### 23rd Putnam 1962

Problem B3

Show that a convex open set in the plane containing the point P, but not containing any ray from P, must be bounded. Is this true for any convex set in the plane?

Solution

Let S be the convex set. Take any ray R from P. We can find X on R not in S. Since S is open we can take a small segment AB perpendicular to R through P with AP = PB which is entirely contained in S. Take A' on the line AX, the opposite side to A, and B' on BX the opposite side to B, with A'B' perpendicular to R. Then no point C' on the segment A'B' can be in S. For if it was, then we could project C'X to meet the segment AB in C and X would lie between two points of S and hence would have to be in S. But the angle A'PB' is greater than zero. So we have found an open interval about the direction PX (measured as a polar angle at P) such that S extends at most a distance AA' for directions within that interval.

The interval [0, 2π] is compact and is covered by these open intervals, so a finite number of them cover it. Hence there is a finite distance d such that we can find a point not in S within a distance d of P in any direction. That means that no point Q a distance greater than d from P can be in S (otherwise all points on the segment QP would be in S and there would be no point not in S within a distance d in that direction). So S lies inside the disk radius d on P and hence is bounded.

Take the slab 0 < y < 1 together with the point (0, 0). It is convex and unbounded but does not contain any ray through the origin.