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...
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) :...
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...
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...
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...
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...
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*. ...
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...
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...
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...
Dear Bernard ... (M) ... the ... simple ... conic ... A1, ... pivot ... B', ... Very good. In fact, it gives (after projectivity) a construction of the common ...
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...
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...
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....
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...
The following paper has been published in Forum Geometricorum. It can be viewd at http://forumgeom.fau.edu/FG2005volume5/FG200513index.html The editors Forum...
ForumGeom
ForumGeom@...
Jul 5, 2005 5:14 pm
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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...
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...
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...
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 +...
... 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...
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...
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...
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...
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 ...
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