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?
Answer: 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 d2θ/dt2 = (2π/15) dr/dt + (dθ/dt) cos θ. Substituting from the first equation, 2 (dr/dt) (dθ/dt) + r d2θ/dt2 = (2π/15) dr/dt. Multiplying through by r and integrating wrt t, we get r2 dθ/dt = (π/15) r2 + C, for some constant C. At t = 0, r = 3/2 and dθ/dt = 2π/15, so C = 3π/20. Thus r2 dθ/dt = (π/15) r2 + 3π/20. But the original equation gives r2 dθ/dt = 2r2π/15 + r sin θ. Hence π r2/15 + 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 r2 = (sin2θ + (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 r2 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
5 Mar 2002
Last corrected/updated 9 July 03