T-Shapes Coursework
We have been asked to investigate the relationship between the T-total and the T-number for our T-shapes coursework. We have been given our first T-shape on a 9 by 9 grid. This T-shape is shown below:
The total of the numbers inside the T-shape is
1 + 2 + 3 + 11 + 20 = 37
this is called the T-total. The T-number for this
T-shape is 20.
T-number
I have now moved the shape over one square on the grid. The new total of the numbers inside the shape is 2 + 3 + 4 + 12 + 21 = 42. Looking at this I can see that 5 has been added to the T-total. This is because when moved over, each number increases by one, and because there are 5 squares in the T-shape, the number increases by 5. Knowing this I can now predict that if I move the T-shape over by another square, the new T-total will be 47.
I have now moved the T-shape over by another square and the new T-total is
3 + 4 + 5 + 13 + 22 = 47. This is what I originally predicted.
T-total of original T-shape.
37
T-total after first translation.
42
T-total after second translation.
47
I have recorded the results of the original T-shape total and the
T-total after each translation in the table above. I have noticed that after each translation 5 has been added to the T-total. This is because when the T-shape is moved across once on the grid, one is added to each number, and because there are 5 numbers in the T-shape, 5 is added all together.
I have now used a different T-shape; this is still on a 9 by 9 grid. The total for this T-shape is 4 + 5 + 6 + 14 + 23 = 52.
Now, I have moved my T-shape down by one on the grid. The new T-total for this shape is 97. This is because when the shape is moved down, 9 is added to each square because it is on a 9 by 9 grid. So, because there are 5 squares in the T-shape, 45 is added (9x5). I can now predict that if I move my T-shape down again, the T-total will equal 142.
I have worked out the T-total of my new T-shape to be 142. My prediction was correct.
T-total of original T-shape.
52
T-total after first translation.
97
T-total after second translation.
42
I have recorded the results of the original T-shape total and the
T-total after each translation in the table above. I have noticed that after each translation 45 has been added to the T-total. This is because when the T-shape is moved down once on the grid, 9 is added to each number, and because there are 5 numbers in the T-shape, 45 is added all together.
Looking at the T-shape on the left I can see that by
using algebra we can work out an expression for each
T-number. When the expressions are added together
they equal 5t-63. "t" is the T-number. So 5t is 5 lots of the T-number.
I have tested my expression to see if it works. In this case, 5t would be 100, and 100-63 is 37. I can now work out the T-total and if it is 37 my expression will work wherever it is put on a 9 by 9grid.
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Looking at the T-shape on the left I can see that by
using algebra we can work out an expression for each
T-number. When the expressions are added together
they equal 5t-63. "t" is the T-number. So 5t is 5 lots of the T-number.
I have tested my expression to see if it works. In this case, 5t would be 100, and 100-63 is 37. I can now work out the T-total and if it is 37 my expression will work wherever it is put on a 9 by 9grid.
+ 2 + 3 + 11 + 20 = 37
I am now using a 10 by 10 grid. The T-total for this shape is 1 + 2 + 3 + 12 + 22 = 40. Knowing this I can now predict that if I move the T-shape across once in the grid 5 will be added to the T-total.
The T-total for my new T-shape is
2 + 3 + 4 + 13 + 23 = 45. my prediction was correct as 5 has been added to my original T-total. This is for the same reason as on a 9 by 9 grid, when the T-shape is moved across on the grid, one is added to each number and because there are 5 squares, 5 is added.
I now know that because I have moved the T-shape over again, 5 will be added to my previous T-total. To my new T-total will be 50. 3 + 4 + 5 + 14 + 24 = 50.
T-total of original T-shape.
40
T-total after first translation.
45
T-total after second translation.
50
I have recorded my results into a table. I can see that when the shape is moved across one on the grid, 5 is added each time.
I have now used a different T-shape, still on a 10 by
10 grid. The T-total for this T-shape is
6 + 7 + 8 + 17 + 27 = 65
I have now moved my T-shape down once on the grid. My new T-total is 16 + 17 + 18 + 27 + 37 = 115. 50 has been added to my previous T-total. This is because each square has moved down on the grid once, and because it is a 10 by 10 grid, 10 has been added to each square. So because there are 5 squares, 50 had been added. I can predict that if I move the T-shape down once again, 50 will be added again.
The T-total for my new T-shape is
26 + 28 + 29 + 38 + 48 = 165. To work out if my theory is correct I will take 50 off my new T-total and see if it equals 115, my previous T-total.
65 - 50= 115. My theory was correct.
T-total of original T-shape.
65
T-total after first translation.
15
T-total after second translation.
65
I have recorded my results into a table. I can see that when the shape is moved down once on the grid, 50 is added each time.
Looking at the T-shape on the left I can see that by using algebra we can work out an expression for each square in the T-shape. When the expressions are added together they equal 5t-70.
I have tested my theory to see if it works. 5t in this case would be 55 x 5 which is 275. 275 - 70 = 205. So if I add all the numbers in my T-shape they should equal 205. 34 + 35 + 36 + 45 + 55 = 205.
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I will now repeat the steps I did for a 9 by 9 and 10 by 10 grid for an 8 by 8 grid.
The T-total for this shape on an 8 by 8 grid is
2 + 3 + 4 + 11 + 19 = 39
I have moved the T-shape over by one square and I
believe that, just the same as the 9 by 9 and 10 by 10 grid, 5 will be added to the T-total. So the T-total of this shape should equal 44.
3 + 4 + 5 + 12 + 20 = 44
I have moved the T-shape over another square and I expect the new T-total to be 49.
4 + 5 + 6 + 13 + 21 = 49
T-total of original T-shape.
39
T-total after first translation.
44
T-total after second translation.
49
I have recorded the results of the original T-shape total and the
T-total after each translation in the table above. I have noticed that after each translation 5 has been added to the T-total. This is because when the T-shape is moved across once on the grid, one is added to each number, and because there are 5 numbers in the T-shape, 5 is added all together.
The T-total for this shape is 94.
13 + 14 + 15 + 22 + 30 = 94
I have now moved the T-shape down one square on the grid. My new T-total is
21 + 22 + 23 + 30 + 38 = 134. 40 has been added to my previous T-total. I now predict that if I move my T-shape down one more square, 40 will be added to my new T-total making it 174.
The T-total for this shape is
29 + 30 + 31 + 38 + 46 = 174 so my prediction was correct.
T-total of original T-shape.
94
T-total after first translation.
34
T-total after second translation.
74
I have recorded my results into a table. I can see that when the shape is moved down once on the grid, 40 is added each time.
Looking at my T-shape I can use algebraic expressions to work out the total of each T-number. When the expressions are added together they equal 5t-56
I have tested to see if the expression I created works. If I add all the T-numbers up in my T-shape up they should equal 5t, which is 155, minus 56, which is 99.
4 + 15 + 16 + 23 + 31 = 99. This shows that my expression works.
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The algebraic expression that can be used for any grid size (g grid) is T-total = 5t - 7g. I know this because for a 10 grid the expression used is 5t - 70, for a 9 grid the expression used is 5t - 63, and for an 8 grid the expression used is 5t - 56. Each time, the number that is taken away from 5t is 7 lots of the grid number. ( 7 x 10 = 70, 7 x 9 = 63 & 7 x 8 = 56). Knowing this I can now test this expression on a random grid size, which I have chosen to be 11 by 11.
30 x 5 = 150
7 x 11 = 77
150 - 77 = 73
The T-total should equal 73
7 + 8 + 9 + 19 + 30 = 73
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By applying the translation the total increases by +5m - 5ng from its original total. I know this because the translation m moves the
T-shape horizontally, and the translation n moves the T-shape vertically. When the T-shape is moved horizontally, 1 is added (or taken away if m < 0) to each square. So 5 is added altogether. So because "m" represents the number that the shape is translated by, 5m shows that the number that it is translated by is times by 5.
5ng is taken away because the shape is moved upwards by the value "n" in the vector. So because there are 5 boxes, each time the translation takes place, 5 lots of the translation are taken away (or added if n < 0.) And because each time n lots of the grid size is taken away and there are 5 boxes it totals 5ng which is the same as 5 lots of the horizontal movement times the grid size. The total algebra for this T-shape will be 5t - 7g + 5m - 5ng.
To demonstrate that this works I will use a basic 10 by 10 grid. The T-shape I will use is on the left. The total for this T-shape is 90. By applying the translation I would expect 5 to be added because of the value "m" and 50 to be taken away because on the value "n" which means 5 times the grid size. So I would expect my new T-total to be 45.
I have worked out the T-total of my new T-shape after the translation.
2 + 3 + 4 + 13 + 23 = 45. So my expression works correctly.
Now, I have carried out a vertical reflection through the T-number. I know the algebraic expression for my T-shape is 5t - 7g .If the shape is reflected, the expression will stay the same because the shape does not move.
I have now carried out a reflection of my T-shape in a
horizontal line through the T-number. The algebra for my original T-shape was 5t - 7g. So, now by reflecting the T-shape through the T- number, the
algebra will be reversed and instead of the grid size (g) being taken away from each box, it will now be added making the algebra 5t + 7g. This is because when the shape is reflected, the squares below the
T-number are used and because they are below the
T-number, they will be increasing by the number of the grid size instead or decreasing.
To demonstrate this I will use the T-shape to my left. I will use a 10 by 10 grid. The numbers in the original T-shape add up to 110. By using the algebra I know that after carrying out a reflection through the T-number my T-total would be 250. This is because 5t has been added (180) and 7g has been added (70.)
36 + 46 + 56 + 55 + 57 = 250. My algebra has been tested and proven correct.
Finally I am going to reflect my T-shape through a consecutive horizontal line and see how this affects my algebra.
T
T-g
T-2g
T-2g+1
T-2g-1
5T-7g
T
T+g
T+2g
T+2g+1
T+2g-1
5T+7g
T+2g
T+2g+1
T+2g-1
T+3g
_T+4g_
5T+13g
T+4g
T+5g
T+6g
T+6g+1
T+6g+2
5T + 27g
I have noticed that each time the shape is reflected to make a consecutive pattern the number that you multiply the grid by increases by one per square. I will now test the algebra on a size 10 grid to see if my theory works.
2
3
2
22
20
To work out if my formula
works I will use my T-number
22 (22) to calculate the T-totals.
32
42
41
43
80
41
42
43
52
62
240
62
72
82
81
83
380
Yellow
T = 22
5t - 7g
10 - 70 = 40
Pink
5t + 7g
10 + 70 = 180
Black
5t + 13g
10 + 130 = 140
Green
5t + 27g
10 + 270 = 380
