Then the circumcircles of the triangles
AO2O3, BO3O1, CO1O2
are concurrent.
Anopolis #850
For a proof I offer the book:
R. G. SANGER: SYNTHETIC PROJECTIVE GEOMETRY (1939)
APH
RE: [EGML] CONIC - LOCUS
Posted By: cesar_e_lozada
Fri Sep 6, 2013 10:28 pm
Dear Antreas:
They are concyclic for all points P.
The center of their circle is the complement of the isotomic of P.
Regards
Cιsar Lozada
De: Anopolis@yahoogroups.com [mailto:Anopolis@yahoogroups.com] En nombre de Antreas Hatzipolakis
Enviado el: Jueves, 05 de Septiembre de 2013 06:28 p.m.
Para: anopolis@yahoogroups.com
Asunto: [EGML] CONIC - LOCUS
Let ABC be a triangle and A'B'C' the cevian triangle of P.
Denote:
Ab, Ac = the intersections of the circles with diameters BC, AA'
(near to B,C, resp.)
Bc, Ba = the intersections of the circles with diameters CA, BB'
Ca, Cb = the intersections of the circles with diameters AB, CC'
For P = H, the six points are concyclic.
For which P's are the six points lying on a conic?
APH
Anopolis #956
In the book:
Marc Barbut, Bernard Monjardet: Ordre et Classification: Algebre et Combinatoire. Tome II. Hachette Universite, Paris, 1970.
are listed some integer sequences.
In pp. 44-46 the following:
1,3,19,219,4231,130023,6129859,.... It is A001035 in OEIS (The On-Line Encyclopedia of Integer Sequences)
1,4,29,355,6942,209527,9535241..... It is A000798 in OEIS
1,2,5,16,63,316,..... NOT FOUND IN OEIS
1,3,9,33,139,.... It is A001930 in OEIS 1,3,13,75,541,4683,47293,... It is A000670 in OEIS
In p. 101 the following:
1,2,5,,15,52,203,876,... It is A056273 in OEIS
0,1,3,7,15,31,63,.... It is A000225 in OEIS
0,0,1,6,25,90,301,... It is A000392 in OEIS
0,0,0,1,20,65,350,... NOT FOUND IN OEIS
0,0,0,0,1,15,140,... NOT FOUND IN OEIS
In pp. 165-166 the following:
1,4,18,166,7579,7828352,2414682040996,... It is A007153 in OEIS
1,3,8,28,208,.... NOT FOUND IN OEIS
1,2,4,12,81,... It is A001206 in OEIS
1,1,2,3,7,30,703,... NOT FOUND IN OEIS
BIBLIOGRAPHY:
For pages 1-78:
For pages 83-166:
This "theorem" (conjecture) is still unproved (quoetd below).
Seiichi Kirikami has computed the coordinates for (P, P*) = (G, K)
Available here: Hyacinthos #21992
I offer the following books for proofs:
1. For an analytic proof by computing the homogeneous coordinates of the concurrence points:
RICHARD HEGER: ELEMENTE DER ANALYTISCHEN GEOMETRIE IN HOMOGENEN COORDINATEN (1872)
2. For a synthetic proof:
FRANK MORLEY & F. V. MORLEY: INVERSIVE GEOMETRY
3. For any other proof (by complex numbers, etc):
C. ZWIKKER: THE ADVANCED GEOMETRY OF PLANE CURVES AND THEIR APPLICATIONS.
Antreas
--- In Anopolis@yahoogroups.com, "Antreas"
>
> Let ABC be a triangle and P,P* two isogonal conjugate points.
> Denote: H1,H2,H3 = the orthocenters of PBC, PCA, PAB, resp. and
> Ha,Hb,Hc = the orthocenters of P*BC, P*CA, P*AB, resp.
>
> The circumcircles of:
> (1) H1HbHc, H2HcHa, H3HaHb
> (2) HaH2H3, HbH3H1, HcH1H2
>
> are concurrent.
>
> Figure: Here
>
> If P = (x:y:z), which are the points of concurrences?
>
> APH
>
Re: RADICAL CENTERS - NPC - OI LINE
Posted By: amontes1949
Wed Aug 21, 2013 4:44 am
[Antreas P. Hatzipolakis]:
Let ABC be a triangle and A'B'C' the cevian triangle of I.
Denote:
(Nab), (Nac) = the NPCs of AIB', AIC', resp.
(Nbc), (Nba) = the NPCs of BIC', BIA', resp.
(Nca), (Ncb) = the NPCs of CIA', CIB', resp.
R = the radical center of (Nbc), (Nca), (Nab)
S = the radical center of (Nba), (Ncb), (Nac)
*** (Trilinear ccordinates)
R = ((a + b - c) (a - b + c) (a^3 b - a b^3 + 2 a^3 c - 2 a b^2 c - b^3 c + a^2 c^2 - 3 a b c^2 - 3 b^2 c^2 - 2 a c^3 - 3 b c^3 - c^4) : (a - b - c) (a + b - c) (a^4 + 2 a^3 b - a^2 b^2 - 2 a b^3 + 3 a^3 c + 3 a^2 b c - b^3 c + 3 a^2 c^2 + 2 a b c^2 + a c^3 + b c^3) : (a - b - c) (a - b + c) (a^3 b + 3 a^2 b^2 + 3 a b^3 + b^4 + a^3 c + 2 a^2 b c + 3 a b^2 c + 2 b^3 c - b^2 c^2 - a c^3 - 2 b c^3)) T = ((a + b - c) (a - b + c) (2 a^3 b + a^2 b^2 - 2 a b^3 - b^4 + a^3 c - 3 a b^2 c - 3 b^3 c - 2 a b c^2 - 3 b^2 c^2 - a c^3 - b c^3): (a - b - c) (a + b - c) (a^3 b - a b^3 + a^3 c + 2 a^2 b c - 2 b^3 c + 3 a^2 c^2 + 3 a b c^2 - b^2 c^2 + 3 a c^3 + 2 b c^3 + c^4) : (a - b - c) (a - b + c) (a^4 + 3 a^3 b + 3 a^2 b^2 + a b^3 + 2 a^3 c + 3 a^2 b c + 2 a b^2 c + b^3 c - a^2 c^2 - 2 a c^3 - b c^3) )
R and S is a bicentric pair, then the midpoint of RS is a triangle center of trilinear ccordinates:
M = ( (a + b - c) (a - b + c) (3 a^3 b + a^2 b^2 - 3 a b^3 - b^4 + 3 a^3 c - 5 a b^2 c - 4 b^3 c + a^2 c^2 - 5 a b c^2 - 6 b^2 c^2 - 3 a c^3 - 4 b c^3 - c^4) : (a - b - c) (a + b - c) (a^4 + 3 a^3 b - a^2 b^2 - 3 a b^3 + 4 a^3 c + 5 a^2 b c - 3 b^3 c + 6 a^2 c^2 + 5 a b c^2 - b^2 c^2 + 4 a c^3 + 3 b c^3 + c^4) : (a - b - c) (a - b + c) (a^4 + 4 a^3 b + 6 a^2 b^2 + 4 a b^3 + b^4 + 3 a^3 c + 5 a^2 b c + 5 a b^2 c + 3 b^3 c - a^2 c^2 - b^2 c^2 - 3 a c^3 - 3 b c^3) )
M lie on the central line X(1)X(3).
Ideal point of line RT is X(513).
This bicentric pair does not appear in the current edition of "BICENTRIC PAIRS OF POINTS" ( Clark Kimberling)
I suggest as a flower name for this bicentric pair: Canarina
Angel Montesdeoca
Anopolis #860
Denote:
A(b,c,d), B(c,d,a),C(d,a,b), D(a,b,c) = the antimedial triangles of (b,c,d), (c,d,a),(d,a,b), (a,b,c), resp.
Oa, Ob, Oc, Od = the circumcenters of A(b,c,d), B(c,d,a),C(d,a,b), D(a,b,c), resp. [=orthocenters of (b,c,d), (c,d,a),(d,a,b), (a,b,c), lying on the Steiner line]
M = the Miquel point of the quadrilateral (= the point of concurrence of the circumcircles of (b,c,d), (c,d,a),(d,a,b), (a,b,c))
Are the orthopoles of MOa, MOb, Moc, MOd with respect A(b,c,d), B(c,d,a),C(d,a,b), D(a,b,c) concyclic?
The orthopoles lie on the NPCs of A(b,c,d), B(c,d,a),C(d,a,b), D(a,b,c) [ = circumcircles of (b,c,d), (c,d,a),(d,a,b), (a,b,c)]
Antreas P. Hatzipolakis, 19 July 2013
Problem by Antreas Hatzipolakis Solution by Francisco Javier García Capitán ETC LISTING OF Q X(72803)
