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WMLized and corrected some of the files.

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t2/MathVentures/bug_square.html

-<HTML>
-<TITLE>Math-Ventures: Bugs in a square</TITLE>
-<BODY BGCOLOR="#FFFFFF">
-
-<CENTER><H2>Bugs in a square</H2></CENTER>
-
-I first encountered this problem in the science journal of a laboratory
-building I used to study physics in. It is rather well known, and I found
-some other solutions to it. One used differential equations which is rather
-high stuff. The other is "intuitive", and so lacks enough support, so it
-didn't satisfy me either. Anyway, I devised my own solution beforehand.<P>
-
-Here's the riddle:<P>
-
-<TABLE BORDER=1>
-<TR><TD>
-
-Four bugs are standing at the corners of a square. At time 0 a bell rings and
-they start moving towards each other. Each bug advances in the direction of the
-bug which was initially at the position clockwise from it.<P>
-
-Assuming the square is 1 meter long, and the bugs are moving in the velocity of
-1 meter per second, how will it take for the bugs to meet in the center of the
-square?
-
-</TR></TD>
-</TABLE>
-
-The solution can be found some space below:
-<BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR>
-<BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR>
-<BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR>
-<BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR>
-
-<H3>Solution:</H3>
-
-
-
-<table border="0"> 
-<tr>
-<td valign="bottom" align="left">
-I'll use the following model to simplify the problem: suppose the bugs don't
-move towards each other infinitesimly, but rather move in a straight line, and
-then stop, and then move again toward their new positions, and so forth. The bugs
-move toward their initial position until they reach a position which is in proportion
-p to the initial distance between them. Then they move again, towards a position which
-is in proportion p to their distance, and so forth.<P>
-
-Thus they form an infinite series of squares inside each other. You can see an
-illustration of this scheme to the right of this text for the proportion p=0.15.<P>
-</td>
-<td valign="bottom">
-<img src="bugs.gif" alt="Bugs Path Illustration" border="0">
-</tr>
-</table>
-
-<br><br>
-
-Now, if the length of a given square is a, then the length of the square inside it is
-(according to pythgoras theorem): square-root((p*a)^2+((1-p)*a)^2) =
-a*square-root(2*p^2-2*p+1). The lengthes of the squares form a decreasing geometrical
-series, with that proportion. Thus the length of the path a bug travel until they meet is:<P>
-
-p * a1 / [ 1 - square-root(2*p^2-2*p+1) ]<P>
-
-This is according to the formula that the sum of an infinite decresing geometric series is
-a1 / (1-q) where a1 is the value of its first item and q is the proportion between
-two consecutive items.<P>
-
-Now, to find the length an infinitesimal bug will travel, we just limit p to 0:<P>
-
-<PRE>
-
-                                                       _____________
-          p*a1                       p*a1           1+V 2*p^2-2*p+1
-  lim ---------------   =  lim  ----------------  * -----------------  =
-  p->0   ____________      p->0    ____________        _____________
-      1-V 2*p^2-2*p+1           1-V 2*p^2-2*p+1     1+V 2*p^2-2*p+1
-
-
-                ____________                    _____________
-       a1*p*(1+V 2*p^2-2*p+1)          a1 * (1+V 2*p^2-2*p+1 )
-  lim  ---------------------  =  lim   ----------------------  = 
-  p->0                           p->0      2 - 2*p
-       1 - 2*p^2 + 2*p - 1
-
-               _____________
-       a1 * (1+V 2*0-2*0+1  )      a1 * (1+1)  
-  lim  ---------------------  =  ------------- = a1
-  p->0     2 - 2*0                     2
-
-
-
-</PRE>
-
-Therefore, the length of a bug's path is equal to the length of the original
-square's side. The time it will take a 1 meter per second fast bugs who stand at
-the corner of a 1*1 meter square to meet is 1/1 = 1 second.<P>
-
-You can find <A HREF="bugs.scm">here</A> a <A HREF="http://www.gimp.org/">Gimp</A> script
-that generates a series of squares inside squares, and optionally marks the path of
-a single bug.<P>
-
-<HR>
-
-<A HREF="./"><H3>Back to the Math-Ventures Web page</H3></A><P>
-<A HREF="../"><H3>Back to Shlomi Fish' Homepage</H3></A>
-
-
-</BODY>
-</HTML>

t2/MathVentures/bug_square.html.wml

+#include '../template.wml'
+
+<subject "Bugs in a Square" />
+
+<p>
+I first encountered this problem in the science journal of a laboratory
+building I used to study physics in. It is rather well known, and I found
+some other solutions to it. One used differential equations which is rather
+high stuff. The other is "intuitive", and so lacks enough support, so it
+didn't satisfy me either. Anyway, I devised my own solution beforehand.
+</p>
+
+<p>
+Here's the riddle:
+</p>
+
+<table border="1">
+<tr><td>
+
+<p>
+Four bugs are standing at the corners of a square. At time 0 a bell rings and
+they start moving towards each other. Each bug advances in the direction of the
+bug which was initially at the position clockwise from it.
+</p>
+
+<p>
+Assuming the square is 1 meter long, and the bugs are moving in the velocity of
+1 meter per second, how will it take for the bugs to meet in the center of the
+square?
+</p>
+
+</td></tr>
+</table>
+
+The solution can be found some space below:
+<longblank />
+
+
+<h2>Solution:</h2>
+
+<img src="bugs.gif" alt="Bugs Path Illustration" style=" float : right ; margin-left : 0.5em ; margin-bottom : 0.5em ; " />
+
+<p>
+I'll use the following model to simplify the problem: suppose the bugs don't
+move towards each other infinitesimly, but rather move in a straight line, and
+then stop, and then move again toward their new positions, and so forth. The bugs
+move toward their initial position until they reach a position which is in proportion
+p to the initial distance between them. Then they move again, towards a position which
+is in proportion p to their distance, and so forth.
+</p>
+
+
+<p>
+Thus they form an infinite series of squares inside each other. You can see an
+illustration of this scheme to the right of this text for the proportion p=0.15.
+</p>
+
+
+
+
+
+<p>
+Now, if the length of a given square is a, then the length of the square inside it is
+(according to pythgoras theorem): square-root((p*a)^2+((1-p)*a)^2) =
+a*square-root(2*p^2-2*p+1). The lengthes of the squares form a decreasing geometrical
+series, with that proportion. Thus the length of the path a bug travel until they meet is:
+</p>
+
+
+<p>
+p * a1 / [ 1 - square-root(2*p^2-2*p+1) ]
+</p>
+
+
+<p>
+This is according to the formula that the sum of an infinite decresing geometric series is
+a1 / (1-q) where a1 is the value of its first item and q is the proportion between
+two consecutive items.
+</p>
+
+
+<p>
+Now, to find the length an infinitesimal bug will travel, we just limit p to 0:
+</p>
+
+
+<pre>
+                                                       
+        p*a1             
+lim ---------------   =  
+p-&gt;0   ____________   
+    1-V 2*p^2-2*p+1      
+
+
+                            _____________
+          p*a1           1+V 2*p^2-2*p+1
+lim  ----------------  * -----------------   =
+p-&gt;0    ____________        _____________
+     1-V 2*p^2-2*p+1     1+V 2*p^2-2*p+1
+ 
+
+ 
+                ____________     
+        a1*p*(1+V 2*p^2-2*p+1)    
+lim     ---------------------  =  
+p-&gt;0                        
+         1 - 2*p^2 + 2*p - 1
+
+
+               _____________
+        a1 * (1+V 2*p^2-2*p+1 )
+lim     ----------------------  = 
+p-&gt;0       2 - 2*p
+
+
+                 _____________
+        a1 * (1+V 2*0-2*0+1  )   
+lim     ---------------------  = 
+p-&gt;0      2 - 2*0           
+
+
+
+   a1 * (1+1)  
+ ------------- = a1
+      2        
+          
+
+</pre>
+
+<p>
+Therefore, the length of a bug's path is equal to the length of the original
+square's side. The time it will take a 1 meter per second fast bugs who stand at
+the corner of a 1*1 meter square to meet is 1/1 = 1 second.
+</p>
+
+
+<p>
+You can find <a href="bugs.scm">here</a> a <a href="http://www.gimp.org/">Gimp</a> script
+that generates a series of squares inside squares, and optionally marks the path of
+a single bug.
+</p>
+
+
+<hr />
+
+<h2><a href="./">Back to the Math-Ventures Web page</a></h2>
+

t2/MathVentures/dodeca.html

-<HTML>
-<TITLE>Math-Ventures: What's the volume of a dodecahedron?</TITLE>
-
-<BODY BGCOLOR="#FFFFFF">
-
-<CENTER><H2>What's the volume of a dodecahedron?</H2></CENTER>
-
-<IMG SRC="dodeca.gif" alt="A Dodecahedron">
-
-<BR>
-The picture above shows a dodecahedron. It's a solid body which has 12 perfect
-pentagons of the same size as its sides. I wondered what is the volume
-of such a body.<P>
-
-<TABLE BORDER=1>
-<TR><TD>
-What is the volume of a dodecahedron with an edge-length of a?
-</TR></TD>
-</TABLE>
-
-<BR>
-
-The solution can be found some space below:
-
-<BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR>
-<BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR>
-<BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR>
-<BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR><BR>
-
-<H3>Solution:</H3>
-
-Since a dodecahedron is a perfect solid with 12 identical sides at the
-same distance from its center, we can divide it into twelve identical
-pyramids. The corner of each pyramid is found at the dodecahedron's
-center, and the base is one of the sides.<P>
-
-The volume of a pyramid is the area of the base multiplied by its
-height divided by 3. We can calculate the height by the angle between
-the base and one of the surface sides. Since two pyramids are adjacent
-at every edge of the dodecahedron, we can find it by taking the angle
-between two adjacent sides of the dodecahedron and dividing it by 2.<P>
-
-To find that, let's take a corner of the dodecahedron, and form a
-triangle at the points which are at a certain distance along the edges.
-We get the following picture:<P>
-
-<IMG SRC="dodeca_corner.gif" alt="Corner of a dodecahedron from two different views">
-
-<BR>
-From point C, which is found somewhere along the edge, let's lower two
-perpendiculars to the edge, down to the edges of the base triangle. We
-get the triangle CAB. Now since the surface sides of this pyramid are
-isoscles triangles, and the angle of a perfect pentagon is equal to
-180&deg;*3/5=108 degrees, then the angle CDA is equal to
-(180&deg;-108&deg;)/2=36&deg; degrees. <P>
-
-Since angle ACD is a right angle, we find that CA (and CB) is equal to AD *
-sin(36&deg;), and since the base triangle is perfect, AB is equal to
-AD. Thus, we know that AE/CA = AB/CA/2 = AD/CA/2 = 1 / (CA/AD) / 2 =
-1 / (2 * sin(36&deg;)). Thus, the angle ACE, which is half the angle
-between two sides of the dodecahedron and the angle we seek, is equal
-to arcsin(1/(2*sin(36&deg;))). (approximately 58.28&deg;).<P>
-
-Now, let's take the look at a side pyramid of a dodecahedron and calculate its
-volume:<P>
-
-<IMG SRC="dodeca_side_pyramid.gif" alt="A Dodecahedron Side Pyramid">
-
-<BR>
-Since the angle of a perfect pentagon is equal to 108&deg;, angle OBA
-which is half of it is equal to 54&deg;. Since AB is <B>a</B>/2, OA is equal to
-tan(54&deg;)*<B>a</B>/2. Now, the base of the pyramid is made of 5 equal
-triangles each of which has a base of length <B>a</B> and a height of
-OA. Thus, the area of the base is 5*(OA*<B>a</B>/2) =
-5*tan(54&deg;)*<B>a</B>/2*<B>a</B>/2 = 5/4*tan(54&deg;)*<B>a</B>^2.<P>
-
-In the previous section we found out that the angle OAD is equal to
-arcsin(1/(2*sin(36&deg;))), and therefore OD is OA *
-tan(arcsin(1/(2*sin(36&deg;)))) =
-1/2 * tan(54&deg;) * tan(arcsin(1/(2*sin(36&deg;)))) * <B>a</B>.
-The volume of the pyramid is the area of its base multiplied by its
-height (OD) divided by 3 and so it is equal to:<P>
-
-5/4 * tan(54&deg;) * <B>a</B>^2 *<BR>
-1/2 * tan(54&deg;) * tan(arcsin(1/(2*sin(36&deg;)))) * <B>a</B> *<BR>
-1/3 =<BR>
-5/24 * tan(54&deg;)^2 * tan(arcsin(1/(2*sin(36&deg;)))) * <B>a</B>^3<P>
-
-Since there are 12 such pyramids in a dodecahedron, its volume is equal
-to this volume multiplied by 12. Thus, we get that the volume of a
-dodecahedron is:<P>
-
-5/2 * tan(54&deg;)^2 * tan(arcsin(1/(2*sin(36&deg;)))) * <B>a</B>^3 =
-~7.66 * <B>a</B>^3.
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-<HR>
-
-<A HREF="./"><H3>Back to the Math-Ventures Web page</H3></A><P>
-<A HREF="../"><H3>Back to Shlomi Fish' Homepage</H3></A>
-
-
-</BODY>
-</HTML>

t2/MathVentures/dodeca.html.wml

+#include '../template.wml'
+
+<subject "What's the Volume of a Dodecahedron?" />
+
+<img src="dodeca.gif" alt="A Dodecahedron" /> <br />
+
+<p>
+The picture above shows a dodecahedron. It's a solid body which has
+12 perfect pentagons of the same size as its sides. I wondered what
+is the volume of such a body.
+</p>
+
+<table border="1">
+<tr>
+<td>What is the volume of a dodecahedron with an edge-length of
+a?</td>
+</tr>
+</table>
+
+<br />
+<p>
+ The solution can be found some space below:
+</p>
+
+<longblank />
+ 
+<h2>Solution:</h2>
+
+<p>
+Since a dodecahedron is a perfect solid with 12 identical sides at
+the same distance from its center, we can divide it into twelve
+identical pyramids. The corner of each pyramid is found at the
+dodecahedron's center, and the base is one of the sides.
+</p>
+
+<p>The volume of a pyramid is the area of the base multiplied by
+its height divided by 3. We can calculate the height by the angle
+between the base and one of the surface sides. Since two pyramids
+are adjacent at every edge of the dodecahedron, we can find it by
+taking the angle between two adjacent sides of the dodecahedron and
+dividing it by 2.</p>
+
+<p>To find that, let's take a corner of the dodecahedron, and form
+a triangle at the points which are at a certain distance along the
+edges. We get the following picture:</p>
+
+<p><img src="dodeca_corner.gif"
+alt="Corner of a dodecahedron from two different views" /><br />
+From point C, which is found somewhere along the edge, let's lower
+two perpendiculars to the edge, down to the edges of the base
+triangle. We get the triangle CAB. Now since the surface sides of
+this pyramid are isoscles triangles, and the angle of a perfect
+pentagon is equal to 180&deg;*3/5=108 degrees, then the angle CDA
+is equal to (180&deg;-108&deg;)/2=36&deg; degrees.</p>
+
+<p>Since angle ACD is a right angle, we find that CA (and CB) is
+equal to AD * sin(36&deg;), and since the base triangle is perfect,
+AB is equal to AD. Thus, we know that AE/CA = AB/CA/2 = AD/CA/2 = 1
+/ (CA/AD) / 2 = 1 / (2 * sin(36&deg;)). Thus, the angle ACE, which
+is half the angle between two sides of the dodecahedron and the
+angle we seek, is equal to arcsin(1/(2*sin(36&deg;))).
+(approximately 58.28&deg;).</p>
+
+<p>Now, let's take the look at a side pyramid of a dodecahedron and
+calculate its volume:</p>
+
+<p><img src="dodeca_side_pyramid.gif"
+alt="A Dodecahedron Side Pyramid" /><br />
+Since the angle of a perfect pentagon is equal to 108&deg;, angle
+OBA which is half of it is equal to 54&deg;. Since AB is
+<b>a</b>/2, OA is equal to tan(54&deg;)*<b>a</b>/2. Now, the base
+of the pyramid is made of 5 equal triangles each of which has a
+base of length <b>a</b> and a height of OA. Thus, the area of the
+base is 5*(OA*<b>a</b>/2) = 5*tan(54&deg;)*<b>a</b>/2*<b>a</b>/2 =
+5/4*tan(54&deg;)*<b>a</b>^2.</p>
+
+<p>In the previous section we found out that the angle OAD is equal
+to arcsin(1/(2*sin(36&deg;))), and therefore OD is OA *
+tan(arcsin(1/(2*sin(36&deg;)))) = 1/2 * tan(54&deg;) *
+tan(arcsin(1/(2*sin(36&deg;)))) * <b>a</b>. The volume of the
+pyramid is the area of its base multiplied by its height (OD)
+divided by 3 and so it is equal to:</p>
+
+<p>5/4 * tan(54&deg;) * <b>a</b>^2 *<br />
+1/2 * tan(54&deg;) * tan(arcsin(1/(2*sin(36&deg;)))) * <b>a</b>
+*<br />
+1/3 =<br />
+5/24 * tan(54&deg;)^2 * tan(arcsin(1/(2*sin(36&deg;)))) *
+<b>a</b>^3</p>
+
+<p>Since there are 12 such pyramids in a dodecahedron, its volume
+is equal to this volume multiplied by 12. Thus, we get that the
+volume of a dodecahedron is:</p>
+
+<p>5/2 * tan(54&deg;)^2 * tan(arcsin(1/(2*sin(36&deg;)))) *
+<b>a</b>^3 = ~7.66 * <b>a</b>^3.</p>
+
+<hr />
+
+<h3><a href="./">Back to the Math-Ventures Web page</a></h3>
+

t2/MathVentures/toggle_squares.html

-<HTML>
-<TITLE>Math-Ventures: Toggling Squares is not that Trivial...</TITLE>
-
-<BODY BGCOLOR="#FFFFFF">
-
-<CENTER><H2>Toggling Squares is not that Trivial...</H2></CENTER>
-
-Well, this page is not ready yet, and will take some time to be written.<P>
-
-For the time being, the Java game, I'm referring to can be found here:
-<A HREF="http://www.ida.net/users/housley/atag.htm">http://www.ida.net/users/housley/atag.htm</A><P>
-
-<p>
-I also write an implementation of it in JavaScript. It can be found here:
-</p>
-
-<p>
-<a href="http://www.iglu.org.il/shlomif/toggle_squares/">
-http://www.iglu.org.il/shlomif/toggle_squares/
-</a>
-</p>
-
-You should experience with it, and try to figure out a methodical way to
-solve it. Afterwards, you can read <A HREF="../toggle.html">this report</A>
-which I prepared for Ken Housley, who programmed the game applet. The report
-describes an algorithm which we figured out to solve the game, and proves
-why this algorithm is working.<P>
-
-I frequently used many Linear Algebra terms there, so you may need to be familiar
-with this field to fully understand it. When I prepare this page, I will
-explain everything with more basic material, but for now, this will have
-to do.
-
-<BR><BR>
-
-<HR>
-
-<A HREF="./"><H3>Back to the Math-Ventures Web page</H3></A><P>
-<A HREF="../"><H3>Back to Shlomi Fish' Homepage</H3></A>
-
-
-</BODY>
-</HTML>

t2/MathVentures/toggle_squares.html.wm

+<HTML>
+<TITLE>Math-Ventures: Toggling Squares is not that Trivial...</TITLE>
+
+<BODY BGCOLOR="#FFFFFF">
+
+<CENTER><H2>Toggling Squares is not that Trivial...</H2></CENTER>
+
+Well, this page is not ready yet, and will take some time to be written.<P>
+
+For the time being, the Java game, I'm referring to can be found here:
+<A HREF="http://www.ida.net/users/housley/atag.htm">http://www.ida.net/users/housley/atag.htm</A><P>
+
+<p>
+I also write an implementation of it in JavaScript. It can be found here:
+</p>
+
+<p>
+<a href="http://www.iglu.org.il/shlomif/toggle_squares/">
+http://www.iglu.org.il/shlomif/toggle_squares/
+</a>
+</p>
+
+You should experience with it, and try to figure out a methodical way to
+solve it. Afterwards, you can read <A HREF="../toggle.html">this report</A>
+which I prepared for Ken Housley, who programmed the game applet. The report
+describes an algorithm which we figured out to solve the game, and proves
+why this algorithm is working.<P>
+
+I frequently used many Linear Algebra terms there, so you may need to be familiar
+with this field to fully understand it. When I prepare this page, I will
+explain everything with more basic material, but for now, this will have
+to do.
+
+<BR><BR>
+
+<HR>
+
+<A HREF="./"><H3>Back to the Math-Ventures Web page</H3></A><P>
+<A HREF="../"><H3>Back to Shlomi Fish' Homepage</H3></A>
+
+
+</BODY>
+</HTML>
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