Prove that the line joining the -circumcenter of a triangle to a
vertex- is perpendicular to the line -joining the foot of altitudes of
adjoining sides.-
Some sort of a HINT please....

>
> 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]

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http://www.Synergeticists.org | "Dare to be Naive" -- Bucky Fuller

dear colleague
There is a very simple response
the orthic triangle is homothetic to the tangential one
The line joining the -circumcenter of a triangle to a
vertex- is perpendicular to the tangent line in that vetex to the
circumcircle, so to the parallel side of the orthic triangle

There are others proves using barycentrics coordinates

Best regards

Michel Garitte



--- In Hyacinthos@yahoogroups.com, "malisundaresan"
<malisundaresan@...> wrote:
>
> Prove that the line joining the -circumcenter of a triangle to a
> vertex- is perpendicular to the line -joining the foot of altitudes
of
> adjoining sides.-
> Some sort of a HINT please....
>

Dear Franηois,

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@...
========================================

--- In Hyacinthos@yahoogroups.com, "malisundaresan"
<malisundaresan@...> wrote:
>
> Prove that the line joining the -circumcenter of a triangle to a
> vertex- is perpendicular to the line -joining the foot of
altitudes of
> adjoining sides.-
> Some sort of a HINT please....
>
Let E, F be the feet of the altitudes through
B, C upon AC, AB respectively,
Let X be the midpoint of AH (H orthocenter)
X is the center of the circle through A, E, H, F.
XE = XF
The nine-point circle (N) passes through E and F
NE = NF (each = R/2)
Follows NX is the perpendicular bisector of EF.
In the triangle AOH :
X, N being the midpoints of AH, OH we have
NX is parallel to OA.
Together with the fact that NX is perpendicular to EF,
it now follows that OA is perpendicular to EF

Vijayaprasad

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

http://forumgeom.fau.edu/FG2007volume7/FG200721index.html

The editors
Forum Geometricorum
=============================================================
Tibor Dosa,  Some triangle centers associated with the excircles,
Forum Geometricorum, 7 (2007) 151--158.

Abstract: We construct a few new triangle centers associated with the
excircles of a triangle.

Dear Alexey
Here how I checked your property with Cabri (and GSP).
I change your notations a little bit, calling G the ABC-centroid, O the
ABC-circumcenter and  K the ABC-X(6) point (Lemoine or Grebe point).
Now let A'B'C' be any triangle such that B'C' is parallel   to the
ABC-median through A, C'A' is parallel to the ABC-median through B and A'B'
is parallel through the ABC-median through C.
It is nearly obvious that the lengths of the sides B'C', C'A', A'B' are
proportionnal to the lengths of the three ABC-medians.

Now ,of course, I label G' the A'B'C'-centroid, O' the A'B'C' -circumcenter
and K' the A'B'C'-X(6) point
and I checked your theorem on these triangles ABC and A'B'C' that is to say
:
<OGK = <O'G'K'
I find this equality astonishing!
I suppose that there exists a proof through some boring and tedious trigo
calculations.
I began this way handling barycentrics of O, G, L and giving up  quickly
courtesy of my lazyness!

I have another idea:
I label f the direct circular map sending A to A', B to B' and C to C'.
It is known that f(G) = G' but with Cabri I find that we have also:
f(O) =K'  and f(K) = O'.

I find that very strange and maybe related to your problem!
Friendly
Francois



On 9/28/07, Alexey.A.Zaslavsky <zasl@...> wrote:
>
>   Dear colleagues!
> Let M, L be the centroid and the Lemoine point of the triangle. Then the
> angle between ML and the Euler line is the same for any triangle and the
> triangle formed by its medians.
> Have anybody the proof?
>
> Sincerely Alexey
> Antivirus scanning: Symantec Mail Security for SMTP.
>
> [Non-text portions of this message have been removed]
>
>
>


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

Dear Alexey
With Cabri, I found 2 other properties of the map f leading, I think,  to a
solution of your problem.
I label I and J' the limit points of the direct circular map f, that is to
say:
f(I) = Infty anf f(Infty) = J'
(Here Infty is the point at infinity of the circular plane.)
Then I and K are inverse wrt the ABC-circumcircle
and J' and K' are inverse wrt the A'B'C' circumcircle.
Friendly
Francois


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

Dear Alexey and Francois
you wrote that if A'B'C' is the triangle
with sides the medians of ABC that
<OGK = <O'G'K'  and that there exists a proof through
some boring and tedious trigo calculations.

You are right.
cot(<OGK) = (GK^2+GO^2-OK^2)/(4(ABC))
= -[(aa-bb)^2+(bb-cc)^2+(cc-aa)^2]/36(aa+bb+cc)(ABC)
where (ABC) is the area of ABC
and this is invariant if instead of
a,b,c we use the medians of ABC.

Best regards
Nikos Dergiades





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Dear Nikos
You are our Savior!
And now what I am going to do with my circular map, the best is still to go
to bed!
Friendly
Francois

On 10/1/07, Nikolaos Dergiades <ndergiades@...> wrote:
>
> Dear Alexey and Francois
> you wrote that if A'B'C' is the triangle
> with sides the medians of ABC that
> <OGK = <O'G'K'  and that there exists a proof through
> some boring and tedious trigo calculations.
>
> You are right.
> cot(<OGK) = (GK^2+GO^2-OK^2)/(4(ABC))
> = -[(aa-bb)^2+(bb-cc)^2+(cc-aa)^2]/36(aa+bb+cc)(ABC)
> where (ABC) is the area of ABC
> and this is invariant if instead of
> a,b,c we use the medians of ABC.
>
> Best regards
> Nikos Dergiades
>
>
>
>
>
> ___________________________________________________________
> 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
>
>
>
>
>
> Yahoo! Groups Links
>
>
>
>


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

Dear Alexey
The existence of the direct circular map f defined in my previous post gives
indeed a synthetic solution of yout theorem:

1° In the sequel I label Z the point at infinity instead of < Infty>.
2° If L nd L' are 2 lines, the notation (L,L') is the signed angle of lines
of L and L'.
3° The notation PQ means the line PQ.
4° The notation PQR means the circle PQR.
Remind PQ = PQZ that is to say, a line is a circle through the poit at
infinity Z.
5° (PQR, P'Q'R') is the signed angle of the circles PQR and PQ'R' at their
intersection point P that is to say the signed angle of lines (T, T') where
T and T' are tangents to PQR and PQ'R' at point P.
6° Remind that an inversion changes a signed angle of circles in its
opposite  and that a direct circular map preserves these angles.
Now let's go
On 10/1/07, Francois Rideau <francois.rideau@...> wrote:
(GO, GK) = (GOZ, GKZ)
(GOZ, GKZ) = (G'K'J', G'O'J')
(G'K'J', G'O'J') = - (G"J'K', G"ZK') via the inversion wrt circle A'B'C';
Here G" is the inverse point of G' in this inversion.
-(G"J'K', G"ZK') = -(G'G", G'K') since points J', K', G', G" are on a same
circle.
-(G'G", G'K') = -(G'O', G'K')
So (GO, GK) = -(G'O', G'K')
And we are done for this synthetic proof.
Friendly
Francois
PS
Now the problem is to prove existence of the map f of course.
So Nikos proof is the only one available for the moment!


Dear Alexey
> With Cabri, I found 2 other properties of the map f leading, I think,  to
> a solution of your problem.
> I label I and J' the limit points of the direct circular map f, that is to
> say:
> f(I) = Infty anf f(Infty) = J'
> (Here Infty is the point at infinity of the circular plane.)
> Then I and K are inverse wrt the ABC-circumcircle
> and J' and K' are inverse wrt the A'B'C' circumcircle.
> Friendly
> Francois
>
>
>


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

I correct some little typos


Dear Alexey
> The existence of the direct circular map f defined in my previous post
> gives indeed a synthetic solution of your theorem:
>
> 1° In the sequel I label Z the point at infinity instead of < Infty>.
> 2° If L and L' are 2 lines, the notation (L,L') is the signed angle of
> lines of L and L'.
> 3° The notation PQ means the line on P and Q.
> 4° The notation PQR means the circle on P, Q, R.
> Remind PQ = PQZ that is to say, a line is a circle through the point at
> infinity Z.
> 5° (PQR, P'Q'R') is the signed angle of the circles PQR and PQ'R' at their
> intersection point P that is to say the signed angle of lines (T, T') where
> T and T' are tangents to PQR and PQ'R' at point P.
> 6° Remind that an inversion changes a signed angle of circles in its
> opposite  and that a direct circular map preserves these angles.
> Now let's go:
>
> (GO, GK) = (GOZ, GKZ)
> (GOZ, GKZ) = (G'K'J', G'O'J') for the map  f presenves angles.
> (G'K'J', G'O'J') = - (G"J'K', G"ZK') via the inversion wrt circle A'B'C';
> Here G" is the inverse point of G' in this inversion.
> -(G"J'K', G"ZK') = -(G'G", G'K') since points J', K', G', G" are on a same
> circle.
> -(G'G", G'K') = -(G'O', G'K')
> So (GO, GK) = -(G'O', G'K')
> And we are done for this synthetic proof.
> Friendly
> Francois
> PS
> Now the problem is to prove existence of the map f of course with the
> described properties
> So Nikos proof is the only one available for the moment!
>
>
> Dear Alexey
> > With Cabri, I found 2 other properties of the map f leading, I think,
> > to a solution of your problem.
> > I label I and J' the limit points of the direct circular map f, that is
> > to say:
> > f(I) = Infty anf f(Infty) = J'
> > (Here Infty is the point at infinity of the circular plane.)
> > Then I and K are inverse wrt the ABC-circumcircle
> > and J' and K' are inverse wrt the A'B'C' circumcircle.
> > Friendly
> > Francois
> >
> >
> >
>


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

Dear Francois!
Thank you for your proof.
I have another proof. Let the sidelines of the triangle A'B'C' are perpendicular
to the medians of ABC and pass through the common points of these medians and
the NPC of ABC.
Then the centroid M of ABC is also the centroid of A'B'C' and the midpoint of MH
is the Lemoine point of A'B'C'. Now the pedal triangle of K is homothetic to
A'B'C' and the homothety center lies on MK. So the circumcenter of A'B'C' also
lies on MK.

Sincerely                                   Alexey

P.S. The line A'B' is the directrix of the parabola passing through A, B and
touching in these points the line AC, BC. The focus of this parabola is the
common point of CK, the circle ABO and the circle with diameter OK.

   Dear Alexey
   The existence of the direct circular map f defined in my previous post gives
   indeed a synthetic solution of yout theorem:

   1° In the sequel I label Z the point at infinity instead of < Infty>.
   2° If L nd L' are 2 lines, the notation (L,L') is the signed angle of lines
   of L and L'.
   3° The notation PQ means the line PQ.
   4° The notation PQR means the circle PQR.
   Remind PQ = PQZ that is to say, a line is a circle through the poit at
   infinity Z.
   5° (PQR, P'Q'R') is the signed angle of the circles PQR and PQ'R' at their
   intersection point P that is to say the signed angle of lines (T, T') where
   T and T' are tangents to PQR and PQ'R' at point P.
   6° Remind that an inversion changes a signed angle of circles in its
   opposite and that a direct circular map preserves these angles.
   Now let's go
   On 10/1/07, Francois Rideau <francois.rideau@...> wrote:
   (GO, GK) = (GOZ, GKZ)
   (GOZ, GKZ) = (G'K'J', G'O'J')
   (G'K'J', G'O'J') = - (G"J'K', G"ZK') via the inversion wrt circle A'B'C';
   Here G" is the inverse point of G' in this inversion.
   -(G"J'K', G"ZK') = -(G'G", G'K') since points J', K', G', G" are on a same
   circle.
   -(G'G", G'K') = -(G'O', G'K')
   So (GO, GK) = -(G'O', G'K')
   And we are done for this synthetic proof.
   Friendly
   Francois
   PS
   Now the problem is to prove existence of the map f of course.
   So Nikos proof is the only one available for the moment!

   Dear Alexey
   > With Cabri, I found 2 other properties of the map f leading, I think, to
   > a solution of your problem.
   > I label I and J' the limit points of the direct circular map f, that is to
   > say:
   > f(I) = Infty anf f(Infty) = J'
   > (Here Infty is the point at infinity of the circular plane.)
   > Then I and K are inverse wrt the ABC-circumcircle
   > and J' and K' are inverse wrt the A'B'C' circumcircle.
   > Friendly
   > Francois
   >
   >
   >

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




Antivirus scanning: Symantec Mail Security for SMTP.

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

Dear Alexey, dear Nikos and dear friends.
Thank you for your  very elementary proof. I will look at it and compare it
with mine.
In your proof, triangles ABC and A'B'C' are orthologic and in mine they are
parallelogic!

Nikos proof is also interesting. In fact, I begin also to calculate the
cotangent but was so scared by the lengths of the calculations that I gave
up cowardly and went to bed. So I am eager to know how Nikos proceeded,
maybe he used Mathematica , Maple or some other math software or may be some
shortcut I did not see.

However , I think my proof  may have some interest from an educational point
of view for I use the circular group often forgotten in triangle geometry.
I remind you parallelogy and orthology are very special cases of
isogonology, a theory known a long time ago as the theory of metapolar four
points, ( "Theorie des metapoles"  or "Theorie des quadrangles metapolaires"
in french.)
First this theory was exposed in the elementary frame of the euclidian
plane, for example in the old Rouche-Comberousse around 1900, but I noticed
some years ago that the circular plane or Moebius plane was more suitable to
explain some subtleties of the theory.

I have written more than ten years ago,  a rather lengthy paper in french on
this topic. As I am unable and too lazy to translate it in english, however
I can always summarize here in english, of course without proof, the main
results of the theory if you are interested. And maybe some of you can give
me some other interesting references on the topic if you have seen them.
Friendly
Francois
PS
Dear Alexey
1°
In fact I proved the existence of the direct circular map f in your problem
but I can't elaborate here for lack of room!
2° What do you call the NPC of ABC? Sorry for such a stupid question!


On 10/2/07, Alexey.A.Zaslavsky <zasl@...> wrote:
>
>   Dear Francois!
> Thank you for your proof.
> I have another proof. Let the sidelines of the triangle A'B'C' are
> perpendicular to the medians of ABC and pass through the common points of
> these medians and the NPC of ABC.
>



Then the centroid M of ABC is also the centroid of A'B'C' and the midpoint
> of MH is the Lemoine point of A'B'C'. Now the pedal triangle of K is
> homothetic to A'B'C' and the homothety center lies on MK. So the
> circumcenter of A'B'C' also lies on MK.
>
> Sincerely Alexey
>
> P.S. The line A'B' is the directrix of the parabola passing through A, B
> and touching in these points the line AC, BC. The focus of this parabola is
> the common point of CK, the circle ABO and the circle with diameter OK.
>
>
> Dear Alexey
> The existence of the direct circular map f defined in my previous post
> gives
> indeed a synthetic solution of yout theorem:
>
> 1° In the sequel I label Z the point at infinity instead of < Infty>.
> 2° If L nd L' are 2 lines, the notation (L,L') is the signed angle of
> lines
> of L and L'.
> 3° The notation PQ means the line PQ.
> 4° The notation PQR means the circle PQR.
> Remind PQ = PQZ that is to say, a line is a circle through the poit at
> infinity Z.
> 5° (PQR, P'Q'R') is the signed angle of the circles PQR and PQ'R' at their
>
> intersection point P that is to say the signed angle of lines (T, T')
> where
> T and T' are tangents to PQR and PQ'R' at point P.
> 6° Remind that an inversion changes a signed angle of circles in its
> opposite and that a direct circular map preserves these angles.
> Now let's go
> On 10/1/07, Francois Rideau
<francois.rideau@...<francois.rideau%40gmail.com>>
> wrote:
> (GO, GK) = (GOZ, GKZ)
> (GOZ, GKZ) = (G'K'J', G'O'J')
> (G'K'J', G'O'J') = - (G"J'K', G"ZK') via the inversion wrt circle A'B'C';
> Here G" is the inverse point of G' in this inversion.
> -(G"J'K', G"ZK') = -(G'G", G'K') since points J', K', G', G" are on a same
>
> circle.
> -(G'G", G'K') = -(G'O', G'K')
> So (GO, GK) = -(G'O', G'K')
> And we are done for this synthetic proof.
> Friendly
> Francois
> PS
> Now the problem is to prove existence of the map f of course.
> So Nikos proof is the only one available for the moment!
>
> Dear Alexey
> > With Cabri, I found 2 other properties of the map f leading, I think, to
>
> > a solution of your problem.
> > I label I and J' the limit points of the direct circular map f, that is
> to
> > say:
> > f(I) = Infty anf f(Infty) = J'
> > (Here Infty is the point at infinity of the circular plane.)
> > Then I and K are inverse wrt the ABC-circumcircle
> > and J' and K' are inverse wrt the A'B'C' circumcircle.
> > Friendly
> > Francois
> >
> >
> >
>
> [Non-text portions of this message have been removed]
>
>
>
>
> Antivirus scanning: Symantec Mail Security for SMTP.
>
> [Non-text portions of this message have been removed]
>
>
>


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

Dear Geometers,

The GREAT book Advanced Euclidean Geometry – Roger A. Jonhson is
available again.
www.amazon.com.
Dover publications
ISBN: 0486462374.

Sincerely,
Deoclecio Gouveia Mota Junior

Dear Alexey and Nikos and dear friends

Here a short summary (without proof) of the theory  of the metapoles:

We start with 2 triangles ABC and A'B'C' and  to have  existence and
unicity in the following theorem , one suppose that they are not inversely
similar.

Theorem 1
There exists a unique point D which is a common center of  similarity or
pivot for the triangles abc circumscribed to ABC, (A on line bc, B on line
ca, C on line ab), and directly similar to A'B'C'.

D is named the metapole of triangle ABC with respect to triangle A'B'C'
Of course we can swap the roles of ABC and A'B'C' and get the point D'
metapole of the triangle A'B'C' with respect to triangle ABC.

Theorem 2
We have the following equalities between signed angles of lines:
<alpha> = (AD, B'C') = (BD, C'A') = ( CD, A'B')    (mod pi)
-<alpha> = (A'D', BC) = (B'D', CA) = (C'D', AB)    (mod pi)

In case: <alpha> = 0  (mod pi), triangles ABC and A'B'C' are parallelogic
and points D and D' are named centers of parallelogy.
In case: <alpha> = pi/2   (mod pi), triangles ABC and A'B'C' are orthologic
and points D and D' named centers of orthology.
In the general case, A'B'C' is named <alpha>-isogonologic with respect to
ABC, so ABC is -<alpha>-isogonologic with respect to triangle A'B'C'.

Theorem 3 (Isogonologic map)
Let f be the affine map sending ABC onto A'B'C'. Then for any triangle LMN,
triangle f(L)f(M)f(N) is <alpha>- isogonologic with respect to triangle
LMN.
So we can tell that <alpha> is the angle of isogonology of the map f or that
f is an <alpha> isogonologic map.
Every affine map f different from an inverse similarity is
<alpha>-isogonologic for some unique <alpha> (mod pi).

Theorem 3 (Affine invariance of the metapoles)
Let f be the affine map sending ABC on A'B'C', then:  D' = f(D).

Theorem 4 (Circular invariance of the metapoles)
Let g be the direct circular map sending  ABC onto A'B'C', then we have also
: D' = g(D).

Theorem 5 (Characterization of the limit points)
Let I be the object limit point of the map g, that is to say g(I) = Infty,
the point at infinity of the circular plane.
Let U be the isogonal  conjugate of the metapole D with respect to triangle
ABC.
Then
1° I is the inverse point of U in the inversion with respect to the
ABC-circumcircle.
2° g(U) = O', the A'B'C'-circumcenter
In using the inverse map g^(-1), we have the following corollary:

Corollary
Let J' be the the image limit point of g, that is to say: g(Infty) = J'.
Let V' be the isogonal conjugate of the metapole D' with respect to triangle
A'B'C'.
Then
1° J' is the inverse point of V' in the inversion with respect to the
A'B'C'-circumcircle.
2° g(O) = V' where O is the ABC-circumcenter.

Theorem 7 (Metapolar four points)
In the pair of four points (A, B, C, D) and (A', B', C', D'), points A and
A' are metapoles of triangles BCD and B'C'D' and so on….
So we can speak of metapolar four points (A, B, C, D) and (A', B', C', D')


In the Alexey orthologic configuration  or in my parallelogic configuration
of triangles ABC and A'B'C', the centroids G and G' are the metapoles.
The isogonal conjugate U of G wrt ABC is the Lemoine point K and so on …

Friendly
Francois

PS
1° I would be happy to get any other references on this topic if any!
2° Of course, there are other theorems that I can't give here for lack of
room!


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

Dear Francois,
You wrote
> Nikos proof is also interesting. In fact, I begin
> also to calculate the
> cotangent but was so scared by the lengths of the
> calculations that I gave
> up cowardly and went to bed. So I am eager to know
> how Nikos proceeded,
> maybe he used Mathematica , Maple or some other math
> software or may be some
> shortcut I did not see.

I used MathCad and complicated distance formulas.
Another more simple method I see now is the
following general.
Let L: px+qy+rz = 0  and  L': p'x+q'y+r'z = 0 be
two lines in barycentrics.
Let S be the double area of ABC and
A = S_A = (bb+cc-aa)/2, B = S_B, C = S_C
we have sin(<B) = S/ac and cot(<B) = B/S .
If the line L meets the sides  BC, CA, AB of ABC
at the points P, Q, R  and forms with BC angle w then
PB = aq/(q-r), RB = cq/(q-p)
sin(B-w)/sin(w) = PB/BR  or
cotw -cotB = PB/BRsinB or
cotw = aa(q-p)/S(r-q) + B/S . (1)
and for L'
cotw'= aa(q'-p')/S(r'-q') + B/S . (2)
If f is the angle of lines L, L' then
cot(f) = cot(w'-w) = (cotwcotw'+1)/(cotw-cotw'). (3)
From (3) if we substitute (1) and (2)
and aa = B+C  and SS = AB+BC+CA
we get
cot(f) = N1/N2  where
N1 = A(q-r)(q'-r')+B(r-p)(r'-p')+C(p-q)(p'-q')
N2 = S[(qr'-q'r)+(rp'-r'p)+(pq'-p'q)]

Best regards
Nikos Dergiades



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Dear friends
By curiosity, I look at the configuration of a pair of metapolar four points
(A, B, C, D) and (A', B', C', D') , (checking Alexey equality each time),
in case D is some special center or another remarkable point of triangle
ABC.

The drawing is very interesting when D is one of the Brocard point of ABC
and I suppose this special  configuration is known for a long time since
Brocard points and metapoles exist.
Friendly
Francois


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

Dear Nikos
Thank you for your explanation.
Of course, we can calculate the distance of any pair of points knowing their
barycentrics
but I am always scared by the length of the calculations.
I noticed F.G-M gives the distance of some pairs of centers (1453d) and also
cites on this topic an italian book: la recente geometria del Triangolo
written in 1900 by Cristoforo Alasia.
But maybe in this damned Kimberling list, there is something to chew ?
Friendly
Francois
PS
With the notations of my little lecture on metapoles, of course we have:
(DO, DU) = -(D'O', D'V')
with the same proof I give for Alexey equality which is then a special case
of the general metapole drawing.
But I also like your proof very much!
The existence of a triangle T' of which the lengths of the sides are the
lengths of the medians of a triangle T is not so obvious and is a nice
little problem for rookies. A little challenge is to calculate the ratio
Area(T')/Area(T) without calculating the lengths of the medians.



On 10/3/07, Nikolaos Dergiades <ndergiades@...> wrote:
>
> Dear Francois,
> You wrote
> > Nikos proof is also interesting. In fact, I begin
> > also to calculate the
> > cotangent but was so scared by the lengths of the
> > calculations that I gave
> > up cowardly and went to bed. So I am eager to know
> > how Nikos proceeded,
> > maybe he used Mathematica , Maple or some other math
> > software or may be some
> > shortcut I did not see.
>
> I used MathCad and complicated distance formulas.
> Another more simple method I see now is the
> following general.
> Let L: px+qy+rz = 0  and  L': p'x+q'y+r'z = 0 be
> two lines in barycentrics.
> Let S be the double area of ABC and
> A = S_A = (bb+cc-aa)/2, B = S_B, C = S_C
> we have sin(<B) = S/ac and cot(<B) = B/S .
> If the line L meets the sides  BC, CA, AB of ABC
> at the points P, Q, R  and forms with BC angle w then
> PB = aq/(q-r), RB = cq/(q-p)
> sin(B-w)/sin(w) = PB/BR  or
> cotw -cotB = PB/BRsinB or
> cotw = aa(q-p)/S(r-q) + B/S . (1)
> and for L'
> cotw'= aa(q'-p')/S(r'-q') + B/S . (2)
> If f is the angle of lines L, L' then
> cot(f) = cot(w'-w) = (cotwcotw'+1)/(cotw-cotw'). (3)
> From (3) if we substitute (1) and (2)
> and aa = B+C  and SS = AB+BC+CA
> we get
> cot(f) = N1/N2  where
> N1 = A(q-r)(q'-r')+B(r-p)(r'-p')+C(p-q)(p'-q')
> N2 = S[(qr'-q'r)+(rp'-r'p)+(pq'-p'q)]
>
> Best regards
> Nikos Dergiades
>
>
>
> ___________________________________________________________
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>
>
>
>
>
> Yahoo! Groups Links
>
>
>
>


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

Dear Francois,

> The existence of a triangle T' of which the lengths
> of the sides are the
> lengths of the medians of a triangle T is not so
> obvious and is a nice
> little problem for rookies. A little challenge is to
> calculate the ratio
> Area(T')/Area(T) without calculating the lengths of
> the medians.
Yes.
Let T the triangle ABC, G the centroid, M the mid
point of BC and G' the reflection of G in M.
GG' = (2/3)ma, BG = (2/3)mb,  BG' = CG = (2/3)mc
If T" is the triangle BG'G then T" is similar to T'
and (T") = (4/9)(T').
Since
(T") = 2(BMG) = (BCG) = (1/3)(T)  we get
(1/3)(T) = (4/9)(T')
or   (T') = (3/4)(T)

Best regards
Nikos Dergiades





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Dear friends,

in the ETC

X(175) is the center of the outer Soddy-circle and is called the
isoperimetric point.
This isoperimetric property is only valid if the following condition
is true:

a + b + c > 4R + r (1)

X(176) is the center of the inner Soddy-circle and is called the
equal detour point.
The ETC states that X(175) becomes a second equal detour point if the
following condition is true:

greatest angle of triangle ABC > 2.arcsin(4/5) (2)

First I thought that the conditions (1) and (2) were identical
although I found that very odd

After some numerical analysis I came to the followong conclusions:

- condition (2) is only valid when the triangle ABC is isosceles.

If angle B = angle C then I found that AB=AC=5.BC
I limited my analysis to triangles in which the longest equal sides
are integer multiples of the shortest side

- if the triangle is not isosceles then condition (2) should be
replaced by condition (1)

If the radii of the A-, B- and C-Soddycirkels have integer values I
found that:
ra = k.m^2
rb = k.(m+1)^2
rc = k.m^2.(m+1)^2 + cyclic permutations
with k and m any integer values

My final conclusions are:
- if a + b + c = 4R + r then the outer Soddy circle becomes a line
touching the A-, B- and C- Soddy-circle;
- if a + b + c > 4R + r then the A-, B- and C- Soddy-circle are all
inside the outer Soddy circle (they touch internally);
- if a + b + c < 4R + r then the A-, B- and C- Soddy-circle are all
outside the outer Soddy circle (they touch externally).

Am I right?
Any comments?

Kind regards

Eric

Dear Eric,
You are right about all your conclusions.
(I did not test your integer value option).

If x, x' are the radii of the inner and outer
Soddy circles and S is the area of ABC then
x = S/[4R + r +(a + b + c)]
x'= S/[4R + r -(a + b + c)]
So we have always 1/x - 1/x' = 4/r
and if the outer Soddy circle is a line
or 4R + r = a + b + c  then  x = r/4.

Best regards
Nikos Dergiades

> Dear friends,
>
> in the ETC
>
> X(175) is the center of the outer Soddy-circle and
> is called the
> isoperimetric point.
> This isoperimetric property is only valid if the
> following condition
> is true:
>
> a + b + c > 4R + r (1)
>
> X(176) is the center of the inner Soddy-circle and
> is called the
> equal detour point.
> The ETC states that X(175) becomes a second equal
> detour point if the
> following condition is true:
>
> greatest angle of triangle ABC > 2.arcsin(4/5) (2)
>
> First I thought that the conditions (1) and (2) were
> identical
> although I found that very odd
>
> After some numerical analysis I came to the
> followong conclusions:
>
> - condition (2) is only valid when the triangle ABC
> is isosceles.
>
> If angle B = angle C then I found that AB=AC=5.BC
> I limited my analysis to triangles in which the
> longest equal sides
> are integer multiples of the shortest side
>
> - if the triangle is not isosceles then condition
> (2) should be
> replaced by condition (1)
>
> If the radii of the A-, B- and C-Soddycirkels have
> integer values I
> found that:
> ra = k.m^2
> rb = k.(m+1)^2
> rc = k.m^2.(m+1)^2 + cyclic permutations
> with k and m any integer values
>
> My final conclusions are:
> - if a + b + c = 4R + r then the outer Soddy circle
> becomes a line
> touching the A-, B- and C- Soddy-circle;
> - if a + b + c > 4R + r then the A-, B- and C-
> Soddy-circle are all
> inside the outer Soddy circle (they touch
> internally);
> - if a + b + c < 4R + r then the A-, B- and C-
> Soddy-circle are all
> outside the outer Soddy circle (they touch
> externally).
>
> Am I right?
> Any comments?
>
> Kind regards
>
> Eric
>
>
>
>
>
> Yahoo! Groups Links
>
>
>
>
>




___________________________________________________________
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Dear friends,

sorry for the error in my previous message

If angle B = angle C then I found that AB=AC=5.BC
I limited my analysis to triangles in which the longest equal sides
are integer multiples of the shortest side

should be

If angle B = angle C then I found that AB/BC = AC/BC = 5/8

Kind regards

Eric

Dear Nikos
yes
and this is why the triangle I have labeled A'B'C' , parallelogic with ABC
with centers of parallelogy their  centroids G and G' has its sides
proportional to the medians of ABC.
Friendly
Francois

On 10/3/07, Nikolaos Dergiades <ndergiades@...> wrote:
>
> Dear Francois,
>
> > The existence of a triangle T' of which the lengths
> > of the sides are the
> > lengths of the medians of a triangle T is not so
> > obvious and is a nice
> > little problem for rookies. A little challenge is to
> > calculate the ratio
> > Area(T')/Area(T) without calculating the lengths of
> > the medians.
> Yes.
> Let T the triangle ABC, G the centroid, M the mid
> point of BC and G' the reflection of G in M.
> GG' = (2/3)ma, BG = (2/3)mb,  BG' = CG = (2/3)mc
> If T" is the triangle BG'G then T" is similar to T'
> and (T") = (4/9)(T').
> Since
> (T") = 2(BMG) = (BCG) = (1/3)(T)  we get
> (1/3)(T) = (4/9)(T')
> or   (T') = (3/4)(T)
>
> Best regards
> Nikos Dergiades
>
>
>
>
>
> ___________________________________________________________
> 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
>
>
>
>
>
> Yahoo! Groups Links
>
>
>
>


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

In message #6452 (30/01/2003), I wrote

"Related with that, the limit max(A,B,C) < 2 arcsin(4/5) is only valid if
the
triangle is isosceles. If the triangle is far to be isosceles the limit for
uniqueness of points with equal detour property go to pi/2"

Best regards,

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


Eric Danneels wrote:
> Dear friends,
>
> in the ETC
>
> X(175) is the center of the outer Soddy-circle and is called the
> isoperimetric point.
> This isoperimetric property is only valid if the following condition
> is true:
>
> a + b + c > 4R + r (1)
>
> X(176) is the center of the inner Soddy-circle and is called the
> equal detour point.
> The ETC states that X(175) becomes a second equal detour point if the
> following condition is true:
>
> greatest angle of triangle ABC > 2.arcsin(4/5) (2)
>
> First I thought that the conditions (1) and (2) were identical
> although I found that very odd
>
> After some numerical analysis I came to the followong conclusions:
>
> - condition (2) is only valid when the triangle ABC is isosceles.
>
> If angle B = angle C then I found that AB=AC=5.BC
> I limited my analysis to triangles in which the longest equal sides
> are integer multiples of the shortest side
>
> - if the triangle is not isosceles then condition (2) should be
> replaced by condition (1)
>
> If the radii of the A-, B- and C-Soddycirkels have integer values I
> found that:
> ra = k.m^2
> rb = k.(m+1)^2
> rc = k.m^2.(m+1)^2 + cyclic permutations
> with k and m any integer values
>
> My final conclusions are:
> - if a + b + c = 4R + r then the outer Soddy circle becomes a line
> touching the A-, B- and C- Soddy-circle;
> - if a + b + c > 4R + r then the A-, B- and C- Soddy-circle are all
> inside the outer Soddy circle (they touch internally);
> - if a + b + c < 4R + r then the A-, B- and C- Soddy-circle are all
> outside the outer Soddy circle (they touch externally).
>
> Am I right?
> Any comments?
>
> Kind regards
>
> Eric
>
>
>
>
>
> Yahoo! Groups Links
>
>
>

Dear Francois,

> The existence of a triangle T' of which the lengths
> of the sides are the
> lengths of the medians of a triangle T is not so
> obvious and is a nice
> little problem for rookies.

We can notice that such a triangle T' is similar to the pedal triangle
(or the circumcevian triangle) of the symedian point K.
If A'B'C' is the circumcevian triangle of K (Brocard triangle
number ??), then A'B'C' and ABC have the same symedians, thus the same
symedian point and, of course, the same circumcenter
As the centroid G' of A'B'C' is X(353) in ETC, the Alex property
<OGK = <OG'K (with opposite orientation) means that the circles OGK and
OG'K are reflected of each other wrt the line OK.
Friendly. Jean-Pierre

Dear Jean-Pierre
As you notice, this is the opposite orientation equality which do the job
and Alexey only ask for an equality with "ordinary" angles.
That's why my proof, (using alas the circular group), is important.
I have not read yet Alexey proof and maybe he proves also the same equality.
Friendly
Francois

On 10/4/07, jpehrmfr <jean-pierre.ehrmann@...> wrote:
>
>   Dear Francois,
>
> > The existence of a triangle T' of which the lengths
> > of the sides are the
> > lengths of the medians of a triangle T is not so
> > obvious and is a nice
> > little problem for rookies.
>
> We can notice that such a triangle T' is similar to the pedal triangle
> (or the circumcevian triangle) of the symedian point K.
> If A'B'C' is the circumcevian triangle of K (Brocard triangle
> number ??), then A'B'C' and ABC have the same symedians, thus the same
> symedian point and, of course, the same circumcenter
> As the centroid G' of A'B'C' is X(353) in ETC, the Alex property
> <OGK = <OG'K (with opposite orientation) means that the circles OGK and
> OG'K are reflected of each other wrt the line OK.
> Friendly. Jean-Pierre
>
>
>


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

Dear Jean-Pierre
Thanks for your help.
I had already made the drawing in this case.
Note that the metapoles are the respective centroids G and G' of ABC and its
circumcevian A'B'C'.
K and G are ABC-isogonal,  K and G' are A'B'C'-isogonal.
The direct circular map f sending ABC onto A'B'C' is involutive
(transposition in french) and its central point (both object and image limit
point) is the inverse of the Lemoine point wrt ABC-circumcircle = X(?).
I think the fixed points of f are the isodynamic points = X(?)
(I only know X(n) for n = 1 and n = 2 !)

Have you already met metapoles in triangle geometry?
I still don't understand why circular geometry plays a role in it!
If you have any references, please, let me know about it!
Friendly
Francois
On 10/4/07, jpehrmfr <jean-pierre.ehrmann@...> wrote:
>
>   Dear Francois,
>
> > The existence of a triangle T' of which the lengths
> > of the sides are the
> > lengths of the medians of a triangle T is not so
> > obvious and is a nice
> > little problem for rookies.
>
> We can notice that such a triangle T' is similar to the pedal triangle
> (or the circumcevian triangle) of the symedian point K.
> If A'B'C' is the circumcevian triangle of K (Brocard triangle
> number ??), then A'B'C' and ABC have the same symedians, thus the same
> symedian point and, of course, the same circumcenter
> As the centroid G' of A'B'C' is X(353) in ETC, the Alex property
> <OGK = <OG'K (with opposite orientation) means that the circles OGK and
> OG'K are reflected of each other wrt the line OK.
> Friendly. Jean-Pierre
>
>
>


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

I correct a little typo

On 10/7/07, Francois Rideau <francois.rideau@...> wrote:
>
> Dear Alexey
> I try to read your proof with some difficulties.
> 1° You have strange notations for your points.
> It is usual to label the centroid G, the circumcenter O, the orthocenter H
> and so on..
> As for the Lemoine point, one label it K, I don't know why? Maybe L is
> kept for the de Longchamps point
> Sure there are not enough letters to label all Kimberling little friends!
> Besides you label L the Lemoine point in your first post and K in your
> proof.
> 2° I find your triangle A'B'C' a little intricate to define.
> It is sufficient to prove your property with any triangle A'B'C' with
> sides proportional to the ABC-medians and as you notice it, the pedal
> triangle of the ABC-Lemoine point K is more suitable to do the job.
>
> So in the sequel, I label G, O, K the centroid, the circumcenter and the
> Lemoine point of triangle ABC.
> A'B'C' is the pedal point of K;


Of course A'B'C' is the pedal triangle of K.

So the centroid of A'B'C' is G' = K.
> The circumcenter O' of A'B'C' is the middle point of GK.
> So if K' is the Lemoine point of A'B'C', your property is equivalent to
> prove that KK' is parallel to the Euler line GO of triangle ABC.
> Friendly
> Francois
>


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