# Area fractal pentagrams II

A simple fractal. How to find the area of it? (only the arms of the star)

Working with pentagrams is quite complicated, I can not solve this.

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"Only the arms of the star" mean what? – GEdgar Nov 4 '12 at 18:08
only the external triangles, disregarding the center of the pentagram – John Smith Nov 4 '12 at 18:16
To deal with the infinitely many stars, consider the large pentagon minus the next smaller one, and compute the proportion of that which is covered by arms. As you find that same proportion at every iteration, you can extend it to the whole area of the large pentagon. To compute a single triangle, you either know some numbers that occur in a pentagram, or you go the hard way and start by taking some known angles, and computing coordinates from them using trigonometric functions. – MvG Nov 5 '12 at 6:09

The easiest method to calculate only the "arms" of the stars is to subtract the areas of all the larger triangles (for lack of better wording) in each iteration from the area of the entire pentagon.

First of all, we calculate the area of the entire pentagon (assuming the side length is 1) to be $5\times\frac{1}{2}\times\left(\frac{sin{54}}{sin{72}}\right)^2\times\sin{72}$

distance from centre to vertex

Since pentagrams are well known for their relationship with the golden ratio ($\phi=\frac{1+\sqrt{5}}{2}$), the side lengths can be calculated easily using these identities. Diagram of values (all angles are in degrees).

The total area of the obtuse triangles in the first iteration is $5\times\frac{1}{2}\times(\phi-1)^2\times\sin{108}$

Next, we need to calculate the ratio between the area of each iteration of obtuse triangles. Since the triangles are similar, the ratio r (smaller/larger) can be simplified to $(2-\phi)^2$ Since r>1, this is identified as a geometric series and therefore the area of all the obtuse triangles can be found using the formula $\frac{a}{1-r}$

$A_{total}=\frac{a}{1-r}$ $=\frac{5\times\frac{1}{2}\times(\phi-1)^2\times\sin{108}}{1-(2-\phi)^2}$

We then subtract this value from the area of the entire pentagon to get $5\times\frac{1}{2}\times\left(\frac{sin{54}}{sin{72}}\right)^2\times\sin{72}-\frac{5\times\frac{1}{2}\times(\phi-1)^2\times\sin{108}}{1-(2-\phi)^2}$

The total area of all the "arms" of the star $\approx0.65716$