archematics/public/rostamian/trisect-raiford.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>
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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 angle trisection</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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<body style="visibility:hidden">
<h1>An angle trisection</h1>
<h4>
William R. Raiford, <i>An approximate trisection</i>,
American Mathematical Monthly,
vol.&nbsp;68, no.&nbsp;9, Nov 1961, p.&nbsp;917.
</h4>
<table class="centered">
<tr><td align="center">
<applet code="Geometry" archive="Geometry.zip" width="450" height="400">
<param name="background" value="ffffff">
<param name="title" value="An angle trisection">
<param name="e[1]" value="O;point;fixed;210,365">
<param name="e[2]" value="A;point;fixed;410,365">
<param name="e[3]" value="pt0;point;fixed;410,0;none;none">
<param name="e[4]" value="li0;line;connect;A,pt0;none;none;green">
<param name="e[5]" value="cir1;circle;radius;O,A;none;none;none;none">
<param name="e[6]" value="B;point;circleSlider;cir1,0,300;red;red">
<param name="e[7]" value="OA;line;connect;O,A;none;none;blue">
<param name="e[8]" value="OB;line;connect;O,B;none;none;blue">
<param name="e[9]" value="arcAB;sector;sector;O,A,B;none;none;blue;none">
<param name="e[10]" value="pt1;point;angleBisector;A,O,B;none;none">
<param name="e[11]" value="C;point;cutoff;O,pt1,O,A">
<param name="e[12]" value="OC;line;connect;O,C;none;none;lightGray">
<param name="e[13]" value="li1;line;connect;B,C;none;none;lightGray">
<param name="e[14]" value="li2;line;extend;B,C,B,C;none;none;lightGray">
<param name="e[15]" value="T;point;intersection;li0,li2">
<param name="e[16]" value="OT;line;connect;O,T;none;none;red">
<!-- angle marker -->
<param name="e[17]" value="p1;point;fixed;240,385;none;none">
<param name="e[18]" value="c1;circle;radius;O,p1;none;none;none;none">
<param name="e[19]" value="l1;line;chord;OA,c1;none;none;none">
<param name="e[20]" value="q1;point;first;l1;none;none">
<param name="e[21]" value="l2;line;chord;O,T,c1;none;none;none">
<param name="e[22]" value="q2;point;first;l2;none;none">
<param name="e[23]" value="s1;sector;sector;O,q1,q2;none;none;black;orange">
<!-- angle marker -->
<param name="e[24]" value="l3;line;chord;OB,c1;none;none;none">
<param name="e[25]" value="q3;point;first;l3;none;none">
<param name="e[26]" value="s2;sector;sector;O,q2,q3;none;none;black;yellow">
</applet>
</td></tr>
<tr><td>
<b>
Drag the point $B$ to change the angle $AOB$.<br>
Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
The red line $OT$ is an approximate trisector of the angle $AOB$.
</b>
</td></tr></table>
<h2>Construction</h2>
<p>
The construction described in the article cited at the top of the page,
is quite straightforward. Consider the angle $AOB$ represented by the
circular arc $AB$ centered at $O$, as shown in the diagram above.
To trisect $AOB$ do:
<ol>
<li>
Erect a perpendicular to $OA$ at $A$ (shown in green).
<li>
Construct the bisector $OC$ of the angle $AOB$.
<li>
Connect $B$ to $C$ and extend to intersect the green line at a point $T$.
</ol>
The line $OT$ is an approximate trisector of the angle $AOB$.
<h2>Error Analysis</h2>
<p>
Let $\alpha$ and $\beta$ be the sizes of the angles $AOB$ and $AOT$,
respectively. One may verify that
\[
\beta
= \arctan \Big( \sin\alpha - (1 - \cos\alpha)
\cot \big( \frac{3}{4}\alpha \big) \Big)
= \frac{1}{3}\alpha + \frac{1}{2^3\cdot3^4} \alpha^3 + O(\alpha^5)
= \frac{1}{3}\alpha + \frac{1}{648} \alpha^3 + O(\alpha^5).
\]
<p>
The error
$
\ds e(\alpha) = \beta - \frac{\alpha}{3}
$
is monotonically increasing in $\alpha$.
The worst error on the interval $0 \le \alpha \le \pi/2$ is
$e(\pi/2)$ = 0.0063 radians = 0.361 degrees.
The worst error on the interval $0 \le \alpha \le \pi$ is
$e(\pi)$ = 0.06 radians = 3.435 degrees.
<p>
<span class="name">Raiford</span>, whose affiliation is given as IBM,
states that he has calculated
the error in increments of one degree in an IBM&nbsp;709. Computers
were novelties when that article was published.
<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 June 14, 2010.
</em>
<p>
<table width="100%">
<tr>
<td valign="top">Go to <a href="index.html#trisections">list of trisections</a></td>
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