Dear Antreas, ... They are not perspective. This is not the way extraversion goes! Under extraversion, the triangle formed by the polars of the points A, B, C...
Dear all, My apologies for sending an incomplete message. [FvL] ... Proof: I calculated the beautiful identity that for a CircleCevian A'B'C' triangle ... ...
Dear Darij ... Well... If A1A2A3 is a triangle, and (I0),(I1),(I2),(I3) are the in/excircles and F0,F1,F2,F3,F4 the Feuerbach Points, then we have the nine...
Dear Antreas, ... ^^^^^^^^^^^^^^ Well, you probably wanted to say F0, F1, F2, F3. There are 4 Feuerbach points, not 5. ... Actually, you have twelve polars!...
Dear Darij ... Sorry for the stupid errors! (I should have delete it and resend a correct one) ... [...] Thanks for your analysis ! Now, how about a...
Dear Antreas, you wrote ... I suppose you mean the common point of the circumcircles of A'B'C", B'C'A" and C'A'B" where A'B'C' and A"B"C" are the circlecevian ...
... Cheers, Eric, for finding such a beautiful theorem! It was indeed too beautiful for me to leave it without a proper synthetic proof, so I have been looking...
Dear Eric ... Quite Interesting! Given now that X(399) lies on the Neuberg cubic, I guess that its "symmetric" lies on the same cubic, as well. If so, then we...
Dear friends, which is the locus of points that are lying on the circumcircle of their cevian triangle? Are there known points with this property? I found that...
Dear Eric and Antreas [Antreas] ... [Eric] ... A'B'C", ... Let's try a little generalization : A'B'C' is the circlecevian triangle of P, A"B"C" is the...
Dear Antreas, Floor and Eric, I wrote ... circlecevian ... I have to apologize because I realize now that Floor proved in #10352 that it was true for any pair...
Dear Nikolaos! ... I receive c^2*xy(x+z)(y+z)(x+y-z)+...=0. Is it the same equation? Sincerely Alexey...
Alexey.A.Zaslavsky
zasl@...
Sep 2, 2004 11:14 am
10365
Dear Alexey, [ND] ... [AZ] ... Yes thank you it is the same if we substitute c^2 = S_A+S_B a^2 = S_B+S_C b^2 = S_C+S_A. Best regards Nikolaos Dergiades...
Let ABC be a triangle and P a point. A' := The second intersection of AP and the circumcircle of PBC B' := The second intersection of BP and the circumcircle...
Dear Antreas, [APH] ... Am I missing something? It seems that A1B1C1 is just the cirlecevian triangle of P*, and thus indeed for CC2 the analogous theorem...
Let ABC be a triangle, P a point and Op, Oc the centers of the pedal, cevian circles of P, resp. Which is the locus of P such that O, Op, Oc are collinear? Any...
Dear Floor ... I think you are right. Quick computations give that ABC, A1B1C1 are perspective at P*, and also that the circumcircles of BCA1, CAB1, ABC1...
Dear Nikolaos! ... This equation can be written in next nice form a^2/x+b^2/y+c^2/z=a^2/(y+z)+b^2/(z+x)+c^2/(x+y) May be there is another interpretation? ...
Alexey.A.Zaslavsky
zasl@...
Sep 3, 2004 9:08 am
10372
Dear Antreas! ... Your locus is the curve with degree <=7. Sincerely Alexey...
Alexey.A.Zaslavsky
zasl@...
Sep 3, 2004 9:11 am
10373
Let 1,2,3 be three parallel lines. To construct a triangles A1A2A3 such that: (i) A1,A2,A3 be lying on the lines 1,2,3 resp. (ii) The reflections of 1,2,3 in...
Dear Antreas, [APH] ... it is sufficient to construct the triangle A1A2A3 such that the Euler line of this triangle is parallel to the lines 1, 2, 3. Then the...
Dear friends, ... (A1B1C1 is the circlecevian triangle of P*). The circles A'A1C and B'B1C are tangent at C. Let Lc be the tangent to both circles in C....
Last Message #10376 (under Floor's name) was posted by APH, and is a correct edition of ##10370, 10375, that have been deleted. When possible, I will try to...
... Let ABC be a triangle, P,Q,R three points, and A1B1C1, A2B2C2, A3B3C3 the circlecevian triangles of P,Q,R, resp. (1,2) := the point of concurrence of the...
Dear friends, If I 'm not mistaken, X, Y, Z are the touch points of (I_a) and BC, (I_b) and CA, (I_c) and AB. Then I think P can't lie on the Euler line of...
Dear Floor, Jean-Pierre, Eric, and Antreas, [APH]: What point is the point of concurrence for the circlecevian triangles of H and O? Is it in ETC? [ED]: ... My...
Dear Floor, Jean-Pierre, Eric, and Paul ... Thanks !! If you don't mind, I would name the new point as HYACINTHOS POINT, to honor our group. By JPE's...
Dear Alexey, [ND] ... [AZ] ... Excellent!! From this it is evident that if the point P = (x:y:z) of this locus lies on the circumcircle of ABC then also the...
Hi Treegoner, Nice to see you at Hyacinthos! ... Actually, they are not. X, Y, Z are defined as the touch points of (I_a), (I_b), (I_c) with the nine-point...
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