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		<title>For any integer n &gt; 0, there exists an Apollonian gasket defined by the following curvatures:</title>
		<link>http://sagepub.wordpress.com/2010/03/10/for-any-integer-n-0-there-exists-an-apollonian-gasket-defined-by-the-following-curvatures/</link>
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		<pubDate>Wed, 10 Mar 2010 17:59:24 +0000</pubDate>
		<dc:creator>Web Ani</dc:creator>
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		<description><![CDATA[For any integer n &#62; 0, there exists an Apollonian gasket defined by the following curvatures: (-n, n+1, (n)(n+1), (n)(n+1)+1). For example, the gaskets defined by (-2,3,6,7), (-3,4,12,13), (-8,9,72,73), and (-9,10,90,91) all follow this pattern. Because every interior circle that is defined by n+1 can become the bounding circle (defined by -n) in another gasket, [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=sagepub.wordpress.com&amp;blog=11141043&amp;post=40&amp;subd=sagepub&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>For any integer <em>n</em> &gt; 0, there exists an Apollonian gasket defined by the following curvatures:<br />
(-<em>n</em>, <em>n</em>+1, (<em>n</em>)(<em>n</em>+1), (<em>n</em>)(<em>n</em>+1)+1).<br />
For example, the gaskets defined by (-2,3,6,7), (-3,4,12,13), (-8,9,72,73), and (-9,10,90,91) all follow this pattern. Because every interior circle that is defined by <em>n</em>+1 can become the bounding circle (defined by -<em>n</em>) in another gasket, these gaskets can be nested. This is demonstrated in the figure at right, which contains these sequential gaskets with <em>n</em> running from 2 through 20.</p>
<p>-wiki</p>
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		<title>Apollonian gaskets</title>
		<link>http://sagepub.wordpress.com/2010/03/10/apollonian-gaskets/</link>
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		<pubDate>Wed, 10 Mar 2010 17:58:47 +0000</pubDate>
		<dc:creator>Web Ani</dc:creator>
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		<description><![CDATA[The figure at left is an Apollonian gasket that appears to have D3 symmetry. The same figure is displayed at right, with labels indicating the curvatures of the interior circles, illustrating that the gasket actually possesses only the D1 symmetry common to many other integral Apollonian gaskets. The following table lists more of these almost-D3 [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=sagepub.wordpress.com&amp;blog=11141043&amp;post=38&amp;subd=sagepub&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>The figure at left is an Apollonian gasket that appears to have D<sub>3</sub> symmetry. The same figure is displayed at right, with labels indicating the curvatures of the interior circles, illustrating that the gasket actually possesses only the D<sub>1</sub> symmetry common to many other integral Apollonian gaskets.</p>
<p>The following table lists more of these <em>almost</em>-D<sub>3</sub> integral Apollonian gaskets. The sequence has some interesting properties, and the table lists a factorization of the curvatures, along with the multiplier needed to go from the previous set to the current one. The factorizations demonstrate an interesting chain of factors throughout the sequence of gaskets, and the multiplier appears to be converging toward <img src="http://upload.wikimedia.org/math/8/3/4/834b0de87818535d0dd36324931b8862.png" alt="\sqrt{3}+2 \approx 3.732050807..." /></p>
<p>-wiki</p>
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			<media:title type="html">\sqrt{3}+2 \approx 3.732050807...</media:title>
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		<title>Symmetry of integral Apollonian circle packings</title>
		<link>http://sagepub.wordpress.com/2010/03/10/symmetry-of-integral-apollonian-circle-packings/</link>
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		<pubDate>Wed, 10 Mar 2010 17:57:48 +0000</pubDate>
		<dc:creator>Web Ani</dc:creator>
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		<description><![CDATA[  Whenever two of the largest five circles in the gasket have the same curvature, that gasket will have D1 symmetry, which corresponds to a reflection along a diameter of the bounding circle, with no rotational symmetry. If two different curvatures are repeated within the first five, the gasket will have D2 symmetry; such a [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=sagepub.wordpress.com&amp;blog=11141043&amp;post=36&amp;subd=sagepub&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p> </p>
<p>Whenever two of the largest five circles in the gasket have the same curvature, that gasket will have D<sub>1</sub> symmetry, which corresponds to a reflection along a diameter of the bounding circle, with no rotational symmetry.</p>
<p>If two different curvatures are repeated within the first five, the gasket will have D<sub>2</sub> symmetry; such a symmetry consists of two reflections (perpendicular to each other) along diameters of the bounding circle, with a two-fold rotational symmetry of 180°. The gasket described by curvatures (-1,2,2,3) is the only Apollonian gasket (up to a scaling factor) to possess D<sub>2</sub> symmetry.</p>
<p>If none of the curvatures are repeated within the first five, the gasket contains no symmetry, which is represented by symmetry group C<sub>1</sub>; the gasket described by curvatures (−10,18,23,27) is an example.</p>
<p>If the three circles with smallest positive curvature have the same curvature, the gasket will have D<sub>3</sub> symmetry, which corresponds to three reflections along diameters of the bounding circle (spaced 120° apart), along with three-fold rotational symmetry of 120°. In this case the ratio of the curvature of the bounding circle to the three inner circles is <img src="http://upload.wikimedia.org/math/b/7/2/b724fb2513714533c8cf59c486a395ae.png" alt="2\sqrt{3}-3" />. As this ratio is not rational, no integral Apollonian circle packings possess this D<sub>3</sub> symmetry, although many packings come close.</p>
<p>-wkii</p>
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			<media:title type="html">2\sqrt{3}-3</media:title>
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		<title>If any four mutually tangent &#8230;</title>
		<link>http://sagepub.wordpress.com/2010/03/10/if-any-four-mutually-tangent/</link>
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		<pubDate>Wed, 10 Mar 2010 17:57:01 +0000</pubDate>
		<dc:creator>Web Ani</dc:creator>
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		<description><![CDATA[If any four mutually tangent circles in an Apollonian gasket all have integer curvature (where the curvature of a circle is defined to be the inverse of its radius) then all circles in the gasket will have integer curvature.[1] The first few of these integral Apollonian gaskets are listed in the following table. The table [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=sagepub.wordpress.com&amp;blog=11141043&amp;post=34&amp;subd=sagepub&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>If any four mutually tangent circles in an Apollonian gasket all have integer curvature (where the curvature of a circle is defined to be the inverse of its radius) then all circles in the gasket will have integer curvature.<sup><a href="http://en.wikipedia.org/wiki/Apollonian_gasket#cite_note-0">[1]</a></sup> The first few of these integral Apollonian gaskets are listed in the following table. The table lists the curvatures of the largest circles in the gasket; a negative curvature indicates that all other circles are tangent to the interior of that circle (that is, it is the bounding circle), while a positive curvature indicates that all other circles are tangent to the exterior of that circle (these are the interior circles). Only the first three curvatures (of the five displayed in the table) are needed to completely describe each gasket &#8211; all other curvatures can be derived from these three.</p>
<p>-wiki</p>
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		<title>Links with hyperbolic geometry</title>
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		<pubDate>Wed, 10 Mar 2010 17:56:12 +0000</pubDate>
		<dc:creator>Web Ani</dc:creator>
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		<description><![CDATA[  The three generating circles, and hence the entire construction, are determined by the location of the three points where they are tangent to one another. Since there is a Möbius transformation which maps any three given points in the plane to any other three points, and since Möbius transformations preserve circles, then there is [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=sagepub.wordpress.com&amp;blog=11141043&amp;post=33&amp;subd=sagepub&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p> </p>
<p>The three generating circles, and hence the entire construction, are determined by the location of the three points where they are tangent to one another. Since there is a <a title="Möbius transformation" href="http://en.wikipedia.org/wiki/M%C3%B6bius_transformation">Möbius transformation</a> which maps any three given points in the plane to any other three points, and since Möbius transformations preserve circles, then there is a Möbius transformation which maps any two Apollonian gaskets to one another.</p>
<p>Möbius transformations are also isometries of the <a title="Hyperbolic plane" href="http://en.wikipedia.org/wiki/Hyperbolic_plane">hyperbolic plane</a>, so in hyperbolic geometry all Apollonian gaskets are congruent. In a sense, there is therefore only one Apollonian gasket, which can be thought of as a <a title="Tessellation" href="http://en.wikipedia.org/wiki/Tessellation">tessellation</a> of the hyperbolic plane by circles and hyperbolic triangles.</p>
<p>The Apollonian gasket is the limit set of a group of Möbius transformations known as a <a title="Kleinian group" href="http://en.wikipedia.org/wiki/Kleinian_group">Kleinian group</a>.</p>
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		<title>Symmetries</title>
		<link>http://sagepub.wordpress.com/2010/03/10/symmetries/</link>
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		<pubDate>Wed, 10 Mar 2010 17:55:25 +0000</pubDate>
		<dc:creator>Web Ani</dc:creator>
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		<description><![CDATA[  If two of the original generating circles have the same radius and the third circle has a radius that is two-thirds of this, then the Apollonian gasket has two lines of reflective symmetry; one line is the line joining the centres of the equal circles; the other is their mutual tangent, which passes through [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=sagepub.wordpress.com&amp;blog=11141043&amp;post=31&amp;subd=sagepub&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p> </p>
<p>If two of the original generating circles have the same radius and the third circle has a radius that is two-thirds of this, then the Apollonian gasket has two lines of reflective symmetry; one line is the line joining the centres of the equal circles; the other is their mutual tangent, which passes through the centre of the third circle. These lines are perpendicular to one another, so the Apollonian gasket also has rotational symmetry of degree 2; the symmetry group of this gasket is D<sub>2</sub>.</p>
<p>If all three of the original generating circles have the same radius then the Apollonian gasket has three lines of reflective symmetry; these lines are the mutual tangents of each pair of circles. Each mutual tangent also passes through the centre of the third circle and the common centre of the first two Apollonian circles. These lines of symmetry are at angles of 60 degrees to one another, so the Apollonian gasket also has rotational symmetry of degree 3; the symmetry group of this gasket is D<sub>3</sub>.</p>
<p>-from wiki</p>
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		<title>An Apollonian gasket can also be constructed by &#8230;</title>
		<link>http://sagepub.wordpress.com/2010/03/10/an-apollonian-gasket-can-also-be-constructed-by/</link>
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		<pubDate>Wed, 10 Mar 2010 17:54:34 +0000</pubDate>
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		<description><![CDATA[An Apollonian gasket can also be constructed by replacing one of the generating circles by a straight line, which can be regarded as a circle passing through the point at infinity. Alternatively, two of the generating circles may be replaced by parallel straight lines, which can be regarded as being tangent to one another at [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=sagepub.wordpress.com&amp;blog=11141043&amp;post=29&amp;subd=sagepub&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>An Apollonian gasket can also be constructed by replacing one of the generating circles by a straight line, which can be regarded as a circle passing through the point at infinity.</p>
<p>Alternatively, two of the generating circles may be replaced by parallel straight lines, which can be regarded as being tangent to one another at infinity. In this construction, the circles that are tangent to one of the two straight lines form a family of <a title="Ford circle" href="http://en.wikipedia.org/wiki/Ford_circle">Ford circles</a>.</p>
<p>The three-dimensional equivalent of the Apollonian gasket is the <a title="Apollonian sphere packing" href="http://en.wikipedia.org/wiki/Apollonian_sphere_packing">Apollonian sphere packing</a>.</p>
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		<title>An Apollonian gasket can be constructed &#8230;.</title>
		<link>http://sagepub.wordpress.com/2010/03/10/an-apollonian-gasket-can-be-constructed/</link>
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		<pubDate>Wed, 10 Mar 2010 17:53:44 +0000</pubDate>
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		<description><![CDATA[An Apollonian gasket can be constructed as follows. Start with three circles C1, C2 and C3, each one of which is tangent to the other two (in the general construction, these three circles can be any size, as long as they have common tangents). Apollonius discovered that there are two other non-intersecting circles, C4 and [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=sagepub.wordpress.com&amp;blog=11141043&amp;post=27&amp;subd=sagepub&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>An Apollonian <a title="Gasket" href="http://en.wikipedia.org/wiki/Gasket">gasket</a> can be constructed as follows. Start with three circles <em>C</em>1, <em>C</em>2 and <em>C</em>3, each one of which is tangent to the other two (in the general construction, these three circles can be any size, as long as they have common tangents). Apollonius discovered that there are two other non-intersecting circles, <em>C</em>4 and <em>C</em>5, which have the property that they are tangent to all three of the original circles &#8211; these are called <em>Apollonian circles</em> (see <a title="Descartes' theorem" href="http://en.wikipedia.org/wiki/Descartes%27_theorem">Descartes&#8217; theorem</a>). Adding the two Apollonian circles to the original three, we now have five circles.</p>
<p>Take one of the two Apollonian circles &#8211; say <em>C</em>4. It is tangent to <em>C</em>1 and <em>C</em>2, so the triplet of circles <em>C</em>4, <em>C</em>1 and <em>C</em>2 has its own two Apollonian circles. We already know one of these &#8211; it is <em>C</em>3 &#8211; but the other is a new circle <em>C</em>6.</p>
<p>In a similar way we can construct another new circle <em>C</em>7 that is tangent to <em>C</em>4, <em>C</em>2 and <em>C</em>3, and another circle <em>C</em>8 from <em>C</em>4, <em>C</em>3 and <em>C</em>1. This gives us 3 new circles. We can construct another three new circles from <em>C</em>5, giving six new circles altogether. Together with the circles <em>C</em>1 to <em>C</em>5, this gives a total of 11 circles.</p>
<p>Continuing the construction stage by stage in this way, we can add 2·3<sup><em>n</em></sup> new circles at stage <em>n</em>, giving a total of 3<sup><em>n</em>+1</sup> + 2 circles after <em>n</em> stages. In the limit, this set of circles is an Apollonian gasket.</p>
<p>The Apollonian gasket has a <a title="Hausdorff dimension" href="http://en.wikipedia.org/wiki/Hausdorff_dimension">Hausdorff dimension</a> of about 1.3057</p>
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		<title>Apollonian gasket</title>
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		<pubDate>Wed, 10 Mar 2010 17:52:53 +0000</pubDate>
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		<description><![CDATA[In mathematics, an Apollonian gasket or Apollonian net is a fractal generated from triples of circles, where any circle is tangent to two others. It is named after Greek mathematician Apollonius of Perga. from wiki<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=sagepub.wordpress.com&amp;blog=11141043&amp;post=25&amp;subd=sagepub&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>In <a title="Mathematics" href="http://en.wikipedia.org/wiki/Mathematics">mathematics</a>, an <strong>Apollonian gasket</strong> or <strong>Apollonian net</strong> is a <a title="Fractal" href="http://en.wikipedia.org/wiki/Fractal">fractal</a> generated from triples of circles, where any circle is <a title="Tangent" href="http://en.wikipedia.org/wiki/Tangent">tangent</a> to two others. It is named after <a title="Greece" href="http://en.wikipedia.org/wiki/Greece">Greek</a> <a title="Mathematician" href="http://en.wikipedia.org/wiki/Mathematician">mathematician</a> <a title="Apollonius of Perga" href="http://en.wikipedia.org/wiki/Apollonius_of_Perga">Apollonius of Perga</a>.</p>
<p>from wiki</p>
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		<title>Conversion from two parametric equations to a single equation</title>
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		<pubDate>Tue, 09 Feb 2010 07:11:04 +0000</pubDate>
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		<description><![CDATA[Conversion from two parametric equations to a single equation Converting a set of parametric equations to a single equation involves eliminating the variable t from the simultaneous equations . If one of these equations can be solved for t, the expression obtained can be substituted into the other equation to obtain an equation involving x [...]<img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=sagepub.wordpress.com&amp;blog=11141043&amp;post=24&amp;subd=sagepub&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p>Conversion from two parametric equations to a single equation</p>
<p>Converting a set of parametric equations to a single equation involves eliminating the variable <em>t</em> from the simultaneous equations <img src="http://upload.wikimedia.org/math/c/3/a/c3a3818c86a7d4bddfb953015b62a028.png" alt="x=x(t),\ y=y(t)" />. If one of these equations can be solved for <em>t</em>, the expression obtained can be substituted into the other equation to obtain an equation involving <em>x</em> and <em>y</em> only. If <em>x</em>(<em>t</em>) and <em>y</em>(<em>t</em>) are rational functions then the techniques of the <a title="Theory of equations" href="http://en.wikipedia.org/wiki/Theory_of_equations">theory of equations</a> such as <a title="Resultant" href="http://en.wikipedia.org/wiki/Resultant">resultants</a> can be used to eliminate <em>t</em>. In some cases there is no single equation in closed form that is equivalent to the parametric equations.<sup><a href="http://en.wikipedia.org/wiki/Parametric_equation#cite_note-0">[1]</a></sup></p>
<p>To take the example of the circle of radius <em>a</em> <a href="http://en.wikipedia.org/wiki/Parametric_equation#Examples">above</a>, the parametric equations</p>
<dl>
<dd><img src="http://upload.wikimedia.org/math/f/9/d/f9d29798b4900f5cd50fcd97097118d3.png" alt="x = a \cos(t)\," /> </dd>
<dd><img src="http://upload.wikimedia.org/math/2/8/6/28638dab1598540a0c1578a79fba6501.png" alt="y = a \sin(t)\," /> </dd>
</dl>
<p>can be simply expressed in terms of <em>x</em> and <em>y</em> by way of the <a title="Pythagorean trigonometric identity" href="http://en.wikipedia.org/wiki/Pythagorean_trigonometric_identity">Pythagorean trigonometric identity</a>:</p>
<dl>
<dd><img src="http://upload.wikimedia.org/math/b/3/f/b3f478983e6bb7b34214fe33b40906f3.png" alt="x/a = \cos(t)\," /> </dd>
<dd><img src="http://upload.wikimedia.org/math/6/a/c/6ac68e3d9374afbd6fda586242837e0f.png" alt="y/a = \sin(t)\," /> </dd>
<dd><img src="http://upload.wikimedia.org/math/1/c/7/1c7ea7b166435fdc8d964e4e0747c63e.png" alt="\cos(t)^2 + \sin(t)^2 = 1\,\!" /> </dd>
<dd><img src="http://upload.wikimedia.org/math/5/e/8/5e8d98c01185b78b0f04f76b3adc4e84.png" alt="\therefore (x/a)^2 + (y/a)^2 = 1," /> </dd>
</dl>
<p>which is easily identifiable as a type of <a title="Conic section" href="http://en.wikipedia.org/wiki/Conic_section">conic section</a> (in this case, a circle).</p>
<p> http://en.wikipedia.org/wiki/Parametric_equation</p>
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		<media:content url="http://1.gravatar.com/avatar/f0288fd4ebbf45dc6b58914ee401d8ea?s=96&#38;d=identicon&#38;r=G" medium="image">
			<media:title type="html">Web Ani</media:title>
		</media:content>

		<media:content url="http://upload.wikimedia.org/math/c/3/a/c3a3818c86a7d4bddfb953015b62a028.png" medium="image">
			<media:title type="html">x=x(t),\ y=y(t)</media:title>
		</media:content>

		<media:content url="http://upload.wikimedia.org/math/f/9/d/f9d29798b4900f5cd50fcd97097118d3.png" medium="image">
			<media:title type="html">x = a \cos(t)\,</media:title>
		</media:content>

		<media:content url="http://upload.wikimedia.org/math/2/8/6/28638dab1598540a0c1578a79fba6501.png" medium="image">
			<media:title type="html">y = a \sin(t)\,</media:title>
		</media:content>

		<media:content url="http://upload.wikimedia.org/math/b/3/f/b3f478983e6bb7b34214fe33b40906f3.png" medium="image">
			<media:title type="html">x/a = \cos(t)\,</media:title>
		</media:content>

		<media:content url="http://upload.wikimedia.org/math/6/a/c/6ac68e3d9374afbd6fda586242837e0f.png" medium="image">
			<media:title type="html">y/a = \sin(t)\,</media:title>
		</media:content>

		<media:content url="http://upload.wikimedia.org/math/1/c/7/1c7ea7b166435fdc8d964e4e0747c63e.png" medium="image">
			<media:title type="html">\cos(t)^2 + \sin(t)^2 = 1\,\!</media:title>
		</media:content>

		<media:content url="http://upload.wikimedia.org/math/5/e/8/5e8d98c01185b78b0f04f76b3adc4e84.png" medium="image">
			<media:title type="html">\therefore (x/a)^2 + (y/a)^2 = 1,</media:title>
		</media:content>
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