S is a non-empty closed subset of the plane. The disk (a circle and its interior) D ⊇ S and if any disk D' ⊇ S, then D' ⊇ D. Show that if P belongs to the interior of D, then we can find two distinct points Q, R ∈ S such that P is the midpoint of QR.
The existence of D imposes strong restrictions on S. In fact, we will show that S must contain the entire perimeter C of D. For suppose it does not contain a point X on C. Then since S is closed, it does not contain a neighbourhood of X. So we can find a small circle C' centre X so that the disk D' enclosed by C' lies entirely outside S. Let C' meet C at A and B. Take a circle C'' through A and B with centre on the same side of AB as the centre of C, but further away (so that it has a larger radius than C). Then C'' contains all points of D - D' and hence all points of S. But it does not contain all points of D. Contradiction.
Finally, note that any point in the interior of D lies on the midpoint of some chord of C.
22nd Putnam 1961
© John Scholes
15 Feb 2002