archematics/public/rostamian/trisect-durham.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>
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<h1>An angle trisection</h1>
<h4>
R. L. Durham, <i>A simple construction for the approximate trisection
of an angle</i>,
American Mathematical Monthly,
vol.&nbsp;51, no.&nbsp;4, April 1944, pp.&nbsp;217&ndash;218.
</h4>
<table class="centered">
<tr><td align="center">
<applet code="Geometry" archive="Geometry.zip" width="450" height="350">
<param name="background" value="ffffff">
<param name="title" value="An angle trisection">
<param name="e[1]" value="O;point;fixed;210,300">
<param name="e[2]" value="A;point;fixed;420,300">
<param name="e[3]" value="cir1;circle;radius;O,A;none;none;none;none">
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<param name="e[5]" value="OA;line;connect;O,A;none;none;blue">
<param name="e[6]" value="OB;line;connect;O,B;none;none;blue">
<param name="e[7]" value="arcAB;sector;sector;O,A,B;none;none;blue;none">
<param name="e[8]" value="pt1;point;angleBisector;A,O,B;none;none">
<param name="e[9]" value="C;point;cutoff;O,pt1,O,A">
<param name="e[10]" value="OC;line;connect;O,C;none;none;lightGray">
<!-- O-pt2 is 1/3 of OA -->
<param name="e[11]" value="pt2;point;fixed;280,300;none;none">
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<param name="e[14]" value="AG;line;connect;A,G;none;none;lightGray">
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<param name="e[16]" value="li1;line;perpendicular;C,O,O,A;none;none;lightGray">
<param name="e[17]" value="li2;line;chord;li1,cir2;none;none;none">
<param name="e[18]" value="T;point;first;li2">;
<param name="e[19]" value="OT;line;connect;O,T;none;none;red">;
<!-- angle markers -->
<param name="e[20]" value="p1;point;fixed;240,300;none;none">
<param name="e[21]" value="c1;circle;radius;O,p1;none;none;none;none">
<param name="e[22]" value="l1;line;chord;OA,c1;none;none;none">
<param name="e[23]" value="q1;point;first;l1;none;none">
<param name="e[24]" value="l2;line;chord;OT,c1;none;none;none">
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<param name="e[26]" value="s1;sector;sector;O,q1,q2;none;none;black;orange">
<param name="e[27]" value="l3;line;chord;OB,c1;none;none;none">
<param name="e[28]" value="q3;point;first;l3;none;none">
<param name="e[29]" 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>
We wish to trisect the given angle $AOB$ represented by the circular arc
$AB$ centered at $O$, as shown in the diagram above.
<ol>
<li>
Draw the bisector $OC$ of the angle $AOB$.
<li>
Draw the chord $AC$ and trisect it at point $F$ so that $CF=\frac{1}{3}CA$.
<li>
Locate point $G$ on the extension of the chord $AC$ so that $GC=CF$.
<li>
Draw a circle (shown in green) centered at $G$ and through point $F$.
<li>
Let $T$ on the green circle be such that $TC$ is perpendicular to $OC$.
</ol>
The line $OT$ (shown in red) is an approximate trisector of the angle $AOB$.
<p>
<span class="name">R. L. Durham</span>
(see the reference at the top of this page) goes one step further.
Using the line $OT$ as a starting point, he produces a substantially
better approximation $OT'$. The construction for this second stage is
complex and not particularly pretty, so I won't go into that here.
<h2>Error Analysis</h2>
<p>
Let $\alpha$ and $\beta $ be the sizes of the angles $AOB$ and $AOT$,
respectively. It is possible to show that
\[
\beta
= \frac{1}{2}{\alpha}
-
\arctan\bigg(
\frac{4}{3}\sin\Big(
\frac{\alpha}{4} -
\arcsin\big(\frac{1}{2}\sin\frac{\alpha}{4}\big)
\Big)\bigg)
= \frac{1}{3}\alpha + \frac{7}{2^7\cdot3^4} \alpha^3 + O(\alpha^5)
= \frac{1}{3}\alpha + \frac{7}{10368} \alpha^3 + O(\alpha^5).
\]
<!-- The first two terms of the series are the same as those
in trisect-baker.html. The third terms are different.
b_baker := 2*arcsin(4/3*sin(a/8));
series(b_baker,a);
b_durham := a/2 - arctan(sin(a/4 - arcsin(sin(a/4)/2))*4/3);
series(b_durham, a);
-->
<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
$e(\pi/2) =$ 0.00265 radians = 0.152 degrees.
The worst error on the interval $0 \le \alpha \le \pi$ is
$e(\pi)$ = 0.0218 radians = 1.252 degrees.
<h2>Calculation details</h2>
<p>
The derivation of the formula for $\beta$ shown above is a straightforward
application of trigonometry. Here are the details.
<p>
The angles $AOC$ and $ACT$ subtend the arc $AB$ of the circle centered
at $O$. Since $AOC = \alpha/2$, then $ACT=\alpha/4$. Consequently,
the angle $GCT$ is $\pi - \alpha/4$.
<p>
We apply the law of sines in the triangle $GCT$. (The edge $GT$
is not shown in the diagram to reduce clutter.) Let us write
$\theta$ for the angle $CGT$. The the angle $CTG$ is
$\alpha/4 - \theta$. The law of sines is:
\[
\frac{\sin\theta}{CT} = \frac{\sin(\alpha/4-\theta)}{CG} = \frac{\sin(\pi-\alpha/4)}{GT}.
\]
But $GT=2CG$, therefore $2\sin(\alpha/4-\theta) = \sin(\pi-\alpha/4) = \sin(\alpha/4)$, whence
\[
\theta = \frac{\alpha}{4} - \arcsin\big(\frac{1}{2}\sin\frac{\alpha}{4}\big).
\]
Going back to the equation of law of since, we compute $CT$:
\[
CT = \frac{\sin\theta}{\sin(\alpha/4)} GT.
\]
But $GT = GF = 2CF = \frac{2}{3} AC
= \frac{2}{3} \big(2OA\sin\frac{\alpha}{4}\big) = \frac{4}{3} OA\sin\frac{\alpha}{4}$.
We conclude that $CT = \frac{4}{3} OA\sin\theta$.
Then $COT = \arctan \frac{CT}{OC} = \arctan \frac{4}{3} \sin\theta$. Finally, the angle
$\beta$, which equals $COA - COT$, is given by:
\[
\beta = \frac{\alpha}{2} - \arctan \frac{4}{3} \sin\theta.
\]
Substituting the expression for $\theta$ calculated above, we arrive at the desired
expression for $\beta$.
<h2>Aside</h2>
<p>
In 1990, the well-known logician
<span class="name">Willard Van Orman Quine</span>
wrote an expository article in
the <em>Mathematics Magazine</em> where he proves that
some angles cannot be trisected by ruler and compass. The proof
is elementary (but not short) and requires
nothing but simple algebra. Here is the full reference:
<h4>
W. V. Quine, <i>Elementary proof that some angles cannot be trisected by ruler
and compass</i>,
Mathematics Magazine,
vol.&nbsp;63, no.&nbsp;2, April 1990, pp.&nbsp;95&ndash;105.
</h4>
<p>
In a prefatory note he refers to <span class="name">Durham</span>, the author of the trisection
described in this web page. He writes:
<blockquote>
<p>
This purely expository paper dates from April 1946.
<span class="name">Robert Lee Durham</span>,
president emeritus of Southern Seminary Junior and College,
had sent me a hundred dollars and asked me to make it clear to him
why an angle cannot in general be trisected by ruler
and compass. He had himself presented a way of almost trisecting
any angle by ruler and compass, to an accuracy for acute angles of
1/720 of a degree. [Here he refers to <span class="name">Durham</span>'s
1944 article cited at the top of this web page.]
<p>
I welcomed the money and the occasion to familiarize myself
with the famous proof. I was guided in large part by
<span class="name">L. E. Dickson</span>,
<em>Why it is impossible trisect to an
angle or construct a regular polygon of 7 or 9 sides by ruler and compass,</em>
Mathematics Teacher, vol.&nbsp;14 (1921), 217&ndash;223.
<p>
<span class="name">Mr. Durham</span>
expressed satisfaction with my report and proposed
paying for publishing it as a pamphlet. With his approval I
submitted instead to a mathematics journal. After waiting
nineteen months for a decision from the journal, I recalled
the paper and dropped the matter.
</blockquote>
<p>
<span class="name">Quine</span>
goes on to explain how this article was eventually published
more than 50 years after it was written.
<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 13, 2010.
</em>
<p>
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