} +10

} +10

} +10

} +10 (we can assume this, looking at the pattern of the difference increasing in increments of 10)

} +10

Possible Rules

One possible rule is taking one away from the length and multiplying by 10 will give the difference. This, written algebraically is: 10(L-1). However, this may not work with a width that isn’t 2 [I’ll look into this further on].

One formula that takes into account the width is: 5W(L-1). This can be written in words as the Length minus one, times by five times the width.

Checking Formulae

10(L-1), when L = 3: 10x(3-1) = 10x(2)=20? YES

5W(L-1), when L=3 and W=2: (5x2)x(3-1)= (10)x(2)=20? YES

3x‘n’ Rectangles, 10x10 Grid

I’m now investigating rectangles with heights of 3 and see whether they follow the same or similar pattern to those with heights of 2. After this I will see whether the formulae I came up with previously apply for these rectangles, and if not, how I can adapt them to.

3x2

2x21=42

1x22=22

42-22=20, D=20

3x3

63x81=5103

61x83=5063

5103-5063=40, D=40

3x4

29x46=1334

26x49=1274

1334-1274=60

3x5

20x36=720

16x40=640

720-640=80, D=80

The differences are still multiples of 10, however, they are going up in increments of 20.

Table of Results 2(and further predictions)

} +20

} +20

} +20

} +20 (we can assume this, looking at the pattern of the difference increasing in increments of 20)

} +20

Checking Previous Formulae

10(L-1), when L = 5: 10x(5-1)= 10x(4)= 40? NO, it should be double that (80).

5W(L-1), when L=5 and W=3: (5x3)x(5-1)= (15)x(4)=60? NO

Adapting The Formulae

I have decided to continue with the formulae 10(L-1) as the other formula doesn’t seem to work for other widths, whereas this one appears to be adaptable. For 10(L-1), the result is half the expected number, so to give the correct answer it can be multiplied by 2: 2x10(L-1) or 20(L-1). E.g. If L=4: 20x(4-1) = 20x(3) = 60? YES

4x‘n’ Rectangles, 10x10 Grid

To help me find a pattern, I’m going to do rectangles with a width of 4:

4x2

2x31=62

1x32=32

62-32=30, D=30

4x3

9x37=333

7x39=273

333-273=60, D=60

4x4

60x87=5220

57x90=5130

5220-5130=90, D=90

4x5

10x36=360

6x40=240

360-240=120, D=120

The differences are still multiples of 10, but now they are going up in 30. The formula can be adapter here to be 3x10(L-1). E.g. If L=4: 30x(4-1)= 30x(3)=90? YES

Finding A General Rule/Formula

I’m now going to draw up an observation table and try to make links between all the patterns seen and the formula I have to try and make a general rule that would apple for any rectangle.

Something interesting I have noticed is that all of the increment sizes are linked to the width of the rectangle by (W-1) x 10 . I will now put this into the previous formula to try and generalize it.

So the formula for 2x‘n’ was 1 x 10(L-1),

And the formula for 3x‘n’ was 2 x 10(L-1),

And the formula for 4x‘n’ was 3 x 10(L-1).

One possible step would be to replace the 1

, 2

and 3

with (W-1).

This would make the general rule: (W-1) x 10 x (L-1) or 10(W-1)(L-1)

Checking the General Rule

2x5

When W=2 and L=5: 10(W-1)(L-1)= 10x(1)x(4)= 40? YES

3x4

When W=3 and L=4: 10(W-1)(L-1)= 10x(2)x(3)= 60? YES

4x4

When W=4 and L=4: 10(W-1)(L-1)= 10x(3)x(3)= 90? YES

5x3?

When W=5 and L=3: 10(W-1)(L-1)= 10x(4)x(2)= 80? …

3x41=123

1x53=43

123-43=80, D=80. YES

Why does the Rule Work?

Taking this grid as an example, whenever you move to the right, it is adding 1. And moving down one will add 10.

+1 +1 +1

Using this, you can take any rectangle and begin to label all the sections within it in regards to one square. We will call this the ‘Expression Grid’:

If we now use these expressions rather than numbers in finding the difference, we should end up with proof of how the rule works:

Proof

(A+(L-1)) x (A+10(W-1)) = A2+(10A(W-1))+ A(L-1) + 10(L-1)(W-1)

A x A+(L-1)+10(W-1) = A2+A(L-1)+(10A(W-1))

A2+(10A(W-1))+ A(L-1) + 10(L-1)(W-1) – (A2+A(L-1)+(10A(W-1)))= A2+(10A(W-1))+ A(L-1) + 10(L-1)(W-1) - A2-(10A(W-1))

A2+(10A(W-1))+ A(L-1) + 10(L-1)(W-1) - A2-A(L-1)-(10A(W-1))= 10(L-1)(W-1) + A2-A2 +(10A(W-1))-(10A(W-1) + A(L-1)-A(L-1)

10(L-1)(W-1) + A2-A2 +(10A(W-1))-(10A(W-1) + A(L-1)-A(L-1) = 10(W-1)(L-1)

Extension: Using Grids Bigger than 10x10

Previously, I showed how, on a 10x10 grid, the numbers increase by 1 when moving right and 10 when moving down. I then used this to create a grid where useful boxes were labeled using expressions in regards to one square (which I called ‘A’). The exact same thing can be done for any sized grid. The only thing that needs adapting is the value the number increases when moving down. For example, on a 13x13 grid, it would increase by 13 when you go down. The fact that the value it increases when moving down is exactly the same as the grids length, we can substitute this into the equation to make a rule that can apply to any grid:

D=(Grid Length)(W-1)(L-1)

Or simply

D=G(W-1)(L-1)

Proof for the New General Rule

This rule is even better, as it takes into account grid length, length of rectangle, and width of rectangle, compared to the previous rule that only took into account length and width of rectangle. To prove this rule, all we have to do it replace the 10 in the expression grid with L , as we know it increased by 10 because of the length of 10. Here is the new expression grid:

These new changes can now be put into the proof.

Proof

(A+(L-1)) x (A+G(W-1)) = A2+(GA(W-1))+ A(L-1) + G(L-1)(W-1)

A x A+(L-1)+G(W-1) = A2+A(L-1)+(GA(W-1))

A2+(GA(W-1))+ A(L-1) + G(L-1)(W-1) – (A2+A(L-1)+(GA(W-1)))= A2+(GA(W-1))+ A(L-1) + G(L-1)(W-1) - A2-(GA(W-1))

A2+(GA(W-1))+ A(L-1) + G(L-1)(W-1) - A2-A(L-1)-(GA(W-1))= G(L-1)(W-1) + A2-A2 +(GA(W-1))-(GA(W-1) + A(L-1)-A(L-1)

G(L-1)(W-1) + A2-A2 +(GA(W-1))-(GA(W-1) + A(L-1)-A(L-1) = G(W-1)(L-1)

Extension: Rotating 90

and 180

(Old values, before rotation, will be marked with a ‘1’. New values, after rotation, will be marked with a ‘2’)

Using our previous rule, working out whether rotation the rectangle by 90, 180 or 270 degrees would affect the difference is easy.

90

Rotating by 90 degrees makes the length into the width and the width into the length. As the formula is L(L-1)(W-1) , the two same values will have one subtracted and then be multiplied. However, it will them be multiplied by what previously was the width rather then the length. So the new difference can be worked out as:

180

Rotating by 180 degrees does not affect the value of the Width or Length, so the difference is still exactly the same.

270

Like with rotating 90 degrees, the width is just swapped with the length vice versa. So to find out the new difference, you do:

360

Rotating a shape by 360 degrees just returns it to its original orientation and position. Obviously, this means the difference would have still stayed the same.

Evaluation

I think that I have completed the aims of this assessment and have reached the goals. I have found multiple patterns in the difference of opposites corners and used this to find individual rules. I then linked together the patterns and individual rules to make a general rule, which can work for any grid, width or length. I also went on to look at how rotating a shape would affect the difference, which was made easier by my previous work in creating a general rule. I also went very detailed in my explanations and proof of everything.

In the future, it would be interesting to see how this would work in 3D, for example, a cuboid inside a cube. This would be interesting as I’d have to take into account a whole new dimension which increases the work and calculations needed to come to any conclusions. It would also be interesting to see how it worked with different shapes like triangles and hexagons, although a specially designed grid would be necessary.

I am pleased with the way that I handled this task and, at the moment, cannot think of any improvements without looking at a more in-depth success criteria or the standard of other pieces of work.