My Hypothesis for Cubes
3 joints – there are always eight 3 joints on a cube because there are always eight vertices on a cube. The 3 joints are only found on the vertices so there are only eight. There is, therefore, no formula for 3 joints
4 joints – I can see that the number of 4 joints is always a simple multiple of 12. The difference between each number of 4 joints is 12 so the co-efficient is 12. To get the right number, though, the constant of n must be (-1) because otherwise the number is 12 too big. If you look at a cube the number of 4 joints on a particular edge is always one less than n so that is the reason for the constant being (-1). There are twelve edges, hence the co-efficient being 12. Therefore the final formula is 12(n-1).
5 joints – By working out the differences between each number, you have a set of numbers whose difference is 12. Therefore the co-efficient is half of 12 because it took two rounds of numbers to get the differences between each the same. For the same reason n and its constant will have to be squared. The constant will be (-1) again to make the formula work. This formula works because there are 6 faces and on each face (n-1) 2 is how many 5 joints there are on that face, so the final formula is 6(n-1) 2
6 joints – This works in the same way as the 5 joints. There are three sets of numbers before the differences are the same therefore the n and the co-efficient will be cubed. To get the right number you have to take away one again from the n. By looking at the cube I can see that the 6 joints make up another cube inside of the bigger one, which is why this formula works. The final formula is (n-1) 3
*A dash represents irrelevant data.
Unit rods – from conducting the investigation I could see that the number of rods was equal to the number of 3 joints times 3 add the number of 4 joints times 4 add the number of 5 joints times 5 add the number of 6 joints times 6. That is not all though, as by doing that each rod was counted twice, therefore after doing the above I will divide the final result by 2. This can be put in a formula by multiplying in the relevant numbers into the formulae that I already have and adding these formulae together as shown below:
(8x3) + 4x12(n-1) + 5x6(n-1) 2 + 6(n-1) 3
2
= 24 + 48(n-1) + 30 (n-1) 2 + 6(n-1) 3
2
= 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)
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.
* Note that, as the cuboids were randomly chosen, the formulae cannot be worked out in the same way as the cubes because there is no relationship between the numbers for each joint.
My Hypothesis for Cuboids
3 joints – there are always eight 3 joints on a cuboid because there are always eight vertices on a cuboid. The 3 joints are only found on the vertices so there are only eight. There is, therefore, no formula for 3 joints
4 joints – just like for cubes, each edge contains the number of 4 joints equal to n-1. With cuboids, though, there are three different sized lengths, therefore instead of it being 12(n-1) it will be 4(n-1) for each different sized length. The three formulae for each sized length is then added together to give the final formula of 4(x-1)+4(y-1)+4(z-1). This can then be simplified in the following way:
= 4x-4+4y-4+4z-4
= 4x+4y+4z-12
= 4(x+y+z-3)
Therefore the final formula is 4(x+y+z-3).
5 joints – like for cubes you have to times one edge by the other for each face. Each edge is x, y or z minus 1 because this corresponds to the correct number of joints for that dimension. You times x-1 by y-1 twice because there are two faces where x and y are the corresponding edges. So that part of the formula is 2(x-1)(y-1). You do the same for the x and z parts and also the y and z parts, so the final formula is 2(x-1)(y-1)+2(x-1)(z-1)+2(y-1)(z-1).
This formula cannot be simplified in a more logical way.
6 joints – each letter (x, y or z) corresponds to a dimension. The 6 joints make up a cube inside the bigger cube. This inner cube is the same side as each dimension minus one, and then multiplied together. Therefore the final formula for 6 joints is
(x-1)(y-1)(z-1).
Unit rods – this formula is worked out in the exact same way as the formula for the rods of the cubes. I noticed that writing out each formula this time alongside each other would be very cramped and might spill over onto two lines. To prevent this I have split up each part of the formula for the corresponding joints. You have to times the joint number by the number of times that joint appeared in the cuboid to get the number of rods that those joints take. Once this has been done for each set of joints, I will add the formulae together to get one single, final formula for the number of unit rods in a cuboid. But in this formula every rod will have been counted twice, so I will divide the final formula by 2.
Unit rods
In 3 joints – 3 x 8
= 24
Unit rods
In 4 joints – 4(4(x+y+z-3))
= 16(x+y+z-3)
= 16x+16y+16z-48
Unit rods
In 5 joints – 5(2(x-1)(y-1)+2(x-1)(z-1)+2(y-1)(z-1))
= 10(x-1)(y-1)+10(x-1)(z-1)+1(y-1)(z-1)
= (10x-10)(y-1)+(10x-10)(z-1)+(10y-10)(z-1)
= 10xy-10y-10x+10+10xz-10z-10x+10+10yz-10z-10y+10
= 10xy+10xz+10yz -20x-20y-20z+30
Unit rods
In 6 joints – 6((x-1)(y-1)(z -1))
= (6x-6)(y-1)(z -1)
= (6xy-6x-6y+6)(z -1)
= 6xyz-6xy-6xz-6yz+6x+6y+6z-6
Grouping all the like terms from all the other formulae of rods, and then dividing the total by 2 will find the total number of rods.
Total rods
In a cuboid - 6xyz+4xy+4xz+4yz+2x+2y+2z
2
= 3xyz+2xy+2xz+2yz+x+y+z
Therefore the final formula for the number of cuboids is 3xyz+2xy+2xz+2yz+x+y+z
Prediction and Justifying Formulae
I can now use my formulae to predict what the results would be in a 3x6x7 cuboid (randomly chosen). If my formulae are correct then the cuboid would have 8 3joints, 52 4 joints, 104 5 joints, 60 6 joints and 556 unit rods.
(See spotty paper for prediction)
I can see that my prediction was correct and therefore my formulae are correct.
Summary and Conclusion
My task was to discover formulae that would represent the number of unit rods, 3 joints, 4 joints, 5 joints and 6 joints in an nxnxn cube and an xxyxz cuboid.
I set about my task by using spotty paper and tables to spot patterns in my results and ultimately find the appropriate formulae. I then predicted the results of a different sized cube/cuboid using my formulae. I presented these results in a table and then drew simple bar charts as another way to present the collected data.
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
Cuboids
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.
