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Hyacinthos · We discuss themes on Triangle Geometry

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  • Members: 3
  • Category: Geometry
  • Founded: Dec 22, 1999
  • Language: English
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Messages 21953 - 21986 of 21989   Oldest  |  < Older  |  Newer >  |  Newest
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21953 Antreas Hatzipolakis
xpolakis Send Email
Apr 14, 2013
7:27 pm
Thanks!!!!! I would like the n-th ratio be something like this: [f(n)/g(n)] * F(a,b,c) with f(n), g(n) integer sequences, which I would send to seqfan...
21954 César Lozada
cesar_e_lozada Send Email
Apr 15, 2013
4:33 am
Dear geometers: May anyone, please, confirm the coordinates of X(5448) in ETC? I get barycentrics ...
21955 Francisco Javier
garciacapitan Send Email
Apr 15, 2013
7:21 am
You're right....
21956 César Lozada
cesar_e_lozada Send Email
Apr 15, 2013
10:28 am
Dear geometers: If P=(p : q : r ) and U=( u : v : w), what’s the geometrical meaning of Q = ( p*u : q*v : r*w) for barycentric and trilinear coordinates, if...
21957 Antreas Hatzipolakis
xpolakis Send Email
Apr 15, 2013
6:37 pm
I do not find a method to construct P*U, but there are simple loci (ie we can find them by simple computations) ... line? aph ... -- [Non-text portions of this...
21958 Bernard Gibert
bernardgibert Send Email
Apr 15, 2013
6:46 pm
Dear César Lozada, ... You may want to consult §1.2.2 in http://bernard.gibert.pagesperso-orange.fr/files/isocubics.html Best regards Bernard [Non-text...
21959 Barry Wolk
wolkbarry Send Email
Apr 15, 2013
6:47 pm
... This was already answered, and is the whole plane. Another way of stating this result is:   The Poncelet points of the quads BCPP*, CAPP*, ABPP* form a...
21960 Antreas
xpolakis Send Email
Apr 15, 2013
6:50 pm
I guess that for P = I (instead of N) the ratio should be simpler.... ie Let A1, B1, C1 be the NPC centers of IBC, ICA, IAB and O1 the circumcenter of A1B1C1. ...
21962 Francisco Javier
garciacapitan Send Email
Apr 15, 2013
7:35 pm
[APH] Which is the locus of P such that the circumcenter of A1B1C1 is on the Euler line (where A1,B1,C1 are the NPC centers of PBC, PCA, PAB)? ... It is a...
21963 rhutson2 Send Email Apr 15, 2013
9:24 pm
For P = O, Op = complement of X(157). In general, Op = complement of the nine-point-center of the antipedal triangle of P. Randy...
21964 Antreas Hatzipolakis
xpolakis Send Email
Apr 16, 2013
9:29 am
... Euler line (where A1,B1,C1 are the NPC centers of PBC, PCA, PAB)? ... I think it is the same with Bernard's Q038: ...
21965 forumgeom forumgeom
ForumGeom@... Send Email
Apr 16, 2013
3:32 pm
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2013volume13/FG201309ndex.html The editors Forum...
21966 rhutson2 Send Email Apr 16, 2013
10:17 pm
Dear friends, Let ABC be a triangle, and P a point. Let A'B'C' be the pedal triangle of P. Let Ba, Ca be the orthogonal projections of A' onto lines CA, AB,...
21967 Antreas
xpolakis Send Email
Apr 16, 2013
10:30 pm
Let ABC be a triangle. A conic intersects sidelines BC at A1, A2, CA at B1, B2 and AB at C1, C2. Theorem: (BA1 / CA1) * (BA2 / CA2) * (CB1 / AB1) * (CB2 / AB2)...
21968 Antreas
xpolakis Send Email
Apr 16, 2013
10:43 pm
Let ABC be a triangle and P a point. Which is the locus of P such that the NPC center of PBC lies on the line AP ? Ceva triangle variation: Let ABC be a...
21969 Bernard Gibert
bernardgibert Send Email
Apr 17, 2013
7:46 am
Dear Randy, ... a quintic with many simple points but only two (I think) ETC centers : X4, X1498. ... seems very difficult... Best regards Bernard [Non-text...
21970 Antreas Hatzipolakis
xpolakis Send Email
Apr 17, 2013
10:03 am
A CIRCLE: Let ABC be a triangle, A'B'C' the cevian triangle of I and N1, N2, N3 the NPC centers of IB'C', IC'A', IA'B', resp. The points I, N1,N2,N3 are...
21971 César Lozada
cesar_e_lozada Send Email
Apr 17, 2013
1:04 pm
Yes, they are concyclic. The center X of the circle has trilinears: 2*cos(A)+4*sin(3*A/2)*cos(B/2-C/2)+ cos(B-C)+2 : : ETC search: 2.387773069046934..,...
21972 César Lozada
cesar_e_lozada Send Email
Apr 17, 2013
2:45 pm
Correction: X lies on line X(I),X(J) for these (I,J): (1,30), (3,81), (5,581), (21,323), (58,5428), (140,3216), (186,2906), (386,549), (511,1385), (550,991),...
21973 yiuatfauedu Send Email Apr 17, 2013
2:58 pm
Dear Randy and Bernard, [RH] Let ABC be a triangle, and P a point. Let A'B'C' be the pedal triangle of P. Let Ba, Ca be the orthogonal projections of A' onto...
21974 Antreas
xpolakis Send Email
Apr 17, 2013
6:16 pm
... The most interesting line is the first one (X1,X30): If I0 is the center of the circle, then the line II0 is parallel to Euler line of ABC. There are three...
21975 rhutson2 Send Email Apr 17, 2013
7:37 pm
Paul, This is interesting: the perspector you mention with ETC search value 5.10435062529 matches the isogonal conjugate of the polar conjugate of X(1073), and...
21976 Paul Yiu
yiuatfauedu Send Email
Apr 17, 2013
8:18 pm
Dear Randy, Yes, S_{AB} is shorthand for (S_A)(S_B) etc. I always use barycentric coordinates unless the context clearly favors trilinear coordinates. Let's...
21977 rhutson2 Send Email Apr 18, 2013
1:33 am
A related locus: Q such that Q and the NPCs of BCQ, CAQ, ABQ are concyclic. This would include X(13), X(14), the bicentric pair PU(5), the circumcircle...
21980 Antreas
xpolakis Send Email
Apr 18, 2013
6:11 am
Let ABC be a triangle and P a point. Denote: L1, L2, L3 = the Euler Lines of PBC, PCA, PAB, resp. R1 = the radical axis of ((NPC_PBC), (O_PBC)) ie the radical...
21981 Francisco Javier
garciacapitan Send Email
Apr 18, 2013
6:30 am
This is the tricircular sextic Q014 - 4 S^2 x y z (x + y + z) (c^2 x y + b^2 x z + a^2 y z) where S=twice the area of ABC. (tricircular = the circular points...
21982 rhutson2 Send Email Apr 18, 2013
8:20 am
Francisco Javier, While Q014 mentions PU(5), it does not contain them, nor X(13) and X(14), which are on this curve. Also, X(80) which is on Q014, is not on...
21983 rhutson2 Send Email Apr 18, 2013
10:58 am
My apologies. I misread Francisco Javier's post below. As he explained privately, he meant the difference: LOCUS = SEXTIC(Q014) - SEXTIC(S 4 S^2 x y z (x + y...
21985 Antreas
xpolakis Send Email
Apr 18, 2013
1:15 pm
Let ABC be a triangle, P a point, and (Op) the cevian circle of P (with center Op). Which is the locus of P such that the Poncelet point of Op (ie the point of...
21986 xpolakis Send Email Apr 19, 2013
6:06 am
1. Hyacinthos is closed. The archive is available. 2. Hyacinthos members may subscribe to ANOPOLIS list: Related Link: http://groups.yahoo.com/group/Hyacinthos...
Messages 21953 - 21986 of 21989   Oldest  |  < Older  |  Newer >  |  Newest
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