1st Putnam 1938 Problem B3 A horizontal disk diameter 3 inches rotates once every 15 seconds. An insect starts at the southernmost point of the disk facing due north. Always facing due north, it crawls over the disk at 1 inch per second. Where does it again reach the edge of the disk? Solution
: at the northernmost point of the disk. Take polar coordinates with r = 3/2, θ = 0 at the start. The equations of motion are dr/dt = - cos θ, r dθ/dt = 2rπ/15 + sin θ. Differentiating the second equation: (dr/dt) (dθ/dt) + r d 2 θ/dt 2 = (2π/15) dr/dt + (dθ/dt) cos θ. Substituting from the first equation, 2 (dr/dt) (dθ/dt) + r d 2 θ/dt 2 = (2π/15) dr/dt. Multiplying through by r and integrating wrt t, we get r 2 dθ/dt = (π/15) r 2 + C, for some constant C. At t = 0, r = 3/2 and dθ/dt = 2π/15, so C = 3π/20. Thus r 2 dθ/dt = (π/15) r 2 + 3π/20. But the original equation gives r 2 dθ/dt = 2r 2 π/15 + r sin θ. Hence π r 2 /15 + r sin θ = 3π/20 (**). Hence if r = ±3/2, we have sin θ = 0 and hence θ = 0 or π. That is not quite enough to show that the insect reaches the edge again at θ = π. But we can treat (**) as a quadratic in r and solve to get r 2 = (sin 2 θ + (π/5) 2 ) 1/2 - sin θ. This shows that r first decreases, but then increases again to ±1 at θ = π. We can rule out -1 because r 2 is always positive, and r starts positive. So by continuity it must always remain positive. Thus the insect next reaches the edge at the northenmost point of the disk. Many thanks to Melih Ozlem for pointing out an error in the original solution. 1st Putnam 1938 © John Scholes jscholes@kalva.demon.co.uk 5 Mar 2002 Last corrected/updated 9 July 03