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CTK Exchange
alexb
Charter Member
1175 posts |
Jan-09-04, 05:41 PM (EST) |
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1. "RE: Thanks; typo"
In response to message #0
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>This will be my first post at this forum. So let me proclame
>my warmest thanks to Alexander for making me familiar with
>geometry Dear Darij, let's gove praise where it is due. I am not sure that even years ago, when I was as young as you are today, I had as vast knowledge of geometry as yours. Your profound posts at the geometry forum and the Hyacinthos would give a credit to a mature mathematician. I would never guess your age just reading your geometry posts. >and, later, for linking to my website in relation
>with the Ceva theorem. I should probably make more links. In fact, I just wrote an applet that illustrates Zaslavsky's theorem about a triangle and its reflection in a point, for which you posted a solution based on Menelaus' theorem at the geometry forum. >On the Geometry page, the link to "Hexagon Parallel to
>Orthic Triangle" should be
>
>https://www.cut-the-knot.org/Curriculum/Geometry/OrthicPursuit.shtml
>
>rather than
>
>https://www.cut-the-knot.org/Curriculum/Geometry/MedialPursuit.shtml
> Many thanks for that. Alex |
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darij
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Jan-10-04, 01:24 PM (EST) |
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2. "RE: Thanks; typo"
In response to message #1
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Dear Alex, Many thanks for your kind words. I have always been enjoying your website and have learned some of my very first parts of Modern Triangle Geometry here. Thanks for the Zaslavsky applet; here is another result you may be interested in (if you don't already know it!): Hyacinthos message #7463
From: Darij Grinberg
Subject: BA, BA & EL Given a triangle ABC, take a point K on the circumcircle. The parallel to BC through K meets the circumcircle again in L. The parallel to CA through L meets the circumcircle again in M. The parallel to AB through M meets the circumcircle again in N. The parallel to BC through N meets the circumcircle again in O. The parallel to CA through O meets the circumcircle again in J. The parallel to AB through J meets the circumcircle again in K'. Warm-up exercises (from Wolfgang Kroll): The points K and K' coincide. The segments KN, OL and MJ are congruent. Loci problems: (1) Let OJ meet MN at X, let MN meet KL at Y, let KL meet OJ at Z. Triangles ABC and XYZ are evidently homothetic. What is the locus of the homothetic center? (2) What is the locus of the circumcenter of triangle XYZ ? (3) Let KN meet OL at X', let OL meet MJ at Y', let MJ meet KN at Z'. What is the locus of the circumcenter of triangle X'Y'Z' ? Note that the hexagon KLMNOJ is a Tucker hexagon for triangle XYZ. Darij Grinberg
Thanks for correcting the link typo, now it works. Sincerely,
Darij Grinberg |
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darij
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Jan-10-04, 06:13 PM (EST) |
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3. "RE: Thanks; typo"
In response to message #2
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I wrote: >> Warm-up exercises (from Wolfgang Kroll): Sorry, I have forgotten to give the reference where Kroll discussed this little theorem: Wolfgang Kroll, Rundwege und Kreuzfahrten, Praxis der Mathematik 32 (1990) Nr. 1, 1-9, Section 2. >> Loci problems:
>>
>> (1) Let OJ meet MN at X, let MN meet KL at Y, let KL
>> meet OJ at Z. Triangles ABC and XYZ are evidently
>> homothetic. What is the locus of the homothetic center?
>>
>> (2) What is the locus of the circumcenter of triangle
>> XYZ ?
>>
>> (3) Let KN meet OL at X', let OL meet MJ at Y', let
>> MJ meet KN at Z'. What is the locus of the circumcenter of
>> triangle X'Y'Z' ? And the solutions of these problems are: (1) The Brocard axis.
(2) The Brocard axis.
(3) The Euler line. (This explains the title of my Hyacinthos message.) Sincerely,
Darij Grinberg |
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