In the star pentagon $A B C D E$ (each side line has 2 vertices on one side from the remaining vertices, and 1 vertex on the other side). What is the sum of the angles $A B C, B C D, C D E, D E A, E A B$?
I. solution. We use the notation of the figure (the intersection point of lines $BC$ and $DE$ is $J$, and so on).

![](https://cdn.mathpix.com/cropped/2024_05_02_ed8321d1a3fabb16dd0bg-1.jpg?height=781&width=761&top_left_y=206&top_left_x=671)

We apply the well-known theorem of the exterior angle of a triangle first to $\triangle BCN$, then to $\triangle NDL$:

$$
\begin{gathered}
\beta+\gamma=\angle ABC + \angle BCD = \angle NBC + \angle BCN = \angle BND = \angle LND, \\
\angle LND + \angle NDL = \angle NLE, \quad \text{where} \quad \angle NDL = \angle CDE = \delta,
\end{gathered}
$$

thus

$$
\beta+\gamma+\delta=\angle NLE = \angle ALE
$$

Adding $\varepsilon = \angle DEA = \angle LEA$ and $\alpha = \angle EAB = \angle EAL$ to this, we get the sum of the angles of $\triangle AEL$, so $\alpha+\beta+\gamma+\delta+\varepsilon=180^{\circ}$.

Eltető Eszter (Budapest, I. István g. I. o. t.)

II. solution. The sum of the angles in question is obtained by subtracting the sum of the interior angles at the vertices $J, K, L, M, N$ of the triangles $BJL, CKM, DNL, EMJ$, and $ANK$ from the sum of the interior angles of these triangles, which is $5 \cdot 180^{\circ}$. Among these 10 angles, 2 are vertical angles and thus equal, and each is an exterior angle of the common pentagon $JKLMN$; hence, the above subtraction is equal to twice the sum of these exterior angles.

The sum of the interior and exterior angles at each vertex of a common pentagon is $180^{\circ}$, so the sum of the exterior angles is obtained by subtracting the sum of the interior angles from $5 \cdot 180^{\circ}$. The sum of the interior angles, as is well known, is $3 \cdot 180^{\circ}$, so the sum of the exterior angles is $2 \cdot 180^{\circ}$, and the above subtraction is $4 \cdot 180^{\circ}$. Therefore, the desired sum, the remainder of the subtraction, is $1 \cdot 180^{\circ} = 180^{\circ}$.

Bárász Péter (Budapest, Fazekas M. gyak. g. I. o. t.)

III. solution. Let's walk around the pentagon in the order of the labeling, starting from a point $K$ on the $EA$ side. In this way, we need to turn at the vertices $A, B, C, D, E$ by the supplementary angles of $\alpha, \beta, \gamma, \delta$, and $\varepsilon$, respectively, and we complete two full turns. Thus,

$$
\begin{gathered}
180^{\circ}-\alpha+180^{\circ}-\beta+180^{\circ}-\gamma+180^{\circ}-\delta+180^{\circ}-\varepsilon=720^{\circ} \\
\alpha+\beta+\gamma+\delta+\varepsilon=180^{\circ}
\end{gathered}
$$

Herneczki István (Sopron, Széchenyi I. g. I. o. t.)

IV. solution. The sum $\alpha+\beta+\gamma+\delta+\varepsilon$ in question can be considered as the change in direction of the segment $AE$ during the following sequence of movements. The segment $AE$

1. a) rotates around $A$ (by an angle less than $180^{\circ}$) to the ray $AB$, its new position is $A_1E_1 = AE_1$;
2. b) (if necessary,) slides (along the line $AB$) so that $E_1$ reaches $B$, its new position is $A_1'E_1' = A_1'B$;
3. a) rotates around $B$ (i.e., $E_1'$) to the ray $BC$: $A_2E_2 = A_2B$;
4. b) (if necessary,) slides to the position $A_2'E_2' = CE_2'$;
5. a) rotates to the position $A_3E_3 = CE_3$;
6. b) (if necessary,) slides to the position $A_3'E_3' = A_3'D$;
7. a) rotates to the position $A_4E_4 = A_4D$;
8. b) (if necessary,) slides to the position $A_4'E_4' = EE_4'$;
9. rotates to $EA$.

The sequence of movements has swapped the positions of $A$ and $E$, and the direction of $AE$ has rotated by $180^{\circ}$; each rotation occurred in the same direction (in the positive direction in the figure), so the value of the sum in question is $180^{\circ}$.