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<h1>An angle trisection</h1>
<h4>Construction due to
<a href="mailto:avniu66@hotmail.com">Avni Pllana</a></h4>
<table class="centered">
<tr><td align="center">
<applet code="Geometry" archive="Geometry.zip" width="450" height="260">
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<param name="title" value="An angle trisection">
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<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">
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<param name="e[17]" value="li3;line;cutoff;O,N,O,A;none;none;red">
<!-- angle marker -->
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</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 is an approximate trisector of the angle $AOB$.
</b>
</td></tr></table>
<h2>Construction</h2>
<p>
This approximate trisection, due to Avni Pllana, was announced
<a href="http://mathforum.org/kb/message.jspa?messageID=1084688">in a
message</a>
in the <code>geometry.puzzles</code> newsgroup on July&nbsp;23, 2003.
Scroll to the bottom of that page to view the related discussion thread.
<p>
Consider the angle $AOB$ given by the circular arc $AB$ centered at $O$,
as shown in the diagram above.
<ol>
<li>
Pick points $C$ and $D$ on the arc $AB$ so that $OC$ bisects the angle $AOB$
and $OD$ bisects the angle $AOC$.
<li>
Let $M$ be the midpoint of the line segment $OC$.
<li>
Let $N$ be the midpoint of the line segment $MD$.
</ol>
The line $ON$ 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 $AON$,
respectively. One may verify that:
\[
\beta
= \arctan \frac
{\sin\frac{\alpha}{2} + 2\sin\frac{\alpha}{4}}
{\cos\frac{\alpha}{2} + 2\cos\frac{\alpha}{4}}
= \frac{1}{3}\alpha - \frac{1}{2^6\cdot3^4} \alpha^3 + O(\alpha^5)
= \frac{1}{3}\alpha - \frac{1}{5184} \alpha^3 + O(\alpha^5).
\]
<em>Hint:</em> Represent the points as complex numbers
in the polar form $re^{i\theta}$.
<p>
The error
$
\ds e(\alpha) = \frac{\alpha}{3} - \beta
$
increases monotonically with $\alpha$.
The worst error on the interval $0 \le \alpha \le \pi/2$ is
$e(\pi/2)$ = 0.000757 radians = 0.0434 degrees.
The worst error on the interval $0 \le \alpha \le \pi$ is
$e(\pi)$ = 0.00630 radians = 0.361 degrees.
That's quite good for such a simple construction.
<h2>An interesting coincidence</h2>
<p>
The angle $\beta$ constructed by this method coincides <em>exactly</em>
with that of <a href="trisect-jamison.html">Lindberg's construction</a>,
where $\beta$ is given as:
\[
\beta
= \frac{1}{4} \alpha + \arctan
\frac{\sin\frac{\alpha}{4}}{2+\cos\frac{\alpha}{4}}.
\]
One way to verify that the seemingly different expressions
for $\beta$ are in fact identical,
is to compare their derivatives. In both cases we have:
\[
\frac{d\beta}{d\alpha} =
\frac{3(1 + \cos\frac{\alpha}{4})}{2(5 + 4\cos\frac{\alpha}{4})}.
\]
<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 10, 2010.
</em>
<p>
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