• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  • Level: GCSE
  • Subject: Maths
  • Word count: 3666

Hidden Faces Investigation

Extracts from this document...

Introduction

Zia Juwel Haque                        Mathematics Coursework: Hidden Faces

image33.png

image00.png

image01.png

image12.png

image15.png

image12.png

image27.png

image29.png

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: image30.png

x, y and z are collectively image31.png

z       z        known as ‘the variables’.

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

image32.png

Title Pageimage02.png

Introductionimage03.png

Abbreviations

Contents

Start of Investigationimage04.png

Proving the general formula for h

Testing the formulae for h, v and T in arrangements:image05.png

With one changing variable

With two changing variablesimage06.png

With three changing variablesimage07.png

Extensionimage08.png

More abbreviationsimage09.png

Conclusionimage10.png

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.image11.png

image13.pngimage13.pngimage13.png

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.

...read more.

Middle

image11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.png

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

image11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.pngimage11.png

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

...read more.

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

...read more.

This student written piece of work is one of many that can be found in our GCSE Hidden Faces and Cubes section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Hidden Faces and Cubes essays

  1. An investigation to look at shapes made up of other shapes (starting with triangles, ...

    formula which will give you the correct value of H, if you know P and D. H=(P+2D-2)/4 can be rearranged to give P=4H+2-2D and D=2H-P/2+1. Previously, when I rearranged a working formula I then tested out the two new formulas all over again, just to make sure.

  2. Borders Investigation Maths Coursework

    -4 +1 2. 2(4) -4 +1 3. 8 - 4 +1 = 5 Correct As you can see the formula is working I am going to try it on some more sequences to be 100% sure that it is fully working and has no errors Sequence 3 1. 2(32)

  1. Cubes and Cuboids Investigation.

    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: No. of painted faces per small cube Prediction Formula (X=5)

  2. Maths-hidden faces

    if a 30-cubed cuboid is x6, you get the answer of 180; this is the total amount of faces on a cuboid made up of 30 cubes.

  1. An experiment to find out if seeing the eyes of a well known persons ...

    I repeated the same procedure with group 2. This time with the pictures of the celebrities with blacked out eyes. Results: A table to show the difference in face recognition between celebrities faces with eyes and faces with the eyes blacked out.

  2. Border coursework

    + third value Thus � 2*1 + 2*5 + 13 � 25 Hence, it can be seen that for an nth term, it is the summation of twice [2n2 -2n+1] for the values of n from 1 to n-1. This can be stated as: n-1 ?

  1. "With reference to theories of visual object recognition outline the ways in which faces ...

    The final stage that Marr believed to be the visual output was the 3d representation. This draws together all of the previous stages giving the perceiver an overall picture of the object and allows them to recognise it from any viewpoint angle.

  2. Shapes Investigation I will try to find the relationship between the perimeter (in cm), ...

    Now I shall test it, just to make sure it works. So where P=14, D=4 and Q=10, Q=P/2+D-1 � Q=7+4-1 � Q=10 C And where P=16, D=6 and Q=13, Q=P/2+D-1 � Q=8+6-1 � Q=13 C And where P=16, D=9 and Q=16, Q=P/2+D-1 � Q=8+9-1 � Q=16 C As I expected,

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work