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When I saw this image I was a little curious. How can I find the area of this fractal?

enter image description here

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Can you sum a geometric series? –  GEdgar Nov 4 '12 at 18:07
    
Yes, and the sequence is infinity –  John Smith Nov 4 '12 at 18:18
    
An infinite geometric series may have a finite limit. –  GEdgar Nov 4 '12 at 18:22
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There area must be finite, as the entire fractal is contained inside the pentagon formed by joining the vertices of the initial star ... –  Old John Nov 4 '12 at 22:19
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3 Answers

up vote 2 down vote accepted

To avoid accidentally confusing the Koch Snowflake and (what we might call) the Koch Pentaflake, let's work in generality.

Consider a segment of length $1$, within which we identify a central segment of length $\alpha$. (In the Koch Snowflake, $\alpha = 1/3$. In the Pentaflake, $\alpha = 1/\phi^3$, with golden ratio $\phi := 1.618...$.) The "wings" of the segment have length $\omega :=(1-\alpha)/2$. We build an isosceles triangle over the central segment, with legs of length $\omega$; the height of this triangle is $\sqrt{\omega^2 - \left(\frac{1}{2}\alpha\right)^2} = \frac{1}{2}\sqrt{1-2\alpha}$, so that its area is $A_0 := \frac{1}{4}\alpha\sqrt{1-2\alpha}$. (Observe that, both geometrically and algebraically, we require $\alpha \le 1/2$.)

Now we have $4$ segments of length $\omega$. Upon each central segment of length $\omega\alpha$, we construct an isosceles triangle ---with legs of length $\omega^2$--- with area $A_1 := \frac{1}{4}\omega^2\alpha\sqrt{1-2\alpha}$.

At this point, we have $16=4^2$ segments of length $\omega^2$, each of which gives rise to an isosceles triangle ---with legs of length $\omega^4=(\omega^2)^2$--- of area $A_2:=\frac{1}{4}(\omega^2)^2\alpha\sqrt{1-2\alpha}$.

In the next (third) iteration, we have $64=4^3$ segments, and so $4^3$ triangles of area $A_3 := \frac{1}{4}(\omega^2)^3\;\alpha\sqrt{1-2\alpha}$.

For iteration $4$, we have $4^4$ triangles of area $A_4 :=\frac{1}{4}(\omega^2)^4\;\alpha\sqrt{1-2\alpha}$.

And so on.

The total area of these triangles is

$$\begin{align} A := A_0 + 4 A_1 + 4^2 A_2 + 4^3 A_3 + 4^4 A_4 + \cdots &= \sum_{k=0}^{\infty}4^k A_k \\ &= \frac{1}{4} \alpha\sqrt{1-2\alpha}\cdot \sum_{k=0}^{\infty}\left(4\omega^2\right)^{k} \\ &= \frac{1}{4} \alpha\sqrt{1-2\alpha}\cdot \frac{1}{1-4\omega^2} \\ &=\frac{1}{4} \alpha\sqrt{1-2\alpha} \cdot \frac{1}{1-(1-\alpha)^2} \\ &=\frac{\sqrt{1-2\alpha}}{4(2-\alpha)} \\ \end{align}$$

For the Koch Snowflake, $\alpha = 1/3$, so that $A = \sqrt{3}/20$, but note that this is simply the area under the "Koch Curve" forming one side of the Snowflake. The Snowflake's area comprises three copies of $A$, plus the area of the central equilateral triangle of side length $1$; that is, $3 A + \sqrt{3}/4 = 2\sqrt{3}/5$. This agrees with the Wikipedia article on the Koch Snowflake (taking side length $s=1$).

For the Pentaflake, $\alpha = 1/\phi^3$. The figure's full area is equal to five copies of $A$, plus the area of the pentagon of side length $\alpha$ (not $1$! The pentagon sits under the central segments on each side).

Edit. No. No, it is not. Five copies of $A$ over-counts the pointy area: arranging five unit-base "PentaKoch Curves" into a pentagram causes some overlap in the constructed triangles. (Triangles built on the "wings" of one initial unit-length segment overlap those built on a leg of the triangle from the neighboring segment.) Rather than describe how to subtract-off the overlap, I'll just revise the sum as it applies to the starry figure itself.

enter image description here

The full area of the PentaFlake --with wingspan $1$-- is given by the area, $P$, of the pentagon of side-length $\alpha=1/\phi^3$, plus: $5$ copies of $A_0$, and $10$ copies of $A_1$, and $40$ copies of $A_2$, and $160$ copies of $A_3$, and, and, and, ...

$$\begin{align} P + 5 A_0 + 10 A_1 + 10 \cdot 4 A_2 + 10 \cdot 4^2 A_3 + \cdots &= P + 5 A_0 + \frac{10}{4} \sum_{k=1}^{\infty} 4^k A_k \\ &=P + \frac{5}{4}\alpha\sqrt{1-2\alpha} + \frac{10}{16}\alpha\sqrt{1-2\alpha} \frac{4\omega^2}{1-4\omega^2} \\ &=P + \frac{5(1-2\omega^2)}{4(1-4\omega^2)}\alpha\sqrt{1-2\alpha} \end{align}$$ which gives $$\frac{\alpha^2}{4}\sqrt{25+10\sqrt{5}} + \frac{5}{8}\frac{1+2\alpha-\alpha^2}{2-\alpha}\sqrt{1-2\alpha}$$

As before the edit, I'll leave it to the reader to express this value in terms of $\phi$ --noting the reduction formula $\phi^2 = \phi+1$-- or in terms of $\sqrt{5}$ (which is equal to $2\phi-1$ and thus also $\alpha+2$).

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Each segment of the pentagram is the initiator of the fractal. Take its length to be 1. Now the generator consists of 2 line segments each of length $\frac{1}{3}$.

Hence on each iteration $n$ the area can be expressed as follows:

$$A_n=10\sum_{k=0}^{n}2^kS_k +S_{p}$$

Where $S_p$ is the area of the regular pentagon and:

$$S_k=\frac{1}{2}\frac{1}{3^{2k}}\sin\frac{\pi}{5}$$

is the area of the "k-th generation" petal.

$$10\sum_{k=0}^{n}2^kS_k=5\pi\sin\frac{\pi}{5}\sum_{k=0}^{\infty}\left(\frac{2}{9}\right)^k=\frac{45\pi}{7}\sin\frac{\pi}{5}$$

The area of the pentagon is:

$$S_p=\frac{t^2\sqrt{25+10\sqrt{5}}}{4}$$ where $t=2\sin\frac{\pi}{10}$

Finally,

$$A=\frac{45\pi}{7}\sin\frac{\pi}{5}+\sqrt{25+10\sqrt{5}}\left(\sin\frac{\pi}{10}\right)^2$$

Or equivalently

$$A=\frac{45\pi\sqrt{2(5-\sqrt{5})}}{28}+\frac{\sqrt{25+10\sqrt{5}}}{4\phi^2}$$

or any other way you wish to think of it.

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ok, so i made the same mistake taking the ratio to be $\frac{1}{3}$ when in fact it is not.. will revisit this in a while –  Valentin Nov 5 '12 at 8:37
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First, $ Area\ of\ Star\ with\ 5\ petals\ = Area\ of\ pentagram\ ->\ step\ 1 $

please refer http://mathworld.wolfram.com/Pentagram.html for area formula of pentagram. because the tricky part is identifying the GP.

Idea is that from each of the 5 triangles (assuming the new triangle has side one third the length of base triangle on which it forms ,as fractals are regular figures) there are 2 triangles coming up .

after iteration 1 added area =area of 10 equilaterla triangles of side $ \frac{a}3 $

after iteration 2 added area =area of 20 equilaterla triangles of side $ \frac{a}{3^2} $

after iteration 3 added area =area of 40 equilaterla triangles of side $ \frac{a}{3^3} $

so as this is infinite as per your problem this goes on and on

Area = Area of STAR (as found in step 1) + area added after infinite iteration (let this be k)

$k\ = 10\times(\sqrt(3)/4)(\frac{a}{3})^2\ +\ 20\times(\sqrt(3)/4)(\frac{a}{3^2})^2\ +\ 40\times(\sqrt(3)/4)(\frac{a}{3^3})^2\ ...$

$k\ = 10\times(\sqrt(3)/4)a^2 [ \frac{1}{3^2} + \frac{2}{3^4} + \frac{2^2}{3^6} +\ ..... ]$

this within square brackets is a infinite GP with common ratio of $\frac{2}{3^2}\ $and first term is $\frac{1}{3^2} $ as summation of GP with infinite series is $\frac{firstterm}{1-commonratio} $

$k\ = 10\times(\sqrt(3)/4)a^2 \times [\frac{\frac{1}{3^2}}{1-\frac{2}{3^2}}]$

$k\ = 10\times(\sqrt(3)/4)a^2 \times [\frac{1}{7}]$

so

$Area = Area\ of\ PENTAGRAM\ + 10\times(\sqrt(3)/4)a^2 \times [\frac{1}{7}]$

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the "petal" triangles are not equilateral –  Valentin Nov 4 '12 at 22:21
    
As no specifications were given I took it to be equilateral . but the idea is the same .In this fractal the idea is that basically there is a pentagon and from each of the 5 sides a triangle starts and thereafter the GP starts in that each triangle gives rise to two other –  Harish Kayarohanam Nov 4 '12 at 22:25
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well, if it is a regular pentagram there are quite many specifications already given. –  Valentin Nov 4 '12 at 22:27
    
So according to your point, Area = Area of pentagram(which has a formula) + 10×((√3)/4)a^2×[1/7] . Is it ok now ? –  Harish Kayarohanam Nov 4 '12 at 22:52
    
The areas of the "petals" are going to involve the golden ratio. See, for instance, contracosta.edu/legacycontent/math/pentagrm.htm . –  Blue Nov 4 '12 at 22:55
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