Pentagons in Pentagons 2

Follow the side length $1$ of the small pentagon in the center. Using isosceles triangles and a parallelgram, it can be readily seen that the side length of the large pentagon is $4$ .

Consider the relationship between a regular pentagon and the regular pentagon formed by the star inside it (the yellow, orange, and red below), and consider the relationship between a regular pentagon and the regular pentagon formed by joining its midsegments (the green and blue below), all labelled by the letters in the picture given. Let the sides of the red pentagon be $x$ , the sides of the yellow pentagon be $y$ , the sides of the green pentagon be $v$ , and the sides of the blue pentagon be $w$ .

Then $CD = BD - BC$ , and since $BD = DE = y$ and $BC = x$ , $CD = y - x$ . By symmetry, $CD = CE = BE = AB = y - x$ . So $AD = AB + BC + CD = (y - x) + x + (y - x) = 2y - x$ .

Since $\triangle AED$ is similar to $\triangle DCE$ , $\frac{DE}{AD} = \frac{CD}{DE}$ , or $\frac{y}{2y - x} = \frac{y - x}{y}$ , and this solves to $y = \frac{3 + \sqrt{5}}{2}x$ , or $y = \phi^2x$ , where $\phi = \frac{1 + \sqrt{5}}{2}$ , the golden ratio. So the sides of a regular pentagon are greater than the sides of the regular pentagon formed by the star inside it by a factor of $\phi^2$ .

Additionally, since $\triangle FHG$ is also similar to $\triangle DCE$ , $\frac{DE}{FG} = \frac{CD}{HG}$ , or $\frac{y}{v} = \frac{y - x}{\frac{1}{2}w}$ , and since $y = \phi^2x$ and $\phi^2 - 1 = \phi$ , this solves to $w = \frac{2}{\phi}v$ . So the sides of a regular pentagon are greater than the sides of the regular pentagon formed by joining its midsegments by a factor of $\frac{2}{\phi}$ .

In the problem, the smallest pentagon is formed by a star in a pentagon formed by the joining the midsegments of a pentagon formed by joining the midsegments of the largest pentagon, so the sides increased by a factor of $\phi^2 \cdot \frac{2}{\phi} \cdot \frac{2}{\phi} = 4$ . Therefore, the areas increased by a factor of $4^2 = \boxed{16}$ .