Solution

We consider all segments with endpoints at the marked points. Since no three of the marked 64 points lie on a single straight line, there will be $64\times63$/2 = 2016 $>$ 1981 such segments. From problem 116573 it follows that among these segments there will be at least one pair of parallel points. A quadrilateral with vertices at the ends of these segments is the desired one $($it cannot be a parallelogram, since there is no parallelogram with vertices at the vertices of the regular 1981-gon$)$.

Hint

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Hint

Each triple of straight lines forms a triangle.

Solution

Nothing to translate.

Answer

Hint

Each triple of straight lines forms a triangle.

Hint

From each vertex of the polyhedron at least three edges emerge.

Solution

If a polyhedron has four vertices, then it is a tetrahedron with six edges. Let the number of vertices of the polyhedron, n, be at least five. At each vertex of the polyhedron, at least three faces converge, and in this way, at least three edges emerge from each vertex of the polyhedron. Hence, the total number of edges is not less than 3n/2 $>$ 7.

Hint

From each vertex of the polyhedron at least three edges emerge.

Solution

Yes, it could be possible $($see the figure$)$.

Hint

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