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 $I_1$, $I_2$, $I_3\ldots$ as follows:

$I_1$ is an equilateral triangle.

$I_2$ is obtained from $I_1$ by adding an equilateral triangle to the middle third of each side of $I_1$, (see the snowflake figures ).

$I_3$ is obtained from $I_2$ by adding an equilateral triangle to the middle third of each side of $I_2$, and in general

$I_{n+1}$ is obtained from $I_n$ by adding an equilateral triangle to the middle third of each side of $I_n$.

The inner snowflake is the set

$$K_I = \bigcup_{n=1}^\infty I_n,$$

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

To construct the outer snowflake, we first construct a family of polygons $O_1$, $O_2$, $O_3\ldots$ as follows:

$O_1$ is a regular hexagon.

$O_2$ is obtained from $O_1$ by removing an equilateral triangle from the middle third of each side of $O_1$, (see the snowflake figures ),

$O_3$ is obtained from $O_2$ by removing an equilateral triangle from the middle third of each side of $O_2$, and in general

$O_{n+1}$ is obtained from $O_n$ by removing an equilateral triangle from the middle third of each side of $O_n$.

The outer snowflake is the set

$$K_O = \bigcap_{n=1}^\infty O_n,$$

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

An isosceles 120$^\circ$ triangle is an isosceles triangle having a vertex angle of 120$^\circ$. Since the sum of the angles of a triangle is two right angles, the base angles of such a triangle will be $${{1\over 2}(180^\circ -120^\circ )=30^\circ}$$.

The following two technical lemmas^2.2 guarantee that in the process of building $I_{n+1}$ from $I_n$ we never reach a situation where two of the added triangles intersect each other, or where one of the added triangles intersects $I_n$, and in the process of building $O_{n+1}$ from $O_n$ 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 $O_n$.

2.38   Lemma. Let $\triangle BAC$ be an isosceles $120^\circ$ triangle with $\angle BAC=120^\circ$. Let $E,F$ be the points that trisect $BC$, as shown in the figure. Then $\triangle AEF$ is an equilateral triangle, and the two triangles $\triangle AEB$ and $\triangle AFC$ are congruent isosceles $120^\circ$ triangles.

Proof: Let $\triangle BAC$ be an isosceles triangle with $\angle BAC=120^\circ$. Construct $30^\circ$ angles $BAX$ and $CAY$ as shown in the figure, and let $E$ and $F$ denote the points where the lines $AX$ and $AY$ intersect $BC$. Then since the sum of the angles of a triangle is two right angles, we have

$$\angle AEB=180^\circ -\angle ABE-\angle BAE=180^\circ -30^\circ -30^\circ =120^\circ .$$

Hence

$$\angle AEF=180^\circ -\angle AEB=180^\circ -120^\circ=60^\circ,$$

and similarly $\angle AFE=60^\circ$. Thus $\triangle AEF$ is an isosceles triangle with two $60^\circ$ angles, and thus $\triangle AEF$ is equilateral. Now $\angle BAE=30^\circ$ by construction, and $\angle ABE=30^\circ$ since $\angle ABE$ is a base angle of an isosceles $120^\circ$ triangle. It follows that $\triangle BEA$ is isosceles and $BE=EA$. (If a triangle has two equal angles, then the sides opposite those angles are equal.) Thus, $BE=EA=EF$, and a similar argument shows that $CF=EF$. It follows that the points $E$ and $F$ trisect $BC$, and that $\triangle AEB$ is an isosceles $120^\circ$ triangle. A similar argument shows that $\triangle AFC$ is an isosceles $120^\circ$ triangle.

Now suppose we begin with the isosceles $120^\circ$ triangle $\triangle BAC$ with angle $BAC=120^\circ$, and we let $E,F$ be the points that trisect $BC$. Since $A$ and $E$ determine a unique line, it follows from the previous discussion that $EA$ makes a $30^\circ$ angle with $BA$ and $FA$ makes a $30^\circ$ angle with $AC$, and that all the conclusions stated in the lemma are valid. $\diamondsuit$

2.39   Lemma. If $T$ is an equilateral triangle with side of length $a$, then the altitude of $T$ has length $${ {{a\sqrt 3}\over 2}}$$, and the area of $T$ is $${ {{\sqrt 3}\over 4}a^2}$$. If $R$ is an isosceles $120^\circ$ triangle with two sides of length $a$, then the third side of $R$ has length $a\sqrt 3$.

Proof: Let $T=\triangle ABC$ be an equilateral triangle with side of length $a$, and let $M$ be the midpoint of $BC$. Then the altitude of $T$ is $AM$, and by the Pythagorean theorem

$$AM=\sqrt{(AB)^2-(BM)^2}=\sqrt{a^2-\left( {1\over 2}a\right)^2}=\sqrt{{3\over 4}a^2}={{\sqrt 3}\over 2}a.$$

Hence

$$\mbox{\rm area}(T)={1\over 2}(\mbox {base})(\mbox{altitude})={1\over 2}a\cdot {{\sqrt 3}\over 2}a={{\sqrt 3}\over 4}a^2.$$

An isosceles $120^\circ$ triangle with two sides of length $a$ can be constructed by taking halves of two equilateral triangles of side $a$, and joining them along their common side of length ${a\over 2}$, as indicated in the following figure.

Hence the third side of an isosceles $120^\circ$ triangle with two sides of length $a$ is twice the altitude of an equilateral triangle of side $a$, i.e., is $${ 2\left( {{\sqrt 3}\over 2}a\right)=\sqrt 3 a}$$. $\diamondsuit$

We now construct two sequences of polygons. $I_1,I_2,I_3,\cdots$, and
$O_1,O_2,O_3,\cdots$ such that

$$I_1\subset I_2\subset I_3\subset\cdots\subset O_3\subset O_2\subset O_1.$$

Let $O_1$ be a regular hexagon with side $1$, and let $I_1$ be an equilateral triangle inscribed in $O_1$. Then $O_1\setminus I_1$ consists of three isosceles $120^\circ$ triangles with short side $1$, and from lemma 2.39, it follows that the sides of $I_1$ have length $\sqrt 3$. (See the snowflake pictures).

Our general procedure for constructing polygons will be:

$O_{n+1}$ is constructed from $O_n$ by removing an equilateral triangle from the middle third of each side of $O_n$, and $I_{n+1}$ is constructed from $I_n$ by adding an equilateral triangle to the middle third of each side of $I_n$. For each $n$, $O_n\setminus I_n$ will consist of a family of congruent isosceles $120^\circ$ triangles and $O_{n+1}\setminus I_{n+1}$ is obtained from $O_n\setminus I_n$ by removing an equilateral triangle from the middle third of each side of each isosceles $120^\circ$ triangle. Pictures of $I_n$, $O_n$, and $O_n\setminus I_n$ 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 $120^\circ$ triangles to a new set of isosceles $120^\circ$ triangles. Note that every vertex of $O_n$ is a vertex of $O_{n+1}$ and of $I_{n+1}$, and every vertex of $I_n$ is a vertex of $O_n$ and of $I_{n+1}$.

Let

$$\begin{eqnarray*} s_n&=&\mbox{length of a side of $I_n$}.\\ t_n&=&\mbox{area of... ...&\mbox{number of sides of $O_n$}.\\ A_n&=&\mbox{area of $O_n$}. \end{eqnarray*}$$

Then

$$\begin{array}{lll} s_{n+1}={1\over 3}s_n, & &S_{n+1}={1\over ... ...a_{n+1}=a_n+m_n t_{n+1} & &A_{n+1}=A_n-M_n T_{n+1}. \end{array}$$

Since an equilateral triangle with side $s$ can be decomposed into nine equilateral triangles of side $${s \over 3}$$ (see the figure),

we have

$$t_{n+1} = {t_n\over 9} \mbox{ and }T_{n+1} = {T_n \over 9}.$$

Also

$$a_1 = \mbox{area}(I_1) = t_1,$$

and since $O_1$ can be written as a union of six equilateral triangles,

$$A_1 = 6 T_1.$$

The following table summarizes the values of $s_n$, $m_n$, $t_n$, $S_n$, $M_n$ and $T_n$:

$$\begin{array}{\vert l\vert l\vert l\vert l\vert l\vert l\vert... ...4\over 9}\right)^{n-2} {A_1\over 9} [2ex] \hline \end{array}$$

Now

$$A_2 = A_1 - M_1T_2 = A_1- {A_1 \over 9},\mbox{{}}$$

$$A_3 = A_2-M_2T_3= A_1 - {A_1 \over 9} - \left({4\over 9}\right){A_1 \over 9}, \mbox{{}}$$

$$\vdots \mbox{{}}$$

$$A_{n+1} = A_{n} - M_{n}T_{n+1} \mbox{{}}$$

$$= A_1 -{A_1\over 9}- \left({4\over 9}\right){A_1\over 9} - \left({4... ...)^2{A_1 \over 9} - \cdots - \left({4\over 9}\right)^{n-1}{A_1 \over 9}\mbox{{}}$$

$$= A_1 - {A_1\over 9}\left(1+{4\over 9}+\left({4\over 9}\right)^2+\cdots +\left({4\over 9}\right)^{n-1}\right).$$ (2.40)

Also,

$$a_2 = a_1 +m_1t_2 = a_1 + {3\over 9}a_1,\mbox{{}}$$

$$a_3 = a_2 + m_2t_3 = a_1 + {3\over 9}a_1 + {3\over 9}\left({4\over 9}\right)a_1,\mbox{{}}$$

$$\vdots \mbox{{}}$$

$$a_{n+1} = a_n + m_nt_{n+1}\mbox{{}}$$

$$= a_1 + {3\over 9}a_1 + {3\over 9}\left({4\over 9}\right)a_1 +{3\ov... ...r 9\right)}^2 a_1 + \cdots +{3\over 9}\left({4\over 9}\right)^{n-1}a_1\mbox{{}}$$

$$= a_1 + {a_1\over 3}\left(1+{4\over 9}+\left({4\over 9}\right)^2+\cdots +\left({4\over 9}\right)^{n-1}\right).$$ (2.41)

By the formula for a finite geometric series we have

$$1+{4\over 9}+({4\over 9})+\cdots +({4\over 9})^{n-1}={{1-({4\... ...}\over {1-{4\over 9}}}={9\over 5}\left[1-({4\over 9})^n\right].$$

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

$$\mbox{\rm area}(O_{n+1}) = A_{n+1} = A_1 - {A_1 \over 5} \left[ 1 - \left({4\over 9} \right)^n \right] \mbox{{}}$$

$$= {4\over 5}A_1 + {A_1 \over 5} \left( {4\over 9} \right)^n,$$ (2.42)

and

$$\mbox{\rm area}(I_{n+1}) = a_{n+1}= a_1 + {a_1 \over 3} \cdot{9 \over 5} \left[ 1-\left({4\over 9}\right)^n\right]$$

$$= {8\over 5}a_1 - {3a_1 \over 5} \left( {4\over 9} \right)^n.\mbox{{}}$$

Now you can show that $a_1 = \displaystyle {A_1 \over 2}$, so the last equation may be written as

$$\mbox{\rm area}(I_{n+1}) = {4\over 5}A_1 -{3a_1 \over 5} \left( {4\over 9} \right)^n.$$ (2.43)

2.44   Exercise. Show that $a_1 = \displaystyle {A_1 \over 2}$, i.e. show that $\mbox{\rm area}(I_1) = \displaystyle {1\over 2}\mbox{\rm area}(O_1).$

2.45   Definition (Snowflakes.) Let $K_I=\displaystyle { \bigcup_{n=1}^\infty I_n}$ and $K_O=\displaystyle {\bigcap_{n=1}^\infty O_n}$. Here the infinite union $${\bigcup_{n=1}^\infty I_n}$$ means the set of all points $x$ such that $x\in I_n$ for some $n$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$, and the infinite intersection $${\bigcap_{n=1}^\infty O_n}$$ means the set of points $x$ that are in all of the sets $O_n$ where $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$. I will call the sets $K_I$ and $K_O$ the inner snowflake and the outer snowflake, respectively.

For all $k$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$, we have

$$I_k\subset \bigcup_{n=1}^\infty I_n =K_I\subset K_O=\bigcap_{n=1}^\infty O_n \subset O_k,$$

so

$$\mbox{\rm area}(I_k)\leq\mbox{\rm area}(K_I)\leq\mbox{\rm area}(K_O)\leq \mbox{\rm area}(O_k).$$

Since $${ \left({4\over 9}\right)^n}$$ can be made very small by taking $n$ large (see theorem 6.71), we conclude from equations 2.43 and 2.42 that

$$\mbox{\rm area}(K_I)=\mbox{\rm area}(K_O)= {4\over 5}A_1 = {4\over 5} \mbox{\rm area}(O_1).$$

We will call $O_1$ the circumscribed hexagon for $K_I$ and for $K_O$. We have proved the following theorem:

2.46   Theorem. The area of the inner snowflake and the outer snowflake are both $${{4\over 5}}$$ 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 $K_O$ and $K_I$ 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 $K_O$ and $K_I$ 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 $K_O$ 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 $\mbox{{\bf R}}^2$?'' 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 $I_n$ and $O_n$ be the polygons described in section 2.6, which are contained inside and outside of the snowflakes $K_I$ and $K_O$.

a)

Calculate the length of the perimeter of $I_n$.

b)

Calculate the length of the perimeter of $O_n$.

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