What is the maximum number of sides a convex polygon can have if its interior angles form an arithmetic sequence with a difference of $d=1^{\circ}$?
Solution. Let the smallest angle of the polygon be $\alpha$. We know that the sum of the interior angles of an $n$-sided polygon is

$$
(n-2) \cdot 180^{\circ}
$$

On the other hand, we can write the sum of the interior angles using the known sum formula for an arithmetic sequence:

$$
S_{n}=\frac{n}{2}(2 \alpha+n-1)=(n-2) \cdot 180^{\circ}
$$

Rearrange the equation:

$$
2 \alpha n+n^{2}-n=360 n-720
$$

From this equation, solve for $\alpha$:

$$
\alpha=\frac{n-2}{n} \cdot 180-\frac{n-1}{2}
$$

A suitable polygon exists if $\alpha$ (the smallest angle of the polygon) is positive, and $\alpha+(n-1)$ (the largest angle of the convex polygon) is less than $180^{\circ}$. Thus,

$$
\begin{gathered}
\frac{n-2}{2} \cdot 180>\frac{n-1}{2}, \quad \text { and } \\
\frac{n-2}{n} \cdot 180-\frac{n-1}{2}+n-1<180, \quad \text { that is } \\
n^{2}-n-720<0
\end{gathered}
$$

The roots of the equation $n^{2}-n-720=0$ are:

$$
n_{1,2}=\frac{1 \pm \sqrt{2881}}{2} \approx 27.33
$$

or negative, so the largest positive integer satisfying (3) is 27. Since 27 also satisfies the inequality (2), the number of sides of the polygon can be at most 27.