# The Koch Snowflake

Extracts from this document...

Introduction

THE KOCH SNOWFLAKE

Q.1)Using an initial side length of 1, create a table that shows the values of Nn, Ln, Pn and An for n = 0, 1, 2 and 3. Use exact values in your results. Explain the relationship between successive terms in the table for each quantity Nn, Ln, Pn and An.

ANS 1)

STAGE | SIDES ( Nn ) | LENGTH ( Ln ) | PERIMETER ( Pn ) | AREA ( An ) |

STAGE 0 | 3 | 1 | 3 | 0.433013 |

STAGE 1 | 12 | 0.3333333333 | 4 | 0.577350 |

STAGE 2 | 48 | 0.1111111111 | 5.3333333333 | 0.641500 |

STAGE 3 | 192 | 0.037037037 | 7.1111111111 | 0.670011 |

At first, let us take a look at the change in the sides of the snowflake with increase in stages. It is soon realized that with increase in stage, the snowflake structure changes and so do the sides consequently. On further observation, it is seen that the snowflake sides increase in geometric progression.

This is seen because:

12/3=48/12=192/48=4:1

4:1 is the constant ratio. The sides increase by four times with each increase in stage.

Next, we have the length of sides of the snowflake. On first look, it seems that the length of sides decrease tremendously with each stage starting from the initial length of 1. However, a closer look reveals another geometric progression!

Let me show you how:

0.33333333/1=0.111111111/0.333333333=0.037037037/0.11111111=0.33333333:1

0.33333333:1 is the constant ratio. The length of the sides changes by this ratio with the progression of stages.

Thirdly, we come to the perimeter.

Middle

At Stage 1,

Area= (√3/4)*side2*number of new triangles + area of old triangle.

= (√3/4)*0.3333332*3+0.433013

= 0.577350

We can use the same idea for the area of the 2nd stage as well.

At Stage 2,

Area= (√3/4)*side2*number of new triangles + area of triangle of Stage 1

= (√3/4)*0.1111112*12+0.577350

= 0.641500

It is similar for stage 3 as well.

At Stage 3,

Area= (√3/4)*side2*number of new triangles + area of triangle of Stage 2

= (√3/4)*0.0370370372*48+0.641500

= 0.670011

The area of the Koch snowflake is a slightly complex matter. However, there is evidence of geometric progression. This is seen from the observation below:

0.577350/0.433013=0.641500/0.577350=0.670011/0.641500=1.333331794:1

Hence 1.333331794:1 is the common ratio. Hence we can assume that the area of the Koch snowflake increases with the general geometric progression formula=

Un=U1*rn

Where,

Un= the nth term,

U1= the 1st term,

R= the common ratio,

n= the stage of the snowflake under discussion

To elucidate my point, we shall investigate:

U0=0.433013*1.3333317940

=0.433013

U1=0.433013*1.333331794

=0.577350

U2=0.433013*1.3333317942

=0.641500

U3=0.433013*1.3333317943

=0.670011

Q.4) Investigate what happens at n= 4. Use your conjectures from step 3 to obtain values for N4, L4, P4, and A4. Now draw a large diagram of one side (that is one side of the original triangle that has been transformed) of the fractal at stage 4 and clearly verify your predictions.

ANS 4)

At stage 4,

Number of sides (N4) = 768

Length (Ln) = 0.012346

Conclusion

Similarly at Stage 3, the numbers of new triangles formed are 48 which coincide with the number of sides in Stage 2 which is 3 as well.

And so on.

This helps me replace the expression “number of new smaller triangles” with “Nn-1”.

Now I have:

AREA FORMULA:

An=30.5/4*Ln2*Nn-1 + An-1

We can generalize the formula to a greater extent. We can substitute Ln and Nn-1 and rewrite the formula as:

An=30.5/4*{1*(1/3)n}2*{3*4(n-1)} + An-1

OR

An= An-1 +30.5/4*{1*(1/3)n}2*{3*4(n-1)}

This is because length gets multiplied by {1*(1/3)n}2 times and number of new truiangles increases by *{3*4(n-1)} times.

A0 + A1 + A2 + A3+…..+An= An-1 +30.5/4*{1*(1/3)n}2*{3*4(n-1)}

0.433012702 + 0.577349981 + 0.641499754 + 0.670010707 +……+An=An-1 +30.5/4*{1*(1/3)n}2*{3*4(n-1)}

For n=0,

A0= A0-1 +30.5/4*{1*(1/3)0}2*{3*4(0-1)}

As A0-1 and {3*4(0-1)} does not exist

A0= 30.5/4*{1*(1/3)0}2

= 0.433012702

For n=1,

A1= A1-1 +30.5/4*{1*(1/3)1}2*{3*4(1-1)}

= (√3/4)*0.3333332*3+0.433013

= 0.577350

For n=2,

A2= A2-1 +30.5/4*{1*(1/3)2}2*{3*4(2-1)}

= (√3/4)*0.1111112*12+0.577350

= 0.641500

For n=3,

A3= A3-1 +30.5/4*{1*(1/3)2}2*{3*4(3-1)}

(√3/4)*0.0370370372*48+0.641500

= 0.670011

We may assume that the formula holds true for all numbers of n including n=k. So the formula holds true for n=k.

Ak= Ak+1 +30.5/4*{1*(1/3)k}2*{3*4(k-1)}

To prove it for n=k+1:

Ak+1= Ak + Uk+1

Where Uk+1 is the number of new triangles formed in the n=k+1 stage.

Ak +30.5/4*{1*(1/3)k+1}2*(3*4k)= Ak+ Uk+1

Ak +30.5/4*{1*(1/3)k*(1/3)}2*(3*4k) = Ak + 30.5/4*{1*(1/3)k+1}2*(3*4k)

30.5/4*{1*(1/3)k*(1/3)}2*(3*4k) = 30.5/4*{1*(1/3)k*(1/3)}2*(3*4k)

LHS=RHS

Hence the formula stands true for AK+1 as well.

This student written piece of work is one of many that can be found in our GCSE Fencing Problem section.

## Found what you're looking for?

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