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• Level: GCSE
• Subject: Maths
• Word count: 4165

# T-totals. I am going to investigate the relationship between the t-total, T, and the t-number, n. The t-number is always the number at the bottom of the t-shape when it is orientated upright.

Extracts from this document...

Introduction

T-Totals

 1 2 3 11 20

I am going to investigate the relationship between the t-total, T, and the t-number, n.  The t-number is always the number at the bottom of the t-shape when it is orientated upright.  Here the t-number would be 20.  The t-total is the sum of the cells inside the t-shape.  Here it would be 37 as 1+2+3+11+20 = 37

I will calculate the t-total for different t-numbers on 9×9 grid.  Working algebraically, I will find a relationship that will express the t-total in terms of the t-number and the grid size.  I will test this generalisation for t-shapes in an 8×8 and 10×10 grid.  I will then transform the t-shape and investigate it’s affect on this relationship.

T-shapes in a 9×9 grid

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 n T T=5n-63 20 37 5(20)-63=37 23 52 5(23)-63=52 26 67 5(26)-63=67 47 172 5(47)-63=172 50 187 5(50)-63=187 53 202 5(53)-63=202 74 307 5(74)-63=307 77 322 5(77)-63=322 80 337 5(80)-63=337

The t-total increases by 135 when the t-number is increased by 27 and by 15 when the t-number increases by 3.  It thus follows that for every unit increase in the t-number there will be an increase of 5 in the t-total.

 n-20 n-18 n-19 n-9 n

The t-shapes in a 9×9 grid can be represented algebraically.  The t-total, T, can therefore be written in terms of the t-number, n, as T= 5n - 63.  Using similar reasoning we can express T in terms of n in an 8×8 and 10×10 grid.

T-shapes in an 8×8 grid

 1 2 3 4 5 6 9 10 11 12 13 14 17 18 19 20 21 22 25 26 27 28 29 30 33 34 35 36 37 38 41 42 43 44 45 46 n T T=5n-56 18 34 5(18)-56=37 21 49 5(21)-56=52 42 154 5(42)-56=67 45 169 5(45)-56=172

The t-total increases by 120 when the t-number is increased by 24 and by 15 when the t-number increases by 3.  It thus follows that for every unit increase in the t-number there will be an increase of 5 in the t-total.

 n-17 n-16 n-15 n-8 n

The t-shapes in an 8×8 grid can be represented algebraically.  The t-total, T, can therefore be written in terms of the t-number, n, as T= 5n - 56.

T-shapes in a 10×10 grid

n

T

T=5n-70

22

40

5(22)-70=37

25

45

5(25)-70=52

28

50

5(28)-70=67

52

190

5(52)-70=172

55

195

5(55)-70=187

58

200

5(58)-70=202

1

2

3

4

5

6

7

8

9

11

12

13

14

15

16

17

18

19

21

22

23

24

25

26

27

28

29

31

32

33

34

35

36

37

38

39

41

42

43

Middle

-2

-3

31

224

5 {31 - 2 + 3(9) } - 7(8) = 224

10×10

1

3

55

75

5 {55 + 1 – 3(9) } - 7(10) = 75

-3

1

66

200

5 {66 – 3 – 1(9) } - 7(10) = 200

3

-2

74

405

5 {74 + 3 + 2(9) } - 7(10) = 405

-2

-2

27

145

5 {27 - 2 + 2(9) } - 7(10) = 145

The t-total for the translated t-number agrees with the results from the formula.  We can therefore say that T = 5 (n +a - bg) -7g will work for any translation of the form  in any grid size.

Rotations

In the same way that we justified T= 5n-7g, we can rotate the t-shape about the t-number and express the other cells in terms of n (t-number) and g (grid size).  Collecting like terms will simplify a formula for the t-total, T, of the rotated t-shape.  We can then validate the formula in different grid sizes.

Rotation of 90º clockwise/ 270º anti-clockwise

 n-2g-1 n-2g n-2g+1 n-g n n+2-g n n+1 n+2 n+2+g

T= (n) + (n +1) + (n + 2) + (n + 2 –g) + (n + 2) + (n + 2 + g)

T= 5n + 7

Rotation of 180º

 n-2g-1 n-2g n-2g+1 n-g n n n+g n+2g-1 n+2g n+2g+1

T= (n) + (n + g) + (n + 2g) + (n + 2g –1) + (n + 2g +1)

T= 5n + 7g

Rotation of 270º clockwise/ 90º anti-clockwise

 n-2g-1 n-2g n-2g+1 n-g n n-2-g n-2 n-1 n n-2+g

T= (n) + ( n – 1) +( n - 2 )+ (n – 2 – g) + (n – 2 + g)

T= 5n + 7g

Validation

 g n 90º clockwise 180º 270º clockwise T T=5n+7 T T=5n+7g T T=5n-7 9×9 31 157 5(31)+7=157 218 5(31)+7(9)=218 143 5(31)-7=143 40 207 5(40)+7=207 263 5(40)+7(9)=263 193 5(40)-7=193 49 252 5(49)+7=252 308 5(49)+7(9)=308 238 5(49)-7=238 8×8 20 107 5(20)+7=107 156 5(20)+7(8)=156 93 5(20)-7=93 28 147 5(28)+7=147 196 5(28)+7(8)=196 133 5(28)-7=133 36 187 5(36)+7=187 236 5(36)+7(8)=236 173 5(36)-7=173 10×10 45 232 5(45)+7=232 295 5(45)+7(10)=295 218 5(45)-7=218 55 282 5(55)+7=282 345 5(55)+7(10)=345 268 5(55)-7=268 65 332 5(65)+7=332 395 5(65)+7(10)=395 318 5(65)-7=318

As the results from the formulae agree with the t-total of the rotated t-shape, we can confirm the following formulae for a rotation about the t-number in any grid size:

• 90º clockwise: T = 5n + 7
• 180º: T = 5n – 7g
• 270º clockwise: T= 5n - 7

Rotation of 90º clockwise/ 270º anticlockwise about an external point

The distance from the t-number to the centre of rotation is described by the column vector  where c is the horizontal distance (positive to the right and negative to the left) and d is the vertical distance (positive being upwards and negative being downwards).

The t-total of any t-shape that has been rotated 90º clockwise or 270º anticlockwise about an external point is given by (T = t-total, n = t-number, g = grid size):

T = 5 ( n + c – dg –d –cg ) + 7

Justification

 n-2g-1 n-2g n-2g+1 n-g centre of rotation n d c

The general t-shape with t-total T= 5n - 7g is clearly shown.  The distance from the t-number, n, to the centre of rotation is described by the vector .

 n-2g-1+c-dg n-2g+c-dg n-2g+1+c-dg n-2g-1 n-2g n-2g+1 n-g+c-dg n-g n+c-dg n d c

Translate the t-number, n, to the centre of rotation.  By using the earlier justification, a translation by the vector  will add (c-dg) to every cell.  The t-number thus becomes:

n + c – dg

 n-2g-1+c-dg-d-cg n-2g+c-dg-d-cg n-2g+1+c-dg-d-cg n-g+c-dg-d-cg n+c-dg-d-cg-d n-2g-1+c-dg n-2g+c-dg n-2g+1+c-dg c n-g+c-dg n+c-dg

The position of t-shape following a rotation of 90º clockwise has been outlined by a dotted line.  The vector  will translate the t-number,    n+ c–dg, to the position that the t-number will take when the original t-shape has been rotated.  By using the earlier justification, a translation by the vector  will add (-d-cg) to every cell.  The t-number thus becomes:

n + c – dg –d -cg

 n-2g-1+c-dg-d-cg n-2g+c-dg-d-cg n-2g+1+c-dg-d-cg n-g+c-dg-d-cg n+c-dg-d-cg+2-g n+c-dg-d-cg n+c-dg-d-cg+1 n+c-dg-d-cg+2 n+c-dg-d-cg+2+g

Rotate the t-shape 90º clockwise about the t-number, n+c-dg-d-cg.  By using the earlier justification that a rotation of 90º clockwise about the t-number, n, has a t-total T=5n + 7, the t-total of a t-shape that has been rotated 90º about an external point becomes:

T= 5 ( n + c – dg –d -cg ) + 7.

Rotation of 180º about an external point

The t-total of any t-shape that has been rotated 180º about an external point is given by (T = t-total, n = t-number, g = grid size):

T = 5 (n + 2c – 2dg )+ 7g

Justification

We can assume that the t-number, n, has been translated to the centre of rotation by the vector .  The t-number thus becomes n+c-dg.

 n-2g-1+2c-2dg n-2g+2c-2dg n-2g+1+2c-2dg n-2g-1+c-dg n-2g+c-dg n-2g+1+c-dg n-g+2c-2dg n-g+c-dg n+2c-2dg n+c-dg d c

The position of the t-shape following a rotation of 180º has been outlined by a dotted line.  The vector  will translate the t-number, n+c–dg, to the position that the t-number will take when the original t-shape has been rotated.  By using the earlier justification, a translation by the vector  will add (c-dg) to every cell.  The t-number thus becomes:  n + 2c – 2dg

 n-2g-1+2c-2dg n-2g+2c-2dg n-2g+1+2c-2dg n-g+2c-2dg n+2c-2dg n+2c-2dg+g n+2c-2dg+2g-1 n+2c-2dg+2g n+2c-2dg+2g+1

Rotate the t-shape 180º about the t-number,  n+2c-2dg.  By using the earlier justification that a rotation of 180º about the t-number, n, has a t-total T=5n + 7g, the t-total of a t-shape that has been rotated 180º about an external point becomes:

T= 5 ( n + 2c – 2dg ) + 7g.

Rotation of 270º clockwise/ 90º anticlockwise about an external point

The t-total of any t-shape that has been rotated 270º clockwise or 90º anticlockwise about an external point is given by (T = t-total, n = t-number, g = grid size):

T = 5 ( n + c – dg +d +cg ) - 7

Justification

We will assume that the t-number, n, has been translated to the centre of rotation by the vector.  The t-number thus becomes n+c-dg.

 n-2g-1+c-dg n-2g+c-dg n-2g+1+c-dg n-g+c-dg n+c-dgd n-2g-1+c-dg+d+cg n-2g+c-dg+d+cg n-2g+1+c-dg+d+cg -c n-g+c-dg+d+cg n+c-dg+d+cg

The position of the t-shape following a rotation of 270º clockwise has been outlined by a dotted line.  The vector  will translate the t-number, n + c –dg, to the position that the t-number will take when the original t-shape has been rotated.  By using the earlier justification, a translation by the vector  will add   (d+ cg) to every cell.  The t-number thus becomes:

n + c –dg + d +cg

 n-2g-1+c-dg+d+cg n-2g+c-dg+d+cg n-2g+1+c-dg+d+cg n+c-dg+d+cg-2-g n-g+c-dg+d+cg n+c-dg+d+cg-2 n+c-dg+d+cg-1 N+c-dg+d+cg n+c-dg+d+cg-2+g

Conclusion

a

b

c

d

2

2

1

1

61

322

5 { 61 + (-1)(-10) - (-1)(12) + 2 – 22} + 7 = 322

2

2

-1

1

61

422

5 { 61 + (-3)(-10) - (-1)(12) + 2 – 22} + 7 = 422

2

2

1

-1

61

442

5 { 61 + (-1)(-10) - (-3)(12) + 2 – 22} + 7 = 442

2

2

-1

-1

61

542

5 { 61 + (-3)(-10) - (-3)(12) + 2 – 22} + 7 = 542

-2

2

1

1

61

102

5 { 61 + (3)(-10) - (-1)(12) - 2 – 22} + 7 = 102

-2

2

-1

1

61

202

5 { 61 + (1)(-10) - (-1)(12) - 2 – 22} + 7 = 202

-2

2

1

-1

61

222

5 { 61 + (3)(-10) - (-3)(12) - 2 – 22} + 7 = 222

-2

2

-1

-1

61

322

5 { 61 + (1)(-10) - (-3)(12) - 2 – 22} + 7 = 322

Translate and rotate 180º

T = 5 {n + 2 (c - a) – 2g (d - b) + a – bg } + 7g

2

-2

1

1

61

162

5 { 61 + (2)(-1) – (22)(3) + 2 +22 } + 77 = 162

2

-2

-1

1

61

142

5 { 61 + (2)(-3) – (22)(3) + 2 +22 } + 77 = 142

2

-2

1

-1

61

382

5 { 61 + (2)(-1) – (22)(1) + 2 +22 } + 77 = 382

2

-2

-1

-1

61

362

5 { 61 + (2)(-3) – (22)(1) + 2 +22 } + 77 = 362

Translate and rotate 270º clockwise

T = 5 { n + (c - a) (1 + g) –  (d - b) (g – 1) + a – bg } – 7

-2

-2

1

1

61

428

5 { 61 + (3)(12) - (3)(10) - 2 + 22} - 7 = 428

-2

-2

-1

1

61

308

5 { 61 + (1)(12) - (3)(10) - 2 + 22} - 7 = 308

-2

-2

1

-1

61

528

5 { 61 + (3)(12) - (1)(10) - 2 + 22} - 7 = 528

-2

-2

-1

-1

61

408

5 { 61 + (1)(12) - (1)(10) - 2 + 22} - 7 = 408

The t-total for the rotated t-shape agrees with the results from the formula.  We can therefore say that our formulae are consistent.

Summary

T-total

T = 5n – 7g

Translation by vector

T = 5 ( n + a - bg ) - 7g

Rotations about an external point that is vector from t-number

• Rotate 90º clockwise                 T = 5 ( n + c – dg –d –cg ) + 7
• Rotate 180º                                 T = 5 (n + 2c – 2dg )+ 7g
• Rotate 270º clockwise                T = 5 ( n + c – dg +d +cg ) – 7

* for rotations about the t-number c and d will equal zero.

Combined transformation: rotation followed by a translation

• Rotate 90º clockwise and translate

T = 5 ( n + c – dg –d –cg + a – bg ) + 7

• Rotate 180º and translate

T = 5 (n + 2c – 2dg + a – bg )+ 7g

• Rotate 270º clockwise and translate

T = 5 ( n + c – dg +d +cg + a – bg ) – 7

Combined transformation: translation followed by a rotation

• Translate and rotate 90º clockwise

T = 5 { n + (c - a) (1-g) – (d - b) (g+1) + a – bg } + 7

• Translate and rotate 180º

T = 5 {n + 2 (c - a) – 2g (d - b) + a – bg } + 7g

• Translate and rotate 270º clockwise

T = 5 { n + (c - a) (1 + g) –  (d - b) (g – 1) + a – bg } – 7

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