archematics/public/inscribed-revived.html
Glen Whitney f1696120dc feat: Improve construction element handling
This change implements several additional construction methods,
  including the first polygon ones. In particular, it now allows
  arbitrary strings as entity names, even ones that are not allowed
  as GeoGebra identifiers, using captions to show the original
  entity names. In addition, line arguments are interpreted as a pair
  of point arguments as needed.

  Resolves #6.
  Resolves #30.
  Resolves #31.
2023-09-24 17:42:02 -07:00

136 lines
5.1 KiB
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<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
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<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7">
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<title>An equilateral triangle inscribed in a rectangle</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">
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</head>
<body style="visibility:hidden">
<h1>An equilateral triangle inscribed in a rectangle</h1>
<table class="centered">
<tr><td align="center">
<applet code="Geometry" archive="Geometry.zip" width="410" height="370">
<param name="background" value="ffffff">
<param name="title" value="An equilateral triangle inscribed in a rectangle">
<!-- the moving mechanism -->
<param name="e[1]" value="O;point;fixed;290,320">
<param name="e[2]" value="U1;point;fixed;510,320">
<param name="e[3]" value="V1;point;perpendicular;O,U1">
<param name="e[4]" value="U;point;angleDivider;U1,O,V1,3">
<param name="e[5]" value="V;point;angleDivider;V1,O,U1,3">
<param name="e[6]" value="circ1;circle;radius;O,U">
<param name="e[7]" value="li1;line;parallel;U,O,U1">
<param name="e[8]" value="li2;line;parallel;V,O,V1">
<param name="e[9]" value="W;point;intersection;li1,li2">
<param name="e[10]" value="VW;line;connect;V,W;0;0;lightGray">
<param name="e[11]" value="@;point;lineSegmentSlider;V,W,0,220;red;red">
<param name="e[12]" value="li3;line;parallel;@,O,U1">
<param name="e[13]" value="li4;line;chord;circ1,li3">
<param name="e[14]" value="X1;point;first;li4">
<!-- the triangle -->
<param name="e[15]" value="A;point;fixed;50,320">
<param name="e[16]" value="V2;point;perpendicular;A,U1">
<param name="e[17]" value="li5;line;parallel;A,O,X1">
<param name="e[18]" value="X2;point;last;li5">
<param name="e[19]" value="X;point;extend;A,X2,A,X2">
<param name="e[20]" value="tri1;polygon;equilateralTriangle;X,A">
<param name="e[21]" value="Y;point;vertex;tri1,3">
<param name="e[22]" value="B;point;midpoint;X,Y">
<param name="e[23]" value="ABC;polygon;equilateralTriangle;A,B">
<param name="e[24]" value="C;point;vertex;ABC,3">
<!-- the rectangle -->
<param name="e[25]" value="D;point;foot;B,A,U1">
<param name="e[26]" value="F;point;foot;C,A,V2">
<param name="e[27]" value="FE;line;parallel;F,A,D">
<param name="e[28]" value="E;point;last;FE">
<param name="e[29]" value="rect;polygon;quadrilateral;A,D,E,F;0;0;black;0">
<param name="e[30]" value="ADB;polygon;triangle;A,D,B;0;0;0;pink">
<param name="e[31]" value="ACF;polygon;triangle;A,C,F;0;0;0;pink">
<param name="e[32]" value="BCE;polygon;triangle;B,C,E;0;0;0;cyan">
</applet>
</td></tr>
<tr><td>
<b>
Slide the &ldquo;@&rdquo; up and down to change the geometry.<br>
Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
Proposition: The blue area equals the sum of the two pink areas.
</b>
</td></tr></table>
<h2>Problem statement</h2>
<p>
The diagram above shows an equilateral triangle inscribed in a rectangle
in such a way that the two have a vertex in common. This subdivides the
rectangle into four disjoint triangles.
The original equilateral triangle is shown in white
in the diagram; the other three are shown in color.
<p>
<b>Proposition</b>
<em>
The area of the blue triangle equals the sum
of the areas of the two pink triangles.
</em>
<p>
The trigonometric proof is quite straightforward. I don't
know of a classical proof <i>a la</i> <span class="name">Euclid</span>.
(Well, actually I haven't tried much.)
If you can think of a neat non-trigonometric proof, let me know. I will
put it here with due credit.
<p>
This problem appeared as a conjecture
<a href="http://mathforum.org/kb/thread.jspa?forumID=129&amp;messageID=1083967">in an article</a>
in the <code>geometry.puzzles</code> newsgroup on March 15, 1997.
<p>
<b>Note added January 8, 2017:</b>
Here is a
<a href="inscribed-equilateral-solution.html">clever solution</a>
that <b>Peter Renz</b> sent me a in December 2016. Thanks, Peter!
<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>
on July 2, 2010.
</em>
<p>
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