Dear Nilolaos, You are right. Many thanks for this correction. In fact I drew a "subfigure" with the only points B, C, A1, A2, Oa, but I wrote A instead C. ...
6093
alex
alex_geom@...
Dec 1, 2002 11:09 am
Dear Nikos! Thank You very much for the calculatiuns. The book of Kimberling I've mention about, is "Triangle Centers and Central Triangles", published in...
6094
alex
alex_geom@...
Dec 1, 2002 11:10 am
Dear Gilles and Nikos! Thank You for detail analusis of that problem. What if we add to all construction the other points of intersectionof our circles with...
easy but i think i know smth wrong about it..could you please explain the three occasions in the question... four lengths of sides of a quadrilateral are given...
... explain ... what ... of course we cannot find what the sum of the diagonals is...but i meant: what interval can we talk about? you can explain it by...
Dear Hyacinthists, Let A'B'C' be a pedal triangle of a point P. Build squares externally on AB',B'C, CA', A'B, BC' and C'A. The squares that have A as a vertex...
Dear Hyacinthists P is a finite point of the plane (not on the sidelines of ABC, not on the circumcircle) The circle PBC intersects again the line AB at Ab and...
Dear Floor and other Hyacinthists, I upload on the site a figure for my answer : Let AB'RQ and C'A'Q'R' the squares with A as vertex, so AQQ' is the A-flank...
Dear Jean-Pierre, ... May I add several little remarks : the locus of point P such that the 6 points Ab, Ac, Bc, Ba, Ca, Cb lie on a conic is B2 = "Second...
6101
ForumGeom
ForumGeom@...
Dec 2, 2002 3:03 pm
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2002volume2/FG200220index.html The Editors Forum...
ABCD is a square E is the midpoint of [BC], draw [AE], F is the midpoint of [CD], draw [EF], G is the midpoint of [DA], draw [FG], H is the midpoint of [AE],...
Dear Floor and Gilles, a similar explanation is: The point P is the circumcenter of triangle AB'C' and it is known that the line AGa is the median of the...
Sorry, the phrase The point P is the circumcenter of triangle must be The mid-point of AP is the circumcenter of triangle ND ... in ... the ... are ... ...
6106
alex
alex_geom@...
Dec 3, 2002 10:38 am
Dear all! Let ABC be an arbitrary triangle, A1,B1,C1 -the points, where incircle is tangent to the sides of ABC, and I is the center of incircle. Let Ta is the...
Mustafa, Symmetric kite-shaped quadrilaterals similar to EFGH recur at NOPQ and each subsequent seventh point. The centroid of these kites converges to W with...
Dear Nikos, Floor and all Hyacinthists, Let A'B'C' be a triangle, inscribed in ABC. Let Ka, Kb, Kc the Lemoine points of the flanks. The lines AGa and AKa are...
Dear Gilles, Thank you very much. Very good proof. My solution is: If A'B'C' is the cevian triangle of P and the lines AGa, BGb, CGc perpendicular to B'C',...
... H, and I? ... exactly ... Dear Paul, Jean-Pierre and other Hyacinthians. Here I'll present some ideas about it. I'm not complete about wether Jean-Pierre'...
Dear Alex, Who posed problem 1? You or somebody else? Igor Sharygin's (clever) construction I think does not give an answer to problem 1, because how many...
Dear Frans, Paul and other Hyacinthists [PY] ... O, ... [FG] ... Lemoine ... triangle ... says ... the ... Using your method, I think that the conjecture is...
From Darij Grinberg: I have read the following geometry problem marked with "Iran, 1997": Let ABC be a triangle and P a varying point on the arc BC of the ...
Dear Paul, ... This point should be X_56 = external center of similitude of the incircle and the circumcircle = isogonal conjugate of the Nagel point. ...
6116
alex
alex_geom@...
Dec 6, 2002 7:08 pm
Dear Nikos! The first problem was my, when I've thought about it. (Of course, somebody have thought about it some times ago). Of course, I've missed " 1) Draw...
Dear Hyacinthists, [Darij Grinberg] ... "Iran, ... of the ... the ... point ... circumcircle of ABC. ... define the ... found (by ... somebody ... ...
6118
hgsfh ghhh
etancheetanche@...
Dec 8, 2002 12:39 pm
I am looking for a proof with Menelaus theorem of the 3 circles of Monge Consider 3 circles with differents radius, then the pairs of common tangents of each...
ABC is a triangle, A-bisector meets [BC] at D, B-bisector meets [AC] at E, C-bisector meets [AB] at F, if the side lengths of the triangle DEF are given, how...
The points are collinear only if the tangents of the three circles are external tangents. If (A,R1), (B,R2), (C,R3) are the three circles and the three points...
6121
ForumGeom
ForumGeom@...
Dec 9, 2002 4:45 pm
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2002volume2/FG200221index.htm The editors Forum...
