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

http://forumgeom.fau.edu/FG2012volume12/FG201214index.html

The editors
Forum Geometricorum

================================================================
Olga Radko and Emmanuel Tsukerman,
The perpendicular bisector construction, isotopic point and Simson line,
Forum Geometricorum, 12 (2012) 161--189.

Abstract.   Given a noncyclic quadrilateral, we consider an iterative procedure
producing a new quadrilateral at each step. At each iteration, the vertices of
the new quadrilateral are the circumcenters of the triad circles of the previous
generation quadrilateral. The main goal of the paper is to prove a number of
interesting properties of the limit point of this iterative process. We show
that the limit point is the common center of spiral similarities taking any of
the triad circles into another triad circle. As a consequence, the point has the
isoptic property i.e., all triad circles are visible from the limit point at the
same angle. Furthermore, the limit point can be viewed as a generalization of a
circumcenter. It also has properties similar to those of the isodynamic point of
a triangle.} We also characterize the limit point as the unique point for which
the pedal quadrilateral is a parallelogram. Continuing to study the pedal
properties with respect to a quadrilateral, we show that for every quadrilateral
there is a unique point (which we call the Simson point) such that its pedal
consists of four points on a line, which we call the Simson line, in analogy to
the case of a triangle. Finally, we define a version of isogonal conjugation for
a quadrilateral and prove that the isogonal conjugate of the limit point is a
parallelogram, while that of the Simson point is a degenerate quadrilateral
whose vertices coincide at infinity.


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

> From: rhutson2 <rhutson2@...>
>
> Dear friends:
>
> The incircle and Steiner inellipse intersect in 4 points, one of which, say P,
> is a triangle center, and the other 3, say A', B', C', form a
> central triangle (similar to the intersections of NPC and Steiner inellipse =
> X(115) + medial triangle).  P is a non-ETC center, search=2.307963780976356,
and
> A'B'C' is perspective to ABC at non-ETC center, say X,
> search=1.626790309962831.
>
> Questions:
>
> 1.) What are the coordinates of P, A', B', C', and X?

Let t1=sqrt((c+a-b)(a+b-c)), t2=sqrt((a+b-c)(b+c-a)), t3=sqrt((b+c-a)(c+a-b)),
with t1,t2,t3 all > 0. Then P=(a-t1, b-t2, c-t3),
A'=(a-t1, b+t2, c+t3), B'=(a+t1, b-t2, c+t3),
C'=(a+t1, b+t2, c-t3) and X=(a+t1, b+t2, c+t3).

>
> 2.) For the ETC reference triangle, A' is the farthest of the 4
> intersections from A, and likewise for B' and C'.  Does this hold in
> general?
>
> 3.) For the ETC reference triangle, P is the closest of the 4 intersections to
> X(11).  Does this hold in general?
>
> 4.) Any other interesting properties?
>
> 5.) Generalizations for other inconics?
>
> Thanks in advance,
> Randy Hutson

--
Barry Wolk

Dear friends,

[Randy Hutson]
> > The incircle and Steiner inellipse intersect in 4
> points, one of which, say P,
> > is a triangle center, and the other 3, say A', B', C',
> > Questions:
> > 1.) What are the coordinates of P, A', B', C', and X?

[Barry Wolk]
> Let t1=sqrt((c+a-b)(a+b-c)), t2=sqrt((a+b-c)(b+c-a)),
> t3=sqrt((b+c-a)(c+a-b)),
> with t1,t2,t3 all > 0. Then P=(a-t1, b-t2, c-t3),
> A'=(a-t1, b+t2, c+t3), B'=(a+t1, b-t2, c+t3),
> C'=(a+t1, b+t2, c-t3) and 
> X=(a+t1, b+t2, c+t3).

[Randy Hutson]
> Generalizations for other inconics?

If the isotomic conjugates of the persectors of two
inconics are interior points of ABC with
barycentric coordinates (pp : qq : rr),  (PP : QQ : RR)
then their intersections are
S = ( (Qr-qR)^2 : (Rp-rP)^2 : (Pq-pQ)^2 )
A'= ( (Qr-qR)^2 : (Rp+rP)^2 : (Pq+pQ)^2 )
B'= ( (Qr+qR)^2 : (Rp-rP)^2 : (Pq+pQ)^2 )
C'= ( (Qr+qR)^2 : (Rp+rP)^2 : (Pq-pQ)^2 )
and the triangles ABC, A'B'C' are perspective at
X = ( (Qr+qR)^2 : (Rp+rP)^2 : (Pq+pQ)^2 )

Best regards
Nikos Dergiades

Thanks Barry and Nikos.  I love it when these problems have such elegant
solutions.

Best regards,
Randy Hutson




>________________________________
> From: Nikolaos Dergiades <ndergiades@...>
>To: Hyacinthos@yahoogroups.com
>Sent: Friday, June 1, 2012 5:33 PM
>Subject: Re: [EMHL] intersections of incircle and Steiner inellipse
>
>Dear friends,
>
>[Randy Hutson]
>> > The incircle and Steiner inellipse intersect in 4
>> points, one of which, say P,
>> > is a triangle center, and the other 3, say A', B', C',
>> > Questions:
>> > 1.) What are the coordinates of P, A', B', C', and X?
>
>[Barry Wolk]
>> Let t1=sqrt((c+a-b)(a+b-c)), t2=sqrt((a+b-c)(b+c-a)),
>> t3=sqrt((b+c-a)(c+a-b)),
>> with t1,t2,t3 all > 0. Then P=(a-t1, b-t2, c-t3),
>> A'=(a-t1, b+t2, c+t3), B'=(a+t1, b-t2, c+t3),
>> C'=(a+t1, b+t2, c-t3) and 
>> X=(a+t1, b+t2, c+t3).
>
>[Randy Hutson]
>> Generalizations for other inconics?
>
>If the isotomic conjugates of the persectors of two
>inconics are interior points of ABC with
>barycentric coordinates (pp : qq : rr),  (PP : QQ : RR)
>then their intersections are
>S = ( (Qr-qR)^2 : (Rp-rP)^2 : (Pq-pQ)^2 )
>A'= ( (Qr-qR)^2 : (Rp+rP)^2 : (Pq+pQ)^2 )
>B'= ( (Qr+qR)^2 : (Rp-rP)^2 : (Pq+pQ)^2 )
>C'= ( (Qr+qR)^2 : (Rp+rP)^2 : (Pq-pQ)^2 )
>and the triangles ABC, A'B'C' are perspective at
>X = ( (Qr+qR)^2 : (Rp+rP)^2 : (Pq+pQ)^2 )
>
>Best regards
>Nikos Dergiades
>
>
>
>------------------------------------
>
>Yahoo! Groups Links
>
>
>
>
>
>

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

Dear colleagues!
Is next fact known?
Let P, Q be two point on the circumcircle of triangle ABC. Then their Simson
lines meet on the altitude from C iff PQ is parallel to AB.

Sincerely                                                Alexey

[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/FG2012volume12/FG201215index.html

The editors
Forum Geometricorum

Albrecht Hess, A highway from Heron to Brahmagupta,
Forum Geometricorum, 12 (2012) 191--192.

Abstract.   We give a simple derivation of Brahmagupta's area formula for a
cyclic quadrilateral from Heron's formula for the area of a triangle.


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

Yes, it is a particular case of S-triangles.

Two triangles ABC and A'B'C' inscribed in the same circle are S-triangles iff
arc(AA')+arc(BB')+arc(CC')=0 (mod 2PI).
The Simson lines of A',B',C', with respect to ABC and  the Simson lines of
A,B,C, with respect to A'B'C' are concurrent in the midpoint of [HH'] where H
and H' are the orthocenters of ABC and A'B'C'.

The triangles CAB and CPQ are S-triangles iff PQ is parallel to AB.
So, the Simson lines of P,Q, and C are concurrent. But the Simson line of C is
the altitude from C.

See
[1] Traian Lalescu, A Class of Remarkable Triangles,
Gazeta Matematica , vol XX, feb. 1915 ,p 213
(in Romanian)

[2] Trajan Lalesco, La geometrie du triangle ,Bucharest,
1937, Paris, Libraire Vuibert

Sincerely
Daniel Vacaretu


--- In Hyacinthos@yahoogroups.com, "Alexey Zaslavsky" <zasl@...> wrote:
>
> Dear colleagues!
> Is next fact known?
> Let P, Q be two point on the circumcircle of triangle ABC. Then their Simson
lines meet on the altitude from C iff PQ is parallel to AB.
>
> Sincerely                                                Alexey
>
> [Non-text portions of this message have been removed]
>

Dear Daniel,
Very good observation and proof.
So we can generalize Alexey's problem
as follows:
If L is a line and S is the isogonal conjugate
of its infinite point and the chord PQ of the
circumcircle of ABC is parallel to L then the
triangles ABC, SPQ are S-triangles.
Hence the Simson lines of P, Q relative to ABC
meet at M' midpoint of HH'. The locus of H' orthocenter
of SPQ is the perpendicular line from S to L.
Hence the locus of M' is the perpendicular line
from the midpoint M of HS to L.

If L is parallel to BC then S=A, M=H.
Best regards
Nikos Dergiades


> Áðü: Daniel <dvacaretu@...>
> Yes, it is a particular case of
> S-triangles.
>
> Two triangles ABC and A'B'C' inscribed in the same circle
> are S-triangles iff arc(AA')+arc(BB')+arc(CC')=0 (mod 2PI).
> The Simson lines of A',B',C', with respect to ABC and 
> the Simson lines of A,B,C, with respect to A'B'C' are
> concurrent in the midpoint of [HH'] where H and H' are the
> orthocenters of ABC and A'B'C'.
>
> The triangles CAB and CPQ are S-triangles iff PQ is parallel
> to AB.
> So, the Simson lines of P,Q, and C are concurrent. But the
> Simson line of C is the altitude from C.

> "Alexey Zaslavsky" <zasl@...> wrote:
> >
> > Dear colleagues!
> > Is next fact known?
> > Let P, Q be two point on the circumcircle of triangle
> ABC. Then their Simson lines meet on the altitude from C iff
> PQ is parallel to AB.

An easier proof would be to just recall that the Simson line bisects the segment
determined by the orthocenter of the triangle and the point, thus if we let X, Y
be the midpoints of HP and HQ (where H is the orthocenter of ABC), the result
follows at once by Desargues' theorem applied to triangles PXM, QYN, where M, N
are the projections of P, Q on AB.

Best,
Cosmin

--- In Hyacinthos@yahoogroups.com, "Alexey Zaslavsky" <zasl@...> wrote:
>
> Dear colleagues!
> Is next fact known?
> Let P, Q be two point on the circumcircle of triangle ABC. Then their Simson
lines meet on the altitude from C iff PQ is parallel to AB.
>
> Sincerely                                                Alexey
>
> [Non-text portions of this message have been removed]
>

> From: Daniel <dvacaretu@...>
>
> Yes, it is a particular case of S-triangles.
>
> Two triangles ABC and A'B'C' inscribed in the same circle are
> S-triangles iff arc(AA')+arc(BB')+arc(CC')=0 (mod 2PI).
> The Simson lines of A',B',C', with respect to ABC and  the Simson
> lines of A,B,C, with respect to A'B'C' are concurrent in the
> midpoint of [HH'] where H and H' are the orthocenters of ABC and
> A'B'C'.

A routine calculation using complex numbers as coordinates gives
the following:   If A,B,C,D,E are on the unit circle, then the Simson
lines of D and of E (wrt ABC) meet at (A+B+C+D+E+ABC/(DE))/2
 
This formula is simpler than expected, even though it was known
in advance that it had to have some symmetry and scaling properties.
And the S-triangle result follows easily from this.
--
Barry Wolk

Dear Daniel, Nikos, Cosmin and Barry!
Yes, I also found these general facts. Thank you for the references. Also the
envelop of Simson lines is the astroid and its center coincides with the center
of NPC. And what is the locus of point P such that the circumcevian triangle of
P is S-triangle? It is clear that this is some cubic, but which?

Sincerely                                        Alexey


   ----- Original Message -----
   From: Barry Wolk
   To: Hyacinthos@yahoogroups.com
   Sent: Thursday, June 07, 2012 11:33 PM
   Subject: [EMHL] Re: Simson lines



   > From: Daniel <dvacaretu@...>
   >
   > Yes, it is a particular case of S-triangles.
   >
   > Two triangles ABC and A'B'C' inscribed in the same circle are
   > S-triangles iff arc(AA')+arc(BB')+arc(CC')=0 (mod 2PI).
   > The Simson lines of A',B',C', with respect to ABC and  the Simson
   > lines of A,B,C, with respect to A'B'C' are concurrent in the
   > midpoint of [HH'] where H and H' are the orthocenters of ABC and
   > A'B'C'.

   A routine calculation using complex numbers as coordinates gives
   the following:   If A,B,C,D,E are on the unit circle, then the Simson
   lines of D and of E (wrt ABC) meet at (A+B+C+D+E+ABC/(DE))/2

   This formula is simpler than expected, even though it was known
   in advance that it had to have some symmetry and scaling properties.
   And the S-triangle result follows easily from this.
   --
   Barry Wolk





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

Dear Alexey,

> Yes, I also found these general facts. Thank you for the references. Also the
envelop of Simson lines is the astroid and its center coincides with the center
of NPC. And what is the locus of point P such that the circumcevian triangle of
P is S-triangle? It is clear that this is some cubic, but which?

see K024 in

http://bernard.gibert.pagesperso-orange.fr/Exemples/kjp.html

Best regards

Bernard

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

Dear Bernard!

see K024 in

http://bernard.gibert.pagesperso-orange.fr/Exemples/kjp.html

Thank you.

Sincerely                             Alexey

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

Dear Hyacinthists,
I have read that a result of Droz-Farny is associated with the name of NOYER.
Who is Noyer? Where can I find some informations about it? his contribution ?
Sincerely
Jean-Louis

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

Dear Friends,

I noticed there are all kind of centers, lines, conics, cubics, etc. in a
Complete Quadrilateral/Complete Quadrangle.
So I started writing a paper on this subject.
Because I discovered so many new items I decided to make a catalogue of them.
More than 150 items are catalogued now.
I gave them all a unique code and name.
* Jean-Louis'  Euler-Poncelet Point (Hyacinthos message  19258) is coded with
QA-P2.
* Jean-Pierre's Homothetic Center (Hyacinthos message 19635) is coded with
QA-P4.
* Apart from the well-known quadrangle centroid QA-P1 there is also a
quadrilateral centroid QL-P12.
I placed them all at my site with references where I found them.
I also found many new items. These you will find without reference.
When there are more references please let me know.

You can find this Encyclopedia of Quadri-Figures (EQF) at:
http://www.chrisvantienhoven.nl/mathematics/encyclopedia.html

The results also can be downloaded in PDF-format at:
http://www.chrisvantienhoven.nl/7-mathematics/191-downloads-eqf.html

Special thanks for Eckart Schmidt who painstakingly checked all items and helped
me to eliminate several typos and mistakes and gave me lots of good advice. We
had a very nice conversation alternately in German and English.

When you know about other Quadrangle/Quadrilateral items please let me know.

Best regards,
Chris van Tienhoven

I have posted a synthetic solution(given by me and Sayan Mukherjee) of the
Neuberg problem here:-
http://www.artofproblemsolving.com/blog/71562

Can you refer to some other synthetic solution of this problem?

With regards,
Chandan.

Dear Hyacinthists,

Barycentric coordinates are used in respect of a triangle ABC. Let P=(u:v:w) be
a point not on a sideline of ABC and let d=(p:q:r) be a line other than the side
lines of ABC. The P-isoconjugate of d is the line

       d'=(qrv^2w^2:rpw^2u^2:pqu^2v^2).

Let PaPbPc be the cevian triangle of P; let D, E, F be the points where d meets
the sidelines BC, CA, AB; let D', E', F' be the points where d' meets the
sidelines BC, CA, AB, then we have the following relations between the
cross-ratios:

     (B C P_a D)=(C B P_a D'),
     (C A P_b E)=(A C P_b E'),
     (A B P_c F)=(B A P_c F').

The isogonal conjugate of d is the X(1)-isoconjugate of d: The lines AD and AD'
are symmetric with respect to the bisector of angle A.

The isotomic conjugate of d is the X(2)-isoconjugate of d: The points D and D'
are symmetric with respect to the midpoints of the side B$.



If the lines d and d' are P-isoconjugate and perpendicular, then the point of
intersection of d and d' is on bicevian conic C(P,H), where H is the orthocenter
of ABC.



   Let Q be a point not on a sideline of ABC. Let d be a line through Q and d'
its P-isoconjugate, then as d varies the locus of the point of intersection of d
and d' is a cubic L(P,Q) through Q (which are double) and vertices of cevian
triangle of P and Q . There are three pairs of P-isoconjugate and perpendicular
lines (d,d') which intersec in the three points of intersection (other than
Pa,Pb and Pc) of C(P,H) and L(P,Q).

   Consider now the point Q on the bicevian conic C(P,H), we denote the point of
intersection (other than Pa, Pb, Pc and Q) of the C(P,H) and L(P,Q) by S[P,Q].
We call this the "sixth intersection" of the bicevian conic C(P,H) with the
cubic L(P,Q).
The line d_1 passing through Q and S[P,Q] is perpendicular to its
P-isoconjugate line d'_1 , and the point of intersection of d_1 and d'_1 is
S[P,Q]. Let Q' be the second point of intersection of d'_1 with C(P,H) then
S[P,Q]=S[P,Q'].
The tangents d_2 and d_3 in Q to L(P,Q) are perpendicular and
P-isonconjugate to each other.


GeoGebra Construction:
http://webpages.ull.es/users/amontes/geogebra/ConjugateAndPerpendicularLinesH.ht\
ml

====================================================
Someone could suggest a construction of "sixth intersection S[P,Q]" of the
bicevian conic C(P,H) with the cubic L(P,Q), if the point Q is  on C(P,H).
=====================================================


Some examples:


-----------------------------------------------------

The nine-point circle: C(X(2),X(4))

S[X(2),X(11)]=S[X(2),X(119)]= ((b-c)^2(b+c-2a)
(a^2(b+c)+(b+c-2a)b*c-(b^3+c^3))): ... : ...) = X(1647)X(908)

S[X(2),X(114)]=S(X(2),X(115))=X(2679)

S[X(2),X(116)]=S[X(2),X(118)]= X(1566)

S[X(2),X(117)]=S[X(2),X(124)]= ((b-c)^2(b+c-a)(b^2+c^2-a^2)
                                (a^2(b+c)-2a*b*c-(b-c)^2(b+c))
   (2a^4-a^3(b+c)-a^2(b-c)^2+a(b-c)^2(b+c)-(b^2-c^2)^2): ... : ...)

S[X(2),X(120)]=S[X(2),Q']= ((b-c)^2(a^2-2a(b+c)+b^2+c^2)
        (2a^2-a(b+c)+(b-c)^2)  (b^2+ c^2-a(b+c)): ...:...)

S{X(2),X(121)]=S{X(2),Q']= ((2a-b-c)(3a-b-c)(b-c)^2
                       (a(b+c)+b^2-4b*c+c^2): ... : ...)

-------------------------------------------------

Bicevian conic C(X(1),X(4))


  S[X(1),X(11)]=S[X(1),Q']= (a^2(b-c)^2
          (a^2*(b+c)-2a*b*c-(b-c)^2(b+c))
     (a^4-a^3(b+c)-a^2(b^2-3b*c+c^2)+a(b+c)(b^2-4b*c+c^2)+b*c(b+c)^2):
                ... : ...)

     Where Q'=(a^2(a-b-c)(a^2*(b+c)-2a*b*c-(b-c)^2(b+c))
        (a^5*(b+c)-a^3(b+c)(2*b^2-3b*c+2c^2)-a^2b*c(b-c)^2+
             a(b-c)^2(b^3+c^3)+b*c(b^2-c^2)^2) : ... : ...)


   S[X(1),X(2310)]=S[X(1),Q']= (a(b+c)(b-c)^2
             (-a+b+c)^3(a+b-c)^2(a-b+c)^2
             (a+b+c)^4(a^3+b^3+c^3-2a^2(b+c)+a*b*c)
             (a^4-a^2(b^2-b*c+c^2)-b*c(b-c)^2): ... : ...)

       Where  Q'=( a(b+c-a)(a^4-a^2(b^2-b*c+c^2)-b*c(b-c)^2
         (a^4(b+c)-2a^3*b*c-2a^2(b-c)^2(b+c)-b*c(b^3+c^3)+b^5+c^5)
           : ... : ...)

-------------------------------------------------------------

Bicevian conic C(X(1),X(6))

    S[X(6),X(115)]=S[X(6),Q']= (a^4(b^2-c^2)^2(a^2(b^2+c^2)-b^4-c^4)
(a^6-2a^4(b^2+c^2)+a^2(b^4-b^2c^2+c^4)+b^2c^2(b^2+c^2)): ...:...)

  Where Q'=(a^4(a^2(b^2+c^2)-b^4-c^4)
       (a^8(b^2+c^2)   -a^6(3b^4 + 2b^2c^2+3c^4)  +
         3a^4(b^6+c^6)-b^2c^2(b^2-c^2)^2(b^2+c^2)   -
         a^2(b^2-c^2)^2(b^4+c^4) )  : ... : ...)

-----------------------------------------------------------



Best regards
Angel

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

http://forumgeom.fau.edu/FG2012volume12/FG201216index.html

The editors
Forum Geometricorum

Debdyuti Banerjee and Nikoloas Dergiades, Alhazen's circular billiard problem,
Forum Geometricorum, 12 (2012) 193--196.

Abstract.   In this paper we give two simple geometric constructions of two
versions of the famous Alhazen's circular billiard problem.


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

Dear all

how can i find the proof of these results,17 circles concure,?
please put the proof or if the proof is already exist here,put the link or
address.

With Regards.

--- In Hyacinthos@yahoogroups.com, Quang Tuan Bui <bqtuan1962@...> wrote:
>
> Dear All Hyacinthians,
>   In my previous message I still missed a lot of circles. We take O1, O2 only
for triangle ABC wrt Cevian point P. In fact, we can take O1, O2 for four
triangles wrt Cevian point as rest point. I denote here all O1, O2 type for more
clear:
>               O1P, O2P (triangle ABC, Cevian point P)
>               O1A, O2A (triangle PBC, Cevian point A)
>               O1B, O2B (triangle PAC, Cevian point B)
>               O1C, O2C (triangle PAB, Cevian point C)
>   So with only two O1, O2 types, we have already eight concurrent circles. If
we include all Cevian, pedal and nine point circles, now we have total 17
concurrent circles:
>               - Cp is Cevian circle (any four points have only one Cevian
circle)
>               - E, OMa, OMb, OMc (nine point circle of each three points from
A, B, C, P)
>               - Op, OHa, OHb, OHc (pedal circle of each point from A, B, C, P
wrt rest points)
>               - O1P, O2P, O1A, O2A, O1B, O2B, O1C, O2C (our new pedal foot
circles of type 1 and type 2)
>   Now we can sure that the number 17 = 2^2^2+1 is proper, logical and
interesting number of concurrent circles.
>   Best regards,
>   Bui Quang Tuan
>
>
> Quang Tuan Bui <bqtuan1962@...> wrote:  Dear All Hyacinthians,
>   In my previous message "Nine Circles Concur with Nine Point Circle" I have
missed one circle. This circle is pedal circle of P wrt ABC. We denote it as Op.
So now we have ten circles concur with nine point circle and total we have
eleven concurrent circles:
>               E, Cp, O1, O2, OMa, OMb, OMc, OHa, OHb, OHc and Op
>   Please note that the concurrence of 8 from these circles is already known.
These circles are
>               E, OMa, OMb, OMc (nine point circle of each three points from A,
B, C, P)
>               Op, OHa, OHb, OHc (pedal circle of each point from A, B, C, P
wrt rest points)
>
>
> Quang Tuan Bui <bqtuan1962@...> wrote:
>
>   Let's P is arbitrary Cevian point of triangle ABC with Cevian triangle
A'B'C'.
>   Points Ab, Ac are perpendicular foots from vertex A to lines PB, PC
respectively. Points Bc, Ba, Ca, Cb are determined cyclically.
>   Ma, Mb, Mc are the midpoints of BC, CA, AB respectively.
>   Ha, Hb, Hc are the altitude foots from A, B, C  respectively.
>   Intersections of lines:
>               - AbBa and CbBc is B1
>               - BcCb and AcCa is C1
>               - CaAc and BaAb is A1
>               (Shortly: three lines AbBa, BcCb, CaAc form triangle A1B1C1)
>               - CaCb and AcAb is B2
>               - AcAb and BcBa is C2
>               - BcBa and CbCa is A2
>               (Shortly: three lines AcAb, BcBa, CaCb form triangle A2B2C2)
>   We denote ten circles as follow:
>               1. E is nine point circles
>               2. Cp is Cevian circle wrt Ceva point P (circumcircle of A'B'C')
>               3. O1 is circumcircle of triangle A1B1C1
>               4. O2 is circumcircle of triangle A2B2C2
>               5. OMa is circumcircle of MaBcCb.
>               6. OMb is circumcircle of MbCaAc.
>               7. OMc is circumcircle of McAbBa.
>               8. OHa is circumcircle of HaAbAc.
>               9. OHb is circumcircle of HbBcBa.
>               10. OHc is circumcircle of HcCaCb.
>   Result: these ten circles are concurrent at one point T.
>
>
> ---------------------------------
> Yahoo! Messenger with Voice. PC-to-Phone calls for ridiculously low rates.
>
> [Non-text portions of this message have been removed]
>

Like Quang Tuan Bui mentioned, all the circles in the message except O1,O2 are
well-known. So here are the proofs for O1,O2.
Its well-known that if V is the center of the rectangular hyperbola l passing
through A,B,C,P, then V is the concurrency point of all the other circles.
I will use the following lemma to prove the problem:-
If the circumcircle of a triangle XYZ passes through V, then the circumcircle of
the triangle formed by the polars of X,Y,Z wrt l passes through V.
It can be easily proved by simple angle-chasing. So I am not adding its proof
here.
Now note that, if A',B',C' are the feet pf the perpendiculars from P to
BC,CA,AB, then AbBa is the polar of C' wrt l and similar for others. So using
the lemma we get that O1 passes through V.
If U is the foot of the perpendicular from B to AP, then note that C'U is the
pole of Ab wrt l. Similarly, if W is the foot of the perpendicular from C to AP,
then B'W is the pole off Ac wrt l. So intersection point of C'U and B'W is the
pole of AbAc. But note that, it lies on the circumcirle of of A'B'C'. So the
circumcircle of the triangle formed by the poles of AbAc,BcBa,CaCb is same as
the circumcircle of A'B'C'. So using the lemma we get that O2 passes through V.

thank you so much Chandan Banerjee, i will be gratify if u put the solution of
the cevian and pedal circle of arbitrary point P and the nine points circle.
since i ddon't know fontene theorem and have no idea about what should i have to
do.

With Regards.

Dear Hyacinthists,
working again on the Droz-Farny line, it would be nice if somme of You have
access to these references
 
1. Droz-Farny A., Question 14111, The Educational Times 71 (1899) 89-90
2. Albert Noyer, Journal de Mathématiques Spéciales 1893 p. 39
3. W. Mantel, Mathesis 1889 p. 219
 
Thank in advance
Sincerely
Jean-Louis

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

Jean-Louis

There are many volumes of Mathesis and Mathematical Questions from the
Educational times at The Internet Archive http://archive.org/ and on Google
books, but unfortunately not the volumes you want.
I have an odd volume of the Education Times for 1891, but again not of much
help.

It's always worth looking for journals (and books) at the DML Digital
Mathematics Library at
http://www.mathematik.uni-bielefeld.de/~rehmann/DML/dml_links.html
but this has no links at all for any of the journals you want

The internet archive has good runs of
*Nouvelles annales de mathe�matiques,* *Journal de mathématiques
élémentaires [et spéciales] *before de Longchamps split it

In the 1894 edition of the latter  (p 162) there is a theorem by Noyer
giving a ref to the separate spéciales of 1893 but to page 44

If you really get nowhere, I could probably get the Mathesis and
Mathematical Questions from the Educational times refs from The British
Library

John S


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

Dear Hyacinthists,

an article concerning “La droite de Danneels’’ or ‘’Danneels’s
line’’ has been put on my website with original proofs.


http://perso.orange.fr/jl.ayme    vol. 20

Sincerely
Jean-Louis

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

Dear Friends,

I got some very nice remarks following my announcement of the Encyclopedia of
Quadri-Figures.
Seiichi Kirikami attended me on a special property of a Complete Quadrangle.
The Orthopoles of a line with respect to the four component triangles of any
complete quadrangle lie on a straight line known as the Orthopolar Line of for
the given complete quadrangle (see also "Orthopolar line" in Mathworld).
Definition Orthopole of a random line wrt some Triangle:
If perpendiculars are dropped on any line from the vertices of a triangle, then
the perpendiculars to the opposite sides from their perpendicular feet are
concurrent at a point called the Orthopole (see also "Orthopole" in Mathworld).

I tried some things with the Orthopolar Line and found these results:
1. The 4 Orthopoles of a line wrt the Component Triangles in a Complete
Quadrangle are collinear. Let's name this the QA-Orthopole Line.
2. The 4 Orthopoles of a line wrt the Component Triangles in a Complete
Quadrilateral are also collinear. Let's name this the QL-Orthopole Line.
3. Every QA-Orthopole Line of a line through QA-P4 (Isogonal Center) is a line
through QA-P2 (Euler-Poncelet Point).
It looks like this is the only case that a pencil of lines through a point is
being transformed into another pencil through a point.
The locus of their mutual crosspoint is a hyperbola through QA-P2 and QA-P4.
Its conic center is the midpoint of QA-P2 and QA-P4.
Both lines are parallel when they are parallel to the asymptotes of this
hyperbola.
4. A QL-Orthopole Line is always a line perpendicular to the Newton Line
(QL-L1). Most Special.

I refer here to some points as coded in the Encyclopedia of Quadri-Figures:
http://www.chrisvantienhoven.nl/mathematics/encyclopedia.html

Are there any (synthetic) proofs for 1., 2., 3. and 4. ?
Are there any more related properties known?

Best regards,

Chris van Tienhoven


--- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" <van10hoven@...> wrote:

> You can find this Encyclopedia of Quadri-Figures (EQF) at:
> http://www.chrisvantienhoven.nl/mathematics/encyclopedia.html
>
> The results also can be downloaded in PDF-format at:
> http://www.chrisvantienhoven.nl/7-mathematics/191-downloads-eqf.html
>

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

http://forumgeom.fau.edu/FG2012volume12/FG201217index.html

The editors
Forum Geometricorum

Dragutin Svrtan and Darko Veljan, Non-Euclidean versions of some classical
triangle inequalities,
Forum Geometricorum, 12 (2012) 197--209.

Abstract.   In this paper we recall with short proofs of some classical triangle
inequalities, and prove corresponding non-Euclidean, i.e., spherical and
hyperbolic versions of these inequalities. Among them are the well known Euler's
inequality, Rouché's inequality (also called ``the fundamental triangle
inequality"), Finsler - Hadwiger's inequality, isoperimetric inequality and
others.


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

Jean-Louis

I have found your reference to Noyer, p 39 of Journal de mathématiques
spéciales 1893

the web link at the Internet archive is
http://archive.org/details/journaldemathma22unkngoog

The Internet archive catalogue is poor. There are a mixture of journals
which are all indexed as
Journal de mathématiques élémentaires
The trick is to search for "Journal de mathématiques" using the advanced
search


The best way to download the PDF is not to do it from the list at the left
but to first choose read online
then at the top right choose the i symbol and you will get a list of
options from PDF to epub and Kindle

The title of the paper is "Sur les triangles autopolaires"

Noyer also has a paper on the theorem of Fregier on p 178
There are also 4 references to Droz Farny and he has a lot of solutions to
problems in other volumes.

Regards

John S


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

[JS]
The best way to download the PDF is not to do it from the list at the left but
to first choose read online then at the top right choose the i symbol and you
will get a list of options from PDF to epub and Kindle

Another way is using the link "All Files: HTTP" on the left, arriving to a
directory with the file in several formats:

http://ia600508.us.archive.org/16/items/journaldemathma22unkngoog/

Francisco Javier.

Dear Mathlinkers,
continuing some investigations on the end of the XIXe century, where can I found
a short biograpy of G. C. Boubals?
 
Sincerely
Jean-Louis

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

Dear  Chris van Tienhoven,

I have a Quadrangle Center that is not currently in EQF.

Let P1, P2, P3, P4 be the defining Quadrangle Points.
Let S1 = P1.P4 /\ P2.P3, S2 = P1.P3 /\ P2.P4 and S3 = P1.P2/\ P3.P4.
Now S1 S2 S3 is the QA-Diagonal Triangle of the Reference Quadrangle.


For each vertex Pi, we take the triangle TjTkTl, where Tj the
intersection of  the sidelinea PkPl with circumcircle of the triangle S1S2S3
(other than S1, S2, S3)
Qi = Perspector of the triangle PjPkPl and TjTkTl

The "unknown" Quadrangle Center is the common intersection point of lines Pi.Qi

1st CT-Coordinate:

p(q+r)(p+2q+r)^2(p+q+2r)^2
   ((1/(p+2q+r)^2)((p+r)(-c^4(p+r)^2(q+r)^2+
        (p+q)^2(a^4(p+r)^2+b^4(q+r)^2))
        ((q+r)^2(2p+q+r)^2SA+(p+r)^2(p+2q+r)^2SB+
        (p-q)^2(p+q)^2SC))+
    (1/(p+q+2r)^2)((p+q)(c^4(p+r)^2(q+r)^2+(p+q)^2(a^4(p+r)^2-
          b^4(q+r)^2))((q+r)^2(2p+q+r)^2SA+
         (p-r)^2(p+r)^2SB+(p+q)^2(p+q+2r)^2SC)))

1st DT-Coordinate:  a^2(c^4p^2q^2 + (b^4p^2 - a^4q^2)r^2)



Construction GeoGebra:

  http://amontes.webs.ull.es/geogebra/EQF_QA_Pn.html



If ABC is the diagonal  triangle of the Quadrangle with a vertex in triangle
center X(n), Then the Quadriangle Center obtained here is the TCC-perspector of
X(n). See the note just before X(1601) in ETC.


Best regards,


Angel Montesdeoca





--- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" <van10hoven@...> wrote:
>
> Dear Friends,
>
> I got some very nice remarks following my announcement of the Encyclopedia of
Quadri-Figures.
> Seiichi Kirikami attended me on a special property of a Complete Quadrangle.
> The Orthopoles of a line with respect to the four component triangles of any
complete quadrangle lie on a straight line known as the Orthopolar Line of for
the given complete quadrangle (see also "Orthopolar line" in Mathworld).
> Definition Orthopole of a random line wrt some Triangle:
> If perpendiculars are dropped on any line from the vertices of a triangle,
then the perpendiculars to the opposite sides from their perpendicular feet are
concurrent at a point called the Orthopole (see also "Orthopole" in Mathworld).
>
> I tried some things with the Orthopolar Line and found these results:
> 1. The 4 Orthopoles of a line wrt the Component Triangles in a Complete
Quadrangle are collinear. Let's name this the QA-Orthopole Line.
> 2. The 4 Orthopoles of a line wrt the Component Triangles in a Complete
Quadrilateral are also collinear. Let's name this the QL-Orthopole Line.
> 3. Every QA-Orthopole Line of a line through QA-P4 (Isogonal Center) is a line
through QA-P2 (Euler-Poncelet Point).
> It looks like this is the only case that a pencil of lines through a point is
being transformed into another pencil through a point.
> The locus of their mutual crosspoint is a hyperbola through QA-P2 and QA-P4.
> Its conic center is the midpoint of QA-P2 and QA-P4.
> Both lines are parallel when they are parallel to the asymptotes of this
hyperbola.
> 4. A QL-Orthopole Line is always a line perpendicular to the Newton Line
(QL-L1). Most Special.
>
> I refer here to some points as coded in the Encyclopedia of Quadri-Figures:
> http://www.chrisvantienhoven.nl/mathematics/encyclopedia.html
>
> Are there any (synthetic) proofs for 1., 2., 3. and 4. ?
> Are there any more related properties known?
>
> Best regards,
>
> Chris van Tienhoven
>
>
> --- In Hyacinthos@yahoogroups.com, "Chris Van Tienhoven" <van10hoven@> wrote:
>
> > You can find this Encyclopedia of Quadri-Figures (EQF) at:
> > http://www.chrisvantienhoven.nl/mathematics/encyclopedia.html
> >
> > The results also can be downloaded in PDF-format at:
> > http://www.chrisvantienhoven.nl/7-mathematics/191-downloads-eqf.html
> >
>