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• Level: GCSE
• Subject: Maths
• Word count: 3666

Hidden Faces Investigation

Extracts from this document...

Introduction

Zia Juwel Haque                        Mathematics Coursework: Hidden Faces

Introduction

This piece of coursework investigates ‘hidden faces’ on objects made up of cubes placed         on a flat surface. This coursework focuses on finding a formula for the number of faces of cubes that are hidden from sight. I will use many diagrams and tables to help me arrive at a conclusion to my investigation.

Abbreviations

In this investigation:

Let x be the length, y the height and z the width, of a cuboid made up of single unit cubes like this:

x, y and z are collectively

x

Let n be the term number of a cube arrangement.

Let h be the number of hidden faces in an arrangement, v the visible faces and T the total number of faces.

abc means the number of a with dimensions bc. For example, hxy means the number of hidden faces with dimensions xy.

Diff. is short for difference, as in 1st Diff. (First difference, for example)

Letters are in bold to distinguish between x (length) and x (multiplication).

Contents

Title Page

Introduction

Abbreviations

Contents

Start of Investigation

Proving the general formula for h

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

With one changing variable

With two changing variables

With three changing variables

Extension

More abbreviations

Conclusion

Start of investigation

My investigation begins by looking at an individual cube from 3 angles, to show how many faces there are, which are hidden and which aren’t.

y        y                           z        y                 z

x        z                        x                x

Front view           Side view          Top view        Normal view

Another look at the cube itself reminds me there are two faces with dimensions xy, yz and xz. And one face, with dimensions xz, is hidden from sight. So there is 1 hidden face and 5 visible faces, a total of 6.

Middle

 n x y z v h T 1 2 1 2 12 12 24 2 2 2 2 20 28 48 3 2 3 2 28 44 72 4 2 4 2 36 60 96

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=6xyzv=2xy+2yz+xzh=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:

 n x y z v h T 1 1 3 1 13 5 18 2 1 3 2 20 16 36 3 1 3 3 27 27 54 4 1 3 4 34 38 72

Let n=2                         T=6xyzv=2xy+2yx+xzh=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.

 n x y z v h T 1 1 1 1 5 1 6 2 1 2 2 14 10 24 3 1 3 3 27 27 54 4 1 4 4 44 52 96

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=6xyzv=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

Conclusion

 n v 1st Diff. h 1st Diff. 1 5 1 6 6 2 11 7 6 6 3 17 13 6 6 4 23 19

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?

 k hk New hidden faces 1 1 6 2 7 6 3 13 6 4 19

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.

September 2001                    Page  of

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