ALG 18

Week 18: In this set of task we demonstrate some of the power, intrigue and utility of algebra, though perhaps not its serious purpose. We use algebraic expressions (involving a variable, u) to describe the position of the vertices of a quadrilateral, and then look at what happens to the quadrilateral as the value of the variable changes.
This week's tasks involve a rare excursion into the use of algebraic symbols. Pupils might need some gentle support to help them read the expressions correctly, though we are not asking pupils to do much with the expressions, other than to evaluate them for different values of the single variable involved.
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Monday: This task sets up the work for the rest of the week. You might need to work through it quite carefully, to remind pupils of various conventions and ideas, for example that A, B and C, etc, refer to points on the given grid, and that they form the 'vertices' or corners of a 4-sided shape - a 'quadrilateral'. Also, when we write a term like 4u, we mean '4 times u' and that if we know the value of u we can find the specific value of the term, or of an expression like 4u–8.
Other, more subtle ideas are also at play here, for example that an expression like 4u–8 can be thought of as a number in its own right, not just a set of instructions for finding a number; moreover, this number varies as u varies - and different expressions may vary at different rates. These ideas need time to develop over many years.
There is another issue here which may puzzle pupils even if they don't raise it explicitly. We may decide to let the issue stay dormant at this stage, but nonetheless it is important that we ourselves are aware of the issue, which concerns the 'status' of the algebraic expressions. In a 'real life', practical situation, the expressions would have a 'purpose'. They would arise out of the nature of the situation and the relationships involved, and they might help us to solve a worthwhile problem. That is not the case here! Rather, the only 'purpose' of our current expressions is to gain experience of working with such expressions and of getting a sense of how the behaviours of different expressions compare. Our hope is to do this in an intriguing and mathematically illuminating way, despite the absence of a vital problem that algebraic expressions could help us to solve. That is for another time (hopefully!).

In part b), we can tell from the diagram that
4u – 8 = 20, and
4u – 6 = 22, and
u + 7 =14.
We can use any one of these sentences to determine that for this diagram, u = 7.
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Tuesday: Part a) refers back to Monday's task. In part b) the value of u is increased by 1. Before embarking on this part, you might want to ask pupils to predict what might happen.

Points A, B and C all move to the right, but not all by the same amount. Why?
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Wednesday: Here we increase the value of u by 1 again. What will happen?

Do any pupils predict that the points move to the right again, and by the same amount as each moved when we previously increased the value of u by 1?  How do they explain this? Can they relate it to the numbers in each expression?
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Thursday: This is quite challenging, though it can obviously be solved by trial and improvement. Given what has happened to the shape for the values of u that we have considered so far, some pupils might realise that value required here will be smaller.

Some pupils might be able to solve the task analytically, and it might be worth discussing such an approach, even if the class first solves the task by trial and improvement.
Pupils might have noticed that side AB is always the same length and at the same angle, regardless of the value of u (Why?). This means that point A is the 'bottom' vertex of the 'diamond' and the fixed point D is the left-most vertex of the 'diamond'. Also, D will be 2 units to the left of B. We know that D is always 10 units from the red line, so A will be 12 units from the red line. So 4u–8 = 12 when the quadrilateral is this 'diamond' shape. What is the value of u when 4u–8 = 12 ?
We get the 'diamond' when u = 5, as shown in the bottom diagram:

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Friday: Again, we can find the values of u by some form of trial and improvement, or in a more analytical way.

When A and E coincide, the distance of A from the red line is 0 units, ie 4u – 8 = 0. What value of u fits this sentence?
When B and D coincide, the distance of B from the red line will be 10, ie 4u – 6 = 10. What value of u fits this sentence?
The resulting shapes of the quadrilateral (ie when u = 2 and when u = 4) are shown below.

 It is useful to think of what happens to the quadrilateral as u increases over the complete range that we have considered (ie from u = 2 to u = 9). You might want to ask pupils to describe what happens.
When u = 2, the side AB is way over to the left, while point C is just to the left of point D (see the above slide). As the value of u increases, side AB and point C move to the right, but AB moves faster (How much faster?). When u = 4, B has reached D, so the quadrilateral has 'collapsed', and C is just to the right of D (see the above slide). When u = 5, the quadrilateral has opened out again, to form a 'diamond', with C being caught by A and ovetaken by B. As u continues to increase, the quadrilatral becomes more and more stretched, with AB moving further and further to the right and away from C as well as D.
How does this geometric description relate to the algebraic expressions? (A question more for us than the pupils!)

We have a movie to go with this, which you might want to show the pupils at this stage. Stills from the movie are shown below.