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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:

          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:

           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:

  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).

'font-size:14.0pt; '>Now I have found out the formula for cubes with 3 or 2 faces painted.

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

'font-size:14.0pt; '>Y=(X-2)²*6

'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 X-2 in the formula. We can also say that as all of the shapes are cubes, the cubes with one painted face are always going to be in a square so we can say that the formula so far is (X-2)². Lastly as there are 6 faces on a cube there is going to be 6 similar squares with the same amount of cubes with one face painted and so the formula for cubes with only one painted face is Y=(X-2)²*6.

'font-size:14.0pt; '>The last formula that must be found out to make the 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 number of cubes with no painted faces:

'font-size:14.0pt; '>Y=(X-2)³

'font-size:14.0pt; '>Again we can say, as none of the cubes in question are on the outside of the cube and so you have to minus 1 from each side of the length, that X-2 is 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 cube and so the cubes in the middle will also form a cube, therefore we can say that the formula is going to be Y=(X-2)³.

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

'font-size:14.0pt; '>It is possible to check that these formulae work because if they are all added up then it should be equal to the total number of small cubes in the large cube. Which has the formula Y=X³

'font-size:14.0pt; '>Here are the added formulae:

'font-size:14.0pt; '>Y=X³-6X²+12X-8+6X²-24X+24+12X-24+8

'font-size:14.0pt; '>Y=X³

'font-size:14.0pt; '>Therefore my formulae must be correct because when added together they equal Y=X³.

'font-size:14.0pt; '>All of the above covers how to find, how many small cubes will have X number of faces painted. Next I am going to look at a formula to work out the total number of painted faces.

'font-size:14.0pt; '>Here is a table to illustrate this:

'font-size:14.0pt; '>P=C*(6/L) when C=L³

'font-size:14.0pt; '>Because each of the cubes has 6 faces and there are L number of cubes in the large cubes length then the formula will have 6/L in it. There are also a total number of C cubes in the large cube, so to find the total number of painted faces you must multiply the two together, giving you the formula P=C*(6/L).

Check.

          I am now going to test all of the above formulae on a different cube measuring 5*5*5,.

          Here is a table to show my predictions, the formulae and the correct answers:

'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:

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:

           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:

  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 :

'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:

'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:

'font-size:14.0pt; '>We are also able to say that as cubes are just special cuboids all of these rules will work for cubes as well. So the formulae in the table above are the only formulae that you actually need.

It is possible to check that these formulae work because if they are all added together  then it should be equal to the total number of small cubes in the large cuboid. Which is the same as multiplying all three lengths together, using the formulae A*B*C

'font-size:14.0pt; '>Here are the added formulae:

Y=  (A-2)*(B-2)*(C-2)+ 2*((A-2)(B-2)+(B-2)(C-2)+(A-2)(C-2))+ 4A-8+4B-8+4C-8+8

'font-size:14.0pt; '>Y=ABC-2AC-2BC-2AB+4A+4B+4C-8+2AB+2BC+2AC-8A-8B-8C+24+4A+4B+4C-24+8

'font-size:14.0pt; '>Y=ABC

'font-size:14.0pt; '>Therefore my formulae must be correct because when added together they equal Y=ABC

Check.

'font-size:14.0pt; '>I am now going to test all of the above formulae on a different cuboid measuring 3*4*6,.

'font-size:14.0pt; '>Here is a table to show my predictions, the formulae and the correct answers:

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

'font-size:14.0pt; '>From this diagram you can see that my formulae are correct.

Exceptions

'font-size:14.0pt; '>There are though times when my formulae will not work. These can be divided into two categories, cuboids with one dimension of one and the other is for cuboids with two dimensions of one. Again as a cube is a special cuboid you can also add a third category of a cube with dimensions of 1. These are the only exceptions that I can find for cuboids.

'font-size:14.0pt; '>Firstly I am going to look at the cuboids with only one dimension of one. Here are my diagrams. They are 1*2*2, 1*3*3 and 1*2*4.

'font-size:14.0pt; '>Unlike other cuboids I immediately noticed that these only have cubes with 4, 3 and 2 painted sides.

          Here is a table to show the cubes with 4 painted sides:

'font-size:14.0pt; '>Y=4

'font-size:14.0pt; '>Immediately we can say that all of these cuboids have 4 cubes with 4 painted faces. This is similar to before where there was 8 cubes with 3 painted faces because again all of these are at the vertices of the cuboid.

'font-size:14.0pt; '>The above deals with cuboids with 4 painted sides here is a table to show 3 painted sides:

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

'font-size:14.0pt; '>As usual we can say because we are only dealing with cubes that are not at a corner that A-2 and B-2 are going to appear in the formula. Also as each face appears twice we must multiply this by 2 and then to get the equation you must add the two together. Therefore the formula is Y=2(A-2)+2(B-2).

'font-size:14.0pt; '>Now there is only cubes with two sides left to do, so here is a table to show the results:

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

'font-size:14.0pt; '>As usual we are only interested in cubes in the middle of the cuboid so to find these lengths you must minus 2 from each length, so A-2 and B-2 are going to be in the formula. Also as the whole shape is a cuboid and we are interested in the cubes one in from the large shape we can also say that the inner shape that we are interested in will be a cuboid. Therefore to complete the equation you must multiply the two together. This gives you the formula Y=(A-2)(B-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:

'font-size:14.0pt; '>Now I am going to look at cuboids with two dimensions of one, using the cuboids 1*1*2, 1*1*4 and 1*1*3.

'font-size:14.0pt; '>Unlike other cuboids I immediately noticed that these only have cubes with 5 and 4 painted sides.

Here is a table to show the cubes with  painted sides:

'font-size:14.0pt; '>Y=5

'font-size:14.0pt; '>Immediately we can say that all of these cuboids have 2 cubes with 5 painted faces. This is similar to before where there was 8 cubes with 3 painted faces because again all of these are at the vertices of the cuboid.

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

'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:

'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.

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