Equipartition of Convex Sets

Abstract

It is well known that a line can be drawn thru any interior point of a convex set' dividing the set into two pieces of equal area. In What is Mathematics, Courant and Robbins (p. 317 et seq.) several theorems of the same type are proved. In particular, it is shown that a convex region can always be divided into four equal parts by two straight cuts, and that furthermore they can be chosen to be perpendicular. We will be concerned here with three cuts; in general they divide a convex region into seven parts. If the cuts are concurrent, -six parts result. We prove that division into six equal parts is always possible, but that although some regions can be divided into seven equal parts, no convex region can be so divided. Thru any point P of the boundary of a convex region there is a unique cut PP' dividing the region in half. At any point N of PP' draw a line QNQ' so that the area QIVP is exactly one sixth of the whole area. (See figure 1) If N is at P' , the area of Q'P'N is zero; if N is at P, this area becomes one half the whole area. There is than a position of N for which Q'P'N has area one sixth, and this position of N is unique since the area of Q'P'N isa monotone continuous function of N on PP'. Now, draw the lines NR and NR' bisecting the remaining regions. We now have the convex set cut into six equal parts by two straight cuts, and one broken cut RNR'. Moreover, our configuration is uniquely determined by the position of the initial point P. We now show that P can be chosen so that RNR' is a straight line.