# Cubes and Cuboids Investigation.

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Introduction

## Cubes and Cuboids Investigation

I am going to investigate the different patterns that occur with different cubes when all the faces are painted of a large cube and then that is separated into smaller cubes and then how many faces of each small cube are still painted. Here are my cubes. They are 2*2*2, 3*3*3 and 4*4*4.

I am going to establish the patterns that recur as the cube gets larger. For example the number of cubes with one face painted, with two faces painted, with three faces painted and the number of cubes with no faces painted when the larger cube is split up. Here is a table:

Length of cube | No. of small cubes | No. of small cubes with X painted faces | |||

X=3 | X=2 | X=1 | X=0 | ||

2 | 8 | 8 | 0 | 0 | 0 |

3 | 27 | 8 | 12 | 6 | 1 |

4 | 64 | 8 | 24 | 24 | 8 |

Immediately I noticed that all of the cubes have 8 cubes with 3 different faces painted when they are separated. All of these 8 are the vertices of the cube and so every cube except that which has a length of one will have 8 cubes with three faces painted.

This can be shown in the table:

cube length (X) | No. of cubes with 3 painted faces (Y) |

2 | 8 |

3 | 8 |

4 | 8 |

Y=8

'font-size:14.0pt; '>The above tells us how many cubes will have three painted faces to find out how many will have two, here is a table:

cube length (X) | No. of cubes with 2 painted faces (Y) |

2 | 0 |

3 | 12 |

4 | 24 |

Y=12(X-2)

'font-size:14.0pt; '>I noticed this formula because as the differences are 12 then the formula must have something to do with 12X. Also if you look at the diagram above you can see that all the cubes with two painted (brown) are one in from either side and so the formula must have X-2 in it which is the cube length minus 2. Therefore the formula is Y=12(X-2).

Middle

3

8

Y=8

8

2

36

Y=12(X-2)

36

1

54

Y=(X-2)2*6

54

0

27

Y=(X-2)3

27

'font-size:14.0pt; '>Here is the diagram showing the correct answers:

'font-size:14.0pt; '>This proves that my formulae are correct.

Exceptions

None of the formulae will work for the cube with the lengths one which looks like this:

'font-size:14.0pt; '>This pattern does not work because there is more than one vertices on one cube and therefore there is not going to be 8 cubes with 3 painted sides. Also the other formulae can’t work because not only does the first formula not work but all of the others are based upon using formulae that are not on the corner of the cube and so involve taking two from the length but as there is only a length of one you can’t minus two away. The only thing that you can say about this size cube is that there is only one cube which has all six painted faces.

Cuboids

'font-size:14.0pt; '>Previously I have only being investigating into the patterns that occur with painted cubes. That only used one variable but now I am going to look at cuboids in which all three lengths are different, thus creating two more variables. I am going to see what patterns I can come up with as I did before. The principle will be the same in which there are cuboids made up out of smaller cubes and then the six faces will be painted and then all the cubes will be separated and I am going to investigate how many faces are painted on each small cube. Here are my cuboids. They are 2*3*4, 2*3*5 and 3*4*5.

I am going to establish the patterns that recur as the cuboid get larger. For example the number of cubes with one face painted, with two faces painted, with three faces painted and the number of cubes with no faces painted when the larger cuboid is split up. Here is a table showing the relevant information for my cuboids:

Length of cuboids | No. of small cubes | Amount of small cubes with X painted faces | |||

X=3 | X=2 | X=1 | X=0 | ||

2*3*4 | 24 | 8 | 12 | 4 | 0 |

2*3*5 | 30 | 8 | 16 | 6 | 0 |

3*4*5 | 60 | 8 | 24 | 22 | 6 |

I immediately noticed that all of the cuboids have 8 cubes with 3 different faces painted when they are separated. All of these 8 are the vertices of the cuboid and so every cuboid except one will have 8 cubes with three faces painted. This can be shown in the table:

cuboid length (A,B,C) | No. of cubes with 3 painted faces (Y) |

2*3*4 | 8 |

2*3*5 | 8 |

3*4*5 | 8 |

Y=8

'font-size:14.0pt; '>The shows us how many cubes will have three painted faces to find out how many will have two, here is the table:

cuboid length (A,B,C) | No. of cubes with 2 painted faces (Y) |

2*3*4 | 12 |

2*3*5 | 16 |

3*4*5 | 24 |

Y= 4(A+B+C)-24

'font-size:14.0pt; '>I noticed this formula because if you look at the diagram above you can see that all the cubes with two painted (brown) are one in from either side and so the formula must have A-2, B-2 and C-2 in it which is the cuboid lengths minus 2. Also as each of the cuboid lengths appear 4 times on the cuboid each must be multiplied by 4 giving the formula Y= 4A-8+4B-8+4C-8

'font-size:14.0pt; '>I have now found the formulae for cuboids with 3 or 2 faces painted.

'font-size:14.0pt; '>Here is the table showing the cubes with one painted face :

cuboid length (A,B,C) | No. of cubes with 1 painted face (Y) |

2*3*4 | 4 |

2*3*5 | 6 |

3*4*5 | 22 |

'font-size:14.0pt; '>Y= 2*((A-2)(B-2)+(B-2)(C-2)+(A-2)(C-2))

'font-size:14.0pt; '>Again we can say that because we want only the cubes with one painted face which are all in the middle (green) it is going to have A-2, B-2 and C-2 somewhere in the formula. We can also say that as all of the shapes are cuboids, the cubes with one painted face are always going to be in a rectangle so we can say that the formula will also be to do with multiplying the side lengths together for each different size of rectangle and as each different rectangle occurs twice the formula will also have multiplying by 2 in it. And so the formula overall for this will be Y= 2*((A-2)(B-2)+(B-2)(C-2)+(A-2)(C-2)).

'font-size:14.0pt; '>The last formula that must be found out to make this set complete is that which tells us the number of cubes with no painted faces.

'font-size:14.0pt; '>Here is a table to show the amount of cubes with no painted faces:

cuboid length (A,B,C) | No. of cubes with 0 painted faces (Y) |

2*3*4 | 0 |

2*3*5 | 0 |

3*4*5 | 6 |

'font-size:14.0pt; '>Y= (A-2)*(B-2)*(C-2)

'font-size:14.0pt; '>Again we can say, as none of the cubes with no painted faces are on the outside of the cuboid, you have to minus 1 from each side of the length, therefore A-2, B-2 and C-2 are going to be in the formula. We also know that the cube or cubes that have no painted faces, are embedded around the shell of the large cuboid and so the cubes in the middle will also form a cuboid, therefore you have to multiply all of the new lengths together to get the formula to be Y= (A-2)*(B-2)*(C-2)

'font-size:14.0pt; '>Here is a table showing all of the formulae that go together to show how the different amounts of faces painted on the small cubes are arranged:

No. of painted faces | Formula |

0 | Y= (A-2)*(B-2)*(C-2) |

1 | Y= 2*((A-2)(B-2)+(B-2)(C-2)+(A-2)(C-2)) |

2 | Y= 4A-8+4B-8+4C-8 |

3 | Y=8 |

Conclusion

'font-size:14.0pt; '>The above deals with cuboids with 5 painted sides here is a table to show 4 painted sides:

cuboid length (A,1,1) | No. of cubes with 4 painted faces (Y) |

1*1*2 | 0 |

1*1*4 | 2 |

1*1*3 | 1 |

'font-size:14.0pt; '>Y=A-2

'font-size:14.0pt; '>As the cubes that we are interested in for this equation are all one in form each side we can simply say that the formula will be Y=A-2.

'font-size:14.0pt; '>Here is a table showing all of the formulae that go together to show how the different amounts of faces painted on the small cubes are arranged:

No. of painted faces | Formula |

4 | Y=A-2 |

5 | Y=2 |

'font-size:14.0pt; '>As a cube is just a special cuboid I am also going to include the cube of 1*1*1 again to illustrate all of the exceptions that can happen with my formulae. Which looks like this:

'font-size:14.0pt; '>Immediately I can say that the only formula for this type of cuboid is that Y=6. I do not think that a table is needed to aid this as it is very simple.

'font-size:14.0pt; '>None of these cuboids with any number of dimensions of one, work because there is more than one vertices on one of the cubes and therefore there is not going to be 8 cubes with 3 painted sides. Also the other formulae can’t work because not only does the first formula not work but all of the others are based upon using formulae that are not on the corner of the cube and so involve taking two from the length but as there is only a length of one you can’t minus two away because then you will be dealing wuth negative numbers and that could not work

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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