Let ABC be a triangle, P a point and PaPbPc the pedal triangle of P. Ab := AB /\ PPa, Ac := AC /\ PPa A' := (perpendicular to AB at Ab) /\ (perpendicular to AC...
Let ABC be a triangle, P a point, and PaPbPc the pedal triangle of P. Ab := AB /\ PPa, Ac := AC /\ PPa A'b := (Parallel from Ab to BC) /\ AC A'c := (Parallel...
Dear Hyacinthians: If we know the equations of two lines in barycentrics, a1 x + b1 y + c1 z = 0, a2 x + b2 y + c2 z = 0, how can we get the equation of the...
Dear Andreas, XRISTOS ANESTH the locus of P is the line A(B-C)x+B(C-A)y+C(A-B)z = 0 that is the Euler line A = b^2 + c^2 -a^2 etc. and the locus of the other...
Dear Francisco Javier, [FJGC]: If we know the equations of two lines in barycentrics, a1 x + b1 y + c1 z = 0, a2 x + b2 y + c2 z = 0, how can we get the...
Dear Andreas and Nik, [APH]: Let ABC be a triangle, P a point, and PaPbPc the pedal triangle of P. Ab := AB /\ PPa, Ac := AC /\ PPa A'b := (Parallel from Ab to...
Let ABC be a triangle, P a point and PaPbPc the pedal triangle of P. Ab := AB /\ (Parallel from P to AC) Ac := AC /\ (Parallel from P to AB) A' := (Parallel...
Dear Paul Yiu: Thank you again for your fast answer I thought that the result would be easily factored as two lines, but it seems that this occur only in...
Dear Antreas, the locus is the Darboux cubic. If we want the triangles to be homothetic then the point P must be also a point of the cubic x(yyS_AB-zzSAC) +...
Dear Francisco Javier, [FJ]: I thought that the result would be easily factored as two lines, but it seems that this occur only in simple cases, such that of...
Dear Francisco, I think that Paul Yiu answered as follows: If the line PQ is a1 x + b1 y + c1 z = 0, and the line PR is a2 x + b2 y + c2 z = 0, the two...
Let ABC be a triangle, P a point and PaPbPc the pedal triangle of P. Ab := PPa /\ AB, Ac := PPa /\ AC A'b := The Orthogonal projection of Ab on AC A'c := The...
Dear Antreas, ... Reckon on .. The Euler line, the 3 altitudes through H and L.inf. For P on the Euler line, the point of concurrence is the complement of P. ...
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2005volume5/FG200510index.html The editors Forum...
ForumGeom
ForumGeom@...
May 10, 2005 6:19 am
11251
Dear coleagues! Let give 4 points A, B, C, D. X, Y, Z are the common points of AB and CD, AC and BD, AD and BC, P - arbitrary point distinct from X, Y, Z. Then...
Alexey.A.Zaslavsky
zasl@...
May 11, 2005 7:54 am
11252
Dear coleagues! Let ABCD is convex quadrilateral, T is such point that the distances d(T,AB)=d(T,CD), d(T,AD)=d(T,BC). Then T is in th Gauss line if and only...
Alexey.A.Zaslavsky
zasl@...
May 11, 2005 7:59 am
11253
Dear friends, the vanishing of the determinant of a 3x3 matrix expresses that three points are collinear. is there a geometrical interpretation of the...
... Hi Diana, This looks like problem 1711 of Mathematics Magazine. If A', B', and C' lie on line segments BC, CA, and AB respectively of triangle ABC; line...
Dear friends, ... I dont know problem 1711 of Mathematics Magazine. If there other points than centroid does anybody can find where is the false in the...
Dear friends. Sorry. The mistake is in the equality x+y+z=1/(xyz) that is not correct. the correct is cotA+cotB+cotC = 1/(tanAtanBtanC) Best regards Nikolaos...
Dear friends, sorry again I must go to sleep. The mistake is in the equality x+y+z=1/(xyz) that is not correct. In the previus message | wrote nonsense and...
Dear Nikolaos, ... The permanent of a 3x3 matrix is obtained in a similar way as the determinant but with "plus" everywhere instead of alternative "plus" and...
Dear Alexey, ... and CD, AC and BD, AD and BC, P - arbitrary point distinct from X, Y, Z. Then the polars of P wrt all conics passing through A, B, C, D have...
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