Glen Whitney
c99b51dafa
Began with incenter.html, the first one alphabetically. Needed one new point construction method, and a new option to see what was going on. Got the planar diagrams on that page working. The next step on #36 will be to get 3D diagrams as the theorem on this page generalizes to 3D. That will be a bigger task, so merging this now. Reviewed-on: #39 Co-authored-by: Glen Whitney <glen@studioinfinity.org> Co-committed-by: Glen Whitney <glen@studioinfinity.org>
142 lines
4.7 KiB
HTML
142 lines
4.7 KiB
HTML
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
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<html>
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<head>
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<!-- fix buggy IE8, especially for mathjax -->
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<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7">
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<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
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<title>Triangles with common base</title>
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<link rel="stylesheet" type="text/css" media="screen" href="style.css">
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<script type="text/javascript"
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src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML,http://userpages.umbc.edu/~rostamia/mathjax-config.js">
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MathJax.Hub.Queue( function() {document.body.style.visibility="visible"} );
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</script>
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</head>
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<body style="visibility:hidden">
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<h1>Triangles with common base</h1>
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<h4>An interesting problem proposed by
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<a href="mailto:stevebg@adelphia.net">Steve Gray</a></h4>
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<table class="centered">
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<tr><td align="center">
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<applet code="Geometry" archive="Geometry.zip" width="650" height="250">
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<param name="background" value="ffffff">
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<param name="title" value="Triangles with common base">
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<param name="e[1]" value="A;point;fixed;50,215">
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<param name="e[2]" value="B;point;fixed;250,215">
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<param name="e[3]" value="T0;polygon;equilateralTriangle;A,B">
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<param name="e[4]" value="C;point;vertex;T0,3">
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<param name="e[5]" value="D;point;free;115,173;red;red">
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<param name="e[6]" value="DC;line;connect;D,C;none;none;orange">
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<param name="e[7]" value="DA;line;connect;D,A;none;none;blue">
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<param name="e[8]" value="DB;line;connect;D,B;none;none;green">
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<param name="e[9]" value="P;point;fixed;300,215">
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<param name="e[10]" value="Q;point;fixed;600,215">
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<param name="e[11]" value="PQ;line;connect;P,Q">
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<param name="e[12]" value="B';point;similar;P,Q,C,A,D">
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<param name="e[13]" value="PB';line;connect;P,B';none;none;green">
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<param name="e[14]" value="QB';line;connect;Q,B';none;none;green">
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<param name="e[15]" value="C';point;similar;P,Q,A,B,D">
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<param name="e[16]" value="PC';line;connect;P,C';none;none;orange">
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<param name="e[17]" value="QC';line;connect;Q,C';none;none;orange">
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<param name="e[18]" value="A';point;similar;P,Q,B,C,D">
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<param name="e[19]" value="PA';line;connect;P,A';none;none;blue">
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<param name="e[20]" value="QA';line;connect;Q,A';none;none;blue">
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<param name="e[21]" value="T1;polygon;triangle;B',C',A';none;none;lightGray">
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<!-- construct the centroid -->
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<param name="e[22]" value="midA'B';point;midpoint;A',B';none;none">
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<param name="e[23]" value="li1;line;connect;C',midA'B';none;none;none">
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<param name="e[24]" value="midB'C';point;midpoint;B',C';none;none">
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<param name="e[25]" value="li2;line;connect;A',midB'C';none;none;none">
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<param name="e[26]" value="M;point;intersection;li1,li2">
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</applet>
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</td></tr>
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<tr><td>
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<b>
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Drag the point $D$.<br>
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Press “r” to reset the diagram to its initial state.<br>
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Proposition: $A'B'C'$ is equilateral and its centroid $M$ is fixed.
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</b>
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</td></tr></table>
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<h2>The construction</h2>
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<p>
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This problem was proposed by
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<a href="mailto:stevebg@adelphia.net">Steve Gray</a>
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in the <code>geometry.puzzles</code> newsgroup
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(<a href="http://mathforum.org/kb/message.jspa?messageID=1088552">see
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the original message</a>) on July 26, 2002.
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Scroll to the bottom of that page for a link to the solution.
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<p>
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Consider an equilateral triangle $ABC$, a line segment $PQ$,
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and an arbitrary point $D$, as seen in the diagram above.
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On the segment $PQ$ construct three triangles
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$PC'Q$,
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$PA'Q$,
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$PB'Q$,
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similar to the triangles
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$ADB$,
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$BDC$,
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$CDA$,
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respectively.
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<p>
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<b>Proposition 1:</b>
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The triangle $A'B'C'$ is equilateral.
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<p>
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<b>Proposition 2:</b>
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The centroid of $A'B'C'$ is independent of $D$.
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<p>
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Steve adds:
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<blockquote>
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<p>
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Now generalize this for a regular $n$-gon. The new points form a
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regular $n$-gon whose centroid is independent of $D$.
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This problem is original so far as I know. I am interested in the
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simplest synthetic solution; no algebra, please.
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</blockquote>
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<hr width="60%">
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<p>
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<em>This applet was created by
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<a href="http://userpages.umbc.edu/~rostamia">Rouben Rostamian</a>
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using
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<a href="http://aleph0.clarku.edu/~djoyce/home.html">David Joyce</a>'s
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<a href="http://aleph0.clarkU.edu/~djoyce/java/Geometry/Geometry.html">Geometry
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Applet</a>
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July 26, 2002.<br>
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Cosmetic revisions on June 17, 2010.
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</em>
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<p>
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<table width="100%">
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<tr>
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<td valign="top">Go to <a href="index.html">Geometry Problems and Puzzles</a></td>
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<td align="right" style="width:200px;">
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<a href="http://validator.w3.org/check?uri=referer">
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<img src="/~rostamia/images/valid-html401.png" class="noborder" width="88" height="31" alt="Valid HTML"></a>
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<a href="http://jigsaw.w3.org/css-validator/check/referer">
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<img src="/~rostamia/images/valid-css.png" class="noborder" width="88" height="31" alt="Valid CSS"></a>
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</td></tr>
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</table>
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</body>
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</html>
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