Dear friends! Last summer I've found one new (as I hope) circle related to a triangle. The construction is as following: Let ABC be an arbitrary triangle and...
20304
Antreas Hatzipolakis
xpolakis
Nov 9, 2011 8:24 am
... Dear Alexei Is the center of the circle a known point? (ie already listed in ETC) Coordinates? APH [Non-text portions of this message have been removed]...
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amyakishev
Nov 9, 2011 9:06 am
Dear Antreas! The coordinates of the center are rather ugly, smth. like that (if Mathematica 5.1 tells the truth^)): ...
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jpehrmfr
Nov 9, 2011 10:08 am
Dear Alexei ... I don't know if the following can help you but If M,N are the isotomic conjugates (wrt ABC) of the Brocard points of A'B'C', it seems that your...
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amyakishev
Nov 9, 2011 11:00 am
Dear Jean-Pierre! Thank you for that wonderful observation. The next question is: Is this fact about isotomic Brocard's points and the corresponding circle...
20308
jpehrmfr
Nov 9, 2011 2:21 pm
Dear Alexei ... In my case, I didn't know this result before your message. In fact, if you put u=1/bb+1/cc-1/aa, v,w cyclically, we have ...
20309
Sung Hyun Lim
bbblow
Nov 9, 2011 2:22 pm
Dear Angel, Firstly a remark: F* is perspector of circumconic passing through P and Q, and also is the intersection of tripolar of P and Q. Secondly, I'll look...
20310
yiuatfauedu
Nov 9, 2011 2:36 pm
Dear Alexei, Bernard has found this circle before in Hyacinthos 5710: [BG, 5710]: a new (?) circle Dear friends, P a point, PaPbPc its cevian triangle. the...
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Bernard Gibert
bernardgibert
Nov 9, 2011 2:45 pm
Dear Paul and friends, I had forgotten all this which also appears in §2.3 of http://bernard.gibert.pagesperso-orange.fr/files/tuckercubics.html Thank you and...
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jpehrmfr
Nov 9, 2011 5:14 pm
Dear Paul, Bernard, Alexei and other Hyacinthists a good advice : before posting a message, we must read carefully the complete works of Bernard. Friendly....
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Paul Yiu
yiuatfauedu
Nov 9, 2011 5:53 pm
Sound advice from Pope J.-P. Best regards Sincerely Paul ... From: Hyacinthos@yahoogroups.com [mailto:Hyacinthos@yahoogroups.com] On Behalf Of jpehrmfr Sent:...
20314
Sung Hyun Lim
bbblow
Nov 9, 2011 7:23 pm
Sorry, I made a typo. I send a revised version: Dear Angel, Firstly a remark: F* is perspector of circumconic passing through P and Q, and also is the...
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amyakishev
Nov 9, 2011 8:28 pm
Dear Paul, thank you. It's really so hard tj invent something new. But the question about geometrical proof is still remain? Best regards, Sincerely yours, ...
20316
amyakishev
Nov 9, 2011 8:54 pm
Dear friends! Another attept to find realy new circle, related to a triangle. That consruction belongs to Pavel Dolgirev from Moscow, student of P-T institute....
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amyakishev
Nov 9, 2011 8:58 pm
Dear Bernard, sorry for the "new" circle. Newertheless, it would be interesting to know (for me) how you have proved it. Best regards, Alex...
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yiuatfauedu
Nov 10, 2011 4:22 am
Dear Alex, Here is another circle constructed in the same spirit. Given triangle ABC, let A'B'C' be the tangential triangle. Consider the inscribed conic (E)...
20319
yiuatfauedu
Nov 10, 2011 4:27 am
Dear Alex, [AM] Another attempt to find realy new circle, related to a triangle. That consruction belongs to Pavel Dolgirev from Moscow, student of P-T...
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amyakishev
Nov 10, 2011 6:22 am
Dear Paul, Thank you. ... smth.new? Or it was found some years ago:)? Best regards, Alex...
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jpehrmfr
Nov 10, 2011 6:32 am
Dear Paul and Alexei ... [Paul] ... And this circle (Paul's circle) is member of the pencil generated by the incircle and the circumcircle which means, with...
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Bernard Gibert
bernardgibert
Nov 10, 2011 6:37 am
Dear Alex, ... That's OK. I'm pretty sure one can find it in older literature ! ... through barycentric calculations. Best regards Bernard [Non-text portions...
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Chris Van Tienhoven
chris.vantie...
Nov 10, 2011 7:31 am
Dear Friends, ... I just found an addition to this property when I was strolling through older Hyacinthos messages. Ce = also on the P-Pedal Triangle Circle ...
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amyakishev
Nov 10, 2011 8:41 am
Dear Bernard, in fact, nominal by barycentric, I've get it:) Now I'm prepareing the article about this circle. Of course, you would be menthion in it.:) But...
20325
Angel
amontes1949
Nov 10, 2011 9:44 am
Dear Sung Hyun, ... orthojoin(X) is the orthopole of the trilinear polar of the isogonal conjugate of X. (see note above X(1512) in ETC), then (homogenous...
20326
amyakishev
Nov 10, 2011 11:28 am
Dear friends! Let in acute (for the simple) triangle ABC angle A<angle B <angle C. And the Euler circle intersects BC in A1,A2 and so on. (B1,B2,C1,C2) So, the...
20327
Sung Hyun Lim
bbblow
Nov 10, 2011 5:08 pm
Dear Alexey! We can restate your result in a following more general form: Consider cevian triangle of centroid GaGbGc and cevian triangle of orthocenter...
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jpehrmfr
Nov 10, 2011 6:02 pm
Dear Chris ... The first proof that I've seen of this result is due to Bobillier in "Les Annales de Gergonne"(1828-1829). See page 356 at ...
20329
Sung Hyun Lim
bbblow
Nov 10, 2011 6:33 pm
I found a simple generalization. Consider bicevian conic B(G,P), where G is centroid and P is an arbitrary point. Let the center of B(G,P) be O. Let cevian...
20330
forumgeom forumgeom
ForumGeom@...
Nov 10, 2011 8:34 pm
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2011volume11/FG201124index.html The editors Forum...
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Quang Tuan Bui
bqtuan1962
Nov 11, 2011 3:59 am
Dear All My Friends, Â Reflect point P in BC, CA, AB of reference triangle ABC we get reflection triangle PaPbPc. Q is isogonal conjugate of P wrt triangle ...
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Antreas Hatzipolakis
xpolakis
Nov 11, 2011 6:35 am
... Which is the locus of P such that P,gR(P), G (or some other fixed point of ABC) are collinear? APH APH [Non-text portions of this message have been...
