Does anybody know how the converse of the Casey theorem can be proven? The theorem itself is shown in several places like http://www.pandd.demon.nl/casey.htm...
... If N is the Nagel point of triangle ABC, then the nine-point circles of triangles BNC, CNA, ANB and ABC also concur at the Feuerbach point of triangle ABC....
Dear Darij and other friends: With respect to Casey`s theorem application. I finded some relations for a triangle with these conditions: Let ABC triangle with...
Some time ago, I came across an interesting problem in the Swiss journal "Elemente der Mathematik". Alas, I don't remember the exact citation, but it was the ...
... Two centers, namely X(667) and X(1083). Thanks to Edward Brisse for his helpful HTML lists. New questions: (4) How long is UP = UP* ? (5) Are P and P* the...
In one of my old triangle geometry files, I have found a cryptographical notice which I made in 2002 while dipping into some old issues of the "Proceedings of...
On 2 May Darij asked how the converse of Casey's Theorem can be proved. R A Johnson,'Advanced Euclidean Geometry' Dover 1960 reprint addresses Casey's...
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Alexey.A.Zaslavsky
zasl@...
May 8, 2003 8:54 am
Hello! I found an interesting fact. Let I is an incenter of nriangle ABC, a', b', c' - the incircles of the triangles BIC, CIA, AIB. Then internal tangent to...
Dear Friends: My friend Hipólito Reyes send me a problem from Perú , (I don´t know this theorem). Could somebody help me with this problem?. : Be a regular...
Hello, many of the properties of the Feuerbach point are special cases of more general theorems that can be found in the literature. Here I will present a new...
Dear Juan, This formula holds if n > 3. If I is the incenter P is the point and the inradius r the parallel to d1 forms an angle w with IP then d1 = r -...
[English translation below] Dorogoj Alexey Zaslavsky, ... Po-moemu, eto bylo dokazano v reshenii zadachi 72 v sbornike D. O. Shkljarskij, N. N. Chenzov, I. M....
Given two triangles ABC and A'B'C', construct the desmic mate triangle A"B"C" defined through A" = BC' /\ CB'; B" = CA' /\ AC'; C" = AB' /\ BA'. Then, we know...
Dear Hyacinthians, In one of Dick Klingens' sites, http://www.pandd.demon.nl/simson2.htm I found the theorem that if P1 and P2 are the endpoints of a diameter ...
... I have just noted that this solution contains (de-facto) the proof of the Mannheim theorem! [For "newcomers": The Mannheim theorem states that if three...
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alex
alex_geom@...
May 10, 2003 9:56 am
Dear Darij, Alexey and others! Here is one question, conserning Alexey's configuration. Let Ia be the incenter of BIC, Ib - of CIA. C1 is the point, where...
Dear Alex ... Ib)-line. ... lines in the ... If I well understand, your three lines should concur at the radical center of the three Malfatti circles, which is...
1)A triangle ABC is inscribed in the circle C(O,R). Let M be the midpoint of the line segment [RC]. Points B',C' are the projections of B,C on CO, BO...
Dear Chris, ... I think you mean "segment [BC]". [In any case, "segment [OC]" would yield the same circumcircle of triangle MB'C', but I take the symmetrical ...
... Later I gave the trilinears of a Lammel point of triangle ABC, i. e. of the incenter of triangle DEF. I have just noted that the orthocenter of triangle...
In Hyacinthos message #4703, Antreas P. Hatzipolakis ... This is, of course, a generalization of the Simson line. Does anybody have a proof? Or does anybody ...
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Alexey.A.Zaslavsky
zasl@...
May 12, 2003 5:47 am
... the ... The common point of La, Lb, Lc is radical centre of Malfatti circles. It's baricentric coordinates are cited in paper of Milorad Stevanovic in FG. ...
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Alexey.A.Zaslavsky
zasl@...
May 12, 2003 5:48 am
Dear Darij! Certainly, you are right and my fact follows from solving of Malfatti problem. (Except Shklyarskij and al. this solving is cited in Adamar's book.)...
Dear friends, I had found two new points given in barrycentrics. 1.P1(3aa-bb-cc:3bb-cc-aa:3cc-aa-bb) This point is isogonal conjugate of centroid G(of triangle...
Dear friends, Let F1 and F2 are the areas of Cevian triangles of points P1(x:y:z) and P2 which is isogonal conjugate of P1 wrt to ABC then we have the result:...
Dear Milorad, your cubic is the one I usually call the 5th Brocard cubic. it's a circular cubic with singular focus the Tarry point. it contains the Steiner...
Dear Milorad Stevanovic, ... This one is X(193), the anticomplement of the anticomplement of the symmedian point K. Nice new construction. I just see that...
7115
alex
alex_geom@...
May 12, 2003 9:10 pm
Dear Milorad! It seems to me that we can get some generalization of that construction -if I dont't mistake, that fact is correct for any central point - so we...
