Glen Whitney
c99b51dafa
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. Reviewed-on: #39 Co-authored-by: Glen Whitney <glen@studioinfinity.org> Co-committed-by: Glen Whitney <glen@studioinfinity.org>
400 lines
14 KiB
HTML
400 lines
14 KiB
HTML
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
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<html>
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<head>
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<!-- fix buggy IE8, especially for mathjax -->
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<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7">
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<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
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<title>An angle trisection</title>
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<link rel="stylesheet" type="text/css" media="screen" href="style.css">
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<script type="text/javascript"
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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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MathJax.Hub.Queue( function() {document.body.style.visibility="visible"} );
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</script>
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</head>
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<body style="visibility:hidden">
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<h1>An angle trisection</h1>
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<h4>Construction by
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<a href="mailto:mhs210@hotmail.com">Mark Stark</a>
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</h4>
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<table class="centered">
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<tr><td align="center">
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<applet code="Geometry" archive="Geometry.zip" width="600" height="600">
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<param name="background" value="ffffff">
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<param name="title" value="An angle trisection">
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<!-- The angle AOB -->
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<param name="e[1]" value="O;point;fixed;300,300">
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<param name="e[2]" value="A;point;fixed;585,300">
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<param name="e[3]" value="C1;circle;radius;O,A;none;none;lightGray;none">
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<param name="e[4]" value="B;point;circleSlider;C1,70,0;red;red">
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<param name="e[5]" value="OA;line;connect;O,A;none;none;blue">
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<param name="e[6]" value="OB;line;connect;O,B;none;none;blue">
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<!-- The inner circle -->
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<param name="e[7]" value="A';point;fixed;395,300">
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<param name="e[8]" value="C2;circle;radius;O,A';none;none;lightGray;none">
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<param name="e[9]" value="L1;line;chord;OB,C2;none;none;none">
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<param name="e[10]" value="B';point;first;L1">
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<param name="e[11]" value="A'B';line;connect;A',B';none;none;green">
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<!-- Points E and D -->
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<param name="e[12]" value="E;point;circleSlider;C2,600,150;red;red">
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<param name="e[13]" value="C3;circle;radius;A',E;none;none;lightGray;none">
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<param name="e[14]" value="L2;line;chord;A'B',C3;none;none;none">
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<param name="e[15]" value="D;point;first;L2">
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<param name="e[16]" value="L3;line;chord;D,E,C1;none;none;none">
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<param name="e[17]" value="G;point;last;L3">
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<param name="e[18]" value="EG;line;connect;E,G;none;none;lightGray">
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<param name="e[19]" value="L4;line;chord;G,O,C1;none;none;none">
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<param name="e[20]" value="T;point;last;L4">
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<param name="e[21]" value="GT;line;connect;G,T;none;none;red">
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<param name="e[22]" value="p1;point;fixed;325,300;none;none">
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<param name="e[23]" value="c1;circle;radius;O,p1;none;none;none;none">
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<param name="e[24]" value="l1;line;chord;OA,c1;none;none;none">
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<param name="e[25]" value="q1;point;first;l1;none;none">
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<param name="e[26]" value="l2;line;chord;O,T,c1;none;none;none">
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<param name="e[27]" value="q2;point;first;l2;none;none">
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<param name="e[28]" value="s1;sector;sector;O,q1,q2;none;none;black;orange">
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<param name="e[29]" value="p2;point;fixed;325,300;none;none">
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<param name="e[30]" value="c2;circle;radius;O,p2;none;none;none;none">
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<param name="e[31]" value="l3;line;chord;O,T,c2;none;none;none">
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<param name="e[32]" value="q3;point;first;l3;none;none">
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<param name="e[33]" value="l4;line;chord;OB,c2;none;none;none">
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<param name="e[34]" value="q4;point;first;l4;none;none">
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<param name="e[35]" value="s2;sector;sector;O,q3,q4;none;none;black;yellow">
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<param name="pivot" value="O">
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</applet>
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</td></tr>
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<tr><td>
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<b>
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Drag the point $B$ to change the angle $AOB$
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(but stay on the upper half of the semicircle).<br>
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Drag the point “$E$” to change the radius of the circle
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centered at $A'$. Note how little $G$ is affected by the choice of $E$.<br>
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Press “r” to reset the diagram to its initial state.<br>
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The red line ($O$T) is an approximate trisector of the angle $AOB$.
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</b>
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</td></tr></table>
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<h2>Construction</h2>
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<p>
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This construction, due to Mark Stark, was announced
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<a href="http://mathforum.org/kb/message.jspa?messageID=1088614">
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in an message</a> in the <code>geometry.puzzles</code> newsgroup on
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Jun 20, 2002.
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Scroll to the bottom of that page to view the related discussion thread.
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<p>
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The construction is unusual
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because one of the steps involves an arbitrary choice. It is interesting
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that the result is quite insensitive to that choice.
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<p>
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Here I have paraphrased Mark's construction but
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differences from the original are cosmetic. The error analysis is mine.
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<p>
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Consider the circular arc $AB$ on the circle $C$ centered at $O$,
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shown in the diagram above.
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Assume the angle $AOB$ is between 0 and 180 degrees.
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To trisect $AOB$, do:
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<ol>
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<li>
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Draw a circle $C'$ centered at $O$ with a radius 1/3 $OA$.
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Mark $A'$ and $B'$ its intersections with the line segments $OA$ and $OB$,
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respectively.
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<li>
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Draw the line $A'B'$ (shown in green).
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<li>
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Draw a circle $C''$ centered at $A'$ with an arbitrary(!) radius.
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The accuracy of the trisection will be affected by the choice
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of the radius, albeit only slightly. Best results are obtained
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when the angle $EA'D$ (see the next step) is close to one third
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of the angle $AOB$.
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<li>
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Let $D$ be the intersection of $C''$ with the line segment $A'B'$.<br>
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Let $E$ be the intersection of $C''$ with the $C'$, as shown.
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<li>
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Draw the line $ED$ and extend to the intersection point $G$ with the the
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circle $C$.
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<li>
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Draw the diameter $GOT$.
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</ol>
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The line $OT$ is an approximate trisector of the angle $AOB$.
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<h2>Error Analysis</h2>
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<p>
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In the diagram below, I have duplicated the previous diagram
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and added the lines $OE$ and $EA'$ which are not needed in
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the construction, but are needed for the error analysis.
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<p>
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You may zoom and translate the diagram to examine its details.
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To zoom in, grab the point $B'$ with the mouse
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and move it away from $O$. To translate,
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grab $O$ and move it around. As always, type “r” to
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reset the diagram to its initial state.
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<table class="centered">
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<tr><td align="center">
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<applet code="Geometry" archive="Geometry.zip" width="600" height="600">
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<param name="background" value="ffffff">
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<param name="title" value="An angle trisection">
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<!-- The angle AOB -->
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<param name="e[1]" value="O;point;fixed;300,300">
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<param name="e[2]" value="A;point;fixed;585,300">
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<param name="e[3]" value="C1;circle;radius;O,A;none;none;lightGray;none">
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<param name="e[4]" value="B;point;circleSlider;C1,70,0;red;red">
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<param name="e[5]" value="OA;line;connect;O,A;none;none;blue">
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<param name="e[6]" value="OB;line;connect;O,B;none;none;blue">
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<!-- The inner circle -->
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<param name="e[7]" value="A';point;fixed;395,300">
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<param name="e[8]" value="C2;circle;radius;O,A';none;none;lightGray;none">
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<param name="e[9]" value="L1;line;chord;OB,C2;none;none;none">
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<param name="e[10]" value="B';point;first;L1">
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<param name="e[11]" value="A'B';line;connect;A',B';none;none;green">
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<!-- Points E and D -->
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<param name="e[12]" value="E;point;circleSlider;C2,600,150;red;red">
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<param name="e[13]" value="C3;circle;radius;A',E;none;none;lightGray;none">
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<param name="e[14]" value="L2;line;chord;A'B',C3;none;none;none">
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<param name="e[15]" value="D;point;first;L2">
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<param name="e[16]" value="L3;line;chord;D,E,C1;none;none;none">
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<param name="e[17]" value="G;point;last;L3">
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<param name="e[18]" value="EG;line;connect;E,G;none;none;lightGray">
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<param name="e[19]" value="L4;line;chord;G,O,C1;none;none;none">
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<param name="e[20]" value="T;point;last;L4">
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<param name="e[21]" value="GT;line;connect;G,T;none;none;red">
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<param name="e[22]" value="p1;point;fixed;325,300;none;none">
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<param name="e[23]" value="c1;circle;radius;O,p1;none;none;none;none">
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<param name="e[24]" value="l1;line;chord;OA,c1;none;none;none">
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<param name="e[25]" value="q1;point;first;l1;none;none">
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<param name="e[26]" value="l2;line;chord;O,T,c1;none;none;none">
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<param name="e[27]" value="q2;point;first;l2;none;none">
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<param name="e[28]" value="s1;sector;sector;O,q1,q2;none;none;black;orange">
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<param name="e[29]" value="p2;point;fixed;325,300;none;none">
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<param name="e[30]" value="c2;circle;radius;O,p2;none;none;none;none">
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<param name="e[31]" value="l3;line;chord;O,T,c2;none;none;none">
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<param name="e[32]" value="q3;point;first;l3;none;none">
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<param name="e[33]" value="l4;line;chord;OB,c2;none;none;none">
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<param name="e[34]" value="q4;point;first;l4;none;none">
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<param name="e[35]" value="s2;sector;sector;O,q3,q4;none;none;black;yellow">
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<param name="e[36]" value="EA;;line;connect;E,A';none;none;lightGray">
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<param name="e[37]" value="OE;;line;connect;O,E;none;none;lightGray">
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<param name="pivot" value="O">
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</applet>
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</td></tr>
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<tr><td>
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<b>
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To zoom in, grab the point $B'$ with the mouse
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and move it away from $O$.<br>
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To translate, grab $O$ and move it around.<br>
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Type “r” to return to the initial state.
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</b>
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</td></tr></table>
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<p>
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Let $\alpha$ and $\beta$ be the sizes of the angles $AOB$ and $AOT$,
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respectively. We will show that $\beta \approx \frac{1}{3}\alpha$.
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<p>
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The construction leaves the size of the circle $C''$ (centered at $A'$)
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unspecified. We parametrize the circle by the position of the point
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$E$ along the arc $A'B'$, or more precisely, by the value $\gamma$ of the
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angle angle $B'A'E$.
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Thus $\gamma=0$ when $E$ coincides with $B'$
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and $\gamma=\alpha/2$ (easy to verify) when E coincides with $A'$.
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<p>
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Since the angles $B'A'E$ and $B'OE$ subtend the arc $B'E$ of the
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circle $C'$, then the angle $B'OE$ is $2\gamma$. Therefore the angle
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$EOA'$ is $\alpha - 2\gamma$. But $EOA'$ is the vertex angle of the
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triangle $EOA'$, therefore the base angle $OEA'$ is $\frac{1}{2}(\pi - \alpha +
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2\gamma)$.
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<p>
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In the isosceles triangle $DA'E$, the vertex angle is $\gamma$, therefore
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the base angle $DEA'$ is $\frac{1}{2}(\pi - \gamma)$.
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<p>
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Putting the assertions of the two previous paragraphs together, we
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calculate the angle $OED$:
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\[
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OED = OEA' - DEA' = \frac{1}{2}(3\gamma - \alpha)
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\]
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<p>
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In the triangle $GOE$, we have just computed the angle at $E$ (because
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$OED$ is the same as $OEG$). Let us write $x$ for the angle at $G$.
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Then $x$ may be computed by applying the law of sines and noting
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that the ratio of the sides $OG$ to $OE$ is 3. We get:
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$3\sin x = \sin\frac{1}{2}(3\gamma-\alpha)$.
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<p>
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The external angle $EOT$ of the triangle $GOE$ equals the sum of the
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remaining internal angles, that is:
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\[
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EOT = x + \frac{1}{2}(3\gamma-\alpha).
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\]
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On the other hand,
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\[
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EOT
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= EOB' - TOB'
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= EOB' - (A'OB' - A'OT)
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= 2\gamma - (\alpha - \beta).
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\]
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We see then $x + \frac{1}{2}(3\gamma-\alpha) = 2\gamma - (\alpha - \beta)$,
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whence $x = \beta + \gamma/2 - \beta/2$. This leads to the equation:
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\[
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3\sin\big(\beta + \frac{1}{2}\gamma - \frac{1}{2}\beta\big)
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=
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\sin\frac{1}{2}(3\gamma-\alpha),
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\]
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which we may solve for $\beta$:
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\[
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\beta =
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\frac{1}{3}\alpha + \frac{1}{6}(\alpha-3\gamma)
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-
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\arcsin\bigg[
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\frac{1}{3}\sin\Big(\frac{\alpha-3\gamma}{2} \Big)
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\bigg]
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\]
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<p>
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As expected, the constructed angle,
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$\beta$, depends on the original angle $\alpha$ we well as the choice
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of $\gamma$. Let us express this dependence as $\beta = \tau(\alpha,\gamma)$.
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Expanding $\tau$ in power series we get:
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\[
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\beta = \tau(\alpha,\gamma)
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= \frac{1}{3}\alpha
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+ \frac{4}{3}\Big(\frac{\alpha-3\gamma}{6} \Big)^3
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- \frac{4}{5}\Big(\frac{\alpha-3\gamma}{6} \Big)^7
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+ O\bigg( \Big(\frac{\alpha-3\gamma}{6}\Big)^9 \bigg).
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\]
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The term with exponent 5 is absent in the series expansion; that's not a typo.
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<h3>On the choice of $\gamma$</h3>
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<p>
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We see that $\tau(\alpha,\alpha/3) = \alpha/3$, that is,
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the construction produces an <em>exact trisection</em>
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with the choice $\gamma=\alpha/3$.
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Of course, constructing such a $\gamma$
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is equivalent to solving the original trisection problem, therefore
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that is not an option. On the other hand, a constructible $\gamma$ that
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comes close to $\alpha/3$ will serve just fine. The function $\tau$ is
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not very sensitive to the variations of $\gamma$ as is evident from:
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\[
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\frac{\partial \tau(\alpha,\gamma)}{\partial \gamma}
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=
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\frac{1}{2} \bigg(
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\frac{3\cos 3x}{\sqrt{9 - \sin^2 3x}} -1 \bigg),
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\]
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where I have let $x=(\alpha-3\gamma)/6$ to simplify the notation.
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As noted above, best
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results are achieved when $\gamma$ is close to $\alpha/3$. Even
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with a not-so-optimal choice of $\gamma=\alpha/4$ we get $x=\alpha/24$.
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With such a choice, the value of partial derivative in the range
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$0 \le \alpha \le \pi/2$ does not exceed 0.01, indicating that
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the value of the function is essentially independent of $\gamma$
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on that range.
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<p>
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<b>An excellent choice</b> for $\gamma$ is obtained as follows.
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In Step 3 of the construction, first select the point $D$ on
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the line segment $A'B'$ such that $A'D = \frac{1}{3} A'B'$.
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Then draw the circle $C''$ with center $A'$ passing through $D$.
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One may verify that this results in an angle $\gamma$ given by:
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\[
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\hat{\gamma} = \frac{1}{2} \alpha
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- \arcsin\Big( \frac{1}{3} \sin\frac{\alpha}{2} \Big)
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= \frac{1}{3} \alpha + \frac{1}{2\cdot3^4} \alpha^3 + O(\alpha^7).
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\]
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Then, the constructed angle is:
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\[
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\beta = \tau(\alpha,\hat{\gamma})
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= \tau\bigg(\alpha, \frac{1}{2} \alpha
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- \arcsin\Big( \frac{1}{3} \sin\frac{\alpha}{2} \Big)
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\bigg)
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= \frac{1}{3} \alpha - \frac{1}{2^4\cdot3^{13}} \alpha^9 +
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O(\alpha^{13}).
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\]
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The construction error, $e(\alpha) = \frac{1}{3}\alpha-\beta$,
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is monotone increasing. Since $e(\alpha) = O(\alpha^9)$,
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we expect it to be very small.
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Indeed, the worst error on the interval $0 \le \alpha \le \pi/2$ is
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the incredibly small
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$e(\pi/2)$ = 0.00000226 radians = 0.00013 degrees.
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The worst error on the interval $0 \le \alpha \le \pi$ is
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$e(\pi)$ = 0.00103 radians = 0.0592 degrees.
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<p>
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Despite its extraordinary accuracy, this is <em>not</em>
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among my favorite trisection methods because the points $D$ and $E$ are
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too close to each other for locating the point $G$ reliably. For practical
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purposes, should there be such a need, I would much rather use a more
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robust, albeit less accurate, method.
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<hr width="60%">
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<p>
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<em>This applet was created by
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<a href="http://userpages.umbc.edu/~rostamia">Rouben Rostamian</a>
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using
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<a href="http://aleph0.clarku.edu/~djoyce/home.html">David Joyce</a>'s
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<a href="http://aleph0.clarkU.edu/~djoyce/java/Geometry/Geometry.html">Geometry
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Applet</a> on
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July 26, 2002.
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<br>The error analysis was thoroughly revised and extensive
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cosmetic changes were made on June 7, 2010.
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</em>
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<p>
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<table width="100%">
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<tr>
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<td valign="top">Go to <a href="index.html#trisections">list of trisections</a></td>
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<td align="right" style="width:200px;">
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<a href="http://validator.w3.org/check?uri=referer">
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<img src="/~rostamia/images/valid-html401.png" class="noborder" width="88" height="31" alt="Valid HTML"></a>
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<a href="http://jigsaw.w3.org/css-validator/check/referer">
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<img src="/~rostamia/images/valid-css.png" class="noborder" width="88" height="31" alt="Valid CSS"></a>
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</td></tr>
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</table>
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</body>
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</html>
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