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>
268 lines
9.3 KiB
HTML
268 lines
9.3 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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R. L. Durham, <i>A simple construction for the approximate trisection
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of an angle</i>,
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American Mathematical Monthly,
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vol. 51, no. 4, April 1944, pp. 217–218.
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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="450" height="350">
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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;210,300">
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<param name="e[2]" value="A;point;fixed;420,300">
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<param name="e[3]" value="cir1;circle;radius;O,A;none;none;none;none">
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<param name="e[4]" value="B;point;circleSlider;cir1,0,0;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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<param name="e[7]" value="arcAB;sector;sector;O,A,B;none;none;blue;none">
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<param name="e[8]" value="pt1;point;angleBisector;A,O,B;none;none">
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<param name="e[9]" value="C;point;cutoff;O,pt1,O,A">
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<param name="e[10]" value="OC;line;connect;O,C;none;none;lightGray">
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<!-- O-pt2 is 1/3 of OA -->
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<param name="e[11]" value="pt2;point;fixed;280,300;none;none">
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<param name="e[12]" value="F;point;proportion;O,A,O,pt2,C,A,C,A">
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<param name="e[13]" value="G;point;extend;A,C,F,C">
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<param name="e[14]" value="AG;line;connect;A,G;none;none;lightGray">
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<param name="e[15]" value="cir2;circle;radius;G,F;none;none;green;none">
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<param name="e[16]" value="li1;line;perpendicular;C,O,O,A;none;none;lightGray">
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<param name="e[17]" value="li2;line;chord;li1,cir2;none;none;none">
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<param name="e[18]" value="T;point;first;li2">;
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<param name="e[19]" value="OT;line;connect;O,T;none;none;red">;
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<!-- angle markers -->
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<param name="e[20]" value="p1;point;fixed;240,300;none;none">
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<param name="e[21]" value="c1;circle;radius;O,p1;none;none;none;none">
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<param name="e[22]" value="l1;line;chord;OA,c1;none;none;none">
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<param name="e[23]" value="q1;point;first;l1;none;none">
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<param name="e[24]" value="l2;line;chord;OT,c1;none;none;none">
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<param name="e[25]" value="q2;point;first;l2;none;none">
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<param name="e[26]" value="s1;sector;sector;O,q1,q2;none;none;black;orange">
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<param name="e[27]" value="l3;line;chord;OB,c1;none;none;none">
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<param name="e[28]" value="q3;point;first;l3;none;none">
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<param name="e[29]" value="s2;sector;sector;O,q2,q3;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 $AOB$.<br>
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Press “r” to reset the diagram to its initial state.<br>
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The red line, $OT$, 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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We wish to trisect the given angle $AOB$ represented by the circular arc
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$AB$ centered at $O$, as shown in the diagram above.
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<ol>
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<li>
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Draw the bisector $OC$ of the angle $AOB$.
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<li>
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Draw the chord $AC$ and trisect it at point $F$ so that $CF=\frac{1}{3}CA$.
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<li>
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Locate point $G$ on the extension of the chord $AC$ so that $GC=CF$.
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<li>
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Draw a circle (shown in green) centered at $G$ and through point $F$.
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<li>
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Let $T$ on the green circle be such that $TC$ is perpendicular to $OC$.
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</ol>
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The line $OT$ (shown in red) is an approximate trisector of the angle $AOB$.
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<p>
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<span class="name">R. L. Durham</span>
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(see the reference at the top of this page) goes one step further.
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Using the line $OT$ as a starting point, he produces a substantially
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better approximation $OT'$. The construction for this second stage is
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complex and not particularly pretty, so I won't go into that here.
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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 $AOT$,
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respectively. It is possible to show that
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\[
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\beta
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= \frac{1}{2}{\alpha}
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-
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\arctan\bigg(
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\frac{4}{3}\sin\Big(
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\frac{\alpha}{4} -
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\arcsin\big(\frac{1}{2}\sin\frac{\alpha}{4}\big)
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\Big)\bigg)
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= \frac{1}{3}\alpha + \frac{7}{2^7\cdot3^4} \alpha^3 + O(\alpha^5)
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= \frac{1}{3}\alpha + \frac{7}{10368} \alpha^3 + O(\alpha^5).
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\]
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<!-- The first two terms of the series are the same as those
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in trisect-baker.html. The third terms are different.
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b_baker := 2*arcsin(4/3*sin(a/8));
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series(b_baker,a);
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b_durham := a/2 - arctan(sin(a/4 - arcsin(sin(a/4)/2))*4/3);
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series(b_durham, a);
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-->
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<p>
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The error $\ds e(\alpha) = \beta - \frac{\alpha}{3}$
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is monotonically increasing in $\alpha$.
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The worst error on the interval $0 \le \alpha \le \pi/2$ is
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$e(\pi/2) =$ 0.00265 radians = 0.152 degrees.
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The worst error on the interval $0 \le \alpha \le \pi$ is
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$e(\pi)$ = 0.0218 radians = 1.252 degrees.
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<h2>Calculation details</h2>
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<p>
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The derivation of the formula for $\beta$ shown above is a straightforward
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application of trigonometry. Here are the details.
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<p>
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The angles $AOC$ and $ACT$ subtend the arc $AB$ of the circle centered
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at $O$. Since $AOC = \alpha/2$, then $ACT=\alpha/4$. Consequently,
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the angle $GCT$ is $\pi - \alpha/4$.
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<p>
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We apply the law of sines in the triangle $GCT$. (The edge $GT$
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is not shown in the diagram to reduce clutter.) Let us write
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$\theta$ for the angle $CGT$. The the angle $CTG$ is
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$\alpha/4 - \theta$. The law of sines is:
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\[
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\frac{\sin\theta}{CT} = \frac{\sin(\alpha/4-\theta)}{CG} = \frac{\sin(\pi-\alpha/4)}{GT}.
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\]
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But $GT=2CG$, therefore $2\sin(\alpha/4-\theta) = \sin(\pi-\alpha/4) = \sin(\alpha/4)$, whence
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\[
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\theta = \frac{\alpha}{4} - \arcsin\big(\frac{1}{2}\sin\frac{\alpha}{4}\big).
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\]
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Going back to the equation of law of since, we compute $CT$:
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\[
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CT = \frac{\sin\theta}{\sin(\alpha/4)} GT.
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\]
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But $GT = GF = 2CF = \frac{2}{3} AC
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= \frac{2}{3} \big(2OA\sin\frac{\alpha}{4}\big) = \frac{4}{3} OA\sin\frac{\alpha}{4}$.
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We conclude that $CT = \frac{4}{3} OA\sin\theta$.
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Then $COT = \arctan \frac{CT}{OC} = \arctan \frac{4}{3} \sin\theta$. Finally, the angle
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$\beta$, which equals $COA - COT$, is given by:
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\[
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\beta = \frac{\alpha}{2} - \arctan \frac{4}{3} \sin\theta.
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\]
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Substituting the expression for $\theta$ calculated above, we arrive at the desired
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expression for $\beta$.
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<h2>Aside</h2>
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<p>
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In 1990, the well-known logician
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<span class="name">Willard Van Orman Quine</span>
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wrote an expository article in
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the <em>Mathematics Magazine</em> where he proves that
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some angles cannot be trisected by ruler and compass. The proof
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is elementary (but not short) and requires
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nothing but simple algebra. Here is the full reference:
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<h4>
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W. V. Quine, <i>Elementary proof that some angles cannot be trisected by ruler
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and compass</i>,
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Mathematics Magazine,
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vol. 63, no. 2, April 1990, pp. 95–105.
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</h4>
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<p>
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In a prefatory note he refers to <span class="name">Durham</span>, the author of the trisection
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described in this web page. He writes:
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<blockquote>
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<p>
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This purely expository paper dates from April 1946.
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<span class="name">Robert Lee Durham</span>,
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president emeritus of Southern Seminary Junior and College,
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had sent me a hundred dollars and asked me to make it clear to him
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why an angle cannot in general be trisected by ruler
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and compass. He had himself presented a way of almost trisecting
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any angle by ruler and compass, to an accuracy for acute angles of
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1/720 of a degree. [Here he refers to <span class="name">Durham</span>'s
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1944 article cited at the top of this web page.]
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<p>
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I welcomed the money and the occasion to familiarize myself
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with the famous proof. I was guided in large part by
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<span class="name">L. E. Dickson</span>,
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<em>Why it is impossible trisect to an
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angle or construct a regular polygon of 7 or 9 sides by ruler and compass,</em>
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Mathematics Teacher, vol. 14 (1921), 217–223.
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<p>
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<span class="name">Mr. Durham</span>
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expressed satisfaction with my report and proposed
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paying for publishing it as a pamphlet. With his approval I
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submitted instead to a mathematics journal. After waiting
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nineteen months for a decision from the journal, I recalled
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the paper and dropped the matter.
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</blockquote>
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<p>
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<span class="name">Quine</span>
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goes on to explain how this article was eventually published
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more than 50 years after it was written.
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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>
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on June 13, 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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