Problem 1414

1. Identify the relationship between the pentagons:
The smaller pentagon formed by the diagonals of the larger pentagon is similar to the larger pentagon. Therefore, the ratio of their areas is the square of the ratio of their corresponding side lengths.

2. Define the side lengths and distances:
Let the side length of the larger pentagon be $a$ and the side length of the smaller pentagon be $b$. Let $h$ be the distance between the center points of the respective (parallel) sides of the pentagons, and $H$ be the distance from a side of the larger pentagon to the opposite vertex.

3. **Use trigonometry to find $h$ and $H$:**
- The distance $h$ can be found using the cosine of the angle $\frac{\pi}{10}$:
$$h = a \cdot \cos\left(\frac{\pi}{10}\right)$$
- The distance $H - h$ can be found using the cosine of the angle $\frac{3\pi}{10}$:
$$H - h = a \cdot \cos\left(\frac{3\pi}{10}\right)$$
- Therefore, the total distance $H$ is:
$$H = (H - h) + h = a \cdot \cos\left(\frac{3\pi}{10}\right) + a \cdot \cos\left(\frac{\pi}{10}\right)$$

4. **Simplify the expression for $H$:**
Using the trigonometric identity for the sum of cosines:
$$\cos\left(\frac{3\pi}{10}\right) + \cos\left(\frac{\pi}{10}\right) = 2 \cos\left(\frac{2\pi}{10}\right) \cos\left(\frac{\pi}{10}\right)$$
Therefore:
$$H = 2a \cdot \cos\left(\frac{2\pi}{10}\right) \cos\left(\frac{\pi}{10}\right)$$

5. **Find the ratio $k$ of the side lengths:**
The ratio $k$ of the side lengths of the smaller pentagon to the larger pentagon is:
$$k = \frac{b}{a} = \frac{H - h}{H}$$
Substituting the values of $H$ and $h$:
$$k = \frac{a \cdot \cos\left(\frac{3\pi}{10}\right)}{2a \cdot \cos\left(\frac{2\pi}{10}\right) \cos\left(\frac{\pi}{10}\right)}$$
Simplifying:
$$k = \frac{\cos\left(\frac{3\pi}{10}\right)}{2 \cos\left(\frac{2\pi}{10}\right) \cos\left(\frac{\pi}{10}\right)}$$

6. Use known trigonometric values:
Using the known values:
$$\cos\left(\frac{\pi}{10}\right) = \sqrt{\frac{5 + \sqrt{5}}{8}}$$
$$\cos\left(\frac{3\pi}{10}\right) = \sqrt{\frac{5 - \sqrt{5}}{8}}$$
$$\cos\left(\frac{2\pi}{10}\right) = \frac{\sqrt{5} - 1}{4}$$

7. **Calculate $k$:**
Substituting these values:
$$k = \frac{\sqrt{\frac{5 - \sqrt{5}}{8}}}{2 \cdot \frac{\sqrt{5} - 1}{4} \cdot \sqrt{\frac{5 + \sqrt{5}}{8}}}$$
Simplifying further:
$$k = \frac{\sqrt{5 - \sqrt{5}}}{\sqrt{5} - 1} \cdot \frac{1}{\sqrt{5 + \sqrt{5}}}$$

8. Find the area ratio:
The area ratio is $k^2$:
$$k^2 = \left(\frac{\sqrt{5 - \sqrt{5}}}{\sqrt{5} - 1} \cdot \frac{1}{\sqrt{5 + \sqrt{5}}}\right)^2 = \frac{7 - 3\sqrt{5}}{2}$$

The final answer is $\boxed{\frac{7 - 3\sqrt{5}}{2}}$.