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Messages 16998 - 17027 of 18447   Oldest  |  < Older  |  Newer >  |  Newest
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16998
... drawing "by ... Lineal.) ... THX for the tip. DLed. Can someone check the following? It doesn't work with actual test values, and I must have made a stupid...
Hauke Reddmann
shokoshu2
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Dec 1, 2008
2:33 pm
16999
... UNCLE OF EDIT: Flunked a factor of 2. Now it works with the circumcenter (pi-@,pi-ß,pi-y). (@,ß,y) does NOT work. Double drat. (2@,2ß,2y) as well as...
Hauke Reddmann
shokoshu2
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Dec 1, 2008
3:05 pm
17000
Hello, This email message is a notification to let you know that a file has been uploaded to the Files area of the Hyacinthos group. File : /cercle...
Hyacinthos@yahoogroup...
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Dec 1, 2008
3:07 pm
17001
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2008volume8/FG200828index.html The editors Forum...
ForumGeom
ForumGeom@...
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Dec 1, 2008
3:22 pm
17002
Hi Hyacinthians, ... [ That is draw on ABC the isosceles triangles BUC, CVA, AWB, with angles U,VW equal to x (similar isosceles triangles) Then AU, BV, CW...
Philippe
chephip
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Dec 1, 2008
6:16 pm
17003
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2008volume8/FG200829index.html The editors Forum...
ForumGeom
ForumGeom@...
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Dec 3, 2008
8:24 pm
17004
... THX, I would have asked that eventually anyway (my memory - x,x,x works - *didn't* fail me this time :-) This leaves (for ending this thread :-) the...
Hauke Reddmann
shokoshu2
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Dec 4, 2008
1:19 pm
17005
http://pagesperso-orange.fr/bernard.gibert/Tables/table14.html An obvious question would be to ask for the point for which the cevians concur (like the...
Hauke Reddmann
shokoshu2
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Dec 4, 2008
1:46 pm
17006
Dear all, Here is a simple poser: In a triangle ABC, D is the midpoint of BC, B' is the foot of the altitude from B upon AC,and Z is the midpoint of CH. ...
vprasad_nalluri
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Dec 5, 2008
11:29 am
17007
Dear Vijayaprasad, I think that as H you mean the orthocenter of ABC. Construct the symmetric point B" of B relative to Z. The parallel from B" to DZ meets the...
Nikolaos Dergiades
ndergiades
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Dec 5, 2008
12:47 pm
17008
Let ABC and A'B'C' be perspective trangles. Can ABC and A'B'C' be similar *without* AB||A'B' etc.? Hauke...
Hauke Reddmann
shokoshu2
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Dec 5, 2008
2:18 pm
17009
Dear Hauke, I think the answer is yes. If we consider an arbitrary convex quadrilateral ABCD, we can construct a point P inwardly to it, such that the...
Kostas Vittas
vittasko
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Dec 5, 2008
2:59 pm
17010
Dear All My Friends, Following geometrical facts are very easy to prove but they are interesting enough. There is also one problem: to find the best (optimal,...
Quang Tuan Bui
bqtuan1962
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Dec 5, 2008
4:48 pm
17011
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2008volume8/FG200830index.html The editors Forum...
ForumGeom
ForumGeom@...
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Dec 5, 2008
4:51 pm
17012
Dear Nikos Dergiades, You mean the symmetric point B'' of B' relative to Z? Alternatively, Let the circles D(DB') and Z(ZB') cut at C Let B be the symmetric...
vprasad_nalluri
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Dec 5, 2008
5:04 pm
17013
Dear Vijayaprasad, ... Yes. You are right. Sorry for the typo. Best regards Nikos Dergiades ___________________________________________________________ ...
Nikolaos Dergiades
ndergiades
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Dec 5, 2008
5:48 pm
17014
Dear Hauke, If I understood your problem we have the following: Let 2p, 2q, 2r be the peak angles of the isosceles triangles A'BC, B'CA, C'AB outside ABC and x...
Nikolaos Dergiades
ndergiades
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Dec 5, 2008
7:09 pm
17015
Let ABC be a triangle, P a point, and A'B'C' the cevian triangle of P. In the complete quadrilateral bounded by the lines (AB, BB',AC, CC'), denote Emelyanov...
Andreas P. Chatzipola...
xpolakis
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Dec 5, 2008
8:01 pm
17016
Dear Hauke ... Could you please be more explicit ? Regards Bernard [Non-text portions of this message have been removed]...
Bernard Gibert
bernardgibert
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Dec 6, 2008
9:34 am
17017
Dear Alexey, thank you very much. You sent your message only to me and not to hyacinthos. So I tried to answer privately without success. My message returned. ...
Nikolaos Dergiades
ndergiades
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Dec 6, 2008
11:45 am
17018
Dear friends As I have nothing else to do to forget subprimes , I was daydreaming this afternoon about Newton line. I remember you the saga. You start with...
Francois Rideau
francoisrideau
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Dec 6, 2008
1:53 pm
17019
Dear Francois, I have a brief comment on something you wrote in your message on Newton's line. ... not ... proof ... Why are CPU proofs not enough? I am not...
Harold Connelly
haconnelly
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Dec 6, 2008
6:45 pm
17020
Dear François and Harold, I think both your are right, Being a mathematician, I am closer to François, since a good proof requires the checking of all cases....
Francisco Javier Garc...
garciacapitan
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Dec 6, 2008
6:58 pm
17021
Dear Harold Of course I was joking just to titillate your curiosity! And by the way, I am very very happy to have GSP too! As I said it in my previous post,...
Francois Rideau
francoisrideau
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Dec 7, 2008
8:09 am
17022
... A bit of computing also affirms that - a triangle similar to ABC can be rotated (1), scaled (1) and shifted (2 free variables), and the perspector just...
Hauke Reddmann
shokoshu2
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Dec 7, 2008
3:02 pm
17023
... Yup, I also dimly remembered there was a classic solution with half-angles. Your arccot formula explains many things, although the cases 2A,2B,2C and...
Hauke Reddmann
shokoshu2
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Dec 7, 2008
3:14 pm
17024
... OK. Let ABC be a triangle, and A'B'C' an equilateral triangle circumscribed around ABC, like in your table 14. Now demand that AA',BB' and CC' concur in O....
Hauke Reddmann
shokoshu2
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Dec 7, 2008
3:20 pm
17025
Hello, This email message is a notification to let you know that a file has been uploaded to the Files area of the Hyacinthos group. File : /Le...
Hyacinthos@yahoogroup...
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Dec 8, 2008
8:35 am
17026
Dear Hauke, Dear Nikolaos, In addition to the solutions (and notation) of Nikolaos here are some extra solutions though not based on isosceles triangles but on...
chris.vantienhoven
chris.vantie...
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Dec 8, 2008
2:30 pm
17027
... Addendum: A numeric solution for the Kimberling triangle suggest the Fermat point #13 (can't be) and alternatively y=1.9094098548 (not in list, but it...
Hauke Reddmann
shokoshu2
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Dec 8, 2008
5:09 pm
Messages 16998 - 17027 of 18447   Oldest  |  < Older  |  Newer >  |  Newest
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