Hidden Faces Investigation

can see that 2xy still looks good although 2 could be replaced by z, because z was 2 in both arrangements. But 2yz is wrong because 2yz=4 when there are 8 hidden faces. There are 4 more faces than before because of the increase in x. So maybe x has an involvement. But 2yz makes 4, and something-x has to be 2 to make it work. x is 3, too big. But x-1 isn’t. x-1 makes 2 so can make the formula work. I will see if 2yz(x-1) works for the first arrangement:

        x=2, y=1, z=2

      hyz=2yz(x-1)

        =2x1x2(2-1)

=2x1x2x1=4,

this is correct because there are 4 hidden faces with dimensions yz. If with a bit of rearrangement I can make a similar formula work for hxy, then I can include this and try again for hxz. Rearranging the expression gives 2xy(z-1):

x=2, y=1, z=2        x=3, y=1, z=2        So far I have

                                                h=(2xy(z-1))+(2yz(x-1))

      hxy=2xy(z-1)        hxy=2xy(z-1)                but I need to find hxz. So I will use

=2x2x1(2-1)                       =2x3x1(2-1)        the same technique to find what

=4x1=4, true        =6x1=6, true.       remains of the formula.

x=2, y=1, z=2         x=3, y=1, z=2

      hxz=2xz(y-1)                hxz=2xz(y-1)

        =2x2x2(1-1)                     =2x3x2(1-1)

        =8x0=0, false             =12x0=0, false.

        The working out shows the rule or hyz to be wrong because hyz isn’t 0 for either arrangement. But I noticed that adding xz makes it correct in both arrangements. I can say that hxz=(2xz(y-1))+xz. In my next arrangement I intend to test this and the other formulae.

In this arrangement x=1, y=3 and z=1. so hxy=0, hyz=0 and hxz=5. I can test my formulae:

hxy=2xy(z-1)                hyz=2yz(x-1)                        hxz=(2xz(y-1))+xz

  =2x1x3(1-1)                     =2x3x1(1-1)                     =(2x1x1(3-1))+1x1

          =6x0=0                     =6x0=0                             =(2x2)+1=4+1=5

        All the above are correct, so I can say that:

h=(2xy(z-1))+(2yz(x-1))+(2xz(y-1))+xz.

Open the brackets and:         h=2xyz-2xy+2xyz-2yz+2xyz-2xz+xz.

Collect the like terms and:        h=2xyz+2xyz+2xyz-2xy-2yz-2xz+xz.

Simplify and I get:                        h=6xyz-(2xy+2yz+xz)

        T        v

As soon as I saw this I was amazed because I noticed that the first part of the rule was the same as the general formula for T (T=6xyz) and the second part was the general formula for v (v=2xy+2yz+xz). The above basically says h=T-v, hidden faces = total – visible faces which is correct. I have proved myself right.

        Now I will test my formulae in sequences of cuboid arrangements where there are one, two and three changing variables.

        Testing the formulae for h, v and T in arrangements:

        With one changing variable

        Arrangements with one changing variable are those where only one of either x, y or z chance according to the term number n. The other variables remain constant, fixed at a certain value. I will try and prove that the general formulae for T, v and h work for these arrangements.

The first sequence: x=n, y=1, z=1

    n=1         n=2                                                 

The changing variable is x so I will check that the formulae are right: Let n=3 

                                T=6xyz                v=2xy+2yz+xz

                                T=6x3x1x1                v=2x3x1+2x1x1+3x1

n=3                        T =18, true                v=6+2+3=11, true

                                h=6xyz-(2xy+2yz+xz)                 The formulae look good

                                h=6x3x1x1-2x3x1+2x1x1+3x1   but more tests will confirm

                                h=18-6+2+3                                 that they work for all

                                h=18-11=7, true.                                 arrangements with one

n=4                                                changing variable.

        This sequence: x=2, y=n, z=2

        n=1

        n=3

        n=2

                                                     n=4

Now that y is the changing variable, I will test the formulae again: Let n=1

T=6xyz                v=2xy+2yz+xz                h=6xyz-(2xy+2yz+xz)

T=6x2x1x2                v=2x2x1+2x1x2+2x2        h=6x2x1x2-2x2x1+2x1x2+2x2

T=24, correct                v=4+4+4=12, correct        h=24-(4+4+4)=12, correct

So the formulae are right again and seem immortal. One more test and I can say that they all definitely work for arrangements with one changing variable.

        This sequence: x=1, y=3, z=n. The diagrams for the sequence are on the next page due to lack of space.         The last changing variable is z, so this is the final test:

Let n=2                         T=6xyz         v=2xy+2yx+xz            h=6xyz-(2xy+2yx+xz)

T=6x1x3x2        v=2x1x3+2x3x2+1x2  h=36-20=16, true

T=36, true        v=6+12+2=20, true
                                                                        After these tests I can

        say that the three rules

        are true for all

arrangements with one

changing variable. But

I will see if they are true

for arrangements with

        n=1        two, and finally three,

          n=2        changing variables.

                        n=3                         n=4

        With two changing variables

        These arrangements have only one constant variable, the other two change according to n. I will show three sequences in which this is the case, with each variable having a turn at being the constant. I will test my formulae to see if they work for arrangements with two changing variables.

        In this sequence, x=1, y=z=n

   n=1

                                                In this         sequence only x is constant so I will

                                                test my formulae: Let n=4

        n=2

                                                T=6xyz        v=2xy+2yz+xz

                                                T=6x1x4x4        v=2x1x4+2x4x4+1x4

                                                T=96, true        v=8+32+4=44, true

                                                        

                                                        h=6xyz-(2xy+2yz+xz)

    n=4                         h=96-44=52, true

        

Again my formulae are correct, but there are two more sequences to go.

   n=3

 

        This sequence: y=1, x=z=n

   

   n=1

        n=2

                                      n=3                                         n=4

So more tests after y becomes the constant: Let n=2

T=6xyz                h=6xyz-(2xy+2yz+xz)

T=6x2x1x2                h=24-12

T=24, true                h=12, true

v=2xy+2yz+xz                One more sequence and then

v=2x2x1+2x1x2+2x2        I can say that these formulae

v=4+4+4=12, true                work for all arrangements with two changing variables.

        This sequence: x=y=n, z=2

   n=1

     n=2

                          n=3

                                                        n=4

Now it’s the turn of z to be the constant. Let n=3

T=6xyz        v=2xy+2yz+xz

T=6x3x3x2        v=2x3x3+2x3x2+3x2

T=108, true        v=18+12+6=36, true

        h=6xyz-(2xy+2yz+xz)

        h=108-36=72, true

So the rules work in all cuboid arrangements with two changing variables.

        With three changing variables

        There is only one sequence this time, and it is made up entirely of cubes instead of the cuboids above. All the variables change according to n.

        This sequence: x=y=z=n                                 

   n=1

                                                                                  The last set of tests: Let n=2

                                                                        

                n=2                                                

                                                        T=6xyz        v=2xy+2yz+xz                                                    n=3                T=6x2x2x2        v=2x2x2+2x2x2+2x2

T=48, true        v=8+8+4=20, true

   n=4

        h=6xyz-(2xy+2yz+xz)

        h=48-20=28, true

        So I can conclude that the formulae:

        T=6xyz,

        v=2xy+2yz+xz and

        h=6xyz-(2xy+2yz+xz)

        all work for all cuboid arrangements.

        In the extension I will experiment with 3 sequences of arrangements that aren’t cuboids or cubes. Because of this I already know that my formulae won’t work for such arrangements. This is because the variables x, y and z will not apply to such arrangements as they don’t have such clear-cut dimensions. So my formulae are going to be different for each sequence, and will be in terms of n rather than in terms of x, y and z. And I will prove each formula for h by a technique called induction.

        More definitions

        c is the number of cubes in an arrangement

        hk is used in induction to represent the number of hidden faces in an arrangement (it’s just how it’s done).

        First sequence: staircase

        n=1        n=3        n=4

        n=2

        Right at the start of the investigation I said that the total number of faces in any arrangement was 6 (faces per cube) times the number of cubes, so in this case I can say that T=6c for all arrangements. To find the formulae for v and h I will need to make tables to find the differences between each value of v and h.

The formula for v begins with n2 because the 2nd difference is 2. Because it’s the 2nd difference it is halved then n is squared. By the same token the formula for h starts with 2n2. Now I will find the rest, by making another table.

Now I add +4n to the formula for v and -n to the formula for h. They are now:

        v=n2+4n

        h=2n2-n

So I will test them both. Let n=2
        v=n2+4n                h=2n2-n                So the formulae work. Now I will use

        v=2x2+4x2                h=2x2x2-2                induction to further prove the formula

        v=4+8=12, true        h=8-2=6, true        for h.

Turn the page

        Proof by induction

        In the first step of induction n is replaced by k.

        Assuming that hk=2k2-k is true for a staircase with step size k, how many new hidden faces are added when a column of height k+1 is added?

There are 4k+1 new faces when a column of height k+1 is added to a staircase. So for a staircase with step size k+1:

hk+1=hk+4k+1  - now I substitute hk with its rule

hk+1=2k2-k+4k+1

hk+1=2k2+3k+1

        I already know that the formula for hk is true when k=1, so now I replace n in the original rule with k+1 to prove that the formula for hk is true when k+1=2, k+1=3, k+1=4, basically every single value of k.

        hk+1=2(k+1)2-(k+1)

        hk+1=2(k+1)(k+1)-k-1 – square k+1

        hk+1=2(k2+2k+1)-k-1 – multiply out the brackets

        hk+1=2k2+4k+2-k-1 – multiply everything in the bracket by 2

        hk+1=2k2+4k-k+2-1 – collect like terms
        hk+1=2k2+3k+1, same as above.

        So now I have proved that the formula for h is true for every value of n.

This sequence: back-to-back stairs

        n=2                                      n=3

     n=1

n=4

So the formula for v begins with 2n2 because the 2nd difference is 4, and the formula for h starts with 4n2 because the second difference is 8. I will find the rest in another table.

        The formula for v now includes +4n and the formula for h includes –4n. Another table should help me find the rest.

        Therefore the formula for v is: v=2n2-4n-1 and the formula for h is: h=4n2-4n+1.

Now I will prove the rule for h by induction.

        Proof by induction

        Assuming that hk=4k2-4k+1 is true for a staircase with step size k, how many new hidden faces are added when a column of height k+1 is added?

There are 8k new faces when a column of height k+1 is added to a staircase. So for a staircase with step size k+1:

hk+1=hk+8k  - now I substitute hk with its rule

hk+1=4k2-4k+1+8k

hk+1=4k2+4k+1

        hk+1=4(k+1)2-4(k+1)+1

        hk+1=4(k+1)(k+1)-(4k+4)+1

        hk+1=4(k2+2k+1)-4k-4+1

        hk+1=4k2+8k+4-4k-4+1

        hk+1=4k2+8k-4k+4-4+1
        hk+1=4k2+4k+1, same as above.

        So now I have proved that the formula for h is true for every value of n.

This sequence: two arms

        n

        

    n=1        n=2        n=3        n=4

The table below tells me that the formulae for v and h both begin with 6n. When 6n is subtracted from v, -1 remains so the  

rule for v is v=6n-1. by the same token the formula for h is h=6n-5. So on the next page I will prove the formula for h by induction.

Turn the page

Proof by induction

        Assuming that hk=6k-5 is true for two arms with arm size k, how many new hidden faces are added when an arm increases length by k+1?

There are 6 new faces when an arm increases length by k+1. So for two arms with arm size k+1:

hk+1=hk+6  - now I substitute hk with its rule

hk+1=6k-5+6

hk+1=6k+1

        hk+1=6(k+1)-5

        hk+1=6k+6-5

        hk+1=6k+1, same as above

So now I have proved that the formula for h is true for             every value of n.

        Evaluation

         To summarise, I used a sole cube to find formulae for the total number of faces (T) and the number of visible faces (v) in an arrangement, which I then tested. I set out to prove that h=T-v. I did this by first stating that the total of h was the number of hidden faces with dimensions of length and height, height and width, and length and width, added together. I found the corresponding formula and simplified to get the same formula as h=T-v, thus proving this formula. Intensive testing proved the three formulae worked for all arrangements shaped like a cuboid or cube. In the extension I found formulae for three different sequences and proved the rules for h using induction.

        During this coursework I have learnt new techniques like induction and improved on old methods like forming formulae. The above arrangement aren’t the only ones made with cubes, there are infinite arrangements bound only by the imagination. But in every arrangement there are faces of the cubes that we can see and faces that are hidden.