Here is a variation that is symmetric in ABC. Given D, for what P do the circumcenters of the six triangles PBC, PCA, PAB, PDA, PDB, and PDC all lie on a...
18917
Bernard Gibert
bernardgibert
May 18, 2010 8:17 pm
Dear Barry, ... I far prefer this more symmetrical approach but I'm just wondering how can you predict this "small" degree ? Anything to do with Lemoyne's...
18918
Luis
luisgeometria
May 18, 2010 8:43 pm
Dear Hyacintists, Triangle ABC has incenter I and A',B',C' are the centers of the inner-Vecten squares against A,B,C. The rays $A'I ,B'I and C'I cut the...
18919
Francisco Javier
garciacapitan
May 18, 2010 9:51 pm
Luis, Both perspectors are in ETC, X1336 for the internal squares and X1123 for the external ones. Francisco Javier....
18920
Francisco Javier
garciacapitan
May 18, 2010 10:16 pm
Both perspectivities hold if we replace I for any point P on the orthocubic K006...
18921
Eric
efn4900
May 19, 2010 8:28 am
Dear friends, let A1, B1 and C1 be the centers of the squares erected outwardly on the sides BC, CA respectively AB of triangle ABC It is well known that AA1,...
18922
Barry Wolk
wolkbarry
May 19, 2010 6:20 pm
... Because I was able to keep reducing the degree by canceling some common factors while attempting (unsuccessfully) to do the calculation. And symmetry meant...
18923
Antreas
xpolakis
May 19, 2010 6:50 pm
... Variations: We can replace circumcenters with orthocenters, or even concyclicity with orthocentricity. That is: For which points P, the orthocenters of of...
18924
Antreas
xpolakis
May 19, 2010 7:12 pm
I think that this can be generalized for any Kiepert triangles. That is: Let ABC be a triangle, and A'BC, B'CA, C'AB three similar isosceles triangles of base...
18925
Antreas
xpolakis
May 19, 2010 7:45 pm
I am not sure if we have discussed this before: Let ABC be a triangle, and P a point. The circumcircle of PBC intersects AB at Ab other than B, and AC at Ac...
18926
Francisco Javier
garciacapitan
May 20, 2010 7:47 am
This is line at infinity, circumcircle and cubic K018 (Brocard second cubic)....
18927
Jean-Louis Ayme
jeanlouisayme
May 20, 2010 7:59 am
Dear Hyacinthists, has someone a little information about h. M. Taylor? What is the signification of H. M. ? Sincerely Jean-Louis [Non-text portions of this...
[APH] ... {Francisco] ... It is equivalent to 7th locus in Bernard's list http://pagesperso-orange.fr/bernard.gibert/Exemples/k018.html (if A' = BAc /\ CAb...
18930
xpolakis
May 22, 2010 6:31 pm
Let ABC be a triangle, P, P* two isogonal conjugate points and PaPbPc the pedal triangle of P. The circle (P*BC) intersects BP at Ab (other than B) and CP at...
18931
Francisco Javier Garc...
garciacapitan
May 23, 2010 4:48 am
1. This is line at infintity, circumcircle and Euler-Morley quartic Q002 2. I found H and O, it seems that locus is complicated. 2010/5/22 xpolakis...
18932
Antreas Hatzipolakis
xpolakis
May 23, 2010 8:58 am
[APH] ... [Francisco] ... I think that for P = O, H the triangles in question are homothetic (homothetic centers?) APH...
18933
shokoshu2
May 24, 2010 9:49 am
Consider a triangle ABC (angles @,ß,y) and a center P. The cevian triangle DEF with respect to P and ABC will have angles @',ß',y' which are generally a mess...
18934
fvlamoenwxs
May 24, 2010 5:52 pm
Dear Hyacinthians, Let A'B'C' be the circumcevian triangle of H. Construct the triangles, tangent to the circumcircle in A', B' and C' respectively and to the...
18935
Francisco Javier
garciacapitan
May 24, 2010 6:48 pm
Dear Floor, if P=(x:y:z) is any point, then A''B''C'' is the cevian triangle of (x/a:y/b:z/c). Best regards, Francisco Javier....
18936
Antreas Hatzipolakis
xpolakis
May 24, 2010 7:10 pm
How about the triangle RaRbRc of the centers of the constructed circles ? [said by Floor as triangles but meant to say circles] Which is the locus of P such...
18937
Francisco Javier
garciacapitan
May 24, 2010 8:09 pm
I get a cubic, through X3, X19, X65, X165, X267, X284, X1030, but I am not able to find in Bernard Gibert site. The equation is -a^4 b c^3 x^2 y - a^3 b^2 c^3...
18938
Antreas Hatzipolakis
xpolakis
May 25, 2010 5:57 am
[APH] ... [Ra := the center of the circle tangent to BC and the circumcircle at A', where A'B'C' is the circumcevian triangle P] [Francisco Javier] ... ...
18939
Antreas Hatzipolakis
xpolakis
May 25, 2010 8:01 am
In triangle ABC, to construct three circles, each of which is tangent to the other two and to the circumcircle at a vertex of ABC. [instead of two sides of...
18940
Alexey Zaslavsky
zasl@...
May 25, 2010 9:02 am
Dear Antreas! In triangle ABC, to construct three circles, each of which is tangent to the other two and to the circumcircle at a vertex of ABC. [instead of...
18941
Bernard Gibert
bernardgibert
May 25, 2010 9:22 am
Dear Floor and Francisco, Unless I misunderstand something, there are two circles tangent at A' to the circumcircle and tangent to BC at A". In one case, the...
18942
Bernard Gibert
bernardgibert
May 25, 2010 9:38 am
Dear Francisco, ... This is pK(X31, S) where the pivot S is 3,19 /\ 5,1839 /\ 6,46 /\ 9,165 etc. It also contains X365, X2939, X3197. When you take the other...
18943
Philippe
chephip
May 25, 2010 12:46 pm
... Dear Antreas and Alexey, Or also : Given any circle and 3 points A, B, C on it. To construct three circles touching the given circle at the given points,...
18944
armpist
May 25, 2010 5:28 pm
Dear friends, We know that when 4 circles are pairwise tangent then radical axis of one pair passes thru similarity center of another. Construct tangential...
18945
Antreas Hatzipolakis
xpolakis
May 25, 2010 8:30 pm
Dear Bernard If A1,B1,C1 are the points of the 1st case and A2, B2, C2 are the points of the 2nd case, then we have three Apollonius - like circles: with...
