Investigate a formula to see how many squares would be needed to make any two-dimensional cross-shape. The second part requires me to extend the investigation into three dimensions.
Assignment
A dark cross-shape has been surrounded by white squares to create a bigger cross-shape. The next cross-shape is always made by surrounding the previous cross-shape with small squares. Investigate to see how many squares would be needed to make any cross-shape built up in this way.
Introduction
For my mathematics coursework I have been asked to do a two part investigation. Part one requires me to investigate a formula to see how many squares would be needed to make any two-dimensional cross-shape. The second part requires me to extend the investigation into three dimensions. Each time the cross shape increases in size, border squares are increased. These border squares cover the outside of the original cross shape thus creating a new, larger one. An example is shown below.
Equipment
In order to investigate the assignment, I will make use of Lego bricks to use them to build my shapes so that I can gather all of the information for finding the formula, which is a required part of this coursework. This will allow me to model the problem and gather preliminary data.
Two dimensional investigation
Investigating the 2D formula
I am going to start by building a figure and adding on the borders gradually. The diagram for these figures and the others that I investigated are shown below. The squares that are having squares added to them are shown in pink; the border squares (the squares that are being added on) are shown in blue.
Figure 1
Figure 3
The order of squares going from left to right are:
Figure 2: 1 + 3 + 1 = 5 (see diagram below for example)
Figure 3: 1 + 3 + 5 + 3 + 1 = 13
In both cases, 2 has been added to the middle number (as shown below):
+ 3 +5 + 7 + 5 + 3 + 1 = 25
All of the above have so far been correct. For the next pattern I predict that the total number of squares will be 41, using the following pattern: 1 + 3 + 5 + 7 + 9 + 7 + 5 + 3 + 1 = 41
Figure 4
My prediction was correct. As well as discovering a correct method of finding the next pattern I noticed that to find the number of pink squares on the next pattern, you use the total number for the previous pattern.
Figure 5
Analysing results
I think that investigating five different figures will give me enough information to find a formula. The data I have collected is shown in the table below.
Blue squares
Pink squares
Total # of squares
0
4
5
8
5
3
2
3
25
6
25
41
20
41
61
From these results I can see a direct relationship with the amount of blue squares. They increment by 4 every time the figure increases in size. I think that this relationship will be important in finding the formula to discover how many blue squares are required to surround the pink square(s) thus making a bigger figure. I think this will mainly be in the form of a multiple of 4.
I discovered that the total amount of squares is always odd. Another relationship with the results is that the total number of pink squares is the same number as the total number of squares on the previous shape. For example, the total number of squares for the third figure is 25 and the number of pink squares for the fourth figure is also 25. The table below displays this:
Pink squares
Total # of squares
5
5
3
3
25
25
41
41
61
Developing a formula
Using the formula an 2 + bn + c (given) and the difference method below. I am able to show a pattern emerging.
Position in sequence: 0 1 2 3 4 5 6
No. Of Squares (c): 1 1 5 13 25 41 61 . . . . . (i)
First Differences (a + b): 0 4 8 12 16 20 . . . . . . (ii)
Second Differences (2a): 4 4 4 4 4 . . . . . . . . (iii)
The bottom row of differences indicates a constant number, which shows there is a pattern. If the fourth row had not indicated a constant number pattern (i.e. 2, 2, 2, 2 or 6, 6, 6, 6) then I would have kept increasing the rows until I found one. If I was struggling to find a constant number, I would gather more information. This is what happened with my 3D investigation.
If n is the position in the sequence, 2a ...
This is a preview of the whole essay
Second Differences (2a): 4 4 4 4 4 . . . . . . . . (iii)
The bottom row of differences indicates a constant number, which shows there is a pattern. If the fourth row had not indicated a constant number pattern (i.e. 2, 2, 2, 2 or 6, 6, 6, 6) then I would have kept increasing the rows until I found one. If I was struggling to find a constant number, I would gather more information. This is what happened with my 3D investigation.
If n is the position in the sequence, 2a = 4 (iii) (therefore) a = 4 / 2 a = 2.
a + b = 0 (ii) and a = 2
We have 2 + b = 0 b = -2 and c = 1 (i).
Substituting the above into the formula a n 2 + b n + c, we have:
an 2 + bn + c
2n 2 - (2n) + 1
2n 2 - 2n + 1
Here we have a formula, 2n 2 - 2n + 1, for calculating the nth term in the sequence.
Testing the formula
I can test the formula by using it on previous shapes, which I have built out of the Lego bricks. I will test five of my results. On the next page, my testing methods and my results are shown.
Single square
an 2 + bn + c
2 x 1 2 - 2 x 1 + 1
Answer: 1
Status: Correct
Figure 3
an 2 + bn + c
2 x 4 2 - 2 x 4 + 1
Answer: 25
Status: Correct
Figure 1
an 2 + bn + c
2 x 2 2 - 2 x 2 + 1
Answer: 5
Status: Correct
Figure 4
an 2 + bn + c
2 x 5 2 - 2 x 5 + 1
Answer: 41
Status: Correct
Figure 2
an 2 + bn + c
2 x 3 2 - 2 x 3 + 1
Answer: 13
Status: Correct
Figure 5
an 2 + bn + c
2 x 6 2 - 2 x 6 + 1
Answer: 61
Status: Correct
So far, all of the figures that I have tested have been correct. This shows that the formula works for these shapes. I cannot, however, prove that it will work for any shape, though by testing some different sized figures, I can discover how many total squares they consist of and then draw them out, to see if the formula was correct. This will still not prove that it works for any number. However, by extending my two dimensional investigation, I will be able to make my investigation more reliable, though not certain.
I have drawn two other figures. They are supposed to consist of 85 and 113 squares. I will now check if is correct.
an 2 + bn + c
2 x 7 2 - 2 x 7 + 1
Answer: 85
Counted Answer: 85
Status: Correct
an 2 + bn + c
2 x 8 2 - 2 x 8 + 1
Formula Answer: 113
Counted Answer: 113
Status: Correct
Evaluation
There is not a way to make sure that the formula will work for every sized shape but I tried to make it more reliable by showing that it works with all the ones tested and the two other figures that I had not investigated before, which were both correct.
The only way to be certain that the formula works with every sized cross shape is by testing it on every possible sized one, which would be very difficult. This is why I feel that I have generated enough information to show that the formula that I have found out, 2n 2 - 2n + 1, gives the expected results.
(3D investigation on next page)
Three Dimensional investigation
Introduction
I will now investigate to find the three dimensional formula. I am going to start by building the first figure and adding on the borders. I again used Lego bricks to make the figures. The larger diagrams for the figures in 3D are too complex to see all the squares on. The figure below is the first one in the sequence with border squares around it and you can see all but the middle square.
Obtaining results
I obtained my results by building this three dimensional figure up as in the two dimensional investigation. The three-dimensional figures are more complicated than the two-dimensional figures, due to the 3D ones having another dimension to draw.
I used Lego bricks again for they helped me to gather my data. This is because they give a visualisation of the problem. This is easier to understand than trying to work it out in my head.
Analysing results
I think that investigating five different figures will give me enough information to find a formula. This is because the information I gathered will give me enough data to analyse and utilise my results. The data I have found out is shown in the table below. I have done one less shape size in this investigation because the 3D shapes required more bricks than I had. So I therefore stopped at the fifth figure.
Border Squares
Inner Squares
Total Squares
0
6
7
8
7
25
38
25
63
66
63
29
From these results I cannot see a direct relationship with the amount of border squares. I will investigate this further as I think that this relationship will be important in finding the formula to discover how many squares are required to surround the inner squares, thus making a bigger figure.
One relationship with the results is that the total number of inner squares is the same number as the total number of squares on the previous shape. For example, the total number of squares for the third figure is 63 and the number of inner squares for the fourth figure is also 63.
Additionally, the total amount of squares is always odd and the amount of border squares is always even with the inner squares always resulting in an odd number.
Finding a formula
I will apply the difference table method to attempt to find a formula for this investigation. The results are shown below:
Position in sequence: 0 1 2 3 4 5
No. Of Squares: -1 1 7 25 63 129 . . . . . . . (i)
First Differences: 2 6 18 38 66 . . . . . . . . (ii)
Second Differences: 4 12 20 36 . . . . . . . . (iii)
Third Differences: 8 8 8 . . . . . . . . . . . . . (iiii)
The bottom row of differences indicates a constant number, which shows there is a pattern. If the fourth row had not indicated a constant number pattern (i.e. 5, 5, 5, 5 or 9, 9, 9, 9) then I would have kept adding rows until I found one. If I were struggling to find a constant number, I would gather more information. This is what happened with my 2D investigation. Due to the 2D investigation being a slightly easier, I only did three tests, those three tests being with the first, second and third figure. This was not enough information for me to discover a direct link, so when I got to several rows down, I decided to gather more information.
We can now use the equation an 3 + bn 2 + cn + d to find the formula. So now we have to find the values for the variables so we can put them into the equation.
We already know the value of n which indicating the position of the figure in the sequence, so we can find out the values of a, b, c and d.
However, from this I am unable to determine a formula; consequently I will look at another method.
I have noticed that the shape is made up from a series of 2D shapes. This is shown in the diagram below:
Figure 2:
This would indicate that I could use some sort of combination of my 2D formula (2n²-bn+c).
This is the standard formula: (2n2 - 2n + 1) + 2
This formula works out the area for the 2D shape (the middle one in the above diagram) and then adds 2 to the answer because every three dimensional shape (apart from the single square) has two end pieces. This makes the correct answer for the shape shown below.
As you can see, two end pieces (the +2 of the formula) are added on to either end to make the shape on the right hand side. This formula works for this shape only (as it has only the 2D layer and the two added pieces on the end).
n = 2:
2D (centre) Layer + 2x End Squares
= (2n2 - 2n + 1) +2
= (2 x 22 - 2 x 2 + 1) +2
= (2 x 4 - 4 +1) +2
= (8 - 4 +1) +2
= (4+1) +2
= 5 +2
= 5 + 2
= 7
This formula would obviously also work for figure 1, when n=1(where there is only one square); without the added '2' on the end of the formula:
n = 1:
2D (centre) Layer
= (2n2 - 2n + 1)
= (2 x 12 - 2 x 1 + 1) Figure 1:
= (2 x 1 - 2 + 1)
= (2 - 2 + 1)
= (0 + 1)
= 1
However, this formula alone will not work for a shape with more layers. So to find the formula for more layers, I have worked out that extra information has to be added to the equation to take account of the additional layers (next page).
The extended formula is shown below:
(2n2 - 2n + 1) + 2 x [(2(n-1) 2 -2(n-1)+1)] + 2
This formula will only work with the 3rd term in the sequence. Below is the formula broken down to show why the mathematics is there:
This works out the 2D shape: (2n2 - 2n + 1)
This works out the second layer: 2 x [(2(n-1) 2 -2(n-1)+1)]
This is the value of the end squares: 2
So when they are joined together they form the formula:
(2n2 - 2n + 1) + 2 x [(2(n-1) 2 -2(n-1)+1)] + 2
So we can test this formula by doing the calculations for the above diagram.
n = 3:
2D (centre) Layer + 2x Second Layers + 2x End Squares
= (2n2 - 2n + 1) + 2 x [(2(n-1) 2 -2(n-1)+1)] +2
= (2 x 32 - 2 x 3 + 1) + 2 x [(2x22 -2x2)+1)] +2
= (2 x 9 - 6 + 1) + 2 x [(8 - 4)+1)] +2
= (18 - 6 + 1) + 2 x (4+1) +2
= (12 + 1) + 2 x 5 +2
= 13 + 10 +2
= 13 + 10 + 2
= 25
This also works out as predicted.
To check further I will calculate one more pattern, when n=4. The diagram of this is shown below.
This time, there will be another term in the calculation to calculate the additional layer. The formula for this will be:
(2n2 - 2n + 1) + 2 x [(2(n-1) 2 -2(n-1)+1)] + 2 x [(2(n-2) 2 -2(n-2)+1)] + 2
Below is the working out for the above diagram. There is another layer this time because the 3D shape's size has increased. So my prediction for this figure is that it will have a total square amount of 63.
n = 4:
2D (centre) Layer + 2x Second Layers + 2x Third Layers + 2x Ends
= (2n2 - 2n + 1) + 2 x [(2(n-1) 2 -2(n-1)+1)] + 2 x [(2(n-2) 2 -2(n-2)+1)] +2
= (2 x 42 - 2 x 4 + 1) + 2 x [(2x32 -2x3)+1)] + 2 x [(2(4-2) 2 -2(4-2)+1)] +2
= (2 x 16 - 8 + 1) + 2 x [(18 - 6)+1)] + 2 x [(2x4 -2x2+1)] +2
= (32 - 8 + 1) + 2 x (12+1) + 2 x (8 -4+1) +2
= 24 + 1 + 2 x 13 + 2 x 5 +2
= 25 + 26 + 10 +2
= 25 + 26 + 10 + 2
= 63
Both the count of the squares and the formula's answer were correct. So from all of my investigation into this formula I now believe that I can predict what the total amount of squares for figure n will be.
Here is an example of how to find the 5th term:
I have calculated the answer as 129, because I built it with the Lego bricks. I will now attempt to calculate it with my formula. Firstly, I will apply the formula 2n-1 to discover how many layers there are; 2 x 5 - 1 = 9 - this shows that there are 9 layers. I can take three layers off this because I know that there will be one 2D layer, the middle one and two end squares. So I will need to work out 6 layers values. Below is the formula:
(2n2 - 2n + 1) + 2 x [(2(n-1) 2 -2(n-1)+1)] + 2 x [(2(n-2) 2 -2(n-2)+1)] + 2 x [(2(n-3) 2 -2(n-3)+1)] + 2
I will now work this formula out and find n = 5:
(2n2 - 2n + 1) + 2 x [(2(n-1) 2 -2(n-1)+1)] + 2 x [(2(n-2) 2 -2(n-2)+1)] + 2 x [(2(n-3) 2 -2(n-3)+1)] + 2
= (2 x 52 - 2 x 5 + 1) + 2 x [(2(5-1) 2 -2(5-1)+1)] + 2 x [(2(5-2) 2 -2(5-2)+1)] + 2 x [(2(5-3) 2 -2(5-3)+1)] + 2
= (2 x 25 - 10 + 1) + 2 x [(2(4) 2 -2(4)+1)] + 2 x [(2(3) 2 -2(3)+1)] + 2 x [(2(2) 2 -2(2)+1)] + 2
= (50 - 10 + 1) + 2 x [(2x16 -2x4+1)] + 2 x [(2x9 -2x3+1)] + 2 x [(2x 4 -2x2+1)] + 2
= (40 + 1) + 2 x (32 -8+1) + 2 x (18 -6+1) + 2 x (8 -4+1) + 2
= 41 + 2 x (24+1) + 2 x (12+1)] + 2 x (4+1) + 2
= 41 + (2 x 25) + (2 x 13) + (2 x 5) + 2
= 41 + 50 + 26 + 10 + 2
= 129
Again, this agrees with my calculation and I now believe that this formula could calculate all the possible sequences.
I have also observed that there may be a method of calculating how many layers there are in the sequence number.
The above diagram shows figure 2 of the 3D investigation; it is built up of three layers. The formula 2n-1, with the value of n being the term in the sequence (in this case 2) so the formula ends up as 2(2)-1, which is 3. This is correct in this case, as there are 3 layers and the formula is the same value.
For the third sequence, I will use the same formula of 2n-1. This time n will be the value of 3. So the formula will become 2(3)-1 which equals to 5. The diagram over shows what the value is.
As the diagram shows, there are five layers. The formula 2n-1 has been successful so far. I will try one more, to make sure the formula is reliable enough. I will try the third figure; there should be 2n-1 layers. So by substituting the value of n for 4 (the term in the sequence is 4), the formula now reads 2(4)-1, which equals 7. So I predict that there will be 7 layers to the third figure. The diagram for this shape is shown below.
This is also the expected answer.
Evaluation
Due to there being too many variables in the equation an 3 + bn 2 + cn + d, I was unable to resolve the problem in this manner. Instead I decided to use a pattern I noticed while investigating the Lego brick models. I think the new method, which I have developed, could be simplified with further development to give a much more concise formula.
There is not a way to make sure that the formula will work for every sized figure but I tried to make it more reliable by showing that it works with the ones I built with Lego bricks and with other figures that I had not investigated before, which were all correct.
The only way to be certain that the formula works with every sized figure is by testing it on every possible sized one, which would be very difficult and tedious. That is why I feel that I have generated enough information to show that the formula I have found gives the expected results.
Alexandra Leckie 11A2 Maths "Borders" Coursework
