Body $A, B, C, D$ and $E$ are vertices of an irregular five-pointed star, see the figure. Determine the sum of the marked angles.

![](https://cdn.mathpix.com/cropped/2024_04_17_fc767241fb97c8703187g-5.jpg?height=528&width=625&top_left_y=1558&top_left_x=721)

Note: the figure is illustrative only.

Hint. What is the sum of the interior angles in a triangle?
Using the marked angles, we can express all other angles in the given polygon. We will start this way and try to find out something about the desired sum. The marked angles and the remaining vertices of the polygon are labeled as follows:[^0]

![](https://cdn.mathpix.com/cropped/2024_04_17_fc767241fb97c8703187g-6.jpg?height=816&width=894&top_left_y=186&top_left_x=587)

The angle $L K O$ is an internal angle of triangle $B K D$, so it can be expressed as $180^{\circ}-\beta-\delta$. The angle $L K A$ is the supplementary angle to angle $L K O$, or the external angle of triangle $B K D$, so it is equal to $\beta+\delta$. Similarly, the angle $K L A$ is equal to $\gamma+\varepsilon$, etc.

The sum of the internal angles in triangle $A K L$ is precisely the desired sum of the marked angles:

$$
\alpha+\beta+\gamma+\delta+\varepsilon=180^{\circ} .
$$

Notes. In the previous solution, we focused on expressing the angles at the tip of the star with vertex $A$. The same result is obtained at any other tip.

Although the marked angles can be very diverse, their sum is always the same. This is a consequence of the somewhat non-obvious statement about the sum of angles in a triangle. This statement will certainly be in the background of any other solution to the problem. For example, it is possible to use the sum of the internal angles of a general polygon: an $n$-sided polygon can be divided (in various ways) into $n-2$ triangles, so the sum of its internal angles is $(n-2) \cdot 180^{\circ}$.

Another solution. The sum of the internal angles of the pentagon $K L M N O$ is equal to $3 \cdot 180^{\circ}=540^{\circ}$. Similarly to the solution mentioned above, the internal angles at vertices $K, L, M, N, O$ can be expressed, respectively, as

$$
180^{\circ}-\beta-\delta, \quad 180^{\circ}-\gamma-\varepsilon, \quad 180^{\circ}-\delta-\alpha, \quad 180^{\circ}-\varepsilon-\beta, \quad 180^{\circ}-\alpha-\gamma .
$$

In total, we get

$$
\begin{aligned}
5 \cdot 180^{\circ}-2(\alpha+\beta+\gamma+\delta+\varepsilon) & =540^{\circ}, \\
2(\alpha+\beta+\gamma+\delta+\varepsilon) & =360^{\circ}, \\
\alpha+\beta+\gamma+\delta+\varepsilon & =180^{\circ} .
\end{aligned}
$$

[^0]:    * Disjoint sets are sets with an empty intersection, i.e., sets without a common element.