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2.6 Area of a Snowflake.

In this section we will find the areas of two rather complicated sets, called the inner snowflake and the outer snowflake. To construct the inner snowflake, we first construct a family of polygons , , as follows:

is an equilateral triangle.

is obtained from by adding an equilateral triangle to the middle third of each side of , (see the snowflake figures ).

is obtained from by adding an equilateral triangle to the middle third of each side of , and in general

is obtained from by adding an equilateral triangle to the middle third of each side of .

The inner snowflake is the set

i.e. a point is in the inner snowflake if and only if it lies in for some positive integer . Observe that the inner snowflake is not a polygon.

To construct the outer snowflake, we first construct a family of polygons , , as follows:

is a regular hexagon.

is obtained from by removing an equilateral triangle from the middle third of each side of , (see the snowflake figures ),

is obtained from by removing an equilateral triangle from the middle third of each side of , and in general

is obtained from by removing an equilateral triangle from the middle third of each side of .

The outer snowflake is the set

i.e. a point is in the outer snowflake if and only if it lies in for all positive integers . Observe that the outer snowflake is not a polygon.

An isosceles 120 triangle is an isosceles triangle having a vertex angle of 120. Since the sum of the angles of a triangle is two right angles, the base angles of such a triangle will be .

The following two technical lemmas2.2 guarantee that in the process of building from we never reach a situation where two of the added triangles intersect each other, or where one of the added triangles intersects , and in the process of building from we never reach a situation where two of the removed triangles intersect each other, or where one of the removed triangles fails to lie inside .

2.38   Lemma. Let be an isosceles triangle with . Let be the points that trisect , as shown in the figure. Then is an equilateral triangle, and the two triangles and are congruent isosceles triangles.

Proof: Let be an isosceles triangle with . Construct angles and as shown in the figure, and let and denote the points where the lines and intersect . Then since the sum of the angles of a triangle is two right angles, we have

Hence

and similarly . Thus is an isosceles triangle with two angles, and thus is equilateral. Now by construction, and since is a base angle of an isosceles triangle. It follows that is isosceles and . (If a triangle has two equal angles, then the sides opposite those angles are equal.) Thus, , and a similar argument shows that . It follows that the points and trisect , and that is an isosceles triangle. A similar argument shows that is an isosceles triangle.

Now suppose we begin with the isosceles triangle with angle , and we let be the points that trisect . Since and determine a unique line, it follows from the previous discussion that makes a angle with and makes a angle with , and that all the conclusions stated in the lemma are valid.

2.39   Lemma. If is an equilateral triangle with side of length , then the altitude of has length , and the area of is . If is an isosceles triangle with two sides of length , then the third side of has length .

Proof: Let be an equilateral triangle with side of length , and let be the midpoint of . Then the altitude of is , and by the Pythagorean theorem

Hence

An isosceles triangle with two sides of length can be constructed by taking halves of two equilateral triangles of side , and joining them along their common side of length , as indicated in the following figure.

Hence the third side of an isosceles triangle with two sides of length is twice the altitude of an equilateral triangle of side , i.e., is .

We now construct two sequences of polygons. , and
such that

Let be a regular hexagon with side , and let be an equilateral triangle inscribed in . Then consists of three isosceles triangles with short side , and from lemma 2.39, it follows that the sides of have length . (See the snowflake pictures).

Our general procedure for constructing polygons will be:

is constructed from by removing an equilateral triangle from the middle third of each side of , and is constructed from by adding an equilateral triangle to the middle third of each side of . For each , will consist of a family of congruent isosceles triangles and is obtained from by removing an equilateral triangle from the middle third of each side of each isosceles triangle. Pictures of , , and are given in the above snowflake figures). Details of these pictures are shown below.

Lemma 2.38 guarantees that this process always leads from a set of isosceles triangles to a new set of isosceles triangles. Note that every vertex of is a vertex of and of , and every vertex of is a vertex of and of .

Let

Then

Since an equilateral triangle with side can be decomposed into nine equilateral triangles of side (see the figure),

we have

Also

and since can be written as a union of six equilateral triangles,

The following table summarizes the values of , , , , and :

Now
 (2.40)

Also,
 (2.41)

By the formula for a finite geometric series we have

By using this result in equations (2.40) and (2.41) we obtain

 (2.42)

and

Now you can show that , so the last equation may be written as

 (2.43)

2.44   Exercise. Show that , i.e. show that

2.45   Definition (Snowflakes.) Let and . Here the infinite union means the set of all points such that for some in , and the infinite intersection means the set of points that are in all of the sets where . I will call the sets and the inner snowflake and the outer snowflake, respectively.

For all in , we have

so

Since can be made very small by taking large (see theorem 6.71), we conclude from equations 2.43 and 2.42 that

We will call the circumscribed hexagon for and for . We have proved the following theorem:

2.46   Theorem. The area of the inner snowflake and the outer snowflake are both of the area of the circumscribed hexagon.

Note that both snowflakes touch the boundary of the circumscribed hexagon in infinitely many points.

It is natural to ask whether the sets and are the same.

2.47   Entertainment (Snowflake Problem.) Show that the inner snowflake is not equal to the outer snowflake. In fact, there are points in the boundary of the circumscribed hexagon that are in the outer snowflake but not in the inner snowflake.

The snowflakes were discovered by Helge von Koch(1870-1924), who published his results in 1906 [31]. Actually Koch was not interested in the snowflakes as two-dimensional objects, but as one-dimensional curves. He considered only part of the boundary of the regions we have described. He showed that the boundary of and is a curve that does not have a tangent at any point. You should think about the question: In what sense is the boundary of a curve?'' In order to answer this question you would need to answer the questions what is a curve?'' and what is the boundary of a set in ?'' We will not consider these questions in this course, but you might want to think about them.

I will leave the problem of calculating the perimeter of a snowflake as an exercise. It is considerably easier than finding the area.

2.48   Exercise. Let and be the polygons described in section 2.6, which are contained inside and outside of the snowflakes and .

a)
Calculate the length of the perimeter of .
b)
Calculate the length of the perimeter of .

What do you think the perimeter of should be? (Since it isn't really clear what we mean by the perimeter of ,'' this question doesn't really have a correct'' answer - but you should come up with some answer.)

Next: 3. Propositions and Functions Up: 2. Some Area Calculations Previous: 2.5 Area Under the   Index
Ray Mayer 2007-09-07