Solution. We prove that there exist two adjacent sides of K which form a sufficiently small triangle.
K is convex and is a subset of the unit square. Therefore its perimeter p is at most 4.
Denote the length of the ith side of K by di and let the angle between the ith and (i+1)th sides be -i. Since
there exists an index i for which .
Consider the triangle determnined by the ith and (i+1)th sides; its area is