Curved square

6328 /www/nrich/html/content/id/6328/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> A square of side length 1 has an arc of radius 1 drawn from each of its corners, as in the following diagram. The arcs intersect inside the square at four points, to create the shaded region. <div style="text-align: center;"><mdo:image src="CurvedSquare.png"></mdo:image></div> What is the area of the largest square that can be completely contained within the shaded region? Is this a good estimate of the actual shaded area? What is the exact area of the central shaded region? How did that compare to your estimate? Can you find more than one method to work out the exact area? <a href="http://nrich.maths.org/public/viewer.php?obj_id=6830&part=">Click here for a poster of this problem.</a> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why do this problem?</h3> <a href="http://nrich.maths.org/public/viewer.php?obj_id=6328&part=">This problem</a> draws together coordinate geometry, equations of circles, and surds, and can also be approached using integration. <h3>Possible approach</h3> <div>Show the <a href="/content/id/6328/CurvedSquare.png">diagram</a> (also available as a <a class="powerpoint" href="/content/id/6328/CurvedSquare.ppt">PowerPoint slide</a>). Give students time to study the diagram, and make notes about what they know and what they can work out. Pose the problem of finding the shaded area, and after some thinking time bring the class together to discuss possible strategies. Two possible strategies are outlined in the following worksheets: <a class="pdflink" href="/content/id/6328/CurvedSquare1.pdf">Sectors Method</a> <a class="pdflink" href="/content/id/6328/CurvedSquare2.pdf">Integration Method</a> You could outline the general methods to the class and give them time to solve the problem for themselves, offering the worksheet as a prompt if they get stuck. Allow time at the end of the lesson for students to compare the different approaches.</div> <h3>Key questions</h3> <div>What information will we need to find the area?</div> <div>What symmetries are present in the diagram? Does the area split up in any obvious ways?</div> <div>Which would be the easiest arcs to work with? Why?</div> <div>How can integration be used to find the area?</div> <h3>Possible extension</h3> <div>Find the areas of the other parts of the diagram.</div> <div>Set up a similar problem using parabolas instead of circles.</div> <h3>Possible support</h3> <div>The worksheets suggest a suitable coordinate grid, and offer prompts for students to follow.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">You can solve this using basic properties of areas of circles, coordinate geometry and Pythagoras's theorem or simple trigonometry. <a href="/content/id/6328/CurvedSquare1.pdf">This worksheet</a> suggests a possible method. You can also solve it using calculus. <a href="/content/id/6328/CurvedSquare2.pdf">This worksheet</a> suggests a possible method. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> The points of intersection are $(\frac{\sqrt{3}}{{2}}, 0.5), (1-\frac{\sqrt{3}}{{2}}, 0.5)$ and the same with x, y interchanged Estimate for the centre area is based on a square From coordinate geometry, the squared side length of the square (and thus the area) linking the points of intersection is $2-\sqrt{3}$ There were several interesting solutions to this problem, using a variety of different approaches, along with many lovely images. We are sorry that we couldn't include all of the solutions in full! The answers were: The area of the shaded area is $$ \frac{\pi}{3}+1-\sqrt{3}\approx 0.31517 $$ The area of the largest square which can be inscribed is $$ 2-\sqrt{3}\approx 0.267949 $$ The error is about $15\%$. InNyeong from Nanjing International School placed the shape on a coordinate grid to express all of the lines in algebraic form and solved for the resulting intersections numerically, with a beautifully written up solution: <mdo:image alt="" height="324" src="Chang1.jpg" width="557"></mdo:image> <mdo:image alt="" height="366" src="Chang2.jpg" width="553"></mdo:image> <mdo:image alt="" height="177" src="Chang3.jpg" width="525"></mdo:image> Ammar Ali followed a similar approach to InNyeon. although his solution was entirely algebraic. He found the equations of the curves and then solved algebraically, finding the coordinates for the points of intersection to be $$ \left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)\quad \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\quad \left(1-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\quad \left(\frac{1}{2}, 1-\frac{\sqrt{3}}{2}\right) $$ The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $$ d=\sqrt{\left(x_1-x_2)^2+(y_1-y_2)^2\right)} $$ By using this formula we can find the squared distance $L^2$ between neighbouring points: $$ L^2 = 2-\sqrt{3} $$ This is the area of the largest possible square that fits in the centre. Ammar then proceeded with calculating the integrals of the curves exactly, finding the correct expression for the areas. We liked this method a lot because it could be generalisable to curves other than circles (such as ellipses). To perform the calculation, Amman used the knowledge of the points of intersection, integrals and the additive properties of areas <mdo:image alt="" height="126" src="Amman.jpg" width="524"></mdo:image> Yike provided an interesting solution based upon the following modified image: <mdo:image alt="" height="300" src="Yike.jpg" width="269"></mdo:image> Trigonometry shows that $\lambda = 60^\circ$ and $\phi = 30^\circ$. The cosine rule on triangle $OP_1P_2$ gives $Q=(P_1P_2)^2=2-\sqrt{3}$, which is also the area of the largest inscribed square (since it is the side-length squared!) To find the exact area $C$, note that the area of triangle $P_1P_2B = 0.5 r^2\sin(\lambda -\phi)=0.5 \sin(30^\circ)=0.25$. So, $$ C= Q+4\left(\frac{(\lambda-\phi)\pi r^2}{360^\circ}-0.25\right) = 2-\sqrt{3}+4\left(\frac{\pi}{12}-0.25\right) =1-\sqrt{3}+\frac{\pi}{3} $$ Bugboy sent a classic geometric proof based on areas of sectors of circles, a very similar approach to that used by Roy from Nanjing International School: The largest square has vertices at the 4 points where two arcs meet within the square. Because the arcs are 1 unit, the same as the side length of the square, the point where arcs meet and the opposite side of the square form an equilateral triangle. Therefore the distance between the point and the far edge via the perpendicular bisector (which also goes through the opposite corner of the inside square) is$\frac{\sqrt{3}}{2}$. The distance between a corner and the opposite side, plus the distance between the opposite corner and the correspondingly opposite side is $2\frac{\sqrt{3}}{2}=\sqrt{3}$. Each $\frac{\sqrt{3}}{2}$ line is made of two parts: the distance between the two corners of the inside square, and then the distance to the side of the outer square. We have two of each in $\sqrt{3}$. The entire distance from one side to another of the outer square is $1$, and this consists of 2 end sections and 1 section between the two corners. Therefore $\sqrt{3} -1$ is just the one diagonal length of the inner square. The diagonal of any square is the side length multiplied by $\sqrt{2}$, so in this case the side is $\frac{\sqrt{3} -1}{\sqrt{2}}$. Square that for the area of the inner square: $\frac{4-2\sqrt{3}}{2} =2-\sqrt{3}$. In the actual shape, we missed out the segments on the edges of the square. The segment they are part of has an angle of $30$ degrees (proof: the angle from the two sides that meet at the corner where the sector is coming from to the points at the end of the segment are $30$ degrees, as the other side of the $90$ degree angle is part of an equilateral, as shown earlier. Hence it's $60$ degrees, and the angles on either side of the sector angle are $30$ degrees, leaving $90-30-30=30$ degrees for the sector angle) so the area of the sector is $\pi\times 1^2\times 30/ 360=\frac{\pi}{12}$. The triangle within the sector that complements the segment to make the sector has two sides length 1 with included angle of $30$. Hence the area is $\frac{1}{2}\times 1\times 1\times \sin30=\frac{1}{4}$. Therefore each segment is $\frac{\pi}{12}-\frac{1}{4}$. There are 4 segments, so the total area is $\frac{\pi}{3}-1$. Therefore the total area of the shape is $\frac{\pi}{3}-1+2-\sqrt{3}=1+\frac{\pi}{3}-\sqrt{3}$. Finally, Scott from Pencoed Comprehensive extracted the key geometrical information very clearly in a good-sized and clear diagram (reduced in size here), so essential to any work in geometry: <mdo:image alt="" height="318" src="scott.jpg" width="310"></mdo:image> </mdoxml> 2 3 0 0 0 0 1 Curved square Can you find the area of the central part of this shape? Can you do it in more than one way? Pre-Calculus and Calculus Integration Pre-Calculus and Calculus Integration by substitution Coordinates and Coordinate Geometry Cartesian equations of circles ajk44 live for solution Stage 5 Core Mapping Document Trigonometry AS Secondary Mapping Document DisplayCabinet sfh10 Teacher Workshop ajk44 Consolidation