1. Circle-polygon problem

Two regular polygons are inscribed in the same circle. The first polygon has 1982 sides and second has 2973 sides. If the polygons have any common vertices, the number of such vertices is ..........

2. Originally Posted by nolanfan
Two regular polygons are inscribed in the same circle. The first polygon has 1982 sides and second has 2973 sides. If the polygons have any common vertices, the number of such vertices is ..........
The central angle of the first polygon is:

$\displaystyle \frac{360^\circ}{1982} = \left( \frac{180}{991}\right)^\circ$

and the central angle of the 2nd polygon is:

$\displaystyle \frac{360^\circ}{2973} = \left( \frac{120}{991}\right)^\circ$

... and now it's your turn!

3. If you have trouble visualising what's going on here, try asking the same question for polygons with fewer sides. For example,
Two regular polygons are inscribed in the same circle. The first polygon is an equilateral triangle and the second is a pentagon (5 sides). If the polygons have any common vertices, the number of such vertices is .......?

Two regular polygons are inscribed in the same circle. The first polygon is an equilateral triangle and the second is a hexagon (6 sides). If the polygons have any common vertices, the number of such vertices is .......?
You should be able to see that in the first case, the triangle and the pentagon will only have the one vertex in common. In the second case there will be three common vertices (each alternate vertex of the hexagon will be a vertex of the triangle).

Now, what might account for those different results? (Hint: think "greatest common divisor".)