What is the maximum number of sides of a convex polygon that has no two adjacent obtuse angles?
We know that the sum of the exterior angles of a convex polygon is $360^{\circ}$, which means there can be at most 4 exterior angles that are not less than $90^{\circ}$, otherwise their sum would exceed $360^{\circ}$.

If the polygon has 4 exterior angles of $90^{\circ}$, then it is a rectangle. This meets the condition, as there are no obtuse angles at all.

![](https://cdn.mathpix.com/cropped/2024_05_02_d138b782641e07f104b9g-1.jpg?height=444&width=426&top_left_y=312&top_left_x=882)

In any other case, the polygon can have at most 3 exterior angles that are not less than $90^{\circ}$, and accordingly, at most 3 interior angles that are less than or equal to $90^{\circ}$. These 3 interior angles must separate the obtuse angles from each other. There can be at most one obtuse angle between two such acute angles, so the number of obtuse angles is also 3.

Therefore, the convex polygon can have at most six sides. Such a polygon is visible in the diagram.

Erdei Zsuzsa (Hajdúszoboszló, Hógyes E. Gymn., 10th grade)