TITLE: Area fractal pentagrams I
QUESTION [6 upvotes]: When I saw this image I was a little curious.
How can I find the area of this fractal?

REPLY [5 votes]: 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.

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 $\alpha+2$).