Dear friends,
--- In Hyacinthos@yahoogroups.com, "yakub.aliyev" <yakub.aliyev@...> wrote:
> 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...

To discuss what is perceived, I prefer to use an observer centered frame. Let
A(0,0) be the observer, and K (c,-t) the centers of the columns (c fixed, t
changing).
Then  BC = 2*r*c*W/(c^2-r^2)  where W^2=c^2-r^2+t^2
as given by Angel Montesdeoca.

If I, the observer, was the sun, BC would be the shadow of the column on some
wall behind the column line. But this quantity cannot be used to describe what
is the human perception of the columns. When t tends to infinity, the shadow
becomes infinite too, but the human perception of the column becomes "too small
to be observed".

We can try to consider the length of the chord that joins the contact points. We
obtain ST= 2*r*sqrt(d^2-r^2)/d where d^2=c^2+t^2, d being the distance from
observer to the center of the column. If radius of columns is 1 meter, and you
are standing 7 meters away from the column line, ST increases of 1%  when t goes
from 0 (the column in front of you) to infinity (the last column in the line).

But perception of size is not obtained by measuring ST. What is measured is
angle (AS,AT) and from **binocular** vision, an estimation of the distance. Does
our embedded telemeter focus on midpoint of ST (inside the column) or on the
proximal point of the column ? In the second case, factor (d+r)/d must be
applied and we obtain 2r*sqrt((d-r)/(d+r)) as estimated size of the column,
instead of the obvious 2r. Using again r=1, c=7, then "estimated" ST ranges from
1.73 when t=0 to 2 when t=inf. This time, effect is huge.

Therefore, the main question is : what are the laws of perception ?

Best regards.

Dear Pierre,
now I found a proof without
restriction for p, q, r.
P + Q = cot(p) + cot(q) = sin(p+q)/sin(p).sin(q)
= -sin(r)/sin(p).sin(q).
Since 0 < sin(p).sin(q) = (1/2)[cos(p-q)-cos(p+q)] <
< (1/2)[1-cos(p+q)] = (1/2)[1-cos(r)] = sin^2(r/2)
we have P + Q <= -sin(r)/sin^2(r/2) = -2.cot(r/2)
Similarly working we get
P + Q + R <= -(cot(p/2) + cot(q/2) + cot(r/2))
<= -3cot[(p+q+r)/6] = -sqr(3)
since the function cot(x) is convex in the
interval (0, pi/2).

Best regards
Nikos Dergiades



> Dear Pierre,
> Sorry I said nonsense that the
> function cot(x) is convex
> and that P + Q + R >= -sqrt(3).
> So my proof is not correct.
> My proof is correct in the direction
> P + Q + R <= -sqrt(3)
> and hence f >= 4S.sqrt(3)
> if p, q, r > pi/2 since the function
> cot(x) is concave in the interval
> (pi/2, pi).
>
> Best regards
> Nikos Dergiades
> 
>





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Dear Pierre,
Sorry I said nonsense that the
function cot(x) is convex
and that P + Q + R >= -sqrt(3).
So my proof is not correct.
My proof is correct in the direction
P + Q + R <= -sqrt(3)
and hence f >= 4S.sqrt(3)
if p, q, r > pi/2 since the function
cot(x) is concave in the interval
(pi/2, pi).

Best regards
Nikos Dergiades





> Dear Pierre
>
> If for a point X inside ABC
> with homogenous barycentrics (x : y : z)
> p = angle (XB, XC)
> q = angle(XC, XA)
> r = angle(XA, XB)
> with cot(p) = P, cot(q) = Q, cot(r) = R
> where p + q + r = 2.pi  (i)
> gives
> P.Q + Q.R + R.P = 1  (ii)
> and S the area of ABC
> from FG200921 we get that
> x = k/(cotA - P)
> y = k/(cotB - Q)
> z = k/(cotC - R)
> where
> cotA = (bb + cc - aa)/4S  . . .
> and from (ii)
> k = (aayz + bbzx + ccxy)/[2S(x+y+z)] .
> Substituting
> x, y, z and k  we get
> P + Q + R = -f/(4S) (iii)
> where f = f(x,y,z) = cycsum( a^2/x*(-x^2+xz+xy+2yz))
> /(x+y+z)
> and since the function cot(x) is convex we have
> by Jensen's inequality
> P + Q + R = cot(p) + cot(q) + cot(r) >= 3.cot[(p+q+r)/3]
> or
> P + Q + R >= -sqrt(3) or from (iii)
> f <= 4S.sqrt(3)
>
> Best regards
> Nikos Dergiades
>
> > Dear friends.
> > The following question is stated in the current
> Monthly:
> > prove that
> > cycsum(a^2)-4sqrt(3)S is not less than
> > 2 cycsum( a^2(x^2-yz)/x)/(x+y+z) for all positive
> x,y,z
> > where a,b,c are the sides of a triangle and S its
> area
> > (1) a more handy form is : define
> > f(x,y,z) = cycsum( a^2/x*(-x^2+xz+xy+2yz)) /(x+y+z)
> and
> > show that
> > f(x,y,z) is not less than 4sqrt(3)S
> > (2) f is homogeneous and thus define a function acting
> over
> > (barycentric) points in the triangle plane
> > (3) using standard numeric values, see that over
> triangle
> > centers in ETC (lying inside ABC), min f is obtained
> for
> > X(13)
> > (4) use formal barycentrics of X(13) and prove that,
> if
> > true, property is optimal since f(X(13))= 4sqrt(3)S
> > (5) read the doc : X(13) minimizes a sum of lengths
> (that
> > would introduce horrific square roots), but also
> satisfies
> > angle equalities.
> > (6) therefore compute g(X) = cycsum(1/tan(XB,XC)) and
> check
> > that
> > g = -f /4/S
> > (7) in the right domain, extremum is obtained when
> angles
> > are equal to 120°, leading to 3*(1/sqrt(3)) as
> needed.
> >
> > Best regards,
> > Pierre.
> >
> >
> >
>
>
>
>      
> ___________________________________________________________
>
> Χρησιμοποιείτε Yahoo!;
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> μηνύματα (spam); Το Yahoo! Mail
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> προστασία κατά των ενοχλητικών
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>
>
>
> ------------------------------------
>
> Yahoo! Groups Links
>
>
>     Hyacinthos-fullfeatured@yahoogroups.com
>
>
>



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Dear Pierre

If for a point X inside ABC
with homogenous barycentrics (x : y : z)
p = angle (XB, XC)
q = angle(XC, XA)
r = angle(XA, XB)
with cot(p) = P, cot(q) = Q, cot(r) = R
where p + q + r = 2.pi  (i)
gives
P.Q + Q.R + R.P = 1  (ii)
and S the area of ABC
from FG200921 we get that
x = k/(cotA - P)
y = k/(cotB - Q)
z = k/(cotC - R)
where
cotA = (bb + cc - aa)/4S  . . .
and from (ii)
k = (aayz + bbzx + ccxy)/[2S(x+y+z)] .
Substituting
x, y, z and k  we get
P + Q + R = -f/(4S) (iii)
where f = f(x,y,z) = cycsum( a^2/x*(-x^2+xz+xy+2yz)) /(x+y+z)
and since the function cot(x) is convex we have
by Jensen's inequality
P + Q + R = cot(p) + cot(q) + cot(r) >= 3.cot[(p+q+r)/3] or
P + Q + R >= -sqrt(3) or from (iii)
f <= 4S.sqrt(3)

Best regards
Nikos Dergiades

> Dear friends.
> The following question is stated in the current Monthly:
> prove that
> cycsum(a^2)-4sqrt(3)S is not less than
> 2 cycsum( a^2(x^2-yz)/x)/(x+y+z) for all positive x,y,z
> where a,b,c are the sides of a triangle and S its area
> (1) a more handy form is : define
> f(x,y,z) = cycsum( a^2/x*(-x^2+xz+xy+2yz)) /(x+y+z) and
> show that
> f(x,y,z) is not less than 4sqrt(3)S
> (2) f is homogeneous and thus define a function acting over
> (barycentric) points in the triangle plane
> (3) using standard numeric values, see that over triangle
> centers in ETC (lying inside ABC), min f is obtained for
> X(13)
> (4) use formal barycentrics of X(13) and prove that, if
> true, property is optimal since f(X(13))= 4sqrt(3)S
> (5) read the doc : X(13) minimizes a sum of lengths (that
> would introduce horrific square roots), but also satisfies
> angle equalities.
> (6) therefore compute g(X) = cycsum(1/tan(XB,XC)) and check
> that
> g = -f /4/S
> (7) in the right domain, extremum is obtained when angles
> are equal to 120°, leading to 3*(1/sqrt(3)) as needed.
>
> Best regards,
> Pierre.
>
>
>




___________________________________________________________
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Dear friends, we received a solution from Angel Montesdeoca using calculus.
See the link:

http://f1.grp.yahoofs.com/v1/AHAXS5dOt8jl27advXRcTwnyIoUlbf5rXME_WkfTQZaFoPqta5x\
ucCVoOSbRKrjKl7sdwuCYPAXZJvgGtOvk9NNaemiwxdgdzQ/Hyacinthos%2520Message%252018465\
%2520%5B1%5D.pdf

It is also possible to present completely synthetic solution for the problem.
Patrick Morton proposed the following correction:

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. As the point A gets closer to w along the fixed line k, the
distance BC decreases, the minimum value of BC being reached when A is on the
perpendicular to k through the center of w.

Thanks to all.
Yakub.

--- In Hyacinthos@yahoogroups.com, "yakub.aliyev" <yakub.aliyev@...> wrote:
>
> 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.
>

Dear friends.
The following question is stated in the current Monthly: prove that
cycsum(a^2)-4sqrt(3)S is not less than
2 cycsum( a^2(x^2-yz)/x)/(x+y+z) for all positive x,y,z where a,b,c are the
sides of a triangle and S its area
(1) a more handy form is : define
f(x,y,z) = cycsum( a^2/x*(-x^2+xz+xy+2yz)) /(x+y+z) and show that
f(x,y,z) is not less than 4sqrt(3)S
(2) f is homogeneous and thus define a function acting over (barycentric) points
in the triangle plane
(3) using standard numeric values, see that over triangle centers in ETC (lying
inside ABC), min f is obtained for X(13)
(4) use formal barycentrics of X(13) and prove that, if true, property is
optimal since f(X(13))= 4sqrt(3)S
(5) read the doc : X(13) minimizes a sum of lengths (that would introduce
horrific square roots), but also satisfies angle equalities.
(6) therefore compute g(X) = cycsum(1/tan(XB,XC)) and check that
g = -f /4/S
(7) in the right domain, extremum is obtained when angles are equal to 120°,
leading to 3*(1/sqrt(3)) as needed.

Best regards,
Pierre.

Thank you Patrick. The first formulation of Problem is correct:

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.

Note that the illusion (that Greek columns which are closer to observer look as
smaller) which I mentioned in my 1st message can also experimented with equal
sized balls (spheres). See also the links:

http://www.istockphoto.com/file_thumbview_approve/6779165/2/
http://lifehackery.com/qimages/5/used-tennis-balls.jpg

Note that the columns/balls in these pictures which are closer to observer look
as smaller with respect to columns/balls in the right and left sides of the
pictures.
Yakub




--- In Hyacinthos@yahoogroups.com, Patrick Morton <patrickmorton289@...> wrote:
>
> 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]
>

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

http://forumgeom.fau.edu/FG2009volume9/FG200928index.html

The editors
Forum Geometricorum
=============================
Shao-Cheng Liu, Trilinear polars and antiparallels,
Forum Geometricorum, 9 (2009) 283--290.

Abstract: We study the triangle bounded by the antiparallels to the
sidelines of a given triangle ABC through the intercepts of the trilinear
polar of a point P other than the centroid G. We show that this triangle is
perspective with the  reference triangle, and also study the condition of
concurrency of  the antiparallels. Finally, we also study the configuration
of  induced GP-lines and obtain an interesting  conjugation of finite
points other than G.

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]

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.

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,

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

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,

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]

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]

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.

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




___________________________________________________________
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---------- 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 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

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 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 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
>





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---------- 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

[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

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 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,

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