VON KOCH’S SNOWFLAKE CURVE
Kuan Chen (Veronica) Li
Introduction
Today we are going to investigate Von Koch’s snowflake curve. In particular we will examine the relationship
between the perimeter and the number of iterations. We will also look at how the iteration number and the
area are related.
We hope to show that these relationships are both sequences or series.
Building the curve
Start with an equilateral triangle.
On the middle third of each of the three sides, build an equilateral triangle. Erase the base of each of the three
new triangles.
On the middle third of each of the twelve sides, build an equilateral triangle again. Erase the base of each of the
twelve new triangles.
Repeat the process with this 48-sided figure. See the likeness to a crystal of snow emerge.
(This image is from http://scidiv.bellevuecollege.edu/Math/Snowflake.html)
We call the first image triangle 1, the second image triangle 2,etc…
Perimeter:
Assume the perimeter of triangle 1 is 3, which means the length of each side is one unit. From triangle 2, we can
see that the length of one side is 4/3, the first image has three sides, therefore triangle 2’s perimeter is 1/3
times more than triangle 1’s, so we can know that the second image has the perimeter of 4. To look at triangle 3,
we use the same idea to calculate it. One side is 1/3 times more than triangle 2, and as total, the perimeter is
1/3 times more than the second one, so the third one has a perimeter of 16/3. And so on, we can calculate
triangle 4’s, triangle 5’s……
These results are displayed in table 1.
n is the number of triangle as well as the number of iterations.
Table 1
n
1
2
3
4
5
perimeter
3
4
16/3
64/9
256/27
We find that each term can be obtained from the previous one by multiplying by 4/3, therefore it is a geometric
sequences.
From the formula of geometric sequence an=a1*r^ (n-1), we use this formula to calculate the perimeter of Von
Koch’s snowflake curve. a1=3, r= 4/3, therefore the general form of the perimeter for n iterations are will be
given by:
P = 3*(4/3)^ n
As n
∞, we find the perimeter of the curve approaches infinity.
Area:
To look at the area, we find that triangle 1 can be divided into 9 little equilateral triangles. We can see from
triangle 2 that three more little equilateral triangles, therefore the area compare to triangle 1, it needs to add
1/3, so we can know triangle 2’s area is 1+1/3= 4/3. To look at triangle 3, it divided each little triangle into nine
much more smaller equilateral triangles. So we find that it is 2/9 larger than a little equilateral triangle,
therefore the area should be triangle 2’s area add 4/9 times 1/3, so it’s 1+1/3(a+4/9)= 40/27. And so on, we can
calculate triangle 4’s, triangle 5’s……
These results are displayed in table 2.
Table 2
n
1
2
3
4
5
area
1
1+1/3
1+1/3(1+4/9)
1+1/3(1+4/9+(4/9)^2)
1+1/3(1+4/9+(4/9)^2+4/9)^3)
area
1
4/3
40/27
376/243
3448/2187
We have learnt that the formula of geometric series is Sn= a1*(1-r^n)/ (1-r), we can use this formula to work out
the snowflake’s area formula. We find that the area is geometric series.
a = 1+ 1/3* (1+ 4/9+ (4/9) ^2+……+ (4/9) ^ (n-2))
a= 1+ 1/3* (1+ (4/9)* (1-(4/9) ^ (n-2))/ (5/9))
a=1+ 1/3 (1+4/5* (1-(4/9) ^ (n-2))
There’s a geometric series formula for n
a= 1+1/3* (1+ (4/9) / (1- 4/9))
a= 1+1/3* (1+ 4/5)
a= 1+ (1/3)*(9/5)
a= 8/5
∞, Sn= a1/ (1/r), and we can calculate the area from this formula.
Therefore as n
∞, the area will get closer and closer to 8/5.
Conclusion
The snowflake curve is connected in the sense that it does not have any breaks or gaps in it. But it’s not smooth
because it has an infinite number of sharp corners in it that are packed together as n increases.
The snowflake encloses a finite amount of area. At each step building the new little triangles adds more than
one unit of length to the curve. To be precise, (4/3) ^ n are added at the nth step, so the length of the snowflake
is infinite. Also, the area after n steps will be close to 8/5.