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Koch Snowflake

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Problem

Define a sequence of figures  recursively as follows. Let  be an equilateral triangle with sides of length .  For , let  be the curve created by removing the middle third of each side of  and replacing it with an equilateral triangle pointing outward. The limiting figure as  is known as Koch’s snowflake.

Project Questions

  1. Find the length of the perimeter of .
  2. Evaluate to find the perimeter of the Koch snowflake.
  3. Find the area of the figure .
  4. Evaluate to find the area of the Koch snowflake.

Solution

Perimeter of Figure

  • Let be the length of the base of the equilateral triangle .

    • Since is equilateral, each side has a side length of .
    • Therefore, for , .

  • In each iteration , each side of the figure divides into sides in the next iteration.

    • However, the length of each side becomes of its original length.

  • Our equations would make sense if we have an equation for the number of sides each iteration.

    • We let be the sides of the figure .
    • Since is an equilateral triangle,
    • As each side of divides into in the next iteration, .
    • Writing the terms of , a conceivable pattern emerges.
  • For subsequent terms of for , each term can be expressed by the explicit function:
  • The side length divides by of its original length per iteration.
    • Since the initial side length is the length of the base , then:

  • To find the perimeter , we count all side lengths from all sides of .
    • Since each side is equal, then the equation for perimeter becomes:

  • From previous equations, and , we can simply substitute this equation to find:
  • This is a geometric sequence with and
    • Since , then this sequence diverges as .
  • Therefore:

  • This indicates that the perimeter of the Koch’s snowflake is infinite.

Area of Figure

  • The expression for the area of a triangle is:
  • Since is equilateral, each side has the same side length .
  • Therefore, can be a function of .
    • Since is the length of the segment from the perpendicular bisector of a side to a point opposite to it, we can simply use the Pythagorean theorem to find an expression for .
  • Substituting this expression to the area, we get:
  • For where , each side grows an equilateral triangle pointing upward with base of the side’s original length.
    • This means that has several components.

  • Here:

    • is the number of sides per iteration .
    • is the length of each side per iteration.
    • The term is the area of the child triangle on one side per iteration .
    • By multiplying to the area of the child triangle, we account for all child triangles formed on each side on one iteration.
    • By computing the infinite series , we add the areas of all triangles formed from this process that forms the Koch snowflake.

  • We begin by substituting and to

Graph View