archematics/public/rostamian/inequality.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

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<!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>A geometric inequality</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>A geometric inequality</h1>
<h4>&hellip;and its solution by
<a href="mailto:haoyuep@aol.com">Dan Hoey</a></h4>
<table class="centered">
<tr><td align="center">
<applet code="Geometry" archive="Geometry.zip" width="400" height="400">
<param name="background" value="ffffff">
<param name="title" value="A geometric inequality">
<param name="e[1]" value="A;point;fixed;100,300">
<param name="e[2]" value="B;point;fixed;300,300">
<param name="e[3]" value="ABC;polygon;equilateralTriangle;A,B">
<param name="e[4]" value="C;point;vertex;ABC,3">
<param name="e[5]" value="P;point;free;180,320;red;red">
<param name="e[6]" value="PA;line;connect;P,A;none;none;green">
<param name="e[7]" value="PB;line;connect;P,B;none;none;green">
<param name="e[8]" value="PC;line;connect;P,C;none;none;magenta">
</applet>
</td></tr>
<tr><td>
<b>
Drag the point $P$.<br>
Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
Proposition: $PC \le PA + PB$.
</b>
</td></tr></table>
<h2>The problem</h2>
<p>
<b>Proposition:</b> <i>Let $ABC$ be an equilateral triangle and $P$
be an arbitrary point in its plane. Then $PC \le PA + PB$.</i>
<p>
This was brought up in
<a href="http://mathforum.org/kb/message.jspa?messageID=1086018">a message </a>
on the <code>geometry.puzzles</code> newsgroup on November&nbsp;11, 2001.
Go to that message and scroll to the bottom of the page to see the discussion
thread.
<p>
On November&nbsp;22, 2001
<a href="mailto:haoyuep@aol.com">Dan Hoey</a>
offered a particularly nice solution. He also commented that he
had learned that problem may be related to the <em>Van Schooten Theorem</em>,
which indeed it is. See
<a href="http://www.cut-the-knot.org/Curriculum/Geometry/Pompeiu.shtml">Van
Schooten's and Pompeiu's Theorems: What are these?</a> for much detail and
historical background.
<h2>The proof</h2>
<p>
Here is Dan Hoey's proof of the proposition as stated above.
<p>
On the line segment $AP$ construct the equilateral triangle
$APD$, as shown in the diagram below, then add the line segment $DC$.
<p>
Let us show that the triangles $APB$ and $ADC$ are congruent. For this,
Let us observe that the sides $AP$ and $AB$ in the triangle $APB$
equal the sides $AD$ and $AC$ in the triangle $ADC$, by the construction.
Moreover, the angles $BAP$ and $CAD$ are equal because each equals
the difference of
a 60 degree angle and the angle $DAB$. Therefore, the triangles $APB$ and
$ADC$ are congruent by the side-angle-side equality.
We conclude, in particular, that $PB = DC$.
<p>
In the triangle $PDC$ we have $PC \le PD + DC$. In this inequality
replace $PD$ and $DC$ by their equivalents $PA$ and $PB$ to arrive at
$PC \le PA + PB$.&nbsp;&nbsp;&nbsp;<b>QED</b>
<table class="centered">
<tr><td align="center">
<applet code="Geometry" archive="Geometry.zip" width="400" height="400">
<param name="background" value="ffffff">
<param name="title" value="A geometric inequality">
<param name="e[1]" value="A;point;fixed;100,300">
<param name="e[2]" value="B;point;fixed;300,300">
<param name="e[3]" value="ABC;polygon;equilateralTriangle;A,B">
<param name="e[4]" value="C;point;vertex;ABC,3">
<param name="e[5]" value="P;point;free;180,320;red;red">
<param name="e[6]" value="PA;line;connect;P,A;none;none;green">
<param name="e[7]" value="PB;line;connect;P,B;none;none;green">
<param name="e[8]" value="PC;line;connect;P,C;none;none;magenta">
<param name="e[9]" value="APD;polygon;equilateralTriangle;A,P;none;none;none">
<param name="e[10]" value="D;point;vertex;APD,3;">
<param name="e[11]" value="AD;line;connect;A,D;none;none;lightGray">
<param name="e[12]" value="PD;line;connect;P,D;none;none;lightGray">
<param name="e[13]" value="CD;line;connect;C,D;none;none;lightGray">
</applet>
</td></tr>
<tr><td>
<b>
Drag the point $P$.<br>
Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
Observation: The triangles $APB$ and $ADC$ are congruent.
</b>
</td></tr></table>
<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 November 23, 2001.<br>
Cosmetic revisions on June 23, 2010.
</em>
<p>
<table width="100%">
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