### 32nd Putnam 1971

**Problem B4**

A and B are two points on a sphere. S(A, B, k) is defined to be the set {P : AP + BP = k}, where XY denotes the great-circle distance between points X and Y on the sphere. Determine all sets S(A, B, k) which are circles.

**Solution**

Answer: S(A, B, k) is a circle iff k = π and A and B are not antipodal, in which case S(A, B, k) is the great circle perpendicular to the great circle through A and B and so that A and B are on the same side of and equidistant from the plane containing it.

Wlog we may take the sphere to have radius 1. Let the great-circle distance between A and B be d. So 0 ≤ d ≤ π. If d = 0, then clearly any k (in the range 0 < k < π) gives a circle. If d = π, then any point P on the sphere has PA + PB = π, so S is either empty or the whole sphere. So let us assume that 0 < d < π.

Let C be the great circle through A and B. Let O be the centre. Let QR be the diameter with angle AOQ = angle BOR. Let B' be the reflection of B in QR. Then AB' is a diameter. Let C' be the great circle through Q and R perpendicular to C. For any point P on the sphere we have AP + PB' = π. If P lies on C', then by symmetry BP = B'P, so AP + PB = π. If P lies in the open hemisphere containing B, then BP < B'P, so AP + BP < π. Similarly, if P lies in the open hemisphere containing B', then BP > B'P, so AP + BP > π. Thus S(A, B, π) is the circle C'.

Let us now assume 0 < k < π. If k < d, then S(A, B, k) is empty, since AP + BP cannot be less than AB. If k = d, then S(A, B, k) is the arc of C between A and B. So suppose d < k < π. Then there are two points on C' in S(A, B, k): Q' on the arc AQ and R' on the arc BR, with AQ' = BR'. Now S(A, B, k) must be symmetrical about the plane containing C', so if it is a circle C'', then it must be the circle diameter Q'R' and perpendicular to C'. It suffices to find a single point X on this circle with AX + BX > k. Let Y be the midpoint of the arc AB, and take X so that the great circle through X and Y is perpendicular to C'. Since we are assuming C'' is a circle we have YQ' = YX = YR'. Since the great circle through X and Y is perpendicular to C'', we must have AX > YX and BX > YX, so AX + BX > 2 YX = 2 YQ' = AQ' + BQ' = k, so Y is not in S(A, B, k) after all.

Finally consider k > π. Let A' be the antipodal point to B. Now for any point P on the sphere, AP = π - B'P and BP = π - A'P, so S(A, B, k) = S(A', B', 2π-k).

32nd Putnam 1971

© John Scholes

jscholes@kalva.demon.co.uk

7 Feb 2001