Dear Hyacinthists looking at the beginning of the very interesting from EH Lemoyne : Suite de theoremes et de resultats concernant la geometrie du triangle ...
... Rephrasing it: Let A1A2A3.... An be a regular polygon whose angle(A3A1A6) = 15 deg. How many diagonals equal to A1A6 has it? Let w be the angle that a side...
A new book, co-authored by the friend and listmember George Baloglou (Associate Professor of Mathematics at the State University of New York, Oswego) An...
In triangle ABC, let P, P* be two isogonal conjugate points, Pa,Pb,Pc the midpoints of PA,PB,PC, resp., and Ma,Mb,Mc the midpoints of P*Pa, P*Pb, P*Pc, resp. ...
Dear Antreas, [APH]: In triangle ABC, let P, P* be two isogonal conjugate points, Pa,Pb,Pc the midpoints of PA,PB,PC, resp., and Ma,Mb,Mc the midpoints of...
Dear Jean-Pierre ... You mean Lemoine (as in the subject line) ... The perpendicular to BC at Ma intersects AB at Ab, and AC at Ac. Define A' = BAc /\ CAb....
... Quoting the first two paragraphs of the web page: ORLANDO, Florida - December 2, 2003 -- Michael Shafer, a 26 year-old volunteer in the Mersenne.org...
Dear Antreas, Thank you very much. Earlier today I was preparing my number theory notes for the coming semester, and was checking the record of Mersenne...
... [PY]: Thank you very much. Earlier today I was preparing my number theory notes ... Of course I have not read it carefully. It was discovered 2 weeks ago. ...
Given a triangle ABC and an angle phi (directed angle modulo 180), we construct the lines through A, B, C which make the angle phi with the lines BC, CA, AB,...
Dear Darij ... Here are some quick elements of answer : As <BA''C = -phi-<BAC+phi = -<BAC mod Pi, A'' lies on the circle BCH. It follows that the Miquel point...
Dear Jean - Pierre ... In other words: Let Ma be *the* point on the line segment BC such that: MaB + MaMab = MaC + MaMac where Mab, Mac are the orthogonal...
Dear Hyacinthians, message 7504 led to some results on orthopoles. In message 7550 I made the following statement: (rephrased) Consider a triangle ABC with H =...
Dear Darij! ... (7) can be proved without using (1)-(6). If the conic is a circle it follows from Ceva theorem and other cases are projectively equivalent. Now...
Alexey.A.Zaslavsky
zasl@...
Dec 4, 2003 5:53 am
8766
Dear Jean-Pierre Ehrmann, Thank you very much for the proofs using pivot point theory. I will add some remarks. ... I don't understand how you draw the last...
Dear Darij [JPE] ... [DG] ... I just use the properties of the Miquel point : UVW is a tringle indcribed in ABC; P (Miquel point) is the common point of the...
Let ABC be a triangle, P a point and PaPbPc its pedal triangle. The parallel La to BC through P, intersects AB at Lab and AC at Lac. Let A' be the reflection...
Fellow Geometers, As some of you might know, I am working on a bibliography of known properties of quadrilaterals. This is a long term project that will mostly...
Eisso Atzema
atzema@...
Dec 4, 2003 9:01 pm
8771
The generalization found by Michael deV which you refer to is a special case of something still more general that I found a few years ago. You may wish to...
Let ABC be a triangle and AA' a given cevian. To construct three CONGRUENT circles (K), (L), (M) such that: (K) is inscribed in angle(BAA') (ie touches AB,...
Dear Sergei Markelov and all who may be interested, I have uploaded a short note named "Anti-Steiner points with respect to a triangle" on my website; this...
Dear Antreas, Let angle BAA' = 2x, angle CAA' = 2y. On the bisector (b1) of angle BAA' we take an arbitrary point K' and let K'' be the reflection of K' in...
... Since Andreas -- to whom I recently sent a copy of my philological book cited above in appreciation of long-term correspondence -- has revived (or rather...
George Baloglou
baloglou@...
Dec 6, 2003 6:32 am
8780
Dear all, I have finished my paper "Generalization of the Feuerbach point". It has 44 pages; for the first two Fontene theorems, the first 10 pages are enough....
Is there a triangle whose the altitudes AA', BB', CC' are bisected by the orthocenter H? (ie AH = HA' and BH = HB' and CH = HC') In general: Let ABC be a...
Let ABC be a triangle and Pa a point. The parallel to AB through Pa intersects BC at Aab The parallel to AC through Pa intersects BC at Aac Let A'ab, A'ac be...
Dear Antreas, ... ^ This should be Pa. Your condition is easily seen to be equivalent to the condition that the B-parallelian and the C-parallelian through Pa...
