From 35678be21369b58d975e119cca56e528d859030b Mon Sep 17 00:00:00 2001
From: Glen Whitney <glen@studioinfinity.org>
Date: Fri, 6 Oct 2023 12:21:48 -0700
Subject: [PATCH] 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.
---
 etc/options.html                              |   7 +-
 public/index.html                             |   1 +
 public/rostamian/incenter.html                | 393 +++++++++++++++++
 public/rostamian/inequality.html              | 156 +++++++
 .../inscribed-equilateral-solution.html       | 131 ++++++
 public/rostamian/inscribed-equilateral.html   | 135 ++++++
 public/rostamian/steve_gray.html              | 142 ++++++
 public/rostamian/trisect-alberts.html         | 403 ++++++++++++++++++
 public/rostamian/trisect-baker.html           | 268 ++++++++++++
 public/rostamian/trisect-cdsmith.html         | 154 +++++++
 public/rostamian/trisect-dudley.html          | 169 ++++++++
 public/rostamian/trisect-dudley2.html         | 155 +++++++
 public/rostamian/trisect-durham.html          | 267 ++++++++++++
 public/rostamian/trisect-jamison-ext.html     | 185 ++++++++
 public/rostamian/trisect-jamison.html         | 219 ++++++++++
 public/rostamian/trisect-pllana.html          | 173 ++++++++
 public/rostamian/trisect-raiford.html         | 152 +++++++
 public/rostamian/trisect-stark.html           | 399 +++++++++++++++++
 src/adapptlet.civet                           |  68 +--
 src/adapptypes.civet                          |   2 +-
 src/options.civet                             |   7 +-
 21 files changed, 3549 insertions(+), 37 deletions(-)
 create mode 100644 public/rostamian/incenter.html
 create mode 100644 public/rostamian/inequality.html
 create mode 100644 public/rostamian/inscribed-equilateral-solution.html
 create mode 100644 public/rostamian/inscribed-equilateral.html
 create mode 100644 public/rostamian/steve_gray.html
 create mode 100644 public/rostamian/trisect-alberts.html
 create mode 100644 public/rostamian/trisect-baker.html
 create mode 100644 public/rostamian/trisect-cdsmith.html
 create mode 100644 public/rostamian/trisect-dudley.html
 create mode 100644 public/rostamian/trisect-dudley2.html
 create mode 100644 public/rostamian/trisect-durham.html
 create mode 100644 public/rostamian/trisect-jamison-ext.html
 create mode 100644 public/rostamian/trisect-jamison.html
 create mode 100644 public/rostamian/trisect-pllana.html
 create mode 100644 public/rostamian/trisect-raiford.html
 create mode 100644 public/rostamian/trisect-stark.html
diff --git a/etc/options.html b/etc/options.html
index 2d982bb..45e7440 100644
--- a/etc/options.html
+++ b/etc/options.html
@@ -7,12 +7,17 @@
 
 <body>
   <h3>Debugging</h3>
-  Trace the following to the JavaScript console:
   <h4>Java Geometry Applets</h4>
+  Trace the following to the JavaScript console: <br/>
   <label for="commands">Commands executed</label>
   <input type="checkbox" id="commands">
   <label for="color">Colors assigned</label>
   <input type="checkbox" id="color">
+  <br/>
+  <br/>
+  Alter execution of the translated applet: <br/>
+  <label for="showall">Show all entities, even hidden ones</label>
+  <input type="checkbox" id="showall">
   <script src="options.js" type="module"></script>
 </body>
 
diff --git a/public/index.html b/public/index.html
index 2767047..cb5c34d 100644
--- a/public/index.html
+++ b/public/index.html
@@ -11,6 +11,7 @@
       <li> <a href="inscribed-revived.html">Revived</a> </li>
       <li> <a href="inscribed-modified.html">After</a> </li>
       <li> <a href="jelts/book1/joyceDefI2.html">Book I Def 2</a> </li>
+      <li> <a href="rostamian/">Dr. Rostamian's unconverted pages</a> </li>
     </ul>
     <h2>WRL Files</h2>
     <ul><li> <a href="wrl_2.html">Target two</a></li>
diff --git a/public/rostamian/incenter.html b/public/rostamian/incenter.html
new file mode 100644
index 0000000..cf80186
--- /dev/null
+++ b/public/rostamian/incenter.html
@@ -0,0 +1,393 @@
+<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
+<html>
+<head>
+<!-- fix buggy IE8, especially for mathjax -->
+<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7">
+
+<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
+<title>The incenter via algebra</title>
+<link rel="stylesheet" type="text/css" media="screen" href="style.css">
+
+<script type="text/javascript"
+  src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML,http://userpages.umbc.edu/~rostamia/mathjax-config.js">
+   MathJax.Hub.Queue( function() {document.body.style.visibility="visible"} );
+</script> 
+
+</head>
+<body style="visibility:hidden">
+
+<h1>A triangle's incenter via algebra</h1>
+<h2>&hellip; and extension to tetrahedra</h2>
+
+<table class="centered">
+<tr><td align="center">
+<applet code="Geometry" archive="Geometry.zip" width="400" height="300">
+<param name="background" value="ffffff">
+<param name="title" value="The incenter via algebra">
+
+<param name="e[1]" value="A;point;fixed;100,250">
+<param name="e[2]" value="B;point;fixed;300,250">
+<param name="e[3]" value="C;point;free;50,50;red;red">
+<param name="e[4]" value="ABC;polygon;triangle;A,B,C">
+<param name="e[5]" value="pt1;point;extend;A,C,C,B;none;none">
+<param name="e[6]" value="C';point;proportion;A,pt1,A,C,A,B,A,B;none;none">
+<param name="e[7]" value="pt2;point;extend;B,A,A,C;none;none">
+<param name="e[8]" value="A';point;proportion;B,pt2,B,A,B,C,B,C;none;none">
+<param name="e[9]" value="pt3;point;extend;C,B,B,A;none;none">
+<param name="e[10]" value="B';point;proportion;C,pt3,C,B,C,A,C,A;none;none">
+<param name="e[11]" value="AA';line;connect;A,A';none;none;magenta">
+<param name="e[12]" value="BB';line;connect;B,B';none;none;magenta">
+<param name="e[13]" value="CC';line;connect;C,C';none;none;magenta">
+<param name="e[14]" value="O;point;intersection;AA',BB'">
+<param name="e[15]" value="H;point;foot;O,A,B;none;none">
+<param name="e[16]" value="IC;circle;radius;O,H;none;none;black;none">
+
+</applet>
+
+</td></tr>
+<tr><td>
+<b>
+Drag $C$ to change the geometry.<br>
+Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
+The incenter lies at the intersection of angle bisectors.
+</b>
+</td></tr></table>
+
+
+<h2>Statement of the problem</h2>
+
+<p>
+It is a fact of elementary geometry that the center of
+a triangle's incircle&mdash;known as its <em>incenter</em>&mdash;lies
+at the point of intersection of the
+triangle's angle bisectors.  This leads to an elegant expression for
+the location of the incenter as a linear combination of the
+triangle's vertices, as stated in:
+
+<p>
+<b>Proposition 1:</b>
+<i>
+Let $a$, $b$, $c$ be the lengths of the
+sides opposite to the vertices $A$, $B$, $C$ of the triangle $ABC$.
+Then the incenter, $O$, is expressed as a linear combination of
+the vertices as:
+\[
+	O = \frac{a}{p}A + \frac{b}{p}B + \frac{c}{p}C,
+\]
+where $p = a + b + c$ is the triangle's perimeter.
+</i>
+
+<p>
+To make sense of this statement, you need to know something about
+the algebra of points.  The following capsule summary is all that's needed:
+
+<blockquote>
+<p>
+<b>The algebra of points:</b>
+Consider the points $A$ and $B$ and a variable $t$ that takes
+values in the range 0 to 1.  
+If $T$ is a point on the segment $AB$
+such that $AT/AB = t$, we write $T = (1-t)A + tB$.  Note that when
+$t=0$ we get $T=A$, and when $t=1$ we get $T=B$.
+As the value of $t$ ranges from 0 to 1, the point $T$ slides
+from $A$ to $B$.
+</blockquote>
+
+
+<h2>A preliminary proposition</h2>
+
+<p>
+The following elementary proposition is needed for the proof of 
+Proposition&nbsp;1.  This one is a standard result and is
+likely to be found in Euclid's <em>Elements</em> but I haven't checked.
+
+<p>
+<i>PS:</i>
+After this web page was written,
+I found out that
+Proposition&nbsp;2 appears in Euclid's <em>Elements</em> as
+<a href="http://aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI3.html">Book&nbsp;VI, Proposition&nbsp;3</a>
+with a simpler and more elegant proof!
+Nevertheless, I am retaining my original proof here as
+a not-so-elegant alternative.
+
+<p>
+<b>Proposition 2:</b>
+<i>
+Let the bisector of the angle $C$ in the triangle $ABC$ meet the side $AB$
+at point $P$.  We have:  
+\[
+\frac{PA}{PB} = \frac{b}{a},
+\]
+where $a$ and $b$ are as in Proposition&nbsp;1.
+</i>
+
+<p>
+<b>Proof:</b>
+From $P$ drop perpendicular $PM$ and $PN$ onto the sides $AC$ and $BC$;
+see the diagram below.  The right triangles $CMP$ and $CNP$ are congruent
+because they share a common hypotenuse and their angles at $C$ are equal.
+Therefore, $PM = PN$.
+
+<table class="centered">
+<tr><td align="center">
+<applet code="Geometry" archive="Geometry.zip" width="400" height="300">
+<param name="background" value="ffffff">
+<param name="title" value="The incenter via algebra">
+
+<param name="e[1]" value="A;point;fixed;50,250">
+<param name="e[2]" value="B;point;fixed;350,250">
+<param name="e[3]" value="C;point;free;100,50;red;red">
+<param name="e[4]" value="ABC;polygon;triangle;A,B,C">
+<param name="e[5]" value="pt1;point;extend;A,C,C,B;none;none">
+<param name="e[6]" value="P;point;proportion;A,pt1,A,C,A,B,A,B">
+<param name="e[7]" value="CP;line;connect;C,P;none;none;magenta">
+<param name="e[8]" value="M;point;foot;P,C,A">
+<param name="e[9]" value="N;point;foot;P,C,B">
+<param name="e[10]" value="PM;line;connect;P,M;none;none;green">
+<param name="e[11]" value="PN;line;connect;P,N;none;none;green">
+<param name="e[12]" value="H;point;foot;C,A,B">
+<param name="e[13]" value="CH;line;connect;C,H;none;none;cyan">
+
+<param name="e[14]" value="AH;line;connect;A,H;none;none">
+<param name="e[15]" value="BN;line;connect;B,N;none;none">
+<param name="e[16]" value="AM;line;connect;A,M;none;none">
+
+</applet>
+
+</td></tr>
+<tr><td>
+<b>
+Drag $C$ to change the geometry.<br>
+Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
+</b>
+</td></tr></table>
+
+<p>
+Let $H$ be the foot of the altitude dropped from $C$. Let us note 
+the area of the triangle $APC$ may be expressed in two different ways
+in terms of the altitudes $CH$ and $PM$:
+\[
+	\frac{1}{2} AP \cdot CH = \frac{1}{2} AC \cdot PM.
+\]
+Similarly, the area of the triangle $BPC$ may be expressed in two
+different way in terms of the altitudes $CH$ and $PN$:
+\[
+	\frac{1}{2} BP \cdot CH = \frac{1}{2} BC \cdot PN.
+\]
+
+Dividing these two equalities and recalling that $PM = PN$,
+we arrive at the desired assertion.&nbsp;&nbsp;&nbsp;<b>QED</b>
+
+<h2>The proof of Proposition 1</h2>
+
+<p>
+The diagram below shows the bisector $CC'$ of the angle $C$.
+It is known from elementary geometry that the incenter $O$ lies
+on the bisector.
+From Proposition 2 we have $C'A/C'B = b/a$.  It 
+follows that $AC'/AB = b/(a+b)$.  In terms of the notation of
+<em>the algebra of points</em> introduced earlier in this page, 
+this is expressed as:
+\[
+	C' = \Big(1 - \frac{b}{a+b} \Big) A + \frac{b}{a+b} B
+	   = \frac{a}{a+b} A + \frac{b}{a+b} B.
+\]
+Since $O = (1-t)C + tC'$ for some $t$, we arrive at:
+\[
+	O = (1-t)C + t\Big[ \frac{a}{a+b} A + \frac{b}{a+b} B \Big].
+\]
+This expresses the triangle's incenter $O$ in terms of its vertices,
+sides lengths, and a yet unknown quantity $t$ that takes values in the range
+0 to 1.
+
+<table class="centered">
+<tr><td align="center">
+<applet code="Geometry" archive="Geometry.zip" width="400" height="300">
+<param name="background" value="ffffff">
+<param name="title" value="The incenter via algebra">
+
+<param name="e[1]" value="A;point;fixed;100,250">
+<param name="e[2]" value="B;point;fixed;300,250">
+<param name="e[3]" value="C;point;free;50,50;red;red">
+<param name="e[4]" value="ABC;polygon;triangle;A,B,C">
+<param name="e[5]" value="pt1;point;extend;A,C,C,B;none;none">
+<param name="e[6]" value="C';point;proportion;A,pt1,A,C,A,B,A,B">
+<param name="e[7]" value="pt2;point;extend;B,A,A,C;none;none">
+<param name="e[8]" value="A';point;proportion;B,pt2,B,A,B,C,B,C;none;none">
+<param name="e[9]" value="pt3;point;extend;C,B,B,A;none;none">
+<param name="e[10]" value="B';point;proportion;C,pt3,C,B,C,A,C,A;none;none">
+<param name="e[11]" value="AA';line;connect;A,A';none;none;none">
+<param name="e[12]" value="BB';line;connect;B,B';none;none;none">
+<param name="e[13]" value="CC';line;connect;C,C';none;none;magenta">
+<param name="e[14]" value="O;point;intersection;AA',BB'">
+<param name="e[15]" value="H;point;foot;O,A,B;none;none">
+<param name="e[16]" value="IC;circle;radius;O,H;none;none;black;none">
+
+</applet>
+
+</td></tr>
+<tr><td>
+<b>
+Drag $C$ to change the geometry.<br>
+Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
+</b>
+</td></tr></table>
+
+<p>
+If we repeat the same calculation by replacing the vertex $C$ by vertex $A$,
+we arrive at:
+\[
+	O = (1-s)A + s\Big[ \frac{b}{b+c} B + \frac{c}{b+c} C \Big],
+\]
+where $s$ is yet another unknown that takes values in the range from 0 to 1.
+
+<p>
+Equating the two expressions for $O$ and collecting the terms, we arrive at:
+\[
+	\Big[\frac{ta}{a+b} + s - 1\Big] A
+	+
+	\Big[\frac{tb}{a+b} - \frac{sb}{b+c} \Big] B
+	+
+	\Big[1 - t - \frac{sc}{b+c} \Big] C = 0.
+\]
+
+In this equations, 
+if the coefficient of $C$ is nonzero, then we may solve for $C$,
+obtaining $C = \alpha A  + \beta  B$ for some $\alpha$
+and $\beta$.  But this would
+mean that $C$ lies along the line $AB$ which would mean the the
+three points $A$, $B$ and $C$ are collinear, therefore the
+triangle $ABC$ is degenerate.  We conclude that if the triangle is
+non-degenerate, then the coefficient of $C$ is zero.  Similar
+arguments show that the coefficients of $A$ and $B$ are zero.
+Thus, the following system of three equations hold:
+\[
+	\frac{ta}{a+b} + s - 1 = 0,
+	\quad
+	\frac{tb}{a+b} - \frac{sb}{b+c} = 0,
+	\quad
+	1 - t - \frac{sc}{b+c}  = 0.
+\]
+Solving this system for the unknowns $t$ and $s$ we obtain:
+\[
+	t = \frac{a+b}{a+b+c}, \quad s = \frac{b+c}{a+b+c}.
+\]
+
+Substituting these in either of the expressions for $O$ we arrive at:
+\[
+	O = \frac{a}{a+b+C}A + \frac{b}{a+b+C}B + \frac{c}{a+b+C}C,
+\]
+which is equivalent to Proposition&nbsp;1's
+assertion.&nbsp;&nbsp;&nbsp;<b>QED</b>
+
+<h2>Extension to tetrahedra</h2>
+
+<p>
+Propositions 1 extends to tetrahedra:
+
+<p>
+<b>Proposition 3</b>
+Let
+$a$, $b$, $c$, $d$
+be the areas of the faces opposite to the vertices
+$A$, $B$, $C$, $D$ of the tetrahedron $ABCD$.  The the tetrahedron's
+incenter $O$ is given by:
+\[
+	O
+	= \frac{a}{\mathcal{A}} A
+	+ \frac{b}{\mathcal{A}} B
+	+ \frac{c}{\mathcal{A}} C
+	+ \frac{d}{\mathcal{A}} D,
+\]
+where $\mathcal{A} = a + b + c + d$ is the tetrahedron's surface area.
+
+<p>
+This is proved with the aid of the following extension of Proposition&nbsp;2:
+
+<p>
+<b>Proposition 4</b>
+<i>
+Let
+$a$, $b$, $c$, $d$
+be the areas of the faces opposite to the vertices
+$A$, $B$, $C$, $D$ of the tetrahedron $ABCD$.
+Let the bisector plane of the (internal) dihedral angle of edge $CD$
+intersect edge $AB$ at point $P$.
+The we have:
+\[
+	\frac{PA}{PB} = \frac{b}{a}.
+\]
+</i>
+
+<table class="centered">
+<tr><td align="center">
+<applet code="Geometry" archive="Geometry.zip" width="400" height="300">
+<param name="background" value="ffffff">
+<param name="title" value="Triangle's incenter">
+
+<param name="e[1]" value="A;point;fixed;100,250,300">
+<param name="e[2]" value="B;point;fixed;300,250,300">
+<param name="e[3]" value="C;point;fixed;350,200,-400">
+<param name="e[4]" value="D;point;free;200,40;red;red">
+<param name="e[5]" value="ABCD;polyhedron;tetrahedron;A,B,C,D">
+<param name="e[6]" value="ABC;plane;3points;A,B,C;none;none;none;none">
+
+<!-- construct the bisector of the dihedral angle with edge DC -->
+<param name="e[7]" value="pl1;plane;perpendicular;D,C;none;none;none;none">
+<param name="e[8]" value="Apl1;point;foot;A,pl1;none;none">
+<param name="e[9]" value="Bpl1;point;foot;B,pl1;none;none">
+<param name="e[10]" value="pt1;point;angleBisector;Apl1,D,Bpl1,pl1;none;none">
+<param name="e[11]" value="tmp1;plane;3points;D,C,pt1;none;none;none;none">
+<param name="e[12]" value="P;point;intersection;A,B,tmp1">
+<param name="e[13]" value="DP;line;connect;D,P;none;none;cyan">
+<param name="e[14]" value="CP;line;connect;C,P;none;none;cyan">
+
+</applet>
+
+</td></tr>
+<tr><td>
+<b>
+Drag $D$ to change the geometry.<br>
+Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
+The plane $CDP$ bisects the dihedral angle of edge $CD$.
+</b>
+</td></tr></table>
+
+<p>
+I leave the proofs to you, the diligent reader.  You will find
+useful information in a
+<a
+href="http://groups.google.com/group/sci.math/browse_thread/thread/56c187e018a85111/436f448dfab35a83">discussion
+in the sci.math newsgroup</a> from January 17, 2006.
+
+
+
+
+
+<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 June 26, 2010.
+</em>
+<p>
+
+<table width="100%">
+<tr>
+<td valign="top">Go to <a href="index.html">Geometry Problems and Puzzles</a></td>
+<td align="right" style="width:200px;">
+<a href="http://validator.w3.org/check?uri=referer">
+<img src="/~rostamia/images/valid-html401.png" class="noborder" width="88" height="31" alt="Valid HTML"></a>
+<a href="http://jigsaw.w3.org/css-validator/check/referer">
+<img src="/~rostamia/images/valid-css.png" class="noborder" width="88" height="31" alt="Valid CSS"></a>
+</td></tr>
+</table>
+
+</body>
+</html>
+
diff --git a/public/rostamian/inequality.html b/public/rostamian/inequality.html
new file mode 100644
index 0000000..49f7824
--- /dev/null
+++ b/public/rostamian/inequality.html
@@ -0,0 +1,156 @@
+<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
+<html>
+<head>
+<!-- fix buggy IE8, especially for mathjax -->
+<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7">
+
+<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
+<title>A geometric inequality</title>
+<link rel="stylesheet" type="text/css" media="screen" href="style.css">
+
+<script type="text/javascript"
+  src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML,http://userpages.umbc.edu/~rostamia/mathjax-config.js">
+   MathJax.Hub.Queue( function() {document.body.style.visibility="visible"} );
+</script> 
+
+</head>
+<body style="visibility:hidden">
+
+<h1>A geometric inequality</h1>
+<h4>&hellip;and its solution by 
+        <a href="mailto:haoyuep@aol.com">Dan Hoey</a></h4>
+
+<table class="centered">
+<tr><td align="center">
+<applet code="Geometry" archive="Geometry.zip" width="400" height="400">
+<param name="background" value="ffffff">
+<param name="title" value="A geometric inequality">
+
+<param name="e[1]" value="A;point;fixed;100,300">
+<param name="e[2]" value="B;point;fixed;300,300">
+<param name="e[3]" value="ABC;polygon;equilateralTriangle;A,B">
+<param name="e[4]" value="C;point;vertex;ABC,3">
+
+<param name="e[5]" value="P;point;free;180,320;red;red">
+<param name="e[6]" value="PA;line;connect;P,A;none;none;green">
+<param name="e[7]" value="PB;line;connect;P,B;none;none;green">
+<param name="e[8]" value="PC;line;connect;P,C;none;none;magenta">
+
+</applet>
+
+</td></tr>
+<tr><td>
+<b>
+Drag the point $P$.<br>
+Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
+Proposition: $PC \le PA + PB$.
+</b>
+</td></tr></table>
+
+<h2>The problem</h2>
+
+<p>
+<b>Proposition:</b> <i>Let $ABC$ be an equilateral triangle and $P$
+be an arbitrary point in its plane.  Then  $PC \le PA + PB$.</i>
+
+<p>
+This was brought up in
+<a href="http://mathforum.org/kb/message.jspa?messageID=1086018">a message </a>
+on the <code>geometry.puzzles</code> newsgroup on November&nbsp;11, 2001.
+Go to that message and scroll to the bottom of the page to see the discussion
+thread.
+
+<p>
+On November&nbsp;22, 2001
+<a href="mailto:haoyuep@aol.com">Dan Hoey</a>
+offered a particularly nice solution.  He also commented that he
+had learned that problem may be related to the <em>Van Schooten Theorem</em>,
+which indeed it is.  See
+<a href="http://www.cut-the-knot.org/Curriculum/Geometry/Pompeiu.shtml">Van
+Schooten's and Pompeiu's Theorems: What are these?</a> for much detail and
+historical background.
+
+<h2>The proof</h2>
+
+<p>
+Here is Dan Hoey's proof of the proposition as stated above.
+
+<p>
+On the line segment $AP$ construct the equilateral triangle
+$APD$, as shown in the diagram below, then add the line segment $DC$.
+
+<p>
+Let us show that the triangles $APB$ and $ADC$ are congruent.  For this,
+Let us observe that the sides $AP$ and $AB$ in the triangle $APB$
+equal the sides $AD$ and $AC$ in the triangle $ADC$, by the construction.
+Moreover, the angles $BAP$ and $CAD$ are equal because each equals
+the difference of
+a 60 degree angle and the angle $DAB$.  Therefore, the triangles $APB$ and
+$ADC$ are congruent by the side-angle-side equality.
+We conclude, in particular, that $PB = DC$.
+
+<p>
+In the triangle $PDC$ we have $PC \le PD + DC$.  In this inequality
+replace $PD$ and $DC$ by their equivalents $PA$ and $PB$ to arrive at
+$PC \le PA + PB$.&nbsp;&nbsp;&nbsp;<b>QED</b>
+
+<table class="centered">
+<tr><td align="center">
+<applet code="Geometry" archive="Geometry.zip" width="400" height="400">
+<param name="background" value="ffffff">
+<param name="title" value="A geometric inequality">
+
+<param name="e[1]" value="A;point;fixed;100,300">
+<param name="e[2]" value="B;point;fixed;300,300">
+<param name="e[3]" value="ABC;polygon;equilateralTriangle;A,B">
+<param name="e[4]" value="C;point;vertex;ABC,3">
+
+<param name="e[5]" value="P;point;free;180,320;red;red">
+<param name="e[6]" value="PA;line;connect;P,A;none;none;green">
+<param name="e[7]" value="PB;line;connect;P,B;none;none;green">
+<param name="e[8]" value="PC;line;connect;P,C;none;none;magenta">
+
+<param name="e[9]" value="APD;polygon;equilateralTriangle;A,P;none;none;none">
+<param name="e[10]" value="D;point;vertex;APD,3;">
+<param name="e[11]" value="AD;line;connect;A,D;none;none;lightGray">
+<param name="e[12]" value="PD;line;connect;P,D;none;none;lightGray">
+<param name="e[13]" value="CD;line;connect;C,D;none;none;lightGray">
+
+</applet>
+
+</td></tr>
+<tr><td>
+<b>
+Drag the point $P$.<br>
+Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
+Observation: The triangles $APB$ and $ADC$ are congruent.
+</b>
+</td></tr></table>
+
+<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 November 23, 2001.<br>
+Cosmetic revisions on June 23, 2010.
+</em>
+<p>
+
+<table width="100%">
+<tr>
+<td valign="top">Go to <a href="index.html">Geometry Problems and Puzzles</a></td>
+<td align="right" style="width:200px;">
+<a href="http://validator.w3.org/check?uri=referer">
+<img src="/~rostamia/images/valid-html401.png" class="noborder" width="88" height="31" alt="Valid HTML"></a>
+<a href="http://jigsaw.w3.org/css-validator/check/referer">
+<img src="/~rostamia/images/valid-css.png" class="noborder" width="88" height="31" alt="Valid CSS"></a>
+</td></tr>
+</table>
+
+</body>
+</html>
+
diff --git a/public/rostamian/inscribed-equilateral-solution.html b/public/rostamian/inscribed-equilateral-solution.html
new file mode 100644
index 0000000..761d1c6
--- /dev/null
+++ b/public/rostamian/inscribed-equilateral-solution.html
@@ -0,0 +1,131 @@
+<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
+<html>
+<head>
+<!-- fix buggy IE8, especially for mathjax -->
+<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7">
+
+<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
+<title>An equilateral triangle inscribed in a rectangle</title>
+<link rel="stylesheet" type="text/css" media="screen" href="style.css">
+
+<script type="text/javascript"
+  src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML,http://userpages.umbc.edu/~rostamia/mathjax-config.js">
+   MathJax.Hub.Queue( function() {document.body.style.visibility="visible"} );
+</script> 
+
+</head>
+<body style="visibility:hidden">
+
+<h1>An equilateral triangle inscribed in a rectangle</h1>
+
+<p>
+The following solution to the
+<a href="http://userpages.umbc.edu/~rostamia/Geometry/inscribed-equilateral.html">inscribed triangle puzzle</a>
+is due to <b>Peter&nbsp;Renz</b>
+who communicated it to me on December&nbsp;2016.
+
+<p>
+For the sake of completeness, let's begin with the statement of the puzzle.
+The figure below depicts a equilateral triangle $AEF$ inscribed in a rectangle
+in such a way that the two share a vertex.  We wish to show that the
+area of the pink triangle ($ECF$) is the sum of the areas of the other two
+colored triangles.
+
+<div class="centered">
+	<img src="./inscribed-equilateral-solution/frame000.png" width=500
+	height=417 alt="[frame000.png]">
+</div>
+
+<p>
+As noted in the referring page (see the link above), a solution with
+the aid of trigonometry is quite straightforward.  The purpose of this
+page is to present a solution in the style of <i>Euclid</i>, without appeal
+to trigonometry.
+
+<p>
+The following animation encapsulates Peter&nbsp;Renz's solution in its
+entirety.
+
+<div class="centered">
+	<img src="./inscribed-equilateral-solution/waterwheel.gif" width=500
+	height=417 alt="[animation.gif]">
+</div>
+
+The animation should be self-explanatory if you stare at it long enough.
+Nevertheless, I will now proceed to point out the reasoning through
+several still images extracted from that animation.
+
+<ol>
+
+<li>
+Construct a circle on the diameter $FE$, and then subdivide its boundary
+into six 60-degree wedges starting at the vertex $C$.  Mark the division
+points $C$, $P$, $Q'$, $C'$, $P'$, and $Q$, as shown.
+
+<div class="centered">
+	<img src="./inscribed-equilateral-solution/frame007.png" width=500
+	height=417 alt="[frame007.png]">
+</div>
+
+<li>
+Rotate the triangle $FDA$ about $F$ by 60 degrees to bring the edge $FA$ to
+coincide with $FE$.
+
+<div class="centered">
+	<img src="./inscribed-equilateral-solution/frame073.png" width=500
+	height=417 alt="[frame073.png]">
+</div>
+
+The key observation is that the rotation moves the vertex $D$ into $P'$.
+
+<li>
+Further rotate the triangle about the circle's center by 180 degrees to
+place it in the $FPE$ position.
+
+<div class="centered">
+	<img src="./inscribed-equilateral-solution/frame111.png" width=500
+	height=417 alt="[frame111.png]">
+</div>
+
+<li>
+Rotate the triangle $EBA$ about $E$ by 60 degrees to bring the edge $EA$ to
+coincide with $EF$.  Then the vertex $B$ will move onto $Q'$ for reasons
+similar to those explained above.
+
+<div class="centered">
+	<img src="./inscribed-equilateral-solution/frame177.png" width=500
+	height=417 alt="[frame177.png]">
+</div>
+
+<li>
+The two red line segments in the figure below are parallel and of equal lengths
+by virtue of being the side and the &ldquo;radius&rdquo; of the
+regular hexagon (not shown) inscribed in the circle.
+
+<div class="centered">
+	<img src="./inscribed-equilateral-solution/frame180.png" width=500
+	height=417 alt="[frame180.png]">
+</div>
+
+The altitudes of the three triangles, dropped from the vertices $C$,
+$P$, and $Q'$ are shown in dashed lines.  It should be clear that the
+altitude dropped from $C$ is equal in length to the sum of those of
+the other two altitudes.  Since the three triangles share a common base,
+the area of one is the sum of the areas of the other two.  Q.E.D.
+
+</ol>
+
+<table width="100%">
+<tr>
+<td valign="top">Go to <a href="index.html">Geometry Problems and Puzzles</a></td>
+<td align="right" style="width:200px;">
+<a href="http://validator.w3.org/check?uri=referer">
+<img src="/~rostamia/images/valid-html401.png" class="noborder" width="88" height="31" alt="Valid HTML"></a>
+<a href="http://jigsaw.w3.org/css-validator/check/referer">
+<img src="/~rostamia/images/valid-css.png" class="noborder" width="88" height="31" alt="Valid CSS"></a>
+</td></tr>
+</table>
+
+</body>
+</html>
+
diff --git a/public/rostamian/inscribed-equilateral.html b/public/rostamian/inscribed-equilateral.html
new file mode 100644
index 0000000..440dde8
--- /dev/null
+++ b/public/rostamian/inscribed-equilateral.html
@@ -0,0 +1,135 @@
+<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
+<html>
+<head>
+<!-- fix buggy IE8, especially for mathjax -->
+<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7">
+
+<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
+<title>An equilateral triangle inscribed in a rectangle</title>
+<link rel="stylesheet" type="text/css" media="screen" href="style.css">
+
+<script type="text/javascript"
+  src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML,http://userpages.umbc.edu/~rostamia/mathjax-config.js">
+   MathJax.Hub.Queue( function() {document.body.style.visibility="visible"} );
+</script> 
+
+</head>
+<body style="visibility:hidden">
+
+<h1>An equilateral triangle inscribed in a rectangle</h1>
+
+<table class="centered">
+<tr><td align="center">
+<applet code="Geometry" archive="Geometry.zip" width="410" height="370">
+<param name="background" value="ffffff">
+<param name="title" value="An equilateral triangle inscribed in a rectangle">
+
+<!-- the moving mechanism -->
+<param name="e[1]" value="O;point;fixed;290,320;0;0;0;0">
+<param name="e[2]" value="U1;point;fixed;510,320;0;0;0;0">
+<param name="e[3]" value="V1;point;perpendicular;O,U1;0;0;0;0">
+<param name="e[4]" value="U;point;angleDivider;U1,O,V1,3;0;0;0;0">
+<param name="e[5]" value="V;point;angleDivider;V1,O,U1,3;0;0;0;0">
+<param name="e[6]" value="circ1;circle;radius;O,U;0;0;0;0">
+<param name="e[7]" value="li1;line;parallel;U,O,U1;0;0;0;0">
+<param name="e[8]" value="li2;line;parallel;V,O,V1;0;0;0;0">
+<param name="e[9]" value="W;point;intersection;li1,li2;0;0;0;0">
+<param name="e[10]" value="VW;line;connect;V,W;0;0;lightGray">
+<param name="e[11]" value="@;point;lineSegmentSlider;V,W,0,220;red;red">
+<param name="e[12]" value="li3;line;parallel;@,O,U1;0;0;0;0">
+<param name="e[13]" value="li4;line;chord;circ1,li3;0;0;0;0">
+<param name="e[14]" value="X1;point;first;li4;0;0;0;0">
+
+<!-- the triangle -->
+<param name="e[15]" value="A;point;fixed;50,320">
+<param name="e[16]" value="V2;point;perpendicular;A,U1;0;0;0;0">
+<param name="e[17]" value="li5;line;parallel;A,O,X1;0;0;0;0">
+<param name="e[18]" value="X2;point;last;li5;0;0;0;0">
+<param name="e[19]" value="X;point;extend;A,X2,A,X2;0;0;0;0">
+
+<param name="e[20]" value="tri1;polygon;equilateralTriangle;X,A;0;0;0;0">
+<param name="e[21]" value="Y;point;vertex;tri1,3;0;0;0;0">
+<param name="e[22]" value="B;point;midpoint;X,Y">
+<param name="e[23]" value="ABC;polygon;equilateralTriangle;A,B">
+<param name="e[24]" value="C;point;vertex;ABC,3">
+
+<!-- the rectangle -->
+<param name="e[25]" value="D;point;foot;B,A,U1">
+<param name="e[26]" value="F;point;foot;C,A,V2">
+<param name="e[27]" value="FE;line;parallel;F,A,D;0;0;0;0">
+<param name="e[28]" value="E;point;last;FE">
+<param name="e[29]" value="rect;polygon;quadrilateral;A,D,E,F;0;0;black;0">
+<param name="e[30]" value="ADB;polygon;triangle;A,D,B;0;0;0;pink">
+<param name="e[31]" value="ACF;polygon;triangle;A,C,F;0;0;0;pink">
+<param name="e[32]" value="BCE;polygon;triangle;B,C,E;0;0;0;cyan">
+
+</applet>
+</td></tr>
+<tr><td>
+<b>
+Slide the &ldquo;@&rdquo; up and down to change the geometry.<br>
+Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
+Proposition: The blue area equals the sum of the two pink areas.
+</b>
+</td></tr></table>
+
+<h2>Problem statement</h2>
+
+<p>
+The diagram above shows an equilateral triangle inscribed in a rectangle
+in such a way that the two have a vertex in common.  This subdivides the
+rectangle into four disjoint triangles.
+The original equilateral triangle is shown in white
+in the diagram; the other three are shown in color.
+
+<p>
+<b>Proposition</b>
+<em>
+The area of the blue triangle equals the sum
+of the areas of the two pink triangles.
+</em>
+
+<p>
+The trigonometric proof is quite straightforward. I don't
+know of a classical proof <i>a la</i> <span class="name">Euclid</span>.
+(Well, actually I haven't tried much.)
+If you can think of a neat non-trigonometric proof, let me know.  I will
+put it here with due credit.
+
+<p>
+This problem appeared as a conjecture
+<a href="http://mathforum.org/kb/thread.jspa?forumID=129&amp;messageID=1083967">in an article</a>
+in the <code>geometry.puzzles</code> newsgroup on March 15, 1997.
+
+<p>
+<b>Note added January 8, 2017:</b>
+Here is a
+<a href="inscribed-equilateral-solution.html">clever solution</a>
+that <b>Peter Renz</b> sent me a in December 2016.  Thanks, Peter!
+
+<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 2, 2010.
+</em>
+<p>
+
+<table width="100%">
+<tr>
+<td valign="top">Go to <a href="index.html">Geometry Problems and Puzzles</a></td>
+<td align="right" style="width:200px;">
+<a href="http://validator.w3.org/check?uri=referer">
+<img src="/~rostamia/images/valid-html401.png" class="noborder" width="88" height="31" alt="Valid HTML"></a>
+<a href="http://jigsaw.w3.org/css-validator/check/referer">
+<img src="/~rostamia/images/valid-css.png" class="noborder" width="88" height="31" alt="Valid CSS"></a>
+</td></tr>
+</table>
+
+</body>
+</html>
+
diff --git a/public/rostamian/steve_gray.html b/public/rostamian/steve_gray.html
new file mode 100644
index 0000000..60e3aee
--- /dev/null
+++ b/public/rostamian/steve_gray.html
@@ -0,0 +1,142 @@
+<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
+<html>
+<head>
+<!-- fix buggy IE8, especially for mathjax -->
+<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7">
+
+<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
+<title>Triangles with common base</title>
+<link rel="stylesheet" type="text/css" media="screen" href="style.css">
+
+<script type="text/javascript"
+  src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML,http://userpages.umbc.edu/~rostamia/mathjax-config.js">
+   MathJax.Hub.Queue( function() {document.body.style.visibility="visible"} );
+</script> 
+
+</head>
+<body style="visibility:hidden">
+
+<h1>Triangles with common base</h1>
+<h4>An interesting problem proposed by
+	<a href="mailto:stevebg@adelphia.net">Steve Gray</a></h4>
+
+<table class="centered">
+<tr><td align="center">
+<applet code="Geometry" archive="Geometry.zip" width="650" height="250">
+<param name="background" value="ffffff">
+<param name="title" value="Triangles with common base">
+
+<param name="e[1]"  value="A;point;fixed;50,215">
+<param name="e[2]"  value="B;point;fixed;250,215">
+<param name="e[3]"  value="T0;polygon;equilateralTriangle;A,B">
+<param name="e[4]"  value="C;point;vertex;T0,3">
+
+<param name="e[5]"  value="D;point;free;115,173;red;red">
+<param name="e[6]"  value="DC;line;connect;D,C;none;none;orange">
+<param name="e[7]"  value="DA;line;connect;D,A;none;none;blue">
+<param name="e[8]"  value="DB;line;connect;D,B;none;none;green">
+
+<param name="e[9]"  value="P;point;fixed;300,215">
+<param name="e[10]" value="Q;point;fixed;600,215">
+<param name="e[11]" value="PQ;line;connect;P,Q">
+
+<param name="e[12]" value="B';point;similar;P,Q,C,A,D">
+<param name="e[13]" value="PB';line;connect;P,B';none;none;green">
+<param name="e[14]" value="QB';line;connect;Q,B';none;none;green">
+
+<param name="e[15]" value="C';point;similar;P,Q,A,B,D">
+<param name="e[16]" value="PC';line;connect;P,C';none;none;orange">
+<param name="e[17]" value="QC';line;connect;Q,C';none;none;orange">
+
+<param name="e[18]" value="A';point;similar;P,Q,B,C,D">
+<param name="e[19]" value="PA';line;connect;P,A';none;none;blue">
+<param name="e[20]" value="QA';line;connect;Q,A';none;none;blue">
+<param name="e[21]" value="T1;polygon;triangle;B',C',A';none;none;lightGray">
+
+<!-- construct the centroid -->
+<param name="e[22]" value="midA'B';point;midpoint;A',B';none;none">
+<param name="e[23]" value="li1;line;connect;C',midA'B';none;none;none">
+<param name="e[24]" value="midB'C';point;midpoint;B',C';none;none">
+<param name="e[25]" value="li2;line;connect;A',midB'C';none;none;none">
+<param name="e[26]" value="M;point;intersection;li1,li2">
+
+</applet>
+</td></tr>
+<tr><td>
+<b>
+Drag the point $D$.<br>
+Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
+Proposition: $A'B'C'$ is equilateral and its centroid $M$ is fixed.
+</b>
+</td></tr></table>
+
+<h2>The construction</h2>
+
+<p>
+This problem was proposed by 
+<a href="mailto:stevebg@adelphia.net">Steve Gray</a>
+in the <code>geometry.puzzles</code> newsgroup
+(<a href="http://mathforum.org/kb/message.jspa?messageID=1088552">see
+the original message</a>) on July 26, 2002.
+Scroll to the bottom of that page for a link to the solution.
+
+<p>
+Consider an equilateral triangle $ABC$, a line segment $PQ$,
+and an arbitrary point $D$, as seen in the diagram above.
+On the segment $PQ$ construct three triangles
+$PC'Q$,
+$PA'Q$,
+$PB'Q$,
+similar to the triangles 
+$ADB$,
+$BDC$,
+$CDA$,
+respectively.
+
+<p>
+<b>Proposition 1:</b>
+The triangle $A'B'C'$ is equilateral.
+
+<p>
+<b>Proposition 2:</b>
+The centroid of $A'B'C'$ is independent of $D$.
+
+
+<p>
+Steve adds:
+<blockquote>
+<p>
+Now generalize this for a regular $n$-gon. The new points form a
+regular $n$-gon whose centroid is independent of $D$.
+This problem is original so far as I know. I am interested in the
+simplest synthetic solution; no algebra, please.
+</blockquote>
+
+
+<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>
+July 26, 2002.<br>
+Cosmetic revisions on June 17, 2010.
+</em>
+<p>
+
+<table width="100%">
+<tr>
+<td valign="top">Go to <a href="index.html">Geometry Problems and Puzzles</a></td>
+<td align="right" style="width:200px;">
+<a href="http://validator.w3.org/check?uri=referer">
+<img src="/~rostamia/images/valid-html401.png" class="noborder" width="88" height="31" alt="Valid HTML"></a>
+<a href="http://jigsaw.w3.org/css-validator/check/referer">
+<img src="/~rostamia/images/valid-css.png" class="noborder" width="88" height="31" alt="Valid CSS"></a>
+</td></tr>
+</table>
+
+</body>
+</html>
+
diff --git a/public/rostamian/trisect-alberts.html b/public/rostamian/trisect-alberts.html
new file mode 100644
index 0000000..fda5356
--- /dev/null
+++ b/public/rostamian/trisect-alberts.html
@@ -0,0 +1,403 @@
+<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
+<html>
+<head>
+<!-- fix buggy IE8, especially for mathjax -->
+<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7">
+
+<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
+<title>An angle trisection</title>
+<link rel="stylesheet" type="text/css" media="screen" href="style.css">
+
+<script type="text/javascript"
+  src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML,http://userpages.umbc.edu/~rostamia/mathjax-config.js">
+   MathJax.Hub.Queue( function() {document.body.style.visibility="visible"} );
+</script> 
+
+</head>
+<body style="visibility:hidden">
+
+<h1>An angle trisection</h1>
+
+<h4>Construction by Chris Alberts</h4>
+
+<p>
+The construction described in this page is due to Chris Alberts, who
+sent it to me in an email on March 15, 2011.
+I have paraphrased and rearranged his construction, but the
+differences from the original are cosmetic.  The error analysis is mine.
+
+<p>
+
+<table class="centered">
+<tr><td align="center">
+<applet code="Geometry" archive="Geometry.zip" width="810" height="600">
+<param name="background" value="ffffff">
+<param name="title" value="An angle trisection">
+<!-- <param name="align" value="left"> -->
+<param name="pivot" value="O">
+
+<!-- The angle AOB -->
+<param name="e[1]"  value="O;point;fixed;400,400">
+<param name="e[2]"  value="A;point;fixed;790,400">
+<param name="e[3]"  value="c;circle;radius;O,A;none;none;lightGray;none">
+<param name="e[4]"  value="B;point;circleSlider;c,720,0;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;530,400">
+<param name="e[8]"  value="c';circle;radius;O,A';none;none;lightGray;none">
+<param name="e[9]"  value="l1;line;chord;OB,c';none;none;none">
+<param name="e[10]"  value="B';point;first;l1">
+
+<!-- the points P, R, M -->
+<param name="e[11]"  value="E;point;midpoint;B',O">
+<param name="e[12]"  value="c'';circle;radius;B',O;none;none;00aa00;none">
+<param name="e[13]"  value="l2;line;parallel;E,A,O;none;none;none">
+<param name="e[14]"  value="l3;line;chord;l2,c'';none;none;none">
+<param name="e[15]"  value="P;point;first;l3">
+<param name="e[16]"  value="l4;line;chord;l2,c';none;none;none">
+<param name="e[17]"  value="R;point;last;l4">
+<param name="e[18]"  value="l8;line;connect;E,R">
+<param name="e[19]"  value="M;point;midpoint;P,R">
+
+<!-- the points N and F -->
+<param name="e[20]"  value="l5;line;chord;B',M,c';none;none;none">
+<param name="e[21]"  value="N;point;last;l5">
+<param name="e[22]"  value="l9;line;connect;B',N;none;none;none;none"> <!-- -->
+<param name="e[23]"  value="l6;line;parallel;B',A,O;none;none;none">
+<param name="e[24]"  value="c1;circle;radius;N,B';none;none;none;none">
+<param name="e[25]"  value="l7;line;chord;l6,c1;none;none;none">
+<param name="e[26]"  value="F;point;last;l7">
+
+<!-- the point G -->
+<param name="e[27]" value="T1;polygon;triangle;B',N,F;none;none;lightGray;ddffcc">
+<param name="e[28]"  value="l10;line;chord;N,F,c;none;none;none">
+<param name="e[29]"  value="G;point;first;l10">
+<param name="e[30]"  value="l11;line;connect;F,G;none;none;cyan">
+<param name="e[31]"  value="l12;line;chord;G,O,c';none;none;none">
+<param name="e[32]"  value="H;point;last;l12">
+<param name="e[33]"  value="l13;line;connect;G,H;none;none;lightGray">
+
+<!-- the points H, D, and J -->
+<param name="e[34]"  value="D;point;last;l1">
+<param name="e[35]"  value="l14;line;connect;O,D">
+<param name="e[36]"  value="l15;line;parallel;D,O,A;none;none;none">
+<param name="e[37]"  value="c2;circle;radius;H,D;none;none;none;none">
+<param name="e[38]"  value="l16;line;chord;l15,c2;none;none;none">
+<param name="e[39]"  value="J;point;last;l16">
+<param name="e[40]" value="T2;polygon;triangle;D,H,J;none;none;lightGray;ddffcc">
+
+<!-- the point U -->
+<param name="e[41]"  value="l17;line;chord;H,J,c;none;none;none">
+<param name="e[42]"  value="K;point;first;l17">
+<param name="e[43]"  value="l18;line;connect;J,K;none;none;cyan">
+<param name="e[44]"  value="c3;circle;radius;A,K;none;none;none;none">
+<param name="e[45]"  value="l19;line;bichord;c,c3;none;none;none">
+<param name="e[46]"  value="T;point;last;l19">
+<param name="e[47]"  value="l20;line;connect;O,T;none;none;red">
+<param name="e[48]"  value="l21;line;connect;K,T;none;none;lightGray">
+
+<!-- labels for the circles -->
+<param name="e[49]"  value="C'';point;circleSlider;c'',150,0;00aa00;none">
+<param name="e[50]"  value="C';point;fixed;255,420;lightGray;none">
+<param name="e[51]"  value="C;point;fixed;120,120;lightGray;none">
+
+<!-- angle markers -->
+<param name="e[52]"  value="po1;point;fixed;445,400;none;none">
+<param name="e[53]"  value="ci1;circle;radius;O,po1;none;none;none;none">
+<param name="e[54]"  value="li1;line;chord;O,T,ci1;none;none;none">
+<param name="e[55]"  value="po2;point;first;li1;none;none">
+<param name="e[56]"  value="li2;line;chord;O,B,ci1;none;none;none">
+<param name="e[57]"  value="po3;point;first;li2;none;none">
+<param name="e[58]"  value="se1;sector;sector;O,po1,po2;none;none;black;yellow">
+<param name="e[59]"  value="se2;sector;sector;O,po2,po3;none;none;black;orange">
+
+</applet>
+</td></tr>
+<tr><td>
+<b>
+Drag the point $B$ to change the angle $AOB$
+(but keep $AOB$ to less than 90 degrees).<br>
+Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
+The red line ($OT$) is an approximate trisector of the angle $AOB$.
+</b>
+</td></tr></table>
+
+<p>
+Consider the circular arc $AB$ on the circle $C$ centered at $O$,
+shown in the diagram above.
+Assume that the angle $AOB$ is between 0 and 90 degrees.
+To trisect $AOB$, do:
+
+<ol>
+
+<li>
+Draw the circle $C'$ centered at $O$ with a radius 1/3&nbsp;$OA$.
+Mark $A'$ and $B'$ its intersections with the line segments $OA$ and $OB$,
+respectively.
+
+<li>
+Draw the circle $C''$ (shown in green) centered at $B'$ through the point $O$.
+
+<li>
+Let $E$ be the midpoint of the line segment $OB'$.
+Draw a line through $E$ parallel to $OA$ and mark its intersections
+$P$ and $R$ with the circles $C''$ and $C'$, as shown.
+
+<li>
+Let $M$ be the midpoint of the line segment $PR$.  Draw the line $B'M$
+and extend to the intersection point $N$ with the circle $C'$.
+
+<li>
+Draw a line through $B'$ parallel to $OA$, and select the point $F$ on
+it so that $NB' = NF$.
+
+<li>
+Extend the line segment $NF$ to intersect the circle $C$ at $G$.
+
+<li>
+Draw the line $GO$ and extend it to the point of intersection $H$
+with the circle $C'$.
+<em>Note:</em> If you look at the diagram closely, you will be able to
+see that the line segments $GO$ and $GN$
+are <em>not</em> collinear.
+
+<li>
+Let $D$ be diagonally opposite the point $B'$ in the
+circle $C'$.  Draw a line through $D$ parallel to $OA$, and
+select the point $J$ on it so that $HD=HJ$.
+<em>Note:</em> Although it is impossible to discern visually,
+the line segments $OH$ and $HJ$
+are <em>not</em> collinear.
+
+<li>
+Extend the line segment $HJ$ to intersect the circle $C$ at $K$.
+
+<li>
+Reflect $K$ about the line $OA$ to get the point $T$.
+
+</ol>
+
+The line $OT$ is an approximate trisector of the angle $AOB$.
+
+<h2>Error analysis</h2>
+
+<p>
+The calculation of the coordinates of all the points that appear
+in the construction is elementary but
+the resulting expressions are massively large, therefore I will refrain
+from putting them on this web page.  (I did the calculations in <em>Maple</em>). 
+<p>
+Let $\alpha$ and $\beta(\alpha)$ be the sizes of the angles $AOB$ and $AOT$,
+respectively.
+Express the trisection error as $e(\alpha) = \frac{\alpha}{3} - \beta(\alpha)$.
+It turns out that $e(0) = e(\pi/2) = 0$.
+In the range $0$ to $\pi/2$ the error is the largest near
+1.22175&nbsp;radians = 70.0013&nbsp;degrees.  The maximum error is
+$2.32\times10^{-18}$&nbsp;radians = $1.33\times10^{-16}$&nbsp;degrees.
+
+<p>
+Expanding $\beta(\alpha)$ in power series we get:
+\[
+	\beta(\alpha)
+	= \frac{1}{3} \alpha +
+	\frac{5^9}{2^{13} \cdot 3^{40}} \alpha^{27} + O(\alpha^{29})
+	= \frac{1}{3} \alpha + 
+	\frac{1953125}{99595595440594360737792} \alpha^{27} + O(\alpha^{29}).
+\]
+
+
+<h2>Alberts' refinement</h2>
+
+<p>
+The extraordinary precision of Chris Alberts' trisection 
+is a result of the application of a refinement technique which
+I will call <em>Alberts' refinement</em>.
+The 10 steps of his trisection procedure, described above,
+consist of three distinct stages:
+<ol>
+<li><em>Stage 1,</em> corresponding to the steps 1&ndash;5 of the construction,
+produces the angle $FB'N$ which is roughly one third of the angle $AOB$.
+The trisection error at this stage may be as large as 0.7 degrees.
+
+<li><em>Stage 2,</em> corresponding to the steps 6&ndash;7 of the construction,
+produces the angle $A'OH$ which is quite close to $\frac13 \alpha$.  The worst
+error is 0.00013791&nbsp;degrees and
+occurs when $AOB$ is near 70 degrees.
+
+<li><em>Stage 3,</em> corresponding to the steps 8&ndash;10 of the construction,
+produces the angle $AOT$ which, as noted above, is within 
+$1.33\times10^{-16}$&nbsp;degrees of the exact trisection.
+</ol>
+
+Each of the stages 2 and 3 consists of one application
+of <em>Alberts' refinement</em> which may be formulated as
+a stand-alone geometric proposition illustrated
+in the diagram below.
+
+<p>
+<table class="centered">
+<tr><td align="center">
+<applet code="Geometry" archive="Geometry.zip" width="450" height="320">
+<param name="background" value="ffffff">
+<param name="title" value="Alberts' refinement">
+
+<!-- the lines L2, L1, L3 -->
+<param name="e[1]"  value="O;point;fixed;40,250">
+<param name="e[2]"  value="B';point;fixed;380,250;none;none">
+<param name="e[3]"  value="L2;line;connect;O,B';black">
+<param name="e[4]"  value="c1;circle;radius;O,B';none;none;none;none">
+<param name="e[5]"  value="A';point;circleSlider;c1,100,10;none;none">
+<param name="e[6]"  value="L1;line;connect;O,A';black">
+<param name="e[7]"  value="s1;sector;sector;O,B',A';none;none;blue;none">
+<param name="e[8]"  value="X;point;circleSlider;c1,450,110;red;red">
+<param name="e[9]"  value="L3;line;connect;O,X;none;none;blue">
+
+<!-- the points A, P, B -->
+<param name="e[10]"  value="p1;point;fixed;160,250;none;none">
+<param name="e[11]"  value="A;point;cutoff;L1,O,p1">
+<param name="e[12]"  value="c2;circle;radius;A,O;none;none;none;none">
+<param name="e[13]"  value="l1;line;chord;O,X,c2">
+<param name="e[14]"  value="P;point;last;l1">
+<param name="e[15]"  value="T1;polygon;triangle;O,A,P;none;none;black;ffddee">
+<param name="e[16]"  value="c3;circle;radius;P,O;none;none;none;none">
+<param name="e[17]"  value="l2;line;chord;O,B',c3">
+<param name="e[18]"  value="B;point;last;l2">
+<param name="e[19]"  value="T2;polygon;triangle;O,P,B;none;none;black;ddffcc">
+
+<!-- the circle of radius 3 -->
+<param name="e[20]"  value="l3;line;extend;O,A,O,A;none;none;none;none">
+<param name="e[21]"  value="l4;line;extend;l3,O,A;none;none;none;none">
+<param name="e[22]"  value="p2;point;last;l4;none;none">
+<param name="e[23]"  value="c4;circle;radius;A,O,p2;none;none;magenta;none">
+
+<!-- the point G -->
+<param name="e[24]"  value="l5;line;chord;P,B,c4;none;none;none">
+<param name="e[25]"  value="G;point;first;l5">
+<param name="e[26]"  value="l6;line;connect;B,G;none;none;cyan">
+<param name="e[27]"  value="l7;line;connect;A,G">
+
+<!-- the horizontal line -->
+<param name="e[28]"  value="l8;line;parallel;A,O,p1">
+<param name="e[29]"  value="L;point;last;l8">
+
+<!-- angle markers -->
+<param name="e[30]"  value="p3;point;fixed;80,250;none;none">
+<param name="e[31]"  value="c5;circle;radius;O,p3;none;none;none;none">
+<param name="e[32]"  value="l9;line;chord;O,P,c5;none;none;none">
+<param name="e[33]"  value="p4;point;first;l9;none;none">
+<param name="e[34]"  value="s2;sector;sector;O,p3,p4;none;none;none;blue">
+
+<param name="e[35]"  value="p5;point;cutoff;l8,O,p3;none;none">
+<param name="e[36]"  value="c6;circle;radius;A,p5;none;none;none;none">
+<param name="e[37]"  value="l10;line;chord;A,G,c6;none;none;none">
+<param name="e[38]"  value="p6;point;first;l10;none;none">
+<param name="e[39]"  value="s3;sector;sector;A,p6,p5;none;none;none;red">
+
+</applet>
+</td></tr>
+<tr><td>
+<b>
+Drag the point $X$.
+Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
+If the blue angle is an approximate trisection of the angle $AOB$,
+then the red angle is a much better trisection.<br>
+Note that the red angle hardly changes as the point $X$ varies.
+</b>
+</td></tr></table>
+
+<p>
+The lines $L_1$ and $L_2$ intersect at the point $O$. Suppose that the
+line $OX$ is a crude trisector of the angle between $L_1$ and $L_2$.
+The rest of the diagram shows a
+straightedge and compass construction
+that produces a much finer trisection.
+
+Here are the details of the construction:
+
+<ol>
+<li> Pick an arbitrary point $A$ (other than $O$) on the line $L_1$.
+<li> Locate the point $P$ on the line $OX$ so that $AO = AP$.
+<li> Locate the point $B$ on the line $L_2$ so that $PO = PB$.
+<li> Draw a circle centered at $A$ and radius equal to three
+times the length of $OA$.  (This circle is shown in magenta.)
+<li> Extend the line segment $PB$ to a intersect that circle at a point $G$.
+(This extension is shown in cyan.)
+<li> Draw a line $AL$ through $A$ and parallel to the line $L_2$.
+</ol>
+
+The angle $LAG$ is quite close to being one third of the angle $AOB$.
+To see this, let us write $\alpha$, $\beta$, and $\beta'$ for
+the measures (in the radian units)
+of the angles $AOB$, $XOB$, and $LAG$, respectively.
+Let $\beta = \frac13 \alpha + \delta$ and $\beta' = \frac13 \alpha + \delta'$.
+Thus, $\delta$ and $\delta'$ measure the trisection errors corresponding to the
+angles $\beta$ and $\beta'$.
+
+<p>
+A quite straightforward calculation, involving an application of the
+<em>law of sines</em> in the triangle $APG$ leads to the equation:
+\[
+	\delta' = \delta - \arcsin\Big(\frac13 \sin3\delta\Big).
+\]
+Expanding this into power series in $\delta$, we obtain:
+\[
+	\delta' = \frac43 \delta^3 - \frac45 \delta^7 + O(\delta^9).
+\]
+This explains the notable efficiency of the refinement.  For instance,
+if the value of $\beta$ has two significant digits after the decimal point,
+the value of $\beta'$ will have six significant digits after the decimal
+point.
+
+<p>
+<i>Remark 1:</i> Move the point $X$ in the diagram and note
+how insensitive the angle
+$LAG$ is to the choice of $X$.  This indicates that even a crude
+initial approximation produces an excellent trisection.
+
+<p>
+<i>Remark 2:</i> If you examine closely Chris Alberts' trisection
+described earlier in
+this page, you will find buried in it two instances of <em>Alberts'
+refinement</em>. 
+
+
+<h2>A final comment</h2>
+
+<p>
+Comparing the precision of the trisection described in this
+page to those of others presented on my website may not
+seem to be quite fair.  After all, <em>any</em> approximate trisection may
+be applied iteratively to refine its own result.
+Nevertheless, I am making an exception in this case because in
+Chris Alberts' trisection, the iterative refinement is an
+inherent feature of the 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
+March 23, 2011.
+</em>
+<p>
+
+<table width="100%">
+<tr>
+<td valign="top">Go to <a href="index.html#trisections">list of trisections</a></td>
+<td align="right" style="width:200px;">
+<a href="http://validator.w3.org/check?uri=referer">
+<img src="/~rostamia/images/valid-html401.png" class="noborder" width="88" height="31" alt="Valid HTML"></a>
+<a href="http://jigsaw.w3.org/css-validator/check/referer">
+<img src="/~rostamia/images/valid-css.png" class="noborder" width="88" height="31" alt="Valid CSS"></a>
+</td></tr>
+</table>
+
+</body>
+</html>
diff --git a/public/rostamian/trisect-baker.html b/public/rostamian/trisect-baker.html
new file mode 100644
index 0000000..bc7e500
--- /dev/null
+++ b/public/rostamian/trisect-baker.html
@@ -0,0 +1,268 @@
+<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
+<html>
+<head>
+<!-- fix buggy IE8, especially for mathjax -->
+<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7">
+
+<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
+<title>An angle trisection</title>
+<link rel="stylesheet" type="text/css" media="screen" href="style.css">
+
+<script type="text/javascript"
+  src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML,http://userpages.umbc.edu/~rostamia/mathjax-config.js">
+   MathJax.Hub.Queue( function() {document.body.style.visibility="visible"} );
+</script> 
+
+</head>
+<body style="visibility:hidden">
+
+<h1>An angle trisection</h1>
+
+<h4>Construction by
+  <a href="mailto:wayne.baker@ed.cna.nl.ca">Wayne Baker</a></h4>
+
+<table class="centered">
+<tr><td align="center">
+<applet code="Geometry" archive="Geometry.zip" width="570" height="360">
+<param name="background" value="ffffff">
+<param name="title" value="An angle trisection">
+
+<!-- the angle AOB -->
+<param name="e[1]" value="O;point;fixed;260,310">
+<param name="e[2]" value="A;point;fixed;520,310">
+<param name="e[3]" value="OA;line;connect;O,A;none;none;blue">
+<param name="e[4]" value="C;circle;radius;O,A;none;none;none;none">
+<param name="e[5]" value="B;point;circleSlider;C,20,0;red;red">
+<param name="e[6]" value="OB;line;connect;O,B;none;none;blue">
+<param name="e[7]" value="z1;sector;sector;O,A,B;none;none;blue;none">
+
+<!-- point P quadrisects the arc AB -->
+<param name="e[8]" value="t1;point;angleBisector;A,O,B;none;none">
+<param name="e[9]" value="t2;point;angleBisector;A,O,t1;none;none">
+<param name="e[10]" value="Ot2;line;chord;O,t2,C;none;black;none">
+<param name="e[11]" value="P;point;first;Ot2">
+<param name="e[12]" value="AP;line;connect;A,P;none;none;green">
+
+<!-- the arc A'OB' -->
+<param name="e[13]" value="A';point;fixed;455,310">
+<param name="e[14]" value="C';circle;radius;O,A';none;none;none;none">
+<param name="e[15]" value="t9;line;chord;O,B,C';none;none;none">
+<param name="e[16]" value="B';point;first;t9">
+<param name="e[17]" value="z2;sector;sector;O,A',B';none;none;orange;none">
+
+<!-- P' -->
+<param name="e[18]" value="t3;circle;radius;A',A,P;none;none;none;none">
+<param name="e[19]" value="t4;line;bichord;t3,C';none;none;none">
+<param name="e[20]" value="P';point;first;t4">
+<param name="e[21]" value="s1;line;chord;O,P',C;none;none;none">
+<param name="e[22]" value="s2;point;first;s1;none;none">
+<param name="e[23]" value="s3;line;connect;O,s2;none;none;red">
+<param name="e[24]" value="A'P';line;connect;A',P';none;none;green">
+
+<!-- angle marker -->
+<param name="e[25]" value="p1;point;fixed;290,310;none;none">
+<param name="e[26]" value="c1;circle;radius;O,p1;none;none;none;none">
+<param name="e[27]" value="l1;line;chord;OA,c1;none;none;none">
+<param name="e[28]" value="q1;point;first;l1;none;none">
+<param name="e[29]" value="l2;line;chord;s3,c1;none;none;none">
+<param name="e[30]" value="q2;point;first;l2;none;none">
+<param name="e[31]" value="S1;sector;sector;O,q1,q2;none;none;black;orange">
+<param name="e[32]" value="l3;line;chord;OB,c1;none;none;none">
+<param name="e[33]" value="q3;point;first;l3;none;none">
+<param name="e[34]" value="S2;sector;sector;O,q2,q3;none;none;black;yellow">
+
+</applet>
+</td></tr>
+<tr><td>
+<b>
+Drag the point $B$ to change the angle $AOB$.<br>
+Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
+The red line is an approximate trisector of the angle $AOB$.
+</b>
+</td></tr></table>
+
+<H2>The Basic Construction</H2>
+
+<p>
+Here is a very simple straightedge and compass
+construction of an approximate angle trisector due to
+<a href="mailto:wayne.baker@ed.cna.nl.ca">Wayne Baker</a>.
+
+<p>
+Let us represent the angle by 
+the circular arc $AB$ centered at $O$; see the diagram above.
+The angle's size may be anything from 0 to 180 degrees.
+To trisect, do:
+
+<ol>
+
+<li>
+Quadrisect the angle $AOB$, that is, divide it into four
+equal parts.  The arc $AP$ in the diagram above represents one quarter
+of the original arc $AB$.  Let $L$ be the length of the chord $AP$ (shown in green).
+
+<li>
+Draw a circular arc (shown in orange)
+centered at $O$ and radius 3/4 of $OA$.  Mark $A'$ and $B'$ its intersections
+with the rays $OA$ and $OB$, respectively.
+
+<li>
+Swing an arc (not shown) of radius $L$ centered at $A'$ and mark $P'$ its intersection with
+the arc $A'B'$, as shown.
+
+</ol>
+
+<p>
+The line $OP'$ is an approximate trisector of the angle $AOB$.
+
+<h2>Error Analysis</h2>
+
+<p>
+Let $\alpha$ and $\beta=\tau(\alpha)$ be the sizes of the angles $AOB$ and $A'OP'$,
+respectively.  It is straightforward to show that
+\[
+	\beta
+	= 2 \arcsin\big(\frac{4}{3}\sin\frac{\alpha}{8}\big)
+	= \frac{\alpha}{3} + \frac{7}{2^7\cdot3^4}\alpha^3 + O(\alpha^5)
+	= \frac{\alpha}{3} + \frac{7}{10368}\alpha^3 + O(\alpha^5).
+\]
+<!-- The first two terms of the series are the same as those
+     in trisect-dunham.html.  The third terms are different.
+     b_baker := 2*arcsin(4/3*sin(a/8));
+     series(b_baker,a);
+     b_durham := a/2 - arctan(sin(a/4 - arcsin(sin(a/4)/2))*4/3);
+     series(b_durham, a);
+-->
+
+<p>
+The error
+$
+	\ds e(\alpha) =  \tau(\alpha) - \frac{\alpha}{3}
+$
+is monotonically increasing in $\alpha$. 
+The worst error on the interval $0 \le \alpha \le \pi/2$ is
+$e(\pi/2)$ = 0.002695 radians = 0.154 degrees.
+The worst error on the interval $0 \le \alpha \le \pi$ is
+$e(\pi)$ = 0.0237 radians = 1.360 degrees.
+
+
+<h2>Iterative Improvement</h2>
+
+<p>
+As we see in the asymptotic expansion shown above, the
+angle $\tau(\alpha)$ is slightly larger than the target value of $\alpha/3$.
+Making three copies of the constructed angle, and putting them
+end-to-end as in arcs $A'P'$, $P'P''$, and $P''P'''$ shown in the diagram below,
+we arrive at the endpoint $P'''$ which is very slightly off the point $B'$,
+and just outside the arc $A'B'$.  The constructible angle $B'OP'''$ is exactly
+three times the error $e(\alpha)$.
+If we were able to trisect $B'OP'''$ exactly, then we
+would know the error, and consequently will have achieved
+the exact trisection of the original angle.
+Of course the exact trisection of $B'OP'''$ is impossible in general, but we
+may repeat the method outlined in the <em>Basic Construction</em> above
+to obtain an <em>approximate</em> trisection of $B'OP'''$,
+which will yield $ \tau\big(3\tau(\alpha) - \alpha\big) $,
+and consequently an improved trisection $\tau_{\mathrm{improved}}(\alpha)$
+of the original angle:
+\[
+   \tau_{\mathrm{improved}}(\alpha) = \tau(\alpha)  - \tau\big(3\tau(\alpha) - \alpha\big)
+	= \frac{\alpha}{3} - \frac{7^4}{2^{28}\cdot3^{13}} \alpha^9 +
+	O(\alpha^{11}).
+\]
+
+The error
+$ \ds  e_{\mathrm{improved}}(\alpha) = \frac{\alpha}{3}  - \tau_{\mathrm{improved}}(\alpha)$
+is monotonically increasing in $\alpha$.  In particular,
+$e_{\mathrm{improved}}(\pi/2) = 1.5\times 10^{-9}$ radians
+$ = 8.6\times10^{-8}$ degrees.
+
+<table class="centered">
+<tr><td align="center">
+<applet code="Geometry" archive="Geometry.zip" width="570" height="360">
+<param name=background value="ffffff">
+<param name=title value="An angle trisection">
+
+<!-- the angle AOB -->
+<param name="e[1]" value="O;point;fixed;260,310">
+<param name="e[2]" value="A;point;fixed;520,310">
+<param name="e[3]" value="OA;line;connect;O,A">
+<param name="e[4]" value="C;circle;radius;O,A;none;none;none;none">
+<param name="e[5]" value="B;point;circleSlider;C,50,0">
+<param name="e[6]" value="OB;line;connect;O,B">
+<param name="e[7]" value="z1;sector;sector;O,A,B;none;none;blue;none">
+
+<!-- point P quadrisects the arc AB -->
+<param name="e[8]" value="t1;point;angleBisector;A,O,B;none;none">
+<param name="e[9]" value="t2;point;angleBisector;A,O,t1;none;none">
+<param name="e[10]" value="Ot2;line;chord;O,t2,C;none;black;none">
+<param name="e[11]" value="P;point;first;Ot2">
+<param name="e[12]" value="OP;line;connect;O,P;none;none;none">
+
+<!-- the arc A'OB' -->
+<param name="e[13]" value="A';point;fixed;455,310">
+<param name="e[14]" value="C';circle;radius;O,A';none;none;none;none">
+<param name="e[15]" value="t9;line;chord;O,B,C';none;none;none">
+<param name="e[16]" value="B';point;first;t9">
+<param name="e[17]" value="z2;sector;sector;O,A',B';none;none;orange;none">
+
+<!-- P' -->
+<param name="e[18]" value="t3;circle;radius;A',A,P;none;none;none;none">
+<param name="e[19]" value="t4;line;bichord;t3,C';none;none;none">
+<param name="e[20]" value="P';point;first;t4">
+<param name="e[21]" value="s1;line;chord;O,P',C;none;none;none">
+<param name="e[22]" value="s2;point;first;s1;none;none">
+<param name="e[23]" value="s3;line;connect;O,s2;none;none;red">
+
+<!-- P'' -->
+<param name="e[24]" value="t5;circle;radius;P',A,P;none;none;none;none">
+<param name="e[25]" value="t6;line;bichord;t5,C';none;none;none">
+<param name="e[26]" value="P'';point;first;t6">
+
+<!-- P'''
+     Note the trailing spaces after P'''.  These become a part of the label!
+-->
+<param name="e[27]" value="t7;circle;radius;P'',A,P;none;none;none;none">
+<param name="e[28]" value="t8;line;bichord;t7,C';none;none;none">
+<param name="e[29]" value="P'''  ;point;first;t8">
+<param name="e[30]" value="OP''';line;connect;O,P'''  ;none;none;black">
+<param name="e[31]" value="u1;sector;sector;O,B',P'''  ;none;none;none;magenta">
+</applet>
+</td></tr>
+<tr><td>
+<b>
+Drag the point $B$ to change the angle $AOB$.<br>
+Press  &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
+The red line is an approximate trisector of the angle $AOB$.<br>
+The arcs $P'P''$ and $P''P'''$ are copies of $A'P'$.  The endpoint $P'''$
+is just slightly off the point $B'$.<br>
+The (very small and nearly indiscernible)
+angle $B'OP'''$ is three times the trisection error.
+</b>
+</td></tr></table>
+
+<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
+May 31, 2010.
+</em>
+<p>
+
+<table width="100%">
+<tr>
+<td valign="top">Go to <a href="index.html#trisections">list of trisections</a></td>
+<td align="right" style="width:200px;">
+<a href="http://validator.w3.org/check?uri=referer">
+<img src="/~rostamia/images/valid-html401.png" class="noborder" width="88" height="31" alt="Valid HTML"></a>
+<a href="http://jigsaw.w3.org/css-validator/check/referer">
+<img src="/~rostamia/images/valid-css.png" class="noborder" width="88" height="31" alt="Valid CSS"></a>
+</td></tr>
+</table>
+
+</body>
+</html>
diff --git a/public/rostamian/trisect-cdsmith.html b/public/rostamian/trisect-cdsmith.html
new file mode 100644
index 0000000..1860b94
--- /dev/null
+++ b/public/rostamian/trisect-cdsmith.html
@@ -0,0 +1,154 @@
+<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
+<html>
+<head>
+<!-- fix buggy IE8, especially for mathjax -->
+<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7">
+
+<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
+<title>An angle trisection</title>
+<link rel="stylesheet" type="text/css" media="screen" href="style.css">
+
+<script type="text/javascript"
+  src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML,http://userpages.umbc.edu/~rostamia/mathjax-config.js">
+   MathJax.Hub.Queue( function() {document.body.style.visibility="visible"} );
+</script> 
+
+</head>
+<body style="visibility:hidden">
+
+<h1>An angle trisection</h1>
+
+<h4>
+Construction by <a href="http://cdsmith.wordpress.com/2009/04/20/old-memories-about-trisecting-angles/">cdsmith</a>
+</h4>
+
+<table class="centered">
+<tr><td align="center">
+<applet code="Geometry" archive="Geometry.zip" width="600" height="450">
+<param name="background" value="ffffff">
+<param name="title" value="An angle trisection">
+
+<param name="e[1]"  value="O;point;fixed;250,340">
+<param name="e[2]"  value="A;point;fixed;450,340">
+<param name="e[3]"  value="cir1;circle;radius;O,A;none;none;none;none">
+<param name="e[4]"  value="B;point;circleSlider;cir1,400,0;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">
+
+<param name="e[7]"  value="AB;line;connect;A,B;none;none;lightGray">
+<param name="e[8]"  value="M;point;midpoint;A,B">
+<param name="e[9]"  value="cA;circle;radius;A,M;none;none;green;none">
+<param name="e[10]" value="cM;circle;radius;M,A;none;none;green;none">
+<param name="e[11]" value="cB;circle;radius;B,M;none;none;green;none">
+<param name="e[12]" value="li1;line;bichord;cA,cM;none;none;none">
+<param name="e[13]" value="C;point;first;li1">
+<param name="e[14]" value="li2;line;bichord;cM,cB;none;none;none">
+<param name="e[15]" value="D;point;first;li2">
+
+<param name="e[16]" value="OC;line;connect;O,C;none;none;red">
+<param name="e[17]" value="OD;line;connect;O,D;none;none;red">
+
+<!-- angle marker -->
+<param name="e[18]" value="p1;point;fixed;285,340;none;none">
+<param name="e[19]" value="c1;circle;radius;O,p1;none;none;none;none">
+<param name="e[20]" value="l1;line;chord;OA,c1;none;none;none">
+<param name="e[21]" value="q1;point;first;l1;none;none">
+<param name="e[22]" value="l2;line;chord;O,C,c1;none;none;none">
+<param name="e[23]" value="q2;point;first;l2;none;none">
+<param name="e[24]" value="s1;sector;sector;O,q1,q2;none;none;black;orange">
+<param name="e[25]" value="l3;line;chord;OB,c1;none;none;none">
+<param name="e[26]" value="q3;point;first;l3;none;none">
+<param name="e[27]" value="s2;sector;sector;O,q2,q3;none;none;black;yellow">
+
+</applet>
+</td></tr>
+<tr><td>
+<b>
+Drag the point $B$ to change the angle $AOB$.<br>
+Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
+The red lines $OC$ and $OD$ are approximate trisectors of the angle $AOB$.
+</b>
+</td></tr></table>
+
+<h2>Construction</h2>
+
+<p>
+To trisect the angle $AOB$ (with $OA=OB$), do:
+
+<ol>
+
+<li>
+Find the midpoint $M$ of the line segment $AB$.
+
+<li>
+Draw circles centered at $A$, $M$, and $B$, each of radius $\frac{1}{2}AB$,
+and mark their intersection points $C$ and $D$, as shown in the diagram
+above.
+
+</ol>
+
+The lines $OC$ and $OD$ are approximate trisectors of the angle $AOB$.
+
+<h2>Error Analysis</h2>
+
+<p>
+Let $\alpha$ and $\beta$ be the sizes of the angles $AOB$ and $AOC$,
+respectively.  One may verify that
+\[
+	\beta =
+	\arctan\bigg(
+	 \frac{
+		\sin\frac{\alpha}{2}
+		\sin\big(\frac{\pi}{6}+\frac{\alpha}{2}\big)
+	}{
+		1 +
+		\sin\frac{\alpha}{2}
+		\cos\big(\frac{\pi}{6}+\frac{\alpha}{2}\big)
+	}
+	\bigg)
+	= 
+	\frac{1}{4}\alpha + \frac{\sqrt{3}}{16} \alpha^2 - \frac{1}{16} \alpha^3
+	+ O(\alpha^4).
+\]
+
+<p>
+This says that $\ds \beta \approx \frac{1}{4}\alpha$ when $\alpha$ is small,
+so small angles are quadrisected, rather than trisected!
+(This is clearly visible in the interactive diagram above.)
+For not-so-small angles,
+the method works reasonably well.  In fact, it produces
+<em>exact trisection</em> for angles $\alpha=\pi/2$ and $\alpha=\pi$.
+
+<p>
+The worst error in
+the range $0 \le \alpha \le \pi$ is
+0.0214 radians = 1.23 degrees.
+This occurs at 
+$\alpha=2\arctan(\sqrt{3}\pm\sqrt{2})$ which corresponds to 
+angles of approximately 35 degrees and 145 degrees.
+
+<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 June 14, 2010.
+</em>
+<p>
+
+<table width="100%">
+<tr>
+<td valign="top">Go to <a href="index.html#trisections">list of trisections</a></td>
+<td align="right" style="width:200px;">
+<a href="http://validator.w3.org/check?uri=referer">
+<img src="/~rostamia/images/valid-html401.png" class="noborder" width="88" height="31" alt="Valid HTML"></a>
+<a href="http://jigsaw.w3.org/css-validator/check/referer">
+<img src="/~rostamia/images/valid-css.png" class="noborder" width="88" height="31" alt="Valid CSS"></a>
+</td></tr>
+</table>
+
+</body>
+</html>
diff --git a/public/rostamian/trisect-dudley.html b/public/rostamian/trisect-dudley.html
new file mode 100644
index 0000000..575478b
--- /dev/null
+++ b/public/rostamian/trisect-dudley.html
@@ -0,0 +1,169 @@
+<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
+<html>
+<head>
+<!-- fix buggy IE8, especially for mathjax -->
+<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7">
+
+<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
+<title>An angle trisection</title>
+<link rel="stylesheet" type="text/css" media="screen" href="style.css">
+
+<script type="text/javascript"
+  src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML,http://userpages.umbc.edu/~rostamia/mathjax-config.js">
+   MathJax.Hub.Queue( function() {document.body.style.visibility="visible"} );
+</script> 
+
+</head>
+<body style="visibility:hidden">
+
+<h1>An angle trisection</h1>
+
+<h4>From page 26 of<br>
+Underwood Dudley, <i>The Trisectors</i>, 2nd edition, 1996.
+</h4>
+
+<table class="centered">
+<tr><td align="center">
+<applet code="Geometry" archive="Geometry.zip" width="600" height="250">
+<param name="background" value="ffffff">
+<param name="title" value="An angle trisection">
+
+<param name="e[1]"  value="O;point;fixed;50,200">
+<param name="e[2]"  value="A;point;fixed;200,200">
+<param name="e[3]"  value="C;point;extend;O,A,O,A">
+<param name="e[4]"  value="D;point;extend;O,C,O,A">
+<param name="e[5]"  value="AD;line;connect;A,D;none;none;lightGray">
+<param name="e[6]"  value="OA;line;connect;O,A;none;none;blue">
+
+<param name="e[7]"  value="c0;circle;radius;O,A;none;none;none;none">
+<param name="e[8]"  value="B;point;circleSlider;c0,70,100;red;red">
+<param name="e[9]"  value="OB;line;connect;O,B;none;none;blue">
+<param name="e[10]"  value="arcAB;sector;sector;O,A,B;none;none;blue;none">
+
+<param name="e[11]" value="L;line;parallel;B,O,D;none;none;black">
+
+<param name="e[12]" value="circDE;circle;radius;C,D;none;none;none;none">
+<param name="e[13]" value="x8;line;chord;L,circDE;none;none;none">
+<param name="e[14]" value="E;point;last;x8">
+<param name="e[15]" value="CE;line;connect;C,E;none;none;lightGray">
+<param name="e[16]" value="arcDE;sector;sector;C,D,E;none;none;lightGray;none">
+
+<param name="e[17]" value="F;point;foot;E,AD">
+<param name="e[18]" value="EF;line;connect;E,F;none;none;lightGray">
+
+<param name="e[19]" value="circFT;circle;radius;O,F;none;none;none;none">
+<param name="e[20]" value="x13;line;chord;L,circFT;none;none;none">
+<param name="e[21]" value="T;point;first;x13">
+<param name="e[22]" value="OT;line;connect;O,T;none;none;red">
+
+<param name="e[23]" value="arcFT;sector;sector;O,F,T;none;none;green;none">
+
+<param name="e[24]" value="circOA;circle;radius;O,A;none;none;none">
+<param name="e[25]" value="x2;line;chord;OT,circOA;none;none;none">
+<param name="e[26]" value="P;point;first;x2;none;none;none">
+
+<param name="e[27]" value="p1;point;fixed;80,200;none;none">
+<param name="e[28]" value="c1;circle;radius;O,p1;none;none;none;none">
+<param name="e[29]" value="l1;line;chord;OA,c1;none;none;none">
+<param name="e[30]" value="q1;point;first;l1;none;none">
+<param name="e[31]" value="l2;line;chord;OT,c1;none;none;none">
+<param name="e[32]" value="q2;point;first;l2;none;none">
+<param name="e[33]" value="s1;sector;sector;O,q1,q2;none;none;black;orange">
+
+<param name="e[34]" value="p2;point;fixed;80,200;none;none">
+<param name="e[35]" value="c2;circle;radius;O,p2;none;none;none;none">
+<param name="e[36]" value="l3;line;chord;OT,c2;none;none;none">
+<param name="e[37]" value="q3;point;first;l3;none;none">
+<param name="e[38]" value="l4;line;chord;OB,c2;none;none;none">
+<param name="e[39]" value="q4;point;first;l4;none;none">
+<param name="e[40]" value="s2;sector;sector;O,q3,q4;none;none;black;yellow">
+
+</applet>
+</td></tr>
+<tr><td>
+<b>
+Drag the point $B$ to change the angle $AOB$
+(but keep it less than 90 degrees).<br>
+Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
+The red line, $OT$, is an approximate trisector of the angle $AOB$.
+</b>
+</td></tr></table>
+
+<h2>Construction</h2>
+
+<p>
+We wish to trisect the given angle $AOB$.  Assume the angle is less than
+90 degrees; see the diagram above.
+
+<ol>
+
+<li>
+Draw a line through $B$ parallel to $OA$.
+
+<li>
+Extend $OA$ and mark off $AC$ and $CD$ along it, each equal to $OA$.
+
+<li>
+Draw  the arc $DE$ with center $C$ and radius $CD$.
+
+<li>
+Drop a perpendicular from $E$ to $OD$ and let $F$ be the foot of the perpendicular.
+
+<li>
+Draw the arc $FT$ with center $O$ and radius $OF$ (shown in green).
+
+</ol>
+
+The line $OT$ is an approximate trisector of the angle $AOB$.
+
+<h2>Error Analysis</h2>
+
+<p>
+Let $\alpha$ and $\beta$ be the sizes of the angles $AOB$ and $FOT$,
+respectively.  It is straightforward to show that
+\[
+	\beta
+	= \frac{\sin \alpha}{2+\cos \alpha}
+	= \frac{\alpha}{3} - \frac{1}{2^2 \cdot 3^3 \cdot 5} \alpha^5 +
+	O(\alpha^7)
+	= \frac{\alpha}{3} - \frac{1}{540} \alpha^5 + O(\alpha^7).
+\]
+The error
+$
+	\ds e(\alpha) =  \frac{\alpha}{3} - \beta 
+$
+is monotonically increasing in $\alpha$.  
+The worst error on the interval $0 \le \alpha \le \pi/2$ is
+$e(\pi/2) =$ 0.0236 radians = 1.352 degrees.
+
+<p>
+It is interesting that the error is $O(\alpha^5)$ rather than $O(\alpha^3)$
+as one might have expected.  Despite this, the method's accuracy
+is not particularly remarkable for angles that are not very close to zero.
+
+<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>
+Cosmetic revisions on June 6, 2010.
+</em>
+<p>
+
+<table width="100%">
+<tr>
+<td valign="top">Go to <a href="index.html#trisections">list of trisections</a></td>
+<td align="right" style="width:200px;">
+<a href="http://validator.w3.org/check?uri=referer">
+<img src="/~rostamia/images/valid-html401.png" class="noborder" width="88" height="31" alt="Valid HTML"></a>
+<a href="http://jigsaw.w3.org/css-validator/check/referer">
+<img src="/~rostamia/images/valid-css.png" class="noborder" width="88" height="31" alt="Valid CSS"></a>
+</td></tr>
+</table>
+
+</body>
+</html>
diff --git a/public/rostamian/trisect-dudley2.html b/public/rostamian/trisect-dudley2.html
new file mode 100644
index 0000000..8d563e3
--- /dev/null
+++ b/public/rostamian/trisect-dudley2.html
@@ -0,0 +1,155 @@
+<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
+<html>
+<head>
+<!-- fix buggy IE8, especially for mathjax -->
+<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7">
+
+<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
+<title>An angle trisection</title>
+<link rel="stylesheet" type="text/css" media="screen" href="style.css">
+
+<script type="text/javascript"
+  src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML,http://userpages.umbc.edu/~rostamia/mathjax-config.js">
+   MathJax.Hub.Queue( function() {document.body.style.visibility="visible"} );
+</script> 
+
+</head>
+<body style="visibility:hidden">
+
+<h1>An angle trisection</h1>
+
+<h4>Construction attributed to A. G. O<br>
+From page 133 of<br>
+Underwood Dudley, <i>The Trisectors</i>, 2nd edition, 1996.
+</h4>
+
+<table class="centered">
+<tr><td align="center">
+<applet code="Geometry" archive="Geometry.zip" width="450" height="260">
+<param name="background" value="ffffff">
+<param name="title" value="An angle trisection">
+
+<param name="e[1]"  value="O;point;fixed;210,225">
+<param name="e[2]"  value="A;point;fixed;420,225">
+<param name="e[3]"  value="cir1;circle;radius;O,A;none;none;none;none">
+<param name="e[4]"  value="B;point;circleSlider;cir1,0,0;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">
+<param name="e[7]"  value="arcAB;sector;sector;O,A,B;none;none;blue;none">
+<param name="e[8]"  value="pt1;point;angleBisector;A,O,B;none;none">
+<param name="e[9]"  value="C;point;cutoff;O,pt1,O,A">
+<param name="e[10]" value="OC;line;connect;O,C;none;none;lightGray">
+
+<param name="e[11]" value="D;point;fixed;280,225">
+<param name="e[12]" value="E;point;cutoff;OB,O,D">
+<param name="e[13]" value="G;point;cutoff;OC,O,D">
+<param name="e[14]" value="F;point;extend;A,O,O,D">
+<param name="e[15]" value="OF;line;connect;O,F;none;none;lightGray">
+<param name="e[16]" value="sec1;sector;sector;O,D,E;none;none;orange">
+<param name="e[17]" value="sec2;sector;sector;O,E,F;none;none;lightGray">
+
+<param name="e[18]" value="li1;line;chord;F,G,cir1;none;none;none">
+<param name="e[19]" value="T;point;last;li1">
+<param name="e[20]" value="FT;line;connect;F,T;none;none;lightGray">
+<param name="e[21]" value="OT;line;connect;O,T;none;none;red">
+
+<!-- angle markers -->
+<param name="e[22]" value="p1;point;fixed;240,225;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;OT,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="l3;line;chord;OB,c1;none;none;none">
+<param name="e[30]" value="q3;point;first;l3;none;none">
+<param name="e[31]" value="s2;sector;sector;O,q2,q3;none;none;black;yellow">
+
+</applet>
+</td></tr>
+<tr><td>
+<b>
+Drag the point $B$ to change the angle $AOB$.<br>
+Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
+The red line, $OT$, is an approximate trisector of the angle $AOB$.
+</b>
+</td></tr></table>
+
+<h2>Construction</h2>
+
+<p>
+We wish to trisect the given angle $AOB$ represented by the circular arc
+$AB$ centered at $O$, as shown in the diagram above.
+
+<ol>
+
+<li>
+Draw the bisector $OC$ of the angle $AOB$.
+
+<li>
+Draw the circular arc $DE$ centered at $O$ so that $OD = \frac{1}{3} OA$.
+Let $G$ be where the line $OC$ intersects the arc $DE$.
+
+<li>
+Locate $F$ on the extension of $OA$ so that $OF=OD$.
+
+<li>
+Connect $FG$ and extend to the intersection point $T$ with
+the arc $AB$.
+
+</ol>
+
+The line $OT$ (shown in red) is an approximate trisector of the angle $AOB$.
+
+<h2>Error Analysis</h2>
+
+<p>
+Let $\alpha$ and $\beta$ be the sizes of the angles $AOB$ and $AOT$,
+respectively.  It is straightforward to show that
+\[
+	\beta
+	= \frac{\alpha}{4} + \arcsin\Big(
+		\frac{1}{3}\sin\frac{1}{4}\alpha \Big)
+	= \frac{1}{3}\alpha - \frac{1}{2^4\cdot3^4} \alpha^3 + O(\alpha^7)
+	= \frac{1}{3}\alpha - \frac{1}{1296} \alpha^3 + O(\alpha^7).
+\]
+The term after $\alpha^3$ is $\alpha^7$.  That's not a typo.
+
+<p>
+The error
+$
+	\ds e(\alpha) =  \frac{\alpha}{3} - \beta 
+$
+is monotonically increasing in $\alpha$.
+The worst error on the interval $0 \le \alpha \le \pi/2$ is
+$e(\pi/2) =$ 0.003 radians = 0.171 degrees.
+The worst error on the interval $0 \le \alpha \le \pi$ is
+$e(\pi)$ = 0.024 radians = 1.367 degrees.
+
+<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>
+Cosmetic revisions on June 13, 2010.
+</em>
+<p>
+
+<table width="100%">
+<tr>
+<td valign="top">Go to <a href="index.html#trisections">list of trisections</a></td>
+<td align="right" style="width:200px;">
+<a href="http://validator.w3.org/check?uri=referer">
+<img src="/~rostamia/images/valid-html401.png" class="noborder" width="88" height="31" alt="Valid HTML"></a>
+<a href="http://jigsaw.w3.org/css-validator/check/referer">
+<img src="/~rostamia/images/valid-css.png" class="noborder" width="88" height="31" alt="Valid CSS"></a>
+</td></tr>
+</table>
+
+</body>
+</html>
diff --git a/public/rostamian/trisect-durham.html b/public/rostamian/trisect-durham.html
new file mode 100644
index 0000000..27c1908
--- /dev/null
+++ b/public/rostamian/trisect-durham.html
@@ -0,0 +1,267 @@
+<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
+<html>
+<head>
+<!-- fix buggy IE8, especially for mathjax -->
+<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7">
+
+<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
+<title>An angle trisection</title>
+<link rel="stylesheet" type="text/css" media="screen" href="style.css">
+
+<script type="text/javascript"
+  src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML,http://userpages.umbc.edu/~rostamia/mathjax-config.js">
+   MathJax.Hub.Queue( function() {document.body.style.visibility="visible"} );
+</script> 
+
+</head>
+<body style="visibility:hidden">
+
+<h1>An angle trisection</h1>
+
+<h4>
+R. L. Durham, <i>A simple construction for the approximate trisection
+of an angle</i>,
+American Mathematical Monthly,
+vol.&nbsp;51, no.&nbsp;4, April 1944, pp.&nbsp;217&ndash;218.
+</h4>
+
+
+<table class="centered">
+<tr><td align="center">
+<applet code="Geometry" archive="Geometry.zip" width="450" height="350">
+<param name="background" value="ffffff">
+<param name="title" value="An angle trisection">
+
+<param name="e[1]"  value="O;point;fixed;210,300">
+<param name="e[2]"  value="A;point;fixed;420,300">
+<param name="e[3]"  value="cir1;circle;radius;O,A;none;none;none;none">
+<param name="e[4]"  value="B;point;circleSlider;cir1,0,0;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">
+<param name="e[7]"  value="arcAB;sector;sector;O,A,B;none;none;blue;none">
+<param name="e[8]"  value="pt1;point;angleBisector;A,O,B;none;none">
+<param name="e[9]"  value="C;point;cutoff;O,pt1,O,A">
+<param name="e[10]" value="OC;line;connect;O,C;none;none;lightGray">
+
+<!-- O-pt2 is 1/3 of OA -->
+<param name="e[11]" value="pt2;point;fixed;280,300;none;none">
+<param name="e[12]" value="F;point;proportion;O,A,O,pt2,C,A,C,A">
+<param name="e[13]" value="G;point;extend;A,C,F,C">
+<param name="e[14]" value="AG;line;connect;A,G;none;none;lightGray">
+<param name="e[15]" value="cir2;circle;radius;G,F;none;none;green;none">
+
+<param name="e[16]" value="li1;line;perpendicular;C,O,O,A;none;none;lightGray">
+<param name="e[17]" value="li2;line;chord;li1,cir2;none;none;none">
+<param name="e[18]" value="T;point;first;li2">;
+<param name="e[19]" value="OT;line;connect;O,T;none;none;red">;
+
+<!-- angle markers -->
+<param name="e[20]" value="p1;point;fixed;240,300;none;none">
+<param name="e[21]" value="c1;circle;radius;O,p1;none;none;none;none">
+<param name="e[22]" value="l1;line;chord;OA,c1;none;none;none">
+<param name="e[23]" value="q1;point;first;l1;none;none">
+<param name="e[24]" value="l2;line;chord;OT,c1;none;none;none">
+<param name="e[25]" value="q2;point;first;l2;none;none">
+<param name="e[26]" value="s1;sector;sector;O,q1,q2;none;none;black;orange">
+
+<param name="e[27]" value="l3;line;chord;OB,c1;none;none;none">
+<param name="e[28]" value="q3;point;first;l3;none;none">
+<param name="e[29]" value="s2;sector;sector;O,q2,q3;none;none;black;yellow">
+
+</applet>
+</td></tr>
+<tr><td>
+<b>
+Drag the point $B$ to change the angle $AOB$.<br>
+Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
+The red line, $OT$, is an approximate trisector of the angle $AOB$.
+</b>
+</td></tr></table>
+
+<h2>Construction</h2>
+
+<p>
+We wish to trisect the given angle $AOB$ represented by the circular arc
+$AB$ centered at $O$, as shown in the diagram above.
+
+<ol>
+
+<li>
+Draw the bisector $OC$ of the angle $AOB$.
+
+<li>
+Draw the chord $AC$ and trisect it at point $F$ so that $CF=\frac{1}{3}CA$.
+
+<li>
+Locate point $G$ on the extension of the chord $AC$ so that $GC=CF$.
+
+<li>
+Draw a circle (shown in green) centered at $G$ and through point $F$.
+
+<li>
+Let $T$ on the green circle be such that $TC$ is perpendicular to $OC$.
+
+</ol>
+
+The line $OT$ (shown in red) is an approximate trisector of the angle $AOB$.
+
+<p>
+<span class="name">R. L. Durham</span>
+(see the reference at the top of this page) goes one step further.
+Using the line $OT$ as a starting point, he produces a substantially 
+better approximation $OT'$.  The construction for this second stage is
+complex and not particularly pretty, so I won't go into that here.
+
+<h2>Error Analysis</h2>
+
+<p>
+Let $\alpha$ and $\beta $ be the sizes of the angles $AOB$ and $AOT$,
+respectively.  It is possible to show that
+\[
+	\beta
+	= \frac{1}{2}{\alpha}
+	-
+	\arctan\bigg(
+	\frac{4}{3}\sin\Big(
+		\frac{\alpha}{4} -
+		\arcsin\big(\frac{1}{2}\sin\frac{\alpha}{4}\big)
+	\Big)\bigg)
+	= \frac{1}{3}\alpha + \frac{7}{2^7\cdot3^4} \alpha^3 + O(\alpha^5)
+	= \frac{1}{3}\alpha + \frac{7}{10368} \alpha^3 + O(\alpha^5).
+\]
+<!-- The first two terms of the series are the same as those
+     in trisect-baker.html.  The third terms are different.
+     b_baker := 2*arcsin(4/3*sin(a/8));
+     series(b_baker,a);
+     b_durham := a/2 - arctan(sin(a/4 - arcsin(sin(a/4)/2))*4/3);
+     series(b_durham, a);
+-->
+
+
+<p>
+The error $\ds e(\alpha) =  \beta - \frac{\alpha}{3}$
+is monotonically increasing in $\alpha$.
+The worst error on the interval $0 \le \alpha \le \pi/2$ is
+$e(\pi/2) =$ 0.00265 radians = 0.152 degrees.
+The worst error on the interval $0 \le \alpha \le \pi$ is
+$e(\pi)$ = 0.0218 radians = 1.252 degrees.
+
+<h2>Calculation details</h2>
+
+<p>
+The derivation of the formula for $\beta$ shown above is a straightforward
+application of trigonometry.  Here are the details.
+
+<p>
+The angles $AOC$ and $ACT$ subtend the arc $AB$ of the circle centered
+at $O$.  Since $AOC = \alpha/2$, then $ACT=\alpha/4$.  Consequently,
+the angle $GCT$ is $\pi - \alpha/4$.
+
+<p>
+We apply the law of sines in the triangle $GCT$.  (The edge $GT$
+is not shown in the diagram to reduce clutter.)  Let us write
+$\theta$ for the angle $CGT$.  The the angle $CTG$ is
+$\alpha/4 - \theta$.  The law of sines is:
+\[
+	\frac{\sin\theta}{CT} = \frac{\sin(\alpha/4-\theta)}{CG} = \frac{\sin(\pi-\alpha/4)}{GT}.
+\]
+But $GT=2CG$, therefore $2\sin(\alpha/4-\theta) = \sin(\pi-\alpha/4) = \sin(\alpha/4)$, whence
+\[
+	\theta = \frac{\alpha}{4} - \arcsin\big(\frac{1}{2}\sin\frac{\alpha}{4}\big).
+\]
+Going back to the equation of law of since, we compute $CT$:
+\[
+	CT = \frac{\sin\theta}{\sin(\alpha/4)} GT.
+\]
+But $GT = GF = 2CF = \frac{2}{3} AC
+= \frac{2}{3} \big(2OA\sin\frac{\alpha}{4}\big) = \frac{4}{3} OA\sin\frac{\alpha}{4}$.
+We conclude that $CT = \frac{4}{3} OA\sin\theta$.
+Then $COT = \arctan \frac{CT}{OC} = \arctan \frac{4}{3} \sin\theta$.  Finally, the angle
+$\beta$, which equals $COA - COT$, is given by:
+\[
+	\beta = \frac{\alpha}{2} - \arctan \frac{4}{3} \sin\theta.
+\]
+Substituting the expression for $\theta$ calculated above, we arrive at the desired
+expression for $\beta$.
+
+<h2>Aside</h2>
+
+<p>
+In 1990, the well-known logician
+<span class="name">Willard Van Orman Quine</span>
+wrote an expository article in 
+the <em>Mathematics Magazine</em> where he proves that
+some angles cannot be trisected by ruler and compass.  The proof
+is elementary (but not short) and requires
+nothing but simple algebra.  Here is the full reference:
+
+<h4>
+W. V. Quine, <i>Elementary proof that some angles cannot be trisected by ruler
+and compass</i>,
+Mathematics Magazine,
+vol.&nbsp;63, no.&nbsp;2, April 1990, pp.&nbsp;95&ndash;105.
+</h4>
+
+<p>
+In a prefatory note he refers to <span class="name">Durham</span>, the author of the trisection
+described in this web page.  He writes:
+
+<blockquote>
+<p>
+This purely expository paper dates from April 1946.
+<span class="name">Robert Lee Durham</span>,
+president emeritus of Southern Seminary Junior and College,
+had sent me a hundred dollars and asked me to make it clear to him
+why an angle cannot in general be trisected by ruler
+and compass. He had himself presented a way of almost trisecting 
+any angle by ruler and compass, to an accuracy for acute angles of
+1/720 of a degree. [Here he refers to <span class="name">Durham</span>'s
+1944 article cited at the top of this web page.]
+
+<p>
+I welcomed the money and the occasion to familiarize myself
+with the famous proof. I was guided in large part by
+<span class="name">L. E. Dickson</span>,
+<em>Why it is impossible trisect to an
+angle or construct a regular polygon of 7 or 9 sides by ruler and compass,</em>
+Mathematics Teacher, vol.&nbsp;14 (1921), 217&ndash;223.
+
+<p>
+<span class="name">Mr. Durham</span>
+expressed satisfaction with my report and proposed
+paying for publishing it as a pamphlet.  With his approval I
+submitted instead to a mathematics journal.  After  waiting
+nineteen months for a decision from the journal, I recalled
+the paper and dropped the matter.
+</blockquote>
+
+<p>
+<span class="name">Quine</span>
+goes on to explain how this article was eventually published
+more than 50 years after it was written.
+
+<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 June 13, 2010.
+</em>
+<p>
+
+<table width="100%">
+<tr>
+<td valign="top">Go to <a href="index.html#trisections">list of trisections</a></td>
+<td align="right" style="width:200px;">
+<a href="http://validator.w3.org/check?uri=referer">
+<img src="/~rostamia/images/valid-html401.png" class="noborder" width="88" height="31" alt="Valid HTML"></a>
+<a href="http://jigsaw.w3.org/css-validator/check/referer">
+<img src="/~rostamia/images/valid-css.png" class="noborder" width="88" height="31" alt="Valid CSS"></a>
+</td></tr>
+</table>
+
+</body>
+</html>
diff --git a/public/rostamian/trisect-jamison-ext.html b/public/rostamian/trisect-jamison-ext.html
new file mode 100644
index 0000000..f3030b4
--- /dev/null
+++ b/public/rostamian/trisect-jamison-ext.html
@@ -0,0 +1,185 @@
+<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
+<html>
+<head>
+<!-- fix buggy IE8, especially for mathjax -->
+<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7">
+
+<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
+<title>An angle trisection</title>
+<link rel="stylesheet" type="text/css" media="screen" href="style.css">
+
+<script type="text/javascript"
+  src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML,http://userpages.umbc.edu/~rostamia/mathjax-config.js">
+   MathJax.Hub.Queue( function() {document.body.style.visibility="visible"} );
+</script> 
+
+</head>
+<body style="visibility:hidden">
+
+<h1>An angle trisection</h1>
+
+<h4>
+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">
+
+<param name="e[1]" value="O;point;fixed;200,200">
+<param name="e[2]" value="A;point;fixed;200,350">
+<param name="e[3]" value="C1;circle;radius;O,A;none;none;lightGray;none">
+<param name="e[4]" value="B;point;circleSlider;C1,280,0;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">
+
+<!-- the points F, D, C -->
+<param name="e[7]" value="x1;point;angleBisector;A,O,B;none;none">
+<param name="e[8]" value="x2;point;angleBisector;x1,O,B;none;none">
+<param name="e[9]" value="x3;point;angleBisector;x1,O,x2;none;none">
+<param name="e[10]" value="F;point;cutoff;O,x2,O,A">
+<param name="e[11]" value="D;point;cutoff;O,x3,O,A">
+<param name="e[12]" value="C;point;extend;F,O,F,O">
+
+<!-- the lines FC, CE, OE -->
+<param name="e[13]" value="FC;line;connect;F,C;none;none;lightGray">
+<param name="e[14]" value="CD;line;connect;C,D;none;none;green">
+<param name="e[15]" value="E;point;extend;C,D,C,F">
+<param name="e[16]" value="DE;line;connect;D,E;none;none;green">
+<param name="e[17]" value="OE;line;connect;O,E;none;none;red">
+
+<param name="e[18]" value="p1;point;fixed;225,200;none;none">
+<param name="e[19]" value="c1;circle;radius;O,p1;none;none;none;none">
+<param name="e[20]" value="l1;line;chord;OA,c1;none;none;none">
+<param name="e[21]" value="q1;point;first;l1;none;none">
+<param name="e[22]" value="l2;line;chord;OE,c1;none;none;none">
+<param name="e[23]" value="q2;point;first;l2;none;none">
+<param name="e[24]" value="s1;sector;sector;O,q1,q2;none;none;black;yellow">
+
+<param name="e[25]" value="p2;point;fixed;225,200;none;none">
+<param name="e[26]" value="c2;circle;radius;O,p2;none;none;none;none">
+<param name="e[27]" value="l3;line;chord;OE,c2;none;none;none">
+<param name="e[28]" value="q3;point;first;l3;none;none">
+<param name="e[29]" value="l4;line;chord;OB,c2;none;none;none">
+<param name="e[30]" value="q4;point;first;l4;none;none">
+<param name="e[31]" value="s2;sector;sector;O,q3,q4;none;none;black;orange">
+
+<!-- needed for the error analysis, not the construction -->
+<param name="e[32]" value="OD;line;connect;O,D;none;none;lightGray">
+
+</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>
+This construction, due to Free Jamison
+(see the reference at the top of this page)
+is a more accurate variant of the construction described in
+<a href="trisect-jamison.html">a simpler construction</a>.
+
+<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> Pick the points $F$ and $D$ on the arc $BA$ such that 
+arc $BF$ = 2/8 of the arc $BA$ and
+arc $BD$ = 3/8 of the arc $BA$.
+
+<li> Extend $FO$ to intersect the circle at a point $C$.
+
+<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 $AOB$. 
+
+
+<h2>Error Analysis</h2>
+
+<p>
+
+<p>
+Let $\alpha$ and $\beta$ be the sizes of the angles $AOB$ and $EOB$, respectively.
+The angle $FOD$ equals $\alpha/8$ by the construction, therefore the
+angle $FCD$, which is half the central angle $FOD$, is equal to
+$\alpha/16$.
+The triangle $DOC$ is isosceles, therefore the angle $ODC$ also equals $\alpha/16$.
+
+<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 $x+y = \alpha/16$.  Let us note, however,
+that the angle $y$ equals $DOB$ minus $EOB$.  Thus $y = 3\alpha/8 - \beta$,
+whence $x = \beta - 5\alpha/16$.
+
+<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,
+$\sin(3\alpha/8 - \beta) = 2 \sin(\beta - 5\alpha/16)$.
+Solving this for $\beta$ we arrive at:
+\[
+	\beta
+	= \frac{5}{16} \alpha + \arctan \frac{\sin(a/16)}{2+\cos(a/16)}
+	= \frac{1}{3} \alpha - \frac{1}{2^{12}\cdot3^4} \alpha^3 + O(\alpha^5)
+	= \frac{1}{3} \alpha - \frac{1}{331776} \alpha^3 + O(\alpha^5).
+\]
+
+<p>
+We see that the trisection error $e(\alpha) = \alpha/3 - \beta$ is given by:
+\[
+   e(\alpha) = \frac{1}{48}\alpha - \arctan \frac{\sin(a/16)}{2+\cos(a/16)}.
+\]
+(This formula is also given in Jamison's article.)
+The function $e(a)$ is monotonically increasing in $\alpha$.
+The worst error on the interval $0 \le \alpha \le \pi/2$ is
+$e(\pi/2)$ = 0.0000117 radians = 0.00067 degrees.
+The worst error on the interval $0 \le \alpha \le \pi$ is
+$e(\pi)$ = 0.000093756 radians = 0.00537 degrees.
+Quite impressive!
+
+<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>
+
+<table width="100%">
+<tr>
+<td valign="top">Go to <a href="index.html#trisections">list of trisections</a></td>
+<td align="right" style="width:200px;">
+<a href="http://validator.w3.org/check?uri=referer">
+<img src="/~rostamia/images/valid-html401.png" class="noborder" width="88" height="31" alt="Valid HTML"></a>
+<a href="http://jigsaw.w3.org/css-validator/check/referer">
+<img src="/~rostamia/images/valid-css.png" class="noborder" width="88" height="31" alt="Valid CSS"></a>
+</td></tr>
+</table>
+
+</body>
+</html>
diff --git a/public/rostamian/trisect-jamison.html b/public/rostamian/trisect-jamison.html
new file mode 100644
index 0000000..d63b275
--- /dev/null
+++ b/public/rostamian/trisect-jamison.html
@@ -0,0 +1,219 @@
+<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
+<html>
+<head>
+<!-- fix buggy IE8, especially for mathjax -->
+<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7">
+
+<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
+<title>An angle trisection</title>
+<link rel="stylesheet" type="text/css" media="screen" href="style.css">
+
+<script type="text/javascript"
+  src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML,http://userpages.umbc.edu/~rostamia/mathjax-config.js">
+   MathJax.Hub.Queue( function() {document.body.style.visibility="visible"} );
+</script> 
+
+</head>
+<body style="visibility:hidden">
+
+<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">
+
+<param name="e[1]" value="O;point;fixed;200,200">
+<param name="e[2]" value="A;point;fixed;200,350">
+<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">
+<param name="e[7]" value="Dtmp;point;midpoint;A,B;none;none">
+<param name="e[8]" value="D;point;cutoff;O,Dtmp,O,A">
+<param name="e[9]" value="OD;line;connect;O,D;none;none;lightGray">
+<param name="e[10]" value="C;point;extend;B,O,B,O">
+<param name="e[11]" value="OC;line;connect;O,C;none;none;lightGray">
+<param name="e[12]" value="CD;line;connect;C,D;none;none;green">
+<param name="e[13]" value="E;point;extend;C,D,B,C">
+<param name="e[14]" value="DE;line;connect;D,E;none;none;green">
+<param name="e[15]" value="OE;line;connect;O,E;none;none;red">
+<param name="e[16]" value="E';point;cutoff;O,E,O,A;none;none">
+
+<param name="e[17]" value="p1;point;fixed;200,225;none;none">
+<param name="e[18]" value="c1;circle;radius;O,p1;none;none;none;none">
+<param name="e[19]" value="l1;line;chord;OB,c1;none;none;none">
+<param name="e[20]" value="q1;point;first;l1;none;none">
+<param name="e[21]" value="l2;line;chord;OE,c1;none;none;none">
+<param name="e[22]" value="q2;point;first;l2;none;none">
+<param name="e[23]" value="s1;sector;sector;O,q2,q1;none;none;black;orange">
+
+<param name="e[24]" value="p2;point;fixed;200,225;none;none">
+<param name="e[25]" value="c2;circle;radius;O,p2;none;none;none;none">
+<param name="e[26]" value="l3;line;chord;OA,c2;none;none;none">
+<param name="e[27]" value="q3;point;first;l3;none;none">
+<param name="e[28]" value="l4;line;chord;OE,c2;none;none;none">
+<param name="e[29]" value="q4;point;first;l4;none;none">
+<param name="e[30]" value="s2;sector;sector;O,q3,q4;none;none;black;yellow">
+
+</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>
+
+<table width="100%">
+<tr>
+<td valign="top">Go to <a href="index.html#trisections">list of trisections</a></td>
+<td align="right" style="width:200px;">
+<a href="http://validator.w3.org/check?uri=referer">
+<img src="/~rostamia/images/valid-html401.png" class="noborder" width="88" height="31" alt="Valid HTML"></a>
+<a href="http://jigsaw.w3.org/css-validator/check/referer">
+<img src="/~rostamia/images/valid-css.png" class="noborder" width="88" height="31" alt="Valid CSS"></a>
+</td></tr>
+</table>
+
+</body>
+</html>
diff --git a/public/rostamian/trisect-pllana.html b/public/rostamian/trisect-pllana.html
new file mode 100644
index 0000000..496b03a
--- /dev/null
+++ b/public/rostamian/trisect-pllana.html
@@ -0,0 +1,173 @@
+<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
+<html>
+<head>
+<!-- fix buggy IE8, especially for mathjax -->
+<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7">
+
+<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
+<title>An angle trisection</title>
+<link rel="stylesheet" type="text/css" media="screen" href="style.css">
+
+<script type="text/javascript"
+  src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML,http://userpages.umbc.edu/~rostamia/mathjax-config.js">
+   MathJax.Hub.Queue( function() {document.body.style.visibility="visible"} );
+</script> 
+
+</head>
+<body style="visibility:hidden">
+
+<h1>An angle trisection</h1>
+
+<h4>Construction due to
+<a href="mailto:avniu66@hotmail.com">Avni Pllana</a></h4>
+
+<table class="centered">
+<tr><td align="center">
+<applet code="Geometry" archive="Geometry.zip" width="450" height="260">
+<param name="background" value="ffffff">
+<param name="title" value="An angle trisection">
+
+<param name="e[1]"  value="O;point;fixed;210,225">
+<param name="e[2]"  value="A;point;fixed;410,225">
+<param name="e[3]"  value="cir1;circle;radius;O,A;none;none;none;none">
+<param name="e[4]"  value="B;point;circleSlider;cir1,0,0;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">
+<param name="e[7]"  value="arcAB;sector;sector;O,A,B;none;none;blue;none">
+<param name="e[8]"  value="pt1;point;angleBisector;A,O,B;none;none">
+<param name="e[9]"  value="C;point;cutoff;O,pt1,O,A">
+<param name="e[10]" value="li1;line;connect;O,C;none;none;lightGray">
+<param name="e[11]" value="M;point;midpoint;O,C">
+<param name="e[12]" value="pt3;point;angleBisector;A,O,C;none;none">
+<param name="e[13]" value="D;point;cutoff;O,pt3,O,A">
+<param name="e[14]" value="li2;line;connect;O,D;none;none;lightGray">
+<param name="e[15]" value="N;point;midpoint;M,D">
+<param name="e[16]" value="MD;line;connect;M,D;none;none;cyan">
+<param name="e[17]" value="li3;line;cutoff;O,N,O,A;none;none;red">
+
+<!-- angle marker -->
+<param name="e[18]" value="p1;point;fixed;240,225;none;none">
+<param name="e[19]" value="c1;circle;radius;O,p1;none;none;none;none">
+<param name="e[20]" value="l1;line;chord;OA,c1;none;none;none">
+<param name="e[21]" value="q1;point;first;l1;none;none">
+<param name="e[22]" value="l2;line;chord;O,N,c1;none;none;none">
+<param name="e[23]" value="q2;point;first;l2;none;none">
+<param name="e[24]" value="l3;line;chord;OB,c1;none;none;none">
+<param name="e[25]" value="q3;point;first;l3;none;none">
+<param name="e[26]" value="s1;sector;sector;O,q1,q2;none;none;black;orange">
+<param name="e[27]" value="s2;sector;sector;O,q2,q3;none;none;black;yellow">
+
+</applet>
+</td></tr>
+<tr><td>
+<b>
+Drag the point $B$ to change the angle $AOB$.<br>
+Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
+The red line is an approximate trisector of the angle $AOB$.
+</b>
+</td></tr></table>
+
+<h2>Construction</h2>
+
+<p>
+This approximate trisection, due to Avni Pllana, was announced
+<a href="http://mathforum.org/kb/message.jspa?messageID=1084688">in a
+message</a>
+in the <code>geometry.puzzles</code> newsgroup on July&nbsp;23, 2003.
+Scroll to the bottom of that page to view the related discussion thread.
+
+<p>
+Consider the angle $AOB$ given by the circular arc $AB$ centered at $O$,
+as shown in the diagram above. 
+
+<ol>
+
+<li>
+Pick points $C$ and $D$ on the arc $AB$ so that $OC$ bisects the angle $AOB$
+and $OD$ bisects the angle $AOC$.
+
+<li>
+Let $M$ be the midpoint of the line segment $OC$.
+
+<li>
+Let $N$ be the midpoint of the line segment $MD$.
+
+</ol>
+
+The line $ON$ is an approximate trisector of the angle $AOB$.
+
+<h2>Error Analysis</h2>
+
+<p>
+Let $\alpha$ and $\beta$ be the sizes of the angles $AOB$ and $AON$,
+respectively.  One may verify that:
+\[
+	\beta 
+	= \arctan \frac
+		{\sin\frac{\alpha}{2} + 2\sin\frac{\alpha}{4}}
+		{\cos\frac{\alpha}{2} + 2\cos\frac{\alpha}{4}}
+	= \frac{1}{3}\alpha - \frac{1}{2^6\cdot3^4} \alpha^3 + O(\alpha^5)
+	= \frac{1}{3}\alpha - \frac{1}{5184} \alpha^3 + O(\alpha^5).
+\]
+
+<em>Hint:</em> Represent the points as complex numbers
+in the polar form $re^{i\theta}$.
+
+<p>
+The error 
+$
+	\ds e(\alpha) =  \frac{\alpha}{3} - \beta
+$
+increases monotonically with $\alpha$.
+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
+$e(\pi)$ = 0.00630 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-jamison.html">Lindberg's construction</a>,
+where $\beta$ is given as:
+\[
+	\beta
+	= \frac{1}{4} \alpha + \arctan
+	\frac{\sin\frac{\alpha}{4}}{2+\cos\frac{\alpha}{4}}.
+\]
+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:
+\[
+	\frac{d\beta}{d\alpha} =
+	\frac{3(1 + \cos\frac{\alpha}{4})}{2(5 + 4\cos\frac{\alpha}{4})}.
+\]
+
+
+<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 June 10, 2010.
+</em>
+<p>
+
+<table width="100%">
+<tr>
+<td valign="top">Go to <a href="index.html#trisections">list of trisections</a></td>
+<td align="right" style="width:200px;">
+<a href="http://validator.w3.org/check?uri=referer">
+<img src="/~rostamia/images/valid-html401.png" class="noborder" width="88" height="31" alt="Valid HTML"></a>
+<a href="http://jigsaw.w3.org/css-validator/check/referer">
+<img src="/~rostamia/images/valid-css.png" class="noborder" width="88" height="31" alt="Valid CSS"></a>
+</td></tr>
+</table>
+
+</body>
+</html>
diff --git a/public/rostamian/trisect-raiford.html b/public/rostamian/trisect-raiford.html
new file mode 100644
index 0000000..1a4536c
--- /dev/null
+++ b/public/rostamian/trisect-raiford.html
@@ -0,0 +1,152 @@
+<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
+<html>
+<head>
+<!-- fix buggy IE8, especially for mathjax -->
+<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7">
+
+<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
+<title>An angle trisection</title>
+<link rel="stylesheet" type="text/css" media="screen" href="style.css">
+
+<script type="text/javascript"
+  src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML,http://userpages.umbc.edu/~rostamia/mathjax-config.js">
+   MathJax.Hub.Queue( function() {document.body.style.visibility="visible"} );
+</script> 
+
+</head>
+<body style="visibility:hidden">
+
+<h1>An angle trisection</h1>
+
+<h4>
+William R. Raiford, <i>An approximate trisection</i>,
+American Mathematical Monthly,
+vol.&nbsp;68, no.&nbsp;9, Nov 1961, p.&nbsp;917.
+</h4>
+
+<table class="centered">
+<tr><td align="center">
+<applet code="Geometry" archive="Geometry.zip" width="450" height="400">
+<param name="background" value="ffffff">
+<param name="title" value="An angle trisection">
+
+<param name="e[1]"  value="O;point;fixed;210,365">
+<param name="e[2]"  value="A;point;fixed;410,365">
+<param name="e[3]"  value="pt0;point;fixed;410,0;none;none">
+<param name="e[4]"  value="li0;line;connect;A,pt0;none;none;green">
+<param name="e[5]"  value="cir1;circle;radius;O,A;none;none;none;none">
+<param name="e[6]"  value="B;point;circleSlider;cir1,0,300;red;red">
+<param name="e[7]"  value="OA;line;connect;O,A;none;none;blue">
+<param name="e[8]"  value="OB;line;connect;O,B;none;none;blue">
+<param name="e[9]"  value="arcAB;sector;sector;O,A,B;none;none;blue;none">
+<param name="e[10]"  value="pt1;point;angleBisector;A,O,B;none;none">
+<param name="e[11]"  value="C;point;cutoff;O,pt1,O,A">
+<param name="e[12]" value="OC;line;connect;O,C;none;none;lightGray">
+<param name="e[13]" value="li1;line;connect;B,C;none;none;lightGray">
+<param name="e[14]" value="li2;line;extend;B,C,B,C;none;none;lightGray">
+<param name="e[15]" value="T;point;intersection;li0,li2">
+<param name="e[16]" value="OT;line;connect;O,T;none;none;red">
+
+<!-- angle marker -->
+<param name="e[17]" value="p1;point;fixed;240,385;none;none">
+<param name="e[18]" value="c1;circle;radius;O,p1;none;none;none;none">
+<param name="e[19]" value="l1;line;chord;OA,c1;none;none;none">
+<param name="e[20]" value="q1;point;first;l1;none;none">
+<param name="e[21]" value="l2;line;chord;O,T,c1;none;none;none">
+<param name="e[22]" value="q2;point;first;l2;none;none">
+<param name="e[23]" value="s1;sector;sector;O,q1,q2;none;none;black;orange">
+
+<!-- angle marker -->
+<param name="e[24]" value="l3;line;chord;OB,c1;none;none;none">
+<param name="e[25]" value="q3;point;first;l3;none;none">
+<param name="e[26]" value="s2;sector;sector;O,q2,q3;none;none;black;yellow">
+
+</applet>
+</td></tr>
+<tr><td>
+<b>
+Drag the point $B$ to change the angle $AOB$.<br>
+Press &ldquo;r&rdquo; to reset the diagram to its initial state.<br>
+The red line $OT$ is an approximate trisector of the angle $AOB$.
+</b>
+</td></tr></table>
+
+<h2>Construction</h2>
+
+<p>
+The construction described in the article cited at the top of the page,
+is quite straightforward.  Consider the angle $AOB$ represented by the
+circular arc $AB$ centered at $O$, as shown in the diagram above.
+To trisect $AOB$ do:
+
+<ol>
+
+<li>
+Erect a perpendicular to $OA$ at $A$ (shown in green).
+
+<li>
+Construct the bisector $OC$ of the angle $AOB$.
+
+<li>
+Connect $B$ to $C$ and extend to intersect the green line at a point $T$.
+
+</ol>
+
+The line $OT$ is an approximate trisector of the angle $AOB$.
+
+<h2>Error Analysis</h2>
+
+<p>
+Let $\alpha$ and $\beta$ be the sizes of the angles $AOB$ and $AOT$,
+respectively.  One may verify that
+\[
+	\beta 
+	= \arctan \Big( \sin\alpha - (1 - \cos\alpha)
+		\cot \big( \frac{3}{4}\alpha \big) \Big)
+	= \frac{1}{3}\alpha + \frac{1}{2^3\cdot3^4} \alpha^3 + O(\alpha^5)
+	= \frac{1}{3}\alpha + \frac{1}{648} \alpha^3 + O(\alpha^5).
+\]
+
+<p>
+The error 
+$
+	\ds e(\alpha) = \beta - \frac{\alpha}{3}
+$
+is monotonically increasing in $\alpha$.
+The worst error on the interval $0 \le \alpha \le \pi/2$ is
+$e(\pi/2)$ = 0.0063 radians = 0.361 degrees.
+The worst error on the interval $0 \le \alpha \le \pi$ is
+$e(\pi)$ = 0.06 radians = 3.435 degrees.
+
+<p>
+<span class="name">Raiford</span>, whose affiliation is given as IBM,
+states that he has calculated
+the error in increments of one degree in an IBM&nbsp;709.  Computers
+were novelties when that article was published.
+
+
+<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 June 14, 2010.
+</em>
+<p>
+
+<table width="100%">
+<tr>
+<td valign="top">Go to <a href="index.html#trisections">list of trisections</a></td>
+<td align="right" style="width:200px;">
+<a href="http://validator.w3.org/check?uri=referer">
+<img src="/~rostamia/images/valid-html401.png" class="noborder" width="88" height="31" alt="Valid HTML"></a>
+<a href="http://jigsaw.w3.org/css-validator/check/referer">
+<img src="/~rostamia/images/valid-css.png" class="noborder" width="88" height="31" alt="Valid CSS"></a>
+</td></tr>
+</table>
+
+</body>
+</html>
diff --git a/public/rostamian/trisect-stark.html b/public/rostamian/trisect-stark.html
new file mode 100644
index 0000000..1dce5a9
--- /dev/null
+++ b/public/rostamian/trisect-stark.html
@@ -0,0 +1,399 @@
+<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
+<html>
+<head>
+<!-- fix buggy IE8, especially for mathjax -->
+<meta http-equiv="X-UA-Compatible" content="IE=EmulateIE7">
+
+<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
+<title>An angle trisection</title>
+<link rel="stylesheet" type="text/css" media="screen" href="style.css">
+
+<script type="text/javascript"
+  src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML,http://userpages.umbc.edu/~rostamia/mathjax-config.js">
+   MathJax.Hub.Queue( function() {document.body.style.visibility="visible"} );
+</script> 
+
+</head>
+<body style="visibility:hidden">
+
+<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 &ldquo;$E$&rdquo; to change the radius of the circle
+centered at $A'$.  Note how little $G$ is affected by the choice of $E$.<br>
+Press &ldquo;r&rdquo; 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&nbsp;$OA$.
+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 &ldquo;r&rdquo; 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 &ldquo;r&rdquo; 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&nbsp;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>
+
+<table width="100%">
+<tr>
+<td valign="top">Go to <a href="index.html#trisections">list of trisections</a></td>
+<td align="right" style="width:200px;">
+<a href="http://validator.w3.org/check?uri=referer">
+<img src="/~rostamia/images/valid-html401.png" class="noborder" width="88" height="31" alt="Valid HTML"></a>
+<a href="http://jigsaw.w3.org/css-validator/check/referer">
+<img src="/~rostamia/images/valid-css.png" class="noborder" width="88" height="31" alt="Valid CSS"></a>
+</td></tr>
+</table>
+
+</body>
+</html>
diff --git a/src/adapptlet.civet b/src/adapptlet.civet
index 78146eb..ca478f2 100644
--- a/src/adapptlet.civet
+++ b/src/adapptlet.civet
@@ -303,6 +303,7 @@ function jToG(
    cmdr
 
 function invisible(cname: string): boolean
+   if adapParams.config?.showall then return false
    cname is '0' or cname is 'none'
 
 function joyce2rgb(cname: string, backgroundRGB?: RGB): RGB
@@ -378,16 +379,6 @@ classHandler: Record<JoyceClass, ClassHandler> :=
       aux := name + 'aUx'
       parts[0].push name
       switch method
-         /free|fixed/
-            commands.push `${name} = (${args.scalar?.join ','})`
-            if method is 'fixed'
-               callbacks.push (api: AppletObject) => api.setFixed name, true
-         'perpendicular'
-            // Note only the two-point option implemented so far
-            unless args.subpoints return
-            [center, direction] := args.subpoints
-            // Note clockwise 90° rotation (3π/2) confirmed in Joyce source
-            commands.push `${name} = Rotate(${direction}, 3*pi/2, ${center})`
          'angleDivider'
             // Note doesn't yet handle plane argument
             unless args.subpoints return
@@ -406,6 +397,27 @@ classHandler: Record<JoyceClass, ClassHandler> :=
                `${aux}4 = Rotate(${start}, ${aux}3/${n}, ${center})`
                `${name} = Intersect(${destination}, Ray(${center}, ${aux}4))`
             auxiliaries.push ...[2..4].map (i) => `${aux}${i}`
+         'extend'
+            unless args.subpoints then return
+            sp := args.subpoints
+            direction .= `UnitVector(Vector(${sp[0]},${sp[1]}))`
+            if args.line and (
+                  not args.point or args.point[0] !== args.subpoints[0])
+               direction = `UnitVector(${args.line[0]})`
+            displacement := `Distance(${sp[2]}, ${sp[3]})*${direction}`
+            commands.push `${name} = Translate(${sp[1]}, ${displacement})`
+         'first'
+            unless args.subpoints then return
+            commands.push `${name} = ${args.subpoints[0]}`
+         /fixed|free/
+            commands.push `${name} = (${args.scalar?.join ','})`
+            if method is 'fixed'
+               callbacks.push (api: AppletObject) => api.setFixed name, true
+         'foot'
+            pt := args.subpoints
+            unless pt then return
+            commands.push
+               `${name} = ClosestPoint(Line(${pt[1]},${pt[2]}), ${pt[0]})`
          'intersection'
             // Checking Joyce source, means intersection of lines, not
             // intersection of line segments
@@ -413,6 +425,9 @@ classHandler: Record<JoyceClass, ClassHandler> :=
             l1 := `Line(${args.subpoints[0]},${args.subpoints[1]})`
             l2 := `Line(${args.subpoints[2]},${args.subpoints[3]})`
             commands.push `${name} = Intersect(${l1},${l2})`
+         'last'
+            unless args.subpoints then return
+            commands.push `${name} = ${args.subpoints.at(-1)}`
          'lineSegmentSlider'
             segment .= args.line?[0]
             unless segment
@@ -424,35 +439,28 @@ classHandler: Record<JoyceClass, ClassHandler> :=
             if args.scalar and args.scalar.length
                callbacks.push (api: AppletObject) =>
                   api.setCoords name, ...args.scalar as XYZ
-         'first'
-            unless args.subpoints then return
-            commands.push `${name} = ${args.subpoints[0]}`
-         'last'
-            unless args.subpoints then return
-            commands.push `${name} = ${args.subpoints.at(-1)}`
-         'extend'
-            unless args.subpoints then return
-            sp := args.subpoints
-            direction .= `UnitVector(Vector(${sp[0]},${sp[1]}))`
-            if args.line and (
-                  not args.point or args.point[0] !== args.subpoints[0])
-               direction = `UnitVector(${args.line[0]})`
-            displacement := `Distance(${sp[2]}, ${sp[3]})*${direction}`
-            commands.push `${name} = Translate(${sp[1]}, ${displacement})`
-         'vertex'
-            commands.push
-               `${name} = Vertex(${args.polygon?[0]},${args.scalar?[0]})`
          'midpoint'
             if args.line
                commands.push `${name} = Midpoint(${args.line[0]})`
             else
                commands.push
                   `${name} = Midpoint(${args.point?[0]},${args.point?[1]})`
-         'foot'
+         'perpendicular'
+            // Note only the two-point option implemented so far
+            unless args.subpoints return
+            [center, direction] := args.subpoints
+            // Note clockwise 90° rotation (3π/2) confirmed in Joyce source
+            commands.push `${name} = Rotate(${direction}, 3*pi/2, ${center})`
+         'proportion'
             pt := args.subpoints
             unless pt then return
+            len := `Distance(${pt[2]},${pt[3]})*Distance(${pt[4]},${pt[5]})`
+               + `/ Distance(${pt[0]},${pt[1]})`
+            direction := `UnitVector(Vector(${pt[6]}, ${pt[7]}))`
+            commands.push `${name} = Translate(${pt[6]}, ${len}*${direction})`
+         'vertex'
             commands.push
-               `${name} = ClosestPoint(Line(${pt[1]},${pt[2]}), ${pt[0]})`
+               `${name} = Vertex(${args.polygon?[0]},${args.scalar?[0]})`
 
    line: (name, method, args) =>
       return := freshCommander()
diff --git a/src/adapptypes.civet b/src/adapptypes.civet
index bfabb13..af3e20c 100644
--- a/src/adapptypes.civet
+++ b/src/adapptypes.civet
@@ -1,4 +1,4 @@
-export const flags = ['commands', 'color'] as const
+export const flags = ['color', 'commands', 'showall'] as const
 export type FlagType = (typeof flags)[number]
 export type ConfigType = Partial<Record<FlagType, boolean>>
 
diff --git a/src/options.civet b/src/options.civet
index 3353449..dcf6809 100644
--- a/src/options.civet
+++ b/src/options.civet
@@ -1,11 +1,8 @@
 import {flags} from ./adapptypes.ts
 
-console.log('arrived')
-boxes := ['commands', 'color']
-cache := await browser.storage.local.get boxes
-console.log('Found', cache)
+cache := await browser.storage.local.get flags
 
-for each box of boxes
+for each box of flags
    checkbox := document.getElementById(box) as HTMLInputElement
    unless checkbox then continue
    checkbox.checked = cache[box] ?? false
