Steel Cables

7760 /www/nrich/html/content/id/7760/ 1 2011-09-16T11:56:22 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h4>If you are a teacher, click <a href="http://nrich.maths.org/7760&part=note">here</a> for a version of the problem suitable for classroom use, together with supporting materials. Otherwise, read on...</h4> Cables can be made stronger by compacting them together in a hexagonal formation. Here is a 'size 5' cable made up of 61 strands: <mdo:image alt="cable cross section, one cable in the middle with four rings of cables around it" src="cable1.png"></mdo:image> How many strands are needed for a size 10 cable? How many for a size n cable? Can you justify your answer? Once you've had a go at the problem, click below to see how some 15 year old students worked on it. Can you explain their reasoning? Group 1 <div class="toggle"><mdo:image alt="student's picture of hexagon split into three quadrilaterals, two 5*5 rhombuses and a 4*4 rhombus" src="student3a.png"></mdo:image> <mdo:image alt="table showing dimensions of the three quadrilaterals and total T for a size 2 up to 10 cable and a size n cable" src="student3b.png"></mdo:image> <mdo:image alt="T=n^2+n^2-n+n^2-2n+1 = 3n^2-3n+1 or 3n(n-1) +1" src="student3c.png"></mdo:image></div> Group 2 <div class="toggle"><mdo:image alt="student's picture of cable with horizontal arrows showing row lengths n, n+1, n+2 up to 2n-1 in the middle and then decreasing back down to n" src="Student1a.png" style="padding-bottom: 220px;"></mdo:image> <mdo:image alt="student's method is adding up pairs of rows to make n pairs of (3n-1), with the 2n-1 row counted twice, giving a total 3n(n-1) +1" src="student1b.png"></mdo:image></div> Group 3 <div class="toggle"><mdo:image alt="student's picture of hexagon split into six triangles" src="student2a.png" style="padding-bottom: 200px;"></mdo:image> <mdo:image alt="student's method is to use the formula for the (n-1)th triangular number n(n-1)/2, multiply by six, and then add 1 for the cable in the centre" src="student2b.png"></mdo:image></div> Group 4 <div class="toggle"><mdo:image alt="student's picture of hexagon showing four rings and one cable in the centre" src="student4a.png"></mdo:image> <mdo:image alt="1, 6*1, 6*2, 6*3, 6*4 ... 6*(n-1). We noticed that the area of each ring followed this pattern. To find the total we needed to add the area of each ring. 1 + 6*1 + 6*2 + 6*3 ... 6(n-1)=1+6(1+2+3+...+(n-1) 1+6(n(n-1)/2) 1+3n(n-1)" src="student4b.png"></mdo:image></div> Which of the four approaches makes the most sense to you? What do you like about your favourite approach? Can you think of any other approaches? <div class="framework">Notes and Background Hexagonal packings are often chosen for strength or efficiency. To read more about packings, take a look at the Plus articles <a href="http://plus.maths.org/content/mathematical-mysteries-keplers-conjecture" target="_blank">Mathematical Mysteries: Kepler's Conjecture</a> and <a href="http://plus.maths.org/content/newton-and-kissing-problem" target="_blank">Newton and the Kissing Problem</a></div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">It was great to see so many different ways of approaching this! Here are some of the responses that we received from students. Calvin, from Ethiopia, gave the following interesting solution: At first I noticed a pattern: the number of strands per cable was going up by multiples of 6 every time. Then I plotted the data for sizes 1-10: <table border="1" cellpadding="1" cellspacing="1" style="width: 500px;"> <tbody> <tr> <td>Size</td> <td>1</td> <td>2</td> <td>3</td> <td>4</td> <td>5</td> <td>6</td> <td>7</td> <td>8</td> <td>9</td> <td>10</td> </tr> <tr> <td>Strands</td> <td>1</td> <td>7</td> <td>19</td> <td>37</td> <td>61</td> <td>91</td> <td>127</td> <td>169</td> <td>217</td> <td>271</td> </tr> </tbody> </table> and graphed them. The result was a parabola, so I found its equation and that was the answer. The answer is s = 3n2 - 3n + 1, where s is the number of strands required and n is the size of the cable. Great! But how did Calvin find that formula? Patrick, from Otterbourne, suggests a way: The sequence starts off: 1, 7, 19, 37, 61, ... If we take the difference between successive terms, we get the sequence 6, 12, 18, 24, ... If we take successive differences again, we get 6, 6, 6, ... This is constant, so we must have a quadratic equation s = an$^2$ + bn + c. In fact, half of 6 is 3, so we know that a = 3. Therefore s = 3n$^2$ + bn + c In order to find b and c, we can substitute values in: (n = 1, s = 1) 1 = 3 + b + c (n = 2, s = 7) 7 = 12 + 2b + c and we can solve these simultaneous equations to get b = -3 and c = 1. Szymon from Priory Academy LSST sent us a similar solution: We know (similar to the above) that s = 3n$^2$ + bn + c. We can tabulate some values: <table border="1" cellpadding="1" cellspacing="1" style="width: 500px;"> <tbody> <tr> <td>Strands (s)</td> <td>1</td> <td>7</td> <td>19</td> <td>37</td> <td>61</td> <td>91</td> </tr> <tr> <td>3n$^2$</td> <td>3</td> <td>12</td> <td>27</td> <td>48</td> <td>75</td> <td>108</td> </tr> <tr> <td>Difference (s - 3n$^2$)</td> <td>-2</td> <td>-5</td> <td>-8</td> <td>-11</td> <td>-14</td> <td>-17</td> </tr> </tbody> </table> Now we can work out the successive differences between these terms as before. These are very nice solutions if you've seen this sort of method before. If not, don't worry: there are other ways of going about it! Matthew, from Saint Peter and Paul's, sent us this interesting 'counting argument': We can view the size 5 hexagon as being made up from 6 triangles of side length 5. Each of these triangles contains 15 strands, so we get 6$\times$15 = 90. But wait - these triangles overlap, so we've overcounted! Adjacent triangles overlap in 5 strands, one of which is the centre, and all of the triangles overlap in this centre strand; so we need to subtract 1 from our total for each strand that we've double-counted, and we've counted the centre 6 times so we must subtract 5 from our total for that: 90 - 6$\times$4 $-$5 = 61 For the size n hexagon, the same thing happens: The hexagon is made from 6 triangles of side length n, so each triangle contains n(n+1)/2 strands (the nth triangular number), so 6 triangles give 6n(n+1)/2 strands, or 3n(n+1) strands. Now we need to adjust our answer for our overcounting: Adjacent triangles overlap in (n-1) strands plus the centre strand, so we need to subtract (n-1) for each pair of adjacent triangles (and there are 6 of them), and then we need to subtract 5 for the centre. So we get 3n(n+1) - 6(n-1) - 5 strands, which is 3n$^2$ - 3n + 1. Brady from Canada sent us a similar solution: <mdo:image src="Cables%20Problem.png"></mdo:image> We can break a size n hexagon into 3 big triangles (of side length n) and 3 small triangles (of side length n-2). Since the 3 big triangles meet in the middle, when we add up the number of strands in each we have to subtract 2 because we have overcounted the middle strand by 2: $\frac{3n(n+1)}{2}$ + $\frac{3(n-2)(n-1)}{2}$ - 2 = 3n$^2$ - 3n + 1 Karl from King's School, Grantham, sent us this slick solution: <mdo:image alt="" src="triangles.png" style="width: 400px; height: 581px;"></mdo:image> The middle row contains 2n-1 strands. The top part can be seen as a large triangle (of side length 2n-2) minus a small triangle (of side length n-1), giving in total: $\frac{(2n-2)(2n-1)}{2}$ - $\frac{(n-1)(n)}{2}$ = $\frac{3n^2 - 5n + 2}{2}$ The bottom part is the same as the top. This gives us a total of (3n$^2$ - 5n + 2) + (2n - 1) = 3n$^2$ - 3n + 1. Crayola, from the British School in Tokyo, knew a short-cut: We have to work out n + (n+1) + (n+2) + ... + (n+x) + (n+x-1) + ... + (n+1) + n, where n+x is the longest row of the hexagon. What is x? Well, there are 2x + 1 terms in this sum, one for each row, and the hexagon has 2n-1 rows, so x = n-1. That is, we have to work out n + (n+1) + ... + (2n - 2) + (2n - 1) + (2n - 2) + ... + (n+1) + n. We can regroup these as follows: 2[n + (n+1) + ... + (2n-2)] + (2n-1). Now, a theorem of Gauss tells us that, to work out n + (n+1) + ... + (2n-2), we simply have to add the first and last number together, multiply the result by the number of terms in the sum, and divide by 2: 2$\times$$\frac{(3n-2)(n-1)}{2}$ + (2n-1) = 3n$^2$ - 3n + 1 Well spotted! Of course, if you didn't know this result, no problem - you can work it out simply by one of the other solutions above. Gauss himself probably figured it out by drawing pictures like Karl's triangles above!</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <h3>Why do this problem?</h3> Many students are accustomed to using number patterns in order to generalise. <a href="http://nrich.maths.org/7760">This problem</a> offers an alternative approach, challenging students to consider multiple ways of looking at the structure of the problem, through making sense of other people's approaches, an important part of working mathematically. The powerful insights from these multiple approaches can help us to derive general formulae, and can lead to students' appreciation of the equivalence of different algebraic expressions. This problem and <a href="/6675">Christmas Chocolates</a> are essentially the same question but presented in a different way. <h3>Possible approach</h3> Start by showing this image: <mdo:image src="cable1.png"></mdo:image> "Cables can be made stronger by compacting them together in a hexagonal formation. Here is a 'size 5' cable. Can you work out, without counting every strand, how many strands it contains?" Give students a short time to consider this and then discuss their ideas in pairs, before bringing the class back together to share their different methods. Hand out <a href="/content/id/7760/steel%20cables%20intro.pdf">this worksheet</a>, together with <a href="/content/id/7760/steel%20cables%20templates.pdf">these templates</a>, and ask: "How many strands are needed for a size 10 cable?" "While you are working on this, keep in mind how your method could be adapted to work out the number of strands needed for any size of cable." While students are working, circulate and observe any interesting methods that students are using. When the class is ready, bring them back together and invite those students with an interesting method to explain what they did to the rest of the group. "Here is some work done by groups of 15 year old students who were asked to find a formula to work out the strands needed for a size n cable." Arrange the class in groups of four, and hand out <a href="/content/id/7760/steel%20cables-groups.pdf">this worksheet.</a> "I'd like each of you to take one of the sheets and make sense of the method used by one of the groups. Then when you are ready, you need to explain the method to the rest of your group. Finally, the group needs to decide which of the four approaches makes the most sense to you, and be prepared to justify your choice to the rest of the class." To finish off, each group can explain to the rest of the class which approach they chose, and why. Here are some key questions that could be used to help groups who are struggling to make sense of the different methods: "Is there a quick way of adding up all the numbers from 1 to n?" "Look at the picture for a size 5 cable. What might the group have drawn for a size 6 cable?" <h3> </h3> <h3>Possible extension</h3> Challenge students to come up with alternative, elegant ways of computing the number of strands in a size n cable. The problems <a href="http://nrich.maths.org/6590&part=">Summing Squares</a> and <a href="http://nrich.maths.org/325&part=">Picture Story</a> lead to formulae for some intriguing sequences through analysis of structure. <h3>Possible support</h3> <a href="http://nrich.maths.org/2290&part=">Seven Squares</a> gives lots of simple contexts where formulae emerge by looking at structure rather than number sequences. <a href="http://nrich.maths.org/2274&part=">Picturing Triangle Numbers</a>, <a href="http://nrich.maths.org/6703&part=">Mystic Rose</a>, and <a href="http://nrich.maths.org/6708&part=">Handshakes</a> introduce students to summing consecutive numbers, which is a feature of several of the solutions used by the groups in the problem. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Do you know a quick way of adding up all the numbers from 1 to n? If not, take a look at the problem <a href="/2274">Picturing Triangle Numbers</a>. If you are finding it difficult to make sense of the different groups' work, try to work out what they might have drawn for a size 6 cable. </mdoxml> 2 3 0 0 0 1 0 Steel Cables Some students have been working out the number of strands needed for different sizes of cable. Can you make sense of their solutions? Using, Applying and Reasoning about Mathematics Generalising Using, Applying and Reasoning about Mathematics Visualising Algebra Creating expressions/formulae Secondary Mapping Document Patterns and sequences US Secondary Mapping Document DisplayCabinet