A solution of the “plank problem.”

Proof. M has at least one chord in the direction v with parallel supporting hyperplanes at its ends. It is obvious that the length k of this chord is not less than m. A length k — hol this chord is contained in the intersection. Let A on the boundary of M+v and B on the boundary of M—v he the ends of this length k — h. Il A is taken as center of a similitude of the ratio (k — h)/k<l, then M+v is carried into a part of itself, and the minimal width of this part is m(k — h)/k'^1m — h. Since this new domain is also obtained from M—v through a similitude of the same ratio and with its center in B,