Exploring Concepts with Cubes A resource book
ACTIVITY 1 Gauss s method Gauss s method is a fast and efficient way of determining the sum of an arithmetic series. Let s illustrate the method using the sum of the first 100 natural numbers, i.e. the series 1 + 2 + 3 + 4 +... + 98 + 99 + 100. Gauss s method involves writing down the sequence to be summed twice, once forwards and once backwards, carefully aligning the terms as follows: 1 + 2 + 3 +... + 98 + 99 100 + 99 + 98 +... + 3 + 2 + 1 + 100 The two terms in each column are then added, giving a sum of 101 for each of the 100 columns. This gives a total tally of 10100. However, this is double the required answer, so the sum of the original sequence is 5050. 1 + 2 + 3 +... + 98 + 99 100 + 99 + 98 +... + 3 + 2 + 1 + 100 101 + 101 + 101 +... + 101 + 101 + 101 Sum of first 100 natural numbers = 101x100 2 = 5050 Use the wooden cubes to investigate Gauss s method in order to develop a visual appreciation for why the method works. Begin with the series 1+2+3+4+5 as shown below, and then investigate on your own.
ACTIVITY 2 Sum of cubes a) Complete the following table: Column A Column B 1 3 + 2 3 = 1 + 2 = 1 3 + 2 3 + 3 3 = 1 + 2 + 3 = 1 3 + 2 3 + 3 3 + 4 3 = 1 + 2 + 3 + 4 = b) Describe in words any observed patterns. c) Use the wooden cubes to investigate the patterns you have observed. An example is shown below to get you started. = d) Determine 1 + 2 + 3 + 4 + K + 98 + 99 + 100 using the S n formula for an arithmetic n series: S n = [ 2a + ( n 1) d]. 2 e) Hence determine 3 3 3 3 3 3 1 + 2 + 3 + 4 + K + 98 + 99 + 100 3 f) Given that i n i= 1 3 = 4 n 4 3 n + 2 2 n n n +, prove that i = i 4 i= 1 i= 1 2 3.
ACTIVITY 3 The painted cube Use 27 small wooden cubes to build a large 3 3 3 cube as shown below. Now imagine you paint the outside of the large cube. Some of the smaller cubes will have paint on 3 of their faces. Some will have paint on only 2 of their faces. Others will only have paint on 1 of their faces, and some won t have any faces painted at all. Use the small wooden cubes to investigate for different sized larger cubes. Use the table below to keep track of your results. Use the final column of the table to generalise your results. Side length of large cube 2 3 4 5 n 0 faces painted 1 face painted 2 faces painted 3 faces painted Extension activities Justify each of your generalised formulas based on the physical structure of the large cube. Graph each of the functions in the final column on the same set of axes (for n 2 ). Do this by hand, and then check your graphs using appropriate graphing software (e.g. GeoGebra). Extend the investigation to n m p rectangular prisms.
ACTIVITY 4 Six block challenge This is a game of strategy for two players. Begin with a pile of 6 cubes. The two players, A and B, take turns in removing either one or two cubes from the pile. The winner is the player who takes the last cube or cubes. If player A makes the first move and each player always plays the best move, who wins the game? Devise a winning strategy for Player B i.e. the player who moves second. Investigate with different numbers of starting blocks. Does your winning strategy for player B always work? Under what conditions would there be a winning strategy for Player A?
ACTIVITY 5 Kayles Kayles, a name derived from the French word for skittles (quilles), is a game of strategy for two players. Kayles is played with a row of tokens (or blocks) which represent bowling pins. The two players, A and B, take turns in removing either a single block (a single bowling pin knocked over by a direct strike), or two adjacent blocks (a ball that strikes two pins and knocks them both down). The winner is the player who takes the last block or blocks. Begin with an arrangement of 12 blocks as shown below there are 11 adjacent blocks and an end block moved one space away from the others. By a careful strategy, player A (i.e. the first player to move) can always win. How? Now construct a row of 12 blocks without any gaps, as shown below. From this new starting arrangement (all blocks adjacent with no initial gaps) player A can once again win every time, irrespective of how long the initial row of blocks is. What is this winning strategy?
ACTIVITY 6 Mirror image The purpose of this activity is to engage students in mathematical communication and to provide an opportunity to use mathematical language in the context of giving instructions to another person. As with all vocabulary building, mathematical terms need to be used repeatedly in context in order for them to become internalised. This activity attempts to provide such a context. The mirror image activity is carried out between pairs of students. The two students in each pair must sit facing one another on opposite sides of a table. A partition is then placed between the two students. One of the students then uses the cubes to build a shape. This shape is built on one side of the partition so that the opposite member of the pair is unable to see it. The shape can be as simple or as complicated as desired. The student who built the shape now needs to describe the shape to the other member of the pair, who in turn must begin to build it on the opposite side of the partition such that the partition acts as a mirror plane. Only verbal communication may be used. If done successfully, the two shapes should ultimately be mirror images of each other Once the process has been completed, the members of each pair can swop roles.
ACTIVITY 7 Planes of symmetry Four cubes are arranged as shown alongside. By adding either one or two additional cubes different structures can be created. Some of these new structures have vertical planes of symmetry. Below are two possibilities. On the left a single additional cube could be added, while two additional cubes could be added to form the structure on the right. The respective planes of symmetry are indicated. When positioning the additional cubes, faces in contact with each other must fit exactly, and all cubes must have at least one face in contact with another cube. Also, cubes cannot be placed on top of one another. If these conditions are adhered to, there are 12 ways of adding either one or two additional cubes to the original 4-cube structure that result in a new structure that has a vertical plane of symmetry. Two solutions are shown above. Can you find the other 10 solutions? Extension activities: What if we allowed the additional cubes to be placed on top of existing cubes? What if we allowed the addition of 3 extra cubes? Investigate similar scenarios but with different starting configurations.
ACTIVITY 8 Sums of odd numbers Have you ever noticed the following? 1 + 3 = 4 1 + 3 + 5 = 9 1 + 3 + 5 + 7 = 16 1 + 3 + 5 + 7 + 9 = 25 4 = 2 9 = 3 16 = 4 25 = 5 2 2 2 2 Starting with 1, will the sum of consecutive odd numbers always be a perfect square? Take 16 cubes and arrange them into a 4 4 square (shown below on the left). Have you ever noticed that you can deconstruct the square into a series of L-shaped sections (shown below on the right)? What do you notice about how many cubes there are in each of these L-shaped sections? Now, based on the above observation, start with 25 cubes arranged as shown below (i.e. 1 + 3 + 5 + 7 + 9) and re-arrange them to form a perfect square. Will this method always work? Investigate!
ACTIVITY 9 Surface area generalisation Let us consider the six faces of each cube as each having an area of 1 unit 2. If 20 cubes are put together end-to-end as shown below, what is the surface area of the resulting shape? What would the surface area be if you put 100 cubes end-to-end? Now come up with a general formula to work out the surface area if you put n cubes end-to-end. What if we had a double row of cubes instead of a single row, as illustrated below? If the double row was 20 cubes long (i.e. 40 cubes in total), what would the surface area of the resulting shape be? If the double row was 100 cubes long (i.e. 200 cubes in total), what would the surface area of the resulting shape be? Now come up with a general formula to work out the surface area if the double row is n cubes long (i.e. 2 n cubes in total). Extension activities: What if we had a triple row of cubes instead of a single or double row? What about a quadruple row of cubes? Use the results of this investigation to determine a formula for the surface area of a single layer of cubes measuring n cubes long by m cubes wide.
ACTIVITY 10 Construction site A 3-dimensional structure is built from cubes. The top view, the front view as well as the view from one of the sides are all shown below. Front view Side view Top view What is the smallest number of cubes needed to build the structure? What is the biggest number of cubes needed to build the structure? In how many different ways could the structure be built so that the above profiles are correct? Extension activity: Build your own 3-dimensional structure out of cubes. Draw profiles from the front, top and side, similar to those shown above. You can decide whether you want to draw the side profile from either the right-hand side or the left-hand side. You can even draw both side profiles if you want. Once you ve drawn your profiles challenge a friend to build the structure using the least (or alternatively the greatest) number of cubes possible.
ACTIVITY 11 Cubes of all sizes Consider the diagram shown below. In this diagram there are 30 squares of varying size. Can you find them all? Let s extend this idea to a 3-dimensional arrangement of wooden cubes. Build a large 3 3 3 cube as shown below. This large cube contains 36 cubes of varying size. Can you find them all? Now let s try to generalise. How many cubes of all sizes are there in a 1 1 1 cube? What about a 2 2 2 cube? And a 4 4 4 cube? Try to find a systematic way of counting each of the different sized cubes. Now, without building a 5 5 5 cube, can you determine the total number of cubes of all sizes that it would contain? Finally, try to form an expression for the total number of cubes of all sizes in a large cube that is n cubes long, n cubes wide and n cubes high. Extension activity: Have you noticed the connection between this activity and Activity 2?
ACTIVITY 12 Traffic lights Take four cubes and cover each one with a strip of coloured paper. Use a different colour for each of the four cubes, for example red, blue, yellow and green. If we take two of the cubes (blue and yellow) and stack them vertically, there are only two possible arrangements: Blue at the bottom and yellow on top. Yellow at the bottom and blue on top. If we take three of the cubes (blue, yellow and red) there are six different ways to stack them vertically. What are these six different ways? One of them is shown alongside. If we now move on to four cubes (blue, yellow, red and green), how many different ways can they be stacked vertically, i.e. one upon the other? One way is shown alongside, but there are many other possibilities. Without writing down every possible arrangement, can you work out how many ways there would be of stacking five differently coloured cubes? What about six differently coloured cubes? Can you generalise for n differently coloured cubes?
ACTIVITY 13 Visible faces Place a single cube on a flat surface. There are five visible faces in total - one horizontal face on the top, and four vertical faces around the sides. We can t see the horizontal face at the bottom, so it isn t counted. Add a second column containing two cubes as shown alongside. There are now 11 visible faces in total - the three horizontal faces on the top, and the eight vertical faces around the sides. As before, we can t see the horizontal faces at the bottom, so they aren t counted. Now add more columns where each column contains one more cube than the previous column. The third column will contain three cubes, the fourth column four cubes, and so on. Complete the following table and try to generalise for a structure containing n columns. COLUMNS 1 2 3 4 5 6 7 n VISIBLE HORIZONTAL FACES 1 3 VISIBLE VERTICAL FACES 4 8 TOTAL NUMBER OF VISIBLE FACES 5 11 The general formula for the total number of visible faces should equal the sum of the general formulas for the visible horizontal and vertical faces respectively. Confirm that this is indeed the case. Can you explain why the number of visible vertical faces increases so regularly?
ACTIVITY 14 Area and volume This activity aims to develop the important conceptual link between volume and surface area. Begin by creating a grid on cardboard. The gap between lines should be 2cm i.e. the same width as the wooden cubes. Once you have created your grid, design a net that can be folded to create a closed 3- dimensional box. Now cut the net out of the grid and score along the lines that need to be folded. side side base side top side Now build the box from the grid (with the grid markings on the outside of the box) and use sticky tape to hold the sides of the box in place. You can now pack the box with cubes and explore the relationship between the volume and the total surface area. Rather than working in cm 2 and cm 3 just work with units 2 and units 3 where a unit is the width of a single cube.
ACTIVITY 15 Constant volume, changing area Take eight cubes and position them in a single layer so that any touching faces fit exactly against one another. Each cube must have at least one face in contact with another cube. Depending on how you position the eight cubes the total external surface area (TSA) will differ. The volume will of course remain constant. A few examples are shown below. Volume = 8 units 3 Volume = 8 units 3 Volume = 8 units 3 TSA = 28 units 2 TSA = 34 units 2 TSA = 32 units 2 Investigate with other arrangements and record your results in a sensible format. What arrangement gives the smallest TSA? What arrangement gives the largest TSA? Extend your investigation if we now allow cubes to be placed on top of one another as well (an example is shown below). Volume = 8 units 3 TSA = 32 units 2
ACTIVITY 16 Slicing the cube A number of interesting cross-sectional faces can be formed by making a single slice through a cube. For example, the slice made below results in a triangular cross-section. We can represent this slice diagrammatically as follows: Using the wooden cubes as a visual aid, but without actually cutting the cubes, investigate how you could slice the cube to form each of the cross-sectional faces listed below. A square An equilateral triangle A rectangle that is not a square A pentagon A hexagon An isosceles trapezium Keep in mind that each side of the cross-section will be formed by slicing through a face of the original cube. Keep track of your different slices by making use of diagrams similar to the ones shown above.
ACTIVITY 17 Tetrominoes and pentominoes Polyominoes are normally made by joining unit squares along their edges, but we can also make them using cubes by restricting ourselves to a single layer of cubes (i.e. cubes may not be placed on top of one another). Tetrominoes are composed of four cubes such that any touching faces fit exactly against one another and where each cube has at least one face in contact with another cube. There are five free tetrominoes. Free polyominoes are considered identical if they can be transformed into one other through any combination of translation, rotation or reflection. The five free tetrominoes are shown below. The two tetrominoes alongside are considered identical since the one is the mirror image of the other, or alternatively because they can each be changed into the other by being flipped over. Notice that some of the tetrominoes have 1, 2 and even 4 vertical planes of reflectional symmetry. Some also have rotational symmetry. Pentominoes are similar constructions composed of five cubes. There are 12 different free pentominoes. See if you can find all 12 different pentominoes. Keep a record of the different structures. Remember that (i) any touching faces must fit exactly against one another, (ii) each cube must have at least one face in contact with another cube, and (iii) cubes can only be placed in a single layer. Characterise each of the 12 different pentominoes in terms of their rotational symmetry and any vertical planes of reflectional symmetry they may have. Extension activity: Extend your investigation to hexominoes. There are 35 different free hexominoes.
ACTIVITY 18 Palindromic sums Have you ever noticed that 1 + 2 + 3 + 4 + 3 + 2 + 1 = 16 and 16 can be written as 4 2, i.e. a square number? We can use cubes to visualise why this works: Will this method work for other palindromic sums? How about 1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 or 1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 + 2 + 1? What about palindromic sums of even numbers e.g. 2 + 4 + 6 + 8 + 6 + 4 + 2? If we use the same visual approach with the cubes as previously used, the result will be a 4 by 8 rectangle. Will palindromic sums of even numbers always result in a rectangle where the sides are in a 1:2 ratio? Why? What about palindromic sums of odd numbers? Try 1 + 3 + 5 + 3 + 1. Notice that this sum gives an answer of 13 and that 13 = 3 2 + 2 2, i.e. the sum can be split into two square numbers. Will this always happen? Use the cubes to investigate. It might be helpful to refer to Activity 8.
ACTIVITY 19 From cubes to cuboids In a 2 2 2 cube there are eight 1 1 1 cuboids. There are also various cuboids of different dimensions. There are twelve 2 1 1 cuboids in three different orientations three of these are shown alongside. There are also six 2 2 1 cuboids in three different orientations three of these are shown alongside. Finally there is a single 2 2 2 cuboid. We can summarise all of the above as follows: Cuboid 1 1 1 2 1 1 2 2 1 2 2 2 TOTAL Tally 8 12 6 1 27 Build a 3 3 3 cube using 27 wooden cubes. Set up a similar table to the one given above and investigate the number of cuboids with dimensions 1 1 1, 3 1 1, 3 3 1 and 3 3 3. Note that although there are also cuboids with other dimensions we are restricting ourselves to these four in particular. Now build a 4 4 4 cube using 64 wooden cubes. Set up another table and investigate the number of cuboids with dimensions 1 1 1, 4 1 1, 4 4 1 and 4 4 4. Summarise your results by completing the table shown below: Cuboid 1 1 1 n 1 1 n n 1 n n n TOTAL n=2 8 12 6 1 27 n=3 n=4 Try to derive an algebraic expression (in terms of n) for each column. Think carefully about how you arrived at each number, i.e. think about how the various cuboids were oriented and positioned in the large cube.
ACTIVITY 20 Double-storey Begin with a 2-by-2 grid like that shown alongside. Place 5 cubes on the grid according to the following criteria: 3 cubes must be placed in any of the 4 cells on the grid, one cube per cell. 2 cubes must then be placed anywhere on top of the original cubes to form a second layer. How many ways are there of doing this? In other words, how many different doublestorey structures can be built according to these criteria? To build the first layer there are clearly four different possibilities: If we take the left-most of these four arrangements then there are three different possibilities for the second level: So, how many different double-storey structures can be built in total? Let s now start with a 3-by-2 grid (as shown alongside) and place 5 cubes in the bottom layer and 4 cubes in the second layer. How many ways are there to do this? Now investigate a scenario where we place 5 cubes in the bottom layer but only 3 cubes in the second layer. How many ways are there to do this? Investigate other scenarios. You may even want to extend the investigation to 3 layers.
ACTIVITY 21 Concentric borders Take a single cube and build a border around it using more cubes. This border will contain 8 cubes. Now build a second border around the outside of the first border. This second border will contain 16 cubes. If we build a third border around the outside of the second border, then this third border will contain 24 cubes. These three scenarios are shown below. How many cubes will you need for the fourth border? What about the fifth border? Can you generalise for the n th border? Extend the investigation by building borders around inner rectangles of different dimensions. Try ultimately to generalise for an inner rectangle with dimensions h by k units. The table below might be helpful to summarise your results. The results from the three diagrams above have already been entered into the table. Border layer 1 2 3 4 5 n 1 by 1 inner 8 16 24 1 by 2 inner 1 by 3 inner 1 by 4 inner 2 by 2 inner 2 by 3 inner 2 by 4 inner 3 by 4 inner h by k inner
ACTIVITY 22 Rectangles and factors Take 12 wooden cubes and arrange them into rectangles. There are three different possible rectangles: 3 4, 2 6 and 1 12. Notice that 3 4 = 12, 2 6 = 12 and 1 12 = 12. Also notice that 3, 4, 2, 6, 1 and 12 are all factors of 12. In fact, these six numbers are the only factors of 12. Instead of using 12 wooden cubes, investigate for other numbers. What about extending the idea to rectangular prisms? For example, if you start with 12 blocks then these can be arranged into a 2 2 3 rectangular prism. Investigate for other starting numbers.
ACTIVITY 23 Noughts & Crosses In a standard game of noughts and crosses, where each player aims to get three noughts (or crosses) in a row, there are eight possible winning lines. The three noughts (or crosses) can lie in one of the three horizontal rows, one of the three vertical columns, or one of the two diagonals. These eight possible winning lines are shown below. 7 4 5 6 8 1 2 3 What if we extend the idea to three dimensions? Instead of playing on a 3 3 grid you would need to play in a 3 3 3 matrix of cubes where each nought or cross is placed inside one of the 27 cubes. In such a 3-dimensional game, how many winning lines would there be? Extension activities: Extend the original 2-dimensional game of noughts and crosses to a 4 4 grid in which you have to get four noughts (or crosses) in a row. How many winning lines would there be? What about a 5 5 grid, or a 6 6 grid? Try to generalise for an n n grid. Now extend the 3-dimensional game of noughts and crosses to a 4 4 4 cube in which you have to get four noughts (or crosses) in a row. How many winning lines would there be? What about a 5 5 5 grid of cubes, or a 6 6 6 grid? Try to generalise for an n n n grid of cubes.
ACTIVITY 24 Approximating a pyramid The formula for determining the volume of a pyramid like that shown alongside is: B.h V = 3 where B is the area of the base and h is the perpendicular height. B h From the formula one can see that if we had a pyramid with a square base, then the volume of the pyramid would be one third of the volume of a cube with the same base. Using wooden cubes we can approximate the shape of a pyramid as shown below. Clearly the greater the number of layers the better the approximation is to a pyramid. Since we know that the pyramid being approximated should have a volume equal to a third of the cube with the same base as the pyramid, we can set up a table to investigate how good the approximation really is. n (layers) Volume of cubic solid Volume of pyramid of cubes 1 1 1 1 2 8 5 0.625 3 27 14 0.518518519 4 64 30 0.46875 (Volume of pyramid of cubes) (Volume of cubic solid) We can see from the final column that the approximation improves as the number of levels increases since the final column should ideally give an answer of 0,3333 Use a spreadsheet to extend the table. For the third column devise a recursive formula that will enable you to determine each cell from the previous one. How many layers does one need before Column 3 divided by Column 2 gives an answer less than 0,35? What about 0,335 or 0,3335? Use the spreadsheet software to draw a graph to show how the approximation improves as n (the number of layers) increases.
ACTIVITY 25 Twin halves Use 16 blocks and set them out in a 4 4 square as shown below. There are a number of ways of splitting the square of blocks into two identical pieces. Two possible ways are shown below. What other ways can you find to split the square into two identical pieces? What if we started with a different size square of blocks? What about starting with a rectangle of blocks? Investigate for different sized squares and/or rectangles.
ACTIVITY 26 Visualising fractions We can use the wooden blocks to help visualise fractions. Consider the following fraction sum: 3 4 + 1 6 =? Begin by creating a rectangle of blocks with dimensions equal to the denominators of the two fractions. Since our denominators are 4 and 6 we must create a 4-by-6 rectangle using a total of 24 blocks, i.e. 4 6 blocks. If we now take the rectangular block as our whole, then each of the four horizontal rows of blocks represent one quarter of the whole. Similarly, each of the 6 vertical columns represent one sixth of the whole. 1 4 1 6 Since 4 1 contains 6 blocks, then 4 3 contains 18 blocks, i.e. three horizontal rows. The 1 3 1 fraction contains 4 blocks. So, + thus contains 18 + 4 = 22 blocks out of the 24 6 4 6 3 1 22 11 blocks of the whole. In other words, + = which we can simplify to. 4 6 24 12 5 1 Use the same rectangular arrangement of blocks to help determine. 6 4 2 1 Create a new rectangular arrangement of blocks to determine + as well as 5 4 4 3. 5 4 Investigate how one could use this method to add three or even four fractions if they each had a different denominator.
ACTIVITY 27 Chequered flag Look at the following pattern of blocks: Figure 1 Figure 2 Figure 3 Figure 4 Use the wooden blocks to build the first four figures as shown above. How many blocks will you need to build Figure 5? How many blocks will you need to build Figure 15? Now find a formula that will determine the number of blocks for the n th figure in the sequence. Try to do this in two different ways: (i) by looking for common structural features in each of the figures, and (ii) by working with the numeric sequence which the sequence of figures represents.
ACTIVITY 28 Be amazed! The image alongside shows a maze built out of blocks. In the top left-hand corner of the image is a black dot. In the bottom right-hand corner is a black star. The object of the exercise is to start at the dot and end at the star by moving either through or around the outside of the maze of blocks. The only condition is that you can only move in two directions: (i) to the right, and (ii) downwards. In other words, you can only move due East or due South. There are a number of ways this can be accomplished. Two possible paths are shown below. How many different paths are there? What strategy did you use to ensure that you counted all possible paths? Did you notice that each path was the same length? Why do you think this is? How many paths would there be if the maze was a 4 4 grid of blocks? What about a 5 5 grid of blocks? Can you generalise for an n n grid of blocks?
ACTIVITY 29 Triangular numbers The first five triangular numbers are 1, 3, 6, 10 and 15. Can you see why they re known as the triangular numbers? Write down the next five triangular numbers. 1 3 6 10 15 Triangular numbers can also be arranged pictorially as follows: 1 3 6 10 15 This pictorial arrangement of the triangular numbers is rather convenient. Notice, by way of example, that if we take two lots of the third triangular number they can be rearranged into a 3 4 rectangular arrangement as shown alongside. Use the wooden blocks to confirm that two lots of the fourth triangular number can be rearranged to form a 4 5 rectangular arrangement of blocks. Use the wooden blocks to confirm that two lots of the fifth triangular number can be rearranged to form a 5 6 rectangular arrangement of blocks. Can you see why two lots of the n th triangular number will always be able to be rearranged into an n by (n+1) rectangular arrangement of blocks? Use these observations to find a formula for the n th triangular number, and confirm that your formula works for the first ten triangular numbers.
ACTIVITY 30 Cross pattern Look at the following sequence of shapes: Shape 1 Shape 2 Shape 3 Notice that in each shape there is an inner square of blocks. On each of the four sides of this central square there is a triangular arrangement of blocks. How do the dimensions of the inner square of blocks relate to the shape number? How do each of the four triangular arrangements relate to the shape number? [Hint: think in terms of triangular numbers and use the formula you derived in Activity 29 for the n th triangular number.] Now determine a formula to determine the number of blocks in the n th shape for the above pattern of crosses. Shape 4