Dear Hyacinthians, ... The perpendiculars from each excenter to the line joining the Feuerbach point with the corresponding vertex of the medial triangle are...
Dear Eric, ... Your conjecture is correct. The locus of the perspector is the Jerabek hyperbola of the excentral triangle. With reference to ABC, this has...
Dear Eric, ... Apart from the OI line, there is also the line containing X(36) and its isogonal conjugate X(80). X(36) is the inversive image of I in the...
... That looks like the A. M. Macbeath who wrote a nice little text-book "Elementary Vector Algebra" in 1964. The title-page of the 1966 reprint calls him...
Ken Pledger
Ken.Pledger@...
Dec 3, 2004 12:52 am
10923
Thanks to Ken Pledger for the information about MacBeath. Professor MacBeath himself send me the following autobiographical note: FWDED MESSAGE:...
... whose ... As a triangle is equilateral if and only if his Steiner circumellipse is a circle, it follows that, if L is the focal axis of the Steiner...
Dear Hyacinthians, the following proposition can easily be proven synthetically: The triangle formed by the radical axes of the Bevan circle and the excircles...
Dear Eric Danneels, Sorry for my ignorance, can you explain me what Bevan circle is? Thank you very much. Yours sincerely, Khoa Lu. ... Do you Yahoo!? The...
Dear Khoa Lu, Perhaps I can answer that. The Bevan circle, name coined by Clark Kimberling in 2003, is the circumcircle of the ex-centers. Centered on X(40),...
Dear colleagues! There is an interesting fact. Let A'B'C' is the cevian triangle of point P wrt triangle ABC. Then the excenters of A'B'C' are in rectangle...
Alexey.A.Zaslavsky
zasl@...
Dec 6, 2004 6:15 am
10930
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2004volume4/FG200424index.html The editors Forum...
ForumGeom
ForumGeom@...
Dec 6, 2004 4:34 pm
10931
... This generalizes. Take any point P, and let A1 B1 C1 be the cevian traces of P. Then the isogonal conjugate of the complement of P lies on the radical axis...
Dear Barry, [MS] ... [BW] ... I knew for that result,and from the general situation I made the special case. The circumcircles of triangles BAA1,PAC1,BCC1 and...
Dear Milorad [MS] ... Note that if u+v+w=0 then the point u^2/(s-a):v^2/(s-b):w^2/(s-c) lies on the incircle. There is a typo in P5 : your point lies on the...
Dear Jean-Pierre, [MS] ... [JPE] ... Thanks for your comment and corrections. First about comment. I know about it and for some other similar relations. ...
... (edited version) ... Not quite - the antipode of X(11) is X(1317) This is your P3 - I entered the coordinates wrongly when searching Best wishes wilson...
Dear Milorad, ... (s-c)[2c^2-c(a+b)+(a-b)^2]^2). ... I think some of the points you have found are in ETC. P1. X(1358) P2. X(1357) P3. X(1317) P4. X(1365) ...
Dear Milorad, ... (s-c)[2c^2-c(a+b)+(a-b)^2]^2). ... I think that P5 is the incircle antipode of P1, X(1358), on lines {X(2), X(11)}, {X(56), X(1292)} &...
Dear Milorad, Jean-Pierre and Peter ... and, ... Perhaps it is worth noting that the incircle antipode of Jean-Pierre's point from u:v:w is that from U:V:W...
Dear Milorad, Jean-Pierre, and Wilson, I reckon that your points P7 - P16 are indeed on the incircle. with P16 = X(1364). P7 is on lines {11, 115}, {12, 114},...
Dear Hyacintheans, From Darij 's Schroeder database, we know this following problem : Let ABC be a triangle and M, N, P be midpoints of BC, CA, AB. Call H the...
How an inequality do the bisectors of a triangle must satisfy to form a triangle with these three bisectors? Same question also for ex-bisectors... ...
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2004volume4/FG200425index.html The editors, Forum...
ForumGeom
ForumGeom@...
Dec 13, 2004 8:46 pm
10946
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2004volume4/FG200426index.html The editors Forum...
ForumGeom
ForumGeom@...
Dec 16, 2004 8:55 pm
10947
Dear Hyacinthians, consider a triangle ABC, its intouch triangle A'B'C' and a point P inside ABC. Let A* be the intersection nearest to A of AP and the...
Having problems with message search?
Fill out this form to ensure your group is one of the first to be migrated to the new message search system.
