B3. The internal angles of a convex hexagon measure whole degrees and form an arithmetic sequence. Let $\alpha$ be the largest angle of this hexagon. What is the greatest possible measure of the angle $\alpha$?
B3. We note that the sizes of the angles form an arithmetic sequence. Let $a_{1}$ be the smallest angle, and we use the formula for the sum of 6 terms of an arithmetic sequence $s_{6}=\frac{n\left(2 a_{1}+(n-1) d\right)}{2}$, along with the information about the sum of the interior angles of a hexagon $6 \cdot 180^{\circ}-360^{\circ}=720^{\circ}$. We write the equation $\frac{6\left(2 a_{1}+5 d\right)}{2}=720$, from which we get $2 a_{1}+5 d=240$ and then $a_{1}=\frac{240-5 d}{2}=120-\frac{5 d}{2}$. The number $d$ is positive, and from the last relationship between $a_{1}$ and $d$, we conclude that it is even. The largest angle of the hexagon is $a_{6}=a_{1}+5 d$. By trial and verification, we determine that the largest angle of such a hexagon can measure $175^{\circ}$.

We can solve the problem differently. In a regular hexagon, all angles are $120^{\circ}$. If each angle of the hexagon measures an integer number of degrees and the angles form an arithmetic sequence, the middle angles by size measure $(120-a)^{\circ}$ and $(120+a)^{\circ}$, where $a$ is a natural number. Then the difference in size between these two angles is $(2 a)^{\circ}$, which is the common difference of the arithmetic sequence. This means that the largest angle measures $(120+a+2 a+2 a)^{\circ}=(120+5 a)^{\circ}$. Since each angle of a convex polygon is less than $180^{\circ}$, the largest angle of such a hexagon measures $175^{\circ}$.

Considering the sum of the arithmetic sequence $s_{6}=\frac{n\left(2 a_{1}+(n-1) d\right)}{2} \ldots \ldots \ldots \ldots \ldots \ldots \ldots 1$ point

Considering the sum of the interior angles of the hexagon ......................................................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_bce9c6e2a6f4f88bb0deg-16.jpg?height=62&width=1630&top_left_y=1054&top_left_x=276)

![](https://cdn.mathpix.com/cropped/2024_06_07_bce9c6e2a6f4f88bb0deg-16.jpg?height=54&width=1639&top_left_y=1115&top_left_x=274)

Considering the conditions $d \in \mathbb{Z}, d>1$ and the conclusion that $d$ is an even number ..............................................................................................

Solution $\alpha=175^{\circ}$.......................................................................................................