I OM - B - Task 12

How many sides does a polygon have if each of its angles is $ k $ times larger than the adjacent angle? What values can $ k $ take?
If each angle of a polygon is $ k $ times larger than the adjacent angle,
then the sum of all $ n $ angles of the polygon, which - as
we know - equals $ (n - 2) \cdot 180^{\circ} $, is also $ k $ times larger
than the sum of the corresponding adjacent angles (i.e., the exterior angles),
which means it is $ k $ times larger than $ 360^{\circ} $. Therefore






from which






Since $ n - 2 $ must be a natural number, then $ 2k $ should
be a natural number, i.e., $ k $ must be half
of a natural number, so $ k = \frac{1}{2}, 1, \frac{3}{2}, 2, \dots $