Let's start by showing that the answer is unique. Consider two equilateral pentagons $A B C D E$ and $A^{\prime} B^{\prime} C^{\prime} D^{\prime} E^{\prime}$, in which $\angle A=\angle A^{\prime}=120^{\circ}$,

$\angle C=\angle C^{\prime}=135^{\circ}$. We can assume that the side lengths of the pentagons are equal to 1. Note that by two sides and the angle between them, triangles $E A B$ and $E^{\prime} A^{\prime} B^{\prime}$, as well as $B C D$ and $B^{\prime} C^{\prime} D^{\prime}$, are equal. Therefore, $B E=B^{\prime} E^{\prime}, B D=B^{\prime} D^{\prime}$, and triangles $B D E$ and $B^{\prime} D^{\prime} E^{\prime}$ are equal by three sides. Thus,

$\angle B D E=\angle B^{\prime} D^{\prime} E^{\prime}$. From the isosceles triangles $B C D$ and $B^{\prime} C^{\prime} D^{\prime}$, we see that $\angle C D B=\angle C^{\prime} D^{\prime} B^{\prime}=22.5^{\circ}$. Therefore,

$\angle C D E=\angle C D B+\angle B D E=\angle C^{\prime} D^{\prime} B^{\prime}+\angle B^{\prime} D^{\prime} E^{\prime}=\angle C^{\prime} D^{\prime} E^{\prime}$, which means the answer is unique.

We will now prove that there exists a pentagon satisfying the condition with $\angle D=90^{\circ}$. First, construct a right triangle $C D E$ with legs $C D$ and $D E$ equal to 1. Then $E C=\sqrt{2}$. Next, construct a point $B$ such that $\angle E C B=90^{\circ}, B C=1$. Then $B E=\sqrt{3}$. Now construct a point $A$ such that $\angle A E B=\angle A B E=30^{\circ}$. Then $\angle B A E=120^{\circ}, A B=A E=1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_476f7e28c93a861c2173g-51.jpg?height=451&width=535&top_left_y=1947&top_left_x=773)

## Answer

$90^{\circ}$.
Obukhov B B.

Given a convex pentagon $A B C D E$, all sides of which are equal. It is known that angle $A$ is $120^{\circ}$, angle $C$ is $135^{\circ}$, and angle $D$ is $n^{\circ}$.

Find all possible integer values of $n$.

#