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# File 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>`

# File 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>`
`+`

# File 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>`

# File 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>`
`+`

# File 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>`

# File 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>`