Dear Hyacinthians, Let ABC be a triangle and call A'B'C' it's intouch triangle and A"B"C" it's circummedial triangle. If F denotes the Feuerbach point, then...
16210
jpehrmfr
Mar 19, 2008 11:48 am
Dear Lennart ... little ... circumcircle ... is ... Poncelet ... projectives ... A_1, ... Q_P. ... A'P; ... K_B ... intersect ... Note that your point Q_P is...
16211
Alexey.A.Zaslavsky
zasl@...
Mar 19, 2008 11:48 am
Dear Jan! What is the circummedial triangle? Are its vertex the common points of circumcircle with the medians of ABC? It seems that in this case your...
16212
Francois Rideau
francoisrideau
Mar 19, 2008 11:48 am
Dear friends After some calculations, I think that the points P, for which Q_P = H the orthocenter, are the symmetrics wrt the circumcenter of the 3 points Q'...
16213
Alexey.A.Zaslavsky
zasl@...
Mar 19, 2008 11:48 am
Dear Lennart! Your point Q_P is the reflection of P in its Simson line. It is known that the Simson line pass through the midpoint of PH. So Q_P=H iff the...
16214
Alexey.A.Zaslavsky
zasl@...
Mar 19, 2008 11:48 am
Dear Lennart! I am sorry, of course three points for which Q_P=H aren't the vertex of equilateral triangle and the curve isn't symmetric. Sincerely...
16215
Francois Rideau
francoisrideau
Mar 19, 2008 11:48 am
I think you have a neat parametrization of your flower using the Morley trick. You choose the ABC-circumcircle as the unit circle in the complex plane. It is...
16216
Francois Rideau
francoisrideau
Mar 19, 2008 11:48 am
Your point Q_P, (what a strange name, Q should be simpler), is also the symmetric of P wrt its ABC-Simson line. Friendly Francois On Tue, Mar 18, 2008 at 6:46...
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Jan Vonk
jan.vonk
Mar 19, 2008 1:11 pm
Dear Alexey, perhaps I used an incorrect name for the triangle A"B"C". It's vertices are the midpoints of the arcs BC,AC,AB opposite A,B,C on the circumcircle...
16218
Francois Rideau
francoisrideau
Mar 19, 2008 3:44 pm
As usual, Jean-Pierre's proof is loveliest as usual and I notice I have said something stupid and unnecessary, to say that the Simson lines through the ...
16219
Jan Vonk
jan.vonk
Mar 19, 2008 4:17 pm
Dear Hyacinthians, I will no longer use the name circummedial triangle for A"B"C", since MathWorld seems to give this name to another triangle than my triangle...
16221
Cosmin Pohoata
pohoata_cosm...
Mar 20, 2008 12:02 am
Dear friends, While reviewing some old notes of mine on mixtilinear incircles, I ran through the following result: Let M be an arbitrary point on the...
16222
Alexey.A.Zaslavsky
zasl@...
Mar 20, 2008 1:10 pm
Dear Lennart, Francois and Jean-Pierre! Also it isn't difficult to receive the equation of this line in polar system with center H and axis HO. This equation...
16223
jpehrmfr
Mar 20, 2008 1:10 pm
Dear Jan ... A"B"C" ... the ... the ... This is true for any point F lying on the incircle. More generally, I suggest you the following : A'B'C' and A''B''C''...
16224
Francois Rideau
francoisrideau
Mar 20, 2008 1:10 pm
Dear Cosmin Your first question is very difficult; as for the second one, I think (without proof for the moment) that the result is the following. I name: O...
16225
Alexey.A.Zaslavsky
zasl@...
Mar 20, 2008 1:11 pm
Dear Jan! Thank you for this explication. Now your result can be obtained from the homothety between A'B'C' and A"B"C". Sincerely...
16226
jpehrmfr
Mar 20, 2008 3:41 pm
Dear Jan I wrote ... concurrent ... This is only due to the fact that A'B'C' and A''B''C'' are directly similar : Consider two triangles A'B'C' and A''B''C''...
16227
lennartmeier
Mar 20, 2008 10:37 pm
Dear Hyacinthians! Thank you very much for your answers. Now I see things more clearly, I think. There is an old paper by Patterson, where he computes the...
16228
Harold Connelly
haconnelly
Mar 21, 2008 2:50 am
Dear Friends, Can the three positions of P for which Q_P = H be found with ruler and compass? Regards, Harold Connelly ... This ... P ... on ... the ... that ...
16229
Alexey.A.Zaslavsky
zasl@...
Mar 24, 2008 2:14 pm
Dear Harold! Of course, no. These points are the roots of cubic equation. Sincerely Alexey Dear Friends, Can the...
16230
ForumGeom
ForumGeom@...
Mar 24, 2008 2:14 pm
The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2008volume8/FG200808index.html The editors Forum...
16231
Francois Rideau
francoisrideau
Mar 24, 2008 11:07 pm
But these points are conic constructible by Jean-Pierre Cabri construction. Francois On Fri, Mar 21, 2008 at 8:06 AM, Alexey.A.Zaslavsky <zasl@...> ...
16232
Nikolaos Dergiades
ndergiades
Mar 26, 2008 10:51 pm
Dear Cosmin for your first remark: If P=(x:y:z) is a point in barycentrics A1B1C1 is the circumcevian triangle of P and the tangents from A1 to the incircle ...
16233
Francois Rideau
francoisrideau
Mar 27, 2008 9:25 am
Dear friends I ask you here a question of terminology. In the euclidian plane, we have 2 pairs of points (A, B) and (A', B'). For the sake of simplicity, I...
16234
michgarl
Mar 28, 2008 11:07 am
Dear Hyacintos I join to this message a file with the calculus and a drawing Giving a ideal point[-n-p:n:p], you draw a line from A cutting (BC) in M = [0:n:p]...
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Nikolaos Dergiades
ndergiades
Mar 28, 2008 11:07 am
Dear Francois, I didn't know that Neuberg had used this line and I don't know if there is a name for this. I had also used this line when I proposed problem...
16236
Francois Rideau
francoisrideau
Mar 28, 2008 4:55 pm
Dear Nikos Thank you for that funny surprising name and this interesting special case. I remember to have seen another name for this line as "droite des...
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Francois Rideau
francoisrideau
Mar 28, 2008 4:55 pm
Dear Michel and dear friends I think this construction or another same construction can be found in Lalesco book: Geometrie du triangle. For Cabri addicts,...
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Jean-Louis Ayme
jeanlouisayme
Apr 1, 2008 12:56 am
Dear Floor and Hyacinthists, a new purely proof has being put on my website: volume (2008) Le cercle de van Lamoen http://perso.orange.fr/jl.ayme Sincerely ...
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Francois Rideau
francoisrideau
Apr 1, 2008 12:56 am
... From: Francois Rideau <francois.rideau@...> Date: Sat, Mar 29, 2008 at 2:02 PM Subject: Re: [EMHL] Neuberg line? To: Nikolaos Dergiades...
