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2011-02-01T00:00:01
<mdoxml version="1.0"><p><br></br>
A triangle PQR, right angled at P, slides on a horizontal floor<br></br>
with Q and R in contact with perpendicular walls. What is the locus<br></br>
of P?</p>
<p> </p></mdoxml>
<mdoxml version="1.0"><br></br>
The locus of the point $P$ is a straight line as the vertices $Q$
and $R$ slide along the walls. <br></br>
<br></br>
<table border="0">
<tbody>
<tr>
<td><mdo:image height="228" width="240" src="sol1.gif" alt=""></mdo:image></td>
<td>
<div><span style="font-weight: bold;">Sue Liu</span> of Madras
College, St Andrew's sent this solution to the problem. Without
loss of generality we can let the length of $QR$ be 1 unit, and
take a coordinate system with the origin at $O$ and axes along $OR$
and $OQ$.</div>
<br></br>
<div>If $\angle PQR = \alpha$, where $0 < \alpha < 90^\circ$,
then $PQ = \cos \alpha$ and $PR = \sin \alpha.$ Let $\angle QRO =
\theta $ where $0 \leq \theta \leq 90^\circ$. Then, from the right
angled triangles $PSQ$ and $PTR$, we have $\angle PRT = \angle QPS
= \alpha - \theta$, and hence we can write down the coordinates of
the point $P$.</div>
</td>
</tr>
</tbody>
</table>
<br></br>
<div>
<div class="math">\begin{eqnarray} x &=& \cos \alpha \cos
(\alpha - \theta) \\ y &=& \sin \alpha \cos (\alpha -
\theta). \end{eqnarray}</div>
<div>We see that $${y\over x}= {\sin \alpha \cos (\alpha -
\theta)\over \cos \alpha \cos (\alpha - \theta)}= \tan \alpha.$$
and so $P$ lies on the straight line $y = x\tan \alpha$.</div>
<br></br>
<div>The position $(x,y)$ depends only on $cos(\alpha - \theta)$,
$\alpha$ being a constant, and $\theta$ a variable. The distance of
the point $P$ from $O$ is given by $$OP^2 = x^2 + y^2 =
\cos^2(\alpha - \theta)(\cos^2\alpha + \sin^2\alpha) =
\cos^2(\alpha - \theta).$$ Hence $OP = \cos(\alpha - \theta)$ which
is a maximum when $\cos(\alpha - \theta) = 1$, that is when $\alpha
= \theta$. This occurs when $OQPR$ is a rectangle as shown in the
diagram.</div>
</div>
<br></br>
<mdo:image src="sol2.gif" alt=""></mdo:image><br></br>
<br></br>
We get an even simpler method of solution by using the fact that
the angles $QOR$ and $QPR$ are both 90 degrees so that $OQPR$ is a
cyclic quadrilateral with $PR$ as a chord. We have $\angle POR =
\angle PQR = \alpha$ because these two angles are subtended by the
same chord of the circle. This shows that $\angle POR$ is constant
and hence that the locus of $P$ is the straight line $y = x \tan
\alpha.$ <br></br>
<br></br>
What can you say about the locus of $P$ if the triangle $PQR$ is
not a right angled triangle? <br></br></mdoxml>
<mdoxml version="1.0"><p>You might look for a cyclic quadrilateral.</p>
<p> </p>
<p>We have constructed this in geogebra to help</p>
<p> </p>
<p> </p>
<div style="float: left;"><mdo:applet height="370" width="450" code="geogebra.GeoGebraApplet" archive="http://www.geogebra.org/webstart/geogebra.jar" datafile=""><param name="filename" value="http://nrich.maths.org/content/99/09/15plus5/setsquare.ggb" ></param><param name="framePossible" value="false" ></param><param name="showResetIcon" value="true" ></param><param name="enableRightClick" value="false" ></param><param name="showMenuBar" value="false" ></param><param name="showToolBar" value="false" ></param><param name="showToolBarHelp" value="false" ></param><param name="showAlgebraInput" value="false" ></param></mdo:applet></div>
<p style="clear: both; padding-top: 20px;"> </p>
<p><span style="font-size: small;">Created with <a href="http://www.geogebra.org" target="_blank">GeoGebra</a></span></p>
<div class="framework"><br></br>
NOTES AND BACKGROUND<br></br>
<p>You might like to download your own free copy of GeoGebra from the link above and draw this dynamic diagram for yourself. You will find it easy to get started on GeoGebra with the <a href="http://www.geogebra.org/cms/index.php?option=com_content&task=blogcategory&id=75&Itemid=61">Quickstart</a> guide for beginners.</p>
</div>
<p><br></br>
<br></br>
</p></mdoxml>
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Set Square
A triangle PQR, right angled at P, slides on a horizontal floor
with Q and R in contact with perpendicular walls. What is the locus
of P?
Information and Communications Technology
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Using, Applying and Reasoning about Mathematics
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Sequences, Functions and Graphs
Loci
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Learning through exploration
2D Geometry, Shape and Space
Circle theorems
2D Geometry, Shape and Space
Quadrilaterals