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Messages 8066 - 8097 of 16603   Oldest  |  < Older  |  Newer >  |  Newest
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8066
Dear Barry and your student, You wrote ... I have the following proof. AA'=BB'=CC'=d. 2d^2=aa+bb+cc+4Dsqrt3,where D=area(ABC). ...
Milorad Stevanovic
yumarince
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Oct 1, 2003
7:27 am
8067
Dear Barry Wolk, ... Another high school student (me) is giving a solution he has found right off. I guess it can be simplified a lot. I will use the POMPEIU...
Darij Grinberg
darij_grinberg
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Oct 1, 2003
8:08 am
8068
Dear Georg and Jean-Pierre, ... The solution I wrote this morning assumed that A'AG shouls be collinear. That is a wrong assumption, so basicly all I wrote was...
Floor en Lyanne van L...
fvlamoenwxs
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Oct 1, 2003
9:08 am
8069
Dear Barry and your student, You wrote ... I have the following proof. AA'=BB'=CC'=d. 2d^2=aa+bb+cc+4Dsqrt3,where D=area(ABC). ...
Milorad Stevanovic
yumarince
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Oct 1, 2003
11:08 am
8070
Dear Hyacinthians, in August 2003 a cluster of new points were added to the ETC thanks to Peter Moses. Some of these points were related to the...
efn4900
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Oct 1, 2003
6:11 pm
8071
Dear Milorad Stevanovic, ... This is X(177), the 1st Mid-Arc point. Indeed, the barycentrics given in Kimberling's ETC: ( (sin A)(cos B/2 + cos C/2) sec A/2 :...
Darij Grinberg
darij_grinberg
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Oct 1, 2003
6:21 pm
8072
Dear friends, I cite Kimberling's ETC (with slight modification): Let A', B', C' be the first points of intersection of the angle bisectors of triangle ABC...
Darij Grinberg
darij_grinberg
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Oct 1, 2003
6:22 pm
8073
Regard a triangle ABC and a point P. If x1, y1, z1 are the distances from P to the sidelines BC, CA, AB (and x1 > 0 for P on the same half-plane of BC as A,...
Darij Grinberg
darij_grinberg
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Oct 1, 2003
6:22 pm
8074
If A'B'C' is the Yff central triangle of a triangle ABC, then A' has trilinear coordinates ( cos(B/2) cos(C/2) ... On the other hand, let I be the center of...
Darij Grinberg
darij_grinberg
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Oct 1, 2003
6:29 pm
8076
Dear Eric, [ED]: I call the triangle formed by the points F'a, F'b and F'c where the ... I have just confirmed your result. Indeed the same is true for the...
Paul Yiu
yiuatfauedu
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Oct 1, 2003
7:14 pm
8077
Given a triangle ABC. We have a triangle XYZ which is perspective to all cevian triangles of triangle ABC. Is XYZ necessarily an anticevian triangle? If yes: ...
Darij Grinberg
darij_grinberg
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Oct 1, 2003
7:25 pm
8078
Dear Paul Yiu, ... Very nice observation, but "happens" sounds as if it would be some unexpected coincidence. In fact, there is a simple reason: After the...
Darij Grinberg
darij_grinberg
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Oct 1, 2003
7:40 pm
8079
... [PY]: I have just confirmed your result. Indeed the same is true for the cevian ... It is quite interesting to note that the same is true also for X(970),...
Paul Yiu
yiuatfauedu
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Oct 1, 2003
9:05 pm
8080
Dear Eric and Paul ... There should be a locus of points with that property. Which is it? ... ... and somebody else wrote once something like that, but about...
Antreas P. Hatzipolakis
xpolakis
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Oct 1, 2003
9:23 pm
8081
Dear Darij ... Consider a non-degenerated triangle U,V,W and suppose that U,V,W don't lie on the sidelines of ABC. In any system of projective coordinates with...
jpehrmfr
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Oct 1, 2003
9:29 pm
8082
Dear Paul and Antreas ... [APH] ... Do you mean that we have to wait about three centuries a new Andrew Wiles to know these coordinates? Friendly. Jean-Pierre...
jpehrmfr
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Oct 1, 2003
9:30 pm
8083
Dear Antreas and Jean-Pierre, [PY]: But the coordinates of the perspector is too complicated to write down here. [APH] ... and somebody else wrote once...
Paul Yiu
yiuatfauedu
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Oct 1, 2003
9:55 pm
8085
Dear Darij ... Obviously, these cubics are exactly the cubics with the sidelines of ABC as "flex asymptots" - sorry, but I don't know the English word: I mean...
jpehrmfr
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Oct 1, 2003
10:26 pm
8086
Dear Paul ... new ... double ... are. I'm very sorry for the time you've lost in this tedious computation. Of course, following Antreas, I was trying a very...
jpehrmfr
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Oct 1, 2003
10:31 pm
8087
Dear Jean-Pierre Ehrmann, ... Many thanks for the solution! I was stupid as I thought there should be a finite number of cevian triangles. Sincerely, Darij...
Darij Grinberg
darij_grinberg
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Oct 2, 2003
6:25 am
8088
Dear friends, Let A1,B1,C1 be the second intersections of AI,BI,CI with three excircles. Then 1.Ttriangles A1B1C1 and medial triangle of triangle ABC are...
Milorad Stevanovic
yumarince
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Oct 2, 2003
8:14 am
8089
Dear friends, Given a segment AB of length c, and another segment of length ell, construct in an elegant way a triangle ABC with a + b = ell and a^3 + b^3 =...
Paul Yiu
yiuatfauedu
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Oct 2, 2003
2:55 pm
8090
... Dear Paul, I don't see what property has the special triangle with a^3 + b^3 = c^3 (to apply it for an elegant solution), but in general: if a + b = l, and...
Antreas P. Hatzipolakis
xpolakis
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Oct 2, 2003
5:00 pm
8091
From an old book (from the old good days, when geometry was .... geometry!): Construct a triangle if are given: a, h_a [=altitude from A], 3b^2 + 2c^2 = min. ...
Antreas P. Hatzipolakis
xpolakis
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Oct 2, 2003
5:09 pm
8092
Dear friends, Let A',B',C' be the first points of intersection of the angle bisectors of ABC with its incircle. Then the centroid G'of triangle A'B'C' has...
Milorad Stevanovic
yumarince
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Oct 2, 2003
5:40 pm
8093
Dear Antreas, If we denote by D the feet of perpendicular from A to BC and take BD=a-x,DC=x then 3(b^2)+2(c^2)=7(x^2)-4ax+2(a^2)+3(h_a)^2 has minimum for...
Milorad Stevanovic
yumarince
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Oct 2, 2003
6:07 pm
8094
I think I have already studied partly the following configuration in an old thread. Let ABC be a triangle, P a point, and PaPbPc its pedal triangle. The line...
Antreas P. Hatzipolakis
xpolakis
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Oct 2, 2003
6:28 pm
8095
Let ABC be a triangle, P a point, and PaPbPc its pedal triangle. The line PPa intersects AB at Ab, and Ac at Ac. Let Mac = midpoint of AcP Mab = midpoint of...
Antreas P. Hatzipolakis
xpolakis
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Oct 2, 2003
6:37 pm
8096
... Dear all, I would like to know the date of birth and the date death of Antoine Gob He was a teacher in Hasselt (Belgium) and published with J. Neuberg a...
Darij Grinberg
darij_grinberg
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Oct 2, 2003
6:41 pm
8097
Dear Milorad ... How did you obtain that equation? I think that if we apply twice the Pythagorean theorem (in the triangles DAB, DAC) we get: 3b^2 + 2c^2 =...
Antreas P. Hatzipolakis
xpolakis
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Oct 2, 2003
7:17 pm
Messages 8066 - 8097 of 16603   Oldest  |  < Older  |  Newer >  |  Newest
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