ABCD is a cyclic quadrilateral with consecutive side lenghts. ... Let the measurement of the angle between the shortest and the longest sides be (x). The area...
Floor van Lamoen discovered the exciting fact that if G is the centroid of triangle ABC and A', B', C' are the midpoints of the sides BC, CA, AB, respectively,...
Dear Darij, Juan Carlos and Ricardo, Thanks Darij for being so thoughtful to share with all your Hyacinthos friends the link to the umich online maths books....
... Dear Darij! Thus result can be obtained as partial case of next theorem. Let the lines AP, BP, CP intersect opposite sidelines of ABC in points A1, B1, C1;...
Dear Darij and all friends of Hyacinthos: This website of books on line did send me my friend Ricardo Barroso from Sevilla, Spain. Thank you very much...
In Hyacinthos message #6196, I cited Antreas ... I have shown long ago that the perpendicular bisectors of segments BaCa, CbAb, AcBc concur at one point T,...
Michail Tyomkyn, member of the German team for the IMO, has found a nice theorem about concyclic points: Let ABCD be a cyclic quadrilateral, E the intersection...
Hi,I am a member of www.matematik.kulubu.com from Turkey.Why we can not reach your archives?It must be open to public.Anyway I ask you a question:In...
I have just discovered a nice German site: http://www.uni-duisburg.de/SCHULEN/STG/jufopage/indexjufo.html In a 2003 work, Janne Schilling gives the following ...
From Henry Baker to Dan Hoey (on another mailing list): I found the incenter of the triangle (x,0,0), (0,y,0), (0,0,z). I then found the distance^2 to the ...
A well-known theorem of Victor Thébault reads: Let AM be a cevian in a triangle ABC, M between B and C. Construct two circles that touch BC, AM and the...
- A well-known theorem of Victor Thébault reads: Let AM be a cevian in a triangle ABC, M between B and C. Construct two circles that touch BC, AM and the...
... We can restate the theorem as follows: Let I be the incenter of a triangle ABC. Through the orthocenter H of triangle ABC, draw parallels to AI, BI, CI...
Dear Darij, ... Indeed. ... I do think it is obvious once you have seen that CB_b:B_bA = cos C : cos A. From B_b we can straightforwardly construct the...
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2003volume3/FG200315index.html The Editors, Forum...
in the 5-gon there are 35 triangles can be create by diagonals.What is the general formula for n gons?There are no tree diagonals can pass the same point and...
Dear friends of Hyacinthos: Let ABC triangle with incenter ( I ), inradius (r) and circumradius (R) prove that: 1/AI^2+1/BI^2+1/IC^2=(2R-r)/(2Rr^2). Is this...
Dear Darij, ... As Ta is barycentric -(b+c)^2/p : c^2/(p-c) : b^2/(p-b) where p = (a+b+c)/2, your point T is barycentric x = (p-a)/a^2 and is the isotomic...
Dear Darij, ... Yes these trilinears are correct. They can also be written as: ( 4cos(A) + cos(3A)/[cosAcos(B - C)] : . . . : . . .) Best regards Nikolaos...
The triangle center X(1157) in Clark Kimberling's ETC, the inverse of the Kosnita point in the circumcircle, is known to be the intersection of some nice...
Dear EMHLians, Take a triangle ABC. The circle touching the b-excircle and the c-excircle externally and the a-excircle internally passes through the Spieker...
Chris, Here is one approach to problems 2 and 3: Problem 2. A', B', C' are the near mid-arc points, and the circumradii thru these are the perpendicular...
1) Let A0,B0,C0 be the foot of the altitudes of an acute-angled triangle ABC. Let M be a point from the circumcircle of ABC, A', B',C' the projections of M...
My name is Jean-louis Ayme and I am teacher in a High School in St- Denis de la Reunion, a little insel of the Indian Ocean. Since several year, my interest...
