Commons Vertices

Let the central angle of the 1982-gon be $\alpha = \dfrac {360^\circ}{1982}$ and that of 2973-gon be $\beta = \dfrac {360^\circ}{2973}$ . We note that $\dfrac \alpha \beta = \dfrac {2973}{1982} = \dfrac 32$ . That is $2\alpha = 3\beta$ which means that if we align a pair vertices of the two polygon, then every $2\alpha$ or $3 \beta$ away, there is a common vertex. Therefore, there are $\dfrac {1982}2 = \dfrac {2973}3 = \boxed{991}$ common vertices.

The process is similar as finding the greatest common divisor $\gcd (1982, 2973) = 991$ .

The central angle between adjacent vertices on the first polygon is $\frac{2\pi}{1982}$ . The central angle between adjacent vertices on the second polygon is $\frac{2\pi}{2973}$ .

Therefore, the central angle between common vertices on the two polygons is $\frac{2\pi n}{1982} = \frac{2\pi m}{2973}$ for two positive integers $n,m$ . Solving this gives the ratio $m:n = 3:2$ , so by setting $m=3k$ or $n=2k$ for some positive integer $k$ , we have that the angle between common vertices is $\frac{2\pi (2k)}{1982} = \frac{2\pi k}{991}$ which, starting from a given common vertex, gives $990$ more common vertices corresponding to $k=1,2,\ldots, 990$ .

Counting the original common vertex, they have $1+990 = \boxed{991}$ common vertices.

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Alternatively, just note $\gcd(2973,1982) = 991$ .