Let us call, for reasons discussed shortly (*), a simple quadrilateral that has its opposite sides (or their extensions) going through two of the vertices of a...
Dear friends, Let ABC be our triangle and P be a point not on the sides of ABC. Construct the circumcircles of PBC, PCA, PAB and let A', B', C' be the mid...
We now consider the general case, of any pair of “perspective parallelograms”, i.e. of any pair of simple quadrilaterals that have their opposite sides (or...
Dear friends, A correction: The mid-arcs A', B', C' are such that the arcs BA'C, CB'A, AC'B have the same direction. Best regards Nikos Dergiades ... ...
Dear friends, given the four points P = p:q:r Qa = 0:a2:a3; Qb = b1:0:b3; Qc = c1:c2:0, I want to construct the two points P1 = - a3 q + a2 r : b3 p - b1 r : -...
Dear Bernard, Since your points can be defined as Qa = 0:k.a2:k.a3; Qb = m.b1:0:m.b3; Qc = n.c1:n.c2:0 the point P1 is P1 = - k.a3q + k.a2r : m.b3p - m.b1r : -...
... Dear Nikos, the centers of circles PAB, PBC, PCA build a triangle C'A'B', whose bisectors are the medial lines of segments PM3, PM1, PM2, Mi:middles of...
Dear Nikolaos , ... yes, of course, but these points are supposed to be suitably normalized. for instance, if QaQbQc is the cevian triangle of some point Q,...
Dear Bernard, if the points Qa, Qb, Qc are the traces of the same point Q the coefficients I used k,m,n must be equal k = m = n. but if the points Qa, Qb, Qc...
Dear Paris, Very good observation. You mean that the mid arc points M1, M2, M3 are the symmetric points of P relative to the bisectors of triangle A'B'C' where...
Exactly! Dear Nikos I got interested in this problem because lately I was involved in a problem on bisectors of angles formed by cevians. You can see the...
Dear Nikolaos , ... you're absolutely right. in fact, I'm not interested in P1, P2 themselves but in their barycentric product P which has some geometric...
Dear friends, ... I have found a construction : let Ra be the trilinear pole of the line through AB /\ PQb and AC /\ PQc, define Rb, Rc similarly. these 3...
Dear All My Friends, If we reflect X(9) in X(11) so we have one point on Feuerbach hyperbola with barycentrics: (b + c - a) / ((b + c - a)^2 - b*c) : : ...
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2007volume7/FG200712index.html The editors Forum...
ForumGeom
ForumGeom@...
May 7, 2007 4:43 pm
15266
Dear All My Friends, What we can say about the non degenerated triangle ABC if exists one point D on sideline AB such that BD = DC = CA = OD, here O is...
Dear Nikos, Thank you very much for your results. I would like to make some small remarks with two first solutions: - CO // DH - CO / DH (or DH / CO) = golden...
Dear friends, Let PA be a chord of a circle (Ca) and let M be the mid point of PA. The circle (P, PM) meets (Ca) at the points B, C. Prove for the sides of...
Dear Nikos, Since PB=PC=PM=radius of second circle, and MA=radius (case 1) or MA= k*radius (case 2) the results follow immediately from the application of ...
Dear Nikos, It is clearly that AP = internal angle bisector of BAC Let C' = reflection of C in PA then C' is on AB and AC' = AC M = midpoint of arc CC' so MB...
Dear Nikos, Sorry for one typo mistake: PA // CC'' (both are perpendicular with CC'' ) FALSE It should be read as PA // CC'' (both are perpendicular with CC'...
Dear Peter, thank you for your short clever proof. I didn't thought of such a short proof. I had in mind that BC disects PM at Q and since M is the incenter of...
Dear All My Friends, Given triangle ABC with circumcenter O, incenter I. Ia, Ib, Ic are incenters of BOC, COA and AOB respectively. A'B'C' is circumcevian...
Dear All My Friends, Given triangle ABC and one point P with Cevian triangle PaPbPc. Suppose X is any point with barycentrics (x : y : z). Ab = intersection of...
Dear friends! One construction: Given tiangle ABC. On BC line we mark two points: A1 and A2, so that AA1=AB and AA2=AC. And perpendicylar bissector to that...
Влекђей ...
alex_geom@...
May 14, 2007 9:51 pm
15278
Dear all, My friend Juan Bosco Romero Mrquez and me have been playing with a configuration starting with a triangle ABC and the cevian triangles of two...
Dear Alexei I think it is only a matter about product of central symmetries along each sides of ABC. For example, look at the side BC. Let A' be the middle of...
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