http://tech.groups.yahoo.com/group/Hyacinthos/ We discuss themes on Triangle Geometry Sat, 04 Jul 2009 21:33:05 GMT xpolakis http://tech.groups.yahoo.com/group/Hyacinthos/message/17961 http://tech.groups.yahoo.com/group/Hyacinthos/message/17961 [APH] ... [ND] ... Dear Nikos It is nice! Thanks. Probably there are also nice results for pedal triangles of other points. For the pedal triangle of I, for Sat, 04 Jul 2009 21:17:11 GMT Nikolaos Dergiades http://tech.groups.yahoo.com/group/Hyacinthos/message/17960 http://tech.groups.yahoo.com/group/Hyacinthos/message/17960 Dear Antreas, ... The locus is the linf + the conic with center X(1112) that passes through the vertices of the cevian triangles of X(4) Orthocenter and X(648) Sat, 04 Jul 2009 19:10:09 GMT xpolakis http://tech.groups.yahoo.com/group/Hyacinthos/message/17959 http://tech.groups.yahoo.com/group/Hyacinthos/message/17959 [APH] ... It is true and simple, since ABC, A'B'C' are homothetic. How about if A'B'C' is the orthic triangle? Which is the locus of P? Antreas Sat, 04 Jul 2009 18:23:09 GMT xpolakis http://tech.groups.yahoo.com/group/Hyacinthos/message/17958 http://tech.groups.yahoo.com/group/Hyacinthos/message/17958 Let ABC be a triangle, A'B'C' its medial triangle and P a point. Let La, Lb, Lc be the reflections of PA', PB', PC' in the bisectors AI, BI, CI, resp. Which is Sat, 04 Jul 2009 11:45:02 GMT xpolakis http://tech.groups.yahoo.com/group/Hyacinthos/message/17955 http://tech.groups.yahoo.com/group/Hyacinthos/message/17955 [APH] ... Francisco verified that it is true for P = O,H,G,N. (is it true for all points on the Euler line ?) He has also computed the coordinates of the Fri, 03 Jul 2009 20:29:37 GMT xpolakis http://tech.groups.yahoo.com/group/Hyacinthos/message/17954 http://tech.groups.yahoo.com/group/Hyacinthos/message/17954 [APH] ... Francisco told me it is not true in general. However, it seems it is true for P = O For P = H, it seems that La,Lb,Lc are concurrent. Antreas Thu, 02 Jul 2009 20:13:01 GMT garciacapitan http://tech.groups.yahoo.com/group/Hyacinthos/message/17953 http://tech.groups.yahoo.com/group/Hyacinthos/message/17953 For P = I the concurrence point is the same point L that in Hyacinthos message #17947 Thu, 02 Jul 2009 15:13:42 GMT xpolakis http://tech.groups.yahoo.com/group/Hyacinthos/message/17952 http://tech.groups.yahoo.com/group/Hyacinthos/message/17952 Let ABC be a triangle and P = (x:y:z) a point. Denote: A* :=(Perpendicular from B to CP) /\ (Perpendicular from C in BP) B* :=(Perpendicular from C to AP) /\ Wed, 01 Jul 2009 20:17:07 GMT Eisso J. Atzema http://tech.groups.yahoo.com/group/Hyacinthos/message/17951 http://tech.groups.yahoo.com/group/Hyacinthos/message/17951 Dear Cosmin, I like your argument too, although I would like to observe that your Lemma 1 (or at least the part that you need for your final statement) is Tue, 30 Jun 2009 23:37:16 GMT xpolakis http://tech.groups.yahoo.com/group/Hyacinthos/message/17949 http://tech.groups.yahoo.com/group/Hyacinthos/message/17949 Let ABC be a triangle, P = (x:y:z) a point and A'B'C' the pedal triangle of P. Denote: A* := (The Reflection of BC in BP) /\ (The Reflection of BC in CP) B* := Tue, 30 Jun 2009 18:12:06 GMT garciacapitan http://tech.groups.yahoo.com/group/Hyacinthos/message/17948 http://tech.groups.yahoo.com/group/Hyacinthos/message/17948 For F* the intersection point has coordinates {a (a^5 b - a^4 b^2 - 2 a^3 b^3 + 2 a^2 b^4 + a b^5 - b^6 + a^5 c - 6 a^4 b c + 8 a^3 b^2 c + 4 a^2 b^3 c - 9 a Tue, 30 Jun 2009 17:36:30 GMT garciacapitan http://tech.groups.yahoo.com/group/Hyacinthos/message/17947 http://tech.groups.yahoo.com/group/Hyacinthos/message/17947 The intersection point L has coordinates {a (a^2 b - b^3 + a^2 c - a b c + b^2 c + b c^2 - c^3), b (-a^3 + a b^2 + a^2 c - a b c + b^2 c + a c^2 - c^3), c Mon, 29 Jun 2009 23:00:31 GMT Cosmin Pohoata http://tech.groups.yahoo.com/group/Hyacinthos/message/17946 http://tech.groups.yahoo.com/group/Hyacinthos/message/17946 Dear Eisso, Your idea is great. As I don't usually work with cubics, it is always a joy for me to see their appearance in such rather easy configurations. ... Mon, 29 Jun 2009 18:04:04 GMT xpolakis http://tech.groups.yahoo.com/group/Hyacinthos/message/17945 http://tech.groups.yahoo.com/group/Hyacinthos/message/17945 Let ABC be a triangle and (Oa), (Ob),(Oc) the circumcircles of IBC, ICA, IAB, resp. Let (Qa), (Qb), (Qc) be the reflections of (Oa), (Ob), (Oc) in BC, CA, AB, Mon, 29 Jun 2009 17:12:20 GMT xpolakis http://tech.groups.yahoo.com/group/Hyacinthos/message/17944 http://tech.groups.yahoo.com/group/Hyacinthos/message/17944 Dear Bernard Thanks [APH] ... [BG] ... I think we can ask similar questions for cubics; pivotal cubics in paricular. Do we know any remarkable pivotal cubics