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    <title>Hyacinthos at Yahoo! Groups</title>
    <link>http://tech.groups.yahoo.com/group/Hyacinthos/</link>
    <description>We discuss themes on Triangle Geometry</description>

    <item>
      <title>Re: Forum Geometricorum</title>
      <pubDate>Mon, 08 Feb 2010 23:39:24 GMT</pubDate>
      <dc:creator>pianissimist</dc:creator>
      <link>http://tech.groups.yahoo.com/group/Hyacinthos/message/18622</link>
      <guid isPermaLink="true">http://tech.groups.yahoo.com/group/Hyacinthos/message/18622</guid>
      <description>Would you please explain your method for proving that &lt;MBN is equal to either &lt;HIH&#39; or &lt;IH&#39;Q?</description>
    </item>
    <item>
      <title>Re: Forum Geometricorum</title>
      <pubDate>Mon, 08 Feb 2010 19:17:45 GMT</pubDate>
      <dc:creator>armpist</dc:creator>
      <link>http://tech.groups.yahoo.com/group/Hyacinthos/message/18621</link>
      <guid isPermaLink="true">http://tech.groups.yahoo.com/group/Hyacinthos/message/18621</guid>
      <description>... Uploaded file  &quot;Sangaku 2.4.2&quot; gives easy solution for this &quot; now easy  yet important&quot; Sangaku. Thank you. M.T.</description>
    </item>
    <item>
      <title>New file uploaded to Hyacinthos </title>
      <pubDate>Mon, 08 Feb 2010 19:13:36 GMT</pubDate>
      <link>http://tech.groups.yahoo.com/group/Hyacinthos/message/18620</link>
      <guid isPermaLink="true">http://tech.groups.yahoo.com/group/Hyacinthos/message/18620</guid>
      <description>Hello, This email message is a notification to let you know that a file has been uploaded to the Files area of the Hyacinthos group. File        : /Sangaku</description>
    </item>
    <item>
      <title>Forum Geometricorum</title>
      <pubDate>Mon, 08 Feb 2010 16:33:40 GMT</pubDate>
      <dc:creator>ForumGeom</dc:creator>
      <link>http://tech.groups.yahoo.com/group/Hyacinthos/message/18619</link>
      <guid isPermaLink="true">http://tech.groups.yahoo.com/group/Hyacinthos/message/18619</guid>
      <description>The following paper has been published in Forum Geometricorum. It can be viewed at http://forumgeom.fau.edu/FG2010volume10/FG201002index.html The editors Forum</description>
    </item>
    <item>
      <title>Re: *Hexagonal* peg in a triangle hole :-)</title>
      <pubDate>Mon, 08 Feb 2010 15:28:53 GMT</pubDate>
      <dc:creator>chris.vantienhoven</dc:creator>
      <link>http://tech.groups.yahoo.com/group/Hyacinthos/message/18618</link>
      <guid isPermaLink="true">http://tech.groups.yahoo.com/group/Hyacinthos/message/18618</guid>
      <description>... Dear Hauke, There are 2 solutions for your question of finding hexagons UVWXYZ in such a way that X,V,Z are perspective with C,B,A. In both cases points</description>
    </item>
    <item>
      <title>Circles with diameter [orthocenter-circumcenter]</title>
      <pubDate>Mon, 08 Feb 2010 14:18:45 GMT</pubDate>
      <dc:creator>Michel</dc:creator>
      <link>http://tech.groups.yahoo.com/group/Hyacinthos/message/18617</link>
      <guid isPermaLink="true">http://tech.groups.yahoo.com/group/Hyacinthos/message/18617</guid>
      <description>Dear Hyacintos Is that that fact well knomn? In the (ABC) triangle, orthocenter is H = [Sbc:Sac:Sab] and circumcenter U = [a² Sa:b² Sb: C² Sc] and the</description>
    </item>
    <item>
      <title>Re: *Hexagonal* peg in a triangle hole :-)</title>
      <pubDate>Mon, 08 Feb 2010 12:57:47 GMT</pubDate>
      <dc:creator>shokoshu2</dc:creator>
      <link>http://tech.groups.yahoo.com/group/Hyacinthos/message/18616</link>
      <guid isPermaLink="true">http://tech.groups.yahoo.com/group/Hyacinthos/message/18616</guid>
      <description>... The computation &quot;says&quot; the point of concurrence is #396 and the hexagon center is the barycenter. The boring proof work is yours :-) (Annoyingly, for the</description>
    </item>
    <item>
      <title>Re: *Hexagonal* peg in a triangle hole :-)</title>
      <pubDate>Mon, 08 Feb 2010 10:52:27 GMT</pubDate>
      <dc:creator>shokoshu2</dc:creator>
      <link>http://tech.groups.yahoo.com/group/Hyacinthos/message/18615</link>
      <guid isPermaLink="true">http://tech.groups.yahoo.com/group/Hyacinthos/message/18615</guid>
      <description>And here are the results of the Numerical Jury: Nope. If you draw a regular hexagon around (0,0) with points (1,0),..., define a random point to be center</description>
    </item>
    <item>
      <title>Re: *Hexagonal* peg in a triangle hole :-)</title>
      <pubDate>Sat, 06 Feb 2010 01:38:56 GMT</pubDate>
      <dc:creator>Nikolaos Dergiades</dc:creator>
      <link>http://tech.groups.yahoo.com/group/Hyacinthos/message/18614</link>
      <guid isPermaLink="true">http://tech.groups.yahoo.com/group/Hyacinthos/message/18614</guid>
      <description>Sorry correct tan(f) = instead of = xyz(x&#43;m)(y&#43;m)(z&#43;m)/[(y-z)(z-x)(x-y)] take = (y-z)(z-x)(x-y)/[xyz(x&#43;m)(y&#43;m)(z&#43;m)] Best regards Nikos Dergiades ... </description>
    </item>
    <item>
      <title>Re: *Hexagonal* peg in a triangle hole :-)</title>
      <pubDate>Sat, 06 Feb 2010 01:26:51 GMT</pubDate>
      <dc:creator>Nikolaos Dergiades</dc:creator>
      <link>http://tech.groups.yahoo.com/group/Hyacinthos/message/18613</link>
      <guid isPermaLink="true">http://tech.groups.yahoo.com/group/Hyacinthos/message/18613</guid>
      <description>Dear Philippe, [Houke] ... [Philippe] ... I don&#39;t think so. If P is the 1st isodynamic point X(15) of ABC and A&#39;B&#39;C&#39; is its pedal triangle then rotate the</description>
    </item>
    <item>
      <title>Re: *Hexagonal* peg in a triangle hole :-)</title>
      <pubDate>Fri, 05 Feb 2010 14:56:10 GMT</pubDate>
      <dc:creator>Philippe</dc:creator>
      <link>http://tech.groups.yahoo.com/group/Hyacinthos/message/18612</link>
      <guid isPermaLink="true">http://tech.groups.yahoo.com/group/Hyacinthos/message/18612</guid>
      <description>... ;-)  X(370) &quot;Jiang Huanxin introduced this notion in 1997&quot; ... Dear Hauke, I&#39;ve not completely solved the problem, just played with my favorite geometry</description>
    </item>
    <item>
      <title>Re: Locus</title>
      <pubDate>Fri, 05 Feb 2010 13:10:45 GMT</pubDate>
      <dc:creator>Nikolaos Dergiades</dc:creator>
      <link>http://tech.groups.yahoo.com/group/Hyacinthos/message/18611</link>
      <guid isPermaLink="true">http://tech.groups.yahoo.com/group/Hyacinthos/message/18611</guid>
      <description>Dear Antreas, [APH] ... [ND] ... Yes. Let L be an arbitrary line and M be a point on L. From a point P we take vectors PA&#39; = (1/2).MA PB&#39; = (1/2).MB PC&#39; =</description>
    </item>
    <item>
      <title>Re: Locus</title>
      <pubDate>Fri, 05 Feb 2010 11:42:41 GMT</pubDate>
      <dc:creator>Nikolaos Dergiades</dc:creator>
      <link>http://tech.groups.yahoo.com/group/Hyacinthos/message/18610</link>
      <guid isPermaLink="true">http://tech.groups.yahoo.com/group/Hyacinthos/message/18610</guid>
      <description>Dear Antreas, [APH] ... I think that there are such examples. Perhaps some cubic in Bernard&#39;s list has such a property. I conjecture that for every line we can</description>
    </item>
    <item>
      <title>*Hexagonal* peg in a triangle hole :-)</title>
      <pubDate>Fri, 05 Feb 2010 10:30:24 GMT</pubDate>
      <dc:creator>shokoshu2</dc:creator>
      <link>http://tech.groups.yahoo.com/group/Hyacinthos/message/18609</link>
      <guid isPermaLink="true">http://tech.groups.yahoo.com/group/Hyacinthos/message/18609</guid>
      <description>Suppose you &quot;inscribe&quot; a regular hexagon UVWXYZ into a triangle ABC such that U lies on AB, W on BC and Y on CA. If the cevians CU,AW,BY intersect, this is the</description>
    </item>
    <item>
      <title>Re: Locus</title>
      <pubDate>Fri, 05 Feb 2010 08:35:25 GMT</pubDate>
      <dc:creator>Antreas</dc:creator>
      <link>http://tech.groups.yahoo.com/group/Hyacinthos/message/18608</link>
      <guid isPermaLink="true">http://tech.groups.yahoo.com/group/Hyacinthos/message/18608</guid>
      <description>Dear Nikos [APH] ... [ND] ... 1. For P = X(4) = H (orthocenter) The triangle A*B*C* is the circumcevian triangle of H in the NPC. ABC is the antipedal triangle</description>
    </item>

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