archematics/public/rostamian/steve_gray.html
Glen Whitney 35678be213 feat: Start implementing Rostamian's pages
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.
2023-10-06 12:21:48 -07:00

143 lines
4.7 KiB
HTML

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