Von Koch

C4
C3
C2
C1
VON KOCH’S SNOWFLAKE CURVE
In this investigation we consider a limit curve named after the Swedish mathematician
Niels Fabian Helge von Koch (1870 - 1924).
To draw Von Koch’sSnowflake curve we:

* start with an equilateral triangle, C1
* then divide each side into 3 equal parts
* then on each middle part draw an equilateral triangle
* then delete the side ofthe smaller triangle which lies on C1.

The resulting curve is C2, and C3, C4, C5, .... are found by ‘pushing out’ equilateral
triangles on each edge of the previous curve as we did with C1 to getC2.
We get a sequence of special curves C1, C2, C3, C4, .... and Von Koch’s curve is the
limiting case when n is infinitely large.
Your task is to investigate the perimeter and area of Von Koch’scurve.

What to do:
1 .Suppose C1 has a perimeter of 3 units. Find the perimeter of C2, C3, C4 and C5.
Hint: becomes so 3 parts become 4 parts.
Remembering thatVon Koch’s curve is Cn, where n is infinitely large, find the
perimeter of Von Koch’s curve.
P=3(43)n-1
-C1 divides in 3 parts 33
-A triangle comes out from the middle divisionso there are 43.
-There are 43 times the 3 sides
* C1=3(43)1-1=3(43)0=3(1)=3u
* C2=3(43)2-1=3(43)1=3(43)=123=4u
* C3=3(43)3-1=3(43)2=3(169)=489=163u
*C4=3(43)4-1=3(43)3=3(6427)=19227=649u
* C5=3(43)5-1=3(43)4=3(25681)=76881=25627u

2 .Suppose the area of C1 is 1 unit2. Explain why the areas of C2, C3, C4 and C5 are

* A2 = 1+ 13 units2

* A3 = 1+ 13 [1 + 49 ]units2

* A4 = 1+ 13[1 + 49 + (49)2] units2

* A5 = 1+ 13 [1 + 49 + (49)2 + (49)3] units2.

Use your calculator to find An where n = 1, 2, 3, 4, 5, 6, 7, etc., giving answers
which areas accurate as your calculator permits.
What do you think will be the area within Von Koch’s snowflake curve?
- We have A1= 1u2
- Each triangle that comes out from another equals a 19
-There are...