archematics/public/rostamian/trisect-jamison.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">
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<title>An angle trisection</title>
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<h1>An angle trisection</h1>
<h4>
Construction attributed to C.&nbsp;R.&nbsp;Lindberg, as reported in<br>
Free Jamison, <i>Trisection Approximation</i>, American Mathematical Monthly,
vol.&nbsp;61, no.&nbsp;5, May 1954, pp.&nbsp;334&ndash;336.
</h4>
<table class="centered">
<tr><td align="center">
<applet code="Geometry" archive="Geometry.zip" width="700" height="400">
<param name="background" value="ffffff">
<param name="title" value="An angle trisection">
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<param name="e[3]" value="C1;circle;radius;O,A;none;none;lightGray;none">
<param name="e[4]" value="B;point;circleSlider;C1,350,0;red">
<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">
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</applet>
</td></tr>
<tr><td>
<b>
Drag the point $B$ to change the angle $AOB$
(but stay on the right half of the circle).<br>
Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
The red line $OE$ is an approximate trisector of the angle $AOB$.
</b>
</td></tr></table>
<h2>The construction</h2>
<p>
According to Jamison (see the reference at the top of this page)
the construction's main idea
comes from an unpublished work by C.&nbsp;R.&nbsp;Lindberg.
<p>
Consider the circular arc $AB$ centered at $O$, shown in the diagram above.
Assume the angle $AOB$ is between 0 and 180 degrees.
To trisect $AOB$, do:
<ol>
<li>
Extend $BO$ to intersect the circle at a point $C$.
<li>
Draw the bisector $OD$ of the angle $AOB$.
<li>
Draw the line $CD$ and extend it to a point $E$ such that $DE$ equals the
circle's diameter.
</ol>
<p>
The line $OE$ is an approximate trisector of the angle $AOB$.
<p>
Here is a heuristic explanation for why the construction works,
as explained by Jamison. The key lies
in the observation that (i) in the triangle $ODE$ the angles $O$ and $E$ are
&ldquo;small&rdquo;, and (ii) the side $ED$ is twice as long as the side $OD$.
Therefore from the law of sines we have $\sin(O)/\sin(E) = ED/OD = 2$,
which implies that the angle $O$ is approximately twice the angle $E$ in
the triangle $ODE$.
<p>
Let the measure of the angle $OED$ be $x$.
Then the triangle's external angle at $D$, that is the angle $ODC$, is the sum of
the internal angles $O$ and $E$, therefore it is approximately $3x$.
Therefore the angle $OCD$ is $3x$. Therefore the angle $BOD$ is $6x$.
Since the angle $E'OD$ is $2x$, we conclude that the angle $BOE'$ is $4x$.
Furthermore, since the angle $BOD$ is $6x$, the angle $BOA$ is $12x$. This shows
that $BOA$ is 3 times $BOE'$, as asserted.
<h2>Error Analysis</h2>
<p>
Let $\alpha$ and $\beta$ be the sizes of the angles $AOB$ and $EOB$, respectively.
The angle $DOB$ is half of $AOB$ by the construction, therefore it is equal to
$\alpha/2$. Consequently the angle $DCB$, which is half the
central angle $DOB$, equals $\alpha/4$.
The triangle $DOC$ is isosceles, therefore the angle $ODC$ also equals $\alpha/4$.
<p>
In the triangle $OED$, let $x$ and $y$ be the sizes of the angles
$OED$ and $EOD$, respectively. Since the sum $x+y$ of the triangle's internal
angles equals the triangle's
external angle $ODC$, we have $x+y = \alpha/4$. Let us note, however,
that the angle $y$ equals $DOB$ minus $EOB$. Thus $y = \alpha/2 - \beta$,
whence $x = \beta - \alpha/4$.
<p>
In the triangle $OED$, the side $DE$ is twice the side $OD$ by the construction,
therefore the law of sines gives $\sin y = 2 \sin x$. Consequently,
$\sin(\alpha/2 - \beta) = 2 \sin(\beta - \alpha/4)$. Solving this for $\beta$
we arrive at:
\[
\beta
= \frac{1}{4} \alpha + \arctan \frac{\sin(a/4)}{2+\cos(a/4)}
= \frac{1}{3} \alpha - \frac{1}{2^6\cdot3^4} \alpha^3 + O(\alpha^5)
= \frac{1}{3} \alpha - \frac{1}{5184} \alpha^3 + O(\alpha^5).
\]
<p>
We see that the trisection error $e(\alpha) = \alpha/3 - \beta$ is given by:
\[
e(\alpha) = \frac{1}{12}\alpha - \arctan \frac{\sin(a/4)}{2+\cos(a/4)}.
\]
(This formula is also given in Jamison's article.)
The function $e(a)$ is monotonically increasing.
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
$e(\pi)$ = 0.0063 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-pllana.html">Pllana's construction</a>,
where $\beta$ is given as:
\[
\beta
= \arctan \frac
{\sin\frac{\alpha}{2} + 2\sin\frac{\alpha}{4}}
{\cos\frac{\alpha}{2} + 2\cos\frac{\alpha}{4}}.
\]
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:
\[
\frac{d\beta}{d\alpha} =
\frac{3(1 + \cos\frac{\alpha}{4})}{2(5 + 4\cos\frac{\alpha}{4})}.
\]
<h2>An extension</h2>
<p>
Although this construction is quite good as is, Jamison proceeds to give
<a href="trisect-jamison-ext.html">an extension of Lindberg's method</a>
which requires a bit more work but is substantially more accurate.
<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 22, 2002.<br>
Cosmetic revisions on June 7, 2010.
</em>
<p>
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