317 /www/nrich/html/content/00/06/15plus2/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"><br></br> <p><em>You may wish to try the problem <a href="http://nrich.maths.org/public/viewer.php?obj_id=312&amp;part=index">Bendy Quad</a> first.</em></p> <p><mdo:image src="bendy%20quad.png" style="float: left;"></mdo:image></p> <p> </p> <p>Four rods are hinged at their ends to form a quadrilateral with fixed side lengths.<br></br> Show that the quadrilateral has a maximum area when it is cyclic.</p> <p> </p> <p> </p> <p> </p></mdoxml> <mdoxml version="1.0"><br></br> <div>Four rods are hinged at their ends to form a convex quadrilateral with fixed side lengths. Show that the quadrilateral has a maximum area when it is cyclic. Sue Liu from Madras College, St Andrews used two methods to solve this.</div> <br></br> <br></br> <table border="0"> <tbody> <tr> <td> <div>${\bf METHOD}$ ${\bf 1}$</div> <br></br> <div>First she applied Brahmagupta's formula for the area of a quadrilateral:</div> <br></br> <div style="text-align: center;">$ \delta = \sqrt{(s - a)(s - b)(s - c)(s - d) - abcd \cos^2 \beta}$</div> <br></br> <div>where $ s = \frac{1}{2}(a + b + c + d)$ and $\beta = \frac{1}{2}(A + C)$ or $ \frac{1}{2}(B + D)$.</div> <br></br> <div>Hence the area is clearly the greatest when $abcd \cos^2 \beta$ is least. Since $\cos^2 \beta$ is always positive, this value is least when $\beta $ is $90$ degrees as $\cos 90^{\circ} = 0$. Hence $\frac{1}{2}(A + C) = 90 $ and so $A + C = 180$ showing that the opposite angles in the quadrilateral add up to $180^{\circ}$ and so the area of a quadrilateral with fixed lengths of sides is greatest when it is cyclic.</div> <br></br> <div>This method gives a proof of the required result but you have to assume Brahmagupta's formula and Sue's second method uses only the formula for the area of a triangle.</div> <br></br> <br></br> ${\bf METHOD}$ ${\bf 2}$<br></br> <br></br> <div>The area of the quadrilateral $ABCD$ can be expressed as the sum of the areas of triangle $ABD$ and $BCD$. Let the area of $ABCD$ be $\Delta$. Then \[ \Delta = \frac{1}{2} ad\sin A + \frac{1}{2} bc \sin C \] Thus $$\frac {d\Delta}{dA} = \frac{1}{2} ad\cos A + \frac{1}{2} bc \cos C \frac {dC}{dA}.\quad (1)$$</div> <br></br> <div>From the Cosine Rule, $$a^2 + d^2 - 2ad\cos A = b^2 + c^2 - 2bc\cos C,$$</div> <br></br> <div>Hence, (differentiating both sides with respect to $A$), $$2ad\sin A = 2bc\sin C \frac {dC}{dA}.\quad (2)$$</div> <br></br> <div>From (1) and (2), $$\eqalign{ \frac{d\Delta}{dA} = \frac{1}{2} ad\cos A + \frac{1}{2} bc \cos C{ad\sin A \over bc\sin C}\\ = \frac {ad\cos A \sin C + \sin A \cos C}{2\sin C}\\ = \left(\frac{ad}{2}\right) \frac{\sin (A + C)}{\sin C} }.$$ Hence, for the maximum area, $\sin (A + C)=0$ and $A+C=\pi$ which makes the quadrilateral cyclic.</div> <br></br> <div>We can show that this gives the maximum and not the minimum value of $\Delta$ by finding the second derivative.</div> </td> <td><mdo:image height="221" width="173" alt="" src="fig1.gif"></mdo:image></td> </tr> </tbody> </table> <br></br> <br></br></mdoxml> <mdoxml version="1.0"><p>How would the area change as one of the angles is changed?</p> <p>You could try splitting the quadrilateral into two triangles. Can you find an expression for the length of the diagonal in two different ways?</p> <p> </p></mdoxml> 2 5 0 0 0 0 1 Four rods are hinged at their ends to form a quadrilateral with fixed side lengths. Show that the quadrilateral has a maximum area when it is cyclic. Sequences, Functions and Graphs Maximise/minimise/optimise Trigonometry Cosine rule 2D Geometry, Shape and Space Quadrilaterals Trigonometry Area formulae