# Investigating T-shapes.

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Introduction

Sheena Robinson Maths Coursework 30/04/2007

Investigative Maths Coursework

We looked at a T-shape drawn on a nine width grid like this:

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 |

The total of the numbers inside the T-shape is called the T-total.

1 | 2 | 3 |

11 | ||

20 |

(1+2+3+11+20=37)

The number at the end of the stem of the T-shape is the T-number. This remains the same even if you rotate the T-shape.

Our first task was to translate the T-shape into different positions on the same sized grid and investigate the relationship between the T-total and T-number.

n = T-number t = T-total

(By ‘difference between’ I mean the amount added or subtracted to get to the next number in the second column)

n | t | Difference between | 5n | 5n - t | Therefore 5n - 63 |

20 | 37 | 100 | 63 | t | |

21 | 42 | 5 | 105 | 63 | t |

22 | 47 | 5 | 110 | 63 | t |

23 | 52 | 5 | 115 | 63 | t |

24 | 57 | 5 | 120 | 63 | t |

So the equation for finding the T-total anywhere on a nine width grid if you only know the T-number is 5n – 63 = t.

You can prove his by using algebra:

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

n-9 | ||

n |

n + (n - 9) + (n – 18) + (n – 17) + (n – 19)

= 5n – (9 + 18 +17 + 19)

= 5n – 63

This is so because each time you move up a row on a grid you take away one grid width and down you add it.

Middle

10 width:

n-21 | n-20 | n-19 |

n-10 | ||

n |

n + (n – 10) + (n – 20) + (n – 19) + (n – 21)

= 5n – (10 + 20 + 19 + 21)

= 5n – 70

11 width:

n-23 | n-22 | n-21 |

n-11 | ||

n |

n + (n – 11) + (n – 22) + (n – 21) + (n – 23)

= 5n – (11 + 22 + 21 + 23)

= 5n – 77

12 width:

n-25 | n-24 | n-23 |

n-12 | ||

n |

n + (n – 12) + (n – 24) + (n – 23) + (n – 25)

= 5n – (12 + 24 + 23 + 25)

= 5n – 84

This can be put into a table to solve the equation for a T-shape for any width grid.

w = width

w | t | Difference between | Therefore 5n – 7w |

8 | 5n – 56 | t | |

9 | 5n – 63 | 7 | t |

10 | 5n –70 | 7 | t |

11 | 5n – 77 | 7 | t |

12 | 5n -84 | 7 | t |

Again we can prove this using algebra:

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

n - w | ||

n |

n + (n – w) + (n - 2w) + (n – 2w + 1) + (n – 2w – 1)

= 5n – (7w + 1 – 1)

= 5n – 7w

Now I have the equation for finding the T-total of this T shape on any grid I can move on to the next and final task. This was to investigate the relationship between the T-total when using a transformation or combination of transformations.

There would be no point in studying both rotation AND reflection because a diagonal reflection is the same as a 90º rotation and horizontal the same as 180º.

For this same reason there is no point in studying reflection AND translation as vertical reflection is a translation right or left.

Conclusion

In full the equation for finding the T-total for a rotation of 90º and THEN a translation of any vector is:

5(n + x – yw) + 7 = t

To prove this I will take a 90º rotation ad then a random translation.

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 |

The translation has a vector of (4)

(-2)

This means that x = 4, y = -2 and n = 10. The grid width (w) is 9.

5(n + x – yw) + 7

= 5(10 + 4 – (-2 x 9) + 7

= 5(10 + 4 + 18) + 7

= (5 x 32) + 7

= 160 + 7

= 167

This can be proved by doing the long calculation for the T-total of the new rotation and translation of the shape.

32 + 33 + 34 + 25 + 43 = 167

In conclusion you have to know what to add on to the 5n or 5(n + x – yw) to get the T-total. This is different for every rotation and there is no way of knowing. You can remember it quite easily though. 7w for 0º and 180º. 7 for 90º and 270º. Whether to add or subtract his is simple to deduce. If you times the T-number by 5 and all the other numbers are less than the T-number you will get an answer too high s you will have to minus something. And visa versa.

Whilst doing this investigation I tried to work systematically and clearly to prove my points thoroughly. I am satisfied with my answers and would like to use the extra knowledge I have gained in years to come.

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

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