Ariel Lin 11B
Math
Introduction:
Von Koch’s Snowflake curve is a special mathematical curve found by the Swedish mathematician Helge von Koch in 1904. To draw a Von Koch’s Snowflake curve:
1.
Draw an equilateral triangle:
(C1)
2.
Divide each side of the equilateral triangle into 3 equal segments:
3.
Draw an equilateral triangle on the middle segment of each side:
4.
Delete the side of the smaller equilateral triangle which lies on the side of the bigger triangle:
5.
The result come out as:
(C2)
6.
If we keep repeating the same steps, we will get:
(C3)  (C4)
We can notice that we get a pattern of shapes which its perimeter and area changes every time in a certain pattern.

Ariel Lin 11B
Math
Investigation 1:
Perimeter:
Assuming triangle C1 has a perimeter of 3 units which means that each side of the triangle is 1 unit.
When we divide each side of the triangle into 3 equal segments, each segment will be 1/3 unit. For each side of the triangle C1, we add two 1/3 unit segments and erase one 1/3 unit segment which means there are 4 segments in total.

So each side of the triangle became 4/3 of the original. If each side of the triangle is 4/3 then the perimeter would be 4/3 of the original as well. For instance:
Assuming the side of the triangle is a.  original perimeter = 3a
Now each side is 4/3 a  perimeter = 3 (4/3 a)  = 4/3 (3a)
So every time the perimeter of the shape becomes 4/3 of the perimeter of the one before it which forms a geometric sequence of a ratio of 4/3. The first shape of the sequence which is the equilateral triangle with each side of 1 unit has perimeter of 1×3 = 3. The first term a = 3, and the ratio r = 4/3, n= nth shape:
C n
= a × r n-1  C n
= 3 × (4/3) n-1
.
Investigation 2:
Area:
When the side of the triangle is 1/3 of the original, the area would be (1/3) 2 , 1/9, of the original. For instance:
Assuming the side of the original equilateral triangle is b unit, the area would be (√ (3) / 4)b 2 square unit. If the other triangle has side of 1/3 b unit, the area would be (√ (3) / 4)(1/3 b) 2 square unit. So when the ratio of the side of the two triangle is 1: 1/3, the ratio of the areas would be: (√ (3) / 4)b 2 : (√ (3) / 4)(1/3 b) 2  b 2 : 1/9 b 2  1 : 1/9.
Assuming triangle C1 has an area of 1 square unit, the new little triangles in shape C2 have areas of 1/9 square unit. There are 3 little triangles on all 3 sides of the triangle C1, so the area of C2 is the area of C1 plus 3 ×
1/9 square unit. Starting from shape C2, the amount of side is 4 times of the amount of side of the shape before it. Also, the little new triangles are 1/9 of the previous shape’s little triangle. So the area of each shape would be the area of the previous shape plus 4/9 (4 × 1/9) of the previous shape’s little triangle.
Area of:
C1: 1 square unit.
C2: 1 + 3×1/9 = 1 + 1/3
C3: 1 + 1/3 + 3×4× 1/ (9×9) = 1 + 1/3 + 4/ (3×9)  1 + 1/3 (1 + 4/9)
C4: 1 + 1/3 + 4/ (3×9) + 3×4×4× 1/ (9×9×9) = 1 + 1/3 + 4/ (3×9) + 4×4/ (3×9×9)  1 + 1/3 [1 + 4/9 + 4×4/
(9×9)]  1 + 1/3 [1 + 4/9 + (4/9) 2 ]
Inside the bracket after 1/3, it forms a geometric series with the first term (a) of 1 and the ratio (r) of 4/9:
S n
= a ( 1 – (4/9) n )/ (1 – r)
 Area of C n
= 1 + 1/3 { 1×[1 – (4/9) n-2 ]/ (1 – 4/9)} = 1 + 1/3 [ (1 – (4/9) n-2 )/ (5/9) ] = 1+ 3/5 (1 – (4/9) n-2 ) = 8/5 +
3/5 (4/9) n-2 = (8 +3(4/9) n-2 )/5

Ariel Lin 11B
Math
As we can see, the ratio of the geometric series in the formula for the area of the shapes is between 0 and 1 which means that the series can get to a certain amount when the number of terms is infinite.
S
∞
= a/ (1 – r) : C
∞
= 1 + 1/3 [ 1/ (1 – 4/9)] = 1 + 1/3 × 9/5 = 1 + 3/5 = 8/5
When n gets infinitely large, the area of the snowflake curve is 8/5 square unit.
Conclusion:
From all the investigation and calculations that we have done, we can see that there are geometric relationships for both perimeter and area. The perimeter for each shape formed a geometric sequence which the general term C n
= 3 × (4/3) n-1
. For the areas of the shapes, each time we add a certain amount of area onto the previous one to get the new area which makes it into a series. Each time the new added little triangle is 1/9 of the triangle in the previous shape. In the formula it formed a geometric series with the first term of 1 and ratio of 4/9 which the formula was (8 +3(4/9) n-2 )/5. In addition, the ratio of the geometric series for the area of the shape is between 0 and 1 which makes that the series have a final result when the number of terms got to infinitely large. When that happens, the area of the snowflake curve would be 8/5 unit square.