foci are on its circumcircle that is to say the nine point circle
> of T
> but orthocenter of the medial triangle that is to say the circumcenter
> of T
> is on both directrix. (Faure Theorem)
>
> So both directrix are also easy to draw.
>
> As I have already said it, the big problem in this construction is to find
> the directions of the axis of the parabolas and it was Newton who
> solved it
> first without knowing any Desargues theorem and any line at infinity.
>
> I can elaborate here in this short post but I advise everybody to read
> Dorie
>
> There is a modern way to look at this Newton theorem. In fact he proves
> without noticing it of course that a parabola has a structure of an affine
> line.
>
> Francisco sends me a very interesting paper on this topic, that is to say
> another « solution » of this construction , named Church's problem as if
> nobody on the Island knew that Isaac had solved it a long time ago!
>
> I nearly got a headache reading this paper and in fact I gave up in the
> middle for I was not able to read the drawing it.
>
> But the idea suggested by the writer, a Mr Hyde, is interesting.
>
> The parabola problem takes place in some plane <P>, so Dr Jekill,
> sorry, Mr
> Hyde, makes a conic projection from the plane <P> onto another plane <P'>.
> So he solves a new problem that he finds easier in <P'>, ( and maybe he is
> the only one to think it this way), and he projects back the solution in
> <P>.
>
> What a nightmare and in fact he also uses a making drawing ( epure in
> French) to do the job!
>
> Clearly Mr Hyde fears nothing nor nobody, even projective geometry!
>
> In fact this parabola problem is only a special case of the following
> problem in projective geometry that is to say construct the conics on 4
> points A, B, C, D and tangent to a line L.
>
> Every time you choose an affine chart of the projective plane, that is to
> say that you choose some line of the projective plane as line at infinity,
> then you get a new problem in the affine plane, image of the chart.
>
> I think that Hyde finds easier to choose line CD as a new line at
> infinity.
>
> So he get the new problem, projectively isomorphic to the old:
>
> To construct hyperbolas on 2 points A and B, knowing the directions of the
> asymptots and also tangent to some line L.
>
> I think this problem is easier for you can take a frame in the affine
> plane
> with both axis parallel to the known directions of the asymptots. In this
> frame, hyperbolas are homographic graphs.
>
> I let GSP users to find a metric solution and Cabri users to find an
> affine
> solution but please, don't send me any making drawing!
>
> I would like to have some discussion following the position of the tangent
> line in the affine plane.
>
> Friendly
>
> Francois
>
> [Non-text portions of this message have been removed]
>
>

--

========================================
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@...
========================================
• Dear Francois, ... I find Epure (I don t know the plural in French) especially helpful, don t you? Sincerely, Jeff
Message 8 of 11 , Sep 28, 2007
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Dear Francois,

> What a nightmare and in fact he also uses a making drawing ( epure in
> French) to do the job!

I find 'Epure (I don't know the plural in French) especially helpful,
don't you?

Sincerely, Jeff
• Dear François, Nikos and Francisco, I tried to post this message earlier, but never showed up. I d be interested in reading Church s paper. Could anyone mail
Message 9 of 11 , Sep 28, 2007
• 0 Attachment
Dear François, Nikos and Francisco,

I tried to post this message earlier, but never showed up. I'd be
interested in reading Church's paper. Could anyone mail this
paper to me as well? Section V of Book I of Newton's Principia (the part
on fitting conic sections) has been lying on my desk for over a year
now, but I still have not had a change to look at all of it. I did look
at a few theorems in Section V and it is amazing work -- especially
since this Section V predates the rest of the Principia by almost a
decade, at which time Newton could not really be considered particularly
well-read in mathematics (in fact, he never really was). My own proof of
Newton's construction is a few lines, but it uses a lemma that requires
proof. The proof of Newton's construction relies heavily on the dual of
version of Steiner's Theorem (sometimes referred to as Chasles'
Theorem), which certainly was known to Halley (even though he did not
grasp the potential of this tool) and (implicitly) to Newton as well (he
seems to use something very similar in a few places in Section V, if I
remember correctly).

The lemma that I need is that for any quadrilateral ABCD with diagonal
points E=AC\capBD, F=AD\capBC, G=AB\capCD, the line connecting the
midpoints of EF and EG is a tangent line to the exparabola of ABCD.
Moreover, the lines connecting the points of intersection of this lines
with the sides of ABCD with E are each parallel to the side opposite the
side they are on. From here the proof of Newton's result is
straightforward. I uploaded a GSP drawing with a sketch of proof (or
sketchy proof if you will) of Newton's construction (the file is called
parabola.gsp). I leave the proof of the lemma to you. Judging from my
own proof, it seems to require the dual of Steiner's Theorem, as well as

best,
Eisso

Francois Rideau wrote:
>
> Dear Eisso, Nikos, Francisco
>
> Eventually this problem of the construction of the parabolas on 4
> points A,
> B, C, D is fascinating and you learn a lot of geometry solving it.
>
> Cheer up for your using the diagonal triangle T of the 4 points A, B, C, D
> after noticing it was self conjugate wrt the parabola.
>
> I just want to make another remark.
>
> As you told us, the medial triangle of T is tangent to the parabolas, so
> both foci are on its circumcircle that is to say the nine point circle
> of T
> but orthocenter of the medial triangle that is to say the circumcenter
> of T
> is on both directrix. (Faure Theorem)
>
> So both directrix are also easy to draw.
>
> As I have already said it, the big problem in this construction is to find
> the directions of the axis of the parabolas and it was Newton who
> solved it
> first without knowing any Desargues theorem and any line at infinity.
>
> I can elaborate here in this short post but I advise everybody to read
> Dorie
>
> There is a modern way to look at this Newton theorem. In fact he proves
> without noticing it of course that a parabola has a structure of an affine
> line.
>
> Francisco sends me a very interesting paper on this topic, that is to say
> another « solution » of this construction , named Church's problem as if
> nobody on the Island knew that Isaac had solved it a long time ago!
>
> I nearly got a headache reading this paper and in fact I gave up in the
> middle for I was not able to read the drawing it.
>
> But the idea suggested by the writer, a Mr Hyde, is interesting.
>
> The parabola problem takes place in some plane <P>, so Dr Jekill,
> sorry, Mr
> Hyde, makes a conic projection from the plane <P> onto another plane <P'>.
> So he solves a new problem that he finds easier in <P'>, ( and maybe he is
> the only one to think it this way), and he projects back the solution in
> <P>.
>
> What a nightmare and in fact he also uses a making drawing ( epure in
> French) to do the job!
>
> Clearly Mr Hyde fears nothing nor nobody, even projective geometry!
>
> In fact this parabola problem is only a special case of the following
> problem in projective geometry that is to say construct the conics on 4
> points A, B, C, D and tangent to a line L.
>
> Every time you choose an affine chart of the projective plane, that is to
> say that you choose some line of the projective plane as line at infinity,
> then you get a new problem in the affine plane, image of the chart.
>
> I think that Hyde finds easier to choose line CD as a new line at
> infinity.
>
> So he get the new problem, projectively isomorphic to the old:
>
> To construct hyperbolas on 2 points A and B, knowing the directions of the
> asymptots and also tangent to some line L.
>
> I think this problem is easier for you can take a frame in the affine
> plane
> with both axis parallel to the known directions of the asymptots. In this
> frame, hyperbolas are homographic graphs.
>
> I let GSP users to find a metric solution and Cabri users to find an
> affine
> solution but please, don't send me any making drawing!
>
> I would like to have some discussion following the position of the tangent
> line in the affine plane.
>
> Friendly
>
> Francois
>
> [Non-text portions of this message have been removed]
>
>

--

========================================
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@...
========================================
• ... I don t see exactly what you mean here.Can you be more explicit? ... As for the sides of the medial triangle of the diagonal triangle of ABCD are ... . ...
Message 10 of 11 , Oct 1, 2007
• 0 Attachment
>
> Dear Eisso
>
> Moreover, the lines connecting the points of intersection of this lines
> with the sides of ABCD with E are each parallel to the side opposite the
> side they are on.
>

I don't see exactly what you mean here.Can you be more explicit?
>

As for the sides of the medial triangle of the diagonal triangle of ABCD are
> tangent to the parabola, it is just a consequence of the fact that for a
> point M having polar line L wrt a parabola <P>, the line homothetic of L in
> the dilation of center M and ratio 1/2 is tangent to the parabola.
>

.
> Friendly
>

Francois

>

[Non-text portions of this message have been removed]
• Dear François, I hoped the drawing that I uploaded would clear that up. Anyway, what I mean is that if the line through the midpoints of EFG intersects AD in
Message 11 of 11 , Oct 1, 2007
• 0 Attachment
Dear François,

I hoped the drawing that I uploaded would clear that up. Anyway, what I
mean is that if the line through the midpoints of EFG intersects AD in
F*, BC in F**, AB in G* and CD in G** then EF* is parallel to BC, EF**
is parallel to AD, EG* is parallel to CD and G** is parallel to AB.

Eisso

Francois Rideau wrote:
>
> >
> > Dear Eisso
> >
> > Moreover, the lines connecting the points of intersection of this lines
> > with the sides of ABCD with E are each parallel to the side opposite the
> > side they are on.
> >
>
> I don't see exactly what you mean here.Can you be more explicit?
> >
>
> As for the sides of the medial triangle of the diagonal triangle of
> ABCD are
> > tangent to the parabola, it is just a consequence of the fact that for a
> > point M having polar line L wrt a parabola <P>, the line homothetic
> of L in
> > the dilation of center M and ratio 1/2 is tangent to the parabola.
> >
>
> .
> > Friendly
> >
>
> Francois
>
> >
>
> [Non-text portions of this message have been removed]
>
>

--

========================================
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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