Spotting the loophole

5812 /www/nrich/html/content/id/5812/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> In 2D, vectors may be thought of as arrows with a fixed length and direction. The place at which the arrow starts is not important, so they may be translated around the plane without affecting the value of the vector. This allows us to add and subtract vectors visually: if the vectors can be translated so as to form a closed circuit then the vector sum is zero. If the vectors cannot be translated so as to form a closed circuit then their vector sum is not zero, as in this diagram in which the vectors on the left have zero sum, whereas the vectors on the right don't have zero sum. <mdo:image alt="Summation of vectors" height="201" src="misMatch.GIF" width="457"></mdo:image> In this problem we are given three grids of vectors. The vectors are only represented visually, but it is assumed that their x and y values are whole numbers (Remember: the x and y values represent the horizontal and vertical offsets from the start of the arrow, not the location of the start of the arrow itself). Your task, in each case is to try to see which subsets of the vectors form closed loops. In each grid can you find a closed loop of vectors? In each case is the closed loop unique? You can prove your assertions using algebra or a convincing visual argument. Try hard to solve using visualisation before resorting to algebra. Grid 1: <mdo:image alt="Grid 1" height="269" src="grid%201.JPG" width="457"></mdo:image> Grid 2 <mdo:image alt="Grid 2" height="368" src="grid2.JPG" width="360"></mdo:image> Grid 3 <mdo:image alt="Grid 3" height="458" src="grid%203.JPG" width="463"></mdo:image> <a href="http://nrich.maths.org/public/viewer.php?obj_id=6822&part=">Click here for a poster of this problem</a>. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> We had some users saying that that there are no closed loops, but there are if you search carefully! We received two answers: one from Joseph from Crewe and, which was very visual and one which made full use of algebra, which was submitted by an anonymous user. We particularly liked the way that the solvers appealed to both algebraic and geometrical considerations to solve the problem. Very well done, whoever you were! FIRST SOLUTION : Each vector will be expressed in the form (x y), where the x-component represents the units moved to the right (negative means left), and the y-component represents the units moved upwards (negative means down). Each vector will also be assigned to a letter from A-G: A: Black B: Dark blue C: Green D: Light blue E: Pink F: Red G: Yellow Grid 1: The direction of each vector is as follows: A: (-6 -3) B: (2 3) C: (3 1) D: (-3 0) E: (4 0) F: (-4 1) G: (-2 4) By considering only the y-direction at first, the only vector with a negative y-value is A. To make the y-component 0, B must be added to A. D and E have a y-component of 0, so adding D and/or E to AB (or even DE by itself) will still keep the y-component of 0. Hence there are only 5 permutations with a y-component of 0: AB, ABD, ABE, ABDE and DE. To make a closed loop, the x-component must also be 0; analysing the x-components of the 4 permutations, only ABE gives an x-component of 0 (-6 + 2 + 4), and so it is the only closed loop. Grid 2: The direction of each vector is as follows: A: (-2 3) B: (2 3) C: (2 0) D: (-2 -2) E: (0 -2) F: (2 -3) G: (-2 0) By visualisation alone, it is easy to see that AF (black and red) and CG (green and yellow) create a closed loop. A combination of ACFG also creates a closed loop. However, these may not be the only closed loops present. Again by considering only the y-direction at first, D and E cannot be used to make a closed loop, since any permutations with D and/or E give y-components of -2 or -4. Adding A and/or B, which are the only vectors with positive y-components, cannot make a y-component of 0. By eliminating D and E, F is the only vector with a negative y-component, which is balanced by addition of A or B. Hence there are only 9 permutations with a y-component of 0: AF, AFC, AFG, ACFG, BF, BFC, BFG, BCFG and CG. To make a closed loop, the x-component must be 0, and so AF, ACFG and CG are the only 3 closed loops. Grid 3: The direction of each vector is as follows: A: (0 1) B: (-2 0) C: (3 0) D: (1 -5) E: (-3 3) F: (4 2) G: (-1 -6) Once again, looking at the y-components, there are only two vectors with negative y-components, D and G. The y-component of D can be made 0 by adding E and F, possibly B and/or C, which have a y-component of 0. The y-component of G can be made 0 by adding E, F and A, possibly B and/or C. There are only 9 permutations with a y-component of 0: AEFG, ABEFG, ACEFG, ABCEFG, DEF, CDEF, BDEF, BCDEF and BC. Only AEFG and BDEF have an x-component of 0, so they are the only two closed loops. SECOND SOLUTION: For Grid one: Only one arrow points downwards (the black one). so the black arrow must be part of the closed loop, if such a loop exists. The x-offset of the black arrow is greater than the x offset of all of the other arrows, thus there must be at least two right facing arrows in the closed loop. Thus two or three of pink, green and dark blue must be present. Both dark blue and green togther would be too high and just pink and green would be too low: thus the loop contains pink and dark blue. These form a closed loop. For Grid two: It is obvious that Black + Red + Green + Yellow make a parallelogram closed loop, and it is obvious that this is the only one. For Grid 3: The coordinates of the vectors are Red (4, 2) Yellow (-1, -6) Purple (-2, 0) Green (3,0) Black (0,1) Light Blue (1, -5) Pink (-3, 3) Looking at the y coordinates of these, I can see that any closed loop must contain Red, Pink, Black and Yellow or Red, Pink and Light Blue. Red + Pink + Black + Yellow = (0, 0) So this is a closed loop. Red + Pink + Light Blue = (2,0) This sorts out the Y-coordinate. I now see that I need to include Purple to make another closed loop. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><h3>Why do this problem?</h3> <div>This problem encourages students to use visualisation to help them to spot a solution which can then be verified using algebra. It shows students the power of using visual representations to solve vector problems, often the quickest route to a solution.</div> <h3>Possible approach</h3> This could form a short introduction to work on vectors. Display the example grids on the board, showing the three vectors forming a closed loop and the four vectors on the right which do not have a zero sum. Then challenge students to find any closed loops in each of the three grids. When they think they have spotted a closed loop by eye, they should verify algebraically that it is indeed a closed loop. <h3>Key questions</h3> <div>What can you say about the horizontal components of the vectors in a closed loop?</div> <div>And what about the vertical components?</div> <h3>Possible extension</h3> <div><a href="http://nrich.maths.org/6572&part=">Vector Walk</a> begins to explore properties of vectors and combining two basic vectors to reach points on the coordinate grid.</div> <h3>Possible support</h3> Students may find it helpful to draw the vectors on squared paper. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> In order to solve this problem you will need to think about what makes a loop closed instead of open. In two dimensions the overall changes in the x direction around the loop must cancel out and the overall changes in the y directions around the loop must also cancel out. You can think about these two directions separately if you cannot quickly spot how to form a closed loop. Visually you can use lots of intuitive shortcuts, such as that 'long' arrows may need multiple 'short' arrows to close up into a loop. Of course, when you think that you have spotted a loop you will need to check carefully to see if you are correct by adding up the vectors exactly. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">For the three grids the coordinates of the vectors are <mdo:image width="420" height="241" src="coordinates.JPG" alt="Coordinates"></mdo:image> Solutions are: Grid 1: Black, Dark blue, pink (only solution) Grid 2:Black, Red, yellow, Green (only solution) Grid 3:Black, yellow, red, pink (first solution) Dark blue, light blue, red, pink (second solution) Example visual proof of uniqueness for grid 1 Only one arrow points downwards (the black one). Thus the black arrow must be part of the closed loop, if such a loop exists. The x-offset of the black arrow is greater than the x offset of all of the other arrows, thus there must be at least two right facing arrows in the closed loop. Thus two or three of pink, green and dark blue must be present. Both dark blue and green togther would be too high and just pink and green would be too low: thus the loop contains pink and dark blue. These form a closed loop. I like this sort of argument. I could imagine Pythagoras talking in such a way. </mdoxml> 2 3 0 0 0 1 0 Spotting the loophole A visualisation problem in which you search for vectors which sum to zero from a jumble of arrows. Will your eyes be quicker than algebra? Using, Applying and Reasoning about Mathematics Visualising Vectors Addition of vectors Admin Short problems Secondary Mapping Document Vectors US