# T-total coursework

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Introduction

Mathematics Coursework 1 – “T-Totals”

Part 1

Investigate the relationship between the T-total and the T-number

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 |

1 | 2 | 3 | ||||||

10 | 11 | 12 | ||||||

19 | 20 | 21 |

The T-number is always the number at the bottom of the T, so in this case (the yellow T) it is 20. The sum of all the numbers in the T is the T-total.

In this case, the T-number, (which will now be called n) is 20, and the T-total, (which will now be called T) is 37.

If the number at the bottom of the T is n, these are the other numbers in terms of n, bearing in mind that the width of the grid is 9 (by 9):

n-19 | n-18 | n-17 |

10 | n-9 | 12 |

19 | n | 21 |

The number above the T-number is (n-9) because this number is exactly one row above (n). The width of the grid is 9, so by moving up 1 cell from (n), I am decreasing the value by the width of the grid (9). The cell above (n-9) is (n-18) as the value has been decreased by the width of the grid again. The cell to the left of (n-18) is (n-19) as it has been decreased by 1. The cell to the right of (n-18) is (n-17) as it is 1 more than (n-18).

When these 5 terms are added together I get:

(n) + (n-9) + (n-17) + (n-18) + (n-19) = 5n – 63

The calculation above shows that the sum of the 5 terms within the T-shape is

5n – 63, therefore I can make a proper formula:

T = 5n - 63 where T is the T-total and

Middle

n-2w+1

10

n-w

12

19

n

21

These are the terms of the numbers in a T-shape when the width of the grid is (w); they are all in terms of (n).

1 | 2 | 3 |

10 | 11 | 12 |

19 | 20 | 21 |

11 | 12 | 13 |

10 | 21 | 12 |

19 | 30 | 21 |

If this T-shape (below left) is moved down by 1, and moved right by 1, and the width of the grid is 9 (w=9), I obtain a new T-shape. The old T-number is 20, and the T-total is 37, while in the new T-shape the T-number is 30 and the T-total is 87. If I rewrite these numbers in terms of

If the vertical translation is 1 (the T-shape is moved down by 1) then y = 1, while the horizontal translation is also 1 (the T-shape is moved right by 1) then x = 1. The diagram below left shows what

n-2w-1 | n-2w | n-2w+1 |

10 | n-w | 12 |

19 | n | 21 |

(n+x+wy-2w)-1 | n+x+wy-2w | (n+x+wy-2w)+1 |

10 | n+x+ wy-w | 12 |

19 | n+x+wy | 21 |

happens to the terms in the T-shape when the T-shape is moved down 1 and right 1.

The terms in the translated T-shape are quite similar to the original T-shape terms; (n), (n-w), (n-2w), (n-2w-1) and (n-2w+1). This means that the formula must have 5n and -7w in it, although they do not have to be together. Every term in the T-shape has had two things added to it; (x) which is horizontal translation, and (wy) which is grid width multiplied by the amount of vertical translation.

Each term has had (x) added to it because each square in the T-shape has been moved horizontally by (x). Because the numbers increase by 1 horizontally each time (e.g. 1 → 2 → 3) adding (x) would not make it move down (or up), just across i.e. it is horizontally translated.

Every term in the T-shape has also had (wy) added to it, and this is because they have been moved down by a number of rows i.e. a multiple of (w). This is vertical translation. When the vertical translation is 1, y = 1, and the grid width in this example is 9, so (wy) is 1 x 9 = 9. This means that the terms in the T-shape have been moved down by 9 squares, not 9 rows, and 9 squares is equal to 1 row. Therefore the T-shape has been moved down by 1 row.

When all of the terms in the translated (moved) T-shape are added together, I obtain this formula:

(n+x+wy) + (n+x+wy-w) + (n+x+wy-2w) + (n+x+wy-2w-1) +

(n+x+wy-2w+1) = 5n + 5x + 5wy – 7w and when this is factorised I get:

5(n + x + wy) - 7w

This formula has only been proven in one type of translation; down and right (1 type of diagonal translation, of which there are four types). This is an example of another type of diagonal translation.

78 | 79 | 80 |

10 | 89 | 12 |

19 | 99 | 21 |

IMPORTANT! | ||

Sign conventions | ||

Mvmnt. | Sign | |

↑ | y = negative | |

↓ | y = positive | |

← | x = negative | |

→ | x = positive | |

To translate a T-shape upwards, y must be a negative whole number, so that (wy) is a negative, and hence the T-shape is moved up- rows have been subtracted. To translate a T-shape to the right, x must be a positive integer; because this is the direction in which the numbers are increasing (they are decreasing to the left, so to move a T-shape to the left, x must be negative). |

14 | 15 | 16 |

10 | 25 | 12 |

19 | 35 | 21 |

T = 14 + 15 + 16 +25 + 35 = 105

T = (5 x 99) + (5 x -4) + (5 x (-6 x 10)) - (7 x 10) = 105

These two examples show that the formula works in at least half of all diagonal translations, so I will keep it:

T = 5(n + x + wy) - 7w where T is the T-total, n is the T-number, w is the grid width, x is the horizontal translation, and y is the vertical translation

NB: The T-shape must stay on the grid for the formula to remain valid; if one of the squares is not on the grid (i.e. an upside-down L shape), the formula will not work, as one of the terms is missing and the total (5n+5x+5wy-7w) is not reached.

Part 3

Use grids of different sizes again. Try other transformations and combinations of transformations. Investigate the relationships between the T-total, the T-number, the grid size, and the transformations.

There are 8 types of reflection overall in this project, but I will only investigate 4 of them.

Horizontal Reflection

n-2w-1 | n-2w | n-2w+1 |

10 | n-w | 12 |

19 | n | 21 |

n+(2h+1)w | ||

n+(2h+2)w | ||

n+(2h+3)w-1 | n+(2h+3)w | n+(2h+3)w+1 |

1 | 2 | 3 |

11 | 12 | 13 |

21 | 22 | 23 |

31 | 32 | 33 |

41 | 42 | 43 |

51 | 52 | 53 |

61 | 62 | 63 |

71 | 72 | 73 |

81 | 82 | 83 |

91 | 92 | 93 |

Conclusion

This is what I get when I add all the terms together:

(n+(2h+1)w+2h+1) + (n+(2h+1)w+2h+2) + (n+(2h+1)w+2h+3) + (n+2hw+2h+3) + (n+(2h+2)w+2h+3) = 5n + 10hw + 10h +5w + 12 and when this is factorised I get:

5(n + 2hw + 2h + w) + 12

In the example, the T-total is 447 (worked out by adding up the numbers). Using the formula, when (n) is 33, (w) is 10, and (h) is 2, the calculated T-total is 447, which means that the formula works.

T = 5(n + 2hw + 2h + w) + 12 where T is the T-total, n is the T-number, w is the grid width, and h is the distance from the horizontal line of reflection

There are three other types of diagonal reflection; using this 45° line, the original T-shape can be put in the bottom-right hand corner, and the formula for this is T = 5n – 10hw – 20w – 10h – 13 (or T = 5(n – 2hw – 4w - 2h) – 13), or I can put the original T-shape in either of the two remaining corners when I use a 45° line that is perpendicular to the one used above (2 diagrams shown in rectangular box); the formula for the first of the diagonal reflections in the rectangular box is T = 5n + 10hw + 5w – 10h – 12 or T = 5(n + 2hw + w – 2h) – 12. The formula for the last of the 8 cases of reflection is T= 5n + 10hw + 20w + 10h + 15 or T = 5(n + 2hw + 4w + 2h + 3).

NB: The (n) (T-number) denotes where the original T-shape is.

JA ~ 25th June 2006 - -

This student written piece of work is one of many that can be found in our GCSE T-Total section.

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## Here's what a teacher thought of this essay

An excellent piece of work taking a simple idea and generating some complex algebraic formulae and expressions. 5 stars.

Marked by teacher Mick Macve 18/03/2012