# I am going to investigate different sized cubes, made up of single unit rods and justify formulae for the number of rods and joints in the cubes.

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Introduction

## Higher Tier Coursework

#### Structures – 2003

##### Owen Gates

###### Introduction

I am going to investigate different sized cubes, made up of single unit rods and justify formulae for the number of rods and joints in the cubes. The cubes are made from single unit rods and are not hollow, meaning that the unit rods are constructed inside the cube making smaller, similar cubes inside of the default one. The only cube not to be made up of smaller cubes, will be the 1x1x1 cube as this is the simplest form of cube and will, therefore not have any unit rods inside it. An example of a 2x2x2 cube is shown below.

The individual unit rods in the structure are held together by a series of different types of joints, as shown below.

3 joints – found on the vertices of the cube and connect three different rods together.

4joints – found on the edges of the cube and connect four different rods together

5 joints – found on the faces of the cube and connect five different rods together

6 joints – found on the inside of the cube and connect six different rods together. Without using diagonals, this is the most amount of rods to join together.

Middle

= 24 + 48n-48 +(30n-30)(n-1) + (6n-6)(n-1)(n-1)

2

= 24 + 48n-48 + 30n2–60n+30 + (6n2-12n+6)(n-1)

2

= 24 + 48n-48 + 30n2 –60n+30 + 6n3-12n2+6n-6n2+12n-6

2

= 6n3+30n2-12n2-6n2+48n-60n+6n+12n+24-48+30-6

2

= 6n3 + 12n2 +6n

2

= 3n3 + 6n2 +3n

=3n(n2+2n+1)

=3n(n+1)(n+1)

=3n(n+1) 2

Therefore the final formula for the number of cubes = 3n(n+1) 2

###### Prediction and Justifying Formulae

I can now use my formulae to predict what the results would be in a 6x6x6 cube. If my formulae are correct then the cube would have 8 3joints, 60 4 joints, 150 5 joints, 125 6 joints and 882 unit rods.

(See spotty paper for prediction)

nxnxn | 3 joints | 4 joints | 5 joints | 6 joints | No. of rods |

1x1x1 | 8 | 0 | 0 | 0 | 12 |

2x2x2 | 8 | 12 | 6 | 1 | 54 |

3x3x3 | 8 | 24 | 24 | 8 | 144 |

4x4x4 | 8 | 36 | 54 | 27 | 300 |

5x5x5 | 8 | 48 | 96 | 64 | 540 |

6x6x6 | 8 | 60 | 150 | 125 | 882 |

I can see that my prediction was correct and therefore my formulae are correct.

=

Cuboids

(All working is on spotty paper at the back of this investigation)

Working and Results

By looking at the spotty paper and counting the number of rods and joints, I have produced the following table.

xxyxz | 3 joints | 4 joints | 5 joints | 6 joints | No. of rods |

1x2x3 | 8 | 12 | 4 | 0 | 46 |

2x3x4 | 8 | 24 | 22 | 6 | 133 |

2x4x6 | 8 | 36 | 46 | 15 | 244 |

1x2x9 | 8 | 36 | 16 | 0 | 124 |

4x5x6 | 8 | 48 | 94 | 60 | 1046 |

Conclusion

I can now state that what I set out to investigate has been successful and the discovered formulae are presented in the following tables.

Cubes

Type of joint/ rods | Formulae |

Number of unit rods | 3n(n+1) 2 |

Number of 3 joints | 8 |

Number of 4 joints | 12(n-1) |

Number of 5 joints | 6(n-1) 2 |

Number of 6 joints | (n-1) 3 |

Cuboids

Type of joint/ rods | Formulae |

Number of unit rods | 3xyz+2xy+2xz+2yz+x+y+z |

Number of 3 joints | 8 |

Number of 4 joints | 4(x+y+z-3) |

Number of 5 joints | 2(x-1)(y-1)+2(x-1)(z-1)+2(y-1)(z-1) |

Number of 6 joints | (x-1)(y-1)(z-1) |

These formulae have been justified as being correct

If I was to improve my investigation I could have used line graphs for each of the discovered formulae and then used the gradient of the line and the point at which it cuts the y-axis to justify my formulae furthermore. This would have only strengthened my argument that my formulae are correct and I would have found any additional evidence. Therefore I can conclude that my investigation was successful and my formulae are accurate.

This student written piece of work is one of many that can be found in our GCSE Hidden Faces and Cubes section.

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