Common Vertices

Assume that an integral number of sides (say $y$ ) of the $2973$ -gon and integral number (say $x$ ) of the $1982$ -gon will subtend an equal angle at the center of the circle. Now, one side of the $2973$ -gon will subtend $\frac{360}{2973}$ degrees, so y sides will obviously subtend $\frac{360y}{2973}$ degrees. Similarly the other polygon will subtend $\frac{360x}{1982}$ degrees.

Now remember what we said about the respective number of sides of the polygon (obviously both regular) subtending equal angles at the center? Yeah. Just equate the angles we just got, that is $\theta=\frac{360y}{2973}=\frac{360x}{1982}$ . You get the least integral solution to be something on the lines of $y=3$ and $x=2$ (so yeah, that's like many of the GCD-solutions in the comments here).

Now we're almost done. Just divide the respective polygons' number of sides by $y$ or $x$ . Either way you get $\frac{2973}{y}=\frac{1982}{x}=991$ . And that's your answer.

I do have one doubt though - won't the last coinciding vertex be the same as the first one? Answer should be $990$ then.

let N1=1982 N2=2973 and here angle at vertex of a triangle formed by centre of the circle is given by Nc1=(360/1982) and Nc2=(360/2973) and by taking the ratio Nc1:Nc2=2:3 which means for every three vertices of larger poygon or for every two sides of smaller polygin there will be common vertex so the answer is (1982/2)=(2973/3)=991

To find the common vertices i just find out the greatest integer of 1982 and 2973 and got the ans without considering the no of common roots and all....nyways good sum!

Until and unless it is given that one vertex is the same for both polygons, the solution shall stand invalid

just 2973-1982= 991