20th Putnam 1959 Problem B5 L 1 is the line { (t + 1, 2t - 4, -3t + 5) : t real } and L 2 is the line { (4t - 12, -t + 8, t + 17) : t real }. Find the smallest sphere touching L 1 and L 2 .
The sphere has diameter PQ, where P(s + 1, 2s - 4, -3s + 5) is on L 1 , and Q(4t - 12, -t + 8, t + 17) is on L 2 . PQ is as short as possible. It can also be characterised as perpendicular to L 1 and L 2 . PQ 2 /2 = 7s 2 + st + 9t 2 + 25s - 52t + 457/2. At the minimum the two partial derivatives must be zero, so 14s + t + 25 = 0, s + 18t - 52 = 0. Solving s = -2, t = 3. This gives P (-3, -12, 11), Q(0, 5, 20), the radius (= PQ/2) as 1/2 √251 and the centre as (-3/2, -7/2, 31/2). As originally set, there was a typo, so that the y-coordinate in L 1 appeared as 2t + 4 instead of 2t - 4. That did not change the principle, but made the arithmetic horrendous! 20th Putnam 1959 © John Scholes jscholes@kalva.demon.co.uk 15 Feb 2002