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  <last_published>2011-02-01T00:00:01</last_published>
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&lt;td&gt;&lt;mdo:applet archive=&quot;http://www.geogebra.org/webstart/geogebra.jar&quot; code=&quot;geogebra.GeoGebraApplet&quot; codebase=&quot;/content/98/03/six6/&quot; datafile=&quot;&quot; height=&quot;289&quot; width=&quot;278&quot;&gt;&lt;param name=&quot;filename&quot; value=&quot;http://nrich.maths.org/content/98/03/six6/Shrink.ggb&quot;&gt;&lt;/param&gt;
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&lt;p&gt;&lt;span style=&quot;font-size: small;&quot;&gt;Created with &lt;a href=&quot;http://www.geogebra.org&quot; target=&quot;_blank&quot;&gt;GeoGebra&lt;/a&gt;&lt;/span&gt;&lt;/p&gt;
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&lt;div&gt;Triangle ABC is any right angled triangle and X is a moveable point on the hypotenuse AC. The points P and Q are the feet of the perpendiculars from X to the sides of the triangle.&lt;/div&gt;
&lt;br&gt;&lt;/br&gt;
&lt;div&gt;Find the position of X which makes the length of PQ a minimum.&lt;/div&gt;
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&lt;div class=&quot;framework&quot;&gt;NOTES AND BACKGROUND&lt;br&gt;&lt;/br&gt;
&lt;p&gt;This dynamic image is drawn using Geogebra, free software and very easy to use. You can download your own copy of Geogebra from &lt;a href=&quot;http://www.geogebra.org/cms/&quot;&gt;http://www.geogebra.org/cms/&lt;/a&gt; together with a good help manual and &lt;a href=&quot;http://www.geogebra.org/cms/index.php?option=com_content&amp;amp;task=blogcategory&amp;amp;id=75&amp;amp;Itemid=61&quot;&gt;Quickstart&lt;/a&gt; for beginners.&lt;/p&gt;
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&lt;td align=&quot;center&quot; valign=&quot;center&quot;&gt;&lt;mdo:image src=&quot;shrinkTri.gif&quot; alt=&quot;Helpful diagram&quot;&gt;&lt;/mdo:image&gt;&lt;/td&gt;
&lt;td align=&quot;left&quot; valign=&quot;top&quot;&gt;
&lt;p&gt;The solution to this problem only uses two very simple
facts:&lt;/p&gt;
&lt;ol&gt;
&lt;li&gt;The diagonals of a rectangle are equal in length.&lt;/li&gt;
&lt;li&gt;For any triangle, the shortest distance from a vertex to the
opposite side is the perpendicular to that side.&lt;/li&gt;
&lt;/ol&gt;
&lt;p&gt;Putting the two together, for any right angled triangle the
length of &lt;strong&gt;P&lt;/strong&gt; &lt;strong&gt;Q&lt;/strong&gt; is a minimum when
&lt;strong&gt;X&lt;/strong&gt; is the foot of the perpendicular from
&lt;strong&gt;A&lt;/strong&gt; to &lt;strong&gt;B&lt;/strong&gt; &lt;strong&gt;C&lt;/strong&gt;. As the
point &lt;strong&gt;X&lt;/strong&gt; moves along the hypotenuse the rectangle
&lt;strong&gt;A&lt;/strong&gt; &lt;strong&gt;P&lt;/strong&gt; &lt;strong&gt;X&lt;/strong&gt;
&lt;strong&gt;Q&lt;/strong&gt; changes but &lt;strong&gt;P&lt;/strong&gt;
&lt;strong&gt;Q&lt;/strong&gt; is always equal in length to &lt;strong&gt;A&lt;/strong&gt;
&lt;strong&gt;X&lt;/strong&gt;.&lt;/p&gt;
&lt;p&gt;As a further challenge, you may like to prove that the position
of X is given by: ( &lt;strong&gt;C&lt;/strong&gt; &lt;strong&gt;X&lt;/strong&gt;/
&lt;strong&gt;X&lt;/strong&gt; &lt;strong&gt;B&lt;/strong&gt;) = ( &lt;strong&gt;C&lt;/strong&gt;
&lt;strong&gt;A&lt;/strong&gt;/ &lt;strong&gt;A&lt;/strong&gt;
&lt;strong&gt;B&lt;/strong&gt;)².&lt;/p&gt;
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  <title>Shrink</title>
  <description>X is a moveable point on the hypotenuse, and P and Q are the feet
of the perpendiculars from X to the sides of a right angled
triangle. What position of X makes the length of PQ a minimum?</description>
  <spec_group>2D Geometry, Shape and Space
    <specifier>Perpendicular lines</specifier>
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  <spec_group>Sequences, Functions and Graphs
    <specifier>Maximise/minimise/optimise</specifier>
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  <spec_group>2D Geometry, Shape and Space
    <specifier>Rectangles</specifier>
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    <specifier>Dynamic geometry</specifier>
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