Mathematical Investigations: A Collaborative Approach to Understanding Precalculus
Name:
Key
Sequences and Series
SNOWFLAKES
In 1904, Helge von Koch created a special example by repeating, or iterating, a simple geometric pattern.
The resulting pattern is a fractal since it resembles itself at any level of magnification. In
other words, if we zoom in on any section, we will see a smaller version of the original.
We begin with a line segment. Remove the middle third of the segment, and replace it by two slanted
segments to form part of an equilateral triangle as shown.

The four segments in the transformation all have equal length. Then the pattern is repeated on each of
those four segments and this process is repeated endlessly.
Now, instead of using a single segment, we begin with an equilateral triangle and transform all sides.
This later came to be known as the Koch Snowflake curve.
Stage 1
Stage 2
Stage 3
Stage 4
We are going to explore both the area and perimeter of each stage of this sequence of figures.
First, we'll examine the total area enclosed by the figure at each stage.
Find the area of the triangle in stage 1, assuming that each side has length 1.
3
A
4
To arrive at stage 2, how many additional triangular regions are added? Find the area of each additional
triangle, and then find the total area that is added in going from stage 1 to stage 2.
2
3 additional triangular regions each with side length 13 so each has area 43  13   363 , giving a total area
added from stage 1 to stage 2 of 3  363 
3
.
12
© 2005 Illinois Mathematics and Science Academy®
S&S 11xb.1
Rev. S06
Mathematical Investigations: A Collaborative Approach to Understanding Precalculus
Name:
Key
To arrive at stage 3, how many additional triangular regions are added? Find the area of each triangle,
and then find the total area that is added in going from stage 2 to stage 3.
2
12 additional triangular regions each with side length 19 so each has area 43  91   3243 , giving a total area
added from stage 1 to stage 2 of 12  3243 
3
.
27
In stage 4, how much area will be added to the total area of stage 3?
4 3
2
48  43  271  
243
To find the total area of the enclosed figure if we increase the number of stages indefinitely, add these
areas above, added at each stage, and place "+ . . ." at the end to represent the sum if the process were
continued indefinitely. Then find the sum.
3
3
3 4 3
Total Area 




4 12 27 243
2

3
3
34
34


 
  
4 12 12  9  12  9 
3
3
2 3
 12 4 
4 1 9
5
In order to determine the area of the Koch snowflake, we considered an infinite series found by adding
the areas of the new triangles at each stage.

We now wish to find the perimeter of the figure. In this case, it is easier to consider the total perimeter
at each stage and look at the sequence of perimeters to see the pattern emerge. Again, we will start with
an equilateral triangle where each side has length 1.
Number of segments
length of each segment
total perimeter
Stage 1
3
1
3
Stage 2
12
1
3
4
Stage 3
48
1
9
16
3
 5 13
What happens to this sequence as n  ?
r  43  1 so as n   , perimeter  3  43 
n1
.
© 2005 Illinois Mathematics and Science Academy®
S&S 11xb.2
Rev. S06
Mathematical Investigations: A Collaborative Approach to Understanding Precalculus
Name:
Key
Now, let’s consider what happens if instead of using an equilateral triangle, a square is used to create a
Koch-type curve. That is, take a square with side = 1. On each segment, remove the middle third and
build a square (three sides) in its place. Repeat the process indefinitely. The first three stages are shown
below.
Stage 1
Stage 2
Stage 3
What happens to the perimeter as the number of stages increases without bound?
Number of segments
length of each segment
total perimeter
Stage 1
4
1
4
Stage 2
20
1
3
20
3
 6 23
Stage 3
100
1
9
100
9
 11 91
What happens to this sequence as n  ?
r  53  1 so as n   , perimeter  4  53 
n1
.
© 2005 Illinois Mathematics and Science Academy®
S&S 11xb.3
Rev. S06