C is a closed plane curve. If P, Q ∈ C, then |PQ| < 1. Show that we can find a disk radius 1/√3 which contains C.
C can be any set of points in the plane. The usual proof uses Helly's theorem. Given any three points of the set we can find a circle radius 1/√3 which contains them. This is fairly obvious. Take the points to be P, Q, R with PQ the longest side of the triangle. Take S on the same side of PQ as R so that PQS is equilateral. Then R must lie inside the bounding circle of PQS (in fact it must lie inside the region bounded PQ, the arc of the circle centre P from S to Q, and the arc of the circle centre Q from S to P, and this region lies entirely inside the bounding circle of PQS).
Another way of expressing this is that the three circles radius 1/√3 and centres P, Q, R have a point in common. We can now apply Helly's theorem to the set of circles radius 1/√3 and with centres at all the points of the set C. Every three of these circles have a point in common (proved above) and hence by Helly's theorem there is a point common to all of them. But this point can be taken as the centre of the required circle.
Comment. This is a well-known result (Jung's theorem). But it is hard to prove if you are not familiar with Helly's theorem.
9th Putnam 1949
© John Scholes
11 Mar 2002