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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                 -  -

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