Dear Jean-Pierre,

Thank you very much.

> > many thanks for your nice mails about this
> problem.
> > You've probably noticed that when the angle of the
> Simson lines is
> > Pi/3, we get a nice trifolium inscribed in the
> circle(N,R) N = NP-
> center
> > With N as pole and a tangent at a cusp of the
> Steiner deltoid as
> polar
> > axis, the trifolium has polar equation rho =
> R.cos(3.theta)
>
> Some remarks about this trifolium :
> The vertices - contact points with the circle(N,R) -
> are the vertices
> of the equilateral triangle bounded by the Simson
> lines of the
> vertices of the circumnormal triangle; these Simson
> lines are the
> common tangents of the Steiner deltoid and the
> NP-circle at the
> points where they touch each other.

*********
These points are found by drawing from N parallells
to the bisectors of Morley's triangle.

> These vertices
> are homothetic of
> the cusps of the Steiner deltoid in (N,2/3)
> The rectangular circumhyperbola through N intersects
> the circle (N,R)
> at 4 points; one of them is the antipode of N on the
> hyperbola; the
> three other ones are the vertices above.

********
Very interesting.

Happy New Year to you and all Hyacinthists.
Best regards
Nikos Dergiades.








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Dear Nikolaos
the locus of the common points of two Simson lines with a given angle
touch the Steiner deltoid at six real points (and at the circular
points at infinity). As both curves are bicircular quartics, they have
no other common point.
The 6 points lie 3 by 3 on two circles centered at N. In the case of
the trifolium the 6 points lie on the circle (N, R.root(3)/2)
Friendly. Jean-Pierre

Dear Jean-Pierre,

> the locus of the common points of two Simson lines
> with a given angle
> touch the Steiner deltoid at six real points (and at
> the circular
> points at infinity). As both curves are bicircular
> quartics, they have
> no other common point.
> The 6 points lie 3 by 3 on two circles centered at
> N. In the case of
> the trifolium the 6 points lie on the circle (N,
> R.root(3)/2)

Wonderful figure!
I didn't thought to sketch the Steiner deltoid.
I think this locus must be called anadeltoid.
If A1,B1,C1 are the points for which the Simson
lines pass through N
(I think that angle(OA,OA1)=2(C-B)/3 )
and we rotate about O
by angles -w/3 and -2w/3 the triangle A1B1C1 we
get six points A2,B2,C2 and A3,B3,C3.
If T is the intersection of the Simson lines
S(P), S(Q) where angle (OP, OQ)=w then I think
when P is on the points A2,A3,B2,B3,C2,C3 we get
from T the 6 points of tangency of this
locus (anadeltoid) with the Steiner deltoid.
So I think that the two circles centered at N
that pass through the 6 points is one circle
with radius R.sqrt(5+4.cos(w))/2.

Best regards
Nikos Dergiades



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Dear Nikolaos
[JPE]
> > the locus of the common points of two Simson lines
> > with a given angle
> > touch the Steiner deltoid at six real points (and at
> > the circular
> > points at infinity). As both curves are bicircular
> > quartics, they have
> > no other common point.
> > The 6 points lie 3 by 3 on two circles centered at
> > N. In the case of
> > the trifolium the 6 points lie on the circle (N,
> > R.root(3)/2)

[ND]
> Wonderful figure!
> I didn't thought to sketch the Steiner deltoid.
> I think this locus must be called anadeltoid.
> If A1,B1,C1 are the points for which the Simson
> lines pass through N
> (I think that angle(OA,OA1)=2(C-B)/3 )
> and we rotate about O
> by angles -w/3 and -2w/3 the triangle A1B1C1 we
> get six points A2,B2,C2 and A3,B3,C3.
> If T is the intersection of the Simson lines
> S(P), S(Q) where angle (OP, OQ)=w then I think
> when P is on the points A2,A3,B2,B3,C2,C3 we get
> from T the 6 points of tangency of this
> locus (anadeltoid) with the Steiner deltoid.
> So I think that the two circles centered at N
> that pass through the 6 points is one circle
> with radius R.sqrt(5+4.cos(w))/2.

You are right.
I didn't notice that, in any case, the 6 points lie on the same
circle centered at N and, of course, I agree with your value of the
radius of this circle.
Many thanks.
Friendly. Jean-Pierre

First Announcement

University of Bucharest and Transilvania University of Brasov

invite you to attend the conference

Riemannian Geometry and Applications

Dedicated to the memory of Prof. Dr. Radu Rosca
Brasov, June 21-26, 2007

Topics:

-- Geometry of Riemannian and Pseudo-Riemannian Manifolds
-- Submanifold Theory
-- Structures on Manifolds
-- Complex Geometry
-- Finsler, Lagrange and Hamilton Geometries
-- Applications to other fields

Organizing and Scientific Committee:

Prof. Dr. Ion Mihai (University of Bucharest)
Acad. Prof. Dr. Radu Miron (Romanian Academy)
Prof. Dr. Leopold Verstraelen (Katholieke Universiteit Leuven)
Prof. Dr. Stere Ianus (University of Bucharest)
Prof. Dr. Emil Stoica (Transilvania University of Brasov)
Assist. Prof. Dr. Bogdan Suceava (California State University at Fullerton)

Secretary:

Prof. Dr. Gheorghe Munteanu (Transilvania University of Brasov)
Dr. Adela Mihai (University of Bucharest)
Drd. Valentin Ghisoiu (University of Bucharest)


Confirmed Invited Speakers:

Prof. Dr. David Blair (Michigan State University)
Prof. Dr. Klaus Buchner (Technical University Munich)
Prof. Dr. Alfonso Carriazo (Universidad de Sevilla)
Prof. Dr. Bang-Yen Chen (Michigan State University)
Prof. Dr. Filip Defever (Katholieke Hogeschool Brugge-Oostende)
Prof. Dr. Franki Dillen (Katholieke Universiteit Leuven)
Prof. Dr. Zhiqin Lu (University of California at Irvine)
Prof. Dr. Koji Matsumoto (Yamagata University)
Prof. Dr. Jean Marie Morvan (Universite Claude Bernard Lyon 1)
Prof. Dr. Seiki Nishikawa (Tohoku University)
Prof. Dr. Bernard Rouxel (Universite de Bretagne Occidentale)
Prof. Dr. Yoshihiko Tazawa (Tokyo Denki University)

Remarks:

There is no registration fee.

Those who are interested in attending the conference and/or presenting a
communication are kindly
invited to send an e-mail to Prof. Dr. Ion Mihai at   imihai@...

Second announcement will be posted in March 2007. The conference is partially
supported by the
Grant CEEX-M3 no. 252/2006 of the Romanian Ministry of Education and Research.
======
Posted for hyacinthos by
Bogdan Suceava

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Dear Nikolaos and Francois
Consider the following facts :
1) if M,M' are the projections of O (or any other point) upon two
parallel lines with respective orthopoles P, P' then Vect(PP') = Vect
(MM')
2) the common points of the Simson lines of M,M' such that <MOM' = w
are the orthopoles of the lines tangent to the circle (O, R cos(w/2))
3) as limit case of 2), the orthopole of the tangent at M to the
circumcircle is the point where the Simson line of M touches the
Steiner deltoid
If we consider a system of rectangular coordinates with N = (0,0) and
a Simson line through N as x-axis, the Steiner deltoid is parametric
x = R cos(t) + R/2 cos(2t)
y = R sin(t) -R/2 sin(2t)
where Vect(OM) = [R cos(t), R sin(t)] or, equivalently [R/2 cos(t),
R/2 sin(t)] is the midpoint of HM
It follows immediately from 1-2-3 that the locus of the orthopoles of
the lines tangent to the circle (O, k.R) is parametric
x = k.R cos(t) + R/2 cos(2t)
y = k.R sin(t) - R/2 sin(2t)
With k = 0, we get the NP-circle
k = 1, the Steiner deltoid
k = 1/2, the trifolium mentioned in a previous mail
k = cos(w/2), the required locus
From this we get easily the properties of this locus (sigularities,
extrema of MP, ...) and the fact that, for k = R.cos(w/2), the locus
touches the deltoid at 6 real points, all of them on the Nikolaos-
circle :
center N, radius R/2 root(5 + 4.cos(w))
Friendly. Jean-Pierre

Dear Hyacinthists,

Dear Nikolaos,

Thank you for your reply. With it I was able to prove
the relation in the german book.

Recalling the problem, I want to construct the triangle
ABC given <a,h_a,d_a>. As the data determine e_a,
the external bisector, we have <a,h_a,d_a,e_a>.

The notation A, H_a, M_a, R, and O is standard. D_a
and E_a are the feet of the internal and external
bisector on the side a. M is the medium point of (D_a,E_a).
(A,D_a) meets the circumcircle at D and (E_a,A)
meets the circumcircle at E.

--------|-------------|-----|--------|---------|-----------
       E_a          M    H_a    D_a     M_a

One wants to show that if  \ell_1=D_aD , then
\ell_1(\ell_1 + p) = q^2
where  p = \frac{d_a^3}{d_a^2-h_a^2} and
q^2 = \frac{a^2 d_a^2}{4(d_a^2-h_a^2)} .

This is part of exercice 148 in a book that I am writing/editing.
Sorry for the heavy LaTeX code that follows. However for those
interested in reproducing the proof and that use LaTeX it will be
useful. For those who just want the .pdf file I can send it in a
private email.
And the little text in portuguese shouldn't pose any difficulty
for this group's readers.

In the process I showed that R is equal to

R=\frac{d^4_{_{\!a}}-2d^2_{_{\!a}}h^2_{_{\!a}}+d^2_{_{\!a}}\sqrt{a^2(d^2_{_{\!a}\
}-
h^2_{_{\!a}})+d^4_{_{\!a}}}}{4h_{_{\!a}}(d^2_{_{\!a}}-h^2_{_{\!a}})}

The key idea in the construction/proof is the result
MM_a^2=(a/2)^2 + MD_a^2 as <B,C> and <D_a,E_a>
are two harmonic pairs.

Similarly one can calculate  \ell_2=E_aE.

Well, the LaTeX code is both ugly and lengthy.
I would rather send the pdf file for those interested.

Best regards,
Luνs


>From: Nikolaos Dergiades <ndergiades@...>
>Reply-To: Hyacinthos@yahoogroups.com
>To: Hyacinthos@yahoogroups.com
>Subject: Θέμα: [EMHL] german translation und triangle construction
>Date: Fri, 22 Dec 2006 22:22:14 +0000 (GMT)
>
>Dear Luνs,
>If AH is the A_altitude of ABC and
>the A_bisector d meets BC at D and the
>circumcircle at E then the x in x(x+p)=q^2
>is x = DE.
>Proof:
>If M is the mid point of BC and
>k = sin(HAD) then
>k = DM/x = (BM-BD)/x = (a/2-ac/(b+c))/x
>      = a(b-c)/2x(b+c)
>p = d/kk
>q = a/2k
>From AD.DE = BD.DC we get
>dx = [ac/(b+c)][ab/(b+c)]
>Hence
>x(x+p)
>= xx+xp
>= [a(b-c)/2k(b+c)]^2 + bc[a/k(b+c)]^2
>= [a/2k]^2
>= q^2.
>
>Best regards
>Nikos Dergiades
>PS. I don't know German.
>For a translation I went to
>http://www.gokastoria.gr/metafrasis.html

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Dear All My Friends,
   Happy New Year to you all and thank you very much for the nice messages about
this nice locus! I am very happy to read your messages.
   I want to share with you here my small results when applying this locus with
special cases to get symmetrical picture.
   Let X is a point on circumcircle (O) of reference triangle ABC. L is a line
connected X and circumcenter O. A', B', C' are reflections of A, B, C through L
respectively. La, Lb, Lc are Simson lines of A', B', C' respectively. Three
lines La, Lb, Lc bound one triangle A*B*C* with angles exactly as angles of ABC.
The locus of A*, B*, C* when X moving on (O) are the special cases of your locus
when locus angles are angles of ABC.
   Moreover, we can also consider to study some loci of incenter, centroid,
circumcenter, orthocenter, nine point center...  of triangle A*B*C* when X
moving on circumcircle (O).
   Thank you again and best regards,
   Bui Quang Tuan



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Dear Tuan,

I want to wish you a Very Happy New Year and offer my thanks for
your message here as well!  I am absolutely astounded by the
incredible variety of properties generated from this problem SooHong
Lee proposed.  Your observation has also clued me into more things I
have not seen and for which I am very grateful for.

I look forward to seeing much more on this subject!

Sincerely,
Best Regards,
Jeff



>
> Dear All My Friends,
>   Happy New Year to you all and thank you very much for the nice
messages about this nice locus! I am very happy to read your
messages.
>   I want to share with you here my small results when applying
this locus with special cases to get symmetrical picture.
>   Let X is a point on circumcircle (O) of reference triangle ABC.
L is a line connected X and circumcenter O. A', B', C' are
reflections of A, B, C through L respectively. La, Lb, Lc are Simson
lines of A', B', C' respectively. Three lines La, Lb, Lc bound one
triangle A*B*C* with angles exactly as angles of ABC. The locus of
A*, B*, C* when X moving on (O) are the special cases of your locus
when locus angles are angles of ABC.
>   Moreover, we can also consider to study some loci of incenter,
centroid, circumcenter, orthocenter, nine point center...  of
triangle A*B*C* when X moving on circumcircle (O).
>   Thank you again and best regards,
>   Bui Quang Tuan

Hello,

This email message is a notification to let you know that
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Dear friends,

What is the proper name for a triangle inversely similar to a given
pedal triangle?

Sincerely, Jeff

Dear Jeff,
   I don't know the name of your triangle but it is not anti pedal. Please refer:
   http://mathworld.wolfram.com/AntipedalTriangle.html
   Best regards,
   Bui Quang Tuan


Jeff Brooks <trigeom@...> wrote:          Dear friends,

What is the proper name for a triangle inversely similar to a given
pedal triangle?

Sincerely, Jeff





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Dear Tuan,

Yes, I realize this now.
I found the definitions in TCCT.  But, I really need a good name for
this triangle since it has so very much to do with what we are
discussing in Simson's line triangle.

Sincerely, Jeff

>
> Dear Jeff,
>   I don't know the name of your triangle but it is not anti pedal.
Please refer:
>   http://mathworld.wolfram.com/AntipedalTriangle.html
>   Best regards,
>   Bui Quang Tuan
>
>
> Jeff Brooks <trigeom@...> wrote:          Dear friends,
>
> What is the proper name for a triangle inversely similar to a
given
> pedal triangle?
>
> Sincerely, Jeff
>
>
>
>
>
>  __________________________________________________
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> Tired of spam?  Yahoo! Mail has the best spam protection around
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>
> [Non-text portions of this message have been removed]
>

Dear Jeff
> What is the proper name for a triangle inversely similar to a given
> pedal triangle?

I don't think that such a triangle has a special name but such a
triangle is directly similar to the pedal triangle of the inverse in
the circumcircle of the given point.
A little remark about Quan locus :
If A'B'C' is inscribed in the circumcircle, the triangle A*B*C* bounded
by the Simson lines of A', B', C' is directly similar with A'B'C'.
Thus, apart the case A* = B* = C*, A*B*C* will be similar with ABC if
and only if ABC and A'B'C' are congruent, hence only in two cases
- A'B'C' is the reflection of ABC wrt a diameter (indirect similitude)
and, in this case, ABC and A*B*C* are perspective
- a rotation with center O maps ABC to A'B'C' (direct similitude); this
gives aother way to draw the three Quan curves
Friendly. Jean-Pierre

Happy new year to all geometers!

   I prove this problem:

   An ellipse intersect the sides of the triangle ABC in points
    A1,A2 Є BC, B1,B2 Є CA, C1,C2 Є AB. Let’s consider
   {A3}=AA2∩B1C1,  {B3}=BB2∩C1A1,  {C3}=CC2∩A1B1,
   {A4}=AA1∩B2C2,  {B4}=BB1∩C2A2,  {C4}=CC1∩A2B2,
   {A5}=BB2∩CC1,    {A6}=BB1∩CC2,   {B5}=CC2∩AA1,
   {B6}=CC1∩AA2,    {C5}=AA2∩BB1,   {C6}=AA1∩BB2,
   {A7}=A1B1∩C2A2, {B7}=B1C1∩A2B2, {C7}=C1A1∩B2C2.
   Prove that the lines A1A3, B1B3, C1C3, A2A4, B2B4, C2C4,
   A5A6, B5B6, C5C6, AA7, BB7 and CC7 are concurrent.


   What do you think about this problem?

   Yours sincerely,
   Petrisor Neagoe

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Dear Jean-Pierre,

>
> Dear Jeff
> > What is the proper name for a triangle inversely similar to a
> > given pedal triangle?
>
> I don't think that such a triangle has a special name but such a
> triangle is directly similar to the pedal triangle of the inverse in
> the circumcircle of the given point.

Happy new year to all geometers!

Problem:
   An ellipse intersect the sides of the triangle ABC in points
    A1,A2,B1,B2,C1,C2 (A1,A2 lies on the line BC,
   B1,B2 lies on the line CA and C1,C2 lies on the line AB).
   Let’s consider:
   A3 is the intersection point of the lines AA2 and B1C1
   B3 is the intersection point of the lines BB2 and C1A1
   C3 is the intersection point of the lines CC2 and A1B1
   A4 is the intersection point of the lines AA1 and B2C2
   B4 is the intersection point of the lines BB1 and C2A2
   C4 is the intersection point of the lines CC1 and A2B2
   A5 is the intersection point of the lines BB2 and CC1
   A6 is the intersection point of the lines BB1 and CC2
   B5 is the intersection point of the lines CC2 and AA1
   B6 is the intersection point of the lines CC1 and AA2
   C5 is the intersection point of the lines AA2 and BB1
   C6 is the intersection point of the lines AA1 and BB2
   A7 is the intersection point of the lines A1B1 and C2A2
   B7 is the intersection point of the lines B1C1 and A2B2
   C7 is the intersection point of the lines C1A1 and B2C2
   Prove that the lines A1A3, B1B3, C1C3, A2A4, B2B4, C2C4,
   A5A6, B5B6, C5C6, AA7, BB7 and CC7 are concurrent.

   What do you think about this problem?

Yours sincerely,
Petrisor Neagoe

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Hello,

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As we can supposse, and check with barycentric coordinates, the result
is true for a general conic.

Best regards,
and Happy new year to all hyacinthians!

---
Francisco Javier Garcνa Capitαn
http://garciacapitan.auna.com

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Friday, January 05, 2007 1:15 PM [GMT+1=CET],
garciacapitan <garciacapitan@...> escribiσ:

> As we can supposse, and check with barycentric coordinates, the result
> is true for a general conic.

I suppose you have observed that the six points A3, B3, C3, A4, B4 and C4
are 'conconics'.


> Best regards,
> and Happy new year to all hyacinthians!

Also my best wishes to all hyacinthians in the new year!

Ignacio Larrosa Caρestro
A Coruρa (Espaρa)
ilarrosa@...

Dear All My Friends,
   The Steiner deltoid is a very interesting geometry object and the objects
based on it can generate also strange geometry facts.
   Another variant of this locus is when locus angle is also variable. I think
you also remark it. I recommend here one special case:
   Let X is a point on circumcircle (O) of reference triangle ABC. Suppose the
Simson line Lx of X cuts circumcircle (O) at two points Y, Z and Ly, Lz are
Simson lines of Y, Z respectively. Three Simson lines Lx, Ly, Lz bound one
triangle TxTyTz which shares with ABC common nine point circle.
   We can see that triangle TxTyTz is reflection of triangle XYZ in midpoint of
XTx.
   We can study here the loci of Tx, Ty, Tz and also incenter, mitten punkt ...
of triangle TxTyTz.
   Dear All My Friends,
   The geometry facts of classical Steiner deltoid and this "New Year 2007 Locus"
are very plentiful. I am very happy if one our specialist can combine all these
facts (old and new, after generalization, classification and selection... under
new light of modern computer triangle geometry) in one complete nice FG paper.
This paper can be our thankfulness to Euler, Steiner, Simson, Wallace, Morley,
Miquel ... our Ancients who worked hard with this deltoid by pencil and paper
only without any calculation facilities.
   This paper is also helpful and comfortable to study Steiner deltoid. Now when
I want to know something related this deltoid, I must read here some facts there
some facts and in most case: there is not detail proof.
   This paper is a big work and requires a lot of special scientific and history
researchs. It can be compare with Clawson paper about Complete Quadrilateral.
   I would like to offer my great thankfulness and high respect to whom who can
do it.
   Thank you and best regards,
   Bui Quang Tuan


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Dear Quang Tuan Bui
in order to generalize the Steiner deltoid, we can look at the locus of
the orthopoles of the lines tangent to a given circle with center O; we
get the cycloidal curves we were talking about.
An other way to generalize is the following :
For M lying on the circumcircle, let A1, B1, C1 lie on the sidelines of
ABC such as the three oriented line angles (BC, MA'), (CA, MB'), (AB,
MC') have the same fixed value w.
Then A1,B1,C1 lie on a same line and this line envelops a deltoid
touching the sidelines of ABC. This deltoid is "centered" at U, lying
on the perpendicular bisector of OH and such as <NUO =w.
The ratio of a similitude mapping the Steiner deltoid to this deltoid
is 1/sin(w)
Some properties of this deltoid and the connection with Mac Beath's
works have been discussed in Hyacinthos.
See, for instance #9934, #9947+ and the file Mac Beath.pdf
Friendly. Jean-Pierre

Dear Petrisor and Ignacio,

Ignacio wrote:

>> I suppose you have observed that the six points
>> A3, B3, C3, A4, B4 and C4 are 'conconics'.

This seems to be not true from some calculations and Cabri figure,

Now, for the sake of completness, if A8=B1C1 intersection with B2C2
then A5, A6, A8 are collinear, and the same cyclically.

Best regards,

Francisco Javier.

Does anyone know any interesting properties of the cevians through the
circumcentre? Specifically, I investigated (without success) if the
points at which they intersect the opposite edges have any interesting
properties, so I'd be interested if you know of anything nontrivial.

I am also not an experienced geometer, so even 'well-known' results I
probably have not yet encountered and would find interesting.

Thanks.

Saturday, January 06, 2007 11:50 AM [GMT+1=CET],
garciacapitan <garciacapitan@...> escribiσ:

> Dear Petrisor and Ignacio,
>
> Ignacio wrote:
>
>>> I suppose you have observed that the six points
>>> A3, B3, C3, A4, B4 and C4 are 'conconics'.
>
> This seems to be not true from some calculations and Cabri figure,

You are right, of course ...

Best regards,

Ignacio Larrosa Caρestro
A Coruρa (Espaρa)
ilarrosa@...

> Now, for the sake of completness, if A8=B1C1 intersection with B2C2
> then A5, A6, A8 are collinear, and the same cyclically.
>
> Best regards,
>
> Francisco Javier.
>
>
>
>
>
>
> Yahoo! Groups Links
>
>
>

Dear Friend (you still not inform us your name),
   As I know, there are a lot of interesting facts related Cevians through the
circumcenter and we can not list only some of them in one message. I can give
you some my experiences as following:
   Find the circumcenter results in some websites
   1. Encyclopedia of Triangle Centers (ETC):
   http://faculty.evansville.edu/ck6/encyclopedia/ETC.html
   2. Forum Geometricorum (FG):
   http://forumgeom.fau.edu/index.html
   3. Paul Yiu books:
   http://www.math.fau.edu/yiu/geometry.html
   An Introduction to the Geometry of the Triangle and other books
   4. The Triangle Web (TWW)
   http://www.xtec.es/~qcastell/ttw/ttweng/portada.html
   5. MathWorld
   http://mathworld.wolfram.com/
   6. Cut-The-Knot
   http://www.cut-the-knot.org/geometry.shtml
   7. Search key word "circumcenter" or "circumcircle" ... in Hyacinthos message
search
   8. Do the same with Google or Yahoo search.
   If you have any problems when reading, visiting these websites please post
here. Hyacinthos members will help you if they can.
   Best regards,
   Bui Quang Tuan


ilthigore1 <ilthigore1@...> wrote:          Does anyone know any
interesting properties of the cevians through the
circumcentre? Specifically, I investigated (without success) if the
points at which they intersect the opposite edges have any interesting
properties, so I'd be interested if you know of anything nontrivial.

I am also not an experienced geometer, so even 'well-known' results I
probably have not yet encountered and would find interesting.

Thanks.





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--- In Hyacinthos@yahoogroups.com, "ilthigore1" <ilthigore1@...> wrote:
>
> Does anyone know any interesting properties of the cevians through
the
> circumcentre? Specifically, I investigated (without success) if the
> points at which they intersect the opposite edges have any
interesting
> properties, so I'd be interested if you know of anything nontrivial.
>
> I am also not an experienced geometer, so even 'well-known' results I
> probably have not yet encountered and would find interesting.
>
> Thanks.
>
One well know property of the cevian from A through O is that is
isogonal of the altitude from A. This means that angles BAHa(Ha foot
of altitude) and OAC are equal.

Hello,Hyacinthos!
  This file is testing of connection.
      Friendly,Stalislav

Respected Hyacinthists!
  Respected Jean-Pierre!
Article's Jean-Pierre in Forum Geometricorum(Volume6,2006) and messages
14503,14505,14506,14507,14520 in Hyacinthos induce me to be divided similar
result.I received their quite other way.
Let P is point inside triangle ABC and BB',CC'-cevians.The quadrileral AC'PB'
have an incircle if and only if:
  tanCBP/2*tanC/2= tanBCP/2*tanB/2. Vertexs P of family similar quadrilerals lie
on the hyperbola with foci B,C.Analogous quadrilerals may be constructed
relatively every vertex ABC. Solving
system of three equations we receive formulies for such point P,that three
quadrilerals AC'PB',BA'PC',CA'PB' will have incircles simultaneity,where
AA',BB',CC'-cevians.Point P is intersection of three hyperbolies going via
vertexs ABC.
   tan(CBP/2)= tanB/2:(1+tanB/2+tanC/2)
   tan(BCP/2)= tanC/2:(1+tanB/2+tanC/2)
   tan(CBP/2)'=tanB/2:(1-tanB/2-tanC/2)
   tan(BCP/2)'=tanC/2:(1-tanB/2-tanC/2)
It is proved that these formulies determine the centers of the inner (sign"+")
and outer(sign"-") Soddy's circles.Received formulies corresponding to
Trilinears Kimberling -X(176),X(175).
Let  S1=tanA/2+tanB/2   S1,S2,S3 determine position of the center
      S2=tanA/2+tanC/2   outer Soddy's circle-inside or outside plane
      S3=tanB/2+tanC/2   triangle ABC.If S1 or S2 or S3 equal 1,then center of
outer Soddy's circle lie on side joined indicated angles.
If either  (tanA/2)^2+tanA/2*tanB/2+(tanB/2)^2=1
        or  (tanA/2)^2+tanA/2*tanC/2+(tanC/2)^2=1
        or  (tanB/2)^2+tanB/2*tanC/2+(tanC/2)^2=1
These formulies determine line Soddy parallel to side triangle ABC
joined indicated angles.
Particular case is degeneration of outer Soddy's circle in straight
line,which touches with of three circles (A,p-a),(B,p-b),(C,p-c),
where "p" is semiperimeter ABC.In this case line Soddy is perpendicu-
late to touch line always,moreover tanA/2+tanB/2+tanC/2=2.
Besides It is proved that incircle ABC is outer Soddy's circle for
three incircles triangles BPC,CPA,APB and were calculated radiuses  of these
incircles.Also were calculated radiuses of incircles the
three quadrilerals AC'PB',BA'PC',CA'PB'.AS a result-sums curves ap-
propriate to pair incircles AC'PB'and BPC,BA'PC'and CPA,CA'PB'and
APB are constant and equal curve of inner Soddy's circle ABC,i.e.:
    1/r1+1/r5=1/r2+1/r6=1/r3+1/r4=1/rs,where
   r1,r2,r3-radiuses of incircles quadrilerals-AC'PB',BA'PC',CA'PB';
   r4,r5,r6-radiuses of incircles triangles-APB,BPC,CPA;
   rs-radius of inner Soddy's circle ABC.
  If either  tanA/2+tanB/2=tanC/2 -condition,when r3=r4=2rs;
         or  tanA/2+tanC/2=tanB/2 -condition,when r2=r6=2rs;
         or  tanB/2+tanC/2=tanA/2 -condition,when r1=r5=2rs.
These corelations determine conditions,when radiuses appropriate to pair
incircles(quadrilerals and triangles) are equal.
In conclusion-points X(176),X(175) are foci of ellipce,on which lie vertexs
triangle ABC.However if one of angles triangle ABC more,than
2arctan4/3,then X(176),X(175) are foci of hyperbola.
Note,that in case ellipce perimeters triangles ASS',BSS',CSS' are
equal,i.e.: AS'+AS=BS'+BS=CS'+CS=rs'+rs,where
    S,S'-centers of inner and of outer Soddy's circles;
    rs,rs'-radiuses inner and outer Soddy's circles.
In case hyperbola-- AS'-AS=BS'-BS=CS'-CS=rs'-rs.

     Exuse me my bad English.     Sincerely,Stalislav



































































































































































































(sign"+") and outer(sign"-") Soddy's circles.Received formulies cor-