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>
219 lines
8 KiB
HTML
219 lines
8 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>
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Construction attributed to C. R. Lindberg, as reported in<br>
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Free Jamison, <i>Trisection Approximation</i>, American Mathematical Monthly,
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vol. 61, no. 5, May 1954, pp. 334–336.
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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="700" height="400">
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<param name="background" value="ffffff">
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<param name="title" value="An angle trisection">
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<param name="e[1]" value="O;point;fixed;200,200">
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<param name="e[2]" value="A;point;fixed;200,350">
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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;C1,350,0;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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<param name="e[7]" value="Dtmp;point;midpoint;A,B;none;none">
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<param name="e[8]" value="D;point;cutoff;O,Dtmp,O,A">
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<param name="e[9]" value="OD;line;connect;O,D;none;none;lightGray">
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<param name="e[10]" value="C;point;extend;B,O,B,O">
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<param name="e[11]" value="OC;line;connect;O,C;none;none;lightGray">
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<param name="e[12]" value="CD;line;connect;C,D;none;none;green">
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<param name="e[13]" value="E;point;extend;C,D,B,C">
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<param name="e[14]" value="DE;line;connect;D,E;none;none;green">
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<param name="e[15]" value="OE;line;connect;O,E;none;none;red">
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<param name="e[16]" value="E';point;cutoff;O,E,O,A;none;none">
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<param name="e[17]" value="p1;point;fixed;200,225;none;none">
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<param name="e[18]" value="c1;circle;radius;O,p1;none;none;none;none">
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<param name="e[19]" value="l1;line;chord;OB,c1;none;none;none">
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<param name="e[20]" value="q1;point;first;l1;none;none">
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<param name="e[21]" value="l2;line;chord;OE,c1;none;none;none">
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<param name="e[22]" value="q2;point;first;l2;none;none">
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<param name="e[23]" value="s1;sector;sector;O,q2,q1;none;none;black;orange">
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<param name="e[24]" value="p2;point;fixed;200,225;none;none">
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<param name="e[25]" value="c2;circle;radius;O,p2;none;none;none;none">
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<param name="e[26]" value="l3;line;chord;OA,c2;none;none;none">
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<param name="e[27]" value="q3;point;first;l3;none;none">
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<param name="e[28]" value="l4;line;chord;OE,c2;none;none;none">
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<param name="e[29]" value="q4;point;first;l4;none;none">
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<param name="e[30]" value="s2;sector;sector;O,q3,q4;none;none;black;yellow">
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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 right half of the circle).<br>
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Press “r” to reset the diagram to its initial state.<br>
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The red line $OE$ 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>The construction</h2>
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<p>
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According to Jamison (see the reference at the top of this page)
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the construction's main idea
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comes from an unpublished work by C. R. Lindberg.
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<p>
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Consider the circular arc $AB$ centered at $O$, 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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Extend $BO$ to intersect the circle at a point $C$.
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<li>
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Draw the bisector $OD$ of the angle $AOB$.
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<li>
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Draw the line $CD$ and extend it to a point $E$ such that $DE$ equals the
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circle's diameter.
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</ol>
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<p>
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The line $OE$ is an approximate trisector of the angle $AOB$.
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<p>
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Here is a heuristic explanation for why the construction works,
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as explained by Jamison. The key lies
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in the observation that (i) in the triangle $ODE$ the angles $O$ and $E$ are
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“small”, and (ii) the side $ED$ is twice as long as the side $OD$.
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Therefore from the law of sines we have $\sin(O)/\sin(E) = ED/OD = 2$,
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which implies that the angle $O$ is approximately twice the angle $E$ in
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the triangle $ODE$.
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<p>
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Let the measure of the angle $OED$ be $x$.
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Then the triangle's external angle at $D$, that is the angle $ODC$, is the sum of
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the internal angles $O$ and $E$, therefore it is approximately $3x$.
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Therefore the angle $OCD$ is $3x$. Therefore the angle $BOD$ is $6x$.
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Since the angle $E'OD$ is $2x$, we conclude that the angle $BOE'$ is $4x$.
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Furthermore, since the angle $BOD$ is $6x$, the angle $BOA$ is $12x$. This shows
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that $BOA$ is 3 times $BOE'$, as asserted.
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<h2>Error Analysis</h2>
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<p>
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Let $\alpha$ and $\beta$ be the sizes of the angles $AOB$ and $EOB$, respectively.
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The angle $DOB$ is half of $AOB$ by the construction, therefore it is equal to
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$\alpha/2$. Consequently the angle $DCB$, which is half the
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central angle $DOB$, equals $\alpha/4$.
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The triangle $DOC$ is isosceles, therefore the angle $ODC$ also equals $\alpha/4$.
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<p>
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In the triangle $OED$, let $x$ and $y$ be the sizes of the angles
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$OED$ and $EOD$, respectively. Since the sum $x+y$ of the triangle's internal
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angles equals the triangle's
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external angle $ODC$, we have $x+y = \alpha/4$. Let us note, however,
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that the angle $y$ equals $DOB$ minus $EOB$. Thus $y = \alpha/2 - \beta$,
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whence $x = \beta - \alpha/4$.
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<p>
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In the triangle $OED$, the side $DE$ is twice the side $OD$ by the construction,
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therefore the law of sines gives $\sin y = 2 \sin x$. Consequently,
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$\sin(\alpha/2 - \beta) = 2 \sin(\beta - \alpha/4)$. Solving this for $\beta$
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we arrive at:
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\[
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\beta
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= \frac{1}{4} \alpha + \arctan \frac{\sin(a/4)}{2+\cos(a/4)}
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= \frac{1}{3} \alpha - \frac{1}{2^6\cdot3^4} \alpha^3 + O(\alpha^5)
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= \frac{1}{3} \alpha - \frac{1}{5184} \alpha^3 + O(\alpha^5).
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\]
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<p>
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We see that the trisection error $e(\alpha) = \alpha/3 - \beta$ is given by:
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\[
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e(\alpha) = \frac{1}{12}\alpha - \arctan \frac{\sin(a/4)}{2+\cos(a/4)}.
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\]
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(This formula is also given in Jamison's article.)
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The function $e(a)$ is monotonically increasing.
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The worst error on the interval $0 \le \alpha \le \pi/2$ is
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$e(\pi/2)$ = 0.000757 radians = 0.0434 degrees.
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The worst error on the interval $0 \le \alpha \le \pi$ is
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$e(\pi)$ = 0.0063 radians = 0.361 degrees.
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That's quite good for such a simple construction.
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<h2>An interesting coincidence</h2>
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<p>
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The angle $\beta$ constructed by this method coincides <em>exactly</em>
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with that of <a href="trisect-pllana.html">Pllana's construction</a>,
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where $\beta$ is given as:
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\[
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\beta
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= \arctan \frac
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{\sin\frac{\alpha}{2} + 2\sin\frac{\alpha}{4}}
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{\cos\frac{\alpha}{2} + 2\cos\frac{\alpha}{4}}.
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\]
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One way to verify that the seemingly different expressions
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for $\beta$ are in fact identical,
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is to compare their derivatives. In both cases we have:
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\[
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\frac{d\beta}{d\alpha} =
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\frac{3(1 + \cos\frac{\alpha}{4})}{2(5 + 4\cos\frac{\alpha}{4})}.
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\]
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<h2>An extension</h2>
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<p>
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Although this construction is quite good as is, Jamison proceeds to give
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<a href="trisect-jamison-ext.html">an extension of Lindberg's method</a>
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which requires a bit more work but is substantially more accurate.
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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 22, 2002.<br>
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Cosmetic revisions 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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</table>
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