I correct a little typo in my last answer. With your notation: OP:PH = k:1-k you mean: OP = k OH so P = G for k=1:3 (and not for k=-2!) Friendly François...
11349
Etienne Rousee
etiennerousee
Jul 1, 2005 8:16 am
... Yes, it's an error of mine. I have forgotten to change the subject. Etienne [Non-text portions of this message have been removed]...
11350
jpehrmfr
Jul 1, 2005 5:49 pm
Dear Jeff ... one ... in ... whose ... If T(P) is the circlecevian triangle of P, there exists one and only one point Q such as T(Q) is inscribed in T(P) :...
11351
Bernard Gibert
bernardgibert
Jul 1, 2005 6:18 pm
Dear Jeff and Jean-Pierre, ... I had found the same results and a strong connection with circular isogonal pK. All these points P, P*, Q, Q*, the vertices of...
11352
Antreas P. Hatzipolakis
xpolakis
Jul 1, 2005 7:31 pm
... --...
11353
Jeff Brooks
jbrooks_tulsa
Jul 2, 2005 3:04 pm
Dear Jean-Pierre and Bernard, Thank you very much for your responses. I did not expect the inverse problem to be quite so involved. Bernard, your analysis...
11354
jpehrmfr
Jul 3, 2005 7:21 am
Dear Jeff and Bernard, may be this is more clear. For a given point, let M* = isogonal conjugate of M T(M) = circlecevian triangle of M with vertices...
11355
Bernard Gibert
bernardgibert
Jul 3, 2005 7:29 am
Dear Jean-Pierre and Jeff, ... How to be sure they are ruler and compass constructible ? In this case, it would be possible to construct the common points of a...
11356
Bernard Gibert
bernardgibert
Jul 3, 2005 7:32 am
Dear Jean-Pierre, ... these are the points such that M lies on the infinite branch of the isogonal circular pK with pivot the infinite point of the line MM*. ...
11360
jpehrmfr
Jul 3, 2005 8:25 am
Dear Bernard [JPE] ... [BG] ... Yes but, except the limit cases, we can split the plane in two parts : if M lies in the first part, there are 4 points P; if M...
11361
jpehrmfr
Jul 3, 2005 10:09 am
Dear Bernard and Jeff ... generally ... ruler ... [BG] ... of ... If I well follow you, the points P such as T(P) circumscribes T(M) are the isogonal...
11362
Bernard Gibert
bernardgibert
Jul 3, 2005 4:58 pm
Dear Jean-Pierre and Jeff, ... here's a construction of the four points : let Q be a point and A1B1C1 its circle cevian triangle. we seek P whose circle cevian...
11363
jpehrmfr
Jul 3, 2005 5:28 pm
Dear Bernard ... (M) ... the ... simple ... conic ... A1, ... pivot ... B', ... Very good. In fact, it gives (after projectivity) a construction of the common ...
11364
fredlangch
Jul 4, 2005 8:16 am
... Hello! Hence, we can construct the four tangents from a point of a pK. Bye. Fred...
11365
Bernard Gibert
bernardgibert
Jul 4, 2005 6:24 pm
Dear Jean-Pierre, ... in fact, it is not necessary to have another pair of points on the pK if it is defined by its pivot P and its isoconjugate P*. to...
11366
Nikolaos Dergiades
ndergiades
Jul 4, 2005 9:23 pm
Dear friends, it is known http://mathworld.wolfram.com/AdamsCircle.html that the parallel lines from the Gergonne point Ge = ( 1/(s-a) , 1/(s-b), 1/(s-c) ) in...
11369
Antreas P. Hatzipolakis
xpolakis
Jul 4, 2005 9:36 pm
Dear Nikos [ND] ... This should be known by Conway's extraversion BTW, and Conway's Circle is also centered at I and has three brother-circles by extraversion....
11370
Jeff Brooks
jbrooks_tulsa
Jul 5, 2005 5:46 am
Dear friends, Recent messages (many thanks to Jean-Pierre and Bernard!) on nested circle cevian triangles led me to take a look at the class of circular...
11371
ForumGeom
ForumGeom@...
Jul 5, 2005 5:14 pm
The following paper has been published in Forum Geometricorum. It can be viewd at http://forumgeom.fau.edu/FG2005volume5/FG200513index.html The editors Forum...
11372
Nikolaos Dergiades
ndergiades
Jul 5, 2005 7:54 pm
Thanks Antreas. [ND] ... [APH] ... If the point P has barycentrics (x : y : z) and A'B'C' is its cevian triangle then the parallel line from P to B'C' meets...
11373
Francois Rideau
francoisrideau
Jul 6, 2005 7:38 am
Dear friends Beeing given 3 lines L_{1}, L_{2}, L_{3} in R^3 with equations: x = a_{k} z + b_{k} y = c_{k} z + d_{k} for k= 1, 2, 3 1° Is there a simple way...
11374
Antreas P. Hatzipolakis
xpolakis
Jul 6, 2005 4:51 pm
If three concurrent chords of a conic are equal, then is the conic a circle (and the chords diameters) ? aph --...
11375
Antreas P. Hatzipolakis
xpolakis
Jul 6, 2005 5:14 pm
... Well... not. Now, if three concurrent chords of a circle are equal, then the three chords are necessarily diameters. Simple proof? aph...
11376
Li Zhou
lizhoupolk
Jul 6, 2005 8:44 pm
Dear Friends, The question is about Math. Magazine Problem 1171 (Feb/2005). Suppose the cevians AA', BB', CC' of triangle ABC meet at M. If M is inside ABC and...
11377
Boutte Gilles
g_bouttefr
Jul 6, 2005 9:18 pm
The chords AA', BB', CC' of the circle (O) are concurrent at M. Let l = AA' = BB' = CC', and p the power of M wrt (O). We have : MA + xMA' = MB + xMB' = MC +...
11378
Francois Rideau
francoisrideau
Jul 7, 2005 5:02 am
... Gilles's proof is very nice. Here another one, not using directly the ... If l<>R, then r<>0, and we have a nonsense with 3 distinct tangents from M to a...
11379
Nikolaos Dergiades
ndergiades
Jul 7, 2005 6:16 am
Dear Antreas, we had very nice proofs for the case of circle but before this you wrote ... and maybe somebody is trying to solve this problem. The answer is no...
11380
Francois Rideau
francoisrideau
Jul 7, 2005 7:39 am
Here what the meaning of a curve? If we begin with a "curve" C and we construct the conchoid C' of C wrt O for a length l, the reunion of C and C' is a new...
11381
Nikolaos Dergiades
ndergiades
Jul 7, 2005 8:52 am
Dear François, I had in mind the cardioide. For example in polar coordinates (p, w) the cardioide with equation p = 1 + cosw has every chord AA' that passes...
11382
Alexey.A.Zaslavsky
zasl@...
Jul 7, 2005 12:14 pm
Dear colleagues! There are some interesting properties of two triangle centers. Let P1, P2 are the Brocard points of triangle ABC, C0 is the common point of ...
