2Question: A mathematician studying algebraic topology explores symmetries of a surface modeled by a regular polygon with rotational symmetry of order $ n $. If the polygon has 12 sides, what is the sum of all distinct rotational symmetries (in degrees) that map the polygon onto itself? - Nelissen Grade advocaten
To solve this problem, we begin by understanding that a regular polygon with 12 sides exhibits rotational symmetry. The rotational symmetry group of a regular $ n $-gon includes all rotations by multiples of $ \frac{360^\circ}{n} $ that map the polygon onto itself.
To solve this problem, we begin by understanding that a regular polygon with 12 sides exhibits rotational symmetry. The rotational symmetry group of a regular $ n $-gon includes all rotations by multiples of $ \frac{360^\circ}{n} $ that map the polygon onto itself.
Given the polygon has 12 sides, the full rotational symmetry is generated by rotations of:
$$
\frac{360^\circ}{12} = 30^\circ
$$
Understanding the Context
Thus, the distinct rotational symmetries are:
$$
0^\circ, 30^\circ, 60^\circ, 90^\circ, 120^\circ, 150^\circ, 180^\circ, 210^\circ, 240^\circ, 270^\circ, 300^\circ, 330^\circ
$$
There are 12 such angles, corresponding to the 12th roots of unity in the complex plane, and they form a cyclic group under composition. The sum of these angles can be computed as:
$$
\sum_{k=0}^{11} \left(30k\right) = 30 \sum_{k=0}^{11} k = 30 \cdot \frac{11 \cdot 12}{2} = 30 \cdot 66 = 1980
$$
Key Insights
Thus, the sum of all distinct rotational symmetries (in degrees) that map the 12-sided polygon onto itself is:
$$
\boxed{1980}
$$
