259 /www/nrich/html/content/99/03/15plus4/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"><p>$A$ and $B$ are two fixed points on a circle and $RS$ is a variable diamater. What is the locus of the intersection $P$ of $AR$ and $BS$?</p></mdoxml> <mdoxml version="1.0"><br></br> <table> <tbody> <tr> <td><mdo:image alt="" src="fig1.gif"></mdo:image></td> <td><mdo:image alt="" src="fig2.gif"></mdo:image></td> </tr> </tbody> </table> <div>$\mathbf{A}$ and $\mathbf{B}$ are fixed points on a circle, while $\mathbf{RS}$ is a variable diameter. $\mathbf{P}$ is the intersection of $\mathbf{AR}$ and $\mathbf{BS}$. We make the conjecture that the locus of $\mathbf{P}$ isa a second circle through the points $\mathbf{A}$ and $\mathbf{B}$.</div> <br></br> <div><span style="font-weight: bold;">Edward Wallace of Graveney School, Tooting</span> sent a very good proof of this conjecture.</div> <br></br> <div>First consider the two cases illustrated in the given diagrams. It is impossible to have $\mathbf{R}$ and $\mathbf{S}$ both on the minor arc of the circle. The other two cases, where $\mathbf{R}$ and $\mathbf{S}$ exchange positions, and $\mathbf{P}$ is outside the original circle, are proved similarly.</div> <br></br> <div style="font-weight: bold;">Case 1</div> <br></br> <div>Both $\mathbf{R}$ and $\mathbf{S}$ are on the major arc $\mathbf{AB}$ so that $\mathbf{P}$ is inside the original circle.</div> <br></br> <div style="font-weight: bold;">Case 2</div> <br></br> <div>$\mathbf{S}$ is on the minor arc $\mathbf{AB}$ and $\mathbf{R}$ on the major arc so that $\mathbf{P}$ is outside the original circle.</div> <br></br> <div>In Case 1, $\angle\mathbf{ARB}$ is unchanging (invariant) as $\mathbf{R}$ moves on the circumference because angles subtended on the circumference of a circle by a fixed arc of the circle are equal. Hence ($\angle \mathbf{ARB} = m^o$ (constant). Also $\angle \mathbf{RBS}=90^o$ because the angle subtended by a diameter is a right angle.</div> <br></br> <div>It follows that $\angle \mathbf{RPB}=(90 - m)^o$ because the angles of the triangle $\mathbf{BRP}$ add up to $180^o$.</div> <br></br> <div>Hence $\angle \mathbf{APB}$ is constant and equal to $(90 + m)^o$ because angles on a line add up to $180^o$. This is a necessary and sufficient condition for P to lie on the circumference of a circle of which $\mathbf{AB}$ is a chord. Let us call this arc $\mathbf{AP_{1}B}$ to emphasise the fact that, in this case, $\mathbf{P_1}$ is inside the original circle.</div> <br></br> <div>In Case 2, exactly the same argument applies, giving $\angle \mathbf{ARB} = m^o$ (the same value as in Case 1) and $\angle\mathbf{RBS}=90^o$, but in this case $\angle\mathbf{APB}$ is the same angle as $\mathbf{RPB}$ so $\angle\mathbf{APB}=(90 - m)^o$ (again constant).</div> <br></br> <div>As before, this is a necessary and sufficient condition for $\mathbf{P}$ to lie on the circumference of a circle of which $\mathbf{AB}$ is a chord. We call this arc $\mathbf{AP_{2}B}$ to emphasise the fact that now $\mathbf{P_2}$ is outside the original circle.</div> <br></br> <div>Note that:</div> <div>$\angle\mathbf{AP_{1}B} + \angle\mathbf{AP_{2}B} = (90 + m)^o + (90 - m)^o = 180^o$ and hence $\mathbf{AP_{1}BP_{2}}$ is a cyclic quadrilateral which confirms that the locus of $\mathbf{P}$ is a single circle (and not two arcs of different circles). </div> <br></br></mdoxml> <?xml version="1.0" encoding="ISO-8859-1"?> <mdoxml version="1.0"><br></br> <p>Think about angle $APB$.</p> <p>You might like to use dynamic geometry software to help with this problem. You can download geogebra for free from geogebra.org. We have pre-drawn a diagram if this is useful to you.</p> <div style="float: left;"><mdo:applet height="370" width="370" code="geogebra.GeoGebraApplet" archive="http://www.geogebra.org/webstart/geogebra.jar" datafile=""><param name="filename" value="https://nrich.maths.org/content/99/03/15plus4/fixingit.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> </mdoxml> <mdoxml version="1.0"><p>dfg<a href="/content/99/03/15plus4/fixingit.ggb">fixingit.ggb</a> hdfgh</p></mdoxml> 2 3 0 0 0 0 1 A and B are fixed points on a circle and PQ is a variable diameter. Make and prove a conjecture about the locus of the intersection of AQ and BP. Information and Communications Technology Dynamic geometry Using, Applying and Reasoning about Mathematics Making and proving conjectures 2D Geometry, Shape and Space Circle theorems