Of course barycentrics give proofs of theorem, although synthetic proofs are
better for all of us.

I this case it is convenient the use of computer algebra to make the
calculations easier. I use Mathematica and the package baricentricas.nb
available in my website.

By the way, if P is your point we have the relation

GP:PH = (2-1/(cosA cosB cosC))/3



2009/11/6 yakub.aliyev <yakub.aliyev@...>

>
>
>
>
> Dear friends Eric and Pierre,
> Please, inform me.
> 1. how you get this barycentric coordinates?
> 2. what do you think is it possible, using these coordinates, to prove that
> this point is on the Euler line?
> 3. Do you want to write a joint article about this problem?
> Regards, Yakub.
>
>
> --- In Hyacinthos@yahoogroups.com <Hyacinthos%40yahoogroups.com>,
> "efn4900" <ERIC.DANNEELS@...> wrote:
> >
> > Dear Yakub,
> >
> > your point is not in the ETC
> >
> > Its barycentric coordinates are
> > ((a^6+a^4(b^2+c^2)
> > -5a^2(b^-c^2)^2
> > +3(b^2+c^2)(b^2-c^2)^2)/(b^2+c^2-a^2)....
> >
> > It also lies on the lines X113-X193, X115-X393 and X253-X339
> >
> > Kind regards
> >
> > Eric
> >
> > --- In Hyacinthos@yahoogroups.com <Hyacinthos%40yahoogroups.com>,
> "yakub.aliyev" <yakub.aliyev@> wrote:
> > >
> > > Dear friends, let H be the orthocenter (intersection point of altitudes
> AH1, BH2, CH3), and and G be the centroid (intersection point of medians
> AM1, BM2, CM3). Draw the parallel to AC through H, which intersects BA and
> BC at C1 and A2, respectively. Analogously, draw the parallel through H to
> BA (and to BC), to find the points A1 and B2 (and B1 and C2). Extend the
> lines C1B2, A1C2, and B1A2 to form a new triangle XYZ (X is opposite to A, Y
> to B, Z to C). Denote midpoints of C1B2, A1C2, and B1A2 by T1, T2 and T3. It
> is known that H1, H2, H3, M1, M2, M3, T1, T2 and T3 are concyclic (Euler
> circle). Let this circle intersect C1B2, A1C2, and B1A2 again at the points
> S1, S2 and S3. I find that the lines XS1, YS2, and ZS3 intersect at a new
> point which is collinear with H, G, Oe (center of Euler cirle) (Euler line).
> Is this point a new triangle center? Is it in the Encyclopedia of
> Kimberling?
> > > Yakub.
> > >
> >
>
>
>



--
---
Francisco Javier Garc�a Capit�n
http://garciacapitan.auna.com


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

<http://groups.yahoo.com/group/Hyacinthos/message/18434;_ylc=X3oDMTJxdHFpbGU3BF9\
TAzk3MzU5NzE1BGdycElkAzM4NjU0MARncnBzcElkAzE3MDUwODMzODYEbXNnSWQDMTg0MzQEc2VjA2R\
tc2cEc2xrA3Ztc2cEc3RpbWUDMTI1NzUwODY3OA-->The
internet archive changed their interface earlier in the year and lots of
PDFs are not available.
If this is so, go to the Open Library and try there. I managed to get the
PDF, but they did not put the author in, so I had to use the Open library:

http://openlibrary.org/

This uses the Internet archive databsse so it is very odd the way the two
work.

another good place to find ebooks is the digital maths library:

http://www.mathematik.uni-bielefeld.de/~rehmann/DML/dml_links.html

although this one is not listed there
John S




Book - Wilhelm Fuhrmann
<http://groups.yahoo.com/group/Hyacinthos/message/18434;_ylc=X3oDMTJxdHFpbGU3BF9\
TAzk3MzU5NzE1BGdycElkAzM4NjU0MARncnBzcElkAzE3MDUwODMzODYEbXNnSWQDMTg0MzQEc2VjA2R\
tc2cEc2xrA3Ztc2cEc3RpbWUDMTI1NzUwODY3OA-->
Posted
by: "deocleciomota" deocleciomota@...
<deocleciomota@...?Subject=+Re%3ABook%20-%20Wilhelm%20Fuhrmann>
   deocleciomota
<http://profiles.yahoo.com/deocleciomota>  Thu Nov 5, 2009 8:18 am (PST)Dear
Friends,
I'm looking for the book Synthetische Beweise Planimetrischer S�tze � Autor:
Wilhelm Fuhrmann.
I wish this book was sent to me, in pdf, via e-mail.
The book is available on the site archive.org and google books, but I can
not download it here in Brazil.
Sincerely,
P.S. - Here in Brazil, it is in the public domain.
P.P.S. � My e-mail: deocleciomota@... <deocleciomota%40google.com>

Deoclecio Gouveia Mota Junior


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

Dear Hyacinthos
I send a file  in french with conway notations expleting the mode of calculus
  The name of the file is "New point on Euler Line"

Amically

Michel Garitte

--- In Hyacinthos@yahoogroups.com, "yakub.aliyev" <yakub.aliyev@...> wrote:
>
>
>
> Dear friends Eric and Pierre,
> Please, inform me.
> 1. how you get this barycentric coordinates?
> 2. what do you think is it possible, using these coordinates, to prove that
this point is on the Euler line?
> 3. Do you want to write a joint article about this problem?
> Regards, Yakub.
>
> --- In Hyacinthos@yahoogroups.com, "efn4900" <ERIC.DANNEELS@> wrote:
> >
> > Dear Yakub,
> >
> > your point is not in the ETC
> >
> > Its barycentric coordinates are
> > ((a^6+a^4(b^2+c^2)
> >     -5a^2(b^-c^2)^2
> >        +3(b^2+c^2)(b^2-c^2)^2)/(b^2+c^2-a^2)....
> >
> > It also lies on the lines X113-X193, X115-X393 and X253-X339
> >
> > Kind regards
> >
> > Eric
> >
> > --- In Hyacinthos@yahoogroups.com, "yakub.aliyev" <yakub.aliyev@> wrote:
> > >
> > > Dear friends, let H be the orthocenter (intersection point of altitudes
AH1, BH2, CH3), and and G be the centroid (intersection point of medians AM1,
BM2, CM3). Draw the parallel to AC through H, which intersects BA and BC at C1
and A2, respectively. Analogously, draw the parallel through H to BA (and to
BC), to find the points A1 and B2 (and B1 and C2). Extend the lines C1B2, A1C2,
and B1A2 to form a new triangle XYZ (X is opposite to A, Y to B, Z to C). Denote
midpoints of C1B2, A1C2,  and B1A2 by T1, T2 and T3. It is known that H1, H2,
H3, M1, M2, M3, T1, T2 and T3 are concyclic (Euler circle). Let this circle
intersect C1B2, A1C2, and B1A2 again at the points S1, S2 and S3. I find that
the lines XS1, YS2, and ZS3 intersect at a new point which is collinear with H,
G, Oe (center of Euler cirle) (Euler line).  Is this point a new triangle
center? Is it in the Encyclopedia of Kimberling?
> > > Yakub.
> > >
> >
>

Hello,

This email message is a notification to let you know that
a file has been uploaded to the Files area of the Hyacinthos
group.

   File        : /New point on Euler line.pdf
   Uploaded by : michgarl <michgarl@...>
   Description : New point of Euler Lne: recents messages

You can access this file at the URL:
http://groups.yahoo.com/group/Hyacinthos/files/New%20point%20on%20Euler%20line.p\
df

To learn more about file sharing for your group, please visit:
http://help.yahoo.com/l/us/yahoo/groups/original/general.htmlfiles

Regards,

michgarl <michgarl@...>

Hello,

This email message is a notification to let you know that
a file has been uploaded to the Files area of the Hyacinthos
group.

   File        : /anewpointeulerline.pdf
   Uploaded by : garciacapitan <garciacapitan@...>
   Description : Another "A new point on Euler line" calculations

You can access this file at the URL:
http://groups.yahoo.com/group/Hyacinthos/files/anewpointeulerline.pdf

To learn more about file sharing for your group, please visit:
http://help.yahoo.com/l/us/yahoo/groups/original/general.htmlfiles

Regards,

garciacapitan <garciacapitan@...>

I also have sent my calculations, with Mathematica and my package
baricentricas.nb

I have included the proof of the relation

GP:PH = (2-1/(cosA cosB cosC))/3

The file I have sent is

http://groups.yahoo.com/group/Hyacinthos/files/anewpointeulerline.pdf

Best regards,

Francisco Javier.


--- In Hyacinthos@yahoogroups.com, "Michel" <michgarl@...> wrote:
>
> Dear Hyacinthos
> I send a file  in french with conway notations expleting the mode of calculus
>  The name of the file is "New point on Euler Line"
>
> Amically
>
> Michel Garitte
>
> --- In Hyacinthos@yahoogroups.com, "yakub.aliyev" <yakub.aliyev@> wrote:
> >
> >
> >
> > Dear friends Eric and Pierre,
> > Please, inform me.
> > 1. how you get this barycentric coordinates?
> > 2. what do you think is it possible, using these coordinates, to prove that
this point is on the Euler line?
> > 3. Do you want to write a joint article about this problem?
> > Regards, Yakub.
> >
> > --- In Hyacinthos@yahoogroups.com, "efn4900" <ERIC.DANNEELS@> wrote:
> > >
> > > Dear Yakub,
> > >
> > > your point is not in the ETC
> > >
> > > Its barycentric coordinates are
> > > ((a^6+a^4(b^2+c^2)
> > >     -5a^2(b^-c^2)^2
> > >        +3(b^2+c^2)(b^2-c^2)^2)/(b^2+c^2-a^2)....
> > >
> > > It also lies on the lines X113-X193, X115-X393 and X253-X339
> > >
> > > Kind regards
> > >
> > > Eric
> > >
> > > --- In Hyacinthos@yahoogroups.com, "yakub.aliyev" <yakub.aliyev@> wrote:
> > > >
> > > > Dear friends, let H be the orthocenter (intersection point of altitudes
AH1, BH2, CH3), and and G be the centroid (intersection point of medians AM1,
BM2, CM3). Draw the parallel to AC through H, which intersects BA and BC at C1
and A2, respectively. Analogously, draw the parallel through H to BA (and to
BC), to find the points A1 and B2 (and B1 and C2). Extend the lines C1B2, A1C2,
and B1A2 to form a new triangle XYZ (X is opposite to A, Y to B, Z to C). Denote
midpoints of C1B2, A1C2,  and B1A2 by T1, T2 and T3. It is known that H1, H2,
H3, M1, M2, M3, T1, T2 and T3 are concyclic (Euler circle). Let this circle
intersect C1B2, A1C2, and B1A2 again at the points S1, S2 and S3. I find that
the lines XS1, YS2, and ZS3 intersect at a new point which is collinear with H,
G, Oe (center of Euler cirle) (Euler line).  Is this point a new triangle
center? Is it in the Encyclopedia of Kimberling?
> > > > Yakub.
> > > >
> > >
> >
>

Dear Yakub,
1 --> when using P= p:q:r instead of barycentrics of H, everything turns easier
(and a generalization --using a nine-points conic-- is obtained as a byproduct).
The details are given in #18421.
The key point is : no need for Conway symbols here since
even letters a,b,c aren't occuring in general formulae.

2 --> in fact, points G, P, Y (image of P) are EVER collinear, since
det(G,P,Y) = 0. Moreover, for all points at infinity Y=P and for all finite
points on K016, Y=G #18423.

3 --> OK. IMHO, remaining questions are :
a) geometrical examination of P = H   (are there special properties due to the
special nature of the conic, namely a circle).
b) special case P = I ??
c) explanation of the results obtained when P is on circumSteiner =CC(G)
c) explanation of the role played by K016 (leading to Y=G)
    and K016-3pqr  (leading to Y at infinity, i.e. parallel lines)
d) checking if any of the "simplest unnamed pairs" have any signifiance
e) counterparts when general cevian conics are used.

Best regards.

--- In Hyacinthos@yahoogroups.com, "yakub.aliyev" <yakub.aliyev@...> wrote:
>
> Dear friends Eric and Pierre,
> Please, inform me.
> 1. how you get this barycentric coordinates?
> 2. what do you think is it possible, using these coordinates, to prove that
this point is on the Euler line?
> 3. Do you want to write a joint article about this problem?
> Regards, Yakub.
>
> --- In Hyacinthos@yahoogroups.com, "efn4900" <ERIC.DANNEELS@> wrote:
> >
> > Dear Yakub,
> > your point is not in the ETC
> > Its barycentric coordinates are
> > ((a^6+a^4(b^2+c^2)
> >     -5a^2(b^-c^2)^2
> >        +3(b^2+c^2)(b^2-c^2)^2)/(b^2+c^2-a^2)....
> >
> > It also lies on the lines X113-X193, X115-X393 and X253-X339
> > Kind regards
> > Eric
> >
> > --- In Hyacinthos@yahoogroups.com, "yakub.aliyev" <yakub.aliyev@> wrote:
> > >
> > > Dear friends, let H be the orthocenter (intersection point of altitudes
AH1, BH2, CH3), and and G be the centroid (intersection point of medians AM1,
BM2, CM3). Draw the parallel to AC through H, which intersects BA and BC at C1
and A2, respectively. Analogously, draw the parallel through H to BA (and to
BC), to find the points A1 and B2 (and B1 and C2). Extend the lines C1B2, A1C2,
and B1A2 to form a new triangle XYZ (X is opposite to A, Y to B, Z to C). Denote
midpoints of C1B2, A1C2,  and B1A2 by T1, T2 and T3. It is known that H1, H2,
H3, M1, M2, M3, T1, T2 and T3 are concyclic (Euler circle). Let this circle
intersect C1B2, A1C2, and B1A2 again at the points S1, S2 and S3. I find that
the lines XS1, YS2, and ZS3 intersect at a new point which is collinear with H,
G, Oe (center of Euler cirle) (Euler line).  Is this point a new triangle
center? Is it in the Encyclopedia of Kimberling?
> > > Yakub.
> > >
> >
>

Dear Yakub,
Do you have a simple proof for this problem?
(I have an algebraic proof not simple,
and no other information)

Best regards
Nikos Dergiades

> Dear friends.
> Please inform me. Who can be the author of the following
> interesting geometric problem.
>
> Let ABCD be a convex quadrilateral. Construct a line
> through C intersecting extensions of AB and AD at M and K,
> respectively, such that (1/(Area(BCM)))+(1/(Area(DCK))) is
> minimal.
>
> The answer is MK||BD.
>
> This fact appeared in one russian journal as a problem of
> anonimous author:
>
> Problem 3393, Mathematics in School (in Russian), 6, 1989,
> 130; Solution: 4, 1990, 70.
>
> Is it possible that editor of geometry problems in this
> journal- Sharygin I.F. was author of this problem? But why
> then he didn't indicate his name?
> Do you familiar with this geometric fact? Have you any
> other sources for this problem?
>
> Y.N. Aliyev (Baku)





___________________________________________________________
�������������� Yahoo!;
���������� �� ���������� �������� (spam); �� Yahoo! Mail
�������� ��� �������� ������ ��������� ���� ��� �����������
��������� http://login.yahoo.com/config/mail?.intl=gr

Dear Nikos Dergiades, yes I have a pure geometric proof which is similar to the
solution of the problem in the problems section of journal Math. in School
(Russian). See

Problem 3393, Mathematics in School (in Russian), 6, 1989,
130; Solution: 4, 1990, 70.

See also my recent paper for the proof

http://f1.grp.yahoofs.com/v1/UPP4Str7mNq-f0xPwnCdMIZZeHHgSFYyUodHyc0qQjXLiGu9hMm\
VJE-NGz9LMROrkTfs40RJSUgcqugDVGo6rr3_CDv12B4bMw/20092516nev.pdf

where the first lemma is about this problem.
Yakub.



--- In Hyacinthos@yahoogroups.com, Nikolaos Dergiades <ndergiades@...> wrote:
>
> Dear Yakub,
> Do you have a simple proof for this problem?
> (I have an algebraic proof not simple,
> and no other information)
>
> Best regards
> Nikos Dergiades
>
> > Dear friends.
> > Please inform me. Who can be the author of the following
> > interesting geometric problem.
> >
> > Let ABCD be a convex quadrilateral. Construct a line
> > through C intersecting extensions of AB and AD at M and K,
> > respectively, such that (1/(Area(BCM)))+(1/(Area(DCK))) is
> > minimal.
> >
> > The answer is MK||BD.
> >
> > This fact appeared in one russian journal as a problem of
> > anonimous author:
> >
> > Problem 3393, Mathematics in School (in Russian), 6, 1989,
> > 130; Solution: 4, 1990, 70.
> >
> > Is it possible that editor of geometry problems in this
> > journal- Sharygin I.F. was author of this problem? But why
> > then he didn't indicate his name?
> > Do you familiar with this geometric fact? Have you any
> > other sources for this problem?
> >
> > Y.N. Aliyev (Baku)
>
>
>
>
>
> ___________________________________________________________
> �������������� Yahoo!;
> ���������� �� ���������� �������� (spam); �� Yahoo! Mail
> �������� ��� �������� ������ ��������� ���� ��� �����������
> ��������� http://login.yahoo.com/config/mail?.intl=gr
>

Dear Yakub,
thank you. But I have not access to this site.

Best regards
Nikos Dergiades

>
> Dear Nikos Dergiades, yes I have a pure geometric proof
> which is similar to the solution of the problem in the
> problems section of journal Math. in School (Russian). See
>
> Problem 3393, Mathematics in School (in Russian), 6, 1989,
> 130; Solution: 4, 1990, 70.
>
> See also my recent paper for the proof
>
>
http://f1.grp.yahoofs.com/v1/UPP4Str7mNq-f0xPwnCdMIZZeHHgSFYyUodHyc0qQjXLiGu9hMm\
VJE-NGz9LMROrkTfs40RJSUgcqugDVGo6rr3_CDv12B4bMw/20092516nev.pdf
>
> where the first lemma is about this problem.
> Yakub.
>
>
>
> --- In Hyacinthos@yahoogroups.com,
> Nikolaos Dergiades <ndergiades@...> wrote:
> >
> > Dear Yakub,
> > Do you have a simple proof for this problem?
> > (I have an algebraic proof not simple,
> > and no other information)
> >
> > Best regards
> > Nikos Dergiades
> >
> > > Dear friends.
> > > Please inform me. Who can be the author of the
> following
> > > interesting geometric problem.
> > >
> > > Let ABCD be a convex quadrilateral. Construct a
> line
> > > through C intersecting extensions of AB and AD at
> M and K,
> > > respectively, such that
> (1/(Area(BCM)))+(1/(Area(DCK))) is
> > > minimal.
> > >
> > > The answer is MK||BD.
> > >
> > > This fact appeared in one russian journal as a
> problem of
> > > anonimous author:
> > >
> > > Problem 3393, Mathematics in School (in Russian),
> 6, 1989,
> > > 130; Solution: 4, 1990, 70.
> > >
> > > Is it possible that editor of geometry problems
> in this
> > > journal- Sharygin I.F. was author of this
> problem? But why
> > > then he didn't indicate his name?
> > > Do you familiar with this geometric fact? Have
> you any
> > > other sources for this problem?
> > >
> > > Y.N. Aliyev (Baku)
> >
> >
> >
> >
> >       
> >
> ___________________________________________________________
>
> > ×ñçóéìïðïéåßôå Yahoo!;
> > ÂáñåèÞêáôå ôá åíï÷ëçôéêÜ
> ìçíýìáôá (spam); Ôï Yahoo! Mail
> > äéáèÝôåé ôçí êáëýôåñç äõíáôÞ
> ðñïóôáóßá êáôÜ ôùí åíï÷ëçôéêþí
> > ìçíõìÜôùí http://login.yahoo.com/config/mail?.intl=gr
> >
>
>
>
>
> ------------------------------------
>
> Yahoo! Groups Links
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
>



___________________________________________________________
Χρησιμοποιείτε Yahoo!;
Βαρεθήκατε τα ενοχλητικά μηνύματα (spam); Το
Yahoo! Mail
διαθέτει την καλύτερη δυνατή προστασία
κατά των ενοχλητικών
μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr

Dear Nikos, use this link:
http://www.qafqaz.edu.az/journal/20092516nev.pdf
See Lemma 2.1. Best regards, Yakub.

--- In Hyacinthos@yahoogroups.com, Nikolaos Dergiades <ndergiades@...> wrote:
>
> Dear Yakub,
> thank you. But I have not access to this site.
>
> Best regards
> Nikos Dergiades
>
> >
> > Dear Nikos Dergiades, yes I have a pure geometric proof
> > which is similar to the solution of the problem in the
> > problems section of journal Math. in School (Russian). See
> >
> > Problem 3393, Mathematics in School (in Russian), 6, 1989,
> > 130; Solution: 4, 1990, 70.
> >
> > See also my recent paper for the proof
> >
> >
http://f1.grp.yahoofs.com/v1/UPP4Str7mNq-f0xPwnCdMIZZeHHgSFYyUodHyc0qQjXLiGu9hMm\
VJE-NGz9LMROrkTfs40RJSUgcqugDVGo6rr3_CDv12B4bMw/20092516nev.pdf
> >
> > where the first lemma is about this problem.
> > Yakub.
> >
> >
> >
> > --- In Hyacinthos@yahoogroups.com,
> > Nikolaos Dergiades <ndergiades@> wrote:
> > >
> > > Dear Yakub,
> > > Do you have a simple proof for this problem?
> > > (I have an algebraic proof not simple,
> > > and no other information)
> > >
> > > Best regards
> > > Nikos Dergiades
> > >
> > > > Dear friends.
> > > > Please inform me. Who can be the author of the
> > following
> > > > interesting geometric problem.
> > > >
> > > > Let ABCD be a convex quadrilateral. Construct a
> > line
> > > > through C intersecting extensions of AB and AD at
> > M and K,
> > > > respectively, such that
> > (1/(Area(BCM)))+(1/(Area(DCK))) is
> > > > minimal.
> > > >
> > > > The answer is MK||BD.
> > > >
> > > > This fact appeared in one russian journal as a
> > problem of
> > > > anonimous author:
> > > >
> > > > Problem 3393, Mathematics in School (in Russian),
> > 6, 1989,
> > > > 130; Solution: 4, 1990, 70.
> > > >
> > > > Is it possible that editor of geometry problems
> > in this
> > > > journal- Sharygin I.F. was author of this
> > problem? But why
> > > > then he didn't indicate his name?
> > > > Do you familiar with this geometric fact? Have
> > you any
> > > > other sources for this problem?
> > > >
> > > > Y.N. Aliyev (Baku)
> > >
> > >
> > >
> > >
> > >       
> > >
> > ___________________________________________________________
> >
> > > ×ñçóéìïðïéåßôå Yahoo!;
> > > ÂáñåèÞêáôå ôá åíï÷ëçôéêÜ
> > ìçíýìáôá (spam); �"ï Yahoo! Mail
> > > äéáèÝôåé ôçí êáëýôåñç äõíáôÞ
> > ðñïóôáóßá êáôÜ ôùí åíï÷ëçôéêþí
> > > ìçíõìÜôùí http://login.yahoo.com/config/mail?.intl=gr
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Dear friends,

it is well known that the Feuerbach point, X(11) in the ETC, lies on the cevian
circle of the incenter.

But I didn't know that X(115) and X(3024) also lie on the cevian circle of the
incenter!
X(115) is the second intersection of the cevian circle of the incenter and the
ninepointcircle
X(3024) is the second intersection of the cevian circle of the incenter and the
incircle

Are there any references of this result?

Kind regards

Eric

Dear Eric,
[ED]
> it is well known that the Feuerbach point, X(11) in the ETC, lies on the
cevian circle of the incenter.
> But I didn't know that X(115) and X(3024) also lie on the cevian circle of
the incenter!
> X(115) is the second intersection of the cevian circle of the incenter and
the ninepointcircle
> X(3024) is the second intersection of the cevian circle of the incenter
and the incircle
> Are there any references of this result?



I don't have a reference to this special result.

But some years ago I did do some research on cevian circles.

Maybe it is of interest for you that:

.         X(11) is also on the cevian circles of  X(7), X(8), X(9)

.         X(115) is also on the cevian circles of  X(4), X(10), X(13),
X(14), X(254)

There are 2 other ETC-points that occur a lot on cevian circles:

.         X(122) occurs on the cevian circles of X(20), X(69), X(253),
X(1032)

.         X(125) occurs on the cevian circles of X(3), X(4), X(6), X(69),
X(253)

X(69) and X(253) have the same cevian circle (cyclocevian conjugates).

Best regards,

Chris van Tienhoven





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

Dear friends,

on the internet one can find several assertions concerning poristic triangles
but I didn't manage to find any references. Can someone help ?

1.  the Simson lines of a fixed point P on (O) wrt all poristic triangles XYZ
passes through a fixed point Q.

2. when P traverses (O), the locus of Q is an ellipse described by Jean-Pierre,
message 8302

3. the locus of the isogonal conjugates of a fixed point P wrt XYZ is a circle.

4. the locus of the Lemoine point of XYZ is an ellipse.

I have all the desired equations but I would like to identify my "predecessors".

Thanks

Best regards

Bernard

Dear Bernard

[BG]
> on the internet one can find several assertions concerning poristic triangles
but I didn't manage to find any references. Can someone help ?
>
> 1.  the Simson lines of a fixed point P on (O) wrt all poristic triangles XYZ
passes through a fixed point Q.

Prove that the Simson LInes of triangles with the same incircles
and circumcircles for a fixed point on the circumference are
concurrent.
Math. Gazette 1918

See a proof :

http://www.mathlinks.ro/viewtopic.php?search_id=754072119&t=30611

Antreas

---------- Forwarded message ----------
From: alexey_zaslavsky <alexey_zaslavsky@...>
Date: Thu, Nov 12, 2009 at 5:08 PM
Subject: Re: Poristic triangles
To: Hyacinthos-owner@yahoogroups.com

Dear Bernard!
>
> 3. the locus of the isogonal conjugates of a fixed point P wrt XYZ is a
circle.
>
This fact was discussed by Hiacinthos some years ago.

> 4. the locus of the Lemoine point of XYZ is an ellipse.
>
There is a paper (in Russian) in "Kvant" N 2, 2003 concerning this and
some similar facts. Also I can note that the locus of two Apollonius
points is the curve with degree 4 and similarly for two Torricelli
points.

Sincerely                          Alexey

Dear Bernard,
I have not the reference you want
but by the way of your question
I have something to notice that
perhaps you just know.
Since the NPC is tangent to incircle
the distance of the Nine point center N
of XYZ from I is d = R/2-r the locus of the N
is the circle (I, d).
The centroid G of XYZ is the homothetic of N
in the homothety (O, 2/3) and the locus of G
is a circle with radius (R-2r)/3.
The Orthocenter H of XYZ is the homothetic of G
in the homothety (O, 3) and the locus of H
is a circle with with center X1482 and radius R-2r
and not R/2-r as is written in the Greek edition
of F.G.-M. Theorem 346 - VII � 1185d.
(I don't know if the French edition is correct)

Best regards
Nikos Dergiades


> Dear friends,
>
> on the internet one can find several assertions concerning
> poristic triangles but I didn't manage to find any
> references. Can someone help ?
>
> 1.� the Simson lines of a fixed point P on (O) wrt all
> poristic triangles XYZ passes through a fixed point Q.
>
> 2. when P traverses (O), the locus of Q is an ellipse
> described by Jean-Pierre, message 8302
>
> 3. the locus of the isogonal conjugates of a fixed point P
> wrt XYZ is a circle.
>
> 4. the locus of the Lemoine point of XYZ is an ellipse.
>
> I have all the desired equations but I would like to
> identify my "predecessors".
>
> Thanks
>
> Best regards
>
> Bernard
>





___________________________________________________________
�������������� Yahoo!;
���������� �� ���������� �������� (spam); �� Yahoo! Mail
�������� ��� �������� ������ ��������� ���� ��� �����������
��������� http://login.yahoo.com/config/mail?.intl=gr

Dear Nikos

[ND]
> The Orthocenter H of XYZ is the homothetic of G
> in the homothety (O, 3) and the locus of H
> is a circle with with center X1482 and radius R-2r
> and not R/2-r as is written in the Greek edition
> of F.G.-M. Theorem 346 - VII � 1185d.
> (I don't know if the French edition is correct)

And in the french edition (in the 5th, that I have) is written
R/2 - r

APH

Dear Bernard I saw today your message. I am interested if your
results relate to X(264) discussed in a short note in my gallery
http://www.math.uoc.gr/~pamfilos/eGallery/problems/Poristic.html
best regards
Paris Pamfilos

On Thu, 12 Nov 2009, Bernard Gibert wrote:

> Dear friends,
>
> on the internet one can find several assertions concerning poristic triangles
but I didn't manage to find any references. Can someone help ?
>
> 1.  the Simson lines of a fixed point P on (O) wrt all poristic triangles XYZ
passes through a fixed point Q.
>
> 2. when P traverses (O), the locus of Q is an ellipse described by
Jean-Pierre, message 8302
>
> 3. the locus of the isogonal conjugates of a fixed point P wrt XYZ is a
circle.
>
> 4. the locus of the Lemoine point of XYZ is an ellipse.
>
> I have all the desired equations but I would like to identify my
"predecessors".
>
> Thanks
>
> Best regards
>
> Bernard
>
>

Dear Friends,
I have been doing some work on the second generation of the Apollonian structure
of the triangle which produces the mixtilinear incircles and the mixtilinear
excircles.  In all the references that I can find, three mixtilinear incircles
and only three mixtilinear excircles are discussed.  There are, in fact, nine
mixtilinear excircles, three associated with each excenter, just as there are
three mixtilinear incircles associated with the incenter.  Can anyone provide me
with a reference that discusses all nine mixtilinear excircles?  Thank you.
Regards,
Harold Connelly

---------- Forwarded message ----------
From: alexey_zaslavsky <alexey_zaslavsky@...>
Date: Fri, Nov 13, 2009 at 4:53 PM
Subject: Re: Poristic triangles
To: Hyacinthos-owner@yahoogroups.com


Dear Bernard!
>
> 1. �the Simson lines of a fixed point P on (O) wrt all poristic triangles XYZ
passes through a fixed point Q.
>
> 2. when P traverses (O), the locus of Q is an ellipse described by
Jean-Pierre, message 8302
>
It is more interesting to consider not Simson lines but the lines
passing through the reflections of P in the sidelines of triangles.
Then the locus of respective points Q is the circle. Its center is the
reflection of O in I, and its radius is equal to OI^2/R.

Sincerely � � � � � � � � � � � � � � �Alexey




--
http://anopolis72000.blogspot.com/

Dear Alexey

[AZ]
> It is more interesting to consider not Simson lines but the
> lines
> passing through the reflections of P in the sidelines of
> triangles.
> Then the locus of respective points Q is the circle. Its
> center is the
> reflection of O in I, and its radius is equal to OI^2/R.

OI^2/R = R - 2r and this locus is the same as the
locus of the orthocenter of XYZ.

Best regards
Nikos Dergiades




___________________________________________________________
�������������� Yahoo!;
���������� �� ���������� �������� (spam); �� Yahoo! Mail
�������� ��� �������� ������ ��������� ���� ��� �����������
��������� http://login.yahoo.com/config/mail?.intl=gr

The following paper has been published in Forum Geometricorum. It can be
viewed at

http://forumgeom.fau.edu/FG2009volume9/FG200927index.html

The editors
Forum Geometricorum
==============================================
John F. Goehl, Jr.,  Pythagorean triangles with square of perimeter equal
to an integer multiple of area,
Forum Geometricorum, 9 (2009) 281--282.

Abstract: We determine all Pythagorean triangles with square on perimeter
equal to an integer multiple of its area.

Hello Paul,.... pdf ...

does not work ..

http://forumgeom.fau.edu/FG2009volume9/FG200928.pdf



greetings

Ricardo






________________________________
De: ForumGeom <ForumGeom@...>
Enviado: lun,16 noviembre, 2009 16:11
Asunto: [EMHL] Forum Geometricorum


The following paper has been published in Forum Geometricorum. It can be
viewed at

http://forumgeom. fau.edu/FG2009vo lume9/FG200927in dex.html

The editors
Forum Geometricorum
============ ========= ========= ========= =======
John F. Goehl, Jr.,  Pythagorean triangles with square of perimeter equal
to an integer multiple of area,
Forum Geometricorum, 9 (2009) 281--282.

Abstract: We determine all Pythagorean triangles with square on perimeter
equal to an integer multiple of its area.







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

Dear friends,

Professor Stephen D Cohen from the University of Glasgow informs me that our
friend Wilson Stothers died in July this year.

I feel sorry since he was a good "cubic friend".

He wrote "Grassmann cubics and desmic structures, Forum Geometricorum, 6 (2006)
117--138."

Rest in peace, my friend.

Bernard



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

On Wed, Nov 18, 2009 at 6:05 PM, Bernard Gibert <bg42@...> wrote:

>
>
> Dear friends,
>
> Professor Stephen D Cohen from the University of Glasgow informs me that
> our friend Wilson Stothers died in July this year.
>
> I feel sorry since he was a good "cubic friend".
>
> He wrote "Grassmann cubics and desmic structures, Forum Geometricorum, 6
> (2006) 117--138."
>
> Rest in peace, my friend.
>
> Bernard
>
>
What sad news !

Wilson Stothers had put in his web site interesting Triangle Geometry
material.   .

http://www.maths.gla.ac.uk/~wws/cabripages/triangle/ninepoint/index.htm

I do not know how long will it be available, so a good idea is to download
these pages.

And from me,  rest in peace, Wilson.

Antreas


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

Dear friends,

in 2003 there were some messages about the "JENKINS" circles
starting with message nr 7292

Here is another variation of this configuration

Consider a triangle with its incircle (Ci) and its excircles (Ca), (Cb) and
(Cc). Now consider the circle tangent externally to (Ci), (Cb) and (Cc) and call
Ta its touchpoint with (Ci). Tb and Tc  are defined similarly using (Ci), (Cc)
and (Ca) respectively (Ci), (Ca) and (Cb).

1) Triangles ABC and TaTbTc are perspective

The perspector is not in the ETC.
The barycentric coordinates are:
( 1/(a^2(b+c-a)) : 1/(b^2(c+a-b)) : 1/(c^2(a+b-c)) )
It is the isotomic conjugate of X(55)

2) The triangle formed by the tangents to the incircle in Ta, Tb and Tc is
perspective with the intouch triangle
The perspector is not in the ETC.
The barycentric coordinates are:
( (a(b+c)-bc)/(a(b+c-a)) : (b(c+a)-ca)/(b(c+a-b))
       : (c(a+b)-ab)/(c(a+b-c)) )
It is the isotomic conjugate of X(2319)

Kind regards

Eric

Very sad new! I always remember Wilson Stothers' advices for me about Cevian
Nest theorem!
Rest in peace, Wilson Stothers!

Bui Quang Tuan

--- On Thu, 11/19/09, Antreas Hatzipolakis <anopolis72@...> wrote:

> From: Antreas Hatzipolakis <anopolis72@...>
> Subject: Re: [EMHL] Wilson Stothers has left us
> To: Hyacinthos@yahoogroups.com
> Date: Thursday, November 19, 2009, 2:33 AM
> On Wed, Nov 18, 2009 at 6:05 PM,
> Bernard Gibert <bg42@...>
> wrote:
>
> >
> >
> > Dear friends,
> >
> > Professor Stephen D Cohen from the University of
> Glasgow informs me that
> > our friend Wilson Stothers died in July this year.
> >
> > I feel sorry since he was a good "cubic friend".
> >
> > He wrote "Grassmann cubics and desmic structures,
> Forum Geometricorum, 6
> > (2006) 117--138."
> >
> > Rest in peace, my friend.
> >
> > Bernard
> >
> >
> What sad news !
>
> Wilson Stothers had put in his web site interesting
> Triangle Geometry
> material.���.
>
> http://www.maths.gla.ac.uk/~wws/cabripages/triangle/ninepoint/index.htm
>
> I do not know how long will it be available, so a good idea
> is to download
> these pages.
>
> And from me,� rest in peace, Wilson.
>
> Antreas
>
>
> [Non-text portions of this message have been removed]
>
>
>
> ------------------------------------
>
> Yahoo! Groups Links
>
>
> � � Hyacinthos-fullfeatured@yahoogroups.com
>
>
>

Dear friends. The well known controversial visual illusion effect that in
pictures the farthest column seems to be greater and closer column seems to be
smaller is related to the following geometrical problem:

Problem: Let k and m be parallel lines and w be a circle not intersecting with
k. Let A be a point on k. The two tangents from A to circle w intersect m at two
points B and C. Prove that the length of BC is lesser if point A is closer to
circle w.

Who knows the solution of this problem? Inform us.
Yakub.

Yakub,

I think there's a problem with the way you pose your question:  Don't you mean
the length of BC is "greater" if point A is closer to circle w?

Patrick



________________________________
From: yakub.aliyev <yakub.aliyev@...>
To: Hyacinthos@yahoogroups.com
Sent: Thu, November 26, 2009 4:33:54 AM
Subject: [EMHL] Greek Columns


Dear friends. The well known controversial visual illusion effect that in
pictures the farthest column seems to be greater and closer column seems to be
smaller is related to the following geometrical problem:

Problem: Let k and m be parallel lines and w be a circle not intersecting with
k. Let A be a point on k. The two tangents from A to circle w intersect m at two
points B and C. Prove that the length of BC is lesser if point A is closer to
circle w.

Who knows the solution of this problem? Inform us.
Yakub.







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