Let ABC be a triangle and A'B'C', A"B"C" the pedal triangles of two isogonal conjugate points P,P*. Denote: Ha := The orthocenter of A'B"C" Hb := The...
Dear Antreas ... ABC and HaHbHc are indirectly similar; hence, they are parallelogic and orthologic (two of the centers are upon the circumcircle of ABC, the...
Let ABC be a triangle, P a point and PaPbPc the pedal triangle of P. We construct the following squares (all externally or all internally the triangle): ...
Dear Jeff, ... your locus is the Orthocubic K006 http://pagesperso-orange.fr/bernard.gibert/Exemples/k006.html Best regards Bernard [Non-text portions of this...
Hi Antreas, ... Collinear for points on the Darboux cubic, circumcircle and L.inf ... L.inf & some quite complicated degree 5 thing that includes X{3,4,64} ...
Dear Hyacintos In a rectangular ellips, choose 2 points P and Q and draw in these points the tangent line to the ellips The tangents have a commun point R For...
Locus: Let (Q) be a circle, A a point on the circle, and BC a fixed line segment. Which is the locus of a point P of ABC, as A moves on the circle? If P = O,...
Dear Peter First, Thanks! [APH} ... [PM] ... Can you imagine which would be my next and natural question? (see at the end of the asterisks) * * * * * * * * * *...
[APH] ... [FGC] ... In fact we have several parallel radical axes. Here is a listing: Let P, P* be two isogonal conj. points, and A'B'C', A"B"C" their pedal...
... Dear Michel, What is a "rectangular" ellipse ? an ellipse is an ellipse, full stop. However your property is true for any conic section (= parabola and...
Let ABC be a triangle, P a point and PaPbPc the pedal triangle of P. We construct the following equilateral triangles (all externally or all internally the...
Hi Antreas, ... Collinear for points on the Darboux cubic, circumcircle and L.inf. A*B*C* are the same as for your squares version (#17432). ... L.inf & some...
Hi Peter, You really know how and where to find things! I already had the idea that this hyperbola should be known. I am curious, how did you find/calculate...
Hi Chris, ... You found the point as {{4,74}/\{146,148}}. Effectively, all I did was work the coords from there. It lies on a few other lines too ... { ...
Let ABC be a triangle, A'B'C' the circumcevian triangle of I wrt ABC and A*B*C* the circumcevian triangle of O wrt A'B'C'. [ie A',B',C' are the second...
Dear friends, Is the following result known? I find it very interesting. Denote by isog_{XYZ}(T) the isogonal conjugate of T wrt. triangle XYZ. Problem: Let...
Small typo: The correct statement is: "Let ABCD be a convex quadrilateral and let P = AC /\ BD, E = AB /\ CD and F = AD /\BC. Prove that AC \perp BD if and...
Dear Hyacinthists, let ABC be a triangle, Vn (Vs) the outer (inner) Vecten point of ABC, Ha the A-altitude of ABC, Mb (Mc) the mediator of CA (AB), Y the...
Hi Jean-Loius, The line VnVs is the line NK (nine point to symmedian). Line, Ha, is {0,SB,-SC} Line Mb is the perpendicular to CA at the midpoint of CA = {b^2,...
Lemma: Let (O) be a circle P a fixed point, and Q,Q' two antipodal points on the circle. Let K be the Lemoine (Symmedian) point of PQQ'. As A moves on the...
Dear jean-Louis, Dear Peter, A small contribution. Peter mentioned perspector X847. I also looked at the tripolar centroid of the 3 points on the Vectenline...
Let ABC be a triangle, P,P* two isogonal conjugate points and A*B*C* the pedal triangle of P*. Denote: A'B'C' := the circumcevian triangle of P wrt ABC (in the...
Yes, I get the McCay cubic, also with the line at infinity, the circumcircle and the cubic with equation 4 S^2 xyz + CyclicSum(a^2 yz(c^2y+b^2z)) = 0 Surely...
Dear Francisco, ... Your cubic seems to be K192 = nK(X6, X6, X110), a focal cubic. http://pagesperso-orange.fr/bernard.gibert/Exemples/k192.html Please, can...
Dear Bernard, No, I see that the cubic with equation 4 S^2 xyz + CyclicSum(a^2 y z(c^2y+b^2z)) = 0 (*) also passes through the centers of Apollonian circles,...
Dear Francisco, ... Your "S" is probably the actual area of ABC and not Conway's S i.e. twice the area. Anyway, your cubic is a nK(X6, X6, ?) without known...
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