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<h1>An angle trisection</h1>
<h4>Construction by
<a href="mailto:mhs210@hotmail.com">Mark Stark</a>
</h4>
<table class="centered">
<tr><td align="center">
<applet code="Geometry" archive="Geometry.zip" width="600" height="600">
<param name="background" value="ffffff">
<param name="title" value="An angle trisection">
<!-- The angle AOB -->
<param name="e[1]" value="O;point;fixed;300,300">
<param name="e[2]" value="A;point;fixed;585,300">
<param name="e[3]" value="C1;circle;radius;O,A;none;none;lightGray;none">
<param name="e[4]" value="B;point;circleSlider;C1,70,0;red;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">
<!-- The inner circle -->
<param name="e[7]" value="A';point;fixed;395,300">
<param name="e[8]" value="C2;circle;radius;O,A';none;none;lightGray;none">
<param name="e[9]" value="L1;line;chord;OB,C2;none;none;none">
<param name="e[10]" value="B';point;first;L1">
<param name="e[11]" value="A'B';line;connect;A',B';none;none;green">
<!-- Points E and D -->
<param name="e[12]" value="E;point;circleSlider;C2,600,150;red;red">
<param name="e[13]" value="C3;circle;radius;A',E;none;none;lightGray;none">
<param name="e[14]" value="L2;line;chord;A'B',C3;none;none;none">
<param name="e[15]" value="D;point;first;L2">
<param name="e[16]" value="L3;line;chord;D,E,C1;none;none;none">
<param name="e[17]" value="G;point;last;L3">
<param name="e[18]" value="EG;line;connect;E,G;none;none;lightGray">
<param name="e[19]" value="L4;line;chord;G,O,C1;none;none;none">
<param name="e[20]" value="T;point;last;L4">
<param name="e[21]" value="GT;line;connect;G,T;none;none;red">
<param name="e[22]" value="p1;point;fixed;325,300;none;none">
<param name="e[23]" value="c1;circle;radius;O,p1;none;none;none;none">
<param name="e[24]" value="l1;line;chord;OA,c1;none;none;none">
<param name="e[25]" value="q1;point;first;l1;none;none">
<param name="e[26]" value="l2;line;chord;O,T,c1;none;none;none">
<param name="e[27]" value="q2;point;first;l2;none;none">
<param name="e[28]" value="s1;sector;sector;O,q1,q2;none;none;black;orange">
<param name="e[29]" value="p2;point;fixed;325,300;none;none">
<param name="e[30]" value="c2;circle;radius;O,p2;none;none;none;none">
<param name="e[31]" value="l3;line;chord;O,T,c2;none;none;none">
<param name="e[32]" value="q3;point;first;l3;none;none">
<param name="e[33]" value="l4;line;chord;OB,c2;none;none;none">
<param name="e[34]" value="q4;point;first;l4;none;none">
<param name="e[35]" value="s2;sector;sector;O,q3,q4;none;none;black;yellow">
<param name="pivot" value="O">
</applet>
</td></tr>
<tr><td>
<b>
Drag the point $B$ to change the angle $AOB$
(but stay on the upper half of the semicircle).<br>
Drag the point “$E$” to change the radius of the circle
centered at $A'$. Note how little $G$ is affected by the choice of $E$.<br>
Press “r” to reset the diagram to its initial state.<br>
The red line ($O$T) is an approximate trisector of the angle $AOB$.
</b>
</td></tr></table>
<h2>Construction</h2>
<p>
This construction, due to Mark Stark, was announced
<a href="http://mathforum.org/kb/message.jspa?messageID=1088614">
in an message</a> in the <code>geometry.puzzles</code> newsgroup on
Jun 20, 2002.
Scroll to the bottom of that page to view the related discussion thread.
<p>
The construction is unusual
because one of the steps involves an arbitrary choice. It is interesting
that the result is quite insensitive to that choice.
<p>
Here I have paraphrased Mark's construction but
differences from the original are cosmetic. The error analysis is mine.
<p>
Consider the circular arc $AB$ on the circle $C$ centered at $O$,
shown in the diagram above.
Assume the angle $AOB$ is between 0 and 180 degrees.
To trisect $AOB$, do:
<ol>
<li>
Draw a circle $C'$ centered at $O$ with a radius 1/3 $OA$.
Mark $A'$ and $B'$ its intersections with the line segments $OA$ and $OB$,
respectively.
<li>
Draw the line $A'B'$ (shown in green).
<li>
Draw a circle $C''$ centered at $A'$ with an arbitrary(!) radius.
The accuracy of the trisection will be affected by the choice
of the radius, albeit only slightly. Best results are obtained
when the angle $EA'D$ (see the next step) is close to one third
of the angle $AOB$.
<li>
Let $D$ be the intersection of $C''$ with the line segment $A'B'$.<br>
Let $E$ be the intersection of $C''$ with the $C'$, as shown.
<li>
Draw the line $ED$ and extend to the intersection point $G$ with the the
circle $C$.
<li>
Draw the diameter $GOT$.
</ol>
The line $OT$ is an approximate trisector of the angle $AOB$.
<h2>Error Analysis</h2>
<p>
In the diagram below, I have duplicated the previous diagram
and added the lines $OE$ and $EA'$ which are not needed in
the construction, but are needed for the error analysis.
<p>
You may zoom and translate the diagram to examine its details.
To zoom in, grab the point $B'$ with the mouse
and move it away from $O$. To translate,
grab $O$ and move it around. As always, type “r” to
reset the diagram to its initial state.
<table class="centered">
<tr><td align="center">
<applet code="Geometry" archive="Geometry.zip" width="600" height="600">
<param name="background" value="ffffff">
<param name="title" value="An angle trisection">
<!-- The angle AOB -->
<param name="e[1]" value="O;point;fixed;300,300">
<param name="e[2]" value="A;point;fixed;585,300">
<param name="e[3]" value="C1;circle;radius;O,A;none;none;lightGray;none">
<param name="e[4]" value="B;point;circleSlider;C1,70,0;red;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">
<!-- The inner circle -->
<param name="e[7]" value="A';point;fixed;395,300">
<param name="e[8]" value="C2;circle;radius;O,A';none;none;lightGray;none">
<param name="e[9]" value="L1;line;chord;OB,C2;none;none;none">
<param name="e[10]" value="B';point;first;L1">
<param name="e[11]" value="A'B';line;connect;A',B';none;none;green">
<!-- Points E and D -->
<param name="e[12]" value="E;point;circleSlider;C2,600,150;red;red">
<param name="e[13]" value="C3;circle;radius;A',E;none;none;lightGray;none">
<param name="e[14]" value="L2;line;chord;A'B',C3;none;none;none">
<param name="e[15]" value="D;point;first;L2">
<param name="e[16]" value="L3;line;chord;D,E,C1;none;none;none">
<param name="e[17]" value="G;point;last;L3">
<param name="e[18]" value="EG;line;connect;E,G;none;none;lightGray">
<param name="e[19]" value="L4;line;chord;G,O,C1;none;none;none">
<param name="e[20]" value="T;point;last;L4">
<param name="e[21]" value="GT;line;connect;G,T;none;none;red">
<param name="e[22]" value="p1;point;fixed;325,300;none;none">
<param name="e[23]" value="c1;circle;radius;O,p1;none;none;none;none">
<param name="e[24]" value="l1;line;chord;OA,c1;none;none;none">
<param name="e[25]" value="q1;point;first;l1;none;none">
<param name="e[26]" value="l2;line;chord;O,T,c1;none;none;none">
<param name="e[27]" value="q2;point;first;l2;none;none">
<param name="e[28]" value="s1;sector;sector;O,q1,q2;none;none;black;orange">
<param name="e[29]" value="p2;point;fixed;325,300;none;none">
<param name="e[30]" value="c2;circle;radius;O,p2;none;none;none;none">
<param name="e[31]" value="l3;line;chord;O,T,c2;none;none;none">
<param name="e[32]" value="q3;point;first;l3;none;none">
<param name="e[33]" value="l4;line;chord;OB,c2;none;none;none">
<param name="e[34]" value="q4;point;first;l4;none;none">
<param name="e[35]" value="s2;sector;sector;O,q3,q4;none;none;black;yellow">
<param name="e[36]" value="EA;;line;connect;E,A';none;none;lightGray">
<param name="e[37]" value="OE;;line;connect;O,E;none;none;lightGray">
<param name="pivot" value="O">
</applet>
</td></tr>
<tr><td>
<b>
To zoom in, grab the point $B'$ with the mouse
and move it away from $O$.<br>
To translate, grab $O$ and move it around.<br>
Type “r” to return to the initial state.
</b>
</td></tr></table>
<p>
Let $\alpha$ and $\beta$ be the sizes of the angles $AOB$ and $AOT$,
respectively. We will show that $\beta \approx \frac{1}{3}\alpha$.
<p>
The construction leaves the size of the circle $C''$ (centered at $A'$)
unspecified. We parametrize the circle by the position of the point
$E$ along the arc $A'B'$, or more precisely, by the value $\gamma$ of the
angle angle $B'A'E$.
Thus $\gamma=0$ when $E$ coincides with $B'$
and $\gamma=\alpha/2$ (easy to verify) when E coincides with $A'$.
<p>
Since the angles $B'A'E$ and $B'OE$ subtend the arc $B'E$ of the
circle $C'$, then the angle $B'OE$ is $2\gamma$. Therefore the angle
$EOA'$ is $\alpha - 2\gamma$. But $EOA'$ is the vertex angle of the
triangle $EOA'$, therefore the base angle $OEA'$ is $\frac{1}{2}(\pi - \alpha +
2\gamma)$.
<p>
In the isosceles triangle $DA'E$, the vertex angle is $\gamma$, therefore
the base angle $DEA'$ is $\frac{1}{2}(\pi - \gamma)$.
<p>
Putting the assertions of the two previous paragraphs together, we
calculate the angle $OED$:
\[
OED = OEA' - DEA' = \frac{1}{2}(3\gamma - \alpha)
\]
<p>
In the triangle $GOE$, we have just computed the angle at $E$ (because
$OED$ is the same as $OEG$). Let us write $x$ for the angle at $G$.
Then $x$ may be computed by applying the law of sines and noting
that the ratio of the sides $OG$ to $OE$ is 3. We get:
$3\sin x = \sin\frac{1}{2}(3\gamma-\alpha)$.
<p>
The external angle $EOT$ of the triangle $GOE$ equals the sum of the
remaining internal angles, that is:
\[
EOT = x + \frac{1}{2}(3\gamma-\alpha).
\]
On the other hand,
\[
EOT
= EOB' - TOB'
= EOB' - (A'OB' - A'OT)
= 2\gamma - (\alpha - \beta).
\]
We see then $x + \frac{1}{2}(3\gamma-\alpha) = 2\gamma - (\alpha - \beta)$,
whence $x = \beta + \gamma/2 - \beta/2$. This leads to the equation:
\[
3\sin\big(\beta + \frac{1}{2}\gamma - \frac{1}{2}\beta\big)
=
\sin\frac{1}{2}(3\gamma-\alpha),
\]
which we may solve for $\beta$:
\[
\beta =
\frac{1}{3}\alpha + \frac{1}{6}(\alpha-3\gamma)
-
\arcsin\bigg[
\frac{1}{3}\sin\Big(\frac{\alpha-3\gamma}{2} \Big)
\bigg]
\]
<p>
As expected, the constructed angle,
$\beta$, depends on the original angle $\alpha$ we well as the choice
of $\gamma$. Let us express this dependence as $\beta = \tau(\alpha,\gamma)$.
Expanding $\tau$ in power series we get:
\[
\beta = \tau(\alpha,\gamma)
= \frac{1}{3}\alpha
+ \frac{4}{3}\Big(\frac{\alpha-3\gamma}{6} \Big)^3
- \frac{4}{5}\Big(\frac{\alpha-3\gamma}{6} \Big)^7
+ O\bigg( \Big(\frac{\alpha-3\gamma}{6}\Big)^9 \bigg).
\]
The term with exponent 5 is absent in the series expansion; that's not a typo.
<h3>On the choice of $\gamma$</h3>
<p>
We see that $\tau(\alpha,\alpha/3) = \alpha/3$, that is,
the construction produces an <em>exact trisection</em>
with the choice $\gamma=\alpha/3$.
Of course, constructing such a $\gamma$
is equivalent to solving the original trisection problem, therefore
that is not an option. On the other hand, a constructible $\gamma$ that
comes close to $\alpha/3$ will serve just fine. The function $\tau$ is
not very sensitive to the variations of $\gamma$ as is evident from:
\[
\frac{\partial \tau(\alpha,\gamma)}{\partial \gamma}
=
\frac{1}{2} \bigg(
\frac{3\cos 3x}{\sqrt{9 - \sin^2 3x}} -1 \bigg),
\]
where I have let $x=(\alpha-3\gamma)/6$ to simplify the notation.
As noted above, best
results are achieved when $\gamma$ is close to $\alpha/3$. Even
with a not-so-optimal choice of $\gamma=\alpha/4$ we get $x=\alpha/24$.
With such a choice, the value of partial derivative in the range
$0 \le \alpha \le \pi/2$ does not exceed 0.01, indicating that
the value of the function is essentially independent of $\gamma$
on that range.
<p>
<b>An excellent choice</b> for $\gamma$ is obtained as follows.
In Step 3 of the construction, first select the point $D$ on
the line segment $A'B'$ such that $A'D = \frac{1}{3} A'B'$.
Then draw the circle $C''$ with center $A'$ passing through $D$.
One may verify that this results in an angle $\gamma$ given by:
\[
\hat{\gamma} = \frac{1}{2} \alpha
- \arcsin\Big( \frac{1}{3} \sin\frac{\alpha}{2} \Big)
= \frac{1}{3} \alpha + \frac{1}{2\cdot3^4} \alpha^3 + O(\alpha^7).
\]
Then, the constructed angle is:
\[
\beta = \tau(\alpha,\hat{\gamma})
= \tau\bigg(\alpha, \frac{1}{2} \alpha
- \arcsin\Big( \frac{1}{3} \sin\frac{\alpha}{2} \Big)
\bigg)
= \frac{1}{3} \alpha - \frac{1}{2^4\cdot3^{13}} \alpha^9 +
O(\alpha^{13}).
\]
The construction error, $e(\alpha) = \frac{1}{3}\alpha-\beta$,
is monotone increasing. Since $e(\alpha) = O(\alpha^9)$,
we expect it to be very small.
Indeed, the worst error on the interval $0 \le \alpha \le \pi/2$ is
the incredibly small
$e(\pi/2)$ = 0.00000226 radians = 0.00013 degrees.
The worst error on the interval $0 \le \alpha \le \pi$ is
$e(\pi)$ = 0.00103 radians = 0.0592 degrees.
<p>
Despite its extraordinary accuracy, this is <em>not</em>
among my favorite trisection methods because the points $D$ and $E$ are
too close to each other for locating the point $G$ reliably. For practical
purposes, should there be such a need, I would much rather use a more
robust, albeit less accurate, method.
<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 26, 2002.
<br>The error analysis was thoroughly revised and extensive
cosmetic changes were made on June 7, 2010.
</em>
<p>
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