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
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
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 .
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
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.
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
We now construct two sequences of polygons.
,
and
such that
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
The following table
summarizes the values of , , , ,
and :
By the formula for a finite geometric series we have
By using this result in equations (2.40) and (2.41) we obtain
Now you can show that
, so the last equation may be
written as
For all in
, we have
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.
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.
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.)