http://groups.yahoo.com/group/Hyacinthos/ We discuss themes on Triangle Geometry Fri, 19 Apr 2013 06:06:53 GMT Hyacinthos-owner@yahoogroups.com http://groups.yahoo.com/group/Hyacinthos/message/21986 http://groups.yahoo.com/group/Hyacinthos/message/21986 1. Hyacinthos is closed. The archive is available. 2. Hyacinthos members may subscribe to ANOPOLIS list: Related Link: http://groups.yahoo.com/group/Hyacinthos Thu, 18 Apr 2013 13:15:58 GMT Antreas http://groups.yahoo.com/group/Hyacinthos/message/21985 http://groups.yahoo.com/group/Hyacinthos/message/21985 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 Thu, 18 Apr 2013 10:58:40 GMT rhutson2 http://groups.yahoo.com/group/Hyacinthos/message/21983 http://groups.yahoo.com/group/Hyacinthos/message/21983 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 Thu, 18 Apr 2013 08:20:27 GMT rhutson2 http://groups.yahoo.com/group/Hyacinthos/message/21982 http://groups.yahoo.com/group/Hyacinthos/message/21982 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 Thu, 18 Apr 2013 06:30:32 GMT Francisco Javier http://groups.yahoo.com/group/Hyacinthos/message/21981 http://groups.yahoo.com/group/Hyacinthos/message/21981 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 Thu, 18 Apr 2013 06:11:49 GMT Antreas http://groups.yahoo.com/group/Hyacinthos/message/21980 http://groups.yahoo.com/group/Hyacinthos/message/21980 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 Thu, 18 Apr 2013 01:33:10 GMT rhutson2 http://groups.yahoo.com/group/Hyacinthos/message/21977 http://groups.yahoo.com/group/Hyacinthos/message/21977 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 Wed, 17 Apr 2013 20:18:39 GMT Paul Yiu http://groups.yahoo.com/group/Hyacinthos/message/21976 http://groups.yahoo.com/group/Hyacinthos/message/21976 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 Wed, 17 Apr 2013 19:37:59 GMT rhutson2 http://groups.yahoo.com/group/Hyacinthos/message/21975 http://groups.yahoo.com/group/Hyacinthos/message/21975 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 Wed, 17 Apr 2013 18:16:16 GMT Antreas http://groups.yahoo.com/group/Hyacinthos/message/21974 http://groups.yahoo.com/group/Hyacinthos/message/21974 ... 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 Wed, 17 Apr 2013 14:58:08 GMT yiuatfauedu http://groups.yahoo.com/group/Hyacinthos/message/21973 http://groups.yahoo.com/group/Hyacinthos/message/21973 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 Wed, 17 Apr 2013 14:45:09 GMT César Lozada http://groups.yahoo.com/group/Hyacinthos/message/21972 http://groups.yahoo.com/group/Hyacinthos/message/21972 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), Wed, 17 Apr 2013 13:04:41 GMT César Lozada http://groups.yahoo.com/group/Hyacinthos/message/21971 http://groups.yahoo.com/group/Hyacinthos/message/21971 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.., Wed, 17 Apr 2013 10:03:45 GMT Antreas Hatzipolakis http://groups.yahoo.com/group/Hyacinthos/message/21970 http://groups.yahoo.com/group/Hyacinthos/message/21970 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 Wed, 17 Apr 2013 07:46:11 GMT Bernard Gibert http://groups.yahoo.com/group/Hyacinthos/message/21969 http://groups.yahoo.com/group/Hyacinthos/message/21969 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