2. Given a convex pentagon $A B C D E$. The diagonals of this pentagon form a pentagon $A_{1} B_{1} C_{1} D_{1} E_{1}$ and a five-pointed star.

a) Determine the sum of the angles of the five-pointed star at the vertices $A, B, C, D$ and E.

b) If the given pentagon $A B C D E$ is regular, determine the ratio of the areas of this pentagon and the pentagon $A_{1} B_{1} C_{1} D_{1} E_{1}$.
Solution. a) The sum of the interior angles of the pentagon $A_{1} B_{1} C_{1} D_{1} E_{1}$ is equal to $3 \cdot 180^{\circ}$, and the sum of the interior angles of the decagon $A A_{1} B B_{1} C C_{1} D D_{1} E E_{1}$ is equal to $8 \cdot 180^{\circ}$. Each angle of the pentagon $A_{1} B_{1} C_{1} D_{1} E_{1}$, when added to the corresponding angle of the decagon, gives $360^{\circ}$ (see the diagram on the right). Therefore, the desired sum of the angles of the star, which is equal to the sum of the angles of the decagon at the vertices $A, B, C, D$, and $E$, is

$$
8 \cdot 180^{\circ} + 3 \cdot 180^{\circ} - 5 \cdot 580^{\circ} = 180^{\circ}
$$

Note. The desired sum can also be calculated by subtracting the difference between the sum of the angles of the triangles $A B A_{1}, B C B_{1}, C D C_{1}, D E D, E A A_{1}$ and the sum of the angles of the pentagon $A_{1} B_{1} C_{1} D_{1} E_{1}$, which is equal to $5 \cdot 180^{\circ} - 3 \cdot 180^{\circ} = 2 \cdot 180^{\circ}$. Finally, the desired sum is equal to $3 \cdot 180^{\circ} - 2 \cdot 180^{\circ} = 180^{\circ}$.

b) If the pentagon $A B C D E$ is regular, then it is symmetric with respect to the bisector of each of its angles, so $A A_{1} = A E_{1} = E E_{1}, \quad A C \| E D$ and $\measuredangle A A_{1} E = \measuredangle A_{1} E D = \measuredangle C E A = \measuredangle E A A_{1}$. Therefore, triangle $A E A_{1}$ is isosceles. Let $A E = a$ and $A_{1} E_{1} = x$. Then, from the similarity of the isosceles triangles $A_{1} A E_{1}$ and $A_{1} E A$ (equal base angles), we get $A A_{1} : A_{1} E_{1} = A_{1} E : A A_{1}$, i.e.,

$$
(a - x) : x = a : (a - x)
$$

Solving the last equation for $x$ and considering that $x < a$, we get $x = \frac{3 - \sqrt{5}}{2} a$. The pentagons $A B C D E$ and $A_{1} B_{1} C_{1} D_{1} E_{1}$ are similar, so the ratio of their areas is

$$
S : S_{1} = \left(\frac{a}{x}\right)^{2} = \left(\frac{2}{3 - \sqrt{5}}\right)^{2} = \left(\frac{\sqrt{5} + 3}{2}\right)^{2} = \frac{7 + 3 \sqrt{5}}{2}
$$