To investigate the patterns generated from using rules in a square grid

Maths Coursework

Opposite Corners

Aim:  My task is to investigate the differences of the products of the diagonally         opposite corners of a rectangle, drawn on a 10x10 grid, with the squares         numbered off 1 to 100.

I will aim to investigate the differences for rectangles that are aligned differently, of different lengths, widths and when they are squares.

I plan to use algebra to prove any rules I discover which I will hopefully discover by analysing my results that I will log in tables.  I will test any rules, patterns and theories I find by using predictions and examples.

I will record any ideas and thoughts I have as I proceed.

Method:  

        To start with I will break up my work into sections, my first is:

2x3 Rectangles

I will concentrate on one particular aspect of at a time.

What is the difference between the products of the corners?
Is the difference the same for a rectangle drawn anywhere in the grid?

I now have several questions:

Is the difference different when the rectangle is aligned so that its shortest sides are at the top and bottom?

What is the same between the two alignments?

What About Other Sizes of Rectangles?

I will now try rectangles, all in the 2 x X series, with different lengths.

I think I can now safely assume that the difference is always constant in relation to the size of the rectangle, therefore I only require one example of each difference.

2 x 4

     

2 x 5

        

I theorise that every time I increase the width by one, the difference increases by 10.

I predict that the 2 x 6 rectangle will have a difference of 50.

2 x 6

        

But can I use this rule to estimate the difference of a 2 x 2 square?  I predict that the trend will continue, as a square is just a special form of rectangle, and the difference shall be 10.

2 x 2

     

From my results so far I can draw a table:

What Do I Notice?

        The area increases by 2 each time.  This is because the length is always being multiplied by the height of 2.

        The difference increases in increments of 10.  Also, each line of the grid is 10 squares wide so the next square vertically straight down the is ten more than the square above it.

Possible Formulas

L=Lenght  H=Height  D=Difference

L-1 x 10

or

The length subtracted by one multiplied by ten.  But from past experience I doubt this rule would work with other sized rectangles without 2 as a side length

  L-1 x 5H

or

The length subtracted by one multiplied by five multiplied by the height.

The lowest corner subtracted from the highest corner subtracted by one, but only when the rectangle is aligned so that the shortest sides are at the top and base.

Do These Rules Work With Other Rectangles?

3 x X

        I will now look at rectangles where one of the sides has a length of 3 squares and see if the same patterns apply and investigate whether the above rules are true for rectangles of any shape.

        I believe I shall require only one example of each and will proceed on this principle.

3 x 3

     

3 x 4

     

3 x 5

     

What Do I Notice?

        The “L-1 x 5H” does not apply.                   But L-1 x 20 does.

        I feel confidant that L-1 is correct but I still need to find how H can be used to calculate what to multiply L-1 by.

For 2 x X                L-1 x 10

For 3 x X                L-1 x 20

Could the universal rule be:                 D = L-1 x 10 x (H-1)?                        

        Subtract one from the height, multiply by 10 and then multiply by the height subtracted by one?

Testing the Rule

4 x X

I will now look at 4 x X and see if my new found rule truly encompasses all rectangles.

4 x 4

     

4 x 5

     

The difference is advancing by 30.  4 x 6 will have a difference of 150.

4 x 6

     

(table calculated with a calculator)

Could the  x 10 in my rule be related to width of my number grid, which is ten squares wide?

        To test this idea I will try my rule on a different sized grid.

                                               

The Answer is forty,  my rule works for any rectangle, on any grid.

L-1 x (Grid Width) x (H-1)

or

Subtract one from the height, multiply by the width of the grid and then multiply by the height subtracted by one.

Using Algebra to Look Closer:

Calculating the rectangle’s values in relation to “X”

If the number in the top left hand corner is X, then the following squares will be....

(G = Grid width)

For a 3 x 2 rectangle when the first square is X the difference is:

   

difference for this size of rectangle is: 2G, the width of the grid multiplied by two.  This is correct as from my previous working on a rectangle of these proportions I know that on a 10 x 10 grid the difference is 20 (10 x 2 !)

For a 4 x 2 rectangle when the first square is X the difference is:

   

difference for this size of rectangle is 3G, the width of the grid multiplied by three.  This is correct as from my previous working I know that on a 10 x 10 grid the difference is 30 (10 x 3 !)

How can I produce an equation from the side lengths of this rectangle to give me 3?

Possible equations:

                                      or

        I will look at the ?G rule in other rectangles to see if any of my above rules also apply and are true for rectangles across the board.

the difference for this size of rectangle is 6G, the width of the grid multiplied by six.  This is correct as from my previous working I know that on a 10 x 10 grid the difference is 60 (10 x 6 !)

None of my rules apply, but what I can now build a table of my results:

               This must be because the H has increased by 1 also.  This tells me that the H must be connected to the difference.  But How?

How can I get 2 from 2 & 3, and 3 from  2 & 4 using the same formula?

If I subtract 1 from each of 2 & 3 I get:          L=1  H=2

1 x 2 = 2.

When I subtract one from each 2 & 4 I get:  L=1  H=3

1 x 3 = 3

The answers to both of these short sums give me the figure to multiply G with to get the difference.

I predict that this will work for 3 & 4:

3 - 1 = 2

4 - 1 = 3                          2 x 3 = 6     So   L-1 x H-1 x G = Difference

This methods works, and it is also another way of writing my previous universal rule. “L-1 x H-1 x G” is the same as “L-1 x (Grid Width) x (H-1)”.

So Summarising:

I have found the rule for any size of rectangle.  I think that I can write down the expression to show this, that dose not require the actual figures for the length:

(X + L - 1) x (X + (G(H - 1))) = XxX + X (G + X(H - 1)) + L-1 X + L-1 (G  (H - 1))_

 X x  (X + (G(H - 1))) + L - 1) = XxX + X(G + X(H - 1)) + X L-1 _________________

                                                                          L-1 (G  (H - 1))

To Test This Expression

L-1 (G  (H-1))

Lets substitute the value of X for 1 in a 2x2 rectangle on a 10x10 grid and check that it calculates the square values correctly.

   So far so good

And now I will use the above expression to find the difference (which should be 10)

L-1 (G  (H-1))

2-1 (10  (2-1))                        

1 x  (10 x 1)

1x 10 = 10

It seems fine so far.

I will now check it with a 3 x 4 rectangle on a 10x10 grid.  I Know the difference should be 60 from previous workings.

L-1 (G  (H-1))

4-1 x (10 (3-1))

3    x (10  (2))

3    x  20 = 60

The expression is true !

L-1 (G  (H-1))  is the universal rule with a the brackets in different positions:

L-1 (G  (H-1)) = L-1 x (Grid Width) x (H-1) = Difference

Evaluation

        I believe I have completed my aims and succeeded in attaining the goals I set myself.  I have found the rule to work out the difference for any rectangle of any size on any grid.  I have done this through progressive investigation and through the use of algebraic methods.

        I have taken this course work as far as I can in the time that I have been allotted.  The only other area that is left to be explored would be drawing a cuboid in a cube.  Though I imagine this would be a lot more complicated,  I am left wondering the possibility of the existence of a rule that would cover any, or even an infinite number of dimensions.

        I am happy with the work that I have done and can not think of anything I could have carried out better or improved upon.