For what integers $n>2$ is the following statement true? "Any convex $n$-gon has a side such that neither of the angles at the endpoints of this side is acute."
Solution. The sum of the exterior angles of any convex polygon is $360^{\circ}$, so at most 3 of these can be obtuse angles. Since an interior angle is acute if and only if the adjacent exterior angle is obtuse, a convex polygon can have at most 3 acute angles. Every angle of the polygon lies on 2 sides, so if $n \geq 7$, then any convex $n$-gon has a side on which neither of the angles is acute. Therefore, if $n \geq 7$, the statement is true.

![](https://cdn.mathpix.com/cropped/2024_05_02_fa1da03a6eb7713af059g-1.jpg?height=276&width=290&top_left_y=350&top_left_x=495)

$n=5$

![](https://cdn.mathpix.com/cropped/2024_05_02_fa1da03a6eb7713af059g-1.jpg?height=282&width=290&top_left_y=350&top_left_x=907)

$n=4$

![](https://cdn.mathpix.com/cropped/2024_05_02_fa1da03a6eb7713af059g-1.jpg?height=319&width=288&top_left_y=350&top_left_x=1317)

If $n=3,4,5$ or 6, the statement is not true. In these cases, there are convex $n$-gons where every side has an acute angle. For example, such a polygon can be obtained by attaching an isosceles triangle with a base angle of $10^{\circ}$ to each of the $n-3$ sides of a regular triangle.