==> geometry/smuggler.s <==
One way G can catch S is as follows (it is not the fastest way).
G waits until he knows that S has traveled for one mile. At that time, both
S and G are somewhere on a circle with radius one mile, and with its center
at the original position of S. G then begins to travel with a velocity that
has a radially outward component equal to that of S, and with a tangential
component as large as possible, given G's own limitation of total speed. By
doing so, G and S will always both be on an identical circle having its
center at the original position of S. Because G has a tangential component
whereas S does not, G will always catch S (actually, this is not proven
until you solve the o.d.e. associated with the problem).
If G can go at 40 mph and S goes at 20 mph, you can work out that it will
take G at most 1h 49m 52s to catch S. On average, G will catch S in:
( -2pi + sqrt(3) ( exp(2pi/sqrt(3)) - 1 )) / 40pi hours,
which is, 27 min and 17 sec.