11.4. (SFRY, 70). The diagonals of a convex pentagon $A B C D E$, intersecting, form a pentagon $A_{1} B_{1} C_{1} D_{1} E_{1}$ and a five-pointed star.

a) Find the sum of the angles of this star at the vertices $A$, $B, C, D, E$.

b) Find the ratio of the area of the pentagon $A_{1} B_{1} C_{1} D_{1} E_{1}$ to the area of the pentagon $A B C D E$ under the condition of the regularity of the latter.
11.4. b) The sum of the interior angles of the pentagon $A_{1} B_{1} C_{1} D_{1} E_{1}$ and the decagon $A A_{1} B B_{1} C C_{1} D D_{1} E E_{1}$ are respectively $3 \cdot 180^{\circ}$ and 8 $\cdot 180^{\circ}$. Moreover, each angle of this pentagon, together with the angle of the decagon at the same vertex, sums to $360^{\circ}$ (Fig. 45). Therefore, we have

$$
\angle A_{1} B B_{\mathbf{i}} + \angle B_{1} C C_{1} + \angle C_{\mathbf{i}} D D_{\mathbf{i}} + \angle D_{1} E E_{1} + \angle E_{1} A A_{1} = 8 \cdot 180^{\circ} + 3 \cdot 180^{\circ} - 5 \cdot 360^{\circ} = 180^{\circ}
$$

b) If $A B C D E$ is a regular pentagon, then by symmetry (with respect to the bisector of any of its angles), we have

$$
A A_{1} = A E_{i} = E E_{1}, \quad A C \parallel E D
$$

from which

$$
\begin{aligned}
& \angle A A_{1} E = \angle A_{1} E D = \angle C E A = \\
& = \angle E A A_{1}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_03a501298610b72904feg-206.jpg?height=424&width=441&top_left_y=1117&top_left_x=601)

Fig. 45

i.e., triangle $A E A_{1}$ is isosceles. Let $a = A E$, $x = A_{1} E_{i}$. Then, from the similarity of the isosceles triangles $A_{1} A E_{1}$ and $A_{1} E A$ (by the common angle at the base), we get

$$
A A_{1} / A_{1} E_{1} = A_{1} E / A A_{1}, \text{ i.e., } (a - x) / x = a / (a - x)
$$

from which

$$
x^{2} - 3 a x + a^{2} = 0
$$

Therefore,

$$
x / a = (3 - \sqrt{5}) / 2 \text{ (since } x < a)
$$

and

$$
S_{A_{1} B_{1} C_{1} D_{1} E_{1}} / S_{A B C D E} = (x / a)^{2} = ((3 - \sqrt{5}) / 2)^{2} = (7 - 3 \sqrt{5}) / 2
$$